What type of person would you rather be like?

Oh I missed the last part :lol:

I think the first option is better than the second...I'm not sure what a military person would say to that though :lol:

I try hard to fall into the second category. I'm not necessarily very self-disciplined though. What self-discipline usually means for me is going somewhere far away with a few pieces of paper to a place with no computers and no math books.

I agree with ComplexZeta.

It's so tempting to fall into the first category sometimes though...so very tempting...

hmm.  I fall into the 2nd catagory but people tell me I am too self-disaplined so I dunno which is better.

I fall into...both categories, I have like, multiple personality disorder, i'm completely different when i'm at home, then when i'm at school, at home, i'm the second type, but when i'm at school, i'm the first type.  Hmm...maybe i should switch that around....

I like the first option although I the only thing I find wrong with that is self concious. I fall under the first catagory all except for that. I could care less about public opinion.

The "overemotional" part in the first option disturbs me.. otherwise.. i think i'd be there..
im definitely studious.. but not so sure about brutally self-disciplined.. lol

I want to be the second one, but I generally fail with respect to "brutally self-disciplined".

self discipline isn't really a good thing.  Discipline is also a bad thing.  Who cares? do what you want to do, be what you want to be.  Be a person of no regrets.  Regret is a horrible feeling and its something no person deserve to live with.

[quote="yixuanyang"]You need more options in the poll.
how do you make a poll?[/quote]

Make a new topic, then scroll down to the bottom.

I'm currently amazed to see there is a two-to-one ratio of people who are "more likely to be a cruel and brutal dictator" than that of a "more socially-acceptable and reasonable" one...course I took them out of a context, a la Jimmy :wink:

Self-control and discipline are infinitely preferable to showy displays of feeling. Then again I've always equated emotion with weakness. Unfortunately I've never been able to be completely self-disciplined.
Is it possible to live without regret? Somehow I think that one who's sensitive and self-conscious is more likely to regret and linger in the past than one who's truly disciplined and forges ahead continually.
Besides, living in the moment has its faults too.

I'm neither.  I guess I practice moderation...

im a mix.  im not self concious but im not brutally self-disciplined either.

Out of those two I'd say its safest to be the second, but I'm not either, and i don't wan to be.  There needs to be a middle ground, people who have empathy for others, ut arn't incompasitated by their emotions; people who are disciplined but not afraid to break loose ocasionally. If the world were made of only those types of people I would say that we'd be doomed.

Bah, humbug.

Positive emotions are for wimps (just kidding, but you probably already knew that).

I'd prefer to be the second, for the sake of order and strength of character.  I can't say the first one doesn't have its merits, but I'd rather have a strong sense of self-discipline (and consequently self-control) than be socially comfortable.

I'm neither, or maybe both.  I'm studious but not too self disciplined.  I get the work done but usually after lots of procrastination.  And while I am not overemotional, I have a tendency to get dramatic.  More the second than the first, but then again that's the stereotypical math person.

HOW AM I THE ONLY ONE VOTING ON THESE POLLS :rotfl:


inequalities
function
algebra proposed
algebra

Let $f(x)$ be a real-valued function, defined on the nonnegative real numbers, such that \[f(x)f(y) \leq y^2 f\left(\frac{x}{2}\right) + x^2 f\left(\frac{y}{2}\right)\] for all $x \geq y \geq 0$ and also \[-M \leq f(x) \leq M\textrm{ for }0 \leq x \leq 1\] where $M$ is a fixed real constant.
Prove that \[f(x) \leq x^2.\]

Actually there is a stronger inequality : $ f(x) \leq \frac{x^2}2$ :D

Let $N_x$ be a positive integer such a that $0 \leq x \leq 2^{N_x}  $.  Manifestly :

 $\displaystyle 0 \leq f(x)\leq  x\sqrt{2f(\frac x2)}\leq  ...\leq  \displaystyle x^ {1+\frac 1{2} +... +\frac 1{2^{n-1}}}(2f(\frac x{2^n}))^{\frac 1{2^n}} \leq $ ${ x^{1+\frac 1{2} +... +\frac 1{2^{n-1}}}(2M)^{\frac 1{2^n}},\forall n\geq N_x}$

The limit of the right part , for $n\rightarrow \infty$, is $x^2$ , so $0\leq f(x)\leq x^2,\forall x\geq 0$

Let $L$ denote $L = \{u\ |\  0\leq f(x) \leq ux^2,\forall x\geq 0\}$. Manifestly $1\in L$ so $L\neq \emptyset$, so there is $ \lambda = \inf_{u\in L}(u)$

So $\  0\leq f(x) \leq \lambda x^2 ,\forall x\geq 0 $. But  $0 \leq f(x)\leq  x\sqrt{2f(\frac x2)}\leq x^2 \sqrt{\frac{\lambda}2}$

Finally, because of the definition of $\lambda$ we have to have $\lambda \leq \sqrt{\frac{\lambda}2}$ so $\lambda\leq \frac 12$
:cool: q.e.d. :cool:


Let a,b,c such that a,b,c >0 and abc=1.Prove that 2/(a+b+c)+1/3>or= 3/(ab+bc+ca)

[quote="duy_nguyen"]Let $ a,b,c$ such that $ a,b,c >0$ and $ abc=1.$Prove that $ \frac{2}{a+b+c}+\frac{1}{3}\geq \frac{3}{ab+bc+ca}.$[/quote]
The following my own inequality  stronger:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc=1.$ Prove that
$ 1+\frac{8}{a+b+c}\ge \frac{11}{ab+bc+ca}$

Maybe i'm wrong but this problem is quite trivial with ABC method.
Solution(maybe wrong):
Using ABC theorem we refer to the only case:
$ a\equal{}b\Rightarrow a^{2}c\equal{}1$.
Hence we must prove that
$ 1\plus{}\frac{8}{2a\plus{}c}\geq\frac{11}{a^{2}\plus{}2ac}\Longleftrightarrow$.
$ 1\plus{}\frac{8a^{2}}{2a^{3}\plus{}1}\geq\frac{11a}{a^{3}\plus{}2}\Longleftrightarrow$
$ (a\minus{}1)^{2}(2a^{4}\plus{}12a^{3}\minus{}7a\plus{}2)\geq 0$.
This is obviously for all $ a\geq0$.
Thus we are done. :)


trigonometry
trig identities
Law of Cosines
geometry
Law of Sines

Let $ p,q,r\geq 0$ be such that $ p^2\plus{}q^2\plus{}r^2\plus{}2pqr\equal{}1$. Show that there are $ A,B,C\in [0,\frac{\pi}{2}]$ such that $ p\equal{}\cos A,q\equal{}\cos B,r\equal{}\cos C$, and $ A\plus{}B\plus{}C\equal{}\pi$

[hide="Solution"] Suppose by way of contradiction that $ p > 1$.  Then

$ q^2 \plus{} r^2 \plus{} (2p)qr \equal{} 1 \minus{} p^2$

where the LHS is greater than $ (q \plus{} r)^2$ but the RHS is negative; contradiction.  Hence $ 0 \le p, q, r \le 1$ so there exist $ A, B, C \in [0, \frac {\pi}{2} ]$ such that $ p \equal{} \cos A, q \equal{} \cos B, r \equal{} \cos C$.

Consider a line segment $ XY$ of length $ 1$, and construct right triangles $ \triangle XPY, XQY$ on opposite sides of $ XY$ so that $ \angle PXY \equal{} A, \angle QXY \equal{} B$.  The Law of Cosines on $ \triangle PXQ$ gives

$ PQ^2 \equal{} \cos A^2 \plus{} \cos B^2 \minus{} 2 \cos A \cos B \cos (A \plus{} B) \equal{} p^2 \plus{} q^2 \plus{} 2pq \cos (180 \minus{} A \minus{} B)$

and the Law of Sines gives

$ \frac {PQ}{\sin (A \plus{} B)} \equal{} 2R \equal{} 1 \Leftrightarrow$
$ PQ \equal{} \sin (A \plus{} B)$

so that $ PQ^2 \equal{} 1 \minus{} \cos^2 (A \plus{} B) \equal{} 1 \minus{} \cos^2 (180 \minus{} A \minus{} B)$, hence $ C \equal{} 180 \minus{} A \minus{} B$ as desired.  (Well, this is sort of the wrong direction; for fixed $ p, q$ we have

$ r \equal{} \frac { \minus{} 2pq \pm \sqrt {4p^2 q^2 \minus{} 4(p^2 \plus{} q^2 \minus{} 1)}}{2}$

and the root that isn't $ \cos C$ is clearly negative.)  [/hide]


geometry
circumcircle
inradius
trigonometry
inequalities
inequalities proposed

$ \boxed {\ \sum\frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c \minus{} a}\ \ge\ s\cdot\frac {R \minus{} 2r}{2R}\ \ge\ 0\ }$ $ \Longleftrightarrow$ $ \boxed {\ \sum \frac {yz}{x}\ \ge\ \left[2 \minus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\right]\cdot\sum x\ \ge\ x\plus{}y\plus{}z\ }$

where $ s$ - [u]semiperimeter[/u], $ R$ - [u]circumradius[/u] ,  $ r$ - [u]inradius[/u] for the triangle $ ABC$ and $ x \equal{} s \minus{} a > 0$ a.s.o.

It is equivalent to
\[ \sum\frac {(a \minus{} b)(a \minus{} c)}{(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)}\ge \frac {R \minus{} 2r}{4R} \equal{} \frac {3R \minus{} 2(R \plus{} r)}{4R}
\]

\[ \Leftrightarrow \frac {(a \minus{} b)(a \minus{} c)}{(b \plus{} c)^2 \minus{} a^2} \equal{} \frac {3}{4} \minus{} \frac {1}{2}\sum\cos A
\]

\[ \Leftrightarrow \sum(\frac {4(a \minus{} b)(a \minus{} c)}{(b \plus{} c)^2 \minus{} a^2} \plus{} 2\cos A) \minus{} 2\ge 1
\]

\[ \Leftrightarrow \sum(\frac {4(a^2 \minus{} ab \minus{} ac \plus{} bc)}{(b \plus{} c)^2 \minus{} a^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{bc}) \minus{} 2\sum\frac {b \plus{} c \minus{} a}{a \plus{} b \plus{} c}\ge 1
\]

\[ \sum(\frac {4(a^2 \minus{} ab \minus{} ac \plus{} bc)}{(b \plus{} c)^2 \minus{} a^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{bc} \minus{} 2\frac {b \plus{} c \minus{} a}{a \plus{} b \plus{} c})\ge 1
\]

\[ \Leftrightarrow \sum\frac {(b^2 \plus{} c^2 \minus{} a^2)^2}{bc((b \plus{} c)^2 \minus{} a^2)}\ge 1
\]
We have
\[ \sum\frac {(b^2 \plus{} c^2 \minus{} a^2)^2}{bc((b \plus{} c)^2 \minus{} a^2)}\ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum bc((b \plus{} c)^2 \minus{} a^2)}\ge 1 \Leftrightarrow \sum a^2(a \minus{} b)(a \minus{} c)\ge 0
\]
which is true  :P

[quote="Virgil Nicula"]$ \boxed {\ \sum\frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c \minus{} a}\ \ge\ s\cdot\frac {R \minus{} 2r}{2R}\ }\ ,$

where $ s$ - [u]semiperimeter[/u], $ R$ - [u]circumradius[/u] and $ r$ - [u]inradius[/u] for the triangle $ ABC\ .$[/quote]
I composed this inequality coming from the interesting identity $ \sum \frac {bc}{b \plus{} c \minus{} a} \equal{} \frac {s^2 \plus{} (4R \plus{} r)^2}{2s}\ (*)$

and the strong geometrical inequality $ R(4R \plus{} r)^2\ge 2s^2(2R\minus{}r)\ (**)$ which is given by the distance

$ \mathrm H\Gamma^2 \equal{} 4R^2\left[1 \minus{} \frac {2s^2(2R \minus{} r)}{R(4R \plus{} r)^2}\right]\ge 0$ , where $ \mathrm H$ is [u]orthocenter[/u] and $ \Gamma$ is the [u][b]Gergonne[/b]'s point[/u].

Thus, $ \sum\frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c \minus{} a} \equal{}$ $ \sum\frac {\left[bc \minus{} a(b \plus{} c \minus{} a)\right]}{b \plus{} c \minus{} a} \equal{}$ $ \sum \frac {bc}{b \plus{} c \minus{} a} \minus{} 2s\stackrel {(*)}{\ \equal{} \ }$

$ \frac {s^2 \plus{} (4R \plus{} r)^2}{2s} \minus{} 2s \equal{}$ $ \frac {(4R \plus{} r)^2 \minus{} 3s^2}{2s}\stackrel {(**)}{\ \ge\ } \frac {\frac {2s^2(2R \minus{} r)}{R} \minus{} 3s^2}{2p} \equal{} s\cdot\frac {R \minus{} 2r}{2R}\ .$

[quote="Virgil Nicula"][quote="Virgil Nicula"]$ \boxed {\ \sum\frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c \minus{} a}\ \ge\ s\cdot\frac {R \minus{} 2r}{2R}\ }\ ,$

where $ s$ - [u]semiperimeter[/u], $ R$ - [u]circumradius[/u] and $ r$ - [u]inradius[/u] for the triangle $ ABC\ .$[/quote]
I composed this inequality coming from the interesting identity $ \sum \frac {bc}{b \plus{} c \minus{} a} \equal{} \frac {s^2 \plus{} (4R \plus{} r)^2}{2s}\ (*)$

and the strong geometrical inequality $ R(4R \plus{} r)^2\ge 2s^2(2R \minus{} r)\ (**)$ which is given by the distance

$ \mathrm H\Gamma^2 \equal{} 4R^2\left[1 \minus{} \frac {2s^2(2R \minus{} r)}{R(4R \plus{} r)^2}\right]\ge 0$ , where $ \mathrm H$ is [u]orthocenter[/u] and $ \Gamma$ is the [u][b]Gergonne[/b]'s point[/u].

Thus, $ \sum\frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c \minus{} a} \equal{}$ $ \sum\frac {\left[bc \minus{} a(b \plus{} c \minus{} a)\right]}{b \plus{} c \minus{} a} \equal{}$ $ \sum \frac {bc}{b \plus{} c \minus{} a} \minus{} 2s\stackrel {(*)}{\ \equal{} \ }$

$ \frac {s^2 \plus{} (4R \plus{} r)^2}{2s} \minus{} 2s \equal{}$ $ \frac {(4R \plus{} r)^2 \minus{} 3s^2}{2s}\stackrel {(**)}{\ \ge\ } \frac {\frac {2s^2(2R \minus{} r)}{R} \minus{} 3s^2}{2p} \equal{} s\cdot\frac {R \minus{} 2r}{2R}\ .$[/quote]

Hi dear Virgil Nicula, this is very interesting relation of the centers in triangle and actually I don't know it, can you show me some others properties of the distances of other points in triangle, maybe with Nagel point, Symmedian point ... ,thank you so much. :)

[quote="Virgil Nicula"]
$ \Longleftrightarrow$ $ \boxed {\ \sum \frac {yz}{x}\ \ge\ \left[2 \minus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\right]\cdot\sum x\ \ge\ x \plus{} y \plus{} z\ }$

where $ s$ - [u]semiperimeter[/u], $ R$ - [u]circumradius[/u] ,  $ r$ - [u]inradius[/u] for the triangle $ ABC$ and $ x \equal{} s \minus{} a > 0$ a.s.o.[/quote]

It's nice but I think it isn't strong.
Let $ \frac {yz}{x} \equal{} a^2$..., it's becomes:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 2$
We obtain an old problem of hungkhtn. It's weaker than mathlink contest's old problem:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 2 \sum\ \frac {ab}{(a \plus{} b)^2} \geq\ \frac {5}{2}$
(Notice that this ineq is stronger than Iran96).
But we can prove it easily by SOS (I have 2 proofs for it)  :)


are you kidding?(post it by you twice....,)[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%5C%5Cfrac%7Ba%7D%7Ba%5E2%2Bbc%7D&t=70283]www.mathlinks.ro/Forum/viewtopic.php?highlight=%5C%5Cfrac%7Ba%7D%7Ba%5E2%2Bbc%7D&t=70283[/url]

[quote="SuperMath"]Let $\large a;b;c > 0$.
$\large \sum \frac{a}{a^2+bc} \leq \sum \frac{1}{a+b}$[/quote]
Easy
$\large \sum \frac{a}{a^2+bc} \leq \sum \frac{1}{a+b}\Leftrightarrow \sum\frac{(a-b)(a-c)}{(b+c)(a^2+bc)}\ge 0$
Let $M_a=\frac{1}{(b+c)(a^2+bc)}$
Assume that $a\ge b\ge c$
We have $M_c\ge 0$ and $M_a(a-c)\ge M_b(b-c)\Leftrightarrow (a^2-c^2)(b^2+ac)\ge(b^2-c^2)(a^2+bc)\Leftrightarrow a^3c+bc^3+a^2c^2\ge b^2c^2+ac^3+b^3c\Leftrightarrow(a-b)(a^2c+ac^2+b^2c+bc^2+abc-c^3)$
 :D
Edit:sorry I don't see your post, zhaobin


Find the restrictions on $ x$ if:

1. $ |ax| \leq |bx \minus{} c |$

2. $ |ax| \geq |bx \minus{} c |$

[hide]for the first one. i think $ c\minus{}bx \le ax \le bx\minus{}c$. looking at the first inequality, we get $ c \le x(a\plus{}b) \Rightarrow \frac{c} {a\plus{}b} \le x$. Fromt the second, we get $ x \le \frac{c} {a\minus{}b}$, so $ \frac{c} {a\plus{}b} \le x \le \frac{c} {a\plus{}b}$[/hide]

[quote="abacadaea"][hide]for the first one. i think $ c \minus{} bx \le ax \le bx \minus{} c$. looking at the first inequality, we get $ c \le x(a \plus{} b) \Rightarrow \frac {c} {a \plus{} b} \le x$. Fromt the second, we get $ x \le \frac {c} {a \minus{} b}$, so $ \frac {c} {a \plus{} b} \le x \le \frac {c} {a \plus{} b}$[/hide][/quote]

You mean: $ \frac{c}{a\minus{}b}\le x \le \frac{c}{a\plus{}b}$?

[quote="Bugi"][quote="abacadaea"][hide]for the first one. i think $ c \minus{} bx \le ax \le bx \minus{} c$. looking at the first inequality, we get $ c \le x(a \plus{} b) \Rightarrow \frac {c} {a \plus{} b} \le x$. Fromt the second, we get $ x \le \frac {c} {a \minus{} b}$, so $ \frac {c} {a \plus{} b} \le x \le \frac {c} {a \plus{} b}$[/hide][/quote]

You mean: $ \frac {c}{a \minus{} b}\le x \le \frac {c}{a \plus{} b}$?[/quote]

You mean: $ \frac{c}{a\plus{}b}\le x \le \frac{c}{a\minus{}b}$?

 :D

Yeah, sorry. But what does abacadea mean?

Hmm...If you looked at my whole solution, you should have been able to figure that out:
$ \frac {c} {a \plus{} b} \le x, x \le \frac {c} {a \minus{} b} \Rightarrow \frac {c} {a \plus{} b} \le x \le \frac {c} {a \minus{} b}$
 :D

For the second one, i think the signs are reversed so $ \frac {c} {a \plus{} b} \ge x, x \ge \frac {c} {a \minus{} b}$. Some thing tells me the 2nd is greater, so $ x \in [\frac{c} {a\minus{}b} , \infty]$.

Probably wrong. i didnt put much thought into that.

what???

$ \frac{c}{a\plus{}b}\le x \le \frac{c}{a\minus{}b} \implies x \in [\frac{c}{a\plus{}b},\frac{c}{a\minus{}b}]$


geometry
rectangle
rhombus
perpendicular bisector
geometry proposed

Consider a rhombus $ ABCD$ with center $ O$. Let $ P$ is a point given inside the rhombus, but not situated on the diagonals. Let $ M, N, Q, R$ be the perpendicular foots of  point $ P$ on the sides $ AB, BC, CD, DA$ respectively. Let perpendicular bisectors of the segments $ MN$ and $ RQ$ meet at point $ S$, let perpendicular bisectors of the segments $ NQ$ and $ MR$ meet at point $ T$. Prove that $ P, S, T, O$ are the vertices of a rectangle.

1.MQ=NR
2.SMQ SNR are congruent
3.PMNSB are concyclic => BS bisects angle MBN => B,S,O are collinear

the rest are trivil

MBNP is cyclic, BP is a diameter of this circle, the perpendicular bisector of MN intersects BD at the midpoint of the arc MN, let it be S', and  $ PS' \perp BD $. Similarly, the perpendicular bisector of RQ intersects BD at S", the midpoint of arc RQ, $ PS" \perp BD$, hence S' = S" = S and $PS \perp BD$.
Similarly $PT \perp AC$, done.

Best regards,
sunken rock


probability
ARML
quadratics
induction
expected value
algebra

Here are a bunch of probability problems I handed out at my workshop on "State Diagrams and Probability" at the ARML practice.  Feel free to post your solutions to any of them.  They are roughly ordered from easiest to hardest, though I also tried to group them by topic.

The solution method I had in mind was to draw a "state diagram", and then for each state define one variable, representing a probability (or expected value) from that state to a final state.  It's hard for me to draw diagrams here, but as an example of the method I give a solution (minus the drawing) to Problem 5.  Of course, you can use any method you want to solve the problems.

Many of the early problems are at ARML level; the later problems are more difficult.

[b]1.[/b]  A fair coin is tossed repeatedly.  What is the average number of tosses required to obtain a head?

[b]2. [/b] A die is rolled repeatedly.  On average, how many times must the die be rolled to get a 6?

[b]3. [/b] A couple decides to keep having children until they have at least one boy and one girl.  What is the average number of children they will have?  (Assume no twins.)

[b]4. [/b] I alternately toss a fair coin and throw a fair die until I either toss a head or throw a 2.  If I toss the coin first, what is the probability that I throw a 2 before I toss a head?

[b]5. [/b] A fair coin is tossed repeatedly.  What is the average number of tosses required to obtain two heads in a row?

[hide="Solution"]Let a be the average number of tosses required to obtain two heads in a row.  Let b be the average number of tosses required to obtain two heads in a row, given a single "pretoss" of heads.  We have the equation $a = 1 + \frac{1}{2} a + \frac{1}{2} b$.  That equation comes from considering the first toss: the "1" is the cost of the first toss, $\frac{1}{2}a$ is the contribution when the first toss is a tail, and $\frac{1}{2}b$ is the contribution when the first toss is a head.  Similarly, we get the equation $b = 1 + \frac{1}{2}a$.  Solving the two equations, we find a = 6 and b = 4.  So the answer is $\boxed{6}$. [/hide]

[b]6. [/b] An apple is located at vertex $A$ of pentagon $ABCDE$, and a worm is located two vertices away, at $C$.  Every day the worm crawls with equal probability to one of the two adjacent vertices.  When it reaches vertex $A$, it stops to dine.  What is the average number of days until dinner?

[b]7. [/b] Five people stand at the vertices of a pentagon, throwing frisbees at each other.  They have two frisbees, initially at adjacent vertices.  At each step, each frisbee is thrown either to the left or to the right with equal probability.  This process continues until one person is the target of two frisbees simultaneously; then the game stops.  What is the average number of steps taken?

[b]8. [/b] Three men, $A$, $B$, and $C$, are to fight a pistol duel (truel?).  $A$'s chance of hitting a target is $0.3$, $C$'s chance is $0.5$, and $B$ never misses.  They are to fire at their choice of target in succession in the order $A$, $B$, $C$, cyclically (but a hit man loses further turns and is no longer shot at) until one man is left.  What should $A$'s strategy be?

[b]9.  [/b]Erica flips a fair coin.  Nate then flips the coin and wins if it matches Erica's flip.  If it doesn't match, then Noah flips the coin and wins if it matches Nate's flip.  If Noah doesn't win, then Erica flips and wins if it matches Noah's flip.  The game continues until there is a winner.  What is the probability that Nate wins the game?

[b]10.[/b]  Two persons, $X$ and $Y$, play with a die an unlimited number of times.  $X$ wins a game if the outcome is 1 or 2; $Y$ wins the game in the other cases.  A player wins the match if he wins two consecutive games.  Determine the probability that $X$ wins the match.

[b]11. [/b] Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent.  Both players start at zero, and points are earned one at a time.  If Daniel has a $60\%$ chance of winning each point, what is the probability that he will win the game?

[b]12. [/b] Coupons are each randomly numbered 1 to 5.  All five numbers are required to win a prize.  On average, how many coupons must be taken to win?

[b]13. [/b] Alice and Bob flip a coin until either HHT or HTT occurs.  Alice wins if the pattern HHT comes first; Bob wins if HTT comes first.  What is the probability that Alice will win?

[b]14.[/b]  A fair coin is tossed repeatedly.  What is the probability that the first occurrence of three heads in a row is earlier than the first occurrence of two tails in a row?

[b]15.[/b]  A fair coin is tossed repeatedly.  What is the probability that the first occurrence of HH will occur before the first occurrence of THTH?

[b]16.[/b]  A fair coin is tossed repeatedly.  What is the average number of tosses required to obtain the sequence THTTH?

[b]17. [/b] Player $M$ has 1 dollar, and Player $N$ has 2 dollars.  Each play gives one of the players 1 dollar from the other.  Player $M$ is enough better than Player $N$ that he wins $\frac{2}{3}$ of the plays.  They play until one is bankrupt.  What is the chance that Player $M$ wins?

[b]18. [/b] Two players, $A$ and $B$, start with $m$ and $n$ dollars, respectively.  A fair coin is repeatedly tossed.  If the result is heads, then $A$ pays $B$ one dollar.  Otherwise, $B$ pays $A$ one dollar.  The game ends when one player goes broke.  What is the probability that $A$ wins?

[b]19.[/b]  A bug is on a number line, starting at 0.  A fair coin is tossed repeatedly.  If the coin comes up heads, the bug goes 1 unit left; otherwise, the bug goes 1 unit right.  What is the probability  that the bug eventually returns to 0?

[b]20.  [/b]From where the drunken man stands, one step toward the cliff would send him over the edge.  He takes random steps, either toward or away from the cliff.  At any step, his probability of taking a step away is $\frac{2}{3}$, of a step toward the cliff $\frac{1}{3}$.  What is his chance of falling over the edge?

[b]21. [/b] A microbe either splits into two perfect copies of itself or dies. If the probability of splitting is $\frac{2}{3}$ (and is independent of the microbe's ancestry), what is the probability that a microbes's descendants die out?

[b]22. [/b] Two players, $A$ and $B$, repeatedly play a game of chance. At each turn, either $A$ wins 1 dollar from B with probability $0.51$, or $A$ loses 1 dollar to $B$ with probability $0.49$. Both $A$ and $B$ have unlimited capital. What is $A$'s expected peak cumulative loss?

[b]23. [/b] A particle moves at each step 2 units to the right or 1 unit to the left, with probability $\frac{1}{2}$ each.  The particle starts at 1.  What is the probability that the particle will ever reach the origin?

[b]24.  [/b]A coin is tossed repeatedly.  What is the probability that at some point, the number of tails will be greater than twice the number of heads?

[b]25.  [/b]Real numbers are chosen at random from the unit interval ($0 \le x \le 1$).  If after choosing the $n$th number the sum of the numbers so chosen first exceeds 1, what is the expected value of $n$?

[b]26. [/b] Real numbers are chosen at random from the unit interval ($0 \le x \le 1$).  If after choosing the $n$th number the sum of the numbers so chosen first exceeds 2, what is the expected value of $n$?

i'll answer one or two per day.
[hide="my solutions 1-4"]1. A fair coin is tossed repeatedly. What is the average number of tosses required to obtain a head?

one time = $\frac{1}2$
two times = TH -> $\frac{1}4$
three times = TTH -> $\frac{1}8$
....

$X = 1*\frac{1}2 + 2*\frac{1}4 + 3*\frac{1}8 + ... = 2$

2. A die is rolled repeatedly. On average, how many times must the die be rolled to get a 6?

Similarly, $X = 1*\frac{1}6 + 2*\frac5{36} + 3*\frac{25}{216} + ... = 6$

3.  A couple decides to keep having children until they have at least one boy and one girl. What is the average number of children they will have? (Assume no twins.)

there is $\frac{1}2$ chance of (actually $\frac{2}4$ -> BG or GB) having 2 number of children. after that, either BB or GG, it is the same problem with 1, until G (or B) is born.

Hence, $\frac{1}2 * 2 + \frac{1}2 * (2+2) = 3$

4. I alternately toss a fair coin and throw a fair die until I either toss a head or throw a 2. If I toss the coin first, what is the probability that I throw a 2 before I toss a head?

let $P_1$ = (T, 2)  = $\frac{1}2 * \frac{1}6$ = $\frac1{12}$ and $P_2$ = (H, not2) = $\frac{1}2 * \frac{5}6$ = $\frac5{12}$

and the probability will be
$P_1 + P_2*P_1 + (P_2)^{2}*P_1 + (P_2)^{3}*P_1 + ...  = \frac{1}7$
[/hide]

[hide="Answer_to_Problem_20"]$\displaystyle \frac{1}{2}$[/hide]
To solve Problem 20, I recommend solving Problem 21 first.  Perhaps surprisingly, the two problems are equivalent.

From where the drunken man stands, one step toward the cliff would send him over the edge. He takes random steps, either toward or away from the cliff. At any step, his probability of taking a step away is , of a step toward the cliff . What is his chance of falling over the edge?

Ravi B what are the steps to solving this probability question? I can't seem to come up with the answer 1/2

[quote="Ravi B"][b]20.  [/b]From where the drunken man stands, one step toward the cliff would send him over the edge.  He takes random steps, either toward or away from the cliff.  At any step, his probability of taking a step away is $\frac{2}{3}$, of a step toward the cliff $\frac{1}{3}$.  What is his chance of falling over the edge?

[b]21. [/b] A microbe either splits into two perfect copies of itself or dies. If the probability of splitting is $\frac{2}{3}$ (and is independent of the microbe's ancestry), what is the probability that a microbes's descendants die out?[/quote]To solve Problem 20, I will actually solve Problem 21.  Do you see how the two problems are equivalent?

In Problem 21, let $p$ be the probability of dying out.  Because each microbe either splits or dies, we get the equation
\[ p = \frac{1}{3} + \frac{2}{3} p^2 \, .  \]
Do you see why?  Solving that quadratic equation shows that either $p = \frac{1}{2}$ or $p = 1$.  Hmm, we have narrowed the answer to two numbers:  Which is it?  We could use general probability principles like the Law of Large Numbers to show that $p \ne 1$, but I will give a more elementary approach below.

Let $P_n$ be the probability of dying out in at most $n$ generations.  By the same logic as above, we get the recurrence
\[ P_n = \frac{1}{3} + \frac{2}{3} P_{n-1}^2 \, ,  \]
with initial condition $P_0 = 0$.  By induction, we can show that $P_n \le \frac{1}{2}$.  Because the $P_n$ converge to $p$, we can't have $p = 1$, and so must have $p = \frac{1}{2}$.

Can somebody explain why Ravi put $ p^2$ and not $ p$?

Because the split is into [b]two[/b] copies.  If there is a split, we are looking for the probability that [b]both[/b] branches die out, which happens with probability $ p \cdot p \equal{} p^2$.  Did that make sense?


A stack of 45 dimes is divided into three piles in the ratio $ \frac{1}{6} : \frac{1}{3} : \frac{1}{4}$. How many dimes are in the pile with the least number of dimes?


All right, so I've been preparing a lot for the AMC's because I feel so mad  :mad: that I didn't even qualify for AIME  :mad:  :mad:  :mad: with my FAILFAILFAIL score of 100.5 on AMC10A this year.

So I just took the AMC10 from 2000. It was actually really easy, and I could've gotten a 150, really. But I did make 5 careless mistakes, resulting in a score of only 120 (well, its still better than a 100.5  :lol: )

I was just wondering, were these older AMC10s relatively easier or harder than the recent ones? Or perhaps,  was 2000 an easy year?

Thanks guys.

actually, this year was a surprisingly easy year. you probably suffered from test anxiety and made some careless mistakes. work hard and you'll make aime next year!

Also note that 2000 was the first year for the AMC 10 so the problems were considerably easier than those on the 2008 or 2007. Similarly, the 1983 AIME was easier than the recent AIMEs. And yes, this year's AMC 10 was very easy.

Also remember that calculators were discontinued 2008 (I think)

Difficulty also varies from person to person because of the different subjects that people are better in.


I've been using the AoPS Wiki guide for using asymptote, and I've gotten to this page.

[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode[/url]

I created the Asymptote tool and typed the code in TeXnicCenter(test.asy), but when I try to use the tool by pressing Alt+A, the following appears in the black box:

[code]error: could not load module 'test.asy'
Press any key to continue . . .
[/code]

What's wrong?

This can also be caused by a missing semicolon (as far as I know), check to see if there are any.

[quote="stevem"]Are you sure your file is called test.asy and not test.asy.tex which is easy to do? Click on File, then Save As, then type test.asy, then set the 'Save as type' field to 'All files'. This should make it a .asy file.[/quote]

I was getting the same error as eavqg.whbeg, then I tried what you said. Unfortunately, I got a new error message:

[b] Cannot execute 'latex'\scrollmode
Please put in config.asy:

Important settings;
textpath = "PATH";

where PATH denotes the correct path to the directory containing your latex engine (latex).

Alternatively, set the environment variable ASYMPTOTE_TEXPATH
or use the command line option -texpath = "PATH"[/b]

:(

It is [b]import settings;[/b], not [b]important settings;[/b] :P

Why don't you do exactly what is suggested? Either add the required lines to the config.asy file (create one if it doesn't exist and put it in the directory where asy.exe is; the PATH is the full path to the latex.exe program) or set up a new environmental variable (to do this go to "properties" in "my computer" and then choose "advanced"; after that it should be self-explanatory).

[quote="fedja"]It is [b]import settings;[/b], not [b]important settings;[/b] :P

Either add the required lines to the config.asy file (create one if it doesn't exist and put it in the directory where asy.exe is; the PATH is the full path to the latex.exe program) or set up a new environmental variable (to do this go to "properties" in "my computer" and then choose "advanced"; after that it should be self-explanatory).[/quote]

I followed the instructions of problem 3 in the following link: [url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help[/url] as it seemed to be the problem I am having. However, I've tried it several times and no luck.

Don't want to sound idiotic, but I'm not 100% sure what you mean by "asy directory" and the environmental variable asks for a variable value that I don't know. I am getting pretty desperate right now  :(

[quote="fedja"]The variable value is just the path to the directory where the file "latex.exe" is. Just search for this file and you'll get something like C:\Somewhere\Something\Somename\bin\latex.exe as a result of your search. C:\Somewhere\Something\Somename\bin is the value you want to set for ASYMPTOTE_TEXPATH. :)[/quote]

YES!!!!! THANK YOU! IT WORKED!!!!!  :D  :D  :D

Never mind. My program is sort of working. When I input a code and get latex to run, an image shows up just fine. However, when I change the code, the same image still appears! For example, I write

draw((0,0)--(100,100));

And I get a straight line. However, when I input

draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$P$",(50,50),S);

I still get just a straight line.  :mad:

Edit: Sorry, I just found a way around this problem. Simply save whenever a change is made to the asy document you are working on.  :blush:


function
real analysis
real analysis solved

Let f,g:R->R be continuous functions such as fog is the identity of R.Prove that also gof is the identity of R.

From fog = id, we deduce that g is injective on |R and f is surjective on |R;
Since g is a continuous and injective function, it follows that g is monotone and that g(|R) = I, where I is an interval.
Suppose that g(x) - > a when x - > +oo, where a is a real number.
Since f is continuous at a, we deduce that fog(x) -> f(a) = b when x -> +oo.

But fog is a bijective and continuous function from |R to |R so that fog is monotone and fog(|R) = |R, thus fog(x) - > +/- oo when x -> +oo. A contradiction.
In a similar way, we prove that g(x) -> a when x -> -oo is impossible. Since g is monotone, we deduce that g(|R) = |R, and then that g is surjective, which gives that g is bijective from |R to |R.
Then f is injective and thus f is bijective from |R to |R too.
From fog = id we deduce that f = g <sup>-1</sup>  so that gof = id, and we are done.

Pierre.


I think it is quite easy
hint
$ \frac{1}{n(n\plus{}1)}\equal{}\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$

[quote="CHAITANYA KRISHNA"]P.T +ve integersa1,a2,a3,..am can always be found such that
1/n =1/a1 +1/a2 +...+1/am for any +ve integers n,m.

i need help using latex![/quote]

I am not sure what do you mean but this is $ \text{\LaTeX}$ code :

[code]$a_1,a_2,...,a_m$[/code]$ \rightarrow a_{1},a_{2},...,a_{m}$

and:

[code]$\dfrac{1}{n}=\dfrac{1}{a_1}+\dfrac{1}{a_2}+...+\dfrac{1}{a_m}$[/code]$ \rightarrow\frac{1}{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{m}}$

If $ m>1$
$ \frac{1}{n(n\plus{}1)}\plus{}....\plus{}\frac{1}{(m\plus{}n\minus{}2)(n\plus{}m\minus{}1)}\plus{}\frac{1}{n\plus{}m\minus{}1}\equal{}\frac{1}{n}$
If $ m\equal{}1$ chose $ a_{1}\equal{}n$


What is the arithmetic mean of 87, 78, 95, 84, 100, 92 and 87?

By definition,
\[ \frac {87 \plus{} 78 \plus{} 95 \plus{} 84 \plus{} 100 \plus{} 92 \plus{} 87}{7} \equal{} \boxed{89}
\]
There is another solution that makes use of the fact that if you let $ x$ be the mean of the numbers $ x_1$, $ x_2$, $ x_3$, ..., $ x_n$, then
\[ \sum_{i \equal{} 1}^{n}(x \minus{} x_i) \equal{} 0
\]
I'll let someone else post this.


not for pregnant women and patients of hypertension

How can you bend your thumb like that?  :o

Nope. Only [b]the chosen ones[/b] can do it.

It's true! I can't do it, but my Bulgarian friend can, and because of that he is much better at throwing a frisbee than I am! (Well there are other factors too, like the fact that I suck.)

:rotfl:

this one is really funny. are you from india? just wondering.

ur only a man to the ones who are still there :P

-jorian

They were only scared off to the bathroom, where I am sure they discussed me excitedly  :)

This is clearly a commercial to Pepsi :)

Not until I get a [i]bit[/i] more famous. Just a [i]bit[/i] more.

it looks more like you got poked in the eye

did you know bubka means grandma in yiddish

DUDE ISN'T SCHOOL SPORTS LIKE THE COOLEST GAME EVER

[quote="ragnarok23"]why are you stargazing during the day[/quote]

Because pointing your binoculars at the sun is REALLY FUN. Try it sometime.
[size=59]
Kids, do not try this at home without parental supervision. Mustafa is not responsible for any loss of eyesight, deafness, or internal bleeding caused by this advice.[/size]

[quote="Go Around the Tree"]you all look like alike  :P[/quote]

lol hahaha. a fmaily of evil overlords

I was pointing them the other way by the way.

ah but they are only some cousins
I have hundreds others

[quote="bubka"]because i am bubka of course  :rotfl:

but seriously, can't everybody do that?  :maybe:[/quote]
only double jointed people can :)

you need to work on your evil smile

thats not very evil

nay....cud have titled it-"The dadi maa's man!" :rotfl:  :rotfl:  :rotfl:

no..bongodesh to be precise
umm....Bankura-r gamcha or Assam? :rotfl:  :rotfl:
and Kishen ji must be ur name,rite? :rotfl:  :rotfl:


In the figure, triangles ABC and BCD are equilateral triangles. What is the value of AD $ \div$ BC when expressed in simplest radical form?

[asy]draw((0,0)--(5,8.7)--(10,0)--cycle);
draw((10,0)--(15,8.7)--(5,8.7));
label("$A$",(0,0),SW);
label("$B$",(5,8.7),N);
label("$C$",(10,0),SE);
label("$D$",(15,8.7),NE);[/asy]

Suppose the equilateral triangles $ ABC$ and $ BCD$ have the edge length of 1.
Law of cosine:
$ AD^2 \equal{} AC^2 \plus{} CD^2 \minus{} 2*CD*AC*cos(120) \equal{} 1 \plus{} 1 \minus{} 2*1*1*(\minus{}1/2) \equal{} 3$
$ AD \equal{} \sqrt3$

$ AD/BC \equal{} \sqrt3$

[quote=RadioActive]Any way to do this w/o trigie?[/quote]

There is a easier solution that avoids using trigonometric laws.

We are given that $ \bigtriangleup ABC $ and $ \bigtriangleup DBC $ are equilateral. We can conclude that $ \bigtriangleup ABC \cong \bigtriangleup DBC $ as they both share side $ BC $. Therefore, all sides of the figure are equal and the lengths of them can be assigned as $x$. Now, we draw line $AD$. We can make the assumption that $ AD $ is the perpendicular bisector of $ BC $ and intersects it at point $M$. Now we can see that  $ \bigtriangleup AMB $ is a 30-60-90 triangle.  $ BM = \frac{1}{2}BC$ or $\frac{1}{2}x$, $ AB = x$, and $AM= \frac{1}{2}x\sqrt{3}$. Because $MD \cong MA$ by CPCTC, $AD = x\sqrt{3}$. $BC = x$ so $AD \div BC = x\sqrt{3} \div x$ or $ \sqrt{3} $.

Hope this helps! I rarely post on the forums and I don't have much experience with LaTeX, so if my formatting is off, feel free to grill me. :)


3D geometry
geometry
euclidean distance
Cyclic
point set
IMO Shortlist
IMO Longlist

In the space $n \geq 3$ points are given. Every pair of points determines some distance. Suppose all distances are different. Connect every point with the nearest point. Prove that it is impossible to obtain (closed) polygonal line in such a way.

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

[quote="grobber"]Am I misunderstanding this? The two points $A,B$ connected by the smallest distance are connected to each other, so we have at most $n-1$ edges in our figure (because the edge going out from $A$ is the same as the edge going out from $B$). However, for a closed polygonal line with $n$ vertices we need $n$ edges.[/quote]

Your solution is right. The problem is that trivial. Indeed the real shortlist problem had a stronger assertion, but unfortunately it was wrong :D .

  Darij


[color=blue][b]Problem.[/b] For all positive real numbers $ a,b,c$ we have the following inequality[/color]
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \ge 3\sqrt[6]{\frac{a^6\plus{}b^6\plus{}c^6}{3}}
\]

There exists a short and nice solution by...... one of Vasc's theorems !! :)

[quote="Vasc"]Using the inequality

$ (\frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a})^2\ge \frac {3(a^{3.5} + b^{3.5} + c^{3.5})}{(abc)^{0.5}}$,

and then (n-1)EV-Theorem, we can prove that

${ \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a} \ge 3(\frac {a^p + b^p + c^p}{3}})^\frac 1{p}$,

for $ p = 5.997$.[/quote]
I will check my solution carefully, maybe I made somes mistakes in computation, because my idea is the same as you.
Thank you for your interest.

[quote="svasao_qb111"]it very hard and nice ?
@ehku: your solution? i want see solotion for you![/quote]
Sorry, my solution is wrong, see Vasc's post.

I know have one solution of Vo Quoc Ba Can very nice give problem for U
@: i stupid english  :|

[quote="can_hang2007"]Use the following
\[ \sum \frac {a^2}{b} \ge \frac {3\sum a^4 \plus{} 13\sum a^3(b \plus{} c) \minus{} \sum a^2b^2 \minus{} \sum a^2bc}{3(\sum a) (\sum ab)}
\]
:)

@Khue: Is your solution the same as me?  :P

[hide]Khue, toi nghi cau nen can than hon trong loi noi cua minh![/hide][/quote]
No, can_hang2007. I haven't any solution to it. :(

[hide]Hì, cảm ơn cậu, tớ sẽ chú ý  :wink: (tính đạo hàm sai ấy mà)[/hide]


what's the proof  of this theorem?
 Given $ABC$ triangle. tangents of circumcircle at $B$ and $C$ intersects at point D. And $AF$ is symmedian. Prove that $A, F, D$ are collinear.

Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=564645#564645]http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 post #2[/url].

  darij


For how many integers  x is square root (81 - x^2) a real number ?

A. 9 B. 10 C. 17 D. 18 E. 19

$ sqrt{81\minus{}x^{2}}$ is real as long as $ 81\minus{}x^{2}\geq 0$, so that $ 81\geq x^{2}$.  There are 19 such values of $ x$ (don't forget to include the negatives and 0).

[quote="JRav"]$ \sqrt{81 \minus{} x^{2}}$ is real as long as $ 81 \minus{} x^{2}\geq 0$, so that $ 81\geq x^{2}$.  There are 19 such values of $ x$ (don't forget to include the negatives and 0).[/quote]

<Slightly fixed LaTeX>.

I almost miscalculated and said 18 (Forgot 0  :) )


function
real analysis
real analysis unsolved

Find all differential 2-forms $ \phi$ such that $ d\phi\equal{}x_1x_2^2x_3\,dx_1\wedge dx_2\wedge dx_3$.

No one seems to be answering this one  :( . I don't know how to find them all, but I can find a lot...

Suppose $ \phi\equal{}Adx\wedge dy\plus{}Bdy\wedge dz\plus{} Cdx\wedge dz$ then $ d\phi\equal{}\left(\frac{\partial A}{\partial z}\plus{}\frac{\partial B}{\partial x}\minus{}\frac{\partial C}{\partial y}\right)dx\wedge dy\wedge dz$.

Thus we need to find all functions $ A(x,y,z)$, $ B(x,y,z)$, and $ C(x,y,z)$ such that $ \displaystyle\frac{\partial A}{\partial z}\plus{}\frac{\partial B}{\partial x}\minus{}\frac{\partial C}{\partial y}\equal{}xy^2z$.

So three classes of 2-forms that work are $ \phi\equal{}\left(\frac{xy^2z^2}{2}\plus{}K\right)dx\wedge dy$, $ \phi\equal{}\left(\frac{x^2y^2z}{2}\plus{}K\right)dy\wedge dz$ and $ \phi\equal{}\left(\frac{\minus{}xy^3z}{3}\plus{}K\right)dx\wedge dz$.

Plus the 4 obvious combinations of these. But I'm sure there are more, and I'm not sure other than the tedious way of trying to solve each of these things one at a time like that of how to do it.

[quote="bombaju"]Find all differential 2-forms $ \phi$ such that $ d\phi \equal{} x_1x_2^2x_3\,dx_1\wedge dx_2\wedge dx_3$.[/quote]

Heuristic Investigation:
What if $ d\phi_1 \equal{} d\phi_2 \equal{} x_1x_2^2x_3\,dx_1\wedge dx_2\wedge dx_3$? Then $ d(\phi_1\minus{}\phi_2) \equal{} 0$, or $ \phi_1\minus{}\phi_2$ is harmonic.

Basically, one we find one solution $ \phi$, then the set of solutions is simply $ \{\phi\plus{}\theta \mid d\theta\equal{}0 \}$. The condition $ d\theta\equal{}0$ is another way of saying that $ \theta$ is a harmonic $ 2$-form.


number theory
prime numbers
number theory unsolved

[quote="zibi"]Thera are given two prime numbers $p$ and $q$ such that for some number $n$ number $n^{2}+q$ is divisible by $p$. Prove that there exist natural numbers $a$, $b$, $k$ such that $a^{2}+qb^{2}=pk$ and $k<q$[/quote]
let $a_{i}=ni(\mod p)$
$i=1,2,\dots,[\sqrt{p}]$
then there exsit two distint $a_{l},a_{k}$ such that $\sqrt{p}\geq  a_{l}-a_{k}>0$
let $b=l-k,a=a_{l}-a_{k}$
then $a^{2}+qb^{2}<p+qp$
and we also have $a^{2}+qb^{2}\equiv (l-k)^{2}(n^{2}+q)\equiv 0(\mod p)$
so $a^{2}+qb^{2}=pt$and $t<q$


geometry
circumcircle
cyclic quadrilateral
geometry open

ABCD is a cyclic quadrilateral. The tangent at A meets the ray CB at K, and the tangent at B meets the ray DA at M, so that BK=BC and AM=AD. Show that the quadrilateral has two sides parallel.

$ X, Y \in BC$ are feet of the internal and external bisectors of the angle $ \angle CAB.$ Tangent at $ A$ to the circumcircle of $ \triangle ABC$ cuts the line $ BC$ at the center $ K$ of the A-Apollonius circle with diameter $ XY.$ Denote $ a = BC, b = CA, c = AB\ (b > c).$

${ YB = \frac{ca}{b-c}},\ BX = \frac{ca}{b+c},\ KX = \frac{YX}{2} = \frac{YB + BX}{2} = \frac{abc}{b^2 - c^2},$
$ a = BC = KB = KX - BX = \frac{c^2a}{b^2 - c^2}\ \Longrightarrow\ b^2 = 2c^2,\ CA = AB \sqrt{2}$

Repeating this for $ \triangle ABD$ with its B-Apollonius circle centered at $ M,$ $ BD = AB \sqrt{2}.$ Therefore, $ BD = CA,$ the cyclic quadrilateral $ ABCD$ has equal diagonals, it is an isosceles trapezoid.


topology
function
real analysis
advanced fields
advanced fields unsolved

(1) Suppose (M,d) is a metric space. Let A and B be two non-empty subsets of M. Define d(A,B) = inf { d(x,y) : x in A, y in B }. If A and B are disjoint and closed and if A is compact, prove that d(A,B) > 0.

(2) Give an example (with justification) of a noncompact space with no compact component.

(3) True or false: (i) If A is not connected and f is injective, then f(A) is not connected.
(ii) If A is a closed proper subspace of X and f is uniformly continuous, then f(A) is bounded in $\mathbb{R}$.

(4) Let A be a subset of a Hausdorff space. Prove that the point x is a limit point of A if and only if every neighbourhood of x contains infinitely many points of A.

(5) Give an example of two compact subspaces of a non-Hausdorff space, whose intersection is not compact.

(6) Let C(X,Y) denote the space of continuous functions from X to Y. if Y is Hausdorff, prove that C(X,Y) is Hausdorff in the compact-open topology.

(7) Suppose that X is a connected space and that there exists a non-constant continuous function $f: X \rightarrow \mathbb{R}$. Show that X is an uncountable set.

[quote="chungyc"](1) Suppose (M,d) is a metric space. Let A and B be two non-empty subsets of M. Define d(A,B) = inf { d(x,y) : x in A, y in B }. If A and B are disjoint and closed and if A is compact, prove that d(A,B) > 0.[/quote]
Let $x\in A$, there exists $r_x$ such that $O(x,r_x)\cap B=\emptyset$ because $B^c$ is open. The
collection consisting of all $O(x,\frac 12r_x)$ ($x\in A$) is a covering of $A$. The compactness of $A$ implies that there exists $x_1,x_2,\dots,x_k$ such that \[ A\subseteq \bigcup_{i=1}^k O\left(x_k, \frac 12r_{x_k}\right). \] Let $r = \min\{\frac 12r_{x_1},\dots,\frac 12r_{x_k}\}$ and $x\in A, y\in B$. Then $x$ is contained in $O(x_j, \frac 12r_{x_j})$ for some $x_j$. On the other hand, $d(y,x_j)\geq r_j$. Hence \[ r_j\leq d(y,x_j)\leq d(y,x)+d(x,x_j)\leq d(y,x)+\frac12 r_{x_j} \] whence $d(y,x)\geq \frac12r_{x_j}\geq r$. Thus $d(A,B)\geq r>0$.

[quote="chungyc"]
(4) Let A be a subset of a Hausdorff space. Prove that the point x is a limit point of A if and only if every neighbourhood of x contains infinitely many points of A.
[/quote]

If every neighbourhood of x contains infinitely many points of A, every neighbourhood specifically contains one non-x point of A, and so x is a limit point.

Now, assume there is a neighbourhood U of x containing only a finite number $y_1,\ldots,y_n$ of A. Then $U-\{y_1,\ldots,y_n\}$ is a neighbourhood of x not containing any point of A, and so x is not a limit point.

[quote]
(7) Suppose that X is a connected space and that there exists a non-constant continuous function $f: X \rightarrow \mathbb{R}$. Show that X is an uncountable set.
[/quote]

What's the topology on X? If X has the discrete topology, every mapping is continuous.

2) $\mathbb{R}$ is not compact and is connected.

3) (i) False (connectedness is a topological property, so this has nothing to do with purely set-theoretic considerations) $[0,1[ \cup [2,3]$ is not connected, but the image of $f(x)=x$ if $x \in [0,1[$, and $f(x)=x-1$ otherwise, is connected.
(ii) False; take $f(x)=x$ and $A=]-\infty;0]$.

5) Consider the set $X = \cup_{k=0}^{+\infty}{[2k,2k+1]}$ endowed with the topology given by the basis $\{\emptyset, X, ([2n,2n+1])_{n \geq 2},[0,1] \cup (\cup_{k=2}^{+\infty}{[2k,2k+1]}),[2,3] \cup (\cup_{k=2}^{+\infty}{[2k,2k+1]})\}$. The two last sets are quasi-compact but their intersection is not.

6) Take $f \neq g$ in $C(X,Y)$. Then $\exists x \in X, f(x) \neq g(x)$ and there are two disjoint open sets $U$ and $V$ in $Y$, $f(x) \in U$ and $g(x) \in V$. Now the two open sets $C(\{x\},U)$ and $C(\{x\},V)$ consisting of those continuous mappings such that $f(\{x\}) \subseteq U$ (resp. $g(\{x\}) \subseteq V$) are disjoint and contain $f$ (resp. $g$). Thus $C(X,Y)$ is Hausdorff.

As an addition to (6), is the converse true?

[quote="Nomendil"]

What's the topology on X? If X has the discrete topology, every mapping is continuous.[/quote]

The topology on X was not specified but I suppose it is not discrete because of the assumption in the question.

[quote="chungyc"]
(7) Suppose that X is a connected space and that there exists a non-constant continuous function $f: X \rightarrow \mathbb{R}$. Show that X is an uncountable set.[/quote]

Since $f$ is continuous and $X$ is connected, then $f(X)$ is connected. But connected subsets of $\mathbb{R}$ are  intervals.  Since $f$ is non constant, so the interval is nondegenerate (not a point) and in this case the cardinality of interval is uncountable. But now $|f(X)|\leq |X|$. So $|X|$ is uncountable.


inequalities
calculus
derivative
logarithms
algebra
polynomial
inequalities proposed

Let $ a,b,c$ be positive real numbers such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}3$. Prove that:
\[ 3\sqrt{8a^{2}\plus{}1}\plus{}3\sqrt{8b^{2}\plus{}1}\plus{}3\sqrt{8c^{2}\plus{}1}\ge 8(a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2})\plus{}3\]
well-come everybody  :D

The following inequality is a very nice one  :D
Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Prove that:
\[ \sqrt{8a^{2}\plus{}1}\plus{}\sqrt{8b^{2}\plus{}1}\plus{}\sqrt{8c^{2}\plus{}1}\le 3(a\plus{}b\plus{}c)\]

[quote="zaizai-hoang"]The following inequality is a very nice :blush:  one  :D
Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that:
\[ \sqrt{8a^{2}\plus{}1}\plus{}\sqrt{8b^{2}\plus{}1}\plus{}\sqrt{8c^{2}\plus{}1}\le 3(a\plus{}b\plus{}c)\]
[/quote]

I can only prove the following one :blush: :
For any positive real numbers $ a,b,c$,
\[ \sqrt{8a^{2}\plus{}ab}\plus{}\sqrt{8b^{2}\plus{}bc}\plus{}\sqrt{8c^{2}\plus{}ca}\le 3(a\plus{}b\plus{}c)\]

[quote="zaizai-hoang"]The following inequality is a very nice one  :D
Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that:
\[ \sqrt{8a^{2}\plus{}1}\plus{}\sqrt{8b^{2}\plus{}1}\plus{}\sqrt{8c^{2}\plus{}1}\le 3(a\plus{}b\plus{}c)\]
[/quote]

This problem is easy for me, but ironically, the first problem is annoying me. :oops:

Here is my solution, which seemes to be in fashion in MathLinks recently, I also have discovered this method several years ago :wink: (but, arqady had have discovered it.), because I make inequality's problem by differentiation, I don't know Muirhead, Shur that means I myself can't use them.

$ f(x) \equal{} 3x\minus{}\sqrt{8x^{2}\plus{}1}\minus{}\frac{1}{3}\ln x\ (x > 0)$, $ f'(x) \equal{}\frac{k(x\minus{}1)P(x)}{A(x)B(x)}$, where $ A(x),\ B(x) > 0$ are the expressions with roorts, $ P(x) > 0$ is a polynomial and $ k$ is a some positive constant number. Therefore $ f(x)\geq f(1), f(1) \equal{} 0$, yielding $ f(x)\geq 0$.
Since $ a > 0,\ b > 0,\ c > 0$, we have $ f(a)\plus{}f(b)\plus{}f(c)\geq 0$
$ \Longleftrightarrow 3(a\plus{}b\plus{}c)\minus{}(\sqrt{8a^{2}\plus{}1}\plus{}\sqrt{8b^{2}\plus{}1}\plus{}\sqrt{8c^{2}\plus{}1})\minus{}\frac{1}{3}\ln abc\geq 0$
$ \Longleftrightarrow\sqrt{8a^{2}\plus{}1}\plus{}\sqrt{8b^{2}\plus{}1}\plus{}\sqrt{8c^{2}\plus{}1}\leq 3(a\plus{}b\plus{}c).\ Q.E.D.$

The calculation of $ f'(x)$ is left for you. :D

I would like to know more basic solution without using differentiation.  :)

zaizai-hoang, this is a present for you!

For positive numbers $ a,\ b,\ c$ with $ abc\equal{}1$, prove that

$ \sqrt{1003a^{2}\plus{}1004}\plus{}\sqrt{1003b^{2}\plus{}1004}\plus{}\sqrt{1003c^{2}\plus{}1004}\leq\sqrt{2007}(a\plus{}b\plus{}c)$.

kunny

OMG, this is also my solution and I think the last inequality can be solve by the same way :wink: I discovered this technique several 6 month ago  :P But can you solve the following by this technique?  :D
Let $ a,b,c,d\ge 0,abcd\equal{}1$ Prove that:
\[ \sum\frac{1}{a^{3}\plus{}a^{2}\plus{}a\plus{}1}\ge 1\]
I think that if you using only this technieque, you won't complete your proof  :wink: well-come

[quote="zaizai-hoang"]OMG, this is also my solution and I think the last inequality can be solve by the same way :wink: I discovered this technique several 6 month ago  :P But can you solve the following by this technique?  :D
Let $ a,b,c,d\ge 0,abcd \equal{} 1$ Prove that:
\[ \sum\frac{1}{a^{3}\plus{}a^{2}\plus{}a\plus{}1}\ge 1\]
I think that if you using only this technieque, you won't complete your proof  :wink: well-come[/quote]

You can use $ \frac{1}{e^{3t}\plus{}e^{2t}\plus{}e^{t}\plus{}1}\geq\minus{}\frac{3}{8}t\plus{}\frac{1}{4}$.


number theory
prime numbers
number theory proposed

1.
a) prove that there are infinitely many prime numbers that can not be writen as the sum of another prime number and a power of 2.
*b) prove that there are infinitely many primes that can be writen in the above form.

I have been able to solve (a) but not part (b). of course b will be immidiate if we assume that the number of twin primes is infinity.

You can read more about the part a) of the problem in the famous number theory book written by Sierpinski.


[quote="chien than"]Find $x;y;z$ such that:
$\sqrt{2x}.y^{2}-z^{4}\geq 7$
and $\sqrt{-x^{2}y^{2}+8xy+9}-\sqrt{x^{2}-4}\geq 2(x+\frac{1}{x}$[/quote]
$x\geq2.$ Hence, $\sqrt{-x^{2}y^{2}+8xy+9}=\sqrt{25-(xy-4)^{2}}\geq2(x+\frac{1}{x})\geq5.$
Thus, $x=y=2$ and $8-z^{4}\geq7.$
I don't think that it's Olympiad's level.


Kate rode her bicycle for $ 30$ minutes at a speed of $ 16$ mph, then walked for $ 90$ minutes at a speed of $ 4$ mph. What was her overall average speed in miles per hour?

$ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$

[hide="Solution"]By the way, the question is wrong. The first 30 minutes, not 40.

For the first section, she went
$(0.5)(16)$ miles, or 8 miles.

For the second section, she went
$(1.5)(4)$ miles, or 6 miles.

She went 8 + 6 miles total, and it took her 0.5 + 1.5 hours total, for a net of

$\frac{14}{2}$ mph, or $\boxed{7}$ mph, which is choice A.
[/hide]

[quote="4everwise"]Kate rode her bycicle for $40$ minutes at a speed of $16$ mph, then walked for $90$ minutes at a speed of $4$ mph. What was her overall average speed in miles per hour?

$\text{(A)}\ 7 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 14$[/quote]

yeah, i remember this from the 10 i think

[hide="solution"]$\frac{1}{2}16+\frac{3}{2}4=14$

$\frac{14}{2}=\boxed{7}$[/hide]

The problem is 30 not 40

so that is very simple to find

Even if it were 40 we would obtain A since it would be the closest answer

oh my i spent like 5 minutes trying to solve this without realizing that the problem is wrong....i'm so sad....

Yeah, I used 40 and I still got the right answer because it's the closest.


[quote="GameBot"]A rectangle having integer length and width has a perimeter of 100 units. What is the number of square units in the least possible area?[/quote]

We want one side to be as long as possible and one side as small as possible. Since the perimeter is 100, the sum of the length and width is 50. Thus the sides are 1 and 49, and the area is $ \boxed{49}$.


I have a long document (a textbook which can be downloaded from www. potto.org) when using fancyhdr package I get the header from the chapter command. When it gets into indexes  (there are two indexes) header remains with the last chapter. how to get it corrected to refer to the index.
Do I have to define the marks (markboth, rightmark etc)? Is the index in section level? How I get it to move to chapter level?

According to [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr's manual[/url] chapter 9 you can redefine \chaptermark. Also read Chapter 18 on where to put the redefinition.

Thanks for the reply.

I have try to define using the definitions from fancyhdr's manual
but they don't seem to tack effect immediately. I found work temporary around of
\chapter{Indexes}
\addtocounter{page}{-1}

and then not printing the "fake" page.

I am using the package  multind which seem to me to demote the index to a section level and not to a chapter level.
I believe that regular  index (makeindex) is in a chapter level.
is there more intelligent solution that my silly work around?
Thanks,
Genick

From the fancyhdr manual, section 18:
[quote]If you want the change to take effect at the next page you must make sure that the current page is finished. In most cases this can be done by issuing a \clearpage command before any of the above mentioned changes. If this is not possible you can use the afterpage package.[/quote]Also the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=memoir]memoir class[/url] supports both headers & footers and multiple indexes.

First thank you,

I have read the package and I seems that your explanation it should work. However, it does not don't do the trick.
here is my file.

\fancyhead[re]{\helv \rightmark}

\printindex{sub}{Subjects index}
\printindex{aut}{Authors index}

I it produce the last chapter title on the right and even pages.
The helv is only a font command.

I have tried the memoir class and it clash with some definitions that I have. Maybe if I resolve the definitions it could work. I will keep you posted.

Thanks,

Genick


It is solved [url=http://wangsblog.com/jeffrey/]here[/url].

edit> nevermind, there's a sign change difference.

Heh. That's actually because I copied the problem off a Putnam exam incorrectly. We can see that

$\frac{\left(x+\frac{1}{x}\right)^{6}-\left(x^{6}+\frac{1}{x^{6}}\right)-2}{\left(x+\frac{1}{x}\right)^{3}+\left(x^{3}+\frac{1}{x^{3}}\right)}= \left(x+\frac{1}{x}\right)^{3}-\left(x^{3}+\frac{1}{x^{3}}\right)$

so we can just minimize that. It reduces to

$3 \left(x+\frac{1}{x}\right) \ge 6$

by AM-GM.


Probability with Combinations Part 3

We solve many probability problems by counting the number of successful outcomes and by counting the number of possible outcomes.  In some of these problems, we can either take order into account when counting outcomes, or we can ignore order entirely.  In this lesson, we see that this does NOT mean that if we have combinations in the denominator, we must also have them in the numerator!

It's nice that Rick is making videos that describe common mistakes as well as correct solutions.  Here, he might have justified the solution further by showing how a wrong solution would have looked (in the case of choosing one D, one R, and one I for a committee) if the number of solutions were actually counted with order.

Why is it called the McCain Obama effect?

I think it's called that because

supposedly

"comparing republicans to democrats"

is analogous to

"comparing regardless of order to not regardless of order"

not sure

There are parts one and two
in fact, there is even a part four

the parts might be in different sections: passed, mastered, or the other part (not sure what it's called...)

[quote="PowerOfPi"]Why isn't there a Part 2 or Part 1? Did Part 3 cover all the questions in the first two parts in a better way?[/quote]

You get Parts 1, 2, and 4 at a different time from Part 3. Apparently, most people get part 3 first, so I am in the rare few that got the other parts first.

[hide="My solution"]
The way I approached this problem, was again:
$\mathbb{P}=\frac{Successes}{Possibilities}$

We have 4 republicans, 5 democrats, and 3 independence.

We have to chose 3 from the 12 of them.
That is $\binom{12}{3}$ total ways.

$\frac{Successes}{\binom{12}{3}}$

$\binom{12}{3}=\frac{12*11*10}{3*2}\implies 4*11*5\implies 20*11\implies 20(10+1)\implies 200+20=220$

$\frac{Successes}{220}$

We have to have one republican from the 4 republicans.  That is $\binom{4}{1}=4$
We have to have one democrat from the 5 democrats.  That is $\binom{5}{1}=5$
We have to have one independence from 3 independence.  That is $\binom{3}{1}=3$

We have:
$\frac{\binom{4}{1}*\binom{5}{1}*\binom{3}{1}}{\binom{12}{3}}\implies \frac{4*5*3}{220}\implies \frac{20*3}{220}\implies \frac{60}{220}\implies \frac{\frac{60}{20}}{\frac{220}{20}}\implies \frac{3}{11}$
$\boxed{\frac{3}{11}}$ is our total probability.  [/hide]
Edit:  Thanks bluecarneal for grammar check =)


Find all pairs of positive integers $ (n,k)$ so that $ (n\plus{}1)^k\minus{}1\equal{}n!$.

if $ n \plus{} 1$ is not prime we have a prime like p that $ p$ divide $ n \plus{} 1$ but $ p|n!$ so contradicts
so $ n \plus{} 1 \equal{} p$ and p is prime.
if $ p > 6$ we have both $ \frac {p \minus{} 1}{2}$ and $ 2$ in factorization of $ (p \minus{} 1)!$ so $ (p \minus{} 1)^2|(p \minus{} 1)!$
so we have $ (p \minus{} 1)^2|p^k \minus{} 1$ so $ p \minus{} 1|k$ so $ p \minus{} 1\leq k$ so we have $ p^{p \minus{} 1} \minus{} 1 \leq (p \minus{} 1)!$but we have the  reverse.
so $ n < 6$

[quote="sumita"]so we have $ (p \minus{} 1)^2|p^k \minus{} 1$ so $ p \minus{} 1|k$[/quote]how come $ (p\minus{}1)^2|p^k\minus{}1$ implies that $ p\minus{}1|k$? :huh:

(P-1+1)^k -1 = (p-1)!

Follows from Binomial Theorem

$ \frac{p^k \minus{} 1}{p \minus{} 1} \equal{} \sum_{i\equal{}0}^{k\minus{}1} p^i \equiv \sum_{i\equal{}0}^{k\minus{}1} 1^i \equiv k \bmod p \minus{} 1$.

A similar problem appeared in Singapore Maths Olympiad Senior Section 2000: [url= Singapore Senior MO 2000] https://artofproblemsolving.com/community/c1068820h2051909p14584003[/url]

If, p is a prime factor of $n+1$  its aslo factor  of $n!+1$. However for the numbers $1,2,...,n$ none of them are factor of $n!+1$. So we get $n+1=p$ thus $n+1$ must be prime. (Actualy if we want to proof $n+1$ must be prime we also can use wilson). For $n+1=2,3,5$, we have solution $(n,k)=(1,1),(2,1),(4,2)$. Let us show that there are no other solutions. Indeed supose $n+1$ is a prime number $\geq7$. Then $n=2m$, $m>2$. Now $2<m<n$, and thus $n^2=2mn$ divides $n!$. Thus $n^2$ divides
\[(n+1)^k-1=n^k+kn^{k-1}+...+\frac{(k-1)k}{2}n^2+nk\]
Thus $n^2|kn$ and in particular, $k\geq n$. It follows that
\[n!=(n+1)^k-1>n^k\geq n^n\geq n!\]
A contradiction, then the solution is  $(n,k)=(1,1),(2,1),(4,2)$


geometry
3D geometry
number theory proposed
number theory

A positive integer is written on each of the six faces of a cube. For each vertex of the cube we compute the product of the numbers on the three adjacent faces. The sum of these products is $1001$. What is the sum of the six numbers on the faces?

Call a, b, c, d, e, f the positive integers written on the six faces of the cube; hereby, assume that the numbers a and b are written on opposite faces, so are the numbers c and d, and so are the numbers e and f. Since all numbers a, b, c, d, e, f are positive integers, they are all $\geq 1$, and thus the numbers a + b, c + d and e + f are positive integers $\geq 2$.

Now, according to the problem, we have

ace + acf + bce + bcf + bde + bdf + ade + adf = 1001.

In other words,

(a + b) (c + d) (e + f) = 1001.

But since the canonical representation of 1001 is $1001 = 7 \cdot 11 \cdot 13$, the only way to represent 1001 as a product of three positive integers $\geq 2$ is $1001 = 7 \cdot 11 \cdot 13$. Since we know that the numbers a + b, c + d and e + f are positive integers $\geq 2$, we thus see that the numbers a + b, c + d, e + f equal to the numbers 7, 11, 13 in some order. Hence, the sum of the six numbers on the faces of the cube is

a + b + c + d + e + f = (a + b) + (c + d) + (e + f) = 7 + 11 + 13 = 31.

What we learn from this problem is that the identity

ace + acf + bce + bcf + bde + bdf + ade + adf = (a + b) (c + d) (e + f)

can sometimes be helpful in problems concerning numbers written on the faces of a cube, and that the canonical representation of 1001 can be of use not only in questions about digits.

This is in fact Darij's solution, re-posted.


Compute the sum of all the roots of $ (2x \plus{} 3)(x \minus{} 4) \plus{} (2x \plus{} 3)(x \minus{} 6) \equal{} 0$.

$ \textbf{(A)}\ 7/2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 13$

[hide]Factor/simplify to get $ (2x\plus{}3)(2x\minus{}1)\equal{}0$, so the sum is 7/2,A.[/hide]

[quote="worthawholebean"]Compute the sum of all the roots of $ (2x \plus{} 3)(x \minus{} 4) \plus{} (2x \plus{} 3)(x \minus{} 6) \equal{} 0$.

$ \text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 13$[/quote]
[hide]Factoring out $ (2x\plus{}3)$ from the equation results in $ (2x\plus{}3)(x\minus{}4\plus{}x\minus{}6)$. Simplifying gets it to $ (2x\plus{}3)(2x\minus{}10)$. So the roots are $ 5$ and $ \minus{}\frac{3}{2}$. The sum is then $ \frac{7}{2}$. [/hide]

[hide=Solution]We can factor to get $(2x+3)(2x-10)=0$, so the roots are $5, -\frac{3}{2}$ for a sum of $\frac{7}{2} \implies \boxed{\text{(A)}}$.[/hide]

[quote=sugar_rush]Hey wasn't this originally in HSM? Why'd it get bumped down to MSM?[/quote]

Because it is MSM content :|

[quote=sugar_rush][quote name="megarnie" url="/community/p22432940"]
[hide=Solution]Expanding gives $4x^2-14x-30=0$ By Vieta's formulas, the sum of the roots is $-(-14)/4=\boxed{(A) 7/2}$. [/hide]
[/quote]

Actually, you don't even need Vieta's, and that makes even more sense considering this is only AMC 12 problem 1.

[hide=solution without Vieta's]The factor $2x+3$ is common to both addends, so factoring it out gives $(2x+3)(x-4+x-6)=0$ or $(2x+3)(2x-10)=0$. By the Zero Product Property, one of $2x+3$ or $2x-10$ equals $0,$ so the roots are $-\displaystyle\frac{3}{2}$ and $5$. Adding them up gives $\boxed{\textbf{(A)} ~\displaystyle\frac{7}{2}}$[/hide][/quote]

I used vieta’s to make a different solution.

Imagine thinking roots were $3/2,4,2/3,6$ rip :(


[quote="matex (L)(L)(L)"]Find all pairs of natural numbers $ (a,b)$ for which:

$ a\geq b$

$ gcd(a,b) \plus{} stm(a,b) \plus{} a \plus{} b \equal{} ab$[/quote]

$ gcd(a,b)\equal{}(a,b)\equal{}k$----> $ a\equal{}nk;b\equal{}mk$---> $ (n,m)\equal{}1$ and $ n\geq m$

$ stm(a,b)\equal{}[a,b]\equal{}nmk$-----> $ k\plus{}nmk\plus{}nk\plus{}mk\equal{}nk*mk$/:k ---> $ 1\plus{}nm\plus{}n\plus{}m\equal{}nmk$----> $ (n\plus{}1)(m\plus{}1)\equal{}nmk$---> $ m\plus{}1: n$---> $ n\equal{}m\plus{}1$----->$ (m\plus{}1,m)\equal{}1$---> $ n\plus{}1\equal{}m\plus{}2: m$---->
$ m\equal{}1;2$

$ m\equal{}1$--> $ n\equal{}2$---> $ k\equal{}3$ --->$ a\equal{}6;b\equal{}3$
$ m\equal{}2$--> $ n\equal{}3$---> $ k\equal{}2$ --->$ a\equal{}6;b\equal{}4$


Between New York City and Boston, there are 6 bus stations (A_1,A_2,...,A_6) and the distance between each station is equal.

In the morning, two buses, bus A and bus B, departed from New York and made their stops among these 6 stations.

In the evening, Bus A stopped at station A_6, and bus B returned to station A_1.  Prove that no matter how these two buses traveled, their total travel distances are not equal.

i'll take it that the busses can't just stop in the middle of two stops and turn around, right?

if so... [hide]say the distance between each station was x. then for a bus to go from station 1 to station 6, it would be a distance of 5x. then, the bus can return to station 6 in any even numbered x, because they must go to somewhere and back, taking the same distance. that would mean the bus leaving station 1 and ending at station 6 would always be an odd number x. the bus leaving and coming back to station 1 would always have an even number x for distance, because of the reason explained earlier.[/hide]

i'm really not too good with proofs. there's also a lot of stuff i don't know how to write that might've helped here.

Good job, and a fine write-up -- it's clear, concise, and well-explained.


I'm looking to take an online college math course this summer. I did the 50H sequence at Stanford during the school year, which included MV calc, lin alg, and ODEs. I'm looking to take an Algebra course. I was wondering which websites would be able to offer a mostly move at your own pace course. I know EPGY, but have never taken an online course there, (or anywhere else other than AoPS.)

I liked EPGY's algebra course.  It pretty much covers everything from Stanford's Math 120 except the Sylow theorems, which I found pretty easy to pick up later.

the main thing i'm unsure of, is whether the class is self paced. i'm going to have a lot of free time this summer, so would it be possible to speed through the material in like, 1.5-2 months? or are the lectures at a set time, and it would take the entire quarter length (which would go into my normal high school starting)

From my experience with EPGY (some high school math and C programming classes), it's at your own pace. Faster you go, more money you save :wink:.

Entirely self-paced.  Sometimes they have a schedule that they don't want you to fall behind on, but getting ahead shouldn't be a problem.

I just finished EPGY's modern algebra course :).  They give you a full 6 months to finish the course but you can go faster than that if you wish.  However, in practice I don't think it should take more than 3 if you work at the recommended pace.  But if you take more than 6 moths they double your tuition...

I found the lectures easy to follow (rather, easy given the complexity of the material) and the text book they use is very good.  Your grade is 50% homework and 50% your final, and I found the grading to be reasonable.


geometry
geometric transformation
rotation
analytic geometry
linear algebra
matrix
trigonometry

If I only rotated x axis or only y axis phi degrees, how will a point P be oriented in comparison to point P on a regular cartesian coordinate system?

If P = (j,k), would it be j units on the rotated x axis, and k units up parallel to the y axis?

I don't know if you can find an equivalent point after rotating only one axis.  If you rotate both together by $\phi$, then your rotation matrix is of course

\[ \begin{bmatrix}\cos\phi & -\sin\phi \\ \sin\phi & \cos\phi\end{bmatrix}. \]


[hide="PROBLEM 1 "]

In $ \triangle ABC$, $ \angle B \equal{} 30^o , \angle C \equal{} 20^o$. If $ D$ is on $ BC$ such that $ BD \equal{} AC$ then $ \angle DAC \equal{} ?$[/hide]

[hide="PROBLEM 2"]

$ ABCD$ is a convex quadrilateral, $ AB \equal{} 2 \sqrt {3} , CD \equal{} 3$ and diagonals intersects at M. If $ \angle CMD \equal{} 30^o$ then find the radius of the circumcircle of $ ABCD$[/hide]

[hide="PROBLEM 3 "]

The lengths of sides of $ ABC$ are $ 5,6$, and $ 7$. $ K$ is the center of inscribed circle and $ G$ is the intersection point of medians then $ KG \equal{} ?$[/hide]

[hide="Problem 3"]Let $ AB \equal{} c \equal{} 6, AC \equal{} 5$. If $ P$ is the triangle area and $ s \equal{} 9$ the semiperimeter, then $ P^2 \equal{} 9\cdot 4\cdot 3\cdot 2\implies P \equal{} 6\sqrt {6}\implies r \equal{} {P\over s} \equal{} {2\sqrt {6}\over 3}$. The altitude $ h_c$ is $ h_c \equal{} {2P\over c} \equal{} 2\sqrt {6}$. Let $ G'$ be the foot of the perpendicular from $ G$ to $ AB$. Then $ GG' \equal{} {h_c\over 3} \equal{} {2\sqrt {6}\over 3} \equal{} r$, therefore $ KG\parallel AB$.

Let $ K'$ be the foot of the perpendicular from $ K$ to $ AB$. Then $ AK' \equal{} {b \plus{} c \minus{} a\over 2} \equal{} 2$.

Let $ H$ be the foot of $ h_c$. Then $ AH^2 \equal{} 5^2 \minus{} (2\sqrt {6})^2 \equal{} 1\implies AH \equal{} 1$. If $ M$ is the midpoint of $ AB$, then $ MH \equal{} 2\implies MG' \equal{} {1\over 3}MH \equal{} {2\over 3}\implies AG' \equal{} AM \minus{} MG' \equal{} {7\over 3}$. Hence $ KG \equal{} K'G' \equal{} {1\over 3}$[/hide]

[quote="ertanrock"][quote="Farenhajt"][hide="Problem 3"]Let $ AB \equal{} c \equal{} 6, AC \equal{} 5$. If $ P$ is the triangle area and $ s \equal{} 9$ the semiperimeter, then $ P^2 \equal{} 9\cdot 4\cdot 3\cdot 2\implies P \equal{} 6\sqrt {6}\implies r \equal{} {P\over s} \equal{} {2\sqrt {3}\over 3}$. The altitude $ h_c$ is $ h_c \equal{} {2P\over c} \equal{} 2\sqrt {6}$. Let $ G'$ be the foot of the perpendicular from $ G$ to $ AB$. Then $ GG' \equal{} {h_c\over 3} \equal{} {2\sqrt {6}\over 3} \equal{} r$, therefore $ KG\parallel AB$.

Let $ K'$ be the foot of the perpendicular from $ K$ to $ AB$. Then $ AK' \equal{} {b \plus{} c \minus{} a\over 2} \equal{} 3$, hence $ K'$ is the midpoint of $ AB$.

Let $ H$ be the foot of $ h_c$. Then $ AH^2 \equal{} 5^2 \minus{} (2\sqrt {6})^2 \equal{} 1\implies AH \equal{} 1$. Hence $ KG \equal{} {1\over 3}HK' \equal{} {2\over 3}$[/hide][/quote]

are you sure $ GG' \equal{} r$ ???[/quote]

Yup, I am :) I did have some typos and other errors, hope all is edited now.

[quote="ertanrock"][hide="PROBLEM 0 "]

$ ABCD$ convex quarilateral, $ AC$ and $ BD$ are diagonals. If $ \angle BAC \equal{} 10^o , \angle BDA \equal{} 50^o , AB \equal{} AD$ and $ AC \equal{} AD$ then $ \angle BCA \equal{} ?$[/hide]
[/quote]

[hide="?????"]
I think $ \angle BCA \equal{} \angle BDA \equal{} 50^\circ$.
[/hide]

Wrong, i thought it's cyclic  :blush: .

[quote="mathson"][quote="ertanrock"][hide="PROBLEM 0 "]

$ ABCD$ convex quarilateral, $ AC$ and $ BD$ are diagonals. If $ \angle BAC \equal{} 10^o , \angle BDA \equal{} 50^o , AB \equal{} AD$ and $ AC \equal{} AD$ then $ \angle BCA \equal{} ?$[/hide]
[/quote]

[hide="?????"]
I think $ \angle BCA \equal{} \angle BDA \equal{} 50^\circ$.
[/hide][/quote]

correction $ AB \equal{} BD$

[hide="Problem 1"]Take $E$, the circumcenter of $\triangle ABC$; $\triangle ACE$ is equilateral, $\triangle ABE$ is isoscels, $40^\circ-70^\circ-70^\circ$, so $\triangle BDE\cong\triangle EAB (s.a.s)$, and $ABED$ is an isosceles trapezoid, by easy angle chase getting $\angle DAC=20^\circ$, done.[/hide]

Best regards,
sunken rock


Let m and n be arbitrary non negative integers.Prove that

                         (2m!)(2n!)/m!n!(m+n)!
is an integer.

It's enough to prove:

$[\frac{2m}{p^{i}}]+[\frac{2n}{p^{i}}] \geq [\frac{n}{p^{i}}]+[\frac{m}{p^{i}}]+[\frac{m+n}{p^{i}}]$

which is kind of easy


Hi. As most of you probably know, I am Young Jeong from Thomas Jefferson HS. More people seem to join AoPS and participate in more math contests, so our school has decided to host an Elementary/Middle School Math Contest (grades 4-8) called Middlementary Math Bonanza. (on saturday, December 2nd)

Basically, these are the tests: Individual and Team. Individual is a single test consisting of 30 FR questions to be solved in 60 minutes. Team tests consist of Mental Math (5 questions per person... we all know how M.M. goes so I dont need to explain in detail), and subject tests such as algebra, geometry, probability and potpourri. It is much like the Blaine tests, except different format (#Q's, level of difficulty, etc.)

Awards will be given to the top 10 individuals in each divisions, and top 5 teams in each division based on the team tests.

The registration fee is $40 for a team, or$15 per individual. In addition, the coach will receive paper and electronic copies of all tests, answers, and solutions.

So for more info, you can PM me, call me at 206-370-0542, or email me at youngsjeong@gmail.com, and I will give you more information for you as well as your school. And, dont forget, spread the words around!!!

Good luck with this, unfortunately I don't really associate w/ middle schoolers (wow, that sounds snobbish), else I would tell everyone I know about this.

update: the testing will be postponed to January. I will keep those of you interested with the latest soon. Or PM.


inequalities
function
inequalities proposed

Let $x, y, z>0$ such that $x+y+z\geq 1$. prove the following inequality:

                           \[ \frac{x\sqrt{x}}{y+z}+\frac{y\sqrt{y}}{z+x}+\frac{z\sqrt{z}}{x+y}\geq\frac{\sqrt{3}}{2} \].

WLOG suppose $x\geq y \geq z$, it's obvious that we just need to consider $x+y+z=1$

by cauchy schwarz we have that
$\displaystyle \sum_{cyc}\frac{x\sqrt x}{y+z}\sum_{cyc}\sqrt x(y+z)\geq (\sum_{cyc}x)^2=1$

but by chebyschev and cauchy  we have that
$\displaystyle \sum_{cyc}\sqrt x(y+z)\leq \frac{\sum_{cyc}\sqrt x \sum_{cyc}y+z}{3}=\frac{2}{3}\sum_{cyc}\sqrt x \leq \frac{2}{3}\sqrt{3\sum_{cyc}x}=\frac{2}{\sqrt3}$

which implies the desired result...

More general  , for $x_1,...,x_n > 0$ and $a \geq 1$ the following inequality holds $\sum_{k=1}^n\frac{x_k^a}{\displaystyle\sum_{i\neq k}x_i} \geq \frac{n^{2-a}}{n-1}(\displaystyle \sum_{k=1}^nx_k)^{a-1}$,
because of the function $\frac{x^a}{S-x}$ is convexe for $0 < x < S$ and $a \geq 1$
 :cool:


We have distributed two hundred balls into one hundred boxes with the restrictions that no box got more than one hundred balls, and each box got at least one. Prove that it is possible to find some boxes that together contain exactly one hundred balls.

[hide="Solution"]
Label the boxes 1,2,3,....,100 and let $ a_i$ be the number of balls in the i'th box
Considering all 100 sums of the first k $ a_i's$ for k ranging through 1 to 100, we see that one of these sums is divisible by 100 or php ensures two sums to be equal mod 100 and the difference of these two is the sum we seek cz the sum is greater than zero and less than 200
[/hide]

[quote="lajanugen"][...] one of these sums is divisible by 100 or php ensures two sums to be equal mod 100 and the difference of these two is the sum we seek cz the sum is greater than zero and less than 200[/quote]

What if it is the last sum that is divisible by $ 100$, in which case it is equal to $ 200$ and none of the other sums are congruent. For example suppose the sums of the first $ k$ boxes are congruent to $ \{1,2,\dots ,98,99,0\}$.

Oops, i didn't rule out that case, thankyou for pointing it out jvenezuela716.

[hide="Solution"]
Consider the sums $ a_1, a_2, a_1 \plus{} a_2, a_1 \plus{} a_2 \plus{} a_3,.....,a_1 \plus{} .... \plus{} a_{199}$. Now if one of these is divisible by 100, then we're done (each box got atleast one ball), if not, two are congruent mod 100 and except for the case when these two are $ a_1$ and $ a_2$, we are done.
Now relabelling the boxes and following the same argument, if we so far do not have a sum divisible by 100, we finally get that all the $ a_i's$ are congruent mod 100 and since each box got atleast one ball, this is only possible when each had 2 balls and this obviously has the given property

Anyone has others solns?
[/hide]


combinatorics unsolved
combinatorics
Chip firing

At a round table there are 1991 girls, playing a game with $ n$ cards. Initially, one girl holds all the cards. In each turn, if at least one girl holds at least two cards, one of these girls must pass a card to each of her two neighbors. The game ends when and only when each girl is holding at most one card. Prove that if $ n \leq 1991,$ then the game must end.

proof is obvious if n<4, if n=4, say person 3 has all the cards at first. persons 2 and 4 both end up with two cards. then, persons 1 and 5 each have one card, and person 3 gets two cards. after that, persons 1,2,4,5 have one card each, and everyone else has none. a similar, though more complex, pattern, occurs for all n <= 1991.

easy analysis: ultimately, through theory of entropy, the cards will end up spread out as thinly as possible, so 1991 is the highest number where everyone will end up with less than two cards each.

[quote="UrInvalid"]through theory of entropy[/quote] The only substance in your solution is hidden in this phrase.  It might be a true statement that one can use the theory of entropy to show this, but you haven't done so.  Moreover, nowhere in your argument have you used any properties of the number 1991, but to do so is necessary since e.g. the result is false for 6 players and 6 cards.

The keywords for the connection to higher mathematics are "sandpile game" or "chip-firing."  Some discussion is available here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=428298&hilit=Sandpile


geometric sequence
algebra unsolved
algebra

Find a1, a2, a3, ..., a10, given that:
i) they are in geometric progression, in this order.
ii) a1, a2, a3, a4, a5, a6 have 4 digists and a10 have 5 digits.

Are they supposed to be positive integers?

Pierre.

Let $a_i=a_1\cdot r^{i-1}$, where $r=a_2/a_1\in\mathbb{Q}$. Since $a_{10}/a_1>1$ we conclude that $r>1$. Suppose $r=p/q$, where $p/q$ is irreducable fraction. Then $a_{10}=\frac{a_1p^9}{q^9}\in\mathbb{N}$. It implies that $q^9\,|\,a_1$. If $q\geq 3$ then $a_1>10^5$ -- contradiction. If $q=1$ then $p\geq 2$ and $a_{10}=a_1p^9\geq 1000\cdot 2^9>10^5$ -- contradiction. Thus $q=2$ $\Rightarrow$ $512\,|\,a_1$. Suppose $a_1\geq 3\cdot 512$. In this case $a_6=a_1(p/2)^6\geq 3\cdot 512\cdot (3/2)^6>10^4$ -- contradiction. Hence $a_1=1024$. We know that $p$ is odd. Suppose $p\geq 5$ $\Rightarrow$ $a_6>10^4$ -- contradiction. So $p=3$. Finally, $a_i=1024\cdot(3/2)^{i-1}$.

Since the problem doesn't say anything about the numbers being positive or not, I think we should also consider the following possibilities:

$a_i = (-1024) \cdot (\frac{3}{2})^{i-1} $

$a_i = (1024) \cdot (- \frac{3}{2})^{i-1} $

and

$a_i = (-1024) \cdot (- \frac{3}{2})^{i-1} $.


inequalities
calculus
derivative
inequalities proposed

find the best value of k to this ineq is truefor all $ a,b,c\geq 0,abc \equal{} 1$
$ k(\sum \frac {a^2 \plus{} b^2}{c^2}) \plus{} \sum \frac {ab}{c^2} \minus{} \frac {3}{4}\geq \sum \frac {c^2}{ab} \plus{} \sum \frac {a \plus{} b}{2c}$
I think it can be killed by pqr tech but it is not a nice proof

[quote="Sunkern_sunflora"][quote="Allnames"]I think it can be killed by pqr tech but it is not a nice proof[/quote]
Now what is "[color=red]pqr technique[/color]"? Is it some sort of substitution? Please post your solution, however ugly it looks![/quote]
Sorry,i only said it was able to killed by pqr tech
My proof is  diffrent from it,i think my proof is quite short

[quote="Sunkern_sunflora"]I wanted to know [b]what is "pqr tech"[/b]??? Is it some sort of substitution? :([/quote]
pqr tech is a tech Where we let $ a\plus{}b\plus{}c \equal{}p,\sum ab\equal{}q ,abc\equal{}r$,and the ineq can be rewrited in the "pqr form" Then we use some ineq of "pqr tech" and try to used them to solved our ineq.I amnt good at pqr so i cant tell you more about it Sorry

First . Think your problem has some mistakes ;) and the real problem like that :
$ a,b,c \ge 0$, find the best value of $ k$ to this ineq true for all $ a,b,c \ge 0$.
$ k ( \sum \frac{a^2+b^}{c^2} -\sum \frac{a+b}{c} ) +\sum \frac{ab}{c^2} \ge \frac{c^2}{ab}$
Ps: For your problem , we try the simplest set of $ (a,b,c)=(1,1,1)$ =>$ k=\frac{5}{8}$

[quote="mitdac123"]First . Think your problem has some mistakes ;) and the real problem like that :
$ a,b,c \ge 0$, find the best value of $ k$ to this ineq true for all $ a,b,c \ge 0$.
$ k ( \sum \frac {a^2 + b^}{c^2} - \sum \frac {a + b}{c} ) + \sum \frac {ab}{c^2} \ge \frac {c^2}{ab}$
Ps: For your problem , we try the simplest set of $ (a,b,c) = (1,1,1)$ =>$ k = \frac {5}{8}$[/quote]
Sorry mitdac123 ,my problem is true,My best value is similar to yours,It is not hard and i think you can solve it.Really!

[quote="Allnames"]find the best value of k to this ineq is truefor all $ a,b,c\geq 0,abc = 1$
$ k(\sum \frac {a^2 + b^2}{c^2}) + \sum \frac {ab}{c^2} - \frac {3}{4}\geq \sum \frac {c^2}{ab} + \sum \frac {a + b}{2c}$
I think it can be killed by pqr tech but it is not a nice proof[/quote]
[quote="Allnames"]find the best value of k to this ineq is truefor all $ a,b,c\geq 0,abc = 1$
$ k(\sum \frac {a^2 + b^2}{c^2}) + \sum \frac {ab}{c^2} - \frac {3}{4}\geq \sum \frac {c^2}{ab} + \sum \frac {a + b}{2c}$
I think it can be killed by pqr tech but it is not a nice proof[/quote]
with $ a=b=c$ we find$ k=\frac {5}{8}$
$ Let a+b+c=p;ab+bc+ca=q;abc=r$. We have:
$ Ineq \leftrightarrow  \frac {5}{8}\sum a^2b^2(a^2+b^2)+ \sum a^3b^3 - \frac {3a^2b^2c^2}{4} \ge\ abc(a^3+b^3+c^3) + \frac {1}{2}\sum ab(a+b)$
 $ \leftrightarrow  \frac {5}{8}(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)+ \sum a^3b^3 - \frac {3a^2b^2c^2}{4} - abc(a^3+b^3+c^3) - \frac {1}{2}\sum ab(a+b)-\frac {15}{8} \ge\ 0$
$ \leftrightarrow  \frac {5}{8}(p^2-2q)(q^2-2p)+(q^3-3pq+3)-(p^3-3pq+3)- \frac {1}{2}(pq-3) -\frac {21}{8}\ge\0$
$ \leftrightarrow \frac {5p^2q^2}{8}-\frac {q^3}{4}-\frac {9p^3}{4}+2pq-\frac {9}{8} \ge\ 0$
Follow Schur : $ r \ge\ \frac { (4q-p^2)(p^2-q)}{6p} \rightarrow 1 \ge\ \frac { (4q-p^2)(p^2-q)}{6p}$
We easy prove last ineq

$ \frac{5}{8}(\sum \frac {a^2 \plus{} b^2}{c^2}) \plus{} \sum \frac {ab}{c^2} \minus{} \frac {3}{4}\geq \sum \frac {c^2}{ab} \plus{} \sum \frac {a \plus{} b}{2c}$
$ \Leftrightarrow 5p^2q^2\minus{}10p^3\minus{}10q^3\plus{}16pq\minus{}9\ge 0$

setting $ f(p)\equal{}5p^2q^2\minus{}10p^3\minus{}10q^3\plus{}16pq\minus{}9$

$ f'(p)\equal{}10pq^2\minus{}30p^2\plus{}16q\ge 0$

$ \Rightarrow f(p)\ge f(\sqrt{3q})\equal{}(\sqrt{q}\minus{}\sqrt{3})(5q^2\sqrt{q}\plus{}5\sqrt{3}q^2\plus{}15q\sqrt{q}\plus{}\sqrt{3}q\plus{}3q\plus{}3\sqrt{3})$

by Am-Gm $ \Rightarrow q\ge 3 \Rightarrow f(p)\ge f(\sqrt{3q}) \ge 0$

$ "\equal{}"\Leftrightarrow (a,b,c)\equal{}(1,1,1)$

[quote="bunhiacovski"]$ \frac {5}{8}(\sum \frac {a^2 \plus{} b^2}{c^2}) \plus{} \sum \frac {ab}{c^2} \minus{} \frac {3}{4}\geq \sum \frac {c^2}{ab} \plus{} \sum \frac {a \plus{} b}{2c}$
$ \Leftrightarrow 5p^2q^2 \minus{} 10p^3 \minus{} 10q^3 \plus{} 16pq \minus{} 9\ge 0$

setting $ f(p) \equal{} 5p^2q^2 \minus{} 10p^3 \minus{} 10q^3 \plus{} 16pq \minus{} 9$

$ f'(p) \equal{} 10pq^2 \minus{} 30p^2 \plus{} 16q\ge 0$

$ \Rightarrow f(p)\ge f(\sqrt {3q}) \equal{} (\sqrt {q} \minus{} \sqrt {3})(5q^2\sqrt {q} \plus{} 5\sqrt {3}q^2 \plus{} 15q\sqrt {q} \plus{} \sqrt {3}q \plus{} 3q \plus{} 3\sqrt {3})$

by Am-Gm $ \Rightarrow q\ge 3 \Rightarrow f(p)\ge f(\sqrt {3q}) \ge 0$

$ " \equal{} "\Leftrightarrow (a,b,c) \equal{} (1,1,1)$[/quote]
Perhaps,it is the proof which is used "pqr tech" that i said>thank your proof
My ineq $ \Leftrightarrow  \sum c^2(a^2\minus{}b^2\minus{}2ac\minus{}2bc)^2\geq 0$ It is not natural,i know that

[hide="to_allnames"]
BÀi bdt của anh không giống như những bài hằng số tốt nhất khác vì bài của anh nếu thử với trường hợp đơn giản nhất $ a\equal{}b\equal{}c$ là thu được $ k\equal{}\frac{8}{5}$. Còn với 1 bài hằng số tốt nhất thường là khi với k là hằng số tốt nhất thì dấu bằng xảy ra tại không chỉ 1 điểm $ (a,b,c)\sim (1,1,1)$ mà còn xảy ra tjai các điểm khác như :$ (a,b,c)\sim (0,1,2)\sim (0,1,1)\sim (1,2,3)...$. Và em nói chữ thường ở đây vì có 1lọai BDT với hằng số tốt nhất dấu bằng chỉ xảy ra tại 1 điểm nào đó khác $ (a,b,c)\sim (1,1,1)$ hoặc chỉ là gần đạt được điểm xảy ra dấu bằng thứ2 thôi ;) .Em ví dụ 1 bài .
Tìm hằng số $ k$ tốt nhất để :
$ k(\sum a)^6\plus{} (ab\plus{}bc\plus{}ca)^3 \ge abc(a\plus{}b\plus{}c)^3$
;) .Anh đừng $ k\equal{}0$ đấy ;)
[/hide][/hide]

[quote="mitdac123"][hide="to_allnames"]
BÀi bdt của anh không giống như những bài hằng số tốt nhất khác vì bài của anh nếu thử với trường hợp đơn giản nhất $ a \equal{} b \equal{} c$ là thu được $ k \equal{} \frac {8}{5}$. Còn với 1 bài hằng số tốt nhất thường là khi với k là hằng số tốt nhất thì dấu bằng xảy ra tại không chỉ 1 điểm $ (a,b,c)\sim (1,1,1)$ mà còn xảy ra tjai các điểm khác như :$ (a,b,c)\sim (0,1,2)\sim (0,1,1)\sim (1,2,3)...$. Và em nói chữ thường ở đây vì có 1lọai BDT với hằng số tốt nhất dấu bằng chỉ xảy ra tại 1 điểm nào đó khác $ (a,b,c)\sim (1,1,1)$ hoặc chỉ là gần đạt được điểm xảy ra dấu bằng thứ2 thôi ;) .Em ví dụ 1 bài .
Tìm hằng số $ k$ tốt nhất để :
$ k(\sum a)^6 \plus{} (ab \plus{} bc \plus{} ca)^3 \ge abc(a \plus{} b \plus{} c)^3$
;) .Anh đừng $ k \equal{} 0$ đấy ;)
[/hide][/hide][/quote]
[hide="to mitdac123"]Nhất trí với em là bài kiểu này ko được xem là tìm hằng số tốt nhất anh sẽ thử bài của em (nhưng ko nhanh được đâu vì anh còn khá nhiều thứ em thông cảm),trước hết em thử xem bài # của anh hoặc bài http://www.mathlinks.ro/viewtopic.php?t=216355 anh ko nghĩ là nos quá dễ đâu.Cảm ơn em[/hide]

[quote="Allnames"][quote="bunhiacovski"]$ \frac {5}{8}(\sum \frac {a^2 \plus{} b^2}{c^2}) \plus{} \sum \frac {ab}{c^2} \minus{} \frac {3}{4}\geq \sum \frac {c^2}{ab} \plus{} \sum \frac {a \plus{} b}{2c}$
$ \Leftrightarrow 5p^2q^2 \minus{} 10p^3 \minus{} 10q^3 \plus{} 16pq \minus{} 9\ge 0$

setting $ f(p) \equal{} 5p^2q^2 \minus{} 10p^3 \minus{} 10q^3 \plus{} 16pq \minus{} 9$

$ f'(p) \equal{} 10pq^2 \minus{} 30p^2 \plus{} 16q\ge 0$

$ \Rightarrow f(p)\ge f(\sqrt {3q}) \equal{} (\sqrt {q} \minus{} \sqrt {3})(5q^2\sqrt {q} \plus{} 5\sqrt {3}q^2 \plus{} 15q\sqrt {q} \plus{} \sqrt {3}q \plus{} 3q \plus{} 3\sqrt {3})$

by Am-Gm $ \Rightarrow q\ge 3 \Rightarrow f(p)\ge f(\sqrt {3q}) \ge 0$

$ " \equal{} "\Leftrightarrow (a,b,c) \equal{} (1,1,1)$[/quote]
Perhaps,it is the proof which is used "pqr tech" that i said>thank your proof
[color=red]My ineq $ \Leftrightarrow \sum c^2(a^2 \minus{} b^2 \minus{} 2ac \minus{} 2bc)^2\geq 0$ It is not natural,i know that[/color][/quote]
it is here ,it not natural,sorry

[quote="Math pro"]the same to vasc 's problem! :blush:[/quote]
I think this Allnames 's problem only from a develop ; don't same  to vasc 's problem!

[quote="mitdac123"][hide="to_allnames"]
BÀi bdt của anh không giống như những bài hằng số tốt nhất khác vì bài của anh nếu thử với trường hợp đơn giản nhất $ a \equal{} b \equal{} c$ là thu được $ k \equal{} \frac {8}{5}$. Còn với 1 bài hằng số tốt nhất thường là khi với k là hằng số tốt nhất thì dấu bằng xảy ra tại không chỉ 1 điểm $ (a,b,c)\sim (1,1,1)$ mà còn xảy ra tjai các điểm khác như :$ (a,b,c)\sim (0,1,2)\sim (0,1,1)\sim (1,2,3)...$. Và em nói chữ thường ở đây vì có 1lọai BDT với hằng số tốt nhất dấu bằng chỉ xảy ra tại 1 điểm nào đó khác $ (a,b,c)\sim (1,1,1)$ hoặc chỉ là gần đạt được điểm xảy ra dấu bằng thứ2 thôi ;) .Em ví dụ 1 bài .
Tìm hằng số $ k$ tốt nhất để :
$ k(\sum a)^6 \plus{} (ab \plus{} bc \plus{} ca)^3 \ge abc(a \plus{} b \plus{} c)^3$
;) .Anh đừng $ k \equal{} 0$ đấy ;)
[/hide][/hide][/quote]
[hide="to mitdac123"]bài này anh chưa thử hết,nhưng anh nghĩ là dùng pqr,chuẩn hóa $ p\equal{}1$ anh nghĩ là đc( bởi anh gặp nhiều bài dạng này rồi )
Nếu ko đúng thì em bỏ qua nha[/hide]

[quote="Allnames"][quote="mitdac123"]Yep ;)
$ k \equal{} \frac {5.\sqrt {5}}{54} \minus{} \frac {11}{54}$;)[/quote]
[hide="to mitdac123"]số đẹp nhỉ cảm ơn em vì đã cho anh 1 kinh nghiệm thank hiều[/hide][/quote]
[hide="to Allnames & mitdac123"]số này mà đẹp àh Allnames; mitdac123 có thể post lời giải ko?[/hide]

It only uses mixing varibles + derivative ;) [/hide]

You guys keep writing posts with help of links, and I'm being unable to see your posts.  :(   Will you please explain it in details mitdac123! :(
Anyway, what happened to your "cat avatar", mitdac123?

I think you see that I'm not good at English :P , so I have lots of sentence what I can't translate to English :P.

My solution for my problem :P
Assume :$ c \equal{} max(a,b,c)$;$ a \plus{} b \plus{} c \equal{} 1$
Setting :$ t \equal{} \frac {(a \plus{} b)}{2} \le \frac {1}{3}$
We have :
$ (ab \plus{} bc \plus{} ca)^3 \minus{} abc \minus{} [ (t^2 \plus{} 2tc)^2 \minus{} t^2c ] \equal{} \minus{} \frac {1}{64}(a \minus{} b)^2(a^4 \plus{} 12a^3c \plus{} 8a^3b \plus{} 48c^2a^2 \plus{} 30a^2b^2 \plus{} 84a^2bc \plus{} 84ab^2c \plus{} 8ab^3 \plus{} 96bc^2a \plus{} 12b^3c \minus{} 16c \plus{} b^4 \plus{} 48b^2c^2) \ge \frac {1}{4}(a \minus{} b)^2(1 \minus{} 3t)(9t^3 \minus{} 9t^2 \plus{} t \plus{} 1) \ge 0$
(Note : $ \\
a^4 \plus{} 12a^3c \plus{} 8a^3b \plus{} 48c^2a^2 \plus{} 30a^2b^2 \plus{} 84a^2bc \plus{} 84ab^2c \plus{} 8ab^3 \plus{} 96bc^2a \plus{} 12b^3c \plus{} b^4 \plus{} 48b^2c^2 \le 48t^4 \plus{} 192t^3c \plus{} 192c^2t^2$)
So :
$ k \plus{} (ab \plus{} bc \plus{} ca)^3 \minus{} abc \ge k \minus{} t^2(1 \minus{} t)(3t \minus{} 1)^3$
So we must find the maximun value of :$ f(t) \equal{} t^2(1 \minus{} t)(1 \minus{} 3t)^3 \, with \, t \in [0,\frac {1}{3}]$
By derivative , we got:
$ max(f(t)) \equal{} \frac { 5\sqrt {5} \minus{} 11}{54}$
 ;)
P/s: I like bamboo :P

I have another a proof  :wink: I prove it by pqr + Lemma + derivative
This is my proof  :P
assume $ p \equal{} a \plus{} b \plus{} c \equal{} 1$ and $ q \equal{} ab \plus{} bc \plus{} ca \equal{} \frac {1}{3}(1 \minus{} t^2), r \equal{} abc$
Lemma  $ p \equal{} a \plus{} b \plus{} c \equal{} 1, q \equal{} ab \plus{} bc \plus{} ca \equal{} \frac {1}{3}(1 \minus{} t^2$) we have
$ r\le \frac {1}{27}(1 \minus{} 3t^2 \plus{} 2t^3)$
[quote="mitdac123"]
$ k(\sum a)^6 \plus{} (ab \plus{} bc \plus{} ca)^3 \ge abc(a \plus{} b \plus{} c)^3$[/quote]
$ \Leftrightarrow k\ge r \minus{} q^3$
we need prove find $ max(r \minus{} q^3)$
apply lemma we have
$ r \minus{} q^3\le \frac {1}{27}(1 \minus{} 3t^2 \plus{} 2t^3) \minus{} \frac {1}{27}(1 \minus{} t^2)^3 \equal{} \frac {1}{27}t^3(2 \plus{} t)(t \minus{} 1)^2$
$ \Leftrightarrow 27(r \minus{} q^3)\le t^3(2 \plus{} t)(t \minus{} 1)^2 \equal{} g(t)$
By derivative , we got:
$ g(t)\le g(\frac {\sqrt {5} \minus{} 1}{2}) \equal{} \frac {1}{64}( \minus{} 1 \plus{} \sqrt {5})^3(3 \plus{} \sqrt {5})( \minus{} 3 \plus{} \sqrt {5})^2$
$ \Rightarrow max(r \minus{} q^3) \equal{} \frac {1}{1728}( \minus{} 1 \plus{} \sqrt {5})^3(3 \plus{} \sqrt {5})( \minus{} 3 \plus{} \sqrt {5})^2 \equal{} \frac {5\sqrt {5}\minus{}11}{54}$

[quote="bunhiacovski"]I have another a proof  :wink: I prove it by pqr + Lemma + derivative
This is my proof  :P
assume $ p \equal{} a \plus{} b \plus{} c \equal{} 1$ and $ q \equal{} ab \plus{} bc \plus{} ca \equal{} \frac {1}{3}(1 \minus{} t^2), r \equal{} abc$
Lemma  $ p \equal{} a \plus{} b \plus{} c \equal{} 1, q \equal{} ab \plus{} bc \plus{} ca \equal{} \frac {1}{3}(1 \minus{} t^2$) we have
$ r\le \frac {1}{27}(1 \minus{} 3t^2 \plus{} 2t^3)$
[quote="mitdac123"]
$ k(\sum a)^6 \plus{} (ab \plus{} bc \plus{} ca)^3 \ge abc(a \plus{} b \plus{} c)^3$[/quote]
$ \Leftrightarrow k\ge r \minus{} q^3$
we need prove find $ max(r \minus{} q^3)$
apply lemma we have
$ r \minus{} q^3\le \frac {1}{27}(1 \minus{} 3t^2 \plus{} 2t^3) \minus{} \frac {1}{27}(1 \minus{} t^2)^3 \equal{} \frac {1}{27}t^3(2 \plus{} t)(t \minus{} 1)^2$
$ \Leftrightarrow 27(r \minus{} q^3)\le t^3(2 \plus{} t)(t \minus{} 1)^2 \equal{} g(t)$
By derivative , we got:
$ g(t)\le g(\frac {\sqrt {5} \minus{} 1}{2}) \equal{} \frac {1}{64}( \minus{} 1 \plus{} \sqrt {5})^3(3 \plus{} \sqrt {5})( \minus{} 3 \plus{} \sqrt {5})^2$
$ \Rightarrow max(r \minus{} q^3) \equal{} \frac {1}{1728}( \minus{} 1 \plus{} \sqrt {5})^3(3 \plus{} \sqrt {5})( \minus{} 3 \plus{} \sqrt {5})^2 \equal{} \frac {5\sqrt {5} \minus{} 11}{54}$[/quote]
Here is my proof,thank bunhiacopxki


$ 0\in \{\sqrt {2},\frac {\sqrt {2}} {2},\frac {\sqrt {2}} {4},\frac {\sqrt {2}} {8},\cdots \}$
$ 0\not\in \{1,\frac {1} {2},\frac {1} {4},\frac {1} {8},\cdots \}$

Is this right ? I feel like related complete axiom (?) , but I can't say  'right' or 'no' explicitly .
Please help me .

$ 0$ is a limitpoint of $ A\equal{}\{\frac{\sqrt{2}}{2^n}|n\in \mathbb{N}\}$ but $ 0\not\in A$. The same applies without the $ \sqrt{2}$.

$ A \equal{} \{1,\frac {1} {2},\frac {1} {4},\frac {1} {8},\cdots \}$ is not open nor closed.$ A$ has a limitpoint $ 0$ (only one).

Lemma.   $ [F$ is closed $ ]$ $ \Leftrightarrow [x_{n} \in F , x_{n} \rightarrow x \Rightarrow x \in F]$

$ \Rightarrow 0\not\in A$

But I don't have the proof that $ A$ is not open nor closed. In fact , the part for not-closed needs . :roll:

Moreover , I am not sure that $ \{\sqrt {2},\frac {\sqrt {2}} {2},\frac {\sqrt {2}} {4},\cdots \}$ is open or closed too.

It is simpler than you think: If $ 0\in A$ then $ 0$ would be of the form $ \frac{\sqrt{2}}{2^n}$ for some $ n$. There is no need for real analysis here.


Find all integer solutions $ (x,y)$ such that

$ 1\plus{}x^2y\equal{}x^2\plus{}2xy\plus{}2x\plus{}y$

I have a solution, but maybe you find a different one  :D

[hide="Solution"] We have

$ y(x^2 \minus{} 2x \minus{} 1) \equal{} x^2 \plus{} 2x \minus{} 1 \implies$
$ y \equal{} \frac{x^2 \plus{} 2x \minus{} 1}{x^2 \minus{} 2x \minus{} 1} \equal{} 1 \plus{} \frac{4x}{x^2 \minus{} 2x \minus{} 1}$.

Computation for $ \minus{}2 \le x \le 6$ gives the solutions $ (0, 1), (1, \minus{}1), (\minus{}1, \minus{}1), (2, \minus{}7), (3, 7)$.  When $ x \le \minus{}3$ or $ x \ge 7$ we have $ |4x| < |x^2 \minus{} 2x \minus{} 1|$ and hence no integer solutions. [/hide]


I have some questions regarding the SAT essay. I'm having trouble writing an effective (and long) essay for practice prompts in the given time limit. The hardest thing for me is thinking of appropriate examples relating to the prompt. I was wondering, since I can think of an example from Naruto that works for just about any prompt :) , will examples from Naruto (or something similar) yield the same points as examples from say, literature or history? (assuming that my examples are relevent)

While the SAT readers will claim that they give equal weight to all examples, keep in mind that they ARE all high school or college English/history teachers, and they'll be inherently biased to something like Pride and Prejudice and The Adventures of Huckleberry Finn, as opposed to Naruto or Bleach.


Does anyone know the written rankings for the last year's National CD (I only know 6)

Also does anyone know the scores of the top 12. I'm just curious.

Also does 5 vs. 12 winner always play 4 like in March Madness or do they re-seed like in NHL?

Sorry if I too many questions, I've never been to NATS before.