geometry rhombus circumcircle parallelogram rectangle conics ellipse

Let $ (O)$ be the circumcircle of $ ABCD$ and $ KLMN$ be an arbitrary quadrilateral (not necessarily a parallelogram), such that $ R \in KM, P \in LN.$ Project the circle $ (O)$ to a circle $ (O')$ and the line $ PR$ to infinity. The cyclic quadrilateral $ ABCD$ goes into a cyclic parallelogram (rectangle) $ A'B'C'D'.$ The lines $ RKM, PLN$ go into parallels $ K'M' \parallel B'C' \parallel D'A'$ and $ L'N' \parallel A'B' \parallel C'D'.$ Let $ K'L', M'N'$ meet $ C'D', A'B'$ at $ U', V'$ and let $ K'L', M'N'$ meet $ A'C'$ at $ X', X''.$ By Menelaus theorem for the $ \triangle A'C'P'$ ($ P'$ is at infinity) cut by the transversals $ K'U'X', M'V'X'',$

$ \frac {\overline{K'P'}}{\overline{K'A'}} \cdot \frac {\overline{X'A'}}{\overline{X'C'}} \cdot \frac {\overline{U'C'}}{\overline{U'P'}} \equal{} 1\ \Longrightarrow\ \frac {\overline{X'C'}}{\overline{X'A'}} \equal{} \frac {\overline{K'P'}}{\overline{K'A' }} \cdot \frac {\overline{U'C'}}{\overline{U'P'}} \equal{} \frac {\overline{U'C'}}{\overline{K'A'}}$

$ \frac {\overline{V'P'}}{\overline{V'A'}} \cdot \frac {\overline{X''A'}}{\overline{X''C'}} \cdot \frac {\overline{M'C'}}{\overline{M'P'}} \equal{} 1\ \Longrightarrow\ \frac {\overline{X''C'}}{\overline{X''A'}} \equal{} \frac {\overline{V'P'}}{\overline{V'A' }} \cdot \frac {\overline{M'C'}}{\overline{M'P'}} \equal{} \frac {\overline{M'C'}}{\overline{V'A'}}$

$ \frac {\overline{X'C'}}{\overline{X'A'}} \cdot \frac {\overline{X''A'}}{\overline{X''C'}} \equal{} \frac {\overline{U'C'}}{\overline{K'A'}} \cdot \frac {\overline{V'A'}}{\overline{M'C'}} \equal{} \frac {\overline{U'C'}}{\overline{M'D'}} \cdot \frac {\overline{V'A'}}{\overline{K'B'}} \equal{} \frac {\overline{L'C'}}{\overline{L'B'}} \cdot \frac {\overline{N'A'}}{\overline{N'D'}} \equal{} 1$

This means that the points $ X' \equiv X''$ are identical, $ K'L', M'N'$ meet at $ X' \in A'C'$ and similarly, $ L'M', N'K'$ meet at $ Y' \in B'D'.$ Therefore, the original lines $ KL, MN$ meet at $ X \in AC$ and $ LM, NK$ meet at $ Y \in BD.$ If $ KLMN$ is a parallelogram, $ KL \parallel MN \parallel AC,$ $ LM \parallel NK \parallel BD$ and $ X, Y$ are originally at infinity.

In the projected case, $ X', Y'$ are finite points outside of the rectangle $ A'B'C'D'.$ WLOG, assume that $ X', Y'$ are on the rays $ (A'C', (B'D'.$  A unique conic $ \mathcal K$ is tangent to the 5 lines $ B'C', D'A', A'C', B'D', X'Y'.$ By our WLOG assumption, the pentagon formed by these 5 lines is convex and this conic is therefore an ellipse. Let $ M''$ be arbitrary point on the segment $ C'D'$ and let $ M''Y', M''X'$ meet the segments $ B'C', D'A'$ at points $ L'', N''.$ Consider the hexagon $ L''C'X'Y'D'N''$ with its main diagonals $ L''X', C'D', X'N''$ concurrent at $ M''$ and its 5 sidelines $ L''C' \equiv B'C', C'X' \equiv A'C', X'Y', Y'D' \equiv B'D', D'N'' \equiv D'A'$ tangent to the ellipse $ \mathcal K.$ By Brianchon theorem, the remaining sideline $ L''N''$ is also tangent to this ellipse. But the ellipse $ \mathcal K$ has a unique tangent $ L'N' \parallel A'B' \parallel C'D'$ intersecting the segments $ B'C', D'A',$ corresponding to a unique position of $ M'$ on the segment $ C'D'.$ Consequently, the quadrilateral $ K'L'M'N',$ such that $ X' \equiv K'L' \cap M'N' \cap A'C',$ $ Y' \equiv L'M' \cap N'K' \cap B'D'$ and $ L'N' \parallel A'B' \parallel C'D',$ $ K'M' \parallel B'C' \parallel D'A'$ is unique. As a result, the original parallelogram $ KLMN,$ such that $ KL \parallel MN \parallel AC,$ $ LM \parallel NK \parallel BD$ and $ P \equiv LN \cap AB \cap CD,$ $ R \equiv KM \cap BC \cap DA$ is also [color=red]unique.[/color]

Let the internal bisectors of the angle $ \angle BPC \equiv \angle DPA$ meet $ BC, DA$ at $ L, N$ and let the internal bisector of the angle $ \angle ARB \equiv \angle CRD$ meet $ AB, CD$ at $ K, M.$ Let the angle bisectors $ PLN, RKM$ meet at a point $ Q.$ With reference to the posted figure, consider the concave quadrilateral $ APQR.$

$ \angle PQR \equal{} 360^\circ \minus{} (360^\circ \minus{} \angle A) \minus{} \frac {\angle PBC}{2} \minus{} \frac {\angle CRD}{2} \equal{}$

$ \equal{} \angle A \minus{} \frac {180^\circ \minus{} (\angle B \plus{} \angle C)}{2} \minus{} \frac {180^\circ \minus{} (\angle C \plus{} \angle D)}{2} \equal{}$

$ \equal{} \angle A \plus{} \angle C \plus{} \frac {\angle B \plus{} \angle D}{2} \minus{} 180^\circ \equal{} 90^\circ.$

This means that the $ \triangle KPM, \triangle LRN$ are both isosceles and $ Q$ is the midpoint of both $ KM, LN.$ The resulting quadrilateral $ KLMN$ has perpendicular diagonals cutting each other at half, hence, it is a rhombus.


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Denote
\[ x^2 \equal{} \frac {a^2 \plus{} b^2}{a \plus{} b},y^2 \equal{} \frac {b^2 \plus{} c^2}{b \plus{} c},z^2 \equal{} \frac {c^2 \plus{} a^2}{c \plus{} a}
\]
where $ x,y,z \ge 0$.
Prove that
\[ xy \plus{} yz \plus{} zx \ge a \plus{} b \plus{} c.
\]
:)[/quote]
One can also prove the inequalities
$ a^2\plus{}b^2\plus{}c^2\leq x^4\plus{}y^4\plus{}z^4\leq 3(\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c})^2$

Dear tran300491,

have you nice solutions for all these very nice hard inequalities?

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Denote
\[ x^2 \equal{} \frac {a^2 \plus{} b^2}{a \plus{} b},y^2 \equal{} \frac {b^2 \plus{} c^2}{b \plus{} c},z^2 \equal{} \frac {c^2 \plus{} a^2}{c \plus{} a}
\]
where $ x,y,z \ge 0$.
Prove that
\[ xy \plus{} yz \plus{} zx \ge a \plus{} b \plus{} c.
\]
:)[/quote]
I think, squaring and Cauchy-Schwartz kills it.

Arqady, have you proved it ?

Thank you very much.

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Denote
\[ x^2 \equal{} \frac {a^2 \plus{} b^2}{a \plus{} b},y^2 \equal{} \frac {b^2 \plus{} c^2}{b \plus{} c},z^2 \equal{} \frac {c^2 \plus{} a^2}{c \plus{} a}
\]
where $ x,y,z \ge 0$.
Prove that
\[ xy \plus{} yz \plus{} zx \ge a \plus{} b \plus{} c.
\]
:)[/quote]
Here is my proof:
$ xy \plus{} yz \plus{} zx \ge a \plus{} b \plus{} c\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b \plus{} c)}\geq(a \plus{} b \plus{} c)\sqrt {(a \plus{} b)(a \plus{} c)(b \plus{} c)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b \plus{} c) \plus{} 2\sum_{cyc}(b^2 \plus{} c^2)\sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)(a \plus{} b)(a \plus{} c)}\geq$
$ \geq(a \plus{} b \plus{} c)^2(a \plus{} b)(a \plus{} c)(b \plus{} c).$
But $ \sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)(a \plus{} b)(a \plus{} c)}\geq(a^2 \plus{} bc)(a \plus{} \sqrt {bc}).$
Id est, it remains to prove that:
$ \sum_{cyc}(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b \plus{} c) \plus{} 2\sum_{cyc}(b^2 \plus{} c^2)(a^2 \plus{} bc)(a \plus{} \sqrt {bc})\geq$
$ \geq(a \plus{} b \plus{} c)^2(a \plus{} b)(a \plus{} c)(b \plus{} c),$ which equivalent to
$ \sum_{sym}(a^3b^2 \minus{} a^3bc \plus{} 2a^{\frac {5}{2}}b^2c^{\frac {1}{2}} \plus{} 2a^{\frac {7}{2}}b^{\frac {3}{2}} \minus{} 4a^2b^2c)\geq0,$ which true by Muirhead.

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Denote
\[ x^2 \equal{} \frac {a^2 \plus{} b^2}{a \plus{} b},y^2 \equal{} \frac {b^2 \plus{} c^2}{b \plus{} c},z^2 \equal{} \frac {c^2 \plus{} a^2}{c \plus{} a}
\]
where $ x,y,z \ge 0$.
Prove that
\[ xy \plus{} yz \plus{} zx \ge a \plus{} b \plus{} c.
\]
:)[/quote]
My solution based on the trivial inequality
\[ \sum \frac{(a^2\plus{}b^2)(a^2\plus{}c^2)}{(a\plus{}b)(a\plus{}c)} \ge \sum a^2\qquad \forall a,b,c \ge 0\]
:)
Indeed, setting $ a\equal{}m^2,b\equal{}n^2,c\equal{}p^2$ $ (m,n,p \ge 0)$, the problem requires us to prove
\[ \sum\sqrt{ \frac{(m^4\plus{}n^4)(m^4\plus{}p^4)}{(m^2\plus{}n^2)(m^2\plus{}p^2)}} \ge \sum m^2\]
By Holder's Inequality, we get
\[ (m^4\plus{}n^4)(m\plus{}n)(m\plus{}n) \ge (m^2\plus{}n^2)^3\]
\[ \Rightarrow \frac{m^4\plus{}n^4}{m^2\plus{}n^2} \ge \frac{(m^2\plus{}n^2)^2}{(m\plus{}n)^2}\]
\[ \Rightarrow \sum \sqrt{ \frac{(m^4\plus{}n^4)(m^4\plus{}p^4)}{(m^2\plus{}n^2)(m^2\plus{}p^2)}} \ge  \sum \frac{(m^2\plus{}n^2)(m^2\plus{}p^2)}{(m\plus{}n)(m\plus{}p)} \ge \sum m^2\]
We have done. :)

[quote="tran300491"]
One can also prove the inequalities
$ a^2 \plus{} b^2 \plus{} c^2\leq x^4 \plus{} y^4 \plus{} z^4\leq 3(\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c})^2$[/quote]

for left inequality we have a clear inequality
\[ (\frac{a^2 \plus{}b^2}{a\plus{}b})^2\geq\frac{a^2\plus{}b^2}{2}\]
and for right inequality <=>
\[ x^4 \plus{} y^4 \plus{} z^4 \minus{}(a^2 \plus{}b^2 \plus{}c^2)\leq 3(\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c})^2 \minus{}(a^2 \plus{}b^2 \plus{}c^2)\]
or
\[ \sum (a\minus{}b)^2 (\frac{a^2 \plus{}b^2 \plus{}c^2}{(a\plus{}b\plus{}c)^2} \minus{}\frac{a^2 \plus{}b^2}{2(a\plus{}b)^2})\geq 0\]
that is true

[quote="Protonios"]Mine

Prove:

$ \sum\frac {(a^2 \plus{} b^2)^3}{(a \plus{} b)^3}\geq\sum a^3$[/quote]

We have this inequality
\[ (\frac{a^2 \plus{}b^2}{a\plus{}b})^3\geq\frac{a^3 \plus{}b^3 }{2}\]


geometry
3D geometry
number theory proposed
number theory

Prove that the equation $ x^x\equal{}y^3\plus{}z^3$ has infinitely many solutions in positive integers $ x,y,z$.

Can you write down how you got it?Also are these the only solutions?

[quote="Cronus"]Can you write down how you got it?[/quote]
have you checked the answar? If not, check now (start from the RHS). Then read the rest(I doubt that you will need to after checking...).

[hide]At first I realized that I need not find all solutions. So I tried to find those easy to find. The first thing to note was If some constant is multiplied to both $ y,z$ then the LHS gets multiplied by its cube. And if that constant is a power of $ x$ then we need not worry about the high and variable power of $ x$ in the LHS(this power is the source of the seeming difficulty of the problem). Now how much power do I reduce? The best optrion is the final power to be 1 because there's obviously infinitely many solutions to the equation $ x \equal{} y^3 \plus{} z^3$. In that case the power of $ x$ in LHS, which is $ x$ itself must be one more than a multiple of $ 3$, also $ x$ must be sum of two cubes Thus I chose $ x$ to be what it is. then $ y \equal{} 3kx^{\frac {x \minus{} 1}{3}}$ and $ z \equal{} (3k \plus{} 1)3kx^{\frac {x \minus{} 1}{3}}$.

This url has a similar solution: http://www.mathlinks.ro/viewtopic.php?t=208805
In that case the RHS could be raised by a power more than the LHS. Thus I chose the initial RHS to be least, i.e. 2, making $ x,y,z$ consicutive. Then adjusted the parity on the LHS.[/hide]

[quote="Cronus"]Also are these the only solutions?
[/quote]
I'm $ 99\%$ sure there are other solutions.


[quote="José"]Find all solutions of triples of integrers $(x,y,z)$ such that $xy+xz+yz-xyz=2$[/quote]
$xy+xz+yz-xyz=2$<=>$(1-x)+(1-y)+(1-z)=(1-x)(1-y)(1-z)$
Let that $1-x=a,1-y=b,1-z=c$=>$a+b+c=abc$ and $a;b;c\in{Z}$
I think it isn't hard to solve lest equation in$Z$ ;) .


Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=222906]this[/url] for rules.  However, when only one person remains, there'll be a final twist to see if he truly does deserve to live!

Ten people to start, don't post here to join.  Instead, PM me your bullet choices. [hide=" "]55[/hide]

The order in which you send me the bullet choices will be the order that you go, so send 'em fast!

I guess I'll cancel this.  I'm not active much anyways. ;)


Can we have topis related to Inequality in the Intermediate Forum. It could help many of us as many will osting problems to be solved which they may have solved or may not have solved.

Give some comments on it, and if it is a good idea I can start it or if anyone else has only done this please tell me.

For now please prove this inequality:
8(abcd+1)>(a+1)(b+1)(c+1)(d+1),  where a,b,c,d all are greater than 1.

I found Inequality marathon in Pre-Olympiad section so my idea is not new. Please carry on solving the question.

\[\begin{aligned} abcd + 1 & = abcd + 1 \\ abcd + 1 & > a + bcd \\ abcd + 1 & > b + cda \\ abcd + 1 & > c + dab \\ abcd + 1 & > d + abc \\ abcd + 1 & > ab + cd \\ abcd + 1 & > bc + da \\ abcd + 1 & > ac + bd \end{aligned}\] and adding these gives the result.

These inequalities are easy to prove: for example, since $a,b,c,d > 1$ we have \[(a - 1)(bcd - 1) > 0 \Rightarrow abcd + 1 > a + bcd\] and the other inequalities can be proved in a similar way.

[quote="Arne"]There are plenty of inequality lessons/marathons in all forums already, look at those.[/quote]

Is there any in the Intermediate section.

I think there was a thread here, but it was probably moved to the Pre-Olympiad forum or something, not sure. Have a look at the thread in Pre-Olympiad :)

The thread Arne is referring did indeed start in Intermediate but it was moved to Pre Olympiad as the content level started to rise.

Should I try all the questions posted in that thread till now to reach that difficulty.
Or suggest me some way to learn how to do inequalities.

It is probably a good idea to go through the whole tread.
Just try to improve your skills by doing all exercises there.


Let $ x,y,z>0$ and $ xy\plus{}4yz\plus{}9zx\plus{}2xyz\equal{}36$ . Show :

$ x\plus{}y\plus{}z\ge 4$ .

[quote="Pain rinnegan"]Let $ x,y,z > 0$ and $ xy \plus{} 4yz \plus{} 9zx \plus{} 2xyz \equal{} 36$ . Show :

$ x \plus{} y \plus{} z\ge 4$ .[/quote]

let $ x\equal{}\frac{4a}{b\plus{}c},y\equal{}\frac{9b}{c\plus{}a},z\equal{}\frac{c}{a\plus{}b}$

I think AM-GM does the trick after this :lol:

[quote="Pain rinnegan"]Let $ x,y,z > 0$ and $ xy \plus{} 4yz \plus{} 9zx \plus{} 2xyz \equal{} 36$ . Show :

$ x \plus{} y \plus{} z\ge 4$ .[/quote]
Assume that for some $ X,Y,Z$ the condition is satisfied and $ X \plus{} Y \plus{} Z < 4$. Let $ t \equal{} \frac {4 \minus{} X \minus{} Y \minus{} Z}{3}$. Then $ (X',Y',Z') \equal{} (X \plus{} t,Y \plus{} t,Z \plus{} t)$ satisfies $ X' \plus{} Y' \plus{} Z' \equal{} 4$ and obviously $ X'Y' \plus{} 4Y'Z' \plus{} 9Z'X' \plus{} 2X'Y'Z' > 36$.
So it is enough to prove that $ f(x,y,z) \equal{} xy \plus{} 4yz \plus{} 9zx \plus{} 2xyz \le 36$ when $ x \plus{} y \plus{} z \equal{} 4$.
Consider:
$ f(x,0,z \plus{} y) \minus{} f(x,y,z) \equal{} 2y(4x \minus{} 2z \minus{} zx)$ So $ f(x,0,z \plus{} y) \ge f(x,y,z) \iff 4x \minus{} 2z \minus{} zx\ge 0$.
$ f(x \plus{} y,0,z) \minus{} f(x,y,z) \equal{} y(5z \minus{} x \minus{} 2xz)$ So $ f(x,0,z \plus{} y) \ge f(x,y,z) \iff 5x \minus{} x \minus{} 2zx \ge 0$.
Since $ (4x \minus{} 2z \minus{} zx) \plus{} (5x \minus{} x \minus{} 2zx) \equal{} 3(x \plus{} y \minus{} zx) \ge$ $ \frac {3}{4}(4(x \plus{} z) \minus{} (x \plus{} z)^2) \equal{} \frac {3}{4}(x \plus{} z)(4 \minus{} x \minus{} z) \ge 0$. At least one of them must holds and we can wlog assume $ y \equal{} 0$.
$ f(x,0,z) \equal{} 9xz \le 9\left ( \frac {x \plus{} z}{2} \right )^2 \equal{} 36$
Equality iff $ (x,y,z) \equal{} (2,0,2)$


AMC
AIME
geometry
3D geometry
quadratics
USA(J)MO
USAMO

:( shoot i only had a 7, i couldve done 11 but for miscounting on #3, not noticing obvious stuff in #6 and #8, and intimidated by #12
im a freshmen, and i missed my chance
only next yr :?

1... thats a new record, down from 4 last year.

[quote="DPopov"]1 on #3!!!...grr....could have had that 14[/quote]

duuuude me too I got 120 instead of 109 on #3... :/ so I got 8 instead of 9. I don't know if I'll make it then :x is there a diff floor value thing for 8th graders?

Same floor....i forgot the cubes for some reason, I thought about it and I just brushed them off so I put 105. :(

More than I didn't...

5 very stupid mistakes(#3, 4, 7, 10), one should-have-known-better mistake (#14)

Two!

 :D

Only 2 mistakes this time around, thankfully.

 On the semicircle problem I found the radius, but forgot to multiply by 2....

I'm still happy with my score though.

Yeah, I'm pretty happy considering the 12 point increase from last year but I wish I would have caught that careless mistake on #3...... :P ....don't we all though

1, on #3.

I also think it was a stupid mistake not to do #1. (But I thought my geometry sucks... Could've gotten it.)

Not so many dumb mistakes for me...
Possibly because I was dumb enough to give up after a few problems and check the ones I knew.
I randomly guessed 15 but was off by 1. Oops.

Oh yes, my score went down from 10 to 4.
Here's a sampling of my mistakes:
-15C2 = 210
- On the semicircle one found radius
-70*2 = 630
- I took the square root wrong in 6.
-And 2 dumb adding/multiplying mistakes.

The reason I didn't take this was because of dumb mistakes on the AMC. I could have easily qualified for the AIME if I just didn't make so many stuped mistakes! :(  But I'm only in 7th grade, so I have a lot more chances.

Wow, lots of ppl got 3 and 11 wrong. I actually got #3 right. I fell for the trap on 11 though...

One mistake, on the semicircle one (put radius instead of diameter).

Eh...3, seems like many people missed #3, put radius instead of diameter, and messed up the quadratic formula on number (i think it was) 6.

For the sake of not completely embarrassing myself, I'll just say I had 3 stupid mistakes and not name them. :-)

4 stupid mistakes on #'s 3, 5, 7, 13 (very stupid on 3/5)

1 very stupid mistake #1 300pi = 628
Don't ask for the rest of them...  :(  my score speaks for itself (2)

3 stupid mistakes
#6: let y=x-1 and found y
#7: had hypotenuse of right triangle as 12 and one side as $\sqrt{3}$ and found that the other side was $\sqrt{147}$
#11: what do you think???

Was it 13? Found number paths to (4,4) instead of (5,5).
And probably adding mistake in 12.

I missed my frosh chance too...

Somehow missed #12 by 1 because I forgot to include 2005  :(

hmm, at least 5 really stupid mistakes

1 mistake, same as last year, hopefully that won't hurt my USAMO chance,.  11 on AIME 240 index

2 stupid mistakes--for one I didn't know how to add and for the other I overlooked something, so it wasn't [i]that[/i] stupid.  Still sad though, it really put me behind this year since it was so easy.  Last year I think I had 2 also but it was much harder, so I had more leeway in terms of stupid mistakes.

Many.  Addition, overlooked cases, other arithmetic, bad drawings (eg my mind switched 'B' with 'C' on one of the problems) , forgetting that a cube has 12 edges, etc.
These brought my score from what should have been around 8 or 9 down to a 3.
Oh well, the other testees at my school all got 3's or lower too.

I could say I had stupid mistakes on all of the problems I missed (because looking back, none of them were completely confounding), but just counting stupid mistakes that I [i]really[/i] should not have made, I think I only had 2 of those (#3 I put 120 and #14 I had a minor algebraic error in an otherwise correct solution). I am happy that I only made this many though. It is a big improvement over last year.

Just 1--missed cubes of primes in #3--but that's not much to be proud of as I only solved 6 (counting #3) anyway.

0, which is suprising cause I always find a way to screw up on something on these tests.

I found out immediately after that I could have done 2 of the ones I didn't REALLY EASILY.  Count those if you like.

JB

just 1, on #13, i drew the diagrams and got 81. i forgot the one that went straight from (0,0) to (5,5). that's the most obvious one. that mistake caused be to leave out one solution.
score: 4 (+1 from last year)

Agh, for 13, I missed 4 obvious ones and got a 79 instead.

And for 8 got the 3rd power polynomial, and everything, but did some weird thingy, and didn't even guess because didn't get an integer.

I did make any computational type of careless mistakes, but I did solve 3 more questions easily after the test(#9, 12, 15, well #9 and #15 I knew exactly what to do during the test, but didn't do them cuz I got mental issues that day, #12 I didn't attempt). So I guess those were just as bad as careless mistakes.

0. I seldom make mistakes, but I have a LOT of problems that i can't solve no matter what i do.


geometry
3D geometry
rotation
geometric transformation
reflection
symmetry
pigeonhole principle

(1)
How many ways can the faces of a cube be colored with six colors?

(2)
How many ways can the cubic faces of a hypercube be colored with 8 colors?

(1) The answer is 30. But, how do you get it?

[quote="Christine"]But, how do you get it?[/quote]

http://www.artofproblemsolving.com/Forum/search.php

(Searching for the words "cube" and "color" works nicely.)

JBL,

The first question is easy to solve. How about the second question?

"How many ways can the cubic faces of a hypercube be colored with 8 colors?"

If by "colored with 8 colors" we meant that each color must appear exactly once, it's just as simple.  All we need to know is how many symmetries a hypercube has.  It has 8 cubic faces, each of which has 6 neighbors, so we can rotate it in $ 8 \cdot 6 \equal{} 48$ ways and the answer is $ \frac {8!}{48} \equal{} 840$.  If we consider two painted cubes the same if they are reflections of each other, this expands the symmetry group by a factor of 2 and shrinks the number of distinct colorings by the same factor.

More generally, for $ n > 2$, $ n$-cube has $ 2n$ $ n \minus{} 1$-dimensional faces, each of which has $ 2n \minus{} 2$ neighbors, so the symmetry group of rotations has $ 2n(2n \minus{} 2) \equal{} 4n(n \minus{} 1)$ members while the symmetry group with reflections has $ 8n(n \minus{} 1)$ members.  Thus the number of different ways to paint an $ n$-cube with $ 2n$ colors (each used once) is $ \frac {(2n)!}{4n(n \minus{} 1)}$ if we are interested in "different up to rotations" and is $ \frac {(2n)!}{8n(n \minus{} 1)}$ if we're interested in "different up to rotations and reflections."

JBL,

Did you use Ramsey's theorem or pigeonhole principle? If not, how can we use Ramsey's?

I can't even imagine how one would apply Ramsey's Theorem or the pigeonhole principle to this problem.

[quote="JBL"]It has 8 cubic faces, each of which has 6 neighbors, so we can rotate it in $ 8 \cdot 6 \equal{} 48$ ways[/quote]

No. Consider two SO(4)-rotations given by matrices
[code]1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1[/code]
and
[code]1 0 0  0
0 1 0  0
0 0 0 -1
0 0 1  0[/code]

The two rotations preserve both of the vectors (1,0,0,0) and (0,1,0,0). In particular, each of them preserves the position of the cubic faces centered at (1,0,0,0) and (0,1,0,0). This shows that specifying position of two neighboring cubic faces is not enough to determine a rotation of the hypercube.

Indeed, (the number of rotation of hypercube) = (the number of cubic faces) X (the number of rotations of cube).
The rotation group of the cube is isomorphic to the stabilizer subgroup of the rotation group of the hypercube at (1,0,0,0), for the same reason that SO(n) is isomorphic to the stabilizer subgroup of SO(n+1) at the north pole.

The number of rotations of the $ n$-dimensional cube, $ n\ge2$, is $ 4\cdot6\cdots(2n\minus{}2)\cdot(2n)$

Yes, of course, that's much better (i.e., not wrong).  An alternative way to count is to note that you may choose one vertex, select its location in $ 2^n$ ways, and arrange the vertices adjacent to it independently ($ n!$ ways).  This counts reflections, so we then have to divde by 2.

another way to count:

An $ n \times n$ matrix with $ \pm 1$ appearing exactly once in each row and in each column determines a rotation (including reflection) of $ n$ dimensional cube. There are $ n!$ ways to determine the places of nonzero entries in the matrix and then there are $ 2^n$ ways to determine the signs (+1 or -1) of the nonzero entries.

nice. My question would be the following:
What is the stabilizer of the vertex (1,1,1...,1,1) ?

cheers


Isogonal conjugate
geometry proposed
radical axis
geometry

Dear Luis,
nice result.
See for historical note
Stevanovic M. R., Symmedian as radical axis, Message Hyacinthos:
http://tech.groups.yahoo.com/group/Hyacinthos/message/10931
http://tech.groups.yahoo.com/group/Hyacinthos/message/10905
http://tech.groups.yahoo.com/group/Hyacinthos/message/10904
Sincerely
Jean-Louis

Thanks for the links Jean-Louis, I found this result very interesting and with lots of applications. There is also a couple of additional things that I would like to highlight.

Let $ P \ (u: v: w) \ , \ A' \ (0: v: w) \ , \ B' (u: 0: w) \ , \ C' (u: v: 0)$

Barycentric equations of circles $\odot (ABB')$ and $\odot (ACC')$  are then:

$ a^2yz \plus{} b^2xz \plus{} c^2xy \minus{} \frac {ub^2z}{u \plus{} w}(x \plus{} y \plus{} z) \equal{} 0$

$ a^2yz \plus{} b^2xz \plus{} c^2xy \minus{} \frac {uc^2y}{u \plus{} v}(x \plus{} y \plus{} z) \equal{} 0$

Thus, their radical axis $ AX$ has equation  $ AX \equiv \frac {c^2y}{u \plus{} v} \minus{} \frac {b^2z}{u \plus{} w} \equal{} 0$

Similarly, equations of $BY, CZ$ are given by:

$ BY \equiv \frac {c^2x}{u \plus{} v} \minus{} \frac {a^2z}{w \plus{} v} \equal{} 0 \ , \ CZ \equiv \frac {a^2y}{v \plus{} w} \minus{} \frac {b^2x}{u \plus{} w} \equal{} 0$

Therefore, $ AX,BY,CZ$ concur at $ J \left(\frac {a^2}{v \plus{} w}: \frac {b^2}{u \plus{} w}: \frac {c^2}{u \plus{} v} \right),$

which is the isogonal conjugate of the complement of $ P$

Additionally, if we consider the three centers of the spiral similarities that  takes the vectors $ \overrightarrow{BP},\overrightarrow{CP},\overrightarrow{AP}$ into the vectors $ \overrightarrow{CA},\overrightarrow{AB},\overrightarrow{BC},$ then the triangle formed by these three centers is perspective with $ \triangle ABC$ and the perspector is $ J.$

Further, the incenter of this triangle is the first Stevanovic point $X_{1130}$ of $\triangle ABC.$

Dear Luis,
sorry, I have enconter no name for that point.
Sincerely
Jean-Louis

My solution:

Let $ \ell_A $ be the bisector of $ \angle BAC $ .
Let $ Q \in AC, R \in AB $ such that $ BQ \parallel CP, CR \parallel BP $ and $ T=BQ \cap CR $ .
Let $ \Psi $ be the composition of inversion $ \mathbf{I} (A, \sqrt{AB \cdot AC})$ and reflection $ \mathbf{R} (\ell_A) $ .

Since $ AB \cdot AC=AB' \cdot AR $ ,
so $ R $ is the image of $ B' $ under $ \Psi $ .
Similarly we can prove $ Q $ is the image of $ C' $ under $ \Psi $ ,
so $ BQ, CR $ is the image of $ \odot (ACC'), \odot (ABB') $ under $ \Psi $, respectively ,
hence $ T  \equiv BQ \cap CR $ is the image of $ X \equiv \odot (ACC') \cap \odot (ABB') $ under $ \Psi $ . [b]...[/b] $ (\star) $

Since $ AT $ pass through the complement $ P^{*} $ of $ P $ WRT $ \triangle ABC $ ,
so from $ (\star) $ we get $ AX $ passes through the isogonal conjugate $ V $ of $ P^* $ WRT $ \triangle ABC $ .

Similarly we can prove $ BY, CZ $ pass through $ V $ ,
so we conclude that $ AX, BY, CZ $ are concurrent at $ V $ .

Q.E.D

Here is my proof without using barycentric coordinates:

Let $D$ be the point forming the parallalelogram $BPCD$ $\Longrightarrow$ $AD$ is the A-cevian of the complement of $P.$ Since $[DCB']=[DCP]=[DBP]=[DBC']$ $\Longrightarrow$ $\text{dist}{(D,AC)}:\text{dist}(D,AB)=BC':CB'.$ But since $X$ is center of the spiral similarity that swaps $\overline{CB'}$ and $\overline{C'B},$ we have $\text{dist}{(X,AB)}:\text{dist}(X,AC)=BC':CB'$ $\Longrightarrow$ $\text{dist}{(D,AC)}:\text{dist}(D,AB)=\text{dist}{(X,AB)}:\text{dist}(X,AC)$ $\Longrightarrow$ $AD$ and $AX$ are isogonals WRT $\angle BAC$ $\Longrightarrow$ $AX$ goes through the isogonal of the complement of $P$ and similarly $BY$ and $CZ.$

Another proof:

Let $ P_A, P_B, P_C $ be the reflection of $ P $ in the midpoint of $ BC, CA, AB $, respectively .

Since $ X $ is the center of spiral similarity that maps $ CP \mapsto AB $ ,
so $ \triangle XCP \sim \triangle XAB \Longrightarrow XC:BP_A=XC:PC=XA:AB $ . ... $ (\star)$
Since $ \angle P_ABA=\angle PCB+\angle CBP+\angle PBA=\angle B'XA+\angle CXB'=\angle CXA $ ,
so combine with $ ( \star ) $ we get $ \triangle CXA \sim \triangle P_ABA \Longrightarrow \angle BAP_A=\angle XAC $ .

Similarly we can prove $ \angle CBP_B=\angle YBA, \angle ACP_C=\angle ZCB $ ,
so $ AX, BY, CZ $ are concurrent at the isogonal conjugate of the complement of $ P $ WRT $\triangle ABC $ .

Q.E.D

Another solution:let $S$ reflection of $P$ in the midpoint of $BC$ then note that $\angle PXB=\angle CC'A=\angle CXA , \angle XBP=\angle XC'B=\angle XCA \longrightarrow \triangle PXB\sim \triangle CXA$ so $\frac{AC}{PB}=\frac{AX}{BX}$ $\longrightarrow \frac{AC}{CS}=\frac{AX}{BX}$ also $\angle ACS=\angle AB'B=\angle AXB$ so $\triangle ACS\sim \triangle AXB \longrightarrow \angle BAX=\angle CAS$ DONE

Let $D$ be the reflection of $P$ across $M$, the midpoint of $BC$, and let $E$ be the intersection of $CX$ and $BD$. Since $PC\parallel BD$,
\[
\angle XEB=\angle XCP=\angle XB'P=\angle XAB
\]
so $AB'XBE$ is cyclic. Now $B'B\parallel DC$, so by Reim's theorem $ACDE$ is cyclic. Then
\[
\angle XAB=\angle XEB=\angle CED=\angle CAD
\]
as desired.

[quote=Luis González]$P$ is an arbitrary point on the plane of $\triangle ABC$ and let $\triangle A'B'C'$ be the cevian triangle of $P$ WRT $\triangle ABC.$ The circles $\odot (ABB')$ and $\odot (ACC')$ meet at $A,X.$ Similarly, define  the points $Y$ and $Z$ WRT $B$ and $C.$ Prove that the lines $ AX,BY,CZ$ concur at the isogonal conjugate of the complement of $P$ WRT $ \triangle ABC.$[/quote]

Let $\Phi_1$ denote the $\sqrt{AB\cdot AC}$ Inversion followed by a reflection along the angle bisector of $\angle BAC$. Let $\Delta M_AM_BM_C$ denote the Medial triangle of $\Delta ABC$ against $A,B,C$ respecticely. Let $\{X^*,Y^*,Z^*\}$ denote the reflections of $P$ over $\{M_A,M_B,M_C\}$  respectively. Notice that $\Phi_1(B')=\overline{CX^*}\cap\overline{AB}$ and $\Phi_1(C')=\overline{BX^*}\cap\overline{AC}$. So, $\Phi_1((\odot(ABB')\cap\odot(ACC'))=\overline{C\Phi_1(B')}\cap\overline{B\Phi_1(C')}=X^*$. Similarly if $\Phi_2$ denote the $\sqrt{BA\cdot BC}$ Inversion with a reflection around the angle bisector of $\angle ABC$ and if $\Phi_3$ denotes the $\sqrt{CA\cdot CB}$ Inversion with a reflection around the angle bisector of $\angle ACB$ then $\Phi_2((\odot(BAA')\cap\odot(BCC'))=Y^*$ and $\Phi_3(\odot(CAA')\cap\odot(CBB'))=Z^*$ and as $\{AX,AX^*\},\{BX,BX^*\},\{CX,CX^*\}$ are Isogonal Conjugates. Hence it suffices to show that $AX^*,BY^*,CZ^*$ are concurrent at the Complement of $P$ WRT $\Delta ABC$. Notice that $\{\Delta ABC,\Delta X^*Y^*Z^*\}$ are Homothetic. Hence, $AX^*,BY^*,CZ^*$ are Concurrent at a point $P^*$ on the Homothetic center of $\{\Delta ABC,\Delta X^*Y^*Z^*\}$. Now the Pairwise Homothetic Perspectors of the triangles $\{\Delta ABC,\Delta M_AM_BM_C,\Delta X^*Y^*Z^*\}$ must be collinear $\implies\{\overline{P-G-P^*}\}$ where $G$ is the Centroid of $\Delta ABC$. Now notice that $\Delta ABC\stackrel{-}{\cong}\Delta X^*Y^*Z^*$. Hence, $P^*A=P^*X^*$; $P^*B=P^*Y^*$ and $P^*C=P^*Z^*$. Now applying Menelaus Theorem on $\Delta PP^*X^*$ cut by the transversal $\overline{AM_A}$ we get that $-1=\frac{AM_A}{M_AG}\cdot\frac{GK}{GL}\cdot\frac{P^*X^*}{X^*A}\implies GP=2GP^*\implies\{P^*\}$ is the Complement of $P$ WRT $\Delta ABC$. $\blacksquare$


ratio
limit
real analysis
real analysis unsolved

I was looking at the following problem from the GRE math subject practice test.
The set of real numbers x for which the series $ \sum_{n = 1}^{\infty}\frac {n!x^{2n}}{n^n(1 + x^{2n})}$ converges is

(A) {0}
(B) {$ x: - 1 < x < 1$}
(C) {$ x: - 1 \leq x \leq 1$}
(D) {$ x: - \sqrt {e} \leq x \leq \sqrt {e}$}
(E) $ \mathbb{R}$

Using the ratio test $ \lim_{n\to\infty}\frac {\frac {(n + 1)!x^{2(n + 1)}}{{n + 1}^{n + 1}(1 + x^{2(n + 1)})}}{\frac {n!x^{2n}}{n^n(1 + x^{2n})}}$ = $ \lim_{n\to\infty}\frac {(n + 1)!n^nx^{2n + 2}(1 + x^{2n})}{n!{n + 1}^{n + 1}x^{2n}(1 + x^{2n + 2})}$ = $ \lim_{n\to\infty}\frac {n^nx^2(1 + x^{2n})}{n^{n + 1}(1 + x^{2n + 2})}$ = ${ \lim_{n\to\infty}(\frac {n}{n + 1})^n\lim_{n\to\infty}\frac {x^2(\frac {1}{x^{2n}} + 1)}{\frac {1}{x^{2n}} + x^2}}$ = $ \lim_{n\to\infty}(\frac {n}{n + 1})^n$ but what is this limit equivalent to?

Good :) The limit corresponds to $ \frac{1}{e}$ I believe.

Seen as $ (\frac{n\plus{}1}{n})^{\minus{}n} \equal{} ((\frac{n\plus{}1}{n})^{n})^{\minus{}1})$ = $ e^{\minus{}1}$.

$ \left(\frac{n}{n\plus{}1}\right)^n\equal{}\left[\left(1\plus{}\frac1n\right)^n\right]^{\minus{}1}.$ Does that help?

Of course, you might want to reconsider how you've handled the powers of $ x$ in that ratio test.

The limit you're looking for is indeed $ \frac {1}{e}$, but it's not the right limit, and in any case using the ratio test in this case isn't really necessary.  Notice that $ 0\leq \frac {x^{2n}}{1 + x^{2n}} < 1$ for all $ x\in \mathbb{R}$ and $ n\in \mathbb{N}$.  So really what the question is asking is whether or not $ \sum _{n = 1}^\infty{\frac {n!}{n^n}}$ converges.  If it does, the answer must be $ x\in \mathbb{R}$.  It's not hard to show that it does indeed converge (for example estimate that for sufficiently large $ n$ you have $ n! < \left( \frac {n}{2}\right)^n$).

It seems the math GRE likes to ask relatively easy questions which, if attacked without further thought using the standard textbook approach, can be very time-consuming.  Since time is an important factor on that test, it might be a good idea to spend a few moments and look to see if a problem could be simplified before expending precious minutes on unnecessary algebra.

You don't have to try a billion cases to check when $ n! < (\frac{n}{2})^n$ either. When $ n\equal{}6$, it is easy to see $ 720 < 3^6 \equal{} 729$. :)

to see that only E is possible put x=2
using the rootcriterium you get
(n!^(1/n) /n) * 4/(1+4^n)^(1/n) < (n!^(1/n) /n) < 1/2


In the triangle below, angle $ C$ is a right angle. What is the number of inches in the length of $ \overline{AB}$ in simplest radical form?
[asy]draw((0,0)--(.866,0)--(.866,.5)--cycle);
label("$30^\circ$",(0,0),3ENE+2E);
label("$A$",(0,0),W);
label("$6''$",(.433,0),S);
label("$B$",(.866,.5),NE);
label("$C$",(.866,0),SE);[/asy]

Use 30-60-90 triangle ratios.  BC=6/sqrt(3)=2sqrt(3), so AB=4sqrt(3).


Solve the inequation:
$ 5x^2\minus{}20x\plus{}26>\frac{4}{x^2\minus{}4x\plus{}5}$                                   (Exam 2006 U.Politehnica,Bucharest)

We are to solve $ 5t\plus{}1>\frac{4}{t}$ for $ t\geq 1$, where $ t\equal{}...$


logarithms
calculus
calculus computations

hello, your sum does converge, it is $ \minus{}\zeta(1,\frac{4}{3})$.
Sonnhard.

For large enough $ n$, say $ N$, $ \ln n < Cn^{\frac{1}{6}}$ so
$ \\ \begin{eqnarray*}
\sum_{n = 1}^{\infty}{\frac{\ln n}{n^{\frac{4}{3}}} &=& \sum_{n = 1}^{N-1}{\frac{\ln n}{n^{\frac{4}{3}}} + \sum_{n = N}^{\infty}{\frac{\ln n}{n^{\frac{4}{3}}}
\\&<& \sum_{n = 1}^{N-1}{\frac{\ln n}{n^{\frac{4}{3}}} + \sum_{n = N}^{\infty}{\frac{Cn^{\frac{1}{6}}}{n^{\frac{4}{3}}}
\\&=& \sum_{n = 1}^{N-1}{\frac{\ln n}{n^{\frac{4}{3}}} + C\sum_{n = N}^{\infty}{\frac{1}{n^{\frac{7}{6}}}
\end{eqnarray*}$

The first sum is finite (the negative term at $ n = 2$ doesn't matter) and the second is a convergent p-series, hence convergent by the comparison test.


The units digit of 3^1 is 3
3^2 is 9
3^3 is 7
3^4 is 1
3^5 is 3
3^6 is 9
3^7 is 1
clearly there is a pattern, it repeats every four
because 2004/4 is 501 r 0, there will be a units digit of 1
Answer: 1


1. In an experiment, there are 3 cards: Card W is white on both sides, card R is red on both sides, card B is red on one side and white on the other side.

If you see one of these cards with a red side showing, what is the probability that it's red on the other side too?

2. A man leaves for work at a random time b/w 7:00 AM and 8:00AM. His paper arrives at his house at a random time b/w 6:30AM and 7:30 AM. What is the probability he has his newspaper before he leaves for work?

Do I use geometric areas for this one? If so, how do I set my region up?

3. At a county fair, a game is played by tossing a dime onto a large table divided into a grid made up of 2 cm squares. If the coin lands entirely inside one of the squares w/o touching any line, you win a prize. What are your chances of winning on your first toss? (the diameter of a dime is 1.8cm)

Is the answer simply the area of the dime divided by the area of a square?

Thanks.

[hide]
1. Well:3 ways to pick side of a card so that it is red. 2 ways for the other side to be red (either side of red card). So probability is $\frac(2/3)$.

2. Well: picture it like this

6:30                 7:00                      7:30                         8:00
_____________________________
l                        l                                l
l____________l________________l
                         ________________________________
                         l                                l                                l
                         l________________l________________l

The probability they'll both arrive in the middle region is 1/4. The probability the newspaper will arrive before the man is 1/2. So answer is 1/8.

3. Think of the locus.[/hide]

sorry if this is too late
[hide="thoughts on #3"]
i think it would help if you thought of the probability in terms of the dime's center. In the 2 cm X 2 cm grid, the region that makes it possible for the coin to satisfy the conditions is only a 0.2 X 0.2 region:
therefore the probability should be $\frac{(0.2)^2}{(2)^2}=\frac{1}{100}$
[/hide]
the image looks better if you click and enlarge, sorry

Thanks for the replies guys, but can someone please explain #2 again? I don't understand the timeline.


quadratics
modular arithmetic
symmetry
complex numbers
algebra
quadratic formula

[hide]There should be an easier way, but this was my way:

If we let $ x=r \cis \theta$, then
\[ r^2 \mathrm{cis} 2 \theta + r \mathrm{cis} \theta y + y^2=0.\]
Using the quadratic formula to solve for $ y$, we have
\[ y=\frac{-r \mathrm{cis} \theta \pm \sqrt{r^2 \mathrm{cis} 2 \theta - 4r^2 \mathrm{cis} 2 \theta}}{2}=\frac{-r \mathrm{cis} \theta \pm \sqrt{3}i r\mathrm{cis} \theta}{2}\]
Further simplifying:
\[ \frac{r \mathrm{cis} \theta}{2} \left(-1 \pm \sqrt{3}i\right)=r \mathrm{cis} \theta \mathrm{cis} \frac{2\pi}{3}=r \mathrm{cis} \left(\theta + \frac{2\pi}{3}\right).\]

Plugging in these values of $ x$ and $ y$ into the expression, we find that
\[ \left(\frac{x}{x+y}\right)^{2008}+\left(\frac{y}{x+y}\right)^{2008}=\boxed{-1}.\][/hide]

[hide="Solution"]Let $ S_n \equal{} a^n \plus{} b^n$. Observe that $ S^n \equal{} (a \plus{} b) S_{n \minus{} 1} \minus{} (ab) S_{n \minus{} 2}$

In the case of $ a \equal{} \frac x {x \plus{} y}$ and $ b \equal{} \frac y {x \plus{} y}$, we have: \[ a \plus{} b \equal{} \frac x {x \plus{} y} \plus{} \frac y {x \plus{} y} \equal{} \frac{x\plus{}y}{x\plus{}y} \equal{} 1
\] \[ ab \equal{} \frac {xy} {(x \plus{} y)^2} \equal{} \frac {xy}{xy \plus{} (x^2 \plus{} xy \plus{} y^2)} \equal{} \frac{xy}{xy} \equal{} 1
\]  \[ \therefore S_n \equal{} S_{n \minus{} 1} \minus{} S_{n \minus{} 2}
\]
Using initial conditions $ S_0 \equal{} 2$ and $ S_1 \equal{} 1$, we find that the sequence repeats like $ 2,1, \minus{} 1, \minus{} 2, \minus{} 1,1, \ldots$ for $ n \equal{} 0,1,2,3,4,5 \ldots$

$ 2008 \equiv 4 \pmod{6} \implies S_{2008} \equal{} S_4 \equal{} \minus{} 1$[/hide]

[hide="Discussion"] The condition and the desired value share the same symmetry, which is homogeneity, so we can divide out by $ x$ and assume without loss of generality that $ x \equal{} 1$.  This gives $ y \equal{} \omega$ (a primitive third root of unity) and

$ \frac{x}{x \plus{} y} \equal{} \frac{1}{1 \plus{} \omega} \equal{} \frac{1}{\minus{}\omega^2} \equal{} \minus{} \omega$

and

$ \frac{y}{x \plus{} y} \equal{} 1 \minus{} \frac{x}{x \plus{} y} \equal{} 1 \plus{} \omega \equal{} \minus{} \omega^2$.

These are primitive sixth roots of unity.  The result follows. [/hide]

[quote="vinskman"]$ x,y$ are non-zero complex numbers and  $ x^2 \plus{} xy \plus{} y^2 \equal{} 0$

Find the value of

$ \left (\frac x {x \plus{} y}\right)^{2008} \plus{} \left(\frac y {x \plus{} y}\right)^{2008}$
 :roll:[/quote]
[hide]
 $ x\not \equal{} y, (x\minus{}y)(x^2 \plus{} xy \plus{} y^2) \equal{} 0\iff x^3\equal{}y^3\iff(\frac x y)^3\equal{}1$
$ (x  \plus{} y)^2 \equal{} x^2 \plus{} xy \plus{} y^2\plus{}xy \equal{} xy$

$ \therefore \left (\frac x {x \plus{} y}\right)^{2008} \plus{} \left(\frac y {x \plus{} y}\right)^{2008}\equal{}\left(\frac{x} {y}\right)^{1004}\plus{}\left(\frac{y} {x}\right)^{1004}$

$ \equal{}\left(\frac{x}{y}\right)^{3\times334\plus{}2}\plus{}\left(\frac{y}{x} \right)^{3\times334\plus{}2}\equal{}\left(\frac{x}{y}\right)^2\plus{}\left(\frac{y}{x} \right)^2\equal{}\minus{}1$

[/hide]


Consider any covering of a $ 6*6$ board with $ 18$ dominoes.Show that for any such layout of dominoes
It is possible to cut the board into two pieces so that  the cut does not pass through a domino

Suppose than each vertical or horizontal cut cuts at least one domino.
Then it must cut at least two, because the number of cut dominoes is
always even. Since we have 10 cuts and less than 20 dominoes this is
impossible.

[quote="shasha"]Then it must cut at least two, because the number of cut dominoes is
always even.[/quote]
When only 1 domino is cut, then the bord is devided into two parts, with both parts having an even number of squares (one of the sides is always 6).
But with 1 halved domino each part has a number of dominoes + 1/2 domino. Each domino covers 2 sqares, so the number of covered squares is odd for both parts. But that number has to be even.
A conflict, so the numbers of cut dominoes can never be 1 (or odd).

http://www.mathlinks.ro/viewtopic.php?t=138743

[hide="my solution"]consider one row. the number of horizontal dominos in each row must be even. there is at least one horizontal domino in row 1-2, row 2-3, ..., row 5-6. so the minimum number of horizontal dominos in each row 1-2, ..., 5-6 is 2, totaled 10 horizontal dominos. similarly, there are at least 10 vertical dominos. so there must be at least 20 dominos. however, there are only $ \frac{6\cdot6}{2\cdot1}\equal{}18$ dominos. contradiction, hence our proof is complete.[/hide]


I invented a new google game!

Maybe you've heard of googlewhack, where you enter two words and try to come up with one result. In "Goggle", you take one word and misspell it in different ways, and see if you can come up with one result.

For example -- "salamander"

Salamander = 2,490,000 results
Salmander = 41,300 results
Salamnder = 1,750 results
Slamander = 523 results
Slmander = 48 results
[b]Slemander = 2 results! [/b](both from the same page)

Try it out, see if you can beat that!

EDIT-- A 1-result search is a "true" goggle if, when the word is spelled correctly in its sentence, it makes sense.
For example, "hecksagone" came from the following sentence: "...Then with a wrench un-screew the main fitting ( it is the hecksagone shaped piece above the hose cover)..."

somebody has too much free time on their hands...

Or you could try to get the typo with the [i]most[/i] results.

For example, protien = 531,000

hhhhhhhhhhhhhhhello: 1

I suppose that's cheating, though :D

hitcthike: 1 result
That was only on my second try  :!:

I think buzzer11 has the only TRUE goggle so far... It wasn't found on a 'wordlist' website (Where they intentionally misspell the word in 200 different ways) and if it WAS spelled right on that page, it would make sense.

Wow this is cool. I've never invented a successful game before. :D

bush - 292 000 000
bushy - 2 100 000
bussy - 1 530 000
buss - 3 760 000
pushy - 1 920 000
push - 121 000 000
puss - 1 850 000
pussy - 10 300 000
I got an octoplet of all above the million :)

btw, kkeyboard as misspell of keyboard has no matches at all :)

833,000 results for "gogle"
923,000 results for "googel"

MithsApprentice - 21,400

*gives up*

 ;)

:P supercalifragilisticexpialidociuus has 2 results from the same page.

[quote="MithsApprentice"]MithsApprentice - 21,400

*gives up*

 ;)[/quote]
there are actually only 11 results.

[quote="jmadsen"]there are actually only 11 results.[/quote]

Lol. Most of them are just links to AoPS posts. :-P But they still count.

[quote="jmadsen"][quote="Treething"]pr0n has over 2 million results.[/quote]
That doesn't count because it's misspelled on purpose.[/quote]

<_<
>_>

some people dont have lives :huuh:  :huh:  :ewpu:

[quote="nonie"]231, 000 results for nonie
284 for my full name
3 relevant ones![/quote]

Really? 3? Last time I looked I found 1 for you...

But my name has no irrelevant results and I have ~150 results for me.

osylloscope- 1 result! Or it says 1- 1 of about 100; I'm not sure why it says that.

refurunce- 1 result!

:D oooooh! I'm good!

1-1 of 100? Weird. Google must have a glitch.

lololololololololololololololololololololololololololololololololololololololololololololololololololololol 2 results.  there's no correlation between number of ol's tacked on to the end and the number of search results...

antidisestablishmentharism (from antidisestablishmentarism...): 0 RESULTS!!!!!

Maybe disqualified, though... :blush:

[quote="samath"]This isn't really relevant, but try searching "failure" on google.  Check out the first item.[/quote]

This is the result of a bug in Google (known as the GoogleBomb), which lets people who know about it make Google display results it ordinarily wouldn't.  In case you're interested, the original GoogleBomb phrase was "Talentless Hack", but I think the GoogleBombed website has been removed from the #1 site now.

insaynly for insanely

Also for the most results of a mispelled word. Wierd - 3,610,000

[quote="samath"]This isn't really relevant, but try searching "failure" on google. Check out the first item.[/quote]
Note the second item as well- both sides are capable of googlebombing. The rest of the top ten includes some legitimate hits (heart failure society!) and a Googleblog entry referring to this very search.

[quote="samath"]Did you expect antidisestablishmentharism to be there?
I wouldn't be surprised if the word spelled correctly was there...[/quote]

No. That's why I put it here.

Eviletes   it works, not miss speled and only one result :ninja:

[quote="Drunken_Math"][quote="buzzer11"]I got 1 result for hecksagone (from hexagon)!!!!!!!!!!!!!!!!!!  :D[/quote]

The funny thing is that NOW there's 2 topics!

The new link is one leading to this topic!  :rotfl:[/quote]

Interesting. I'm trying rubbecs (Rubik's), which currently has no hits. Maybe it will have one after I post this.


integration
combinatorics unsolved
combinatorics

Hi does any one know how to solve the following equation?

y(x)={1/(Sqrt[x] Exp[x])} -{ Erfc[Sqrt[x]]}

Solve for x in terms of y

Can we do it by using some software like Mathematica?

What is Erfc¿

And if I look at the first term, I think only numerical solutions are possible.

[quote="ZetaX"]What is Erfc¿[/quote]
Mathworld helps.  $\text{Erfc}(z)=\frac{2}{\sqrt{\pi}}\int_{z}^\infty e^{-t^{2}}\text{ d}t$.


trigonometry
limit
calculus
calculus computations

sigma 1 to infinite (1-cos(1/k))
any ideas?

$ 1 \minus{} \cos x$ looks like what power of $ x$ for $ x$ near zero? That is, can you find an exponent $ m$ such that $ \lim_{x\to0}\frac {1 \minus{} \cos x}{x^m}$ exists and is not zero?

If you can answer that, then you'll have a limit comparison test argument for the convergence or divergence of the series.

Can you answer the first part of the hint?

well i graphed 1- cos x
and it kind of looked like x^2 near zero

You can confirm that by finding $ \lim_{x\to0}\frac{1\minus{}\cos x}{x^2}.$

[hide="Answer."]$ \frac12$[/hide]
Now that you have that, you know that $ 1\minus{}\cos\left(\frac1k\right)$ looks like a constant times $ \frac1{k^2}$ for large $ k.$ What can you do with that?

sorry, i'm still confused on why we found the limit of 1-cosx/ x^2 and i am confused on your last post also.

Gmac525, Professor Kent is trying to show you that you can use a convergence test known as limit comparision test. As an hint, you were asked to find

$ \lim_{x \to 0} \frac {1 \minus{} \cos x}{x^2} \equal{} \frac {1}{2}$.

Now if you make $ x \equal{} \frac {1}{k}$ you get

$ \lim_{k \to \infty} \frac {1 \minus{} \cos\left(\frac {1}{k}\right)}{\frac {1}{k^2}} \equal{} \frac {1}{2}$.

Note that in this last limit we are in fact comparing your series with the series of general term 1/k^2. Now, according to the limit comparision test I referred above, as the sum $ \sum_{k \equal{} 1}^{\infty} \frac {1}{k^2}$ converges and as the limit above is a number, then also your series converges.


Hola,

Veran, usualmente cuando alguien propone un problema de ecuaciones como esta $2^m=3^n+5$, pues casi siempre hay solo una solucion. Sin embargo en este caso eso no es asi (traten y veran, hay dos muy claras), sin embargo, mi pregunta es algo distinta, y no creo que sea facil (honestamente no se la solucion)

Dados tres enteros $p,q,r$ todos primos relativos tal que la ecuacion $p^n=q^m+r$ tiene al menos una solucion, bajo que condiciones tiene exactamente una solucion.

Es solo algo que se me ocurrio. Alguna idea?

Ha sido casi un mes, y aun parece que nadie tiene una sola idea al respecto! Alguien lo quiere intentar?


Denote $\zeta=e^{\frac{2\pi i}3}$. Then
\[e^{x}=u+v+w \]

\[e^{\zeta x}=u+\zeta v+\zeta^{2}w \]

\[e^{\zeta^{2}x}=u+\zeta^{2}v+\zeta w \]
I'm not sure, but maybe
\[1=e^{x}\cdot e^{\zeta x}\cdot e^{\zeta^{2}x}\]
is what U need.
P.S. Yes, it is.

Differentiate $u,\ v,\ w$ with respect to $x,$ we have $u'=w,\ v'=u,\ w'=v.$ Thus using this, we are to prove that $\frac{d}{dx}(u^{3}+v^{3}+w^{3}-3uvw)=0.$

This is in Stewart's calculus book, in one of the "Problems Plus" sections. He probably had what kunny is suggesting in mind.


geometry
ratio
calculus
calculus computations

if tangent at point P on y=x^3 insersects the curve at Q also, let A be area of region bounded by curve and line segment PQ.  Let B be the area of the region defined in the same way starting with Q instead of P.  Show how are A and B related.

kunny was right; the answer is $\frac AB=\frac1{16}$.
Here's a way to get that answer, without even calculating $A$ or $B$.

Let $P=(p,p^3)$, $Q=(q,q^3)$, $R=(r,r^3)$, where $R$ is the other intersection of the tangent from $Q$. Because the curve $y=x^3$ has no $x^2$ term, the three roots of $x^3=y=ax+b$ must add to zero, and therefore $2p+q=0$ and $2q+r=0$.
Consider the linear transformation $(x,y)\mapsto(-2x,-8y)$. The curve $y=x^3$ is mapped to itself, and tangents are mapped to tangents by the linearity. Also, $P\mapsto Q$ and $Q\mapsto R$, so region $A$ goes to region $B$. It is easy to see that this transformation multiplies areas by a factor of 16, so the area of $B$ is 16 times the area of $A$.

If you were to calculate the actual areas, the area of $A$ would be $c\cdot p^4$ for some constant $c$, and the area of $B$ would be $c\cdot q^4$. The ratio of 16 comes from the fact that $q=2p$.

nice dude
thanks
algebra has always been  a nemesis


geometric inequality
geometry
convex quadrilateral
IMO
IMO 1989

Let $ ABCD$ be a convex quadrilateral such that the sides $ AB, AD, BC$ satisfy $ AB \equal{} AD \plus{} BC.$ There exists a point $ P$ inside the quadrilateral at a distance $ h$ from the line $ CD$ such that $ AP \equal{} h \plus{} AD$  and $ BP \equal{} h \plus{} BC.$ Show that:
\[ \frac {1}{\sqrt {h}} \geq \frac {1}{\sqrt {AD}} \plus{} \frac {1}{\sqrt {BC}}
\]

Let $\omega$ be a circle of radius $h$ with center $P$.  Let $\omega_1$ be a circle of radius $AD$ with center $A$.  Let $\omega _2$ be a circle of radius $BC$ with center $B$.  Note that all three circles are externaly tangent to each other.  Also, $\omega$ is tangent to $CD$.  Let this tangency point be $E$.
In order to satisfy the problem, there must exist a tangent to $\omega$ through $E$ such that the tanget to $\omega$ at that point intersects both $\omega _1$ and $\omega _2$ thus forming $D$ and $C$.
The smallest radius of $\omega _1$ occurs when $CD$ is tangent to both $\omega _1$ and $\omega _2$.  If $\omega _1$ is any smaller, then when tangent $EC$ is drawn, it is always below $\omega _1$, contradiction.
Using Pythagorean's Theorem, we can find that $\frac{1}{\sqrt{h}}=\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}$.
Since $AD$ is greater than or equal to the already determined value for it, we get the desired inequality.


What is the sum of the reciprocals of all the positive divisors of
8? Express your answer as a mixed number.

Yeah, I got that too. It said it was wrong... It said 17/8 huh


How do I initialize  a signed long long array in C?

Thanks

what do u mean by init? do u want to use memset? if not then u can do it by for loop!

I mean to declare the variable into memory so it can be used as an array in the program.

long long a[];

but it will not work on VC++.

and it will not work on all the compilers as well. if u have cygwin or gcc compiler then it is OK.

another ez tip, use long at ur computer and while submitting to the judge just paste another long ;)

hmm, well that was what i had but it wasnt the problem, how do i format it for output?
fprintf(out, "%lli\n", flag[val]);
This truncates the value down and
fprintf(out, "%lld\n", flag[val]);
doesnt work either

%lld will work. but as i said it will not work in all the compilers. but the judge computer will get it right.

Or why not use the "evil" iostream? :)

It should be OK if you don't print/read very much... :P

Otherwise you can check if __GNUC__ is defined, like:

#ifdef __GNUC__
printf("%lld\n", number);
#else // VC++
printf("%I64d\n", number);
#endif


MIT
college
email
analytic geometry
Stanford
Harvard
Princeton

Hey :)  Any mathlinkers/AoPSers that have applied to MIT? A friend told me that the decisions are given next saturday:-SS U know something different?
Moreover, all the decisions will be given in a few weeks...I am a little scared :maybe: What about u?

You can find on their site http://www-math.mit.edu/
I know at least one member learning at that :D

I'd imagine they'll be released in the next week since they're usually out on Pi Day, or just about.  They only say "mid to late march" on their site, so the date's flexible.

They say they're sending out letters March 16th and emails are to be sent out the 17th.

The MIT Online Decisions site is on.  In 24 hours, more or less, we will find the decisions... :| I am extremely anxious :roll: I think I'll make a hole in the floor :rotfl:

Hey guys,

I am as anxious as you are(maybe even more :blush: ); they say that decisions will come out on 17 march(tomorrow) at noon 12:00 Eastern time :huh: .

Could anyone explain what means exactly "Eastern time"? Maybe time in Tokyo? :blush:

Thanks!

"Eastern time" means the time observed in eastern cities in the United States such as New York City or Washington, DC. The United States has several different time zones: from east to west they are Eastern, Central, Mountain, Pacific, and a few more that exist only in Alaska (with one including Hawaii). Usually announcements of national events in the United States are given for one time zone, naming the time zone, and people who live in other places convert the time themselves. Eastern Time is Greenwich Mean Time (GMT) minus five hours.

Good luck Tiks !:) Let's hope we will get good news :)  BTW, I think you will find this helpful http://www.time.gov/timezone.cgi?Eastern/d/-5/java

[quote="DPatrick"][quote="tokenadult"]Eastern Time is Greenwich Mean Time (GMT) minus five hours.[/quote]Right now I believe it is GMT -4 hours.  (We moved up an hour last weekend, but GMT never changes.)[/quote]

I was trying to remember that adjustment for daylight saving time before I posted. I see from a convenient Web page

http://www.timetemperature.com/tzus/time_zone_codes_us.shtml

that of course you are correct. I was misled by my computer's clock-setting program. Thanks for making sure that AoPS participants in Europe and elsewhere know when they can access admission results online.

I am waiting for the MIT decision as well :).

The whole household is nervous for my brother's big decision..

[quote="MIT decisions website"]
We know that applying to college is stressful, and that the closer you get to receiving your admissions decisions, the easier it becomes to let that stress consume you. We encourage you to acknowledge it, embrace it, and then let it go. This is your last semester of high school, and your primary responsibility is to enjoy every remaining minute of this journey before you embark on the next.

Think about how you answered question 3a on our application (“what do you do purely for fun?”). Try to fill your days with that, and decisions will be here before you know it.[/quote]

I'll be on a distance run tomorrow when it comes.  I'm planning on joggling it, and calling my brother to check it for me.  Wish everyone the best!

9 hours left :ninja:  I think I'll enjoy the morning for the moment :)

Hey Ciprian, do you know anyone else who has applied to MIT from Romania, besides you?Maybe any member of Romanian team for IMO 2006?

And good luck to you too and to other Mathlinkers awaiting the decisions :) !

P.S. 8 hours and 8 minutes left :D .

[quote="SuperThread"]
1) At MIT, how socially exclusive are the fraternities? Do you think that joining a frat is necessary?
[/quote]

From my experience, fraternities here aren't very socially exclusive at all.  As far as open parties go, they definitely want a good turnout so there should be little problem with that respect.  For closed parties (i.e. open to just people in the fraternity and their friends) you'll have to know someone in the frat to be able to attned.  But parties are a relatively small part of what makes fraternities appealing to live at.  I personally had never attended a party at a fraternity before I joined.  I find it unlikely that someone would hang out at a certain fraternity on a consistent basis and not end up joining it.  But anyway, it's not necessary to join...

[quote="SuperThread"]2) For those of you who got into top-tier colleges: what was your GPA from high school? How much do you think that USAMO achievement, etc, helped you?[/quote]
a) At my high school, we had a weighted GPA system where you'd get +1.0 in any honors or AP class so long as you got a C or better. On that system, my GPA at graduation was something like 4.3 or 4.4, and I was 9th or 13th or something in a class of just under 500.
b) I never qualified for USAMO, and I still got into Princeton. But I still think it helps that I was active in competitive academic extracurricular activities, and I am sure "USAMO achievement" (whatever that means) can't hurt.

I mean, how much do you think (say) an honorable mention on the USAMO would be worth? How about just Blue MOP?

Also -- is it feasible for me to get into MIT with a 3.8ish GPA? My school, for some insane reason, only gives +.3 for Honors and AP (and it doesn't start "Honors" until a way up the ladder -- you need to be in Precalc or above to get "Honors" for math, for example).

That last post by PTynan89 was helpful, because it addresses a lot of issues that students worry about.

Another point to consider: often, when colleges (especially selective ones) look at students' high school GPA's, the first thing they generally do is to un-weight them.  Therefore, your school's weighting system should not have any impact on whether or not you get in.

Having just gone through the college admissions process, most of it felt very arbitrary.  My best friend got rejected by Yale, but accepted by Harvard.  We both got rejected by a safety school (same school).  All my teachers were shocked when I was waitlisted by MIT, and there was no one reason that it happened, that's just the way it happened.

The lesson I learned was don't pin your hopes in one place.  Find lots of schools you really like.  I only found four, and only got into one.  Just one rejection away from a disaster, but it all turned out all right in the end.  Rejection stings, but almost everyone at my school ended up someplace they're happy with, including me.

[quote="SnowStorm"][quote="PTynan89"]Dude, I never made USAMO.

I was very close to the real cutoff, though, last and this year, and also 2nd highest index in DE last year, and 3rd this year.

But... I did make Physics Team last and this year.

And silver medal in USAMTS last year, gold this year.

My school also only does .3 weights for AP, .2 honors.

So, my GPA was between 4.0 and 4.1.

I got deferred from MIT and Caltech EA, but got into both regular.

I also got into Princeton, and was initially waitlisted by Harvard, but just found out today that I got in.

My class rank is 36.

However, I was the only one from my school to get into MIT, and also the only one to get into Princeton.

One other guy got into Caltech from my school, and one girl from my school also got into Harvard, so there were 2 acceptances for each.

Thus, despite being 36th out of about 230 in my class, I got into all 4 of those schools, whereas no one else from my high school got into even 2.

I hope this doesn't sound braggy, but I just want to make sure that I truly emphasize this point.

Although, I should admit, that my school does a really funny ranking system.

So taking a half year study hall in 9th grade, and then a half-year one last year together probably killed my rank by ~ 30 places.

Additionally, I should note that my grades had an UPWARD trend, whereas the majority of the people ahead of me had either a constant or downward trend.

I hope this was helpful. =)[/quote]

I know this is kind of taboo.......but
Just curious, what is your ethnicity?[/quote]

Oh, I'm white.

For the record, I think Physics Team (the physics equivalent to making MOSP) is what got me in.

Also, I got rejected by UPenn, which I considered to be one of my safeties.

It seems like the ideal student for safety schools (like state schools) is different from the ideal student for elite schools. I have noticed that elite schools seem to focus more on getting people that are very passionate and accomplished in a few areas, whereas other "safety" schools are more about very well-rounded students.
In my case I was accepted to MIT and Caltech, whereas I didn't get a presidential scholarship at Georgia Tech, which was a fair deal easier (percentage wise) than getting admitted to MIT or Caltech.

Can anybody confirm my observation, or is it not entirely true?

There may be some truth to drunner's observation but I wouldn't generalize too far.  Put yourself in the admissions officer's shoes at your 'safety' school.  You have some nice scholarship money to give to a group of "elite" students at your institution.  Do you try to get the students with the best possible academic credentials to fill these slots? Not necessarily.

In your admissions officer role, suppose you see an application that screams "I'm really a good fit at a place like MIT or Caltech and would really like to go to one of those places but I've also applied to you because they might not accept me."  Do you offer this person one of your scholarships?

Remember, your job is to maximize the desirability of the incoming class subject to many constraints that might be imposed.  Desirability might be defined as a function of average test scores, average HS GPAs, acceptance rates, yield (percentage of accepted students that enroll,) future retention & college graduation rates, geographic/ethnic diversity, and other criteria used to market the institution.  It will also be subject to many constraints that might be imposed by the state or individual programs (music, athletics,...) at the institution.  Yield, retention, and graduation rates are functions of how happy the incoming students are likely to be with the services that your institution can provide.  If there's any hint in the application that a student views the institution as a 'safety' and may not be happy at the institution, why admit them - or give them free rides?  They'll probably just lower your yield, retention, and graduation rates.

One of the worst mistakes an admissions officer can make is to recruit top students with full-ride scholarships who end up so disappointed that they bad-mouth the institution and make future recruiting difficult.  Admissions officers are acutely aware of the limitations of their schools and learn quickly to avoid such mistakes.  It might be a useful exercise to think about ways that potential admissions mistakes are identified.

College applicants often assume that there is a "peck-order" among institutions and that if one is accepted by a school high on the peck-order, then he or she will be accepted by all "lesser" colleges.  College admissions just don't work this way.  drunner's experience is more the norm than the exception.  It's unwise to use any school with restrictive admission standards (say, less than 30% acceptance rates) as a "safety."

When you enter the job market, you'll find that employers use similar logic.  Being "over-qualified" for a position does not bode well for your chances of being hired.  Employers go to great lengths to avoid "hiring" mistakes.  Think about how this can be done.  Predicting how happy someone will be at the job is critical to these processes.

Excellent post GT59!!!

[b][i]GT59 QUOTE: “but I wouldn't generalize too far”[/i][/b]

Why stop short? Let me be controversial – I TOTALLY WOULD!

[b][i]GT59 QUOTE: “If there's any hint in the application that a student views the institution as a 'safety' and may not be happy at the institution, why admit them - or give them free rides? They'll probably just lower your yield, retention, and graduation rates”[/i][/b]

My feeling is that the above is the only relevant truth! I’ll be brutal and even suggest that the eventual happiness of the enrollees does not factor much in the admissions officers’ decision.

Hence the term “Tufts Syndrome”, which has been made especially (in)famous by Washington University at St Louis.

[b][i]GT59 QUOTE: “College applicants often assume that there is a "peck-order" among institutions and that if one is accepted by a school high on the peck-order, then he or she will be accepted by all "lesser" colleges. College admissions just don't work this way. drunner's experience is more the norm than the exception. It's unwise to use any school with restrictive admission standards (say, less than 30% acceptance rates) as a "safety.”[/i][/b]

Take the above as the golden rule -- EDITED/ADDED: except as [i][b]noted by jmerry below, "a lot of state schools use purely formula-driven admissions"[/b][/i].

Many LACs (e.g. Harvey Mudd) are in that category.
------

Further Takes:

A way to increase one’s odd is by applying Early Decision (ED), which obviously limits one's option completely. Or one can apply Early Action (EA) if available, but that may not be good enough to sway the decision of those “safety” schools’ admission decision! Perhaps a very personalized application that shows tremendous intent to enroll if admitted, just falling short of ED, might sway such strategic rejection. That again might only help if the college in question only offers EA but not ED.

Also, though I cannot prove this (yet), I also suspect that the method of “Wait-Listing” is used by even the top-top colleges to improve their chance to land the most desirable applicants without jeopardizing their yield rates. I am saying that, if a most desirable applicant gets wait-listed by a top-top college, an immediate and vigorous protest may very well get the applicant off the wait-list right away. When one does that, I assume one truly wants to go to that wait-listed college.

My experience: I was waitlisted by Princeton- and I'm pretty sure that was because I said in the interview that my first choice was Berkeley. If I hadn't said that- they would have admitted me, and I would have rejected them. As it was, they left me out of the admit list, but gave me a chance to reconsider.

In the realm of "safety schools", a lot of state schools use purely formula-driven admissions, at least in the first pass. When I went there, Berkeley chose about half of its freshman class by simply taking the top scores in their grade/SAT based index. Those places won't reject you unless you have a real hole somewhere, and they won't mind too much if an overqualified student goes elsewhere.


oh use these for the last 2 steps
6th step
R U2' R2' U' R2 U' R2' U2 R (only cross)
R U2 R' U' R U R' U' R U' R' (onyl cross)
R U R' U R U2' R' (jims fishy)
R U2 R' U' R U' R' (anti-sune)
R2' D R' U2 R D' R' U2 R' (theone that pisses me off)
l' U' L U R U' r' F (opposite above)
# l' U' L' U R U' L U (x') (double fish)

last layer
R2' U R U R' U' R' U' R' U R'
R2 U' R' U' R U R U R U' R
R U' R' U D R' D U' R' U R D2' F
r2 R2' U r2 R2' U2 r2 R2' U r2 R2

managed to get down to just under a minute last summer.  im gonna try and learn fridrichs F2L when I have a few weeks off  :D

59 seconds... NICE. im averaging about 47

I was inspired by this topic to go find my rubiks cube and spent a day or two practicing and finally broke a minute... 10 times and brought my average to 65, pr of 55!!

cool! 1 min isn't bad. took me a long time

well 1 min decent, but i think its easy to get to if you know a couple tricks. I get there and it takes me 20 seconds to place the first 4 edges...!!! (i know it is awful)

55-60 s

been cubing for a little over 4 weeks.

[quote="NeverOddOrEven"]i average in the 30-40s.[/quote]

wow nice. i got 37 when i was  lucky... skipped 1 step

uh basic? with some more intermediate algos

f2L with crappy LL. (like 3 algs not inc mirrors)

im working on the f2l algos... but i know the pll stuff a little bit


Where could I download a free (if impossible I can just use my pirate skills) program that has a range of equations that I can do? I'm looking for advanced year 7 to standard year 8 (First and second years of high school for Aus) mathematics.

I've searched but cannot find anything :(

Can you explain what you mean? You just want practice equations?

You could trying googling for some algebra worksheets.


Yes, it is. Why do you ask?.. it is trivial to show.


Find the maximum "k" so that the following inequality holds for $ a,b,c \geq 0$:
\[ a^3\plus{}b^3\plus{}c^3\plus{}3abc \cdot \left(\frac{3abc}{a^3\plus{}b^3\plus{}c^3}\right)^k \geq a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)
\]

I am just posting this to learn how to find best constant. I hope someone will solve this one.

Thanks

 :)

[quote="shaaam"]Find the maximum "k" so that the following inequality holds for $ a,b,c \geq 0$:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \cdot \left(\frac {3abc}{a^3 \plus{} b^3 \plus{} c^3}\right)^k \geq a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)
\]

[/quote]
If $ b\equal{}c\equal{}1$ then, $ k\leq0.$ :wink:

[quote="arqady"]
If $ b \equal{} c \equal{} 1$ then, $ k\leq0.$ :wink:[/quote]

Can you please explain why $ b \equal{} c \equal{} 1$ forces $ k$ to be negative? I have also tried counterexamples.
for $ k \equal{} 1$ it is obviously wrong. $ k \equal{} \frac {1}{2}$ is also wrong. What if $ k \equal{}\frac{1}{10}$? I mean, if $ k$ is very small positive number, then inequality might be true. And my question is: Is there such fixed constant?

Thank you for the attention.

[quote="shaaam"][quote="arqady"]
If $ b = c = 1$ then, $ k\leq0.$ :wink:[/quote]

Can you please explain why $ b = c = 1$ forces $ k$ to be negative? [/quote]
If $ b=c=1$ then,
$ a^3+b^3+c^3+3abc \cdot \left(\frac{3abc}{a^3+b^3+c^3}\right)^k \geq a^2(b+c)+b^2(c+a)+c^2(a+b)\Leftrightarrow$
$ \Leftrightarrow a^3+2+3a\left(\frac{3a}{a^3+2}\right)^k\geq2a^2+2a+2\Leftrightarrow a^2+3\left(\frac{3a}{a^3+2}\right)^k\geq2a+2,$
which is wrong for $ k>0.$ Try $ a=0.$ :wink:

Thanks Arqady. It was great explanation. I should have noticed it before  :oops:.

How about the following?
Let $ A \equal{} a^2b \plus{} b^2c \plus{} c^2a$ and $ B \equal{} ab^2 \plus{} bc^2 \plus{} ca^2$. Find the maximum "$ k$" , so that the following inequality holds:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq \left(\frac {A}{B}\right)^k \cdot A \plus{} \left(\frac {B}{A}\right)^k \cdot B
\]
For $ k \equal{} 0$ we get Schur's inequality. For $ k \equal{} 1$ we get Can_Hang's Inequality at [url]http://www.mathlinks.ro/viewtopic.php?t=234161[/url]

[quote="shaaam"]
Let $ A \equal{} a^2b \plus{} b^2c \plus{} c^2a$ and $ B \equal{} ab^2 \plus{} bc^2 \plus{} ca^2$. Find the maximum "$ k$" , so that the following inequality holds:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq \left(\frac {A}{B}\right)^k \cdot A \plus{} \left(\frac {B}{A}\right)^k \cdot B
\]
[/quote]
Try $ c \equal{} 0$ and you will see easily by AM-GM that it holds iff $ \minus{} 2\le k\le 1$.  :)
(it all boils down to $ x^3 \plus{} 1\ge x^{2 \plus{} k} \plus{} x^{1 \minus{} k}$ :) , and by this you can also see, putting $ f(k): \equal{}\left(\frac {A}{B}\right)^k \cdot A \plus{} \left(\frac {B}{A}\right)^k \cdot B$, that $ f(1)\equal{}f(\minus{}2)\ge f(k)$ for $ k\in[\minus{}2,1]$)


to find the value of x in e^2x-4e^x-12=0

don't tell me how to do it, like give me some hints, like do i do the FOIL method or may be move the 12 to the other side

You have two variables and one equation so I don't think you can solve completely... You can find integer values and what not...

[hide]

factorise the quadratic

[/hide]

i believe hes refering to e as the number not e as a variable

[hide="BIG hint"]Solve the quadratic for e^x then use natural logs[/hide]


Srry, I posted this in the Suggestions forum.

Anyhow...

Sometimes when I post reply's and don't put any smilies in my reply (my sig doesn't have smilies), the server tells me that I have posted too many smilies when I didn't even post any. I don't know if it's just me or has this happened to someone else. Also, this happen atleast once a day...

I hope you can tell me whats wrong...

Does this just happen if you answer to posts by moderators¿
Their smiley limit is deactivated.

No, it happens also to people who post problems. It will say "Error. Too many smileys in this post. Maximum allowed in this post: 3 You have:5"

I'm curious as to what are you trying to post exactly. Can you email me the contents of the post that's being rejected?

okay, i sent it to you. I hope you can tell me whats going on...

Until further notice this is classified as pilot error.


Given an arbitrary triangle, can it be translated, rotated and scaled such that all three vertices lie on the rectangular hyperbola xy=k??

can it be only translated and rotated such that all three vertices lie on the rectangular hyperbola xy=k??

having no idea how to do this, i'm curious to see a solution to this.. anyone?

[quote="bugzpodder"]Given an arbitrary triangle, can it be translated, rotated and scaled such that all three vertices lie on the rectangular hyperbola xy=k??[/quote]

Are we to consider both arcs of that equation, or just the part in the first quadrant?


How to prove
$ \frac {\pi^2}{6} \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2}}$
in many ways?

Have a look  here : [url]http://pictor.math.uqam.ca/~plouffe/OEIS/archive_in_pdf/zeta2.pdf[/url]

This is Basel Problem.

http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf


vector
MATHCOUNTS
AMC
AIME
Pythagorean Theorem
geometry

Hi, people.

Here are some super-duper beginning problems.

A motor boat travels 10 km/h relative to the water. The water below it flows also at a rate of 10 km/h. What is the boat's velocity relative to the shore when it heads upstream? Downstream?

Another one:

How much horsepower does a 2200-pound car generate that runs at a velocity of 22 feet/second? NOTE: Horsepower formula (Distance traveled (feet) X weight of object (pounds) / (Time (seconds) X 550 (constant))

Last one (Bit difficult. Has to do with vectors)

A boat travels between to banks of a river at a sped of 10 km/h relative to the water. The river itself also flows at a rate of 10 km/hr. What 's the velocity of the boat relative to its starting point on shore? (Remember to name the velocity, NOT THE SPEED! Velocity adds one new concept)

Requires some physics knowledge here, though. Are you "intelligent" enough?

Just being an ignoramus like usual, but what's the difference between speed and velocity?

[quote="LynnelleYe"]but what's the difference between speed and velocity?[/quote]

Speed is scalar ie it only has magnitude (eg 60km/h). Velocity is a vector ie it needs both direction and magnitude to be described (eg 60km/h going north).

[quote="marucho8d7"]Hi, people.

Here are some super-duper beginning problems.

A motor boat travels 10 km/h relative to the water. The water below it flows also at a rate of 10 km/h. What is the boat's velocity relative to the shore when it heads upstream? Downstream?[/quote]

Upstream - 20 km/hr
Downstream - 0 km/hour

[quote="marucho8d7"]Another one:

How much horsepower does a 2200-pound car generate that runs at a velocity of 22 feet/second? NOTE: Horsepower formula (Distance traveled (feet) X weight of object (pounds) / (Time (seconds) X 550 (constant))[/quote]

88 horsepower

[quote="tokenadult"]The last one belongs on the Intermediate forum. It's an interesting problem.[/quote]

i guess so. i thought the pythagorean theorem wasn't so intermediate.

[quote="marucho8d7"]Last one (Bit difficult. Has to do with vectors)

A boat travels between to banks of a river at a sped of 10 km/h relative to the water. The river itself also flows at a rate of 10 km/hr. What 's the velocity of the boat relative to its starting point on shore? (Remember to name the velocity, NOT THE SPEED! Velocity adds one new concept)

Requires some physics knowledge here, though. Are you "intelligent" enough?[/quote]

in one hour it will have travelled 10 km perpendicular to the river and 10km down the river. The total displacement is :sqrt:(102 + 102) = 10 :sqrt: 2

So it's velocity is 10 :sqrt: 2 km h-1 at an angle of pi/4c to the river.

Haha, just don't reply to his post anymore.

I tutor him for our Jr. High's MathCounts program....

TOTAL PAIN IN THE ***

[quote="h_s_potter2002"][quote="soulzmischief"]STOP POSTING YOUR SCIENCE HOMEWORK FOR OTHER PEOPLE TO SOLVE FOR YOU.

Don't attempt to act "intelligent" or anything by posting problems straight from your science textbook.[/quote]

Man, I feel so stupid now.  I didn't realize that this was the guy who also posted the "Sig Fig" problems.[/quote]

are you saying that im stupid?

[quote="tokenadult"]The last one belongs on the Intermediate forum. It's an interesting problem.[/quote]

Really.  Unfortunately, I think that the Intermediate forum has been constantly eroded over the space of several months now.

This is really a trivial problem.  There's nothing complex about it, you draw a diagram, you add vectors tip to tail, you're done.

If this keeps going on, the Getting Started forum will only be suitable for middle schoolers and below, the Intermediate Forum will only be suitable for non-advanced high schoolers, the Pre-Olympiad forums will continue to get random, generally non-interesting (not non-difficult, note) and students who are fairly advanced (the kind who score ~120 on the AMC 12 and ~5 on the AIME, who make up a considerable amount of the users on this forum) will have to go to the Olympiad forums.

I say this because it seems like the Pre-Olympiad forums just doesn't get good problems posted in it very often.  I found that the old Intermediate Forum, as it was about 3-4 months ago, had much more interesting problems in it, that were suitable for most high schoolers.

[quote="marucho8d7"]sorry. and that wasnt hw

so ive figured your true identity. your sister told me

really? i suppose that's because v***** is in the club. you cant just say that im a pain. im not (well, sorta. i get a bit annoying on forums. ive got to stop.

someone delete this topic. im going to die. STOP[/quote]

You CAN delete in this forum

Now I know how you got so many posts so fast...

WWMIII

[quote="h_s_potter2002"][quote="marucho8d7"]Hi, people.

Here are some super-duper beginning problems.

A motor boat travels 10 km/h relative to the water. The water below it flows also at a rate of 10 km/h. What is the boat's velocity relative to the shore when it heads upstream? Downstream?[/quote]

Upstream - 20 km/hr
Downstream - 0 km/hour[/quote]

Shouldn't that be 0km/h and 20km/h for Upstream and Downstream, respectively?


I've seen people use that balls and urns thing a lot...what is it? Or rather, what is it used for and how is it derived?

Huh?   :huh: Yo no comprendo.  Could you explain it in caveman terms for me?

let X's be balls and |'s be urns

then you have ||XXXXXXXXXX. 10 balls and 2 urns to separate them into 3 groups. with a total of 12 spots, 12C2 is the total number of solutions.

:what?: OMG!!!  How did you get 2 urns?  Aren't there 3 variables? :?:

His wording is bad, | doesn't represent an urn, it represents an urn divider. For example you have like

URN|URN|URN, which has 3 urns, and we need two dividers in order for them to be seperated.

[quote="CircleSquared"]Balls and urns is used when you have know the sum, and want to find some ordered pairs.  For example:

Find all ordered pairs of non-negative integers (a,b,c) such that a+b+c=10.
Let each number be represented by a number of balls.  The total number of balls will be 10.  Because of that, we can just insert 2 dividers into there to separate the "numbers".  Now the number of ordered pairs is just the number of ways we can insert 2 dividers amongst 10 balls, so it's $\binom{12}{2}$.

Hope that's clear.[/quote]
Wouldn't it be $\binom{11}2$ because there are 11 places for the dividers to go?

[quote="i_like_pie
Wouldn't it be $\binom{11}2$ because there are 11 places for the dividers to go?[/quote]

Well, there are 12 objects, 10 balls, and 2 dividers.

Out of those 12 objects, there are $\binom{12}{2}$ ways we can choose 2 to be the dividers.

Hope that clarifies everything.  Either that, or I've misunderstood balls and urns for a very long time.

Oh... Yes, that makes sense. Thanks for clarifying. :)

[quote="Math Geek"]What if you were to have a coin problem that asked how many ways could you use any amount and type of coins to get a certain amount of money.  Could you use balls and urns for that, too?

For example, how many combinations of [u]coins [/u]could you have to get $5.35? ($1 coins don't count).[/quote]
I suggest that you don't use such a flamboyant font and color.
Anyways, you can't use balls and urns for that because of complications such as coin value and not knowing the number of urns.
However, you can use [url=http://www.artofproblemsolving.com/Wiki/index.php/Generating_functions]generating functions[/url]

[quote="Math Geek"]What if you were to have a coin problem that asked how many ways could you use any amount and type of coins to get a certain amount of money.  Could you use balls and urns for that, too?

For example, how many combinations of [u]coins [/u]could you have to get $5.35? ($1 coins don't count).[/b][/i][/color][/size][/quote]

I think that in MATHCOUNTS, the numbers are small enough so that you could just count all of the cases.

[quote="diophantient"][quote="Math Geek"][size=150][color=blue][i][b]What if you were to have a coin problem that asked how many ways could you use any amount and type of coins to get a certain amount of money.  Could you use balls and urns for that, too?

For example, how many combinations of [u]coins [/u]could you have to get $5.35? ($1 coins don't count).[/quote]
I suggest that you don't use such a flamboyant font and color.
Anyways, you can't use balls and urns for that because of complications such as coin value and not knowing the number of urns.
However, you can use [url=http://www.artofproblemsolving.com/Wiki/index.php/Generating_functions]generating functions[/url][/quote]
wait, I dont really get it, could you do it with generating function... or would this be high school level?

Edited out all that size 18 bold blue font.

1. Blue is my color.
2. Don't use other colors, fonts or all-bold text unless there is a VERY specific purpose.  (y'know, like, when people do the [color=red]Orange[/color] [color=blue]White[/color] [color=indigo]Yellow[/color][color=green]Blue[/color] thing...but that should be saved for G&FF anyways).

Thanks.


limit
trigonometry
ratio
function
calculus
calculus computations

It oscillates. More formally,

$ \lim _{n \to \infty} \sin n \equal{} \lim _{m \to \infty} \sin \left(m \pi \plus{} \frac {\pi}{2} \right)$

For successive integer values of $ m$, the ratio is $ \minus{} 1$, so it fails the ratio test.

$ \lim_{x\to \infty}(\sin(x))$ can be transformed into: $ \lim_{t\to 0}(\sin(\frac{1}{t}))$

Let $ f(t): = \sin(\frac{1}{t})$ for $ t \neq 0$.
Define $ t_{n}: = \frac{1}{n\pi}$ and $ s_{n}: =\frac{1}{\frac{\pi}{2} + 2\pi n}$.
Then $ \lim(t_{n}) = \lim(s_{n}) = 0$
But $ \lim(f(t_{n})) = 0 \neq \lim(f(s_{n})) = 1$. Hence, $ \lim_{t\to \0}(\sin(\frac{1}{t}))$ does not exist. By Divergence Criteria Theorem.

[quote="math227"]prove $ \lim_{n\rightarrow\infty} \sin{n}$ does not exist.[/quote]

Since $ \sin$ is a periodic function, then limit as $ n$ approaches $ \plus{}\infty$ apparently does not exists.

I think the intention of the original problem was that $ n$ be an integer.  In that case, we need a different proof from the ones given so far.

I think I assumed $ n$ to be an integer in my proof, is there something wrong with that proof?

Yes: your $ n$ is not the same as the original poster's $ n$.

Let $ I_1, I_2, \cdots$ be intervals given by

\begin{eqnarray*}
I_1 & = & \left[ \tfrac{\pi}{4}, \tfrac{3\pi}{4}\right]\\
I_2 & = & \left[ \tfrac{5\pi}{4}, \tfrac{7\pi}{4}\right]\\
I_3 & = & \left[ \tfrac{9\pi}{4}, \tfrac{11\pi}{4}\right]\\
I_4 & = & \left[ \tfrac{13\pi}{4}, \tfrac{15\pi}{4}\right]\\
\vdots & & \vdots
\end{eqnarray*}

Since the length of each interval is strictly greater than 1, we can find an integer $ n_k \in I_k$ for all $ k = 1, 2, \cdots$.

Then $ \sin n_k \geq \frac{1}{\sqrt{2}}$ for all odd $ k$ but $ \sin n_k \leq -\frac{1}{\sqrt{2}}$ for all even $ k$.

Therefore, $ \lim_{n\to\infty} \sin n$ does not exists.

In fact, the set $ \{ \sin n \, | \, n \in \mathbb{N}\}$ is dense in $ [-1, 1]$. For any $ r \in [-1, 1]$, we can choose a strictly increasing sequence $ n_k$ of positive integers so that $ \sin n_k$ converges to $ r$.


linear algebra
matrix
linear algebra unsolved

Suppose $ T \in L(V)$ is self-adjoint, $ \lambda \in F$, and $ \epsilon \succ 0$. If there exists $ v \in V$ such that $ \parallel{}v\parallel{} \equal{} 1$ and $ \parallel{}Tv\minus{} \lambda v \parallel{} \prec \epsilon$ then T has an eigenvalue $ \lambda'$ such that $ |\lambda \minus{} \lambda'| < \epsilon$.

We assume that $ V$ is an euclidean space (in particular a real space of finite dimension and then $ \lambda\in\mathbb{R}$) and $ \parallel{}.\parallel{}$ is it's euclidean norm; Let $ \mu$ be a $ T$-eigenvalue such that $ \forall{\alpha_i}\in{spectrum(T)},|\mu\minus{}\lambda|\leq|\alpha_i\minus{}\lambda|$. In a certain orthonormal basis $ (e_i)_i$ the $ T$-matrix is $ diag(\alpha_1,\cdots,\alpha_n)$; if $ v\equal{}\sum{x_ie_i}$ then $ \sum(\alpha_i\minus{}\lambda)^2x_i^2<\epsilon^2$. Therefore  $ \sum(\mu\minus{}\lambda)^2x_i^2<\epsilon^2$ and $ |\mu\minus{}\lambda|<\epsilon$.  QED


Show that if $ X_{0}$ is a particular solution of the system $ AX\equal{}B$. Then $ X$ can be written as: $ X\equal{}X_{0}\plus{}Y$, where Y is a solution of $ AX\equal{}0$.

[quote="danilorj"]Show that if $ X_{0}$ is a particular solution of the system $ AX \equal{} B$. Then $ X$ can be written as: $ X \equal{} X_{0}\plus{}Y$, where Y is a solution of $ AX \equal{} 0$.[/quote]

Let $ X_{1}$ and $ X_{0}$ be particular solutions. Then, $ AX_{0}\equal{}B$ and $ AX_{1}\equal{}B$. So $ AX_{0}\minus{}AX_{1}\equal{}0\implies A(X_{0}\minus{}X_{1})\equal{}0$. But $ X_{0}\minus{}X_{1}\equal{}Y$ because they solve the homogenous equation so $ X_{0}\equal{} X_{1}\plus{}Y$.

This thinking is explicitly taught concerning linear differential equations. It's exactly the same principle: the linear operator is a differential operator rather than a matrix, but it's all the same argument.


calculus
derivative
calculus computations

find the derivative of
$ y \equal{} \sqrt{1 \plus{} x \plus{} \sqrt{2x \plus{} \sqrt{3x}}}$

$ y' \equal{} \left(\frac{1}{2\sqrt{1 \plus{} x \plus{} \sqrt{2x \plus{} \sqrt{3x}}}}\right)\left(1 \plus{} \left(\frac{1}{2\sqrt{2x \plus{} \sqrt{3x}}}\right)\left(2 \plus{} \frac{3}{2\sqrt{3x}}\right)\right)$

thanks to anyone who reviews this


induction
inequalities
function
inequalities unsolved

seems to be a good candidate for induction :D
i'm trying anyway :)

Good problem. And we no need induction to solve it!

First of all rewrite inequality as
(a1-a2)/(1+a1a2)+...+(an-1-an)/(1+an-1an) \leq (a1-an)/(1+a1an).

Each item can be represented as (1/ak+1-1/ak)*1/(1+1/akak+1).

Let bk be (1/ak+1-1/ak)/(1/an-1/a1). Then \sum bk=1 and bk is nonenegative (k=1,2,...,n-1).

Function 1/(1+1/x) is concave function. So
\sum bk*1/(1+1/akak+1) \leq 1/(1+1/(\sum bkakak+1)) = 1/(1+1/a1an) !!!

Another possible approach. Let a_k=tg x_k, with 0 \leq x_n \leq ... \leq x_1 \leq  \pi /2. Then the inequality can be written tg(x_1-x_2+x_2-x_3+...+x_n-1-x_n) \geq tg(x_1-x_2)+...+tg(x_n-1-x_n), which is true since tga+tgb \leq tg(a+b) for 0 \leq a,b \leq  \pi /2

Both the solutions are splendid .
I have some developments of this problem :

1. Let a 1  \geq a2 \geq a3 \geq ... \geq an>0
Prove that :
a1/(a2+1/a1)+a2/(a3+1/a2)+a3/(a4+1/a3)+...+an/(a1+1/an) \leq a1/(an+1/a1)+a2/(a1+1/a2)+a3/(a2+1/a3)+...+an/(an-1+1/an)                 (1)

2.Let a1 \geq a2 \geq a3 \geq ... \geq an \geq  \sqrt 3/3
Prove that :
 \sqrt a1/(a1+1/a2)+ \sqrt a2/(a2+1/a3)+ \sqrt a3/(a3+1/a4)+...+ \sqrt an/(an+1/a1)  \geq   \sqrt a1/(a1+1/an)+ \sqrt a2/(a2+1/a1)+ \sqrt a3/(a3+1/a2)+...+ \sqrt an/(an+1/an-1)              (2)

I must use induction to solve these problem.
I hope somebody will find another way to solve them.

Hello, Tung Lam!
Try to use html tags like n next time.

Well here is the version I'm using: $\sum w_{i}f(x_{i})\leq f\left(\sum w_{i}x_{i}\right)$ under the condition $\sum w_{i}=1$ is equivalent to:
\[\sum y_{i}f(x_{i})\leq \left(\sum y_{i}\right) f\left(\frac{\sum y_{i}x_{i}}{\sum y_{i}}\right)\] for suitable $y_{i}$.
Notice $\sum (a_{i+1}-a_{i})=a_{n}-a_{1}$

[quote="me@home"]Well here is the version I'm using: $\sum w_{i}f(x_{i})\leq f\left(\sum w_{i}x_{i}\right)$ under the condition $\sum w_{i}=1$ is equivalent to:
\[\sum y_{i}f(x_{i})\leq \left(\sum y_{i}\right) f\left(\frac{\sum y_{i}x_{i}}{\sum y_{i}}\right) \]
for suitable $y_{i}$.
Notice $\sum (a_{i+1}-a_{i})=a_{n}-a_{1}$[/quote]

any abitrary $y_{i}$? or $\sum y_{i}=1$


Find the integral solutions to 19x^3-84y^2=1984

[hide="One line"] We have $ y^{2}\equiv-1 \bmod 19$, but $ \left( \frac{-1}{19}\right) =-1$.  No solution. [/hide]


let $n$ be a natural number and is prime number such that
$p \mid n^3 -1$ and $n \mid p-1$
prove that :
$4p-3$ is a perfect square.

i'm pretty sure i've seen this posted before


induction
inequalities
function
logarithms
algebra unsolved
algebra
floor function

[quote="Beat"]$x>0$ and $n$-natural number. Prove that $[nx]\geq \sum_{i=1}^{n} \frac{[nx]}{n}$, where $[t]$ denotes the biggest integer number, which is $\leq t$[/quote]

But that's $1 \geq 1$. I think you mistyped it... $i$ must appear in the $RHS$

[quote="Beat"]$x>0$ and $n$-natural number. Prove that $[nx]\geq \sum_{i=1}^{n} \frac{[nx]}{n}$, where $[t]$ denotes the biggest integer number, which is $\leq t$[/quote]
I think it's $[nx]\geq \sum_{i=1}^{n} \frac{[ix]}{n}$.
Maybe we can use the formula $[x_1+x_2+\cdots+x_n]\geq [x_1]+[x_2]+\cdots+[x_n]$ to prove this. :D

[quote="leepakhin"][quote="Beat"]$x>0$ and $n$-natural number. Prove that $[nx]\geq \sum_{i=1}^{n} \frac{[nx]}{n}$, where $[t]$ denotes the biggest integer number, which is $\leq t$[/quote]
I think it's $[nx]\geq \sum_{i=1}^{n} \frac{[ix]}{n}$.[/quote]
It should be $[nx]\geq \sum_{i=1}^{n} \frac{[ix]}{i}$,isn't it?Since $n\times[nx]\geq \sum_{i=1}^{n} {[ix]}$ seem trivial...
It seem that I saw this question before,in a set of book called "xiao cong shu"(leepakhin should know what I mean),but I forget the detail of the proof,I only remember it is using Mathematical Induction

No, leepakhin's version should wrong, since it's also trivial...
The only one remaining seems to be $[nx]\geq \sum_{i=1}^{n} \frac{[ix]}{i}$.

Sorry, it`s :$[nx]\geq \sum_{i=1}^{n} \frac{[ix]}{i}$

And leepakhin why is this true:$[x_1+x_2+\cdots+x_n]\geq [x_1]+[x_2]+\cdots+[x_n]$ and where can I find more formulas and lemas like this one for [t].

Sorry for guessing the R.H.S. wrongly without thinking. :oops:
For the correct version, induction kills it immediately. :)

[quote="Beat"]And leepakhin why is this true:$[x_1+x_2+\cdots+x_n]\geq [x_1]+[x_2]+\cdots+[x_n]$ and where can I find more formulas and lemas like this one for [t].[/quote]
This is a generalization of the inequality $[x+y]\geq [x]+[y]$. The proof goes as follows.
$[x+y]=[[x]+[y]+\{x\}+\{y\}]=\begin{cases}[x]+[y]+1, &\{x\}+\{y\}\geq 1 \\ [x]+[y], &\{x\}+\{y\}<1\end{cases}$
(I think you know $\{x\}: =x-[x]$)

This is one of the basic properties of the Gaussian function $[x]$. I'm sorry that I don't know where to find them, except in Chinese books, so I am posting some of them below.

[u][b]Properties[/b][/u]
$[x+y]\geq [x]+[y]$
$\{x+y\}\leq \{x\}+\{y\}$
$[n+x]=n+[x]$ where $n$ is an integer
$[-x]=\begin{cases}-[x]-1, \text{when x is not an integer} \\ -[x], \text{when x is an integer}\end{cases}$
$[\frac{x}{n}]=[\frac{[x]}{n}]$ where $n$ is an integer

[u][b]Theorem[/b][/u]
Let $n$ be a positive integer and $p$ a prime. The highest power that $p$ divides $n!$ is
$\sum_{k=1}^\infty [\frac{n}{p^k}]$
(In fact, the exact bound is $[\log_{p} n]$)


Fie 2 nr a<b si o functie derivabila f:[a;b]->R.
Sa se arate ca exista si e unica o functie polinomiala de grad mai mic sau egal decat 3 si are proprietatea ca functia F: R->R, definita prin

                     F(x)=    f(x) , daca x apartine [a;b]
                                 P(x) , daca x nu apartine [a;b]
este derivabila.
Ramane adevarat daca gradul <3 ?

.......................

luand cazuri particulare pt f(x) no problem
in general please help

Fie $p(x) = p x^{3}+q x^{2}+r x+s$.

Noi trebuie sa rezolvam un sistem de forma $p(a) = u, \, p(b) = v, \, p^\prime (a) = w, \, p^\prime (b) = t$, i.e.:

$p a^{3}+q a^{2}+r a+s = u$;

$p b^{3}+q b^{2}+r b+s = v$;

$3 p a^{2}+2 q a+r = w$;

$3 p b^{2}+2 q b+r = t$.

Considerand acest sistem in variabilele $(p,q,r,s)$, matricea asociata lui este \[A = \left( \begin{matrix}a^{3}& a^{2}& a & 1 \\ b^{3}& b^{2}& b & 1 \\ 3 a^{2}& 2a & 1 & 0 \\ 3b^{2}& 2b & 1 & 0 \end{matrix}\right) .\] Ca sa-l calculam pe $\det \, A$, scadem prima linie din a doua, apoi desfacem dupa ultima coloana. Pana la urma obtinem (cred) \[\det \, A =-(a-b)^{4},\] care e $\neq 0$ pentru $a \neq b$, deci $A$ e inversabila.

Rezultă că $\left( \begin{matrix}p \\ q \\ r \\ s \end{matrix}\right) = A^{-1}\cdot \left( \begin{matrix}u \\ v \\ w \\ t \end{matrix}\right)$.

[quote="abedhu"]Ramane adevarat daca gradul <3 ?[/quote]

Bineinteles ca nu. Consideram un polinom oarecare de grad $3$ si potrivim o functie ca-n problema. Daca ar mai exista si o alta solutie (de grad $\leq 2$), atunci $\textrm{dim}\left( \ker \, A \right) > 0$, contradictie cu faptul ca $A$ e inversabila.

problema consta in a gasi un polinom de grad cel mult 3 cu proprietatile :
$P(a)=f(a)$
$P(b)=f(b)$
$P'(a)=f'(a)$
$P'(b)=f'(b)$.
acest fapt este adevarat deoarece :
pentru inceput, evident $P(a)=f(a)$, altfel $F$ nu e continua in $a$.
fie $x$ real, diferit de $a,b$. atunci evident $F$ e derivabila in $x$, din constructia lui $F$.
pt $x=a$ avem
$\lim_{x\nearrow a}\frac{F(x)-F(a)}{x-a}=\lim_{x\nearrow a}\frac{P(x)-P(a)}{x-a}=P'(a)$. ($P$e polinom deci e functie derivabila)$\lim_{x\searrow a}\frac{F(x)-F(a)}{x-a}=\lim_{x\searrow a}\frac{f(x)-f(a)}{x-a}=f'(a)$. ($f$ e derivabila din ipoteza).
conditia ca $F$ sa fie derivabila este ca aceste doua derivate sa fie egale, adica chiar $P'(a)=f'(a)$.
fie acum $P=ux^{3}+vx^{2}+wx+t$. (unii coeficienti pot fi $0$).
atunci conditiile de mai sus sunt un sistem liniar in $(u,v,w,t)$, cu 4 ecuatii si 4 necunoscute.
cum functia $f$ apare doar in coloana termenilor liberi, determinantul sistemului este doar in functie de $a$ si $b$.
se arata ca determinantul este nenul, si deci solutia este unica. (l`am calculat pe foaie, dar nu mai are rost sa scriu)
daca $P$ ar fi de grad <3, atunci sistemul ar avea doar trei necunoscute si patru ecuatii, deci nu ar avea intotdeauna solutie.
deci $P$ trebuie sa fie de grad $\leq 3$, si solutia e unica.

edit : perfect_radio mi`a luat`o inainte, dar a postat in timp ce scriam, si am observat abia dupa ce am postat.

:blush:  multumesc mult nu ma astepatam sa fiti atat de promti


Divide through $z^2$, then you are looking for the rational solutions of $x^2+xy+y^2=1$.
Now assume $y=ax+1$ with rational $a$ and solve for $x,y$, giving solutions.
Geometrically we intersected the line with the circle, and since the first intersection point of the line is $(0,1)$, the second one will be rational too. Conversly, every rational point $\neq (0,-1)$ belongs to such $a$. Thus we found all solutions.

Sorry I revive this, but is there anway to relate this to this problem? http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=436572

At least not directly, but to the equivalent version I posted on the same thread.

Hardly so, since yours is $\displaystyle x^2+xy+y^2=z^3$, and it's [b]not[/b] homogeneous (of degree $2$).

[quote="Br1"]Hi  :)

To find all integers that satisfy the equation: $\displaystyle x^2+xy+y^2=z^2$.

(Excuse me, but have you already resolved this problem in this forum?)

'Bye[/quote]

this problem has a infinity of solution ..
 (0,0,0) is a solution  .
  now , if we put x=z ===> y=-x so (k,-k,k) is  a solution

                        x=0 ===>  y=z or y=-z  so, (0,k,k) and (0,-k,k) is solutions  too

please,  Correct me if i'm wrong


analytic geometry
quadratics
algebra
function
domain
quadratic formula

Finding the $ x$ coordinate of the graphs $ y=|x^{2}-1|$ and $ y=x(x-2)$ gives [hide="Answer"]$ x=\frac{1-\sqrt{3}}{2}$[/hide]

[hide="a bit more algebraic"]consider the cases $ x^{2}-1\ge 0$ and $ x^{2}-1<0$. In the first case, our equation is $ \frac{x^{2}-1}{x-2}=x$ from which we get $ x=\frac{1}{2}$. But this doesn't satisfy $ x^{2}-1\ge 0$.

If $ x^{2}-1<0$, we get $ \frac{1-x^{2}}{x-2}=x$. That is $ 2x^{2}-2x-1=0$. Solving the quadratic we get $ x=\frac{1\pm\sqrt{3}}{2}$. But since $ x^{2}-1<0$, we only get the solution $ x=\frac{1-\sqrt{3}}{2}$.[/hide]

Perfect, but you should reffer to $ x\neq 2$.

Answer

$ x^{2}-1 = (x-1)(x+1)$

when $ x \leq-1$ and $ x \geq 1$: $ x^{2}-1 \geq 0 \Longrightarrow |x^{2}-1| = x^{2}-1$

$ \Longrightarrow \frac{x^{2}-1}{x-2}= x$
$ \Longrightarrow x^{2}-1 = x^{2}-2x$
$ \Longrightarrow-1 =-2x$
$ \Longrightarrow x = \frac{1}{2}$

But this lies outwith the domain $ x \leq-1$ and $ x \geq 1$ so no solutions lie in this domain.

when $ -1 \leq x \leq 1$:  $ x^{2}-1 \leq 0 \Longrightarrow |x^{2}-1| = 1-x^{2}$

$ \Longrightarrow \frac{1-x^{2}}{x-2}= x$
$ \Longrightarrow 1-x^{2}= x^{2}-2x$
$ \Longrightarrow 2x^{2}-2x-1 = 0$

Applying quadratic formula: $ x = \frac{1 \pm \sqrt{3}}{2}$

but $ \sqrt{3}> 1 \Longrightarrow x = \frac{1+\sqrt{3}}{2}$ is outwith domain

$ \sqrt3 < 3 \Longrightarrow x = \frac{1-\sqrt{3}}{2}$ lies in the domain and is the only solution.

Edit: nayel beat me!