erm, why do Anderson Wang of Pennsylvania and Kevin Wang of Vermont have the same picture??

my teammate and i found this out one night when we were up longer than we were supposed to be... :P

yeah I noticed that too. I think kevin wang might have just pulled a prank, because that picture is obviously Anderson.

also, what does the front of the yearbook mean?? (I am talking about the $ 1.702243^{e\minus{}13} LY$ thing on the billboard) Its also on the shirt...

hmm did anyone notice there was a typo that said something like 1+2+2+1=8 in the team round? it was the question with how many squares contain each unit square etc.

[quote="Yongyi781"]Yeah, that's why they gave it to everybody. That's also why EVERY team seemingly got #5 correct in the team round, even if they left it blank.

Apparantly, the original problem had a 5 by 5 square, for which 8 would've been correct in the statement. Then they changed it at the last minute to a 4 by 4 square and forgot to change the 8 to a 6. Must've caused a lot of confusion![/quote]

Wait... .so everyone got that right? That doesnt seem fair... I worked/got that problem!!! (If the answer was 104...)

I think that's unfair too; they should've accepted 2 answers for both the 5x5 and 4x4 cases...

The organizers gave the following note about Team Round #5 to each coach.

[quote="MathCounts"]Although no formal protests were submitted and 51 of the 57 teams reported the desired answer, every team was given credit for Team Round #5.  With or without this accommodation, the ranking of the top teams remains the same.  This decision was made due to the fact that the wording was not completely consistent throughout the problem.  The original problem included a 5-by-5 square.  When the problem was edited to a 4-by-4 situation, not all of the necessary information in the problem was changed accordingly.[/quote]

[quote="Quickster94"]they made a mistake with the picture, b/c both of the ppl have the same last name.  They have already apologized and offered to send the VT team a set of corrected yearbooks[/quote]

Theres also a Stephen Wang of Tenn., but they got him right.

[quote="dynamo729"][quote="Yongyi781"]LY represents light-years and that many light-years equals one mile. Since Denver is the "Mile-high City."[/quote]

Thanks. By chance.... how did you know that??? We tried to figure it out for so long....[/quote]

my team figured it out too... We figured out that LY meant light years and changed it to feet. We saw it was 5280, so we assumed it was for the "mile high city"

Yeah, I remember I was so surprised when I saw my picture twice. It was really funny when one of my teammates and I met the real Kevin Wang.  :lol:


If xy=a, xz=b, yz=c and none of these quantities is 0 then x squared + y squared + z squared equals what?

could you make it more clear how u got this result?

[hide="solution"]
$x=\frac{a}{y}$ from $xy=a$ also,
$x=\frac{b}{z}$ from $xz=b$
So,
$x^2=\frac{ab}{yz}$
Now,
$yz=c$
So, applying this to $y$ and $z$ we get the answer
[/hide]

[hide="answer"]$ab=x^2yz$
$bc=xyz^2$
$ca=xy^2z$

dividing out the terms that arnt squared from the equation, u get

${ab}/c=x^2$
${bc}/a=z^2$
${ca}/b=y^2$

thus, $x^2+y^2+z^2={ab}/c+{ca}/b+{bc}/a$[/hide]


AMC
USA(J)MO
USAMO
geometry
3D geometry
tetrahedron

[quote="Arvind_sn"]I'm starting to do some olympiad problems, and I was doing the USAMO problems when I found something weird, could you guys tell me what's going on?

2.  A tetrahedron has opposite sides equal. Show that all faces are acute-angled.

I think i found a counterexample to this. 0.o
If you draw out the diagram and label congruent sides, you find that the tetrahedron's faces are just congruent triangles.

Well what if it was the congruent triangle with sides 1, 1, and the square root of 3? That's a 120-30-30 triangle, and it seems to contradict the theorem.

Help plz!  :D[/quote]

We know that the tetrahedron's faces are congruent. We have not proved that it can be any congruent triangle.

Yes, but I was just pointing out that the triangle I gave violated the theorem. Can you explain that? Like what you said? I'm not sure I understand. 0.o

You can't actually construct a tetrahedron made out of four of those triangles. If you have two $ 1, 1, \sqrt{3}$ triangles attached together at the $ \sqrt{3}$ side, there's no way to get the two free vertices a distance $ \sqrt{3}$ apart.

A tetrahedron which has opposite sides equal can be put in a cuboid,and the sides of the tetrahedron are the cuboid's surface diagonals.


Here are some thoughts, that might make a solution by the time I'm done:

[hide]So, first off lets assume one of the variables is 1.  Let c=1, then b=1/a.  So

a + b + c + ab + bc + ca = a + 1/a + 1 + 1 + 1/a + a =

= (2 + 2a) + 2/a

3 + ab + bc +ca = 3 + a + 1/a^2 + a = (2 + 2a) + (1 + 1/a^2)

It follows by AM-GM that the second is greater than or equal to the first.

I'm not sure if that was necessary.  Anyhow, lets go back to having a, b, and c, and lets consider the expression that the RHS becomes if we move everything over there.  In other words,

3 + ab + bc +ca - (a + b + c + ab + bc + ca) =

= (a:^2:b - ab - a + 1) + (b:^2:c - bc - b + 1) + (c:^2:a - ca - c + 1) = (ab - 1)(a - 1) + (bc - 1)(b - 1) + (ca - 1)(c - 1).

I tried to keep going from here, but I don't see it.[/hide]

I also found that the inequality - to - prove is equivalent to
(ab - 1)(a - 1) + (bc - 1)(b - 1) + (ca - 1)(c - 1) >= 0.

I tried Chebyshef and at first sight it worked, but it does not :cry:

Probably there exists a brute force solution using homogenizing and "bunching" (Muirhead inequality). But I'm not used to prove inequalities this way so maybe someone can help me out ?

This actually looks like two inequalities stuck together. The first half is an AM-GM inequaltiy, a + b + c :le: 3. Then the remaining terms form another inequality, ab + bc + ca :le: a:^2:b + b:^2:c + c:^2:a.

[color=cyan]But the first half doesn't work -- a + b + c >= 3, not less that or equal.[/color]

doah. That's true.

The only other thing I see is that the left side can also be written as (a+1)(b+1)(c+1) - 2. But I don't know if that is useful.

[color=cyan]I was thinking about trying something like that . . . it's really bugging me that it isn't symmetric, though.  And the only tricks I know to make it symmetric aren't reversible.[/color]

[color=cyan]Can you prove that for me?[/color]

I'll post the proof. Here's some info about Schur's inequality :

http://planetmath.org/encyclopedia/SchursInequality.html

Proof of Schur's inequality.

Suppose x, y, z > 0. We will prove that
x + y + z + 3xyz >= xy(x + y) + xz(x + z) + yz(y + z)
or equivalently that
x(x - y)(x - z) + y(y - x)(y - z) + z(z - x)(z - y) >= 0.
Suppose WLOG that x >= y >= z.
Then the former inequality can be rewritten as
(x - y)[x(x - z) - y(y - z)] + z(x - z)(y - z) >= 0
which is obviously true. Nice, isn't it ?

[quote="Arne"]Oh I almost forgot this very helpful trick ! 8-)

Since abc = 1 we can substitute a = x/y, b = y/z, c = z/x (with x, y, z > 0).
The inequality - to - prove becomes
x + y + z + 3xyz >= xy(x + y) + yz(y + z) + zx(z + x).
But this is well - known as Schur's inequality. :D[/quote]

Now, I'm quite stupid, could you please show the steps involved in moving from the original inequality to the new one after the substitution.

Doing that substitution gives x/y + y/z + z/x + y/x + z/y + x/z :le: 3 + x:^2:/yz + y:^2:/xz + z:^2:/xy..

Multiply everything by xyz..

x:^2:z + x:^2:y + y:^2:x + y:^2:z + z:^2:x + z:^2:y :le: 3xyz + x:^3: + y:^3: + z:^3:.

(and then you can factorise the left hand side as given).

[quote="Arne"]Prove that if a, b and c are positive reals with abc = 1 then
a + b + c + ab + bc + ca <= 3 + ab + bc +ca.[/quote]

What I don't get is that you can just cancel ab+bc+ca, to get the wrong inequality $a+b+c \le 3$.

Did the OP edit it to something different?

The problem was
[quote="Arne"]Prove that if $a, b$ and $c$ are positive reals with $abc = 1$ then
\[a + b + c + ab + bc + ca \le 3 + a^2b + b^2c +c^2a\][/quote]


Find all real numbers $x$ such that

$\lim_{n \rightarrow \infty}{\frac{n!}{n^{n-x}}}$ is a positive real.

[hide="Solution"] Once more I will appeal to Stirling's:  $n! \approx \sqrt{ 2 \pi n }\left( \frac{n}{e}\right)^{n}$

So

$\lim_{n \to \infty}\frac{n!}{n^{n-x}}\approx \sqrt{ 2 \pi }n^{x+\frac{1}{2}}e^{-n}$

Hard luck.  You've got a power function in $n$ multiplied by $e^{-n}$.  This will always tend to zero. [/hide]

um you forgot to put what value $n$ tends to.

Hmm, you can use L'hopetal's Rule but that's a huge pain, so I don't think you can do it  :lol: at least not easily.


There are 67 people in a tennis tournament. A player is eliminated from the tournament after losing 4 matches. What is the maximum number of matches that can be played so that exactly 5 people are left in the tournament?

This is how I did it:
[hide]There can be 62 matches before there are 5 people remaining.
There can be a total of 7 matches before a player is eliminated because someone could lose 3 times and win 4.

So the maximum number is 62x7.[/hide]

However, that's way off.  Can someone tell me which step I totally messed up on?

Thanks.  :)

Here's where you messed up -- you counted total number of wins AND losses. However, a match always contains a winner and a loser, so if you want to count total matches, just count losses. Also, you forgot to factor the losses that the other 5 players can have.

[hide]The maximum number of losses that can occur between all losing players is $62*4$, because all 62 lose four times. That amount is $248$. Then, you add the losses the current players have. The maximum they can have is 3. $3*5$ is $15$. $15+248 = 263$ [/hide]

This is how i approached it.
[hide] Yes, I did the solution the exact same way as fanatic. This problem is just a variant of the problem where you try to find the number of matches played before the winner was decided, so you approach the problem the same way by finding the number of people and count the number of losses because you are double counting a match when you count the win and loss as two different matches which would be four. Then you must count the maximum number of losses that someone can have without being eliminated and multiply that by the number of remaining players. [/hide]

http://www.gomath.com/Questions/question.php?question=18713

says something else...

They clearly didnt understand the question and thought that you could only play an opponent once./


I added AMC 12 class today and have read the transcript but can't see the message board.  Even the link from my classes does not work.

If you signed up close to the registration date, then it might not be updated (see my topic about the Advanced MATHCOUNTS message board)

o. okay. tht makes me a little less worried. thanks  :lol:

[quote="rrusczyk"]You should be able to see the message board now.[/quote]

where is it

Only students that are registered for the AMC 12 class can see the message board.


use this lemma : if ABCD be a cyclic quadrilateral and H=(AC,BD),X=(AD,BC)and Y=(AB,DC) .then H is the pole of the line XY. :)

[quote="Nemion"]Consider $ ABC$ a triangle with ortocenter $ H$, the circle with diameter $ BC$ intersects $ AH$ in $ F$ and $ G$, prove $ AHFG$ is harmonic.[/quote]
I guess you mean to prove that $ (AH,FG)\equal{}\minus{}1$.
The proof of the fact that you need is given in my blog,the last entry.
In fact,there is given the generalization of your problem,which was described by M.A as a hint,see above.
Link for my blog is given below(look at my signature) :wink:


[color=cyan]Andrew flips a fair coin and rolls a standard six-sided die.  If the coin comes up heads, he gets as many points as the number showing on the die.  If it comes up tails, he rolls the die again and gets as many points as the larger of the two rolls.  What is the expected value for the number of points Andrew will get in one turn?[/color]

o man

expected value, never thought i would hear that phrase again

maybe shoulda paid attention to those lectures....

[hide] 287/72?

The value he expects to get on the dice roll if he flips heads on the coin is 7/2... the value he expects to get if he flips tails is 1(1) + 3(2) + 5(3) + 7(4) + 9(5) + 11(6) / 36 = 161/36 if I did my calculations right (the above is because he could only obtain a total of 1 if both his dice rolls were 1, he could get a 2 if he rolled 1,2; 2,1; 2,2 ,a 3 if he rolled 1,3; 2,3; 3,2; 3,1; 3,3; the number of ways he could roll each value n is equal to 2(n-1) + 1) ... so we add 1/2(7/2) + (1/2) (161/36) since he has an equal probability of flipping either heads or tails. [/hide]

By the way, thanks for all the problems!

I liked this because the answer was just a tiny bit less than [hide]4[/hide], which amused me for some reason.  Anyhow, the problems are my pleasure -- I already have them typed up anyhow, so it's not even work to put them up.

Could someone please inform me of what an expected value is?

[color=cyan]What expected value means is, literally, the value you can expect to get on any given turn.  What that does [i]not[/i] mean is "the most likely of all possible options."  What it [i]does[/i] mean is, "if you repeat this procedure many many many (say, approaching infinitely many) times, what will my average score get closer and closer to?  Or in other words, "on average, what can I expect to get?"  Actually, that second way is nicer.[/color]

So you're saying that if there is a game where if a person tosses heads, he/she gains 10 points, but loses 5 points if he/she tosses tails. So after n turns, the number of points the person will have will be a multiple of 5, while the expected value is an average and may not be a multiple of 5, right?

I know that game was a bad example, but you get my idea of the being a multiple of 5 and all hopefully.

Right. The expected value in your problem would be 2.5n (1/2 (10+ (-5)) but the number of points would always be a multiple of 5.

Yea, there are some really nice expected value problems involved with gambling, using expected values to determine how good a strategy is for the "patient gambler" and the "anxious gambler"

Hmm.. on the topic of expected values..

You pay $x to play this game.. you keep tossing a coin until it comes up tails.. if this is on the first turn, you win $2, on the second you win $4, on the third you win $8, doubling each time. How much would you be willing to pay to play this game?

I'll play this game with you for any amount of money you want provided I can play it as many times as I want...

What about if you can only play it once?

Expected value is still infinite, but I'm too lazy to figure out what the probability that I lose money is if I offer a certain amount.

But isn't it that when u calculate expected value vs. probabiltiy of winning, the odds favor your opponent more and more the more you play?

I'm not exactly sure what you mean there.. could you explain that?

Ok, what i am blabbering about is the following:

(notice, I HAVE ABSOLUTELY no idea wat im talking about)

Let's say we have a game. i start off with 2 dollars, and u start off with 10 dollars. in this game, i have a 60% chance of winning, and u have a 40% chance of winning. Each time someone loses, he ahs to give $1 to his opponent. Whoever loses all their money first loses.

In a case like this, i vaguely remember the lecturer saying that the more games u play, the higher chance ur opponent has of winning, or something like that

Anyway, i now realize this has nothing to do with expected value....


function
real analysis
real analysis solved

Prove that there exist functions f:R->R derivable on R wich are not monotonous on every interval I included in R

Have you a 'simple' solution for this one? It seems to me that is it a hard problem, which have been for the first time by Denjoy in the half of the last century...

Pierre.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=244776[/url]. God knows how many times this appeared on the forum. I'm glad we posted a complete proof at last and I can safely retire all these threads to "Solved Problems"  :D


There is a big cylinder of radius R and 2 small cylinders of radius $ R_{1}, R_{2}$. The rotating pivot is the center of mass of the bar. Find the total kinetic energy as function of $ \theta$. Use the "rolling-only condition" for the small cylinders inside the big one.

Pic:
[img]http://www.df.unipi.it/~cella/daily/070202/070202-img8.png[/img]

My solution:

Let's use cylincric coordinates centered in the CM of the bar and versors $ \{\hat\theta,\hat\rho,\hat z\}$.
Let:

$ \omega_{i}\hat z$ the angular speeds of the small cylinders
$ v_{C,i}\hat\theta$ the speed of the center of mass of each cyl.
$ C_{i}\equal{} (R\minus{}R_{i})\hat\rho$ the position of the center of mass of the cyls.
$ \omega_{b}\equal{}\dot\theta$ theangular speed of the bar.
then $ v_{C,i}\equal{}\omega_{i}R_{i}$ because che cyls only roll.

It is $ v_{C,i}\equal{} (R\minus{}R_{i})\dot\theta$ then $ \omega_{i}\equal{}\frac{R\minus{}R_{i}}{R_{i}}\dot\theta$

The kinetic energy of each body is:
for the cyls:
$ K_{i}\equal{}\frac{1}{2}\left ( m_{i}v_{i,C}^{2}\plus{}I_{i}\omega_{i}^{2}\right ) \equal{}\frac{1}{2}(R\minus{}R_{i})^{2}\dot\theta^{2}\left ( m_{i}\plus{}\frac{I_{i}}{R_{i}^{2}}\right )$

for the bar:

$ K_{b}\equal{}\frac{1}{2}I_{b}\dot\theta^{2}$

then the total is $ K_{1}\plus{}K_{2}\plus{}K_{b}$

right??


How can I show that the projection matrix $ P \equal{} \frac{aa^T}{a^Ta}$ is equal to its own square?

I wrote out the following $ PP \equal{} \frac{aa^T}{a^Ta} \frac{aa^T}{a^Ta}$, but I don't know how to cancel things from here.

Try thinking about the [[associative property]].  (It's not like there's anything else to be done with that expression.)


Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[a^{2}+b^{2}+c^{2}\ge 2(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}).\]

[quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[a^{2}+b^{2}+c^{2}\ge 2(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}). \]
[/quote]
SOS kills it enough easy. But I'll look for nice proof.

Yeah... Your proof is very nice, Arqady. I think it is hard for SOS to solve this problem (or it will be complicated, as a lot of expanding).

How do you prove the last inequality?

Very clear explanation, Arqady. Thank you,

In the same article, I proposed the following stronger inequality

$\bigstar$ (Hungkhtn's, SI inequalities, Chapter I).
Let $a,b,c$ be non-negative real numbers with sum $2$. Prove that
\[a^{2}+b^{2}+c^{2}\ge 2(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}+4a^{2}b^{2}c^{2}).\]

This inequality is very sharp, Arqady. Let you kill it  :lol:


Let $ A_1,A_2,....,A_n$ be arithmetic progressions of positive integers, each of $ k$ terms, such that every two progressions have at least two common terms. If $ b$ of the APs have common difference $ d_1$ and the others have common difference $ d_2$, where $ 0<b<n$, prove that \[ b \leq 2 ( k \minus{} \frac{d_2}{gcd(d_1,d_2)}) \minus{}1\]

This is just a idea, I think we can build a rigorous solution.

Let us suppose we have a AP ($ A_1$ wlog) of $ k$ terms with common difference $ d_1$. Excepting the leftmost and the rightmost terms of this AP we can pick every term and construct a AP towards left and right. So $ b \minus{} 1\leq 2k \minus{} 2$.

Similarly we shall try building the rest $ n \minus{} b$ APs starting from a term of $ A_1$. On the scale of AP with cd $ d_1$, two terms of a AP with cd $ d_2$ differ by $ lcm(d_1,d_2)$.
We cannot construct sequences like in the picture.

So we have

$ b \minus{} 1\leq 2\big( k \minus{} \frac {lcm(d_1,d_2)}{d_1}\big) \minus{} 2$

$ \Rightarrow b \minus{} 1\leq 2\big( k \minus{} \frac {d_2}{gcd(d_1,d_2)}\big) \minus{} 2$


Θεωρούμε τον γραμμικό μετασχηματισμό Τ του $ C^3$ του οποίου ο πίνακας ως προς την κανονική βάση {(1,0,0), (0,1,0), (0,0,1)} του $ C^3$ είναι ο πίνακας
$ \left(
\begin{array}{c c c}
3 & 1 & 1 \\
0 & 1 & 0 \\
6 & 2 & 4
\end{array}
\right)$
θέλουμε να βρούμε την S και N μορφή του

Θα ήθελα να συμπληρώσω κάποιες διευκρνιστικές πληροφορίες γιατί ίσως δεν έγινε κατανοητό αυτό που ζητάει η άσκηση.
θέλουμε την ανάλυση του γραμμικού μετασχηματισμού Τ σε S+Ν όπου S διαγωνιοποιήσιμος πίνακας  και ο Ν μηδενοδύναμος.
Σχετική θεωρία είναι τα"
ΛΗΜΜΑ. Αν Τ και S μηδενοδύναμοι και TS=ST τότε T+S μηδενοδύναμος
ΘΕΩΡΗΜΑ. Αν Τ γραμμικός μετασχηματισμός τότε υπάρχει μοναδικός S(διαγωνιοποιήσιμος) και Ν (μηδενοδύναμος) ώστε
(α) Τ=S+Ν
(β) SN=NS

Nomizo oti den xreiazete na kaneis tipota allo apo to na pareis tin Jordan normal form tou pinaka.


AMC
AIME
trigonometry
geometry
3D geometry

The equation $ z^6\plus{}z^3\plus{}1\equal{}0$ has roots with argument $ \theta$ in the complex plane between $ 90$ degrees and $ 180$ degrees. What is the degree measure of $ \theta$?

how do we define argument? I'm stuck on this

All of the roots can be written as: $ r cis{\theta}\equal{} r(\cos{\theta} \plus{} i\sin{\theta})$. The question is asking what $ \theta$ is.

[hide="Solution"]
Let $ x \equal{} z^3$. Then the equation transforms to:

\[ x^2 \plus{} x \plus{} 1 \equal{} 0\]

That has the roots which are the cube roots of unity (excluding 1). They are:

$ x \equal{} cis{120}$ or $ x \equal{} cis{240}$.

So therefore, $ z^3 \equal{} cis{120}$ or $ z^3 \equal{} cis{240}$.

So the solutions are:

$ z \equal{} cis40$ , $ cis 160$, $ cis 280$ (from the first part) and $ z \equal{} cis80$, $ cis 200$, and $ cis 320$. (from the second part)

So the only $ \theta$ between 90 and 180 is 160, which is our final answer.

[/hide]

[hide="another way"]Another thing you can do is to change the equation into $ \frac {z^9 \minus{} 1}{z^3 \minus{} 1} \equal{} 0$

Use roots of unity 0, 40, 80... (in degrees)

then, take out the cube roots of unity, because the equation doesn't evaluate to 0 at those places.

The only argument in the second quadrant left is 160

But I guess my way is more cheap  :P [/hide]

When you say to define argument, the argument is just the angle the complex point makes with the x-axis.


Can you suggest some good magazine high school student should read very often? Thanks

That's kind of a vague question.  What kind of magazine?

essay writing is improved by writing essays.
vocabulary is improved by reading challenging novels.

[quote="simo14"]vocabulary is improved by reading challenging novels.[/quote]  Yes, because there is some magical quality novels possess that improves vocabulary that is not shared with any other form of written language.   :roll:

I subscribe to and greatly enjoy the New Yorker.  Much of its content is available free on its website, [url]http://www.newyorker.com/[/url].

Read some kind of scienc-y magazine, those have tons of huge words.  My mom always reads stuff like Discover and Nat'l Goegraphic.  Reader's digest also has this vocab section in it.

If you don't like to read the nerdy stuff, read something like...uh...i dunno. I read Game Informer, but that probably doesn't help.

[quote="Big M"]Any good magazine can improve essay writing and improve the vocabulary.[/quote]
You'd rather read  books for that...

Magazines help.  Good news magazines, like Newsweek and Time certainly help.  And don't forget newspapers.  The New York Times is huge and is bound to help you (assuming you trust the "liberal media" :wink: ).

I second JBL.  If your goal is to improve your reading skills and vocabulary, and to see models of what good writing is, you can do no better than The New Yorker.  It's the most well-written magazine in wide circulation.  Unlike most other magazines, it contains a rare combination of news, politics, essays, criticism, literature, and frivolous nonsense.  However, the magazine is extremely New York-centered, which is a bit off-putting if you're not from New York, but you can always skip over all the New York bits.

A less New York-centered alternative, which is also very well-written, is Harper's.  It skews a bit more toward literature.  I should mention that both magazines are a bit left-leaning.  There aren't really any right-leaning counterparts to The New Yorker and Harper's.

Another place to find very good writing is The New York Review of Books.  It has less to do with both "New York" and "book reviews" than one might expect.  Many of the articles are written by world-famous intellectuals, but they sometimes place serious demands on the reader's level of background knowledge.

Many of the serious magazines in wide circulation are heavily political, with The New Republic and The Nation on the left, and the Weekly Standard and The National Review on the right.  I guess that leaves The Atlantic somewhere in between (?).   (Sidenote: I never read any of these.)

Newsweek and Time are respectable, but a typical high schooler with a decent SAT verbal score is not going to improve her verbal skills by reading them.  In any case, I think that The Economist is a better substitute (but it's British).

Also, while reading novels is a great way to improve your command over the English language, novels do not provide models for what a good essay should look like.  But really, when it comes to reading, you should treat it the same way that you should treat learning math:  Do whatever seems most natural and enjoyable, but always make sure to challenge yourself.

Just wanted to correct myself.  By happenstance I was reading The Atlantic yesterday, and it's really not comparable to those political magazines I mentioned.  I think I just assumed it was political because of Andrew Sullivan (who writes a very popular blog for The Atlantic).


(i) Use the method of differences to show that

$ \sum_{r\equal{}1}^{n} {(r\plus{}1)^{4} \minus{} r^{4}} \equal{} (n\plus{}1)^{4} \minus{}1.$

(ii) Show that

$ (r\plus{}1)^{4} \minus{} r^{4} \equal{} 4r^{3} \plus{} 6r^{2} \plus{}4r \plus{}1$

(iii) Hence show that

$ 4\sum_{r\equal{}1}^{n} r^{3}\equal{} n^{2}(n\plus{}1)^{2}$.

[hide="(i)"]
expand it ... the alternate terms cancel out

$ \sum_{r\equal{}1}^n (r\plus{}1)^4 \minus{} r^4 \equal{} 2^4 \minus{} 1^4 \plus{} 3^4 \minus{} 2^4 \plus{} \cdots n^4 \minus{} (n\minus{}1)^4 \plus{} (n\plus{}1)^4 \minus{} n^4$

leaving you with 2 terms

$ (n\plus{}1)^4 \minus{} 1$

[/hide]

[hide="(ii)"]

By Binomial Theorem

$ (r\plus{}1)^4 \equal{} r^4 \plus{} 4r^3 \plus{} 6r^2 \plus{} 4r \plus{} 1$

[/hide]

[hide="(iii)"]

 $ \sum_{r\equal{}1}^n (r\plus{}1)^4 \minus{} r^4 \equal{} \sum_{r\equal{}1}^n (4r^3 \plus{} 6r^2 \plus{} 4r \plus{} 1) \equal{} (n\plus{}1)^4 \minus{} 1$

^ using results (1) and (ii)

So,

$ (n\plus{}1)^4 \minus{} 1 \equal{} 4 \sum_{r\equal{}1}^n r^3 \plus{} 6 \sum_{r\equal{}1}^n r^2 \plus{} 4 \sum_{r\equal{}1}^n r \plus{} \sum_{r\equal{}1}^n 1$

$ 4 \sum_{r\equal{}1}^n r^3 \equal{} n^4 \plus{} 4n^3 \plus{} 6n^2 \plus{} 4n \minus{} n(n\plus{}1)(2n\plus{}1) \minus{} 2n(n\plus{}1) \minus{}n$

$ \equal{} n^4 \plus{} 2n^3 \plus{} n^2 \equal{} [ n(n\plus{}1) ]^2$

[/hide]

[quote="AndrewTom"]Thanks, Nora.91.

How do you hide solutions when posting?[/quote]

you are welcome !  :)

well, we have the option HIDE

syntax is [b]hide=click[/b] { inside square brackets [ ] } code ...then  [b]/hide [/b]inside [ ]


Let $ x, y$ be nonnegative integers such that $ xy \plus{} x \plus{} y$ is a perfect square. Prove that there is a nonnegative integer $ z$ for which the seven numbers

$ xy \plus{} z, yz \plus{} x, xz \plus{} y, yz \plus{} y \plus{} z, zx \plus{} z \plus{} x, xy \plus{} yz \plus{} zx, xy \plus{} yz \plus{} zx \plus{} x \plus{} y \plus{} z$

are all squares.

[ [b]Kvant 1799[/b] ]

Hints:
1) If $ ab \minus{} 1$ is a square, then it is possible to write $ a \equal{} p^2 \plus{} q^2$ and $ b \equal{} r^2 \plus{} s^2$ such that $ ps \minus{} qr \equal{} 1$.
2) Let $ x \plus{} 1 \equal{} p^2 \plus{} q^2$ and $ y \plus{} 1 \equal{} r^2 \plus{} s^2$ such that $ ps \minus{} qr \equal{} 1$.  Choose $ z \equal{} (p\minus{}r)^2\plus{}(q\minus{}s)^2$.
[hide]
Hence,
\[ xy \plus{} x \plus{} y \equal{} (pr \plus{} qs)^2
\]

\[ yz \plus{} y \plus{} z \equal{} \left(r^2 \plus{} s^2 \minus{} pr \minus{} qs\right)^2
\]

\[ zx \plus{} z \plus{} x \equal{} \left(p^2 \plus{} q^2 \minus{} pr \minus{} qs\right)^2
\]

\[ xy \plus{} z \equal{} (pr \plus{} qs \minus{} 1)^2
\]

\[ yz \plus{} x \equal{} \left(r^2 \plus{} s^2 \minus{} pr \minus{} qs \minus{} 1\right)^2
\]

\[ zx \plus{} y \equal{} \left(p^2 \plus{} q^2 \minus{} pr \minus{} qs \minus{} 1\right)^2
\]

\[ xy \plus{} yz \plus{} zx \equal{} \left(p^2 \plus{} q^2 \plus{} r^2 \plus{} s^2 \minus{} pr \minus{} qs \minus{} 2\right)^2
\]

\[ xy \plus{} yz \plus{} zx \plus{} x \plus{} y \plus{} z \equal{} \left(p^2 \plus{} q^2 \plus{} r^2 \plus{} s^2 \minus{} pr \minus{} qs \minus{} 1\right)^2
\]
[/hide]

Could you please give a detailed explanation to Hint 1

I don't know how to show this without using Gaussian numbers.  Does anyone know?

[hide]

   Suppose $ ab \equal{} n^2 \plus{} 1 \equal{} (n \minus{} i)(n \plus{} i)$.  Let $ p \minus{} qi \equal{} \gcd(a,n \plus{} i)$ and $ r \minus{} si \equal{} \gcd(b,n \minus{} i)$.  Then, we must have $ p \plus{} qi \equal{} \gcd(a,n \minus{} i)$ and $ r \plus{} si \equal{} \gcd(b,n \plus{} i)$.  Therefore, $ a \equal{} (p \plus{} qi)(p \minus{} qi) \equal{} p^2 \plus{} q^2$, $ b \equal{} (r \minus{} si)(r \plus{} si) \equal{} r^2 \plus{} s^2$, and, for some unit $ u$,
\[ n \plus{} i \equal{} u(p \minus{} qi)(r \plus{} si) \equal{} u \left((pr \plus{} qs) \plus{} (ps \minus{} qr)i\right)\,.
\]
If $ u \equal{} 1$, we are done; else, you need to do some minor changes.

EXAMPLE: $ a \equal{} 25$, $ b \equal{} 41$, and $ n \equal{} 32$.
\[ 25\cdot41 \equal{} 1025 \equal{} (32 \plus{} i)(32 \minus{} i)
\]

\[ \gcd(25,32 \plus{} i) \equal{} 4 \minus{} 3i
\]

\[ \gcd(41,32 \minus{} i) \equal{} 5 \minus{} 4i
\]
Thus, $ p \equal{} 4$, $ q \equal{} 3$, $ r \equal{} 5$, and $ s \equal{} 4$.  (Check $ 32 \plus{} i \equal{} (4 \minus{} 3i)(5 \plus{} 4i)$.)

If $ x \plus{} 1 \equal{} a$ and $ y \plus{} 1 \equal{} b$, then $ z \plus{} 1 \equal{} (p \minus{} r)^2 \plus{} (q \minus{} s)^2 \equal{} 2$.
Hence, $ xy \plus{} x \plus{} y \equal{} 32^2$, $ yz \plus{} y \plus{} z \equal{} 9^2$, $ zx \plus{} z \plus{} x \equal{} ( \minus{} 7)^2$, $ xy \plus{} z \equal{} 31^2$, $ yz \plus{} x \equal{} 8^2$, $ zx \plus{} y \equal{} ( \minus{} 8)^2$, $ xy \plus{} yz \plus{} zx \equal{} 32^2$, and $ xy \plus{} yz \plus{} zx \plus{} x \plus{} y \plus{} z \equal{} 33^2$.

[/hide]

[quote="Batominovski"]Choose $ z \equal{} (p \minus{} r)^2 \plus{} (q \minus{} s)^2$.[/quote]

Congratz! I tried something of the form $ z \equal{} \sum p^2 \plus{} \cdots$ but I eventually got bored. Anyway, the lemma that gets things going was already [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150421]here[/url] and [b]does[/b] have an elementary solution (I didn't check it, however it's along the lines).

It seems that the lemma arises naturally by considering the equation

$ A^2 \plus{} I \equal{} 0$

over $ \mathcal{M}_2\left(\mathbb{Z}\right)$.

thank you. I havent advanced as far as Gaussian numbers. I hope I understand your proof when I get there!

Quite curious, however by letting $ x \plus{} y \plus{} xy \equal{} t^2$, $ z : \equal{} x \plus{} y \plus{} 2t \plus{} 1$ is a solution as well (thanks to [b]mateivld[/b] for pointing this out).

[quote="nguyenvuthanhha"][i]Haha , Frankly speaking , I was deeply impressed by Batominovski's solution but I think he had made some mistakes in his solution

    Firstly , with $ x = p^2 + q^2 -1 ; y = r^2 + s^2 -1 ; ps - qr = 1 ; z = (p-r)^2 + ( q-s)^2 $

 then $ xy + z = (pr+qs-1)^2 +1$ [b]not[/b] $ (pr+qs-1)^2$[/i][/quote]

Well, you are right.  I wanted to say $z+1=(p-r)^2 + ( q-s)^2$.


can anybody tell me some more formulas for the "agonal numbers" in geometry

**FOR EXAMPLE the pentagonal number FORMULA is \[ \frac{n(3n\minus{}1)}{2}\], or the hexagonal number formula is $ n(2n\minus{}1)$**

$ p^n_r \equal{} \frac12r\left[(n \minus{} 2)r \minus{} (n \minus{} 4)\right]$
where $ p^n_r$ is the $ r^{th}$ $ n$-gonal number,take a look at:
[url]http://mathworld.wolfram.com/PolygonalNumber.html[/url]


Consider a sequence $ (u_n)$, $ (u_1\equal{}1)$, $ u_{n\plus{}1}\equal{}\frac{16u^3_n\plus{}27u_n}{48u^2_n\plus{}9}$.
Find the largest integer wich is smaller the sum S of 2008 summands
$ S\equal{}\frac{1}{4u_1\plus{}3}\plus{}\frac{1}{4u_2\plus{}3}\plus{}...\plus{}\frac{1}{4u_{2008}\plus{}3}.$

To Mods: Please lock this topic, this is a problem from M&Y, issue 1,,2008, and we can't discuss it now


conics
geometry
circumcircle
USAMO
acute
acute triangle

Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on its side $BC$. Construct a point $C_{1}$ in such a way that the convex quadrilateral $APBC_{1}$ is cyclic, $QC_{1}\parallel CA$, and $C_{1}$ and $Q$ lie on opposite sides of line $AB$. Construct a point $B_{1}$ in such a way that the convex quadrilateral $APCB_{1}$ is cyclic, $QB_{1}\parallel BA$, and $B_{1}$ and $Q$ lie on opposite sides of line $AC$.  Prove that the points $B_{1}$, $C_{1}$, $P$, and $Q$ lie on a circle.

Myth has already solved some of them, so I guess that if there's any harm to be done here then it's already done :).

Assume we have shown that $A,B_1,C_1$ are collinear. Then the result would follow, since $\angle B_1C_1P=\angle AC_1P=\angle ABP=\angle B_1QC=\pi-\angle B_1QP$.

Now, in order to show that $A,B_1,C_1$ are collinear, we can work in reverse: given $B_1,C_1$ on the circles s.t. $B_1,A,C_1$ are collinear, we construct the paralleles through $C_1$ to $AC$ and through $B_1$ to $AB$ so that they cut $AB,AC$ respectively. Let these intersect at $Q$. From a quick angle chase we find $BC_1\|CB_1$, so the hexagon $ABC_1QB_1C$ is isncribed in a conic. Since this conic passes through $A,B_1,C_1$ which are collinear, and also through $B,C$, it must be the union of $B_1,C_1,BC$, so $Q\in BC$. Now, if we make $B_1C_1$ the tangent to $(APB)$, then $Q=B$, and if we make it the tangent to $(APC)$, then $Q=C$, so $Q$ will go along the entire edge $BC$.

So the following is not true?

Let $\mathcal C_1,\mathcal C_2$ be two circles intersecting in $U,V$. The tangent to $\mathcal C_1$ in $U$ cuts $\mathcal C_2$ again in $S$. Let $M,N$ be points on $\mathcal C_1,\mathcal C_2$ respectively, collinear with $V$. Then $UM\|NS$.

I really think it is, and I also have a proof, but I'll let you check it (and check the fact that it's equivalent to what you have quoted) :).

Sorry,   my ABC  was not acute enough.

   Maj.  Pestich

Er...how much would you say the showing collinearity part is worth?

[quote="SJS"]My intuition tells me this problem can be solved by inversion.  [/quote]

Be sure, it may be solved by inversion (in point $P$), we get an easy angle-counting problem.

Construct $B_2$ on the circumcircle of $APB$ so that $B_1$, $A$, and $B_2$ are collinear, and take $R$ the intersection of lines $QC_1$ and $PB_2$. Observe that $RPQB_1$ is cyclic and that $\angle{PRB_1} \cong \angle{PB_2B_1}$. It must be that $R = B_2 = C_1$!

(Thanks to Darij for pointing out the [now corrected] typo in the cyclic quadrilateral.)

Consider $M$ the second intersection of the two circles $APBC_1$ and $AQCB_1$. Let $PC_1$ intersect $AM$ at $T$. Then $MPQT$ is cyclic. Hence $TQ // AB$. Therefore $T, Q, B_1$ are collinear. Hence $PC_1B_1Q$ is cyclic.

Alas, I prevented myself from the joy of solving this problem, when I repeatedly misread the problem text and drew wrong sketches, what finally irritated me to an extent such that I gave up and read the solution. Before I comment on the solutions, let me say a few words about the problem and the characteristic "classical triangle geo": The problem is [i]not[/i] classical. Actually, in the problem, two points ($B_1$ and $C_1$) are defined as intersections of lines and circles, so their coordinates (cartesian, let's say) contain quadratic irrationalities; on the other hand, the assertion of the problem, the concyclicity of the points $B_1$, $C_1$, P and Q, yields that the points $B_1$ and $C_1$ are rationally dependent, i. e. if one of them is given, then the other one can be constructed by rational operations only (e. g., if you know $B_1$, then you can find $C_1$ as the point of intersection of the circumcircles of triangles APB and $B_1PQ$ apart from the point P). As a consequence, the conditions that the points $C_1$ and Q lie on opposite sides of the line AB and that the points $B_1$ and Q lie on opposite sides of the line AC are really important as conditions (to distinguish between the two different points of intersection of lines with circles). This makes the problem different from the usual kind of triangle geometry problems, which either don't contain any irrationalities at all, or the irrationalities are independent (i. e., each line-circle intersection can be replaced by the other intersection without that the problem would become wrong).

Anyway, enough of rambling for now, let's come to the solution of the problem. If there is a shortest solution for the problem, then it's Mildorf's solution in post #9, although it contains a typo:

[quote="Mildorf"]Observe that $RPQB$ is cyclic[/quote]

This should be "Observe that $RAPB$ is cyclic" in my opinion.

Here is also a simpler (or, at least, more elementary) way to state Grobber's solution from post #2:

Let the line $AB_1$ intersect the circumcircle of triangle APB at a point $B_2$ (apart from the point A). We will show that $B_2=C_1$.

We will work with directed angles modulo 180°. Since the points A, P, C and $B_1$ lie on one circle, $\measuredangle AB_1C=\measuredangle APC$. Since the points A, P, B and $B_2$ lie on one circle, $\measuredangle AB_2B=\measuredangle APB$. But < APC = < APB, and thus $\measuredangle AB_1C=\measuredangle AB_2B$. Since the point $B_2$ lies on the line $AB_1$, this equality yields $CB_1\parallel BB_2$. In other words, the point of intersection of the lines $BB_2$ and $CB_1$ is an infinite point.

Also, since $QB_1\parallel BA$, the point of intersection of the lines $QB_1$ and BA is an infinite point.

Now, by the Pappus-Pascal theorem, applied to the collinear point triples (B; C; Q) and ($B_1$; $B_2$; A), we see that the point of intersection of the lines CA and $QB_2$, the point of intersection of the lines $QB_1$ and BA, and the the point of intersection of the lines $BB_2$ and $CB_1$ lie on one line. Since we know that the point of intersection of the lines $BB_2$ and $CB_1$ and the point of intersection of the lines $QB_1$ and BA are infinite points, this line is the line at infinity, so that the point of intersection of the lines CA and $QB_2$ must also lie on the line at infinity, i. e. it is an infinite point. In other words, $QB_2\parallel CA$. On the other hand, $QC_1\parallel CA$. Thus, both points $B_2$ and $C_1$ lie on the parallel to the line CA through the point Q. Also, both of these points lie on the circumcircle of triangle APB, and, more exactly, on the arc AB of this circumcircle (the point $C_1$ lies on this arc because the quadrilateral $APBC_1$ is convex and cyclic and the points $C_1$ and Q lie on opposite sides of the line AB). Since the parallel to the line CA through the point Q intersects the arc AB of the circumcircle of triangle APB at only one point, the points $B_2$ and $C_1$ must therefore coincide. Since we knew that the points $B_1$, $B_2$ and A are collinear, we thus obtain that the points $B_1$, $C_1$ and A are collinear.

Now, the rest is easy: We have $\measuredangle C_1PQ=\measuredangle C_1PB$. On the other hand, $\measuredangle C_1PB=\measuredangle C_1AB$, since the points $C_1$, B, P and A lie on one circle. Also, $\measuredangle C_1AB=\measuredangle C_1B_1Q$ since $QB_1\parallel BA$ (and since the points $B_1$, $C_1$ and A are collinear). Thus, $\measuredangle C_1PQ=\measuredangle C_1B_1Q$. And it follows that the points $B_1$, $C_1$, P and Q lie on one circle. This solves the problem.

  Darij

[quote="darktreb"]Mildorf, what did you think when you saw this problem (or more specifically, when you got the solution)?

Because I sure wasn't expecting that to work :(.[/quote]

Actually, this was probably one of the (I'm guessing) two problems I lost points on due to sloppy writing. I found that solution with literally a couple minutes remaining. I constructed $B_2$ because I knew that it had to be $C_1$, then, after observing that $\angle{B_1PB_2} \cong \angle{BAC} \cong \angle{B_1QC_1}$, it was logical to construct $R$ and proceed to show that three points are the same.

If I could prove that $ BC_1 \parallel{} B_1C$, then I'd be done. Unfortunately, I can't. Darn.

[quote="Fedor Petrov"][quote="SJS"]My intuition tells me this problem can be solved by inversion.  [/quote]

Be sure, it may be solved by inversion (in point $ P$), we get an easy angle-counting problem.[/quote]

I hear a lot of "you just invert, then angle chase" and I never see a concrete solution.
Is anyone willing to provide an actual formal inversive solution?
Thanks

We first show that for points $B'$ on $B_1Q,C'$ on $C_1Q$, we have
\[A\in B'C'\iff BC'\parallel CB'.\]
We present two proofs of this result.

By Pappus: If $A\in B'C'$, $BC'\parallel B'C$ by Pappus on $BQCB'AC'$. Similarly, if $BC'\parallel B'C$, then $A\in B'C'$ by Pappus on $BQC\infty_{AC}\infty_{BC'}\infty_{AB}$.

By Moving Points: Let $\ell$ denote the line at infinity. Consider the projective maps
\[C'\mapsto C'A\mapsto C'A\cap B_1Q, C'\mapsto C'B\mapsto C'B\cap\ell=P_\infty\mapsto CP_\infty\mapsto CP_\infty\cap B_1Q.\]
It suffices to check these maps are the same. Note both of these maps are linear, so we need only check two values of $C'$ (if we needed to check a third, we would check the point at infinity along line $C_1Q$). At $C'=Q$, the result is trivial and $B'=C'=Q$. At $C'=C_1Q\cap AB$, $B'$ is the point at infinity along line $B_1Q$. Hence, these maps are the same and the result holds.

Now, define $B'=C_1A\cap B_1Q$. By the claim, we have $BC_1\parallel CB'$. Then, note that
\[\measuredangle APC=\measuredangle APB=\measuredangle AC_1P=\measuredangle B'C_1P=C_1B'C=\measuredangle AB'C,\]
so $B'=B_1$. Finally, note that
\[\measuredangle PQB_1=\measuredangle PBA=\measuredangle PC_1A=\measuredangle PC_1B_1,\]
implying the desired.

We will use [b]mOviNG pOiNtS[/b] !!

[b][color=#0f0]Claim :[/color][/b] $A$ , $B_{1}$ , $C_{1}$ are collinear.

[b][color=#0f0]Proof:[/color][/b]  Fix $P$ and move $Q$ on the line $BC$ , let $AC_{1} \cap (APC) = B_{1} '$. Note that the mappings -
$$ Q \mapsto Q\infty_{AC} \mapsto Q\infty_{AC} \cap (APB) = C_{1} \mapsto AC_{1} \mapsto AC_{1} \cap (APC) = B_{1}'$$ , and $$ Q \mapsto  Q\infty_{AB} \mapsto Q\infty_{AB} \cap (APC) = B_{1}$$ are projective. Hence it suffices to prove the claim for $3$ values of $Q$.

First when $Q = B , C$ we are trivially done. When $Q = P$ , we are done by Reim's Theorem. $\blacksquare$.

Now we will finish from a simple angle chase ---
Note that \[\measuredangle C_{1}B_{1}Q =\measuredangle C_{1}AB=\measuredangle C_{1}PB =\measuredangle C_{1}PQ,\]. And we are done. $\blacksquare$

[quote=Rg]Invert when nothing else makes sense[/quote]
[quote=SJS]My intuition tells me this problem can be solved by inversion.   I wish I had a better handle on this topic (one that seems strange, elusive, and interesting all at the same time).[/quote]

This 2 line proof is for you  :D

Perform a $\sqrt{bc}$ inversion with a reflection along the angle bisector of $\angle BAC$
Note that $P',Q' \in \odot(ABC)$, $B_1' \in P'C$ such that $\odot(B_1'AQ)$ is tangent to $AB$ respectively and let $B_1'A \cap PB' = C_1"$. We want to show that $C_1'' = C$. Now $$\measuredangle C_1''B_1'Q' = \measuredangle AB_1'Q' = \measuredangle  BAQ'= \measuredangle  BP'Q'= \measuredangle  C_1''P'Q' \implies \odot(P'Q'B_1'C_1'')$$ $$\measuredangle  Q'C_1''B_1'= \measuredangle  Q'P'B_1'= \measuredangle  Q'P'C= \measuredangle  Q'AC \implies \odot(C_1''AQ')\ \text{is tangent to } AC \implies C_1''=C_1' \implies \odot(B_1C_1QP) \ \blacksquare$$

Let $B_2$ be the second intersection of $AC_1$ with $(APC)$. Then $\angle B_2PQ=\angle B_2AC=\angle B_2C_1Q$, so $PQB_2C_1$ is cyclic. Now $\angle C_1B_2Q=\angle C_1PB=\angle C_1AB\implies B_2Q\parallel AB\implies B_1=B_2$, so we're done.

[quote=kootrapali]We begin with the following claim:

[b]Claim:[/b] $A,B_1,C_1$ are collinear.
[b]Proof:[/b] We have $\measuredangle AB_1C=\measuredangle APC=\measuredangle APB=\measuredangle AC_1B$, so $BC_1\parallel CB_1$. By Reim’s, this implies $B_1C_1$ passes through $A$, as desired.

Now, we have $\measuredangle B_1PQ=\measuredangle B_1AC=\measuredangle B_1C_1Q$, so $PQB_1C_1$ is cyclic, and we are done.[/quote]

Unfortunately, this is a circular reasoning.

Dunno if this is right.

Let $B_1'$ be the point such that $B_1'C\parallel C_1B$, $B_1'Q\parallel AB$, and $B_1'\in AC_1$ (which must exist by Pappus).

Now we can just angle chase to show that $B_1'\equiv B_1$ and we're done.

(Idea is to use phantom points to turn weak claim -> strong claim into strong claim -> weak claim)

We prove the following restated problem.

[quote=Restated problem]
Let $ABC$ be a triangle and $P$ lie on segment $\overline{BC}$. Let $E,F$ be selected on $(ABP)$ and $(ACP)$ such that $A,E,F$ are collinear. Then the line through $E$ parallel to $\overline{AC}$, the line through $F$ parallel to $\overline{AB}$, and $\overline{BC}$ concur.
[/quote]

To prove this, note that by Reim's theorem $\overline{BE} \parallel \overline{CF}$. Let $\infty_1$ be the point at infinity along $\overline{AB}$ and $\overline{CF}$, $\infty_2$ be the point at infinity along $\overline{AC}$ and $\overline{BE}$, and $\infty_3$ the point at infinity along $\overline{BE}$ and $\overline{CF}$. The problem then follows by Pascal's on degenerate hyperbola $\infty_1A\infty_2E\infty_3F$.

Now note that if we vary $E$ along minor arc $AB$ of $(ACP)$, $F$ varies along minor arc $AC$ of $(ACP)$ as well, and the concurrency point can lie anywhere on $\overline{BC}$ (consider two limiting cases, use continuity). Thus in the original problem we always have $B_1,C_1,A$ collinear. Then $\measuredangle C_1QB_1=\measuredangle BAC$, and
$$\measuredangle C_1PB_1=\measuredangle BPB_1+\measuredangle C_1PC=\measuredangle BAB_1+\measuredangle C_1AC=\measuredangle BAC,$$
proving the desired concyclicity. $\blacksquare$

Well, I wonder if I fakesolved my solution isn't as involved as most others.

Simply let $B_1' = (APC) \cap \overline{AC_1}$ and $Q' = (C_1PB_1') \cap \overline{BC}$. Then, one notices that
\[\measuredangle BQ'C_1 = \measuredangle PQ'C_1 = \measuredangle PB_1'C_1 = \measuredangle PB_1'A = \measuredangle PCA = \measuredangle BCA\]
Thus, $C_1Q' \parallel AC$ and we must have $Q'=Q$. Further,
\[\measuredangle B_1'QC = \measuredangle B_1'CP = \measuredangle AC_1P = \measuredangle ABP =\measuredangle ABC\]
Thus, we also have that $B_1'Q \parallel AB$ and thus $B_1'= B_1$ and indeed we have that points $P,Q,B_1$ and $C_1$ lie on a circle.

:coolspeak:

[b][color=#f00]Claim.[/b][/color] $A, B_1, C_1$ are collinear.

[i]Proof.[/i] Note that $\angle AC_1P = B$ and $\angle AB_1P = C$, so $\angle C_1PB_1 = A$. Then $\angle C_1AB + \angle B_1AC + \angle A = 180^\circ$. $\blacksquare$

Now let $R = \overline{B_1C_1} \cap \overline{BC}$. Then note $RQ \cdot RA = RC_1 \cdot RC$, so $$RC_1 \cdot RB_1 = RP \cdot RQ \iff RA \cdot RC_1 \cdot RB_1 = RC_1 \cdot RC \cdot RP$$ which is obviously true.

Let $B'_1=AC_1\cap (APC)$.
[color=#00f]Claim:[/color]
$A,B_1,C_1$ are all collinear.
[i]Proof:[/i]
It suffices to show that $B'_1=B_1$, or that $QB'_1\parallel AB$. Note that $\triangle PB'_1C_1\sim \triangle ACB$ by AA. Therefore:
\[\angle C_1QP=180-\angle C_1QB=180-\angle ACB=180-\angle C_1B'_1P,\]
so $C_1QPB'_1$ is cyclic.\\

This implies that:
\[\angle C_1B'_1Q=\angle C_1PQ=\angle C_1AB,\]
showing that $AB\parallel QB'_1$, or $B_1=B'_1$ $\square$\\

We have already shown that $C_1QPB'_1$ is cyclic, so $C_1QPB_1$ is cyclic, as desired $\blacksquare$.

We consider $B_1$ as define and $C' = B_1A \cap (APB)$

Now $\angle AC'P = \angle ABC = \angle B_1QC$ hence $C',B_1,P,Q$ are cyclic

Now from $$\angle ACP = \angle AB_1P = \angle C'B_1P = \angle C'QP$$
which give us $C'Q \parallel AC$ so $C'=C_1$ $\blacksquare$

Notice that

\begin{align*}
\text{Points } A, B_1, C_1 \text{ are collinear} & \iff \angle B_1C_1P = \angle AC_1P \\
& \iff \angle B_1C_1P = \angle ABP \\
& \iff \angle B_1C_iP = \angle B_1QC \\
& \iff B_1C_1PQ \text{ is cyclic.}
\end{align*}

Thus, it suffices to prove that points $A$, $B_1$, and $C_1$ are collinear.

Denote the intersection of $\overline{AC_1}$ and $(ACP)$ as $B_2$. We have

\[\angle AB_2P = \angle ACP = \angle C_1QP,\]

meaning $B_2C_1PQ$ is cyclic. Then,

\[\angle CQB_2 = \angle AC_1P = \angle ABP,\]

so $\overline{AB} \parallel \overline{B_2Q}$. Hence, $B_1 \equiv B_2$, so we are done. $\square$

Okay. Note that $\angle AB_1P = \angle C$ and $\angle AC_1P = \angle B$. Also $\angle B_1QC_1 = \angle A$. Then our next claim finishes the problem.

[color=#00f][b]Claim:[/b][/color] The points $A$, $B_1$ and $C_1$ are collinear.
[i]Proof.[/i] Let $\overline{C_1A}$ meet $(ACP)$ at $X$. We claim $X \equiv B_1$. Note \[\angle AXP = \angle ACP = \angle C_1QB\] Thus $(C_1PQX)$ is cyclic. However then \[\angle XQC = \angle XC_1P = \angle B\] so $\overline{XQ} \parallel \overline{AB}$ and hence $X \equiv B_1$. $\square$

Then we finish as we find $\angle B_1 PC_1 = A$ and hence $B_1C_1 P Q$ is cyclic. $\blacksquare$

We use a phantom point argument by defining $C_2 = B_1A \cap (APB)$. We first note that
\[\measuredangle B_1QP = \measuredangle B_1QC = \measuredangle ABP = \measuredangle B_1C_2P,\]

so $B_1C_2PQ$ is cyclic. We're finished if we can show $C_1 = C_2$, which follows from
\[\measuredangle PCA = \measuredangle PB_1A = \measuredangle PQC_2 \implies QC_2 \parallel CA. \quad \blacksquare\]

nice
Let $B_2 = \overline{AC_1} \cap (APC)$. We have
$$\measuredangle B_2PQ = \measuredangle B_2PC = \measuredangle B_2AC = \measuredangle B_2C_1Q,$$
so $B_2C_1PQ$ is cyclic. But
$$\measuredangle B_2QP = \measuredangle B_2C_1P = \measuredangle AC_1P = \measuredangle ABP = \measuredangle ABQ,$$
which implies that $\overline{BA} \parallel \overline{B_2Q}$, so $B_2 = B_1$ and we are done.


The sequence 1,000,000; 500,000; 250,000 and so on, is made by repeatedly dividing by 2. What is the last integer in this sequence?

Notice that

$ 250,000 \equal{} 5^6 * 2^4$

After dividing by $ 2$ four times, we won't be able to divide by 2 anymore. Hence, the last integer in the sequence is $ 5^6 \equal{} \boxed{3125}$.


geometry
trapezoid
trigonometry
number theory
relatively prime
trig identities
Law of Cosines

Source: 1996 AHSME #30

A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides of length 5.  The chord of the circle that divides the hexagon into two trapezoids with three sides of lengths 3 and 5, respectively, has length equal to m/n, where m and n are relatively prime integers.  Find m + n.

(A) 309
(B) 349
(C) 369
(D) 389
(E) 409

right, this is what i'm thinking...[hide]consider the cyclic isosceles trapezoids mentioned in the problem and use ptolemy. we let our chord (that we're trying to find) be x, and if ABCDEF is the hexagon with AB=BC=CD=3 and DE=EF=FA=5, then let angle BCD be :theta:. Note that angle DEF will then be 240-:theta:. We find AC=BD by the law of cosines to be 18-18cos:theta: and DF=EA to be 50-50cos(240-:theta:). Okay, we use ptolemy and get some expressions. Hmm bit bulky though.[/hide]

what I'm not getting is how this connects to biology...oh well. :)

All angle measures in my solution are in degrees.
[hide]
Let the first trapezoid be $ABCD$, where $AB=BC=CD=3$.  Then the second trapezoid is $AFED$, where $AF=FE=ED=5$.  We look for $AD$.

Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$, if we drew $AC$, we would see $\angle BCA=\angle BAC$.  Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ (I couldn't find a better symbol; $\widehat{AB}$ means arc AB).  Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$.

Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$.  Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$.  Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$.  $\angle CDE=120$ as well.

Now I focus on triangle $FAB$.  By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$, so $BF=7$.  Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$, we can now use the Law of Sines to get:
\[\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.\]

Now I focus on triangle $AFD$.  $\angle AFD=3\phi$ and $\angle ADF=\theta$, and we are given that $AF=5$, so
\[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.\]
We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$, but we need to find $\sin{3\phi}$.  Using various identities, we see
\begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\
&=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\
&=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\
&=\sin{\phi}(3-4\sin^2{\phi})\\
&=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343}
\end{align*}
Returning to finding $AD$, we remember \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.\]
Plugging in and solving, we see $AD=\frac{360}{49}$; $\boxed{49}$; $\boxed{\mathbb E}$.
[/hide]


geometry
circumcircle
function
geometry unsolved

Let $ \triangle ABC$ be a right-angled triangle with $ \angle C\equal{}90^\circ$. $ CD$ is the altitude from $ C$ to $ AB$, with $ D$ on $ AB$. $ \omega$ is the circumcircle of $ \triangle BCD$. $ \omega_1$ is a circle situated in $ \triangle ACD$, it is tangent to the segments $ AD$ and $ AC$ at $ M$ and $ N$ respectively, and is also tangent to circle $ \omega$.
(i) Show that $ BD\cdot CN\plus{}BC\cdot DM\equal{}CD\cdot BM$.
(ii) Show that $ BM\equal{}BC$.

(i) is the Casey theorem for the 'circles' $ D, B, C$ and $ \omega_1$, all tangent to $ \omega$.
(ii) Let's note $ AM=AN=x$, $ a, b$ and $ c$ the sides of the rightangled triangle, verifying $ c^2=a^2+b^2$, calculate all segments (except CD) as function of $ a, b, c$ and $ x$, replace in the relation (i), keep in mind that $ a\cdot b=c\cdot CD$ (*) and get $ CD\cdot (c-a-x)=\frac{a}{c}\cdot (a+c) \cdot (c-a-x)$. Since $ CD$ has the value from (*), it follows that $ x=c-a$, i.e. $ BM = BC$.

Best regards,
sunken rock


logarithms
function
floor function
algebra

Solve for $ x$: $ [\log_\frac {1}{2} (2x \minus{} 3)] \equal{} 1$.

where $ []$ represents the Greatest Integer Function
here's my partial solution, i can't seem to figure it out...
[hide]
$ \log_\frac {1}{2} (2x \minus{} 3) \equal{} 1 \plus{} h$ where h=fractional part.
so, $ (\frac {1}{2})^{1 \plus{} h} \equal{} 2x \minus{} 3$
and,$ 2x \equal{} 3 \plus{} (\frac {1}{2})^{1 \plus{} h}$ which is $ x \equal{} \frac {3}{2} \plus{} (\frac {1}{2})^{2 \plus{} h}$
[/hide]...

There is not a distinct value of $ x$ that works, but a range of $ x$ values. So now, what values can $ h$ take?
[hide="Ans"]$ 0 \leq h < 1$, so now you can just plug it in[/hide]
[b]Summary of solution:[/b]

$ \lfloor \log _{1/2} (2x\minus{}3) \rfloor \equal{} 1$

$ 1 \leq \log_{1/2} (2x\minus{}3) < 2$
$ \minus{}1 \geq \log_2 (2x\minus{}3) > \minus{}2$
$ \frac{1}{2} \geq 2x\minus{}3 > \frac{1}{4}$
$ \frac{7}{2} \geq 2x > \frac{13}{4}$

$ \boxed{\frac{13}{8} < x \leq \frac{7}{4}}$

oh right...i just lacked the fact that $ 0\leq h < 1$. thanks anyways, for that great solution:)


Two creatures, one very fast, and one very slow, decide to hold a race. The faster one, confident he would win, allowed the slower one to decide where both of their starting locations were, and the units that the race would be measured in. They both, however, had to travel the same distance in the units that the slowed one decided. The slower animal, however, extremely easily won the race. How?

(there were no obstacles in the way of either creature)

do the units have to be units of distance?

what are the dimensions of both creatures?

That's strange... I'm not normally good at this, but today I just... know. And, I'm pretty sure it's the answer Temperal intends.

I don't want to spoil. I distrust hides. So, here's a hint...

[hide="A hint"]
Can the slower animal make use of some natural force?
[/hide][/hide]

do the units have to be units of distance? [b]No[/b]
what are the dimensions of both creatures? [b]Not a yes/no question[/b]
Can the slower animal make use of some natural force? Please clarify.

[hide]Maybe the slower guy rented a crazy fast landspeeder thing and hired a driver..? [/hide]

Maybe the slower guy rented a crazy fast landspeeder thing and hired a driver..? [b]Uh, no.[/b]

This was probably answered in your post, but did the slower creature start earlier/in a different timezone? [b]No[/b]

what I meant in my post is I *think* (= am 97% sure) I know the answer. That's not a yes/no question...

but now, it looks like it favors the 3%...

I'll try to confirm first.

Are they allowed to go in any direction they please?
Do they have to go in the same direction?
Does the distance have to be measured in a straight line? (i.e. Can they run in a circle with the circumference equal to the distance?)

is the slower creature significantly longer than teh faster creature?
is the faster creature significantly longer than teh slower creature?
''''''' taller"""
"""" wider"""
""""deeper"""
[b]No to all[/b]

Are they allowed to go in any direction they please? [b]No[/b]
Do they have to go in the same direction? [b]Yes[/b]
Does the distance have to be measured in a straight line? (i.e. Can they run in a circle with the circumference equal to the distance?) [b]Well, technically since they're running on the earth, they're running an arc in three-dimensional space, but whatever. Yes.[/b]
Are the units energy? [b]No[/b]
Was the landscape ice, and the slower animal that would normally be on earth be slow? (like a penguin) [b]Whoa, nice one. ONE of the animals' landscape was ice[/b]
Assuming the slower guy could swim, maybe he jumped into a river that traveled faster than the faster guy? [b]Nice guess, but no[/b]

[hide]
Well, to turn it in a new direction, how about measuring their travelled distance in degrees of longitude and letting both go 5 degrees east? Is it allowed?
[/hide]

Well, to turn it in a new direction, how about measuring their travelled distance in degrees of longitude and letting both go 5 degrees east? Is it allowed?  [b]Yes[/b]

Okay then...
[hide]
The slower animal was on the polar ice cap (either permafrost to the north or antarctica will do) and the faster animal was at the equator. Both must run 180 degrees to the east.
[/hide]

[quote="math_explorer"]Oh that means I was in a completely wrong direction... I thought it was they had to "travel" downwards. The slow guy could start in mid-air, the fast guy was forced to start in a really big continent. Dig!

Not that the slow guy would survive...[/quote]

Penguins and polar bears?

Um, what? Not necessarily, as long as the slow guy doesn't create too much air resistance...

and is heavy enough.

Hmm... dense enough.


1.) Compute all values of $x$ which satisfy:

$(x!)^2 =25(x!) - 24$

2.) $D$ is the midpoint of hypotenuse $AC$ in right triangle $ABC$. Line segment $BD$ is drawn. If $AD = \sqrt{29}$, and $AB = 4$, then compute the area of $\triangle{BCD}$.

[hide=""Hint""]
1) Solve the 'quadratic' equation.

2) Point $D$ is equidistant from the vertices. Use Pythagorean Theorem.
[/hide]

[Edit] For #2, that triangle doesn't even exist because $AB>AC$ and $AC$ is the hypotenuse. Please make the correction.

Whoops, you are right  :blush:, the second problem has been corrected ($AB = 4$). As far as the first problem goes, if we were to solve the quadratic:

Let $x! = x$

$(x!)^2 =25(x!) - 24$
$(x)^2 -25x + 24 = 0$
$(x-24)(x-1) = 0$
$x = 24, 1$

This means that $x! = 24$ and $x! = 1$, which means $x = 4$ and $x = 1$. Is this the right answer?

What other number is a solution to x! = 1?

(As a style comment, you should use a different variable, i.e. n = x!, not x=x!).

[quote="prime_lover"]1.) Compute all values of $x$ which satisfy:

$(x!)^2 =25(x!) - 24$
[/quote]

This is how you would solve it:

Let $x! = n$

$(x!)^2 =25(x!) - 24$
$n^2 -25n + 24 = 0$
$(n-24)(n-1) = 0$
$n = 24, 1$

This means that $x! = 24$ and $x! = 1$, which means $x = 4$ and $x = 1$. However, there is another solution because $0! = 1$ as well. This means that the three solutions to this equation are $(0,1,4)$.

[quote]2.)  $D$ is the midpoint of hypotenuse $AC$  in right triangle $ABC.$  Line segment $BD$ is drawn.
If $AD = \sqrt{29}$, and $AB = 4$, then compute the area of $\Delta BCD.$[/quote][hide]
$\text{Since }M\text{ is the midpoint of }AC:\; CD = \sqrt{29}\;\;\Rightarrow\;\;AC = 2\sqrt{29}.$

$\text{From Pythagorus: }BC^2 + AB^2 = AC^2\;\;\Rightarrow\;\;BC^2 + 4^2 = (2\sqrt{29})^2$
$.\qquad\text{Hence: }BC = 10.$

$\Delta ABC\text{ has area: }\frac{1}{2}(10)(4) = 20\text{ units}^2.$

$\text{A median divides the triangle into two regions of equal area.}$
$.\qquad\text{Therefore: area of }\Delta BCD = 10\text{ units}^2.$[/hide]


inequalities algebra polynomial calculus

[hide="A very poor solution"]Add $ 2(ab\plus{}ac\plus{}bc)$ to both sides of the condition. This gives $ (a\plus{}b\plus{}c)^2 \equal{} 4(ab\plus{}bc\plus{}ca)$.

Consider the monic cubic polynomial with roots $ a,b,c$. If this polynomial is $ x^3 \minus{} px^2 \plus{} qx \minus{} r$, then we know $ p^2 \equal{} 4q$ and we want to prove that $ p^3 \geq 54r$. Equivalently, we want to figure out how high the constant term can be for the polynomial to still have $ 3$ real roots.

Using the condition, we consider the polynomial $ x^3 \minus{} px^2 \plus{} p^2x/4 \minus{} r$. Take the derivative and set it equal to $ 0$. We get $ 3x^2 \minus{} 2px \plus{} p^2/4 \equal{} 0$, which has solutions $ p/6$ and $ p/2$. Since we have a positive leading coefficient, we get a local maximum at the lower critical point $ p/6$. We need the polynomial at this local maximum to be greater than or equal to $ 0$ in order to have $ 3$ real roots. So $ (p/6)^3 \minus{} p(p/6)^2 \plus{} p^2(p/6)/4 \minus{} r \geq 0$, or $ p^3/216 \minus{} p^3/36 \plus{} p^3/24 \geq r$. When simplified, we have $ p^3/54 \geq r$ as desired.[/hide]

see this post in here : http://mathysc.com/comment/1385#comment-1385

اگر z مینیمم باشه

it means : if $ z = min \{ x,y,z \} $

[quote=arqady]Let $ a,$ $ b$ and $ c$ are non-negative numbers for which
 $ a^2\plus{}b^2\plus{}c^2\equal{}2(ab\plus{}ac\plus{}bc).$ Prove that
\[ a\plus{}b\plus{}c\geq3\sqrt[3]{2abc}\][/quote]

[img]http://s6.sinaimg.cn/middle/006ptkjAzy76MgrIwHr55&690[/img]

We can subsitute $a=x^2,$ $b=y^2,$ $c=(x+y)^2$ and the inequality becomes:
$$2(x-y)^{2}(2x+y)^{2}(x+2y)^{2} \geq 0$$


A ball thrown in a game of cricket leaves the bowler's hand at a height of 2 metres above a level playing field, reaches a height of 3 metres and lands at point 20 metres from the feet of the bowler
.
What was the speed of delivery?

[color=blue]Upward direction as positive
The two components of the ball's velocity in the horizontal and vertical direction
$v_x=v_0\cos\theta$
$v_y=v_0\sin\theta -gt$
where $v_0$ is the initial velocity of the ball and $\theta$ is the angle between the velocity vector and the horizontal.
The two equations of the ball's movement in that two directions
$x=v_xt$
$y=h_0+v_yt-\frac{gt^2}{2}$
where $h_0$ is the initial height of the ball.
When tha ball reaches the maximum height $H$, we have
$v_0^2\sin^2\theta=2g(H-h_0)$
When the ball lands, we also get
$v_0^2\cos^2\theta=\frac{gL^2}{2(h_0+L\tan\theta)}$
where $L$ is the horizontal displacement of the ball
From the two latter equations, we can get the initial velocity $v_0$ and the angle $\theta$[/color]


geometry
point set
Sequence
IMO Shortlist
IMO Longlist

Given a segment $AB$ of the length 1, define the set $M$ of points in the
following way: it contains two points $A,B,$ and also all points obtained from $A,B$ by iterating the following rule: With every pair of points $X,Y$ the set $M$ contains also the point $Z$ of the segment $XY$ for which $YZ = 3XZ.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


inequalities
quadratics
algebra
quadratic formula
algebra proposed

Let $a\geq0$.Prove that:
      $\sqrt{a+\sqrt{a+\sqrt{a+...}}}<\frac{1+\sqrt{4a+1}}{2}$.

Hint: square it and take $a$ from left to the right, continue the process - just see how the right side behaves after squaring and adding $-a$.

I suppose $a>0$ and I note for any $n\in N^*,$

$a_n=\sqrt{a+\sqrt{a+{\sqrt{a+\cdots +\sqrt{a+\sqrt{a+\sqrt a}}}}}}$,

where the symbol $\sqrt{\ldots}$ appears by $n$ times. Thus,

$a_1<a_2$ and $(\forall )n\in N^*,\ (a_{n+2}-a_{n+1}).s.s.(a_{n+1}-a_n)$

$\Longrightarrow\sqrt a=a_1<a_2\ldots <a_n<a_{n+1}<\ldots,$

i.e the sequence $a_n,\ n\in N^*$ is increased. Therefore, $a_{n+1}>a_n\Longrightarrow$

$a^2_{n+1}>a^2_n\Longrightarrow a^2_n-a_n-a>0\Longrightarrow \sqrt a\le a_n<\frac{1+\sqrt{1+4a}}{2}.$

[u]Remark.[/u] $X.s.s.Y$ ($X,Y$ have the [u][b]s[/b][/u]ame [i][b]s[/b][/i]ign)$\Longleftrightarrow XY>0\ \vee\ X=Y=0$.

[b]Another exercises.[/b] For any $n\in N^*,\ n>1$ there are the following inequalities:

$\blacksquare\ 1^{\circ}.\ \sqrt{1+\sqrt{2+\sqrt{3+\ldots +\sqrt{(n-2)+\sqrt{(n-1)+\sqrt n}}}}}<4;$

$\blacksquare\ 2^{\circ}.\ \sqrt{2+\sqrt[3] {3+\sqrt [4] {4+\ldots +\sqrt [n-2] {(n-2)+\sqrt [n-1] {(n-1)+\sqrt [n] n}}}}}<2,$

(the symbol $\sqrt{\ldots}$ appears in the exercises $1^{\circ}$ , $2^{\circ}$ by $n$, $n-1$ times respectively).

$\Let S=\sqrt{a+\sqrt{a+\sqrt{a+...}}}$
$S^2=a+\sqrt{a+\sqrt{a+...}}$
$S^2-a=\sqrt{a+\sqrt{a+...}}$
$S^2-a=S$
$S^2-S-a=0$
By quadratic formula
$S= \frac{1 \displaystyle ^+_- \sqrt{1+4a}}{2} = \sqrt{a+\sqrt{a+\sqrt{a+...}}}$
Wouldn't it follow that it is equal, and the inequality is wrong, in all cases except where a=0?
$\sqrt{a+\sqrt{a+\sqrt{a+...}}} \ \displaystyle ^?_=\frac{1+\sqrt{4a+1}}{2}$
Please tell me if I'm doing something wrong here...


geometry calculus trigonometry derivative rectangle perimeter

Find the maximum area of a right triangle whose hypotenus is 5.

ABSOLUTELY NO CALCULUS, IT TRIVIALIZES THE PROBLEM

[hide]
Let $x$ and $y$ be the legs of the triangle.
Then we seek to maximize the value of $xy$ given that $x^2+y^2=25$ (by Pythagorean).
By AM-GM, $\frac{x^2+y^2}{2}\ge\sqrt{(x^2)(y^2)}$ and $\frac{25}{2}\ge xy$ with equality at $x=y=\frac{5\sqrt{2}}{2}$.
Then the maximum area is $\frac{xy}{2}$ or $\frac{25}{4}$.
[/hide]

Wow, that's funny. Each post seems to build a little more insight to the solution... :P And yet they were posted independently of each other.

[quote="chess64"][quote="paladin8"]Wow, that's funny. Each post seems to build a little more insight to the solution... [/quote]

except yours  :)[/quote]

I don't quite understand, but ok.

Man, I gotta start remembering about AM GM.  I don't think I've ever used it... the only place I've seen it is on these forums.
I solved by figuring the max would either be when the legs are equal or when one is infinitely large, and deduced the former would give a larger area (since the latter would be 0).  But just in case, I found the area of a 30-60-90 triangle with hypotenuse 5, and found it's area to be less than that of the 45-45-90 one.  So I proved to myself what the largest one.  Too bad that doesn't count in competitions...

Yeah!  Logic power!

How would you do this calc way?

A more geometric approach.
The right triangle must be inscribed in a semicircle with the hypothenuse as the diameter. Since the base is constant, the area is maximized at the greatest height, which is directly above the center. This can be more rigorously shown with calculus, but only basic trig is necessary. So the height is half the hypothenuse and A=(1/2)(5/2)(5)=25/4.

[quote="towersfreak2006"][hide="To deej21"]
Let the legs be $x$ and $\sqrt{25-x^2}$ (from Pythagorean again).
We want the maximum of $x*\sqrt{25-x^2}$.
Differentiating and setting equal to zero, we have $\left(\frac{1}{2}\right)(50x-4x^3)\left(\frac{1}{25-x^2}\right)=0$.
Then $50x=4x^3$ or $x=\frac{5\sqrt{2}}{2}$ since $x\neq0$.
So then the maximum area is $\left(\frac{1}{2}\right)*\frac{5\sqrt{2}}{2}*\sqrt{25-\frac{25}{2}}$.
Or $\frac{25}{4}$.
[/hide][/quote]

You could also do it implicitly... for some reason I like implicit differentiation :P

[hide]
$x^2+y^2 = 25$
Differentiating, we have
$2x+2y\frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{x}{y}$

We have $A = \frac{xy}{2}$ so $\frac{dA}{dx} = \frac{1}{2}(y+x\frac{dy}{dx})$, which, upon substitution, yields $\frac{dA}{dx} = \frac{1}{2}(y-\frac{x^2}{y})$.

Solving this for $0$ to determine the relative maximum, we have $\frac{1}{2}(y-\frac{x^2}{y}) = 0 \Rightarrow y^2-x^2=0 \Rightarrow y^2=x^2 \Rightarrow y = x = \frac{5}{\sqrt{2}}$ and $xy = \frac{25}{2}$.

Thus the minimum area is $A = \frac{xy}{2} = \frac{\frac{25}{2}}{2} = \frac{25}{4}$.[/hide]

25/4
You can say that you have a circle/semicircle with diameter 5.  Then you're trying to find the point on the circle that will give the greatest area for a triangle, given that one side of the triangle is the diameter of 5.  Well, clearly, if you let 5 be the base of the triangle, the area will be maximized when the height is maximized.  That is done when you make it an isosceles right triangle.  Then the height is clearly 5/2, and the area is 25/4.

EDIT: Ah, Bictor717 posted this solution already...

[quote="shobber"][quote="isaacchao"]i'll just say double up the triangle to make a rectangle..
area is maximized with square configuration..
side is 5/sqrt2

area of triangle is 25/4[/quote]
i guess you also have to prove that square has the maximized area.[/quote]
[hide]A rectangle with sides x+y and x-y where y is some positive number has the same perimeter as a square with side x, but it has area x^2-y^2 which is obviously less than x^2, the area of the square.[/hide]

or you can do LaGrange multipliers to make the calc solution even faster:

[hide]

Maximize $\frac{xy}2$ subject to the constraint that $x^2+y^2=25$

The gradients are $(\frac{y}2,\frac{x}2)$ and $(2x,2y)$  So, skipping some steps, $x=y$

Thus, $x=y=\frac{5}2$, so the Area is $\frac{25}4$

[/hide]


function
trigonometry
algebra unsolved
algebra

if X is a real number and f:R-R, f(f(x))=f(x)
f(x)=?

$ f(x)$ is not uniquely defined. You must gives other constraints, for example, $ f(x)$ is continuous function $ \plus{}$ other constraints (only continuous is not enough still).

$ f(f(x))\equal{}f(x) \Rightarrow f(x)\equal{}x$

[quote="peine"]$ f(f(x)) \equal{} f(x) \Rightarrow f(x) \equal{} x$[/quote]
Not if f is not given to be bijective. :!:

The only conclusion you can make is that there exist disjoint subsets $ A, B \subseteq \mathbb{R}$ with $ A \cup B \equal{} \mathbb{R}$ such that $ f$ is the identity on $ B$ and $ f$ maps $ A$ (edit:) to $ B$.  Intuitively, $ f$ "projects onto $ B$."

Why does $ f$ have to map $ A$ injectively onto $ B$?  Doesn't something like
$ A \equal{} \mathbb{R} \minus{} \{0\}$
$ B \equal{} \{0\}$
$ f(x) \equal{} 0$
work, despite not being injective?

but there is no more info. and if f(x)=a, and f(a)=a f(x) can't be always any "a" for example sometimes it can't be f(x)=t

So if we receive a problem that has the reals as its range, the function does not actually have to have a output into every real number?

Of course not.  For example, $ f(x) \equal{} x^2$ doesn't map into the negative reals.  By the way, anuonhu, I'm not sure what you're objecting to - can you clarify?

What's $ E(x)$?  We're given that $ f : \mathbb{R} \to \mathbb{R}$, aren't we?

Let $ f(\mathbb{R})\equal{}A$. For all $ x\in \mathbb{R}$ we have $ f(x)\in A$. Let $ y\equal{}f(x)$ and we have $ f(y)\equal{}f(f(x))\equal{}f(x)\equal{}y$ hence $ f(x)\equal{}x$ for all $ x\in A$. So $ f$ is defined
$ f(x)\equal{}x$ for $ x\in A$ and $ f(x)\equal{}z$ where $ z\in A$ for $ x\in \mathbb{R}\backslash A$.

[quote="anuonhu"]sorry but we don't know that the f(x) is continuously!!!
If f(x) is continuously that is easy problem!!![/quote]

The answers of t0rajir0u and limes123 are quite correct.

And continuity does not change anything.

Take for example :

$ f(x) \equal{} x$ $ \forall x$ such that $ |x|\leq 1$
$ f(x) \equal{} \sin(\frac {\pi x}{2})$ $ \forall x$ such that $ |x| > 1$

This is a continuous function which respects the equation.


geometry
Euler
circle
perpendicular
IMO Shortlist
IMO Longlist

Well, that's easy using polars. Let E be the point of intersection of the tangents to the circle at the points C and D, and let F be the point of intersection of the lines AC and BD. We then have to prove that $EF \perp AB$.

Well, let O be the center of our circle, i. e. the midpoint of the diameter AB, and let P be the point where this diameter AB meets the chord CD. By a well-known property of polars, for any four points $A_1$, $A_2$, $A_3$, $A_4$ on a circle, the point of intersection of the lines $A_1A_2$ and $A_3A_4$ lies on the polar of the point of intersection of the lines $A_2A_3$ and $A_4A_1$ with respect to the circle. Applying this to our points A, C, D, B which lie on our circle, we see that the point of intersection of the lines AC and DB lies on the polar of the point of intersection of the lines CD and BA with respect to our circle. In other words, the point F lies on the polar of the point P with respect to our circle. The point E also lies on the polar of the point P, since the point E is actually (par definitionem) the pole of the line CD, and as the line CD passes through the point P, the pole E of the line CD lies on the polar of the point P.

Thus, we see that both points F and E lie on the polar of the point P with respect to our circle. Hence, the line EF is the polar of point P with respect to our circle. By the definition of a polar, this yields $EF \perp OP$, or, in other words, $EF \perp AB$. Problem solved.

  Darij

[quote="orl"]Given four points $A,$ $B,$ $C,$ $D$ on a circle such that $AB$ is a diameter and $CD$ is not a diameter. Show that the line joining the point of intersection of the tangents to the circle at the points $C$ and $D$ with the point of intersection of the lines $AC$ and $BD$ is perpendicular to the line $AB.$[/quote]

Let K=AC^BD and H=AD^BC. As H is the orthocenter of ABK, KH is orthogonal to AB.

Let O be the point of intersection of the tangents to the circle at the points $C$ and $D$.  It holds that <ODH = <ODB = <DAB = <DHK, as both complementary of the same angle <DBA. Since by similar arguments one can prove that <OCH = <CHK, we ha ve that O is the center of c(DHCK), i.e. O lies on HK, then we are done.

[quote="pestich"]We have another 'Polish joke' here.

     AB being a diameter makes points C and D feet of altitudes from intersection
     points.  According to Archimedes Theorem  all three of the altitudes will
     be concurrent at point A.

     Pestich.[/quote]

Does anybody understand this?

  Darij

[quote="pestich"]
AB being a diameter makes points C and D feet of altitudes from intersection
points.  According to Archimedes Theorem  all three of the altitudes will
be concurrent at point A.
[/quote]

Pestich thought that we are interested in the intersection of AC and BD, where problem says the intersection of the tangents at C and D.

There's one thing that we have to be careful when working with this kind of problem - we are not given the order which the points A, B, C, D are in. As a result, different configurations must be considered.

For instantance, sprmnt21's angles don't make very much sense on the diagram that I drew.

Darij's method seems to avoid this problem. Or we could use directed angles if we were to approach it in a euclidean fashion.

[quote="billzhao"]
For instantance, sprmnt21's angles don't make very much sense on the diagram that I drew.
[/quote]

or, [i]viceversa[/i], the diagram you drew doesn't make very sense wrt the angles I referred in the proof  :D .

Of course  I'm joking. I completely agree with you about the fact "  that we have to be careful when working with this kind of problem".

But in this case ... yes, as you remember, "we are not given the order which the points A, B, C, D are in" but it is quite usual, if not specified,  consider the clockwise order or (also working in this case) the counterclockwise order.

I agree with you that in an official contest is a must to consider all the "different configurations", but I, usually, don't give a complete solution to the problems. In most of cases, I try to propose  an idea or, at the best, a sketch.

Ah well, let's stop doing banalities. Of course, most of the proofs using non-directed angles refer to one possible arrangement and work analogously for the other possible arrangements. Sprmnt's very nice solution of the Polish problem is one example of this rule. By the way, he could have completely avoided the need to distinguish different arrangements by using directed angles modulo 180 and thus tackling all possible arrangements at once.

[quote="pestich"]Darij,

    Do not pretend  not understanding me.  I considered configuration with points
    C and D on opposite sides of AB.   But [b]proof[/b] is still the same for all possible
    arrangements of points C and D.  All that's needed is the Archimedes theorem
   of altitudes concurrency in a triangle.

    Pestich.[/quote]

Pestich, I really don't see any "proof" in your post. I have asked you about the actual crux of your proof rather than about bagatelles like different arrangements. The problem asks for a proof of the fact that "the line joining the point of intersection of the tangents to the circle at the points C and D with the point of intersection of the lines AC and BD is perpendicular to the line AB". In your post, the point of intersection of the tangents to the circle at the points C and D isn't even mentioned.

  Darij

[quote="pestich"]I know that the intersection point of the tangents at C and D was not
    mentioned.   There is a good reason for it : it does not change the picture.
    In the orthocentric configuration this point will be a foot of a median or one
    of Euler points on altitude. Do you agree?

     Pestich.[/quote]

Yes. But you didn't mention it, and you didn't prove it (although the proof is quite simple). Anyway, once written down completely, this solution is exactly Sprmnt's solution.

  Darij

Let $O$ be the center of $\odot (ABCD)$, Let $AC \cap BD=Z$ and Let $AD \cap BC=Y$, let $DD \cap CC=X$, from Pascal's Theorem on $ACCBDD$, we have, $X-Y-Z$, Now, let $AB \cap CD =R$, $\implies$ $R \in AB$, from Brocard's Theorem, $YZ$ is the polar of $R$, but since, $X \in YZ$, hence, $XZ$ is also the polar of $R$ $\implies$ $XZ \perp AB$

Solution without polars :)

Let $F’$ Be the intersection of $(AOC)$ and $(BOD)$. Observer that $F’$ is the image of F after inversion with respect to $(ABCD)$. So $O,F,F’$ are collinear and $OD^2=OC^2=OF.OF’ (*)$.

By angle chasing, $F’,D,F,C$ are concyclic. From $(*)$, we get that $E$ is the circumcenter of $(CFD)$. Now we can finish by angle chasing.

It's just a Pascal on $ACCBDD$ (note that $AD\cap BC$ - orhocenter of $[AC\cap BD]AB$).

Let $P=CC \cap DD ; R=AD \cap BC ; Q=AC \cap BD ; T=AB \cap CD$
By Brocard theorem: $RQ \perp AB$ , $RQ \equiv t$
By polars: $DT \equiv p \implies P \equiv t$
$\implies P,Q,R$ are collinear in the polar of $T$.
$\implies PQ \perp AB$.$\blacksquare$


Two different natural numbers are selected from the set $ \{1, 2, 3, \ldots , 6\}$. What is the probability that the greatest common factor of these two numbers is one? Express your answer as a common fraction.

ernie, you miscounted.

The correct answer is $ \frac{11}{15}$.

There are 11 ways to have a GCF = 1 or have the two numbers be relatively prime - all the sets containing 1 (there are 5), plus (2,3), (2,5),(3,4),(3,5),(4,5),(5,6) for a total of 11.  That gives [b]11/15[/b] since as mentioned about there are 15 possible ways to choose a set of 2 numbers.  I think about the 15 this way.  There are 6 ways to choose the first number and 5 ways to choose the second number, then divide by 2 to eliminate the sets that where chooses with the members in reverse order.


Alright here is my question. I cannot find a formula related to the number of sides in a polygon to the number of total diagonals in that figure. I have the following pattern given to me to make my formula from... 3 sides no diagonals, 4 sides 2 diagonals, 5 sides 5 diagonals, 6 sides 9 diagonals, 7 sides 14 diagonals. My formula would need to be "n" the number of sides equals the total amount of diagonals. Thanks in advance!!  :)

I may very well be wrong but isn't this a classic textbook case to demonstrate that inductive reasoning isn't a proof method?? (because in this specific example inductive reasoning does not lead to the right answer if I remember correctly.)

Induction is a perfectly valid "proof method." There are ways of employing it that are wrong...

nono, not mathematical induction. inductive reasoning. mathematical induction is a form of deductive reasoning. (I'm assuming he's talking about inductive reasoning and not mathematical induction from the thread title?)

Inductive reasoning (going from specific cases to the formula) works in this case b/c the formula is only of second degree (and is easily guessable or curve-fittable), but the formula isnt hard to think of either.

[hide="Here it is:"]
There are $_{n}C_{2}$ different ways to draw a line between two vertices, but $n$ of those will be sides; thus the formula is $\frac{(n-1)(n)}{2}-n=\boxed{\frac{n^{2}-3n}{2}}$[/hide]

yup, you are correct that inductive reasoning will work. I looked and what I was thinking of was the formula for finding the number of  regions that are formed for every new chord added to a circle (assuming the chords are placed in a way that allows for the most regions to be formed). You know... lines through shapes, it's almost the same thing  :P.


Does the base of the logarithm actually matter when computing the Shannon index? My biology teacher does not like the fact that my computation of the Shannon index yields a value greater than $ 1$. I originally thought that the actual value means relatively little in absolute terms, but only gains significant meaning in relative terms when considering the indices of other considered populations.

Yeah, I think it totally depends.  It would be much easier to compare between two indices.  However, imo, although I haven't tried it yet, theoretically, changing the base shouldn't detract from the comparison of two series of data.

Well, I'm relatively sure that, regardless of the base of the logarithm, one can always construct a hypothetical set of data that yields an index which is greater than $ 1$. Is there a source that definitively identifies the base of the logarithm?

I teach environmental science and have always used ln (log base e) to calculate the Shannon Index.  That's what I was taught in college and all of the ecology sources I've found also use ln.  The actual number is relative, so if you were only comparing values you calculated, I guess the base of the logarithm wouldn't matter, but if you wanted to compare it to H' values others calculate, you should use ln.

Base certainly matters, at least in any sort of biological calculation. In this case, the base is [i]e[/i], the base of ln(x), also known as the [i] natural[/i] log, should be used. Why? Because it is called the natural log for a reason, which is that [i]e[/i] is represented in a variety of natural processes like growth or decay (number of atoms, size of population) and various other occurrences in nature. For this reason it is standard for biologists to utilize [i]e [/i]as the base or exponent in their calculations.

Therefore, it really only makes sense to uses [i]e[/i], and no other base. This would also be especially useful when comparing data to other types of indices.

The base of the logarithm only matters in terms of the units of measurement.  Using base-2 corresponds to units of "bits" (binary digits) whereas using base-e corresponds to units of "nats" (natural digits). Really, there is no fundamental difference between the two bases (it's like converting from centimeters to inches).  However, it IS important to be consistent in using the same base for comparisons, so I would suggest following whatever base is commonly used in your field of study.

-------

BTW, all this is from the original paper by Claude Shannon in 1948, which established the foundation of a fascinating field called Information Theory. The concept of Shannon Entropy (what's called Shannon Index by Enviro-Scientists) was first invented to study communication theory (phone lines, modems, cell phones, etc) but has also been used in various diverse disciplines ranging from economics to biology.

http://en.wikipedia.org/wiki/Claude_Shannon


How many full seven-day weeks are there in seven consecutive
years? Assume the first day of the first year is the first day of
the week.

365...bcuz
there are 365  times 7 days
7 days in a week, so 365 weeks! :lol:

Well there are 365 days in a year and 7 days in a week. so in 7 years there is 365*7 days, which is equal to 2555. You divide this number by 7 because there are 7 days in a week giving you 365, so essentially the number of weeks is just the number of days in a year because you are multiplying by 7 then dividing by 7 and they cancel out.

Answer: 365


Show that

\[ 3\equal{}\sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{1\plus{}4\sqrt{1\plus{}5\sqrt{\cdots}}}}}.\]

induction shows that
$ 3\equal{}\sqrt{1\plus{}2 \sqrt{1\plus{}3\sqrt{ \ldots \plus{}n\sqrt{1\plus{}(n\plus{}1)(n\plus{}2)}}}}$


Euler
arithmetic sequence
number theory unsolved
number theory

Prove that in an infinite arithmetic progression of integers, there is always an infinite subset of terms having the same prime divisors

Suppose that $ a,a\plus{}b,...$ is the AP mentioned. we can suppose that $ \gcd(a,b)\equal{}1$ and $ b>0$.obviously we can suppose that $ a>1$. by Euler theorem: $ a^{\phi(b)} \equiv 1 \mod {b}$. if $ n$ is a positive integer then $ a^{n \phi(b)} \equiv 1 \mod {b}$.
Denote by $ s_{n}$ the number $ \frac{a^{n \phi(b)\plus{}1} \minus{} a}{b},n\equal{}1,2,...$.then u can see that
$ a\plus{}bs_{n}\equal{}a(a^{\phi(b)})^{n},n\equal{}1,2,...$ but we had $ a>0$ it follows that $ 0 \leq s_1<s_2<...$.
So we can conclude that the infinite set $ a\plus{}bs_1,a\plus{}bs_2,...$ is what we wanted.Note that it is also a geometric progression.

[hide]Let $ (a_n \equal{} a \plus{} nr)_{n\geq 0}$ be the arithmetic progression. The case $ r\equal{}0$ is trivial, and we may assume $ r>0$, then $ a>r$ (otherwise start with a large enough $ a_n$). Take $ d \equal{} \gcd(a, r)$, $ a\equal{}d\alpha$, $ r\equal{}d\rho$, so $ \alpha$ and $ \rho$ are relatively prime. It is enough if we show there exists $ n>0$ such that $ a\plus{}nr$ has same prime divisors with $ a$, since then we can generate infinitely more, step by step, by the same method.

From Euler's indicator theorem we have $ \rho \mid \alpha^{\varphi(\rho)} \minus{} 1$. Take $ m$ their quotient, so $ 1\plus{}m\rho \equal{} \alpha^{\varphi(\rho)}$. Now take $ n \equal{} m\alpha$, so $ a\plus{}nr \equal{} d(\alpha \plus{} n\rho) \equal{} d\alpha(1 \plus{} m\rho) \equal{} d\alpha^{\varphi(\rho)\plus{}1}$, having the same prime divisors with $ a \equal{} d\alpha$.[/hide]


Let $ X$ be a set of $ 2k$ elements and $ F$ a family of subsets of $ X$ each of cardinality $ k$ such that each subset of $ X$ of cardinality $ k \minus{} 1$ is contained precisely in one member of $ F$. Show that $ k \plus{} 1$ is a prime.

Finally solved it!

Let's say $ X \equal{} \{1, 2, 3, \ldots, 2k \}$.

Now suppose such family $ F$ exists.  Then:

the number of sets in $ F$ containing $ \{1, 2, 3, \ldots, k\minus{}r \}$ is precisely $ (k\plus{}r)(k\plus{}r\minus{}1) \ldots (k\plus{}2)/r!$ for (obviously) $ 2 \leq r < k$.

Proof. There are $ 2k \minus{} (k\minus{}r) \equal{} k \plus{} r$ elements remaining in $ \{1, 2, 3, \ldots, 2k \}$ if i disregard $ \{1, 2, \ldots, k\minus{}r \}$.  There are $ k\plus{}r$ elements we can use to add to $ \{1, \ldots, k\minus{}r \}$.  After we have added that element, call it $ a_1$, there are $ k\plus{}r\minus{}1$ remaining elements we can use to add to $ \{1, \ldots, k\minus{}r, a_1 \}$.  Then continue until you get an enlarged set consisting of $ k\minus{}1$ elements, which is by the condition on $ X$ and $ F$ uniquely in some set in $ F$.  So there were $ (k\plus{}r)(k\plus{}r\minus{}1)\ldots(k\plus{}2)$ ways to enlarge the set $ \{1, \ldots, k\minus{}r \}$ to a set having cardinality $ k\minus{}1$, which is uniquely contained in some set in $ F$.  Since we added $ r$ more elements to the original set $ \{1, 2, \ldots, k\minus{}r \}$ to get an element of $ F$, there are total $ (k\plus{}r)(k\plus{}r\minus{}1)\ldots(k\plus{}2)/r!$ elements in $ F$ that contains the set $ \{1, 2, \ldots, k\minus{}r \}$.

So we see that $ (k\plus{}r)\ldots (k\plus{}2)/r!$ is always an integer for $ 2 \leq r < k$.  In particular, it's an integer for every prime $ r$ in that range.  But this means every prime $ r$ with $ 2 \leq r < k$ divides one of $ k\plus{}r, k\plus{}r\minus{}1, \ldots, k\plus{}2$.  In particular, it cannot divide $ k\plus{}1$.  In other words, $ k\plus{}1$ is a prime.


well ill give you a clue, consider a polynomial..... :D

OK! I understood! Nice problem.

Consider polynomial $p(x)=a_1x+a_2x^2+...+a_{49}x^{49}$. Let $z_{k,i}$ (i=1,...,k-1) be roots of $z^k-1=0$ except 1.
If $k$ is an odd prime number then:
sequence is k separable ==> $p(z_{k,i})=0$ (since $\sum\limits_{j=1..k} (z_{k,i})^j =1+\sum z_{k,i}$ and $1+\sum z_{k,i}=0$ because of Viete's theorem).
So our polynomial p(x) has at least 2+4+6+10+12+16=50 roots ==> $p(x)\equiv 0$.


I'm sure that there exists something like this.

\[x_1^{x_2}x_2^{x_3}...x_n^{x_1}\leqslant\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{x_1+x_2+\cdots+x_n}\]

What is the name of this inequality and what conditions must be satisfied in order for it to be true?

I think you are looking for weighted amgm?

No, not exactly. In this case weighted AM-GM states:
\[x_1^{\frac{x_2}{x_1+x_2+\cdots+x_n}}x_2^{\frac{x_3}{x_1+x_2+\cdots+x_n}}\dots x_n^{\frac{x_1}{x_1+x_2+\cdots+x_n}}\leqslant \frac{x_1x_2+x_2x_3+\cdots+x_nx_1}{x_1+x_2+\cdots+x_n}\]
which is not the same expression.


This is not a physical/real world problem, but more like a thought experiment.
You are ignoring gravity and the motions of bodies in the universe.

Suppose you have one billion cubic blocks that are 10" on a side.  You are to put them in a row so that, EXCEPT for the first block and the last block in the line,... [FORMER]... each block is touching exactly one block to one side and also one to its other side.[FORMER]  [NEWER]... each block has one block on each side of it (as they are in a line) making direct contact with it
(touching it). [NEWER]

The distance to compare it to will be that of 238,857 miles, according to one source, which is Wikipedia.  (Even if this is in question, please use this value
for this problem.  This number is supposed to be the average number of miles
between the centers of the Earth and the Moon.)

Question:  Is it possible to have the one billion 10" blocks in a line with the requisite touching between and among them so that they will reach from the
Earth's center to AT LEAST the Moon's center (average distance)?

Can you show some minimal work toward it (arithmetic maybe)?

**  EDIT in place

How is this a puzzle?  It just looks like unit conversion to me.

$238857 \text{ miles}= 1.5134... \times 10^{10}\text{ inches}$

Whereas a billion ten-inch blocks only has a length of $1.0 \times 10^{10}$ inches.

Let me show you.

You failed to "think outside the box," so to speak.

I stated the boxes had to be touching each other, not HOW they had to be touching each other in relation to one another.

THAT, is the essence behind the puzzle.

Please try again.

[quote="box of rocks"]You are to put them in a row so that, except for the first block and the last block in the line, each block is touching exactly one block to one [b]side[/b] and also one to its other [b]side[/b].[/quote]

This is unclear.  If you meant

- [b]edges[/b]:  Your length is $1.414... \times 10^{10}$ inches, which is still insufficient.
- [b]faces[/b]:  This is the original case.

But I couldn't possibly interpret "side" to mean [b]vertex[/b], which would give a sufficient length of $1.732... \times 10^{10}$.

Although I DID make an edit for a supposed person's clarity,

I would have still stood by my original wording, because

the blocks are stated to be in a line, and so they are side by side,

not referring to the "sides"/faces/edges of a cube or a block.

So when a block is stated to be to one side or the other side,

then it is in relation to the line of blocks, not in relation to a

face or an edge of a block.

It is possible when the touching areas between boxes are very small squares, say, $\le3.87 \text{ inch}^{2}$.

Or, you could use the billion where it = 10^12
(I remember someone recently posted a link about something like this)

I had not stated it, but I intended the American version of
a "billion" all along.  Good point.  It must be stated to keep out
the ambiguity.


Let set $ S\equal{}\{1,2,...,n\}$ and set $ T$ be the set of all subsets of $ S$ (including $ S$ and the empty set). One tries to choose three (not necessarily distinct) sets from the set $ T$ such that either two of the chosen sets are subsets of the third set or one of the chosen set is a subset of both of the other two sets. In how many ways can this be done?


Hey everybody. I'm doing my Algebra II homework and I'm not sure how to do this one:

8 = 3 * 5^x

Solve for x

Let me know if you know how to work it out ASAP.
thanks a lot.

Taylor


vector
analytic geometry
calculus
integration
number theory
combinatorial geometry
number theory proposed

I really enjoyed this problem ,Its marvelous.

let $w_1,...,w_k$ be different real numbers such that $\sum w_i \neq 0$.

Prove that there exist integers numbers like $n_1,...,n_k$  that

$\sum n_iw_i > 0$ and for every permutation of $w_i$s like $\pi(\pi \neq 1)$ ,

$\sum n_1w_{\pi(i)}  < 0$

there were great mistakes in the problem and it was nonsense with them , now i have edited it.
There were pointed out by a friend (i have been told not to mention his name  :) ).So thanks.

I have a solution which makes this look as anything but a number theory problem :). I hope I can explain it properly.

Consider the $ n! $ vectors of the form $ v_{\pi}=(w_{\pi(1)},\ldots,w_{\pi(n)})^T,\ \pi\in S_n $ (all vectors are column vectors here). Suppose we find a vector $ v\in\mathbb R^n $ s.t. $ (v,v_{\pi}) $ (the usual scalar product in $ \mathbb R^n $) is $ >0 $ for $ \pi=1 $ and $ <0 $ for other $ \pi $. Then, since the set of vectors $ v $ for which $ (v,v_1)>0 $, as well as the sets of $ v $ for which $ (v,v_{\pi})<0,\ \pi\ne 1 $ are all open, it means that the set of $ v $ which satisfy our conditions is non-empty and open. This means that we can find such a $ v $ with rational coordinates, and we can scale it to one with integral coefficients, as desired.

The above means that all we need to show is that there must be a vector $ v\in\mathbb R^n $ with $ (v,v_{\pi})>0 $ if $ \pi=1 $ and $ <0 $ if $ \pi\ne 1 $. This is equivalent to finding a hyperplane in $ \mathbb R^n $ which separates $ v_1 $ from $ \{v_{\pi}\ |\ \pi\ne 1\} $ and also passes through the origin $ 0 $. Consider the affine hyperplane $ \mathcal H $ consisting of those points whose coordinates add up to $ \sum w_i $, and let $ P_{\pi}\in\mathcal H $ be the point corresponding to $ v_{\pi} $. It's easy to deduce from the given conditions that $ P_1 $ does not belong to the convex hull of $ \{P_{\pi}\ |\ \pi\ne 1\} $, so there is a hyperplane (of dimension $ n-2 $) $ \mathcal H' $ in $ \mathcal H $ which separates $ P_1 $ from $ \{P_{\pi}\ |\ \pi\ne 1\} $. The $ n-1 $-dimensional hyperplane generated by $ 0 $ and $ \mathcal H' $ is the required hyperplane passing through $ 0 $ and separating $ v_1 $ from $ \{v_{\pi}\ |\ \pi\ne 1\} $.


function
inequalities
induction
real analysis
real analysis solved

Say that $f$ is continuous on $\mathbb{R}$ and that for every $a,b\in\mathbb{R}$,

$f\left(\frac {a+b}{2}\right)\leq \frac {f(a)+f(b)}{2}$

Prove that $f$ is convex (that is, for every $\lambda\in (0,1)$, $f(\lambda a+(1-\lambda)b)\leq \lambda f(a)+(1-\lambda)f(b)$)

Nice! I wait the proof of an other member of MathLinks  :blush:  :P .

You just show the desired inequality for $\lambda$ being a dyadic rational (i.e. with denominator a power of $2$) by induction, and then use the continuity of $f$ and the density of the set of dyadic rationals in $\mathbb R$ to conclude what you want.

As pointed by grobber, the following inequality holds for $n\in \mathbb{N}$ inductively.

\[f\left(\sum_{k=1}^n \frac{x_k}{2^k}\right)\leqq \sum_{k=1}^n \frac{f(x_k)}{2^k}\]

Let $0\leqq \lambda \leqq 1$, we can write  $\lambda =\sum_{k=1}^{\infty} \frac{\lambda _k}{2^k}$   where $\lambda _k=0,1$

Therefore

$f(\lambda  a+(1-\lambda )b)=f\left(\sum_{k=1}^{\infty}\frac{\lambda _k a+(1-\lambda _k)b}{2^k}\right)$

$\leqq \sum_{k=1}^{\infty} \frac{f(\lambda _k a+(1-\lambda _k)b)}{2^k}$

$=\sum_{k=1}^{\infty} \frac{\lambda _k f(a)+(1-\lambda _k)f(b)}{2^k}$

$=\lambda f(a)+(1-\lambda)f(b)$, which complets the proof.

Denote the following property by (P)
$f\left(\frac{x_1+\cdots+x_n}{n}\right)\leq \frac{f(x_1)+\cdots+f(x_n)}{n}$

We prove that if it holds (P) for $n=k+1$, then it holds (P) for $n=k$.

Let $A=\frac{x_1+\cdots+x_k}{k}$ then $x_1+\cdots+x_k=kA$, hence
$A=\frac{x_1+\cdots+x_k+A}{k+1}.$
Since it holds (P) for $n=k+1$ hence
$f(A)=f\left(\frac{x_1+\cdots+x_k+A}{k+1}\right)\leq \frac{f(x_1)+\cdots+f(x_k)+f(A)}{k+1}$
Then it follows that
$f(A)\leq \frac{f(x_1)+\cdots+f(x_k)}{k}$
Hence (P) holds for $k$.

Since (P) holds for all $n=2^m$, we know that (P) holds for all positive integer $n$.

Denote $f(\lambda x_1+(1-\lambda)x_2)\leq \lambda f(x_1)+(1-\lambda)f(x_2)$ by (Q).

First we prove that (Q) holds for rational $\lambda=\frac{m}{n}$.
$f(\lambda x_1+(1-\lambda)x_2) = f\left(\frac{m}{n}x_1+(1-\frac{m}{n})x_2 \right)=f\left(\frac{mx_1+(n-m)x_2}{n}\right) \leq \frac{mf(x_1)+(n-m)f(x_2)}{n} = \lambda f(x_1)+(1-\lambda)f(x_2)$

Then by the continuity of $f$, we know that (Q) holds for all $\lambda\in (0,1)$

[quote="blahblahblah"]
Actually, what grobber showed is continuous + $\displaystyle f\left(\frac {a+b}{2}\right)\leq \frac {f(a)+f(b)}{2}\Longrightarrow$ convex

It's easy to show from there that convex $\Longrightarrow f(\sum \lambda _ix_i)\leq \sum \lambda_i f(x_i)$ by induction[/quote]

I didn't show that, blahblahblah, you did.  ;)

I have just edited,but same as liyi,blahblahblah.

kunny