There are 6 blacks spheres and 7 white spheres in a box. Find the probability that 3 spheres taken casualy are from the same colour .

[hide="Solution"]
There are two cases that must be considered, the three spheres are black or the three spheres are white.

Case 1 All three spheres are Black.

The probability is $ \frac{6}{13}\cdot\frac{5}{12}\cdot\frac{4}{11}\equal{}\frac{10}{143}$

Case 2 All three spheres are White.

The probability is $ \frac{7}{13}\cdot\frac{6}{12}\cdot\frac{5}{11}\equal{}\frac{35}{286}$

So the probability is $ \frac{10}{143}\plus{}\frac{35}{286}\equal{}\frac{55}{286}\equal{}\boxed{\frac{5}{26}}$
[/hide]

EZ.
the chance of 3 black are
6/13 x 5/12 x 4/11

the chance of 3 white is

7/13 x 1/2 x 5/11

p1 + p2 = 5/26  :D

Now, consider the probability if the spheres are replaced... :lol:

Is that an extension?

[hide]
$ \left(\frac{6}{13}\right)^3\plus{}\left(\frac{7}{13}\right)^3\equal{}\boxed{\frac{43}{169}}$
[/hide]

Can you please explain how you got to that equation? thanks

[b]ThetaPi[/b] — If you're referring to hunter34's last post, remember that the probability of picking a black sphere is $ \frac {6}{13}$, and the probability of picking a white one is $ \frac {7}{13}$. Since the spheres are replaced in the box, the probability is the same each time the person picks a sphere. The probability of the person picking a black sphere three times in a row would then be $ \frac {6}{13} \times \frac {6}{13} \times \frac {6}{13}$, and that of picking a white sphere, $ \frac {7}{13} \times \frac {7}{13} \times \frac {7}{13}$. Which color is picked doesn't matter, so we simply add up the probabilities.

Ok: “Which color is picked [i]dependent[/i] ...” :P

There are $ 2$ cases to consider.

Case 1: All spheres are White.

The probability of getting a white sphere on the first pick is

$ \frac{7}{13}$, because there are $ 7$ white and $ 13$ total.

The probability of getting a white sphere on the second pick is

$ \frac{1}{2}$, because now there are $ 6$ white and $ 12$ total.

The probability of getting a white sphere on the third pick is

$ \frac{5}{11}$, because now there are $ 5$ white and $ 11$ total.

Multiplying these gives $ \frac{7}{13} \cdot \frac{1}{2} \cdot \frac{5}{11}\equal{}\frac{35}{286}$                                    (1)

Case 2: All spheres are Black

The probability, following the same steps as above, is

$ \frac{6}{13} \cdot \frac{5}{12} \cdot \frac{4}{11}\equal{}\frac{20}{286}$                                                   (2)

Adding $ (1)$ and $ (2)$ gives us

$ \frac{55}{286}\equal{}\boxed{\boxed{\boxed{\boxed{\frac{5}{26}}}}}$

[quote="metafor"]Which color is picked doesn't matter, so we simply add up the probabilities.[\quote]

[quote="hunter34"]The word that would fit more would be dependent instead of doesn't matter.[/quote][/quote]

I think we have a confusion of vocabulary here. The correct wording would be, "since getting 3 black spheres and 3 white spheres are [i]mutually exclusive[/i] events (i.e. cannot occur simultaneously), we add their probabilities".

Mathematically this is written $ P(A \ or B) \equal{} P(A) \plus{} P(B)$ iff A and B are mutually exclusive events.

Dependent means something completely different. Two events are dependent if the occurrence of one would "affect" the probability of the other's occurrence. It doesn't indicate when you can add probabilities, it indicates when you can (or can't) multiply probabilities.

OK here is another kind of solving by combinatirics:

$ P(A) \equal{} \frac {\binom{7}{3} \plus{} \binom{6}{3}}{\binom{13}{3}}$  :)

BOGTRO: You're a little late there. Try to read the full thread before posting  :P

I will just expand to solve:

$ \frac {\binom{7}{3} \plus{} \binom{6}{3}}{\binom{13}{3}}$

$ \equal{} \frac {\frac {7!}{3!(7 \minus{} 3)!} \plus{} \frac {6!}{3!(6 \minus{} 3)!}}{\frac {13!}{3!(13 \minus{} 3)!}}$

$ \equal{} \frac {\frac {7!}{3!4!} \plus{} \frac {6!}{3!3!}}{\frac {13!}{3!10!}}$

$ \equal{} \frac {\frac {7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot4\cdot3\cdot2\cdot1} \plus{} \frac {6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot3\cdot2\cdot1}}{\frac {13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}$

$ \equal{} \frac {35 \plus{} 20}{286}$

$ \equal{} \frac {55}{286}$

$ \equal{}\frac{5}{26}$

*whew!*

I hope that solution was correct...

(I'm sure there are more efficient ways to do it, but this was my solution.)

Yes, well I haven't spent too much time with combinatorics and things of that nature... :P


function
complex analysis
complex analysis unsolved

Show that an analytic function $f$ from the open unit disk $D(0,1)$ into itself has at most one fixed point, or is the identity.

Nice problem.

Of course, it should say 'where $f$ is not the identity function'

Say that $f$ fixes $a$ and $b$, with $a,b$ distinct.  Put
\[ \varphi(z)=\frac{a-z}{1-\overline{a}{z}}  \]
Then $\varphi$ is an automorphism of the unit disk which sends $0$ to $a$.  Let $\psi$ denote the function obtained by conjugating $f$ with $\varphi$, e.g, $\psi=\varphi\circ f\circ \varphi^{-1}$.  We have $\psi(0)=0$ and $\psi(\varphi^{-1}(b))=\varphi^{-1}(b)$.  Now we can apply the Schwarz lemma to get that $\psi(z)=\lambda z$ for some $\lambda$ with modulus $1$.  Since $\psi(\varphi^{-1}(b))=\varphi^{-1}(b)$, $\lambda=1$.  Hence $\psi(z)=z$.  But then $\text{id}=\varphi\circ f \circ \varphi^{-1}$.  Right 'multiplying' by $\varphi$ gives $\varphi = \varphi \circ f$, and left 'multiplying' by $\varphi^{-1}$ gives $f=\text{id}$, as desired.

A similar one (and easier one). Suppose that $f: U\to \mathbb{C}$ is analytic where $U$ is a neighborhood of the closed unit disk. Show that if $|f(z)|<1$ on the unit circle then $f$ has exactly one fixed point in the unit disk.


MATHCOUNTS
geometry
3D geometry
modular arithmetic
probability
quadratics
conics

[quote="MathCounts_gi"]dont they have this on the MC website only problems of the week?[/quote]

Uh, I guess, but most of those problems are the same difficulty, around chapter-level. Also you get your (user)name posted here if you solve it.  :D

[quote="xpmath"]What level will the questions be?
I mean, by "easiest" on Monday, do you mean around school/chapter level?[/quote]

Pretty much.

[quote="minicon"][quote="13375P34K43V312"]
Also you get your (user)name posted here if you solve it.  :D[/quote]
Umm, what does that have to do with anything?  Its not really going to make you [i]Famous[/i] or anything.
 :?:[/quote]

Notice the :D.

[quote="13375P34K43V312"]So tomorrow I guess I'll start posting problems of the day, hopefully up until Nationals [u]if people keep answering them[/u]. The Monday ones will probably be the easiest, and will increase in difficulty until the Friday one, and will start over on Monday.

If you solve the problem, pm me your answer and the next day, when I post the next problem, I'll provide a list of people who solved the problem, as well as a (hidden) solution to the problem.

[b]No signing up is needed, and please try to keep this thread as clean as possible (coming from me...). Nothing is mandatory, and again, I'll keep this running as long as people are still actively doing it. No keeping score or any of that fun stuff.[/b].[/quote]
do we go to weekends too?

That would be nice, that's when I have the most time to sit down and do math. Also, will we be able to suggest questions?

Whoops, I accidentally posted on my lil' sis's account.

No weekends.

Yeah, proposing problems is a great idea, so long as you don't answer to your own problem. I will ALWAYS give credit to the original source if the problem is not mine.

[size=200]DAY 1[/size]

Let's start off easy, shall we? BE CAREFUL ON THIS ONE, I got it wrong, as did a former red MOPper and former top 20 Mathcounts finisher.

If $ \sqrt{1\cdot9\cdot9\cdot1}\equal{}z^{2}$, find all possible real values of $ z$.

Can a mod delete the above post please (by minicon)? If you have not answered, please do not click the hide tag, if you do, it's not like I'll ever know, but you're on your honor.

Also, alot of people are pming me solutions, that's great, but not necessary.

Mihail911,
He said [i]Not[/i] to post solutions.  Please stop posting them because other people may copy it and cheat.  If you would be so kind, please edit your post (Like I did.  I made the same mistake! :D)
[i]-minicon[/i]

[quote="Rafaelloaa"]Very easy :rotfl:[/quote]
Don't say things like that. For some people, they might not be so easy, and also, leet said it would be the easiest problem.

DAY 1 RESULTS:

[hide="Submissions"]11[/hide]
[hide="Correct Solvers"] 6:
CheeseIsGood
CircleSquared
xpmath
Identity
abacadaea
minicon
[/hide]

Yeah, it was easy, but tricky.

[hide="solution"] We have $ \sqrt {1\cdot9\cdot9\cdot1} \equal{} 9 \equal{} z^{2}$, thus $ z^{2} \minus{} 9 \equal{} 0\Leftrightarrow (z \minus{} 3)(z \plus{} 3) \equal{} 0\Leftrightarrow z \equal{} \boxed{\pm3}$.[/hide]

[hide="common incorrect answers"] $ \pm9$, 3[/hide]

[hide="source"] My HS Math Team Tryout (Problem 1)[/hide]

[size=200]Day 2[/size]

I'll post these rules at the beginning of every problem.

1. No collaboration or public discussion until the next problem is posted.
2. PM me answers, solutions not necessary.
3. Calculators are not permitted unless otherwise noted.
4. Answers will be accepted up until the posting of the next problem.

A two-digit number $ AB$, where $ A$ and $ B$ are its digits, is equal to 5 less than three times the sum of its digits. What is the number?

[u]Again, please, please, please refrain from posting anything that is not [b]completely[/b] necessary for the purpose of the thread.[/u]

Out of 3 submissions for 87 and 88, CircleSquared, virtuoso, and abacadaea solved 87 correctly, and CircleSquared and virtuoso had correct answers for 88. 0/2 answers were correct on 89.

[hide="87"] $ \dfrac{3\sqrt {3}}{4}$ or $ \dfrac{4\sqrt {3}}{9}$, depending on whether you did square/triangle or triangle/square[/hide]

[hide="88"]21[/hide]

[hide="89"]1275/2[/hide]

If you guys want to post solutions to 87 and 88, you would be ballers. Preferably non-brute force on 88. I'll post some cool stuff on problems related to 88 when I have time. In the mean time,

[size=200]Day 90[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Let $ a_{1},a_{2},a_{3},a_{4},b_{1},b_{2},b_{3},b_{4}$ be arbitrary nonnegative integers, with the restriction that $ 0\le a_{i}\le i$ for $ i \equal{} 1,2,3,4$. Find the number of ordered octuples $ (a_{1},a_{2},a_{3},a_{4},b_{1},b_{2},b_{3},b_{4})$ such that $ a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} a_{4} \plus{} 2b_{1} \plus{} 3b_{2} \plus{} 4b_{3} \plus{} 5b_{4} \equal{} 19$.

[size=200]Day 91[/size]

For how many positive integers $ n$ is $ 9 < \sqrt {n} < 10$?

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[quote="CatalystOfNostalgia"]Let $ a_{1},a_{2},a_{3},a_{4},b_{1},b_{2},b_{3},b_{4}$ be arbitrary nonnegative integers, with the restriction that $ 0\le a_{i}\le i$ for $ i \equal{} 1,2,3,4$. Find the number of ordered octuples $ (a_{1},a_{2},a_{3},a_{4},b_{1},b_{2},b_{3},b_{4})$ such that $ a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} a_{4} \plus{} 2b_{1} \plus{} 3b_{2} \plus{} 4b_{3} \plus{} 5b_{4} \equal{} 19$.[/quote]

Where did you get this problem?

Because my Dad is a member of CMS(Canadian Mathematics Society) and I get a little booklet with problems, and I send in the solutions. I remember seeing this problem in there... And I don't think the deadline has passed for sending solutions to that problem...

That would be unfortunate, seeing as this was on HMMT.

Day 90: No submissions

[hide="Solution"] Note that $ 2b_{1}$ is an arbitrary nonnegative even integer, and that $ a_{1}$ is either 0 or 1. Thus, each nonnegative integer can be expressed uniquely as $ a_{1} \plus{} 2b_{1}$, for nonnegative $ b_{1}$ and $ a_{1}$ being 0 or 1. Similarly, we find that $ a_{2} \plus{} 3b_{2},a_{3} \plus{} 4b_{3},a_{4} \plus{} 5b_{4}$ are just arbitrary nonnegative integers.

Now, let $ c_{1} \equal{} a_{1} \plus{} 2b_{1}$, and define $ c_{2},c_{3},c_{4}$ similarly. We have $ c_{1} \plus{} c_{2} \plus{} c_{3} \plus{} c_{4} \equal{} 19$, which has $ \dbinom{22}{3} \equal{} \boxed{1940}$ solutions by Balls and Urns (if you haven't heard of it, use Google. Make sure you don't just memorize the formula).[/hide]
[hide="Source"] HMMT 2008, its awesomeness was pointed out by our always-friendly moderator kyyuanmathcount (who stunningly finished 4th on combinatorics and was denied one of those beautiful abaci), whom I beat on the AMC.[/hide]

91:

[hide="Submissions"]7[/hide]
[hide="Correct Answers"]6:
mz94
MrRubix
Ihatepie
lotsofmath
abacadaea
infinity4ever[/hide]
[hide="Solution"] Everything is positive, so we can square both sides to get $ 81 < n < 100$, which clearly has $ 100 \minus{} 81 \minus{} 1 \equal{} \boxed{18}$ integral solutions.[/hide]
[hide="Source"]Mandelbrot[/hide]

[size=200]Day 92[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

In calculus, if $ f(x)$ is a function whose graph lies completely above the $ x$-axis for $ a\le x\le b$, then $ \int_{a}^{b}f(x)dx$ is defined as the area bounded by $ f(x)$, the $ x$-axis, and vertical lines through $ a$ and $ b$ on the $ x$-axis. Find $ \int_{3}^{7}(2x \minus{} 5)dx$.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[quote="cognos599"]is this really mc level, you may want to take this to the AMC forum, it actually ain't that hard but it just looks intimidating.[/quote]

cognos, theres a reason why he wrote it like that.

[quote="perfect628"]A) How was that necessary?
B) The AMC forum is locked.
C) Try the problem before assessing its difficulty.
D) Use proper English. Ain't?[/quote]

A) The post you made about the necessity of my post in unnecessary.
B) I didn't mean right away, and i certainly didn't mean the whole topic.
C) I simply said it looked intimidating, i never said it was impossible.
D) I don't use proper grammar unless i am in school so calm down.
 :)

[quote="CatalystOfNostalgia"]
[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

 :rotfl:  :rotfl:  :rotfl:  :rotfl: at the last 3 posts and this one.

[hide="Submissions"]3[/hide]
[hide="Correct Answers"]3:
abacadaea
tenniskidperson3
CircleSquared[/hide]
[hide="Solution"] We see that $ 2x \minus{} 5\ge0\Leftrightarrow x\ge\dfrac{5}{2}$, so $ f(x) \equal{} 2x \minus{} 5$ is above the $ x$-axis in $ [3,7]$. Thus, we want to compute the area between $ y \equal{} 2x \minus{} 5$, $ x \equal{} 3,x \equal{} 7$, and the $ x$-axis. Note that $ y \equal{} 2x \minus{} 5$ is a line, and intersects $ x \equal{} 3,x \equal{} 7$ at $ (3,1),(7,9)$, respectively. We thus see that this is a trapezoid with bases 1,9 and height 4, and therefore, have an area of $ (1 \plus{} 9)2 \equal{} \boxed{20}$.[/hide]
[hide="Source"]own[/hide]

[size=200]Days 93,94,95[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

93. One leg of a right triangle with integer side lengths has length 6. How many possible lengths are there for the hypotenuse?

94. In a row of 20 seats, four must be painted yellow so that no two adjacent seats are yellow. In how many ways can this be done?

95. A long fence has an opening with length 10 feet in the middle. The endpoints of this opening are A and B. The point C is 20 feet away from B such that BC and BA are perpendicular. Saml is attached to a leash, which is attached to point C. If the leash is 20 feet long, what is the area, in square feet, of the region in which Saml can move?

Note that 93,94,95 are not necessarily in order of difficulty.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

That don't involve my being on a leash...

[quote="CatalystOfNostalgia"]Because of a decline in the number of responses, as well as my lack of time, I'm going to officially suspend this. Unless someone wants to take this up with non-sucky problems.[/quote]

Whatever you decide to do, I'd like to thank you for having one of the most educational (and useful) threads in the entire MathCounts forum.


geometry
circumcircle
trapezoid
parallelogram
geometric transformation
rotation
ratio

Outside the triangle $ ABC$ consider the triangles $ BMC$,$ CNA$,$ APB$ such that $ \angle BCM\equal{}\angle ACN\equal{}30^\circ$,$ \angle MBC\equal{}\angle NAC\equal{}40^\circ$,$ \angle PAB\equal{}\angle PBA\equal{}20^\circ$.Prove that $ PM\equal{}PN$ and $ \angle MPN\equal{}30^\circ$.

Should be: Prove that $ PM \equal{} PN$ and $ \angle MPN \equal{} 80^\circ.$

Let $ P_a, P_b$ be circumcenters of the oppositely similar $ \triangle ACN \sim \triangle BCM.$ Let $ \mathcal N, \mathcal M$ be spiral similarities centered at $ N, M$ that carry $ C$ into $ M, N,$ and $ A, B$ into $ A', B',$ respectively. The transformed $ \triangle A'MN \equiv \mathcal N(\triangle ACN) \cong \mathcal M(\triangle BCM) \equiv \triangle B'NM$ with the common corresponding side $ MN \equal{} NM$ are oppositely congruent, $ A'B'MN$ is an isosceles trapezoid with the circumcenter $ P\equiv \mathcal N(P_a) \equiv \mathcal M(P_b).$ The $ \triangle MPN$ is isosceles with $ PM \equal{} PN,$ and $ \angle MPN \equal{} 2 \angle MA'N \equal{} 2 \angle MB'N \equal{} 80^\circ.$ The $ \triangle A'AN \cong \triangle MBB'$ are directly congruent, because they are both spirally similar to the $ \triangle MCN$ and $ A'N \equal{} MB'.$ Therefore, $ A, B$ are corresponding points of the (directly) congruent isosceles $ \triangle A'PN \cong \triangle MPB'$ and $ PA \equal{} PB.$ The angle $ \angle A'PN \equal{} 2 \angle A'MN \equal{} 60^\circ$ and similarly, the angle $ \angle MPB' \equal{} 60^\circ.$ (The congruent isosceles $ \triangle A'PN \cong \triangle MPB'$ are even equilateral.) Consequently, $ \angle APB \equal{} \angle NPB' \equal{} \angle NPM \plus{} \angle MPB' \equal{} 80^\circ \plus{} 60^\circ \equal{} 140^\circ$ and the base angles of the isosceles $ \triangle APB$ are $ \angle PAB \equal{} \angle PBA \equal{} 20^\circ.$

[quote="Inequalities Master"]Outside the triangle $ ABC$ consider the triangles $ BMC$,$ CNA$,$ APB$ such that $ \angle BCM \equal{} \angle ACN \equal{} 30^\circ$,$ \angle MBC \equal{} \angle NAC \equal{} 40^\circ$,$ \angle PAB \equal{} \angle PBA \equal{} 20^\circ$.Prove that $ PM \equal{} PN$ and $ \angle MPN \equal{} 30^\circ$.[/quote]
[hide="Solution"]
Consider point $ Q$ such that $ \triangle CNQ\sim \triangle CMB$ and $ R$ such that $ \triangle CNA\sim \triangle CMR$.  Thus, a spiral similarity maps $ \triangle CBQ\sim \triangle CMN\sim \triangle CRA$.  Also, $ CN \equal{} CN$, $ \angle NCQ \equal{} \angle MCB \equal{} 30 \equal{} \angle ACN$, and $ \angle NQC \equal{} \angle MBC \equal{} 40 \equal{} \angle NAC$, so $ \triangle NAC\cong \triangle NQC$ and it follows that $ AC \equal{} CQ$.  Since $ \triangle CBQ\sim \triangle CRA$ and $ AC \equal{} CQ$, we have that $ \triangle CRA\cong \triangle CBQ$, which means that $ AR \equal{} BQ$.  Since $ \angle ACQ \equal{} \angle ACN \plus{} \angle QCN \equal{} 30 \plus{} 30 \equal{} 60$, we have that $ \triangle ACQ$ is equilateral, as is $ \triangle CBR$.  Now, notice that $ \angle MBR \equal{} \angle CBR \minus{} \angle MBC \equal{} 60 \minus{} 40 \equal{} 20 \equal{} \angle PBA$.  Since $ MB \equal{} MR$ because of the fact that $ \triangle CMB\cong \triangle CMR$, we find that $ \angle MRB \equal{} 20 \equal{} \angle PAB$.  It follows that $ \triangle BPA\sim \triangle BMR$ and thus, a spiral similarity maps $ \triangle BMP\sim \triangle BRA$.  Similarly, $ \triangle APN\sim \triangle ABQ$.  It follows that $ \frac {PN}{BQ} \equal{} \frac {AP}{AB}$ and, noting that $ \angle PAB \equal{} 20 \equal{} \angle PBA\implies PA \equal{} PB$, we have that
\[ \frac {PM}{AR} \equal{} \frac {PB}{AB} \equal{} \frac {AP}{AB} \equal{} \frac {PN}{BQ} \equal{} \frac {PN}{AR}\implies PM \equal{} PN
\]
Now, notice that $ \triangle MNC\sim \triangle BQC\implies \angle MNC \equal{} \angle BQC$ and $ \triangle ANP\sim \triangle AQB\implies \angle AQB \equal{} \angle ANB$.  It follows that $ \angle ANP \plus{} \angle MNC \equal{} \angle AQB \plus{} \angle BQC \equal{} \angle AQC \equal{} 60$.  Thus,
\[ \angle PNM \equal{} \angle ANC \minus{} (\angle ANP \plus{} \angle MNC) \equal{} (180 \minus{} \angle NAC \minus{} \angle NCA) \minus{} 60 \equal{} 180 \minus{} 30 \minus{} 40 \minus{} 60 \equal{} 50
\]
Similarly, $ \angle PMN \equal{} 50$, so $ \angle MPN \equal{} 180 \minus{} \angle PMN \minus{} \angle PNM \equal{} 80$. [/hide]

Denote Q and R the circumcenters of the triangles BMC & CNA, see that angle <RAC = <QBC = 20 degs and the triangles BMQ and ARN are equilateral (*)
See that PQCR is parallellogram, a well known property which can be easily proved as follows: a rotation of 20 degs around A followed by a dilatation of ratio 2*cos(20) centered A maps P to B and R to C, so PR=CQ and both make the same 20 degs angle with BC (**).
<QPR = <QCR = <C - 40 degs ( 1 ).
Triangles PRN and MQP are congruent (S.A.S), due to: a) PQ = CR = RN and PR = CQ = MQ (both resulting from (*) and (**) and
<PRN = <MQP = 60 + <C ( 2 ) , consequently PM = PN and <RPN = <QMP ( 3 ).
From (2) & (3) we get <MPQ + <RPN = 120 degs - <C ( 4 ). Adding (1) and (4) we conclude <MPQ+<QPR+<RPN=80 degs.

Best regards,
sunken rock

Let $ A'$ be the point for which $ BAC$ and $ BA'M$ are directly similar. That is, a spiral similarity about $ B$ maps $ AC$ to $ A'M$. Note $ \angle A'BM\equal{}\angle B$, so $ \angle ABC\equal{}\angle A'BM\iff \angle CBM\equal{}\boxed{\angle ABA'\equal{}40}$.

Furthermore, that spiral similarity means that a spiral similarity about $ B$ maps $ AA'$ to $ CM$. Thus $ \boxed{\angle A'AB\equal{}\angle MCB\equal{}30}$.

Let $ A''$ be the point in $ \triangle ABC$ such that $ PA''B$ is equilateral. Then $ \boxed{\angle A''BP\equal{}40}$, and $ \angle APA''\equal{}(180\minus{}20\minus{}20)\minus{}60\equal{}80$. Note that $ A''P\equal{}PB\equal{}PA$, so $ \angle A''AP\equal{}50$, so $ \boxed{\angle A''AB\equal{}30}$.

Now by comparing the boxed conditions and betweenness, $ A'$ and $ A''$ are the same point.

Let $ \mathcal{R}$ be a rotation center $ P$ of $ 80$ degrees that maps $ A$ to $ A'$.

Observe that $ \angle PA'M\equal{}\angle PAN\equal{}60\plus{}\angle A$.

It follows that $ \mathcal{R}(N)\equal{}N'$ lies on $ A'M$. It suffices to show that $ AN\equal{}A'M$. This is a straightforward calculation with the law of sines. Perhaps I will find a better way to finish this proof.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=45770]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61873]here[/url] .


Functional Analysis
real analysis
real analysis unsolved

Let $L$ - be a Banach space , and $L'$ - it's conjugate (space of all continuous functional). If $L'$- is separable, proove that $L$- is separable.

$f_n$ a dense sequences in $L'$ and $x_n$ a sequence of unit vectosr in $L$ such that $\frac{||f_n||}{2} < |f_n(x_n)|$. Then the set of linear combinations with rational coefficients of the $x_n$ is countable and dense. Of course it suffices to show that the vectorial space $V$ generated by the $x_n$ is dense. To do this, one have to show that if $f \in L'$ vanishes on $V$, then $f$ vanishes everywhere. Take an $\epsilon >0$ and $f_n$ such that $||f-f_n|| < \epsilon$. Then $\frac{||f_n||}{2} < |f_n(x_n)| \leq |(f_n-f)(x_n)|+|f(x_n)|=|(f_n-f)(x_n)| < \epsilon$. This is why $||f|| \leq ||f-f_n||+||f_n|| \leq \epsilon + 2 \epsilon$.


linear algebra
matrix
geometry
algebra
system of equations
analytic geometry

Let $ (a_1,b_1),(a_2,b_2),$ and $ (a_3,b_3)$ be three noncollinear points. Show that the equation of the circle passing through these three points is:

$ \begin{vmatrix}x^2\plus{}y^2 & x & y & 1\\ a_{1}^{2}\plus{}b_{1}^{2} & a_1 & b_1 & 1\\ a_{2}^{2}\plus{}b_{2}^{2} & a_2 & b_2 & 1\\ a_{3}^{2}\plus{}b_{3}^{2} & a_3 & b_3 & 1\end{vmatrix}\equal{} 0$

Such a circle exists because non-colinear points define a plane and such points further form the vertices of a triangle and all triangles have a circumcircle.

The general form of the equation of a circle is given by:

$ x^2 \plus{} y^2 \plus{} xB \plus{} yC \plus{} D \equal{} 0$

To get three other equations like the one above just substitute each of the three points given in the problem. So you will have four equations and then write them in determinant form and you are done.

[quote="akech"]...So you will have four equations and then write them in determinant form and you are done.[/quote]

I'm not sure what you mean by that - wouldn't $ B, C,$ and $ D$ be constants in this case meaning they would be the coefficients written in the matrix?

Could you please elaborate?

You can write the system of equations as:

$ Bx \plus{} Cy \plus{} D \equal{} \minus{} (x^2 \plus{} y^2)$ [substitute the given points to get the rest of the equations, that is, you will have a system of four equations with three unknowns].

Can you write out what the coefficients matrix is? Its rank must be 3 since the three given points are non-collinear.

Now we just write down the augmented matrix and this will have a solution for B, C, D if and only if its determinant is zero.

I feel as though that this is moving more into Linear Algebra's territory.

[quote="akech"]...Its rank must be 3 since the three given points are non-collinear.

Now we just write down the augmented matrix and this will have a solution for B, C, D if and only if its determinant is zero. [/quote]

I lost you around there, and I'm not sure whats wrong with learning some Linear Algebra along the way - if you were wondering this problem was found under an Analytic Geometry section in my textbook.

I was able to prove it is true by using Cramer's Rule to solve for $ B, C$ and $ D$, but I'm sure there is a better way (especially one that doesn't involve as much tedious expansion).

[quote="happyme"]Let $ (a_1,b_1),(a_2,b_2),$ and $ (a_3,b_3)$ be three noncollinear points. Show that the equation of the circle passing through these three points is:

$ \begin{vmatrix}x^2 \plus{} y^2 & x & y & 1 \\
a_{1}^{2} \plus{} b_{1}^{2} & a_1 & b_1 & 1 \\
a_{2}^{2} \plus{} b_{2}^{2} & a_2 & b_2 & 1 \\
a_{3}^{2} \plus{} b_{3}^{2} & a_3 & b_3 & 1\end{vmatrix} \equal{} 0$[/quote]

Expand the determinant by the first row.  Then you get something of the form required.


[quote="math92"]It takes one man one day to dig a 2m x 2m x 2m hole. How long does it take 3 men working at the same rate to dig a 4m x 4m x 4m hole?[/quote]

[hide]

$2^3=8$ and $4^3=64$
one day, one man:  8 cubic feet
one day, 8 men: 64 cubic feet
To keep the proportion equal you do:
1 * 8/3 days, 8 * 3/8 men: 64 cubic feet
2 days and 16 hours *3 = 64 cubic feet
2 days and 16 hours
[/hide]

[quote="math92"]It takes one man one day to dig a 2m x 2m x 2m hole. How long does it take 3 men working at the same rate to dig a 4m x 4m x 4m hole?[/quote]

[hide]One man in one day can dig 8 cubic meters.

3 men in one day can dig 24 cubic meters.

You want to see how long it takes 3 men to dig 64 cubic meters.

It would take them $\frac{64}{24}=\frac{8}{3}$

So it would take them $2\frac{2}{3}$ days to dig it.[/hide]

[quote="math92"]It takes one man one day to dig a 2m x 2m x 2m hole. How long does it take 3 men working at the same rate to dig a 4m x 4m x 4m hole?[/quote]
[hide]
1 man x 1 day = 8 cubic meters

3men x ?days = 64 cubic meters

it should take 2 and 2/3 days

[/hide]


Suppose I have in one page

[code]\begin{enumerate}
\item[P1] write the problem here
\end{enumerate}[/code]

and in other page

[code]\begin{enumerate}
\item[S1] write the solution here
\end{enumerate}[/code]

I want to click on P1 so that it takes me to S1, and in the same fashion for S1.

I know I can use hyperlink and hypertarget, but didn't work. :(

[quote="Null"]I know I can use hyperlink and hypertarget, but didn't work. :([/quote]
Sigh! What does [i]didn't work[/i] mean? Give details of any errors and, in particular, a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] so we can see what is happening.

[code]\begin{enumerate}
\item[\hypertarget{S1}{}\hyperlink{P1}$\mathcal P1$] write the problem here
\end{enumerate}[/code]

and in other page

[code]\begin{enumerate}
\item[\hypertarget{P1}{}\hyperlink{S1}$\mathcal S1$] write the solution here
\end{enumerate}[/code]

That's what I did, and the errors say that mathcal is allowed to use in math mode, and that there're \$ missing.

Hyperlink (like hypertarget) requires [b]two[/b] parameters
[code]\hyperlink{name}{text}
\hypertarget{name}{text}[/code]
You've forgotten to put {} round \$\mathcal P1\$ and \$\mathcal S1\$

(Now you see why your original post couldn't be answered as no-one could have guessed at the problem.)

Oh, perfect, I almost done.

I tried to locate the hypertarget a little bit above so that when clicking on a hyperlink I can see the whole writting.

How can I fix that?

I tried to make

[code]\begin{enumerate}
\item[\hypertarget{S1}{}

\hyperlink{P1}{$\mathcal P1$}] write the problem here
\end{enumerate}[/code]

and

[code]\begin{enumerate}
\item[\hypertarget{P1}{}

\hyperlink{S1}{$\mathcal S1$}] write the solution here
\end{enumerate}[/code]

But that gave me an error.

You can't start a new paragraph in enumerate. Lift the target using \raisebox
[code]\newcommand{\hypertargetraise}[2]{\raisebox{4ex}[0pt]{\hypertarget{#1}{#2}}}[/code]then replace \hypertarget by \hypertargetraise
Change the 4ex to whatever suits.

Damn stevem! You are definitely the best!


function
irrational number
calculus
calculus computations

I am wondering if this question involves intermediate value theorem.

Question: Let function f be continuous on the closed interval (a,b) and suppose that f has only rational values. Show that f is constant on the closed interval (a,b)

Suppose that f is non-constant. Then, for any irrational k (can we do this?)between f(a) and f(b) there exists a value c such that f(c) = k, and hence contradicting the fact that f has only rational values.

That almost works; however, we could have $ f(a)\equal{}f(b)$. But if $ f$ were not constant, then you could find $ p\in [a,b]$ such that $ f(p)\neq f(a)$ and go from there.

That's true.

There is also one thing that I would like to confirm.

We know that Q is dense in R (that is, there exists a rational number between any two real numbers)
But is it true that there exists an irrational number between any two rational numbers?

Certainly: if $ a,b$ rationals with $ a<b$, then $ a\plus{}\frac{b\minus{}a}{\sqrt{2}}$ works.

In fact, "most" numbers between two rationals (or irrationals) are irrational.

Consider also that a rational interval $ (a, b)$ can be translated to the interval $ (0, b \minus{} a)$ and correspondingly the interval $ (0, 1)$ while preserving the structure of the rational and irrational numbers, and I hope you can convince yourself that there exist irrational numbers in $ (0, 1)$!


Game 9 is starting. Hooray!

Here is who's playing and the roles:

windslicer
GoBraves confirmed
chesspro confirmed
bubala
dinoman
math92 confirmed
jmadsen confirmed
yif man12
Sly Si confirmed
CLee
someperson01 confirmed
LadyKn1ght confirmed
Peter VDD confirmed
elvenchamp777 confirmed
DarkKnight confirmed
zeb
h_s_potter2002 confirmed
baseballfreak2 confirmed
jli confirmed
G-UNIT confirmed
SuperNova confirmed
L_Li confirmed
ficklefudge
blueywinkle confirmed

If I missed anyone or if anybody doesn't want to play who is on the list, correct me.

I think we will have the following roles with explanations of how they will work based on some confusions I have noticed:

6 mafia  The mafia will be allowed to communicate during the day

2 angels The angels will know who eachother are, but will only be allowed to communicate at night and only 1 person saved per night.

3 masons No day communication

1 inspector

6 citizens  Citizens get to do nothing!!!!!!

1 vigilante

1 fairy

1 prosecutor If the vigilante is killed, then the prosecutor will still be allowed to search for the miller

1 miller  The miller thinks that he/she is a citizen

1 priest

1 devil

As soon as the pm's are sent, we will start.

[quote="baseballfreak2"]Isn't it game 9? :D[/quote]

yup its game 9 :P

dang it. I feel like an idiot. I did it on basically all the pms too. Ok well everyone just know that it is 9

All pms for game [b][i][u]9[/u][/i][/b] have been sent. if you didn't get one, then you are a citizen. everyone please confirm that you are playing.

No problem. :) And I just felt asleep...  :sleeping:

[quote="chesspro"]Present. 24 people!  :D

The vigilante is innocent, right?[/quote]How subtle! :D

Don't forget it this time, windslicer dude :P

Okay folks, if you guys fail to kill someone in the next week, this game will be stopped. :)

No kills were made during the night. Neither Mafia nor the vigilante submitted anything.

[quote="MithsApprentice"]No kills were made during the night. Neither Mafia nor the vigilante submitted anything.[/quote] :huh:

OK I think that was it... maybe you would better stop it right away.

Um..can I quit..this game is really boring, and besides I'm very busy with school and all.

Apparently, there is no longer interest in this game. In that case, Mafia automatically win. The roles:

RC-7th mafia
SuperNova mafia
GoBraves mafia
LadyKn1ght mafia

Peter VDD angel

WindSlicer inspector

elvenchamp777 fairy

baseballfreak2 vigilante

yif man12 miller

Those of you interested may join Game 10. There are still open spots.

I suspected gobraves and ladyknight... I had no idea about the other 2 though.

[quote="zeb"]I realized gobraves was mafia shortly after I died (I didn't want to say anything). Seriously, did anyone read his arguments? Also, I'm pretty certain that the mafia would have won.[/quote]

You couldnt have said anything anyways after you died. And what was wrong with my arguments?

The let's lynch inactives part of the game was pretty funny, since we all decided to try to be as active as possible.  :)

Well, gobraves, what was with the "he voted correctly that blah was evil so I say lynch him!!!" That was you, right?

I had also suspected Peter of being the inspector (cause I thought the inspector had inspected me first night, and only Sly Si and Peter supported me without hearing my defense).

[quote="zeb"]Well, gobraves, what was with the "he voted correctly that blah was evil so I say lynch him!!!" That was you, right?

[/quote]

So..? It worked didnt it? :P


Εαν δοθεντων:

1) δεν ειναι αληθες οτι : 1>0 and 2+2=5

2) εαν ~1>0, τοτε $ 1\leq 0$ ....... "~" ειναι, οχι

3) εαν $ 2\plus{}2\neq 5$ τοτε, 2+3=6

4) $ 2\plus{}3\neq 6$

.......να αποδειχθη οτι: $ 1\leq 0$

[quote="τρικλινο"][quote="thatin"][hide]
! (2+3=6) => 2+2=5
οπότε ![(1>0) AND (2+2=5)] <=> !(1>0)
συνεπώς 1<=0
QED
[/hide]
cheers![/quote]

Thatin ,για να κανω τον συνηγορο του διαβολου ,εαν καποιος σου ελεγε οτι η αποδειξη σου ειναι λαθος τι θα εκανες??[/quote]

τίποτα ιδιαίτερο. η εκφώνηση μου φαίνεται αρκετά, στην καλύτερη περίπτωση, ασαφής ώστε να ασχοληθώ παραπάνω δίχως να μου πουν περισσότερα. όπως γιατί είναι λάθος.

o.k λοιπον ας συνεχισουμε αυτο το παιγνιδι.

Εχεις γραψει οτι ¨

...........!(1>0 και 2+2=5) <=> !(1>0).........................................................................(1)

Αλλα  1>0 και 2+2=5 ειναι ψευδες αρα .........  !(1>0 και 2+2=5) ειναι αληθες και επειδη !(1>0) ειναι ψευδες δεν μπορουμε να εχουμε ενα αληθες να συνεπαγεται λογικως ενα ψευδες ,αρα η (1) δεν ειναι αληθης

[quote="thatin"]huuh?
το !(1>0) δεν είναι αυτό που θέλουμε να αποδείξουμε? :maybe:[/quote]

Σωστα αλλα δεν μπορουμε να το κανουμε με λανθασμενο τροπο¨.

Επισης τελικα θελουμε να αποδειξουμε $ 1\leq 0$

Συγνωμη που αργησα να σου απαντησω ,αλλα η καταραμενη γριππη με εχει διαλυσει

[quote="τρικλινο"]
Αλλα  1>0 και 2+2=5 ειναι ψευδες αρα .........  !(1>0 και 2+2=5) ειναι αληθες και επειδη !(1>0) ειναι ψευδες δεν μπορουμε να εχουμε ενα αληθες να συνεπαγεται λογικως ενα ψευδες ,αρα η (1) δεν ειναι αληθης[/quote]

τι ακριβώς εννοείς εδώ?
γιατί:
1)φαίνεται να καταλήγεις στο συμπέρασμα ότι η 1 δεν είναι αληθής θεωρώντας πως η 1 είναι ψευδής πράγμα που δεν έχει πολύ νόημα

2)στο τελευταίο ποστ λες  πως δεν μπορούμε να καταλήξουμε με λάθος τρόπο.
όμως λες πως ο τρόπος μου είναι λάθος διότι η σχέση που καταλήγω (και θέλουμε να αποδειχτεί) είναι λάθος

Εννοω:

!( 1>0 kai 2+2=5) ειναι αληθες,αλλα !(1>0) ειναι ψευδες.

Ετσι εχουμε Α===>ψ το οποιον ειναι ψευδες γιατι δεν μπορουμε να εχουμε ενα αληθες να συνεπαγεται ενα ψευδες.

Επομενως η αποδειξη σου !(1>0 και 2+2=5)<===>!(1>0) ειναι λαθος.

Ο πινακας της υποθετικης προτασεως ( conditional  p-------->q) μας λεει οτι A------>Ψ, ειναι ψευδες

[quote="thatin"]και γιατι το !(1>0) είναι ψευδές?
τότε η άσκηση είναι λάθος[/quote]

Το 1>0 ειναι αληθες, επομενως το !(1>0) ειναι ψευδες.

Αλλα που ειναι η ασκηση λαθος??

[quote="τρικλινο"]

.......να αποδειχθη οτι: $ 1\leq 0$[/quote]

Δηληδη ο συλλογισμος:

Ολοι οι σκυλοι εχουν ουρα 3 μετρα, ο jack ειναι σκυλος ,αρα ο jack εχει ουρα 3 μετρα.

Ειναι λαθος??

όχι. αυτό το συλλογισμό έχω κάνει στην παραπάνω απόδειξη
εσύ είπες πως ο συλλογισμός είναι λάθος διότι a priori οι σκύλοι δεν έχουν 3 ουρές aka 0<1

[quote="thatin"]όχι. αυτό το συλλογισμό έχω κάνει στην παραπάνω απόδειξη
[/quote]

Σε ποιο σημειο της αποδειξεως σου εχεις κανει αυτο τον συλλογισμο.

Αν μπορουσες να γραψεις μια ποιο λεπτομερη αποδειξη ,δικαιλογοντας καθε βημα σου,ισως αυτο να εξαφανιζε καθε διαφωνια μας.

Η συντομια στα μαθηματικα καμμια φορα δημιουργει πορεξηγησεις

στο τελευταίο βήμα μου μου φαίνεται.
θεωρώ πως η απόδειξή μου είναι αρκετά λεπτομερής και αυτό δεν είναι μόνο δική μου άποψη. αντίθετα η εκφώνηση του προβλήματος απαιτεί περισσότερες διευκρινήσεις καθώς μπορεί να ερμηνευτεί με πολλούς τρόπους αφού αν χρησιμοποιήσουμε την καθιερμένη χρήση των συμβόλων τα δεδομένα που δίνει είναι λάθος (πχ η πρόταση 3) και ζητά να αποδειχθεί κάτι λάθος.
όμως πες μου σε ποιο σημείο υπάρχει κατά τη γνώμη σου πρόβλημα
όπως ξαναείπα το επιχείρημά σου ότι καταλήγω σε κάποιο αποτέλεσμα που δεν ισχύει επομένως ο συλλογισμός μου έιναι λάθος δεν ισχύει διότι δεν μπορείς να πεις ότι αυτό που καταλήγω δεν ισχύει (1<=0) εφόσον κάτι τέτοιο δε δίνεται στην άσκηση (και τα σύμβολα 1,2 κτλ δεν έχουν τις συνιθησμένες ιδιότητες) και μάλιστα εφόσον καταλήγω σε αυτό που πρέπει να αποδείξω.
το αποτέλεσμα προφανώς θα είανι αντίθετο προς την κλασσική αριθμητική εφόσον έχουμε αντιθετα προς την κλασσική αριθμητική δεδομένα.

τελικά αυτό που χρειάζεται είναι να διευκρινήσεις τι εννοείς με τα σύμβολα 1,2,3 κτλ καθώς και με τα +, <, <=, = αν δηλαδή τα χρησιμοποιείς με την κανονική ή κάποια άλλη έννοια ώστε να δω μήπως έχω ερμηνεύσει κάτι λάθος.
όπως λες,
η συντομία στα μαθηματικά καμμιά φορά δημιουργεί παρεξηγήσεις;-)

Tα συμβολα εδω εχουν την κανονικη τους εννοια και π.χ ,1>0 ειναι αληθες 1=<0 ειναι ψευδες κ.λ.π

Τι πραγματι συμβαινει εδω ειναι το εξης.

Εαν βαλουμε ..........(1>0)=p, (2+2=5)=q, (1=<0)=r, (2+3=6)=s ,τοτε εχουμε το εξης επιχειρημα.

1) ~( p & q)....................................

2) ~p----------->r...................................

3) ~q---------->s

4) ~s
....................................................................
Nα αποδειχθη r

Η Αποδειξη ειναι αποδειξη στον προτασιακο λογισμο και ειναι γνωστες οι αποδειξεις στον προτασιαλο λογισμο

Τωρα καθε αντικατασταση  των μεταβλητων του επιχειρηματος με οποιαδηποτε αληθεις η ψευδεις προτασεις μας δινει ενα καινουργιο επιχειρημα π.χ εαν βαλουμε.

.........(3>4)=p, (0=0)=q , (London is in Paris)=r, (Athens is in Greece)=s..........το καινουργιο επιχειρημα ειναι.

 1)Δεν ειναι σωστο οτι (3>4 και 0=0)

 2)Εαν το 3 δεν ειναι μεγαλυτερο του 4 ,~(3>4), τοτε το Λονδινο ειναι στο Παρισι

 3)Εαν $ 0\neq 0$ τοτε η Αθηνα ειναι στην Ελλαδα

 4) Ομως η Αθηνα δεν ειναι στην Ελλαδα

 Να αποδειχθη οτι το Λονδινο ειναι στο Παρισι.

 Απειρες ειναι οι αντικαταστασεις ,μερικες με νοημα και αλλες παραξενες και καπως χωρις νοημα.

 Ολες ομως εχουν την ιδια αποδειξη στον προτασιακο λογισμο.

 Μια αντικατασταση ειναι καπως περιεργη γιατι η τιμη των προτασεων ( αληθης,ψευδης) δεν ειναι εμφανης,π.χ λεμε

 καθε τεταρτη παω σινεμα. Η ΑΛΗΘΕΙΑ της προτασεως εξαρταται  απο το αν πραγματι καθε τεταρτη παω σινεμα.

 Σε αυτη την περιπτωση η τιμη των p,q,r,s δεν ειναι σιγουρη αυτο ομως δεν επηρεαζει την αποδειξη,γιατι ειναι η μορφη του επιχειρηματος που παιζει ρολο στην αποδειξη του και οχι η εννοια καθε αντικαταστασης

 Τωρα εχουμε δυο ειδων αποδειξεις στον προτασιακο λογισμο,την σιμαντικη (semantical) και την συντακτικη(syntactical)

 Στην πρωτη αποδεικνυουμε με την βοηθεια  των πινακων αληθειας οτι (για οποιαδηποτε τιμες των μεταβλητων ,p,q,r,s e.t.c) το επιχειρημα ειναι παντοτε ταυτοτητα ,δηλαδη εχει την τιμη αληθες.

 Στην δευτερη αποδεικνυουμε οτι το επιχειρημα ειναι ενα θεωρημα που αποδεικνυεται απο τα αξιωματα του προτασιακου λογισμου και αλλων βασικων θεωρηματων π,χ M.Ponens ,addition introduction, addition elimination ,proof by cases e.t.c

 Τα δυο μερη ενωνονται με το κεντρικο θεωρημα του προτασιακου λογισμου ,και το οποιο θεωρημα αποδεικνυει επισης την πληροτητα του προτασιακου λογισμου.Το θεωρημα ειναι:

Ενα επιχειρημα ειναι σωστο (valid) εαν και μονο εαν ειναι θεωρημα εαν και μονο εαν ειναι ταυτολογια.

Δηλαδη εαν ενα επιχειρημα ειναι ταυτολογια τοτε αποδεικνυεται με την βοηθεια βασικων θεωρηματων ,κανων αποσπασεως κ,λπ

Το παραπανω επιχειρημα ειναι ταυτολογια οπως αποδεικνυεται με τους πινακες αληθειας,και μια οριζοντια γραμμη στον πινακα αληθειας του επιχειρηματος ειναι η παραπανω αντικατασταση

ενδιαφέρον.
άρα τι ακριβώς θέλουμε να δούμε σε αυτή την άσκηση?

Οτι υπαρχει αποδειξη της ασκησης στον προτασιακο λιγισμο κατα την οποια αποδειξη καθε σκαλι της αποδειξεως δικαιολογειται και ετσι αποφευγεται καθε αμφιβολια.

Τωρα εσυ εγραψες μια αποδειξη την οποια θεωρεις λεπτομερη και σωστη.

Θα γραψω λοιπον και εγω μια αποδειξη στο ιδιο στιλ και θα σε ρωτησω εαν ειναι σωστη η λαθος

Η αποδειξη ειναι:

2+2$ \neq 5$ ====> (2+3=6)====>[ (2+3=6) & (2+3$ \neq 6$)] ====>$ 1\leq 0$.

αλλα ~(1>0)--------->$ 1\leq 0$  . Επομενως $ 1\leq 0$.

Kαι σε συμβολα:

Assume ~q

~q====>s=====>( s&~s)===>r. But ~p------>r .Επομενως r

[quote="τρικλινο"]

2+2$ \neq 5$ ====> (2+3=6)====>[ (2+3=6) & (2+3$ \neq 6$)] ====>$ 1\leq 0$.

αλλα ~(1>0)--------->$ 1\leq 0$  . Επομενως $ 1\leq 0$.

Kαι σε συμβολα:

Assume ~q

~q====>s=====>( s&~s)===>r. But ~p------>r .Επομενως r[/quote]

όχι, δε μου φάινεται ορθή.
1) δε δίνεται στην εκφώνηση ότι 2+2$ \neq 5$  ( Assume ~q)
2)στο τρίτο βήμα έχουμε καταλήξει να δεχόμαστε μια πρόταση και την αντίθετή της (αν και δεν κατάλαβα πως)
3)επίσης δεν κατάλαβα πως φτάσαμε στο τέταρτο βήμα

όμως το κύριο πρόβλημα είναι ότι, εφόσον δεχόμαστε τον παραδοσιακό συμβολισμό, τα δεδομένα της άσκησης είανι λάθος διότι καταλήγουμε σε αντιφάσεις.
δεν έχει νόημα ούτε πρακτικό κυρίως αλλά ούτε και θεωρητικό να εφαρμόζουμε κανόνες λογικής αποδεχόμενοι αντικρουόμενα δεδομένα. εκτός αν υπάρχει κάποιο που δεν μπορώ να δω και γι αυτό σε ρώτησα τι είχες κατά νου:-)

Η Παρακατω αποδειξη ειναι μια λεπτομερης αποδειξη της παραπανω αποδειξεως, η οποια αποδειξη ειναι σωστη.

Δοθεντων.

1)  ~(p & q)

2)  ~p--------->r

3)  ~q--------->s

4) ~s

Τωρα προτου αρχισουμε την αποδειξη πρεπει να πουμε για το θεωρημα παραγωγης (Deduction theorem) που χρησιμοποιω στην παραπανω αποδειξη,και το οποιο θεωρημα αποτελει το σημαντικοτερο θεωρημα της μεταγλωσσας εις τον προτασιακο και κατηγορηματικο λογισμο

Εαν στην υποθεση μιας αποδειξις προσθεσουμε μια υποθεση P και κατοπιν αποδειξουμε Q,τοτε εχουμε αποδειξει P------>Q

Kαι σε συμβολα εχουμε.  εαν $ \Delta,P\vdash Q$ Τοτε  $ \Delta\vdash P\rightarrow Q$ ,οπου Δ ειναι η υποθεση.

Συμφωνα με τα παραπανω εχουμε την παρακατω αποδειξη.

5) Υποθετουμε ~q.............................................................................................................και αρχιζουμε την χρησιμοποιηση του θεωρηματος παραγωγης.

6) s................................................................................................................................απο (3) και (5) και χρησιμ.τον κανονα αποσπασεως (Μ.Ponens).

Kαι επειδη απο την γραμμη (4) εχουμε ~s ,εχουμε καταληξη σε μια αντιφαση (s & ~s ).Pαρακατω θα δειξουμε πως μια αντιφαση μπορει να μας οδηγηση σε αυτο που θελουμε να αποδειξουμε.

7) svr...............................................................................................................................απο (6) και χρησιμ. τον κανονα προσθετικης διαζευξεως ( Disjuction introduction, p===>p v q)

8) ~s ------->r..................................................................................................................απο (7) και χρησιμ.τον κανονα αντικαταστασεως ( material implication ,p v q <====>~p------>q)

9) r..................................................................................................................................απο (4) και (8) και χρησιμ. τον κανονα αποσπασεως.

Ετσι βλεπουμε πως μια αντιφαση μπορει να οδηγηση σε αυτο που θελουμε να αποδειξουμε.

10) Και χρησιμ. το θεωρημα παραγωγης απο τις γραμμες (5) εως (9) μπορουμε να συμπερανουμε: ~q------->r....................

11) ~p v ~q --------->r .......................................................................................................απο (2) και (10) και χρησιμ.τον κανονα της αποδειξεως των δυνατων περιπτωσεων( proof by cases, [(p------>r)& (q------>r)]-------->(p v q------>r))

12) ~p v ~q .........................................................................................................................απο (1) και χρησιμ. τον κανονα De Morgan

13) r ....................................................................................................................................απο (11) και (12) και χρησιμ. τον κανονα αποσπασεως.

Ετσι λοιπον αποδειξαμε το ζητουμενο ............................r.........................................................

Kαι συνεπως η αποδειξη που σου παρεθεσα σωστη .Αλλα η συντομια ειναι πολλες φορες απαραδεκτη γιατι οδηγει σε σφαλματα.

Υπαρχουν 3 ,4  αλλες αποδειξεις αρκετα ενδιαφερουσες εκτος της μιας συντομης και απλης.

οκ.
υποθέτεις κάτι αυθαίρετα, καταλήγεις σε μια αντίφαση και από αυτή λύνεις την άσκηση.
τι να πω. κάνετε μήπως παρέα με τον κύριο Βέιλ από τους Άγριους Αριθμούς?

Γιατι εαν θελουμε να χρησιμοπ. την ατοπο απαγωγη,δεν πρεπει να υποθεσουμε ................~r.....................και απο εκει να καταληξουμε στην αντιφαση.................s & ~s ...........και απο εκει στο ...............r..................??

Ας  υποθεσουμε ~r...........και

~r====>p===>~q===>( s & ~s)====>r. Επομενως ............r................

O τροπος αυτος δεν ειναι ακριβως ο ιδιος με αυτον που εχρησιμοπ. εγω??


power of 2
Integer partitions
combinatorics
counting
IMO
IMO 1997

For each positive integer $ n$, let $ f(n)$ denote the number of ways of representing $ n$ as a sum of powers of 2 with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, $ f(4) \equal{} 4$, because the number 4 can be represented in the following four ways: 4; 2+2; 2+1+1; 1+1+1+1.

Prove that, for any integer $ n \geq 3$ we have $ 2^{\frac {n^2}{4}} < f(2^n) < 2^{\frac {n^2}2}$.

I'm confused.  Doesn't the lower bound here contradict the result [url=http://en.wikipedia.org/wiki/Integer_partition#Rademacher.27s__series]here[/url]?  We should have $ f(n) \le p(n)$ if I haven't misunderstood the problem.

It appears that the question has a typo and is supposed to read $ 2^{\frac{n^2}{4}} < f(2^n) < 2^{\frac{n^2}{2}}$.  At least, this is what Negative_3's solution appears to prove :)

Oh, wonderful.  Then $ f(2^n) \equal{} [x^{2^n}] \frac{1}{(1 \minus{} x)(1 \minus{} x^2)(1 \minus{} x^4)...(1 \minus{} x^{2^{n\minus{}1}})} \plus{} 1$ and the leading polynomial term has degree $ n\minus{}1$ and leading coefficient $ \frac{1}{(n\minus{}1)! 2^{ n(n\minus{}1)/2} }$, so I would expect the asymptotic $ \frac{1}{(n\minus{}1)!} 2^{ n(n\minus{}1)/2}$, which is comfortably in-between the two given bounds although much closer to the upper bound.  It should be possible from here to bound the rest of the polynomial terms as well as the periodic terms.

Another thought towards a more accurate asymptotic is to let $ F(x) \equal{} \sum_{n \ge 0} f(n) x^n$ and $ P(x) \equal{} \sum_{n \ge 0} p(n) x^n$ and use the identity

$ P(x) \equal{} F(x) F(x^3) F(x^5) ...$

along with the well-known asymptotic for $ p(n)$.

The typo JBL signaled is still on the main IMO resource page :
[url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1997]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1997[/url]

Edit: Fixed by orl.

There's a very nice way of seeing that the upper limit holds:

A sum $\displaystyle\sum_{i=0}^n k_i 2^i = 2^n$ can be expressed as $\displaystyle\sum_{i,j=0}^n \sigma_{ij} 2^{i+j}$, where $k_i = \displaystyle\sum_{j=0}^n \sigma_{ij}2^j$ and $\sigma_{ij} = 0$  or  $1$.

The value of $\sigma_{k0}$ is determined by the values of the other $\sigma_{ij}$'s with $i+j \leq k$, since
$\displaystyle\sum_{i,j=0}^{i+j=k} \sigma_{ij} 2^{i+j}$ must be a multiple of $2^{k+1}$ for $k < n$.

Excluding the $\sigma_{ij}$'s with $i+j=n$, there are $\dfrac {n(n-1)} 2$ bits that can be either 0 or one.  So there are less than $2^{n(n-1)/2}$ partitions resulting from those bits.  The partitions with $i+j=n$ add $n+1$ more partitions.

So, $f(2^n) < 2^{n(n-1)/2}+n+1$, which is $\leq 2^{n^2/2}$ for $n\geq 3$.

The $\sigma_{ij}$ matrix can also be used to prove the lower limit:

If all the $\sigma_{ij}$'s are 1, then $\displaystyle\sum_{i,j=0}^{i+j=m} \sigma_{ij} 2^{i+j} = \sum_{k=1}^m (k+1) 2^k = m2^{m+1}$.

So if $m2^{m+1}\leq 2^n$ and $\dfrac {m(m+1)} 2 > \dfrac {n^2} 4$ for some $m$, then $f(2^n) > 2^{n^2/4}$.

Since $\displaystyle\lim_{m \rightarrow \infty}\dfrac {\frac {m(m+1)} 2}{(\text{log}_2(m2^{m+1}))^2}  = \dfrac 1 2$,

$\displaystyle\lim_{n \rightarrow \infty}f(2^n)> 2^{n^2/4}$ for large enough $n$.

I think for $n\geq 16$ such $m$ can be found, but the bound would have to be checked in some other way for smaller $n$.

Maybe the $\sigma_{ij}$ matrix could be used to yield a better proof  for the lower bound.

Full solution :-

There exists a bijection between representations in the given form of $2k$ and $2k+1$ for $k= 0,1,2.....$
Adding $1$ to every representation of $2k$, we get $2k+1$, and every representation of $2k+1$ contains atleast one $1$, which can be removed. So, $f(2k+1)= f(2k)$.

Now consider all representations of $2k$. The number of those that contain atleast one $1$ is equal to $f(2k-1)= f(2k-2)$, while the number of those not containing a $1$ equals to $f(k)$.
So, $f(2k)= f(2k-2) + f(k)$.
Summing up those equations over $k=1,2,3....,n$ we get $f(2n) = f(0)+f(1)+....... +f(n)$
By applying induction, we get $f(2^{n+1})\leq 2^nf(2^n)\le 2^n2^{\frac{n^2}{2}} \le 2^{\frac{(n+1)^2}{2}}$ for each $n\ge 3$.

The lower bound can be derived from the first equation. Note that the above given map $f(x+2)-f(x)$ is increasing.
By recursion, for all $m,k\le m$ we have $f(2m+2k)+f(2m-2k)> 2f(2m)$.
Adding all these, we obtain
$f(0) + f(2) + ..... +f(4m) \ge (2m+1)f(2m)$
But $f(2)=f(3),f(4)=f(5), f(5)= f(6), f(6)= f(7) .....$

So, $f(1)+f(3)+.........+ f(4m-1)> (2m-1)f(2m)$

Combining with the above gives
$f(8m)= f(0)+f(1)+........+f(4m)\ge 4mf(2m)$
Atlast we have the inequality, $f(2^n)>2^{\frac{n^2}{4}}$ and for large enough $n$ we have,

$f(2^n)> 2^{n-1}f(2^{n-2})> 2^{n-1+\frac{(n-2)^2}{4}}= 2^{\frac{n^2}{4}}$
So   $\boxed{2^{\frac{n^2}{4}}< f(2^n)<2^{\frac{n^2}{2}}}$

From [i]Twitch Solves ISL[/i]:

It's clear that $f$ is non-decreasing. By sorting by the number of $1$'s we used, we have the equation \[ f(N) = 	f\left( \left\lfloor \frac N2 \right\rfloor \right) 	+  f\left( \left\lfloor \frac N2 \right\rfloor -1 \right) 	+  f\left( \left\lfloor \frac N2 \right\rfloor -2 \right) 	+ \dots 	+ f(1) + f(0). \quad (\bigstar) \]
[b]Upper bound.[/b] We now prove the upper bound by induction. Indeed, the base case is trivial and for the inductive step we simply use $(\bigstar)$: \[ f(2^n) = f(2^{n-1}) + f(2^{n-1}-1) + \dots 	< 2^{n-1} f(2^{n-1}) 	< 2^{n-1} \cdot 2^{\frac{(n-1)^2}{2}} 	= 2^{\frac{n^2}{2} - \frac{1}{2}}. \]
[b]Lower bound.[/b] First, we contend that $f$ is convex. We'll first prove this in the even case to save ourselves some annoyance:

[b][color=red]Claim:[/color][/b]  	[$f$ is basically convex] 	If $2 \mid a+b$ then 	we have $f(2a) + f(2b) \ge 2 f\left( a+b \right)$.
[i]Proof.[/i]  	Since $f(2k+1) = f(2k)$, we will only prove the first equation. 	Assume WLOG $a \ge b$ and use 	$(\bigstar)$ on all three $f$ expressions here; 	after subtracting repeated terms, the inequality then rewrites as 	\[ \sum_{(a+b)/2 \le x \le a} f(x) 		\ge \sum_{b \le x \le (a+b)/2} f(x). \] 	This is true since there are an equal number of terms on each side 	and $f$ is nondecreasing. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	For each $1 \le k < 2^{n-1}$, we have 	\[ f(2^{n-1} - k) + f(k+1) \ge 2f(2^{n-2}) \]
[i]Proof.[/i]  	Use the fact that $f(2t+1)=f(2t)$ for all $t$ 	and then apply convexity as above. $\blacksquare$
Now we can carry out the induction: \[ f(2^n) = f(2^{n-1}) + f(2^{n-1}-1) + \dots 	> 2^{n-1} f(2^{n-2}) + f(0) 	> 2^{n-1} 2^{\frac{(n-2)^2}{4}} = 2^{\frac{n^2}{4}}. \]

Observe that $f(2n) = f(2n+1)$ for all $n$. First, we have the recursion $$f(n) = f(1) + f(2) +\cdots + f(\lfloor n/2 \rfloor)$$ for all $n$ by considering possible groupings of $1$'s in the representation of $n$. In particular, starting from $n = 1+1+\cdots + 1$, we may group any $k \leq \lfloor n/2 \rfloor$ groups of two $1$'s into $k$ $2$'s. We then equivalently must construct a partition for $k$; the remaining $1$'s cannot be used by definition of the grouping.

Now this becomes routine bounding. Notice first that $$f(2^n) = f(1)+f(2)+\cdots+f(2^{n-1}) < 2^{n-1} f(2^{n-1}),$$ so multiplying recursively yields $$f(2^n) < 2^{n-1} \cdot 2^{n-2} \cdots 2^1 \cdot 2^0 = 2^{n(n-1)/2} < 2^{n^2/2}.$$ On the other hand, observe that $f$ is essentially convex because $f(2a)+f(2b) \geq 2f(a+b)$ for all $f$ (simply by expanding each term and noticing that $f$ is nondecreasing). Hence we also have $$f(2^n) = f(1)+f(2)+\cdots+f(2^{n-1}) > 2^{n-1} f(2^{n-2}).$$ Iterating this yields $f(2^n) > 2^{n^2/4}$, as needed.


A new school has exactly 1000 lockers and 1000 students. On the first day of school, the first student enters the school and opens all the lockers. The second student then enters and closes every locker with an even number. The third student will 'reverse' every third locker (if closed, he will open it and if open, he will close it). The fourth student will 'reverse' ever fourth locker and so on, untill all 1000 students have entered and reversed the proper lockers. Which lockers will be open at the end? (source: MAO 1992)

[hide]For a locker to be open at the end, it must have been opened and closed an odd number of times. For a locker to be closed at the end, it must have been opened and closed an even number of times.
Consider the nth student where n<1000. This student will open every locker number that is a multiple of n. Thus, for a locker to be open at the end, it must have an odd number of factors. The only numbers with an odd number of factors are perfect squares, so our answer is {1, 4, 9, ... 961}[/hide]

Locker number $n$ is open/closed by student number $d$ if and only if $d|n$. Hence locker number $n$ will be open/closed exactly $\tau(n)$ times,  where $\tau(n)$ is the number of positive integer divisors of $n$. Locker number $n$ is open at the end if and only if $\tau(n)$ is odd, i.e. $n$ is a square.

I have done this problem before.
[hide="answer"]The only lockers open will be the ones that are perfect squares (1,4,9,16,etc).[/hide]

yes, all correct!!!

 :first:

[hide]Only perfect square lockers will remain open.[/hide]

I have seen this problem many times...
[hide=":D "]Only the perfect square numbered lockers will be open at the end of the "student inquisition"  :D [/hide]


Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$.  As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that the point $\, E \,$ traces the arc of a circle.

[hide]

Let $ a=BC, b=CA, c=AB, d=CD, e=AD, f=DB$
The incircle of $ \triangle ACD$ is tangent to AB at F and to CD at H. The incircle of $ \triangle BCD$ is tangent to AB at G and to CD at J.
The common external tangent to both incircles (besides AB) is tangent to the incircle of $ \triangle ACD$ at X and is tangent to the incircle of $ \triangle BCD$ at Y.

$ FD=HD=\frac{d+e-b}{2}$
$ GD=JD=\frac{d+f-a}{2}$

By symmetry, we see that $ XY=FG$

Also $ XY=XE+YE=HE+JE=$
$ (DE-HD)+(DE-JD)=$
$ (DE-FD)+(DE-GD)=$
$ 2DE-FG$

Thus $ FG=2DE-FG\implies 2FG=2DE\implies FG=DE$

Now $ CE=d-DE=d-FG=d-FD-GD=d-\frac{d+e-b+d+f-a}{2}=$

$ -\frac{e+f-a-b}{2}=\frac{a+b-c}{2}$

Thus CE is constant regardless of the position of D, so all possible positions of E are equidistant from C. Thus they all lie on a circle centered at E.

QED

[/hide]


What's the exact value of:

$ \sqrt{32\plus{}10\sqrt{7}}\plus{}\sqrt{32\minus{}10\sqrt{7}}$

And how to do this ?

I do this question, but my answer was 10. But I like to see your method

$ \sqrt {32\pm10\sqrt {7}} \equal{} \sqrt {(25 \plus{} 7)\pm2\sqrt {25\times7}} \equal{} \sqrt {25} \pm \sqrt {7} \equal{} 5 \pm \sqrt {7}$.

Thus, $ \sqrt {32 \plus{} 10\sqrt {7}} \plus{} \sqrt {32 \minus{} 10\sqrt {7}} \equal{} (5 \plus{} \sqrt {7}) \plus{} (5 \minus{} \sqrt {7}) \equal{} \boxed{10}$.

I doubt this would be Pre-Olympiad, but eh whatever. Yeah all the solutions are correct.


Hello, we are two high school students who intend on creating a Math Club at a local middle school in North Dakota. We are not quite sure where to start, we've presented a mission statement to the teachers at the school, and now we have to actually plan what we want to do  . We have been considering enrolling the students in contests such as Math Olympiad, Math League, and AMC. We would also like to reach beyond contest math, and fulfill our true goal of simply having fun with math.

We are currently looking for ideas of what we can do --including games or other mathematical activities. Any help or advice would be greatly appreciated. Also, we would like to get a sponsorship or otherwise attain money (legally) and are not quite sure what the best way to do so is. Again, any advice would be more than welcome.

Thank you very much,

Greg Brockman and William Hall

OMG it's greg (without a hat)

Here's some tips:

1) Talk to Farmer Joe. Tell him you can't pay rent this month and use the money to fund your club.
2) Consider buying AoPS volume 1 and use it in the club.  It's a great book.
3) BUY LOTS AND LOTS OF PIZZA!!!
4) Read Jeff Boyd's article on how to start a successful MATHCOUNTS program. The idea between his club and your club is pretty much the same.
5) Read my blog.  I am starting a math culture in my local high school as you already know.  You can find all the details on my blog.  Hey, if it works for me, it might work for you.
6) I'm glad you realize that education is important.  Try not to drop out of school.  If you want, I'll write Farmer Joe a letter for you.  Surely he can live with less rent.  Once you finish school, your worthy activities with this math club might earn you a college scholarship :)
7) Don't hit on/abuse the middle school girls, Greg!! They're innocent children.

i cant hear a high squeaky voice, but you still sound like david rush.  was i right?


$a.b.c.d$are natural numbers satisfying $a^2+ab+b^2=c^2+cd+d^2$
prove $a+b+c+d$ is not prime.

Let $P=a+b+c+d$ is a prime numbre=>$d=P-a-b-c$=>$P^2+P(c-2a-2b-2c)+(c+a)(c+b)=0$=>$(c+a)(c+b)$divided $P$=>$c+a$divided$P$or $c+b$divided$P$,but $P>a+b+c$=> $P>c+a$and$P>b+c$ :(  :rotfl:

Yes dashmiz youre right,now I try to direct it.

but  mahmoud we can not assume $a\geq b\geq c\geq c\geq d$

Now it is true:
we can get$(a-b-c+d)(a-b+c-d)=3(c+d-a-b)(c+d+a+b)$
$\Rightarrow(a+b+c+d)|(a-b-c+d)or(a-b+c-d)$

but we have:
$|a+b+c+d|>|a-b-c+d|and|a-b+c-d|$
and it is contradiction.


search
combinatorics unsolved
combinatorics

The following facts are known on a mathematical contest:
(i) There were $ n\geq 4$ problems;
(ii) Each problem was solved by exactly four contestants;
(iii) For any two problems there is exactly one contestant who solved both
problems.
Assuming that there were at least $ 4n$ contestants, find the minimum value of $ n$ for which there always exists a contestant who solved all the problems.

Stop spamming the forums: find a less annoying way to deal with the fact that not every problem you post is going to get some response.  (In particular, you posted this before: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1154909815&t=228688 )  This is made particularly obnoxious by the fact that some of the problems you're posting or multi-posting have already appeared, e.g., this one: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1154909815&t=164264 .  It took me under 1 minute to discover this -- why couldn't you have done it yourself?


A dime is tossed.   It is a fair, unbiased coin.

a.  What is the sample space?

b.  What is the probability that a head will appear?

c.  What is the probability that a tail will appear?

[hide]Theres only two options heads or tails. it is $\frac{1}{2}$ for both problems.[/hide]

the sample space is:

[hide]{heads, tails}[/hide]

lol can't there be some remote possibility that it would land straight up?

[quote="Interval"]So, did you answer a through c?

What is a sample space?[/quote]

there are tons of information all over the web.  two great websites are mathworld.com and wikipedia.com


Solve over the complex number:

$x^{2}-x\sqrt{2}-1=0$

When I use the quadractic formula, I get the right answer. But when solve it this way, I get the wrong answer. Would anyone mind checking my work?  Thank you.

$x^{2}-x\sqrt{2}-1=0$

$x^{2}-1=x\sqrt{2}$

$(x^{2}-1)^{2}=(x\sqrt{2})^{2}$

$x^{4}-2x^{2}+1=2x^{2}$

$x^{4}-4x^{2}+1=0$

let $x^{2}=y$

$y^{2}-4y+1=0$

plugging it into the quadractic formula:
$\frac{4\pm{\sqrt{16-4(1)(1)}}}{2}$

solving I get:
$\frac{4\pm{2\sqrt{3}}}{2}=y$

$2\pm{\sqrt{3}}=y=x^{2}$

$x=\sqrt{2\pm{\sqrt{3}}}$

but the correct answer is
 $\frac{\sqrt{2}\pm\sqrt{6}}{2}$

Thanks  :P

[quote="i_like_pie"]You messed up a little with the quadratic formula. It's $y=\frac{4\pm{2\sqrt{2}}}{2},$ not $y=\frac{4\pm{2\sqrt{3}}}{2}.$[/quote]
Actually, it is $y=\frac{4\pm{2\sqrt{3}}}{2}$.  :wink:

@myc: Your answer actually is correct. Try squaring the two expressions: \[\left(\frac{\sqrt2\pm\sqrt6}{2}\right)^{2}=\frac{8\pm4\sqrt3}{4}=2\pm\sqrt3=\left(\sqrt{2\pm\sqrt3}\right)^{2}\]

[quote="nebula42"][quote="i_like_pie"]You messed up a little with the quadratic formula. It's $y=\frac{4\pm{2\sqrt{2}}}{2},$ not $y=\frac{4\pm{2\sqrt{3}}}{2}.$[/quote]
Actually, it is $y=\frac{4\pm{2\sqrt{3}}}{2}$.  :wink:[/quote]
I thought it was $\sqrt{8},$ not $\sqrt{12}.$ :blush:

Why not just use quadratic formula on the original equation????

You get the answer.

He did. He said,

[quote]When I use the quadractic formula, I get the right answer.[/quote]

He wanted to know why he got the "wrong" answer using the other method.

[quote="myc"]$\frac{4\pm{\sqrt{16-4(1)(1)}}}{2}$

solving I get:
$\frac{4\pm{2\sqrt{3}}}{2}=y$

$2\pm{\sqrt{3}}=y=x^{2}$

$x=\sqrt{2\pm{\sqrt{3}}}$

but the correct answer is
 $\frac{\sqrt{2}\pm\sqrt{6}}{2}$

Thanks  :P[/quote]

It's pretty good up to this point. You should obtain 4 solutions since you made the original quadratic into a quartic by squaring. The 4 solutions should be:

$\pm \sqrt{2 \pm \sqrt{3}}$ two of these had be invalid. One more thing, notice the fact that:

$\frac{\sqrt{2}+\sqrt{6}}{2}= \sqrt{2+\sqrt{3}}$ and

$\frac{\sqrt{2}-\sqrt{6}}{2}=-\sqrt{2-\sqrt{3}}$

So both methods yield the same solutions, it's just that when you made the original quadratic into a quartic, you forgot the fact that you had obtain 4 solutions instead of 2.

how do I know which two of the four solution is invalid?  Do i plug it back into the original equation?

Thanks for all the help! :)

plugging it into the quadractic formula:
\frac{4\pm{\sqrt{16-4(1)(1)}}}{2}

that does equal x, not x^2


$ a$,$ b$ and$ c$ are real numbers such that:
$ \left|a\right|<1$, $ \left|b\right|<1$, $ \left|c\right|<1$
prove that
$ ab\plus{}bc\plus{}ac\plus{}1>0$
from moroccan TST 2005

Let $ c$ have the greatest absolute value.

$ ab\plus{}bc\plus{}ca\plus{}1\equal{}(c\plus{}a)(c\plus{}b)\plus{}(1\minus{}c^2)>0$

[quote="SplashD"]Let $ c$ have the greatest absolute value.

$ ab \plus{} bc \plus{} ca \plus{} 1 \equal{} (c \plus{} a)(c \plus{} b) \plus{} (1 \minus{} c^2) > 0$[/quote]
Why? Can be $ (c \plus{} a)(c \plus{} b)<0.$  :wink:
I think, $ (1\minus{}a)(1\minus{}b)(1\minus{}c)\plus{}(1\plus{}a)(1\plus{}b)(1\plus{}c)>0$ better.


Let a,b,c positive integers  and $a \geq b \geq c$
prove that
$a^2\frac{b}{c}+b^2\frac{c}{a}+c^2\frac{a}{b} \geq a^2+b^2+c^2$

By taking $x=ab, y=ac, z=bc$ we will have that $x\geq y\geq z$ will also be positive integers and the desired inequality will become
$(*)$$\frac{x^2}{z}+\frac{z^2}{y}+\frac{y^2}{x}\geq\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x} \Leftrightarrow x^3y+y^3z+z^3x\geq x^2y^2+y^2z^2+z^2x^2$ and by using that $(x-y)(x-z)(y-z)(x+y+z)\geq 0 \Leftrightarrow x^3y+y^3z+z^3x\geq xy^3+yz^3+zx^3$ and Cauchy-Swartz we get that $(x^3y+y^3z+z^3x)^2\geq (x^3y+y^3z+z^3x)(xy^3+yz^3+zx^3) \geq (x^2y^2+y^2z^2+z^2x^2)^2$ and hence $(*)$ is proven.

Akhrif i think you are studying Hojoo Lees Inequality book thats great book ...  :)

Davron

i m studing this book but this pobleme is from another source :P


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

I am having a problem finding the intermediate fields between $ \mathbb{Q}$ and splitting field of $ x^4\minus{}2$; here's what I have done.

First, $ x^4\minus{}2\equal{}(x\plus{}\sqrt[4]{2})(x\minus{}\sqrt[4]{2})(x\plus{}i\sqrt[4]{2})(x\minus{}i\sqrt[4]{2})$ so the splitting field is $ \mathbb{Q}[\sqrt[4]{2},i]$. Next the degree of $ \mathbb{Q}[\sqrt[4]{2},i]$ over $ \mathbb{Q}$ is 8, so the galois group of $ \mathbb{Q}[\sqrt[4]{2},i]$ over $ \mathbb{Q}$ is of order 8. The image of an element in  $ \mathbb{Q}[\sqrt[4]{2},i]$ is completely determined by the images of $ \sqrt[4]{2}$ and $ i$. Since $ x^4\minus{}2$ and $ x^2\plus{}1$ are irriducible over $ \mathbb{Q}$ we have that $ i$ maps to either $ i$ or $ \minus{}i$ and $ \sqrt[4]{2}$ maps to either $ \sqrt[4]{2},\minus{}\sqrt[4]{2},i\sqrt[4]{2},\minus{}i\sqrt[4]{2}$. We then have 8 automorphisms $ a,b,c,d,e,f,g,h$ which are defined as:
$ a(i)\equal{}i, a(\sqrt[4]{2})\equal{}\sqrt[4]{2}$
$ b(i)\equal{}i, b(\sqrt[4]{2})\equal{}\minus{}\sqrt[4]{2}$
$ c(i)\equal{}i, c(\sqrt[4]{2})\equal{}i\sqrt[4]{2}$
$ d(i)\equal{}i, d(\sqrt[4]{2})\equal{}\minus{}i\sqrt[4]{2}$
$ e(i)\equal{}\minus{}i, e(\sqrt[4]{2})\equal{}\sqrt[4]{2}$
$ f(i)\equal{}\minus{}i, f(\sqrt[4]{2})\equal{}\minus{}\sqrt[4]{2}$
$ g(i)\equal{}\minus{}i, g(\sqrt[4]{2})\equal{}i\sqrt[4]{2}$
$ h(i)\equal{}\minus{}i, h(\sqrt[4]{2})\equal{}\minus{}i\sqrt[4]{2}$

This group is isomorphic to the dihedral group of order 8, and it has 8 subgroups- $ (a,b),(a,e),(a,f),(a,g),(a,h),(a,b,c,d),(a,b,e,f),(a,b,g,h)$
I associated these subgroups to intermediate fields-
$ (a,b)\to\mathbb{Q}[\sqrt[4]{2},i]$
$ (a,e)\to\mathbb{Q}[\sqrt[4]{2}]$
$ (a,f)\to\mathbb{Q}[i\sqrt[4]{2}]$
$ (a,b,c,d)\to\mathbb{Q}[i]$
$ (a,b,e,f)\to\mathbb{Q}[\sqrt{2}]$
$ (a,b,g,h)\to\mathbb{Q}[i\sqrt{2}]$

The problem I have is with $ (a,g)$ and $ (a,h)$. For both of them I get a fixed field of $ \mathbb{Q}[i\sqrt{2}]$, but I heed 2 fields not already found. Can anyone help?

One fault: $ (a,b)\to\mathbb{Q}[\sqrt{2},i]$
and $ (a,g)\to\mathbb{Q}[\sqrt[4]{2}(1\plus{}i)]$
$ (a,h)\to\mathbb{Q}[\sqrt[4]{2}(1\minus{}i)]$


Let $ a,b,c \geq 0$ , so that $ a\plus{}b\plus{}c\equal{}1$ and $ n \in \mathbb {N}$. Find maximal value of the following:
$ \frac {(ab)^n}{1\minus{}ab}\plus{}\frac {(ab)^n}{1\minus{}ab}\plus{}\frac {(ab)^n}{1\minus{}ab}$

1) n>1 , maximum is $ \frac{1}{3\cdot4^{n\minus{}1}}$

2) n=1 , maximum is $ \frac{3}{8}$

1. If n>1:  We have $ ab,bc,ca\leq\frac{1}{4}$ and so $ \sum {\frac{(ab)^n}{1\minus{}ab}}\leq \frac{4}{3}\sum {(ab)^n}$

Be cause $ a^n\plus{}b^n\plus{}c^n\geq ab^{n\minus{}1}\plus{}bc^{n\minus{}1}\plus{}ca^{n\minus{}1}$ we presume that $ a^n\geq bc^{n\minus{}1}$

 Then we have $ \frac{1}{4^n}\geq a^n(1\minus{}a^n)\equal{}a^n(b\plus{}c)^n\geq a^nb^n\plus{}a^nc^n\plus{}na^nb^{n\minus{}1}c\geq a^nb^n\plus{}b^nc^n\plus{}c^na^n$

 For $ a\equal{}b\equal{}\frac{1}{2}  ,c\equal{}0$ maximum is attained.


Although first International Mathematical olympiad was held in 1959 by Romania, I do believe IMO was not the first mathematical competition of sorts. I vaguely seem to recall that Leonhard Euler participated and won several times in Paris Academy Prize Problem competition in 1700s.

What were some of the popular mathematical competitions that existed before IMO and how can I have access to the problems posed in those competitions?

The earliest competition that I know of right now is the AHSME, which started in 1951.

The Putnam exam is much older than the IMO. It has been held continuously since 1938 and all of the questions are easily available in books (as well as online). Attached is an article on the history of the putnam exam, including the questions from the very first exam.

And, though it isn't formally a recreational competition, the Cambridge Mathematical Tripos exams are probably the oldest (and most difficult) math exams that I know of. They date back to at least 1780. The modern exams can be found online, and many of the old exams can be found in various sources. The article at Wikipedia refers to a source with the 1841 and 1842 exams (on pp. 201 and 405 respectively), and many of the problems from the Tripos exam are also found as exercises in Whittaker and Watson's 1902 book, [u]A Course of Modern Analysis[/u].

Hungary has held a high school math competition since the late 19th century -- I think it's gone under a couple of different names, including the $ \textrm{E\"otv\"os}$ competition.  Many if not all of the questions are available in book form -- look for "Hungarian Problem Books."  I believe this topic has been discussed before on the forum -- it's probably worthwhile to try a search, look in the "Books and Journals" section of the forum, etc.


function
algebra
polynomial
complex analysis
complex analysis unsolved

Thank you for your answer.

What would be our answer if we dont have any poles :

Find if there is any the meromorhic function g satisfying g'/g = G ,

when G= a_n z^n + .... +a_0  (a polynomial of deg n)

and G= e^z.

Thanks.

It would be nice to answer in the original topic http://www.mathlinks.ro/Forum/viewtopic.php?t=87104 ;)


geometry
geometric transformation
rotation
reflection

In how many ways can you place eight identical rooks on a standard eight-by-eight chessboard so that no two of the rooks attack each other (i.e., so that no two rooks are in the same row or in the same column)?

[hide]The answer should be $8!$. If you start with 1 square, then it is just 1. 2x2 squares is 2. 3 squares is 6 because there's always a 2x2 square in the corner times 3 because you can move the square in the top row to 3 different spots. The same goes for everything else.[/hide]

[hide]I think that tarquin is wrong--you can do it 2 ways diagonally.  I say $2 \times 8!$[/hide]

[hide]  No, mathgeek I think tarquin is correct. If you consider each rook the same, and only there allingment different, there is (8!)^2/8!= 8! different combinations.  However is you consider each rook different there is, is not only 8!*2 but 8! * 8! different combinations.  In general considering only allignment different there is (n!)^2/n!=n! combinations adn considering each rook different there is n! * n!=(n!)^2 different combinations.  I am not including rotations, if that is of any importance  I can explain that if i got it correct..........(probably not)  :P  [/hide]

[hide]

$8!$, because going by rows, the rook on the first row can be on 8 squares, the second rook can be on 7 squares, etc.
[/hide]

G-unit is right and I will take it upon myself to pose two similar problems. For the first one I know the answer:

A Latin Square of size n is an n-by-n array with n appearances of each integer 1 through n such that no two identical elements are in the same row or column. How many 8*8 Latin Squares are there?

For example, this is a 4*4 Latin Square:
4213
3142
2431
1324
This is not a 4*4 Latin Square for two reasons (1--there are two instances where ones are in the same row/column; 2--there are five 1s):
1423
4231
1342
2114

The second problem I do not know and would appreciate the answer:

How many ways are there to arrange rooks on a chessboard such that they can't attack each other and no two of the arrangements are reflections or rotations of each other?

8!. 2(8!) doublecounts, because it's just a reflection.

[hide]Most people derive at the answer 8!.  There reasoning as follows:  The first rook can be put into any of 8 columns, the second into any of 7, teh third into any of 6, etc.  This can be done using rows instead of columns.  Howver if you do this method only considering one aspect of the board, (either the column or the row) you are forgetting that the rooks must not occupy both rows and columns and attack a piece on the other aspect.  What do I mean?  For example lest say a chessboard is numbered from 1-64.  We move horizontally starting from the very upper-left square.  According to this method, we can put rooks on numbers 1, 2 ,3 ,4 5, 6, 7, 8.  But wait!  These rooks are all on the first row, contradicting the given conditions of the problem.  What method should be approached is this:  There are 8*8= 64 squares that the first rook can be put on.  But since each rook takes up a whole column and row, there is only (8-1)*(8-1) or 7*7=49 remaining squares for the second.  Since teh second rook takes up 1 column and 1 row, there is only (7-1)*(7-1)=6*6=36 squares left for the third rook.  This can be repeated until there is only 1 spot left for the last rook.  As you can see we are simplying finding the product of all the squares from 1-8, which is (8!)^2.  But since we want to find how many ways this can be done out of 8 rooks, we divide it by 8!.  Which is (8!)^2/8!=8!.  The other problem, if you consider each rook different and each rook, distinct, is a simple matter of multiplying each arrangement: 8! by the ways we can change it which is 8!.  So altogether we have (8!)^2.  The formulas if you would like to know where n=the side of a square chess-board are:  (n!)^2/n!= n!  and considering each item distinct (n!)^2.    Hope you find this helpful,  :P [/hide]

[hide]

How many ways are there to arrange rooks on a chessboard such that they can't attack each other and no two of the arrangements are reflections or rotations of each other?

I believe the answer is n!/n, where n= this side of a chessboard.  Consider a 3*3 chessboard.  The piece on the diagnol can be flipped to the right, down, and diagnol.  The pieces not on the diagnols can be flipped, aup, down, and left.  This varys from piece but alwas remains 3.  This is consistent for all chessboards, and thus the answer is (n!)/n,  I think.....  :? [/hide]

Yes, i just wrote it as a fraction bar...


Show that there is no continuous function $f:R\rightarrow R$ satisfying $f(x-f(x))=\frac x 2$ for all real numbers $x$.

Any ideas?

Let $g(x)=x-f(x)$. $g$ is a continuous injective function, so it's either strictly increasing or strictly decreasing. We have $g(g(x))=g(x)-\frac x2$. Since $g\circ g$ is strictly increasing, we find $g$ to be strictly increasing (because $g(x)-\frac x2=g\circ g$ is strictly increasing). Since $g$ is increasing and $g(x)-\frac x2$ is increasing, we find that $\frac{g(x)-x}2=g(g(x))-\frac{g(x)}2$ is increasing, so $g(x)-x$ is increasing. From here we find that $-\frac x2=g(g(x))-g(x)$ is increasing, which is absurd.

how do you get that g(g(x)) is increasing?  before that you said g must be either increasing or decreasing, but then you just neglect the latter case.

it looks fishy.  or maybe i'm missing something

[quote]how do you get that g(g(x)) is increasing?  before that you said g must be either increasing or decreasing, but then you just neglect the latter case. [/quote]

Regardless of whether g(x) is increasing or decreasing, g(g(x)) will be increasing.

Yakko


linear algebra
matrix
vector
combinatorics proposed
combinatorics

A $ 4\times 4$ table is divided into $ 16$ white unit square cells. Two cells are called neighbors if they share a common side. A [i]move[/i] consists in choosing a cell and the colors of neighbors from white to black or from black to white. After exactly $ n$ moves all the $ 16$ cells were black. Find all possible values of $ n$.

[hide="Hint"]Even parity makes casework very simple.[/hide]

By considering parity, $ n$ must be even. If $ a$ is a possible value of $ n$, then $ a\plus{}2$ is also a possible value of $ n$, because we can choose the same square for the $ a\plus{}1$-th and the $ a\plus{}2$-th move. If $ a$ is the least possible value of $ n$, then the answer is $ a,a\plus{}2,a\plus{}4,\ldots$. Now, we have to find the least possible value of $ n$. I have yet to find this least value. It is quite hard for me. :oops:

By the way, I played this kind of game in my computer, and I finished it after about 80 moves. :P

There's a (tedious if written out) solution to this problem that goes as follows:

We can view this problem as a matrix equation $ Ax\equal{}b$ over the field of two elements, where $ A$ is the $ 16 \times 16$ incidence matrix representing which buttons effect which lights, and $ b$ is the vector where every entry is 1.  Gaussian elimination reveals that $ A$ has rank 12, and we can explicitly write down all 16 solutions over $ F_2$.  A  button sequence works iff it corresponds to one of these 16 modulo 2.  The answer is therefore the collection of all numbers of the form $ m\plus{}2n$, where $ m$ is the number of nonzero entries in one of our 16 finite field solutions, and $ n$ is any nonnegative integer.

The actual computation of the 16 solutions in question I leave to the bored reader.

Hmn. I just used very simple casework. All deduction really.

If anyone interested:
[hide="Partial Solution"]
Label each square $ a_{ij}$, where $ i$ denotes the $ i$th row and $ j$ denotes the $ j$th column. We can merely consider only the squares where $ i \plus{} j$ is even because each distinct move can only change the colors of either even squares or odd squares. The case where $ i \plus{} j$ is odd is identical.

Without loss of generality, let each distinct move occur at most once (two of the same move = identity). Consider the top left square. It has two adjacent squares.  We must pick exactly one of these squares in order to successfully change its color. Without loss of generality, pick $ a_{12}$ as a move. $ a_{21}$ cannot be a move.

Now consider $ a_{22}$. We have two cases:

Case 1: One adjacent square is chosen.

We already have that $ a_{12}$ changes the color of $ a_{22}$. Thus $ a_{23}$ and $ a_{32}$ cannot be chosen as moves. Consider $ a_{13}$. As its color is already black, we cannot choose $ a_{14}$. Consider $ a_{24}$. By similar reasoning, $ a_{34}$ must be chosen as a move. Consider $ a_{31}$. $ a_{41}$ must be a move. $ a_{43}$ cannot be a move. We have $ 3$ moves total in this case. (I attached a picture for those who do not want to draw it out)

Case 2: Three adjacent squares are chosen.

Tedious, but rather short. It comes out to be $ 5$ moves.

We see that both of these cases, we have an odd number of moves. Because we must perform the set of moves again on the odd parity squares, we see $ n$ cannot be odd. The smallest amount of even moves possible is $ 3 \plus{} 3 \equal{} 6$. Thus $ n$ could be any even number greater than or equal to $ 6$.
[/hide]

[quote="kevinatcausa"]The question as posted is rather unclear...I had been assuming that the selected square also changes color.[/quote]I think your assumption is correct. :wink:

[quote]
A move consists in choosing a cell and [changing] the colors of neighbors from white to black or from black to white.
[/quote]

The sentence seems to be logically missing that verb, which means that the selected square does not change color. Can the poster clarify?

It changes selected square and neighbors squares also.

[color=darkblue]Yes.. Sorry for my missed word.
"A move" consists in choosing a cell and [u]changing the colors of this cell and its neighbors[/u] from white to black or from black to white."[/color]

[color=darkblue] [b]By a move we can cover the table with 3,4 or 5 cells.

Let us prove that we need at least 4 moves.[/b]

[hide="Proof"]Suppose contrary... Then we can max cover $ 3\cdot 5 \equal{} 15$ cells. But we need to cover 16 cells. Contradiction.[/hide]
[b]Then next step is to prove that we can color the table in 4 moves.[/b]

[hide="Proof"]Look at the picture.
[img]http://i051.radikal.ru/0806/c9/a190539811c5.png[/img] [/hide]
[b]Next step to prove that we can color the table in $ 2k$ moves, where $ k \ge 2$.[/b]

[hide="Proof"]Color the table in black in 4 steps. Then take any cell and change it color twice. So by 2 moves you obtained again a table colored in black. You can repeat this procedure and obtain a black table in $ 4,4 \plus{} 2,4 \plus{} 4,...,4 \plus{} 2p$ moves [/hide]
[b]To take the last 5 points you need to prove that you can't obtain the black table in an odd number of moves.[/b]

[hide="Officiall solution"]Consider the figure as in the picture
[img]http://i073.radikal.ru/0806/22/81975e819a97.png[/img]
And now how you will not do a move you will always cover an odd number of cells from this figure. From this after an odd numbers of moves oyu will color an odd number of cells, but it is even, so you can't.[/hide]
[hide="Solution of an turkish student"]Consider the 4 cells as in the picture.
[img]http://i074.radikal.ru/0806/bf/8afd22324726.png[/img]
How you will not do a move you will always cover only one of this cells. So after an odd number of moves you won't color all 4. [/hide][/color]

i think you missread the problem or i missread the problem:)! The MOVE i think it consist in changing just the neighbours colors! the cell you have chosen is not included!
This new Problem is more difficult that this trivial one:)

If you do not change the color of the chosen cell then the minimum is $ 6$
by chosing the select cells:

1001
1001
0000
0110

In order to paint black the corners at least 4 are required from the following eight cells:
0110
1001
1001
0110
 and you do not paint all either by picking exactly four from those eight cells or by picking an additional one to those four.

try the last problem with dimension of table 2nx2n:)

[quote]The MOVE i think it consist in changing just the neighbours colors! the cell you have chosen is not included! [/quote]
No, the move consists in a color changing of the cell too... I missed this word in the 1 post. So the problem isn't easy at all.

I heard this problem was proposed by Dušan Milijančević from Serbia (took a silver on this year's IMO).

I like this problem but it is very easy and basic :lol:

I agree you Ufuk! When i in the comptetion i solved this problem in five minutes. I think this is too easy for Junior Balkan. I think this problem should send to "Baby Balkan"  :lol:

did you solve it in five minutes!!!!!
when I saw this problem I solved it in twenty-thirty seconds  :rotfl:  :rotfl:

i don't understand why n must be even, can you explain me ?

[hide=hidden for length][quote=Ahiles][color=darkblue] [b]By a move we can cover the table with 3,4 or 5 cells.

Let us prove that we need at least 4 moves.[/b]

[hide="Proof"]Suppose contrary... Then we can max cover $ 3\cdot 5 \equal{} 15$ cells. But we need to cover 16 cells. Contradiction.[/hide]
[b]Then next step is to prove that we can color the table in 4 moves.[/b]

[hide="Proof"]Look at the picture.
[img]http://i051.radikal.ru/0806/c9/a190539811c5.png[/img] [/hide]
[b]Next step to prove that we can color the table in $ 2k$ moves, where $ k \ge 2$.[/b]

[hide="Proof"]Color the table in black in 4 steps. Then take any cell and change it color twice. So by 2 moves you obtained again a table colored in black. You can repeat this procedure and obtain a black table in $ 4,4 \plus{} 2,4 \plus{} 4,...,4 \plus{} 2p$ moves [/hide]
[b]To take the last 5 points you need to prove that you can't obtain the black table in an odd number of moves.[/b]

[hide="Officiall solution"]Consider the figure as in the picture
And now how you will not do a move you will always cover an odd number of cells from this figure. From this after an odd numbers of moves oyu will color an odd number of cells, but it is even, so you can't.[/hide]
[hide="Solution of an turkish student"]Consider the 4 cells as in the picture.
How you will not do a move you will always cover only one of this cells. So after an odd number of moves you won't color all 4. [/hide][/color][/quote]

[quote=MK4J]Another solution for eliminating the case $n$ odd:

Consider the quadrominos of the following form:

|o|x|o|
|x|x|x|   where I denoted with “x” a square of the quadromino and “o” an other square of the table.

We can cover the table with $4$ quadrominos of this type (we can turn them), for example when we put their centers in the $4$ cells we applied the operation for the case $n=4$.

Now, suppose that we can cover the table with an odd number of opperations ($n$ odd). Then, we can delete the pairs of opperations which were performed on the same cell. So we obtain $n_0 \leq n$, $n_0$ odd, such that each cell was performed at most one time ($n_0$ being the number of cells on which was performed the operation and in the same time the number of operations performed).

It is not hard to see that the order of which operation was performed first does not change the result.

But there can’t be an odd number of cells in all the four quadrominos formed, because in that case $n_0$ will be even.

So there is a quadromino from that $4$ which has an even number of cells with operation performed.

Hence, the cell in the center of that quadromino will have its changed an even number of times.

So that cell will be white at the end, contradiction!

In conclussion, $n_0$ must be even, so $n$ must be as well.[/quote]
[/hide]
Hmm, I didn’t quite understand the last parts. I got that you can color the table in 2k moves, k$\geq$2, but nothing else about not being able to obtain the black table in an odd number of moves. I tried parity and didn’t get much. Any alternate/more clarifying solutions would be helpful, although I really want to make more progress on this myself, so maybe like 2-3 progressive hints in succession would be really preferred.


function algebra domain conics parabola trigonometry inequalities

Find the range of the function:

      $f(x)=x+\sqrt{1-x}$

         any hint would be helpful thanks :lol:

EDIT: Texed by mod.

[hide]
The range of the function $f(x)$ is the same as the domain of the function $f^{-1}(x)$.

To find the inverse, let $f(x)=y=x+\sqrt{1-x}$ and switch $x$ and $y$.
Now we have $x=y+\sqrt{1-y}$, and we need to solve for $y$.
After some algebra, we obtain $y=\frac{2x-1\pm\sqrt{5-4x}}{2}$.
So $f^{-1}(x)=\frac{2x-1\pm\sqrt{5-4x}}{2}$.

We must have $5-4x\ge 0$, so the domain of $f^{-1}(x)$ is $(-\infty, \frac{5}{4})$, which is the range of $f(x)$.
[/hide]

You really should post these problems in the "Getting Started" forum, since they do not relate to the AMC.  :)

[quote="paladin8"][quote="JSteinhardt"][quote="kapag"]find the range of the function:

      f(x)=x+squareroot(1-x)

         any hint would be helpful thanks :lol: [/u][/quote]

A slightly nicer solution than that presented above:

[hide]
Let $x = cos^2(\theta)$.

Then the function becomes:

$cos^2(\theta) + sin(\theta) = 1 + sin(\theta) - sin^2(\theta)$

This can be treated as a parabola with vertex at $sin(\theta) = \frac{1}{2}$, or in other words the maximum value f can attain is $1 + \frac{1}{2} - \frac{1}{4} = \frac{5}{4}$. It is obvious that the range extends down to $-\infty$, so the range is

$(-\infty,\frac{5}{4}]$ and we are done.
[/hide][/quote]

You might want to state that you are assuming that the maximum occurs when $0 \le x \le 1$, which is graphically obvious. Otherwise $\theta$ doesn't exist. For instance, the minimum of $1+\sin{\theta}-\sin^2{\theta}$ is definitely not $-\infty$...[/quote]

Well the function is quite obviously increasing on $(-\infty,0)$, as $x$ increases faster than $\sqrt{x+1}$ for all positive $x$, and the function does not exist on $(1,\infty)$. You're right though, I probably should have stated it for clarity.

consider the intersection with $y=k$, the max value will give one solution for this

$k^2-2kx+x^2=1-x$
$x^2+(1-2k)x+(k^2-1)=0$
since there is one solution, the discriminant is 0
$1-4k+4k^2-4k^2+4=0$
$k=\frac{5}{4}$

it is clearly unbounded below, so we have $\left(-\infty,\frac{5}{4}\right]$

Is that necessary?  :huh:   Couldn't you just use trivial inequality but first verify that there is some x-value which makes the expression in the square zero?


Let $ a,b,c$ positive reals. Prove that

$ \sum\frac{2a^2\plus{}3b^2}{2a^3\plus{}3b^3}\plus{}\sum\frac{2b^2\plus{}3a^2}{2b^3\plus{}3a^3}\leq\frac{4}{a\plus{}b}\plus{}\frac{4}{b\plus{}c}\plus{}\frac{4}{c\plus{}a}\leq2(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

The RHS inequality is obtained by using AMHM to prove $ \frac{1}{a}\plus{}\frac{1}{b} \geq \frac{4}{a\plus{}b}$, and then adding cyclically.
For the LHS inequality, we have $ \frac{4}{a\plus{}b}\geq\frac{2a^2\plus{}3b^2}{2a^3\plus{}3b^3}\plus{}\frac{2b^2\plus{}3a^2}{2b^3\plus{}3a^3}$
$ \iff 4(6a^6\plus{}6b^6\plus{}13a^3b^3)\geq(a\plus{}b)(12a^5\plus{}12b^5\plus{}13a^3b^2\plus{}13a^2b^3)$
$ \iff 12a^6\plus{}12b^6\plus{}26a^3b^3 \geq 12a^5b\plus{}12ab^5\plus{}13a^4b^2\plus{}13a^2b^4$
$ \iff 12(a\minus{}b)(a^5\minus{}b^5) \geq 13a^2b^2(a\minus{}b)^2$ which is implied by $ 12(a^4\plus{}a^3b\plus{}a^2b^2\plus{}ab^3\plus{}b^4)>13a^2b^2$
which is implied by $ 12ab(a^2\plus{}b^2)>a^2b^2$ which is implied by $ a^2\plus{}b^2\geq2ab$.
We then add these inequalities up cyclically.


Prove for $ a,b,c$ positive reals,

$ \left(\frac{a}{a\plus{}2b}\right)^2\plus{}\left(\frac{b}{b\plus{}2c}\right)^2\plus{}\left(\frac{c}{c\plus{}2a}\right)^2\ge\frac{1}{3}$

Divide by 3 and square root, by QM-AM it suffices to show that

$ \frac{a}{a\plus{}2b}\plus{}\frac{b}{b\plus{}2c}\plus{}\frac{c}{c\plus{}2a} \ge 1$ Which is true since:

$ \frac{a^2}{a^2\plus{}2ab}\plus{}\frac{b^2}{b^2\plus{}2bc}\plus{}\frac{c^2}{c^2\plus{}2ac} \ge$

$ \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2ac\plus{}2bc} \equal{} 1$

Where we used Engel's form of Cauchy-Schwartz.

[quote="rofler"]Divide by 3 and square root, by QM-AM it suffices to show that

$ \frac {a}{a \plus{} 2b} \plus{} \frac {b}{b \plus{} 2c} \plus{} \frac {c}{c \plus{} 2a} \ge 1$ Which is true since:

$ \frac {a^2}{a^2 \plus{} 2ab} \plus{} \frac {b^2}{b^2 \plus{} 2bc} \plus{} \frac {c^2}{c^2 \plus{} 2ac} \ge$

$ \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc} \equal{} 1$

Where we used Engel's form of Cauchy-Schwartz.[/quote]

I think you mean Andreescu. Sorry if im wrong :)

I'm pretty sure both are ok? it doesn't really matter >.>

We can scale a, b, and c such that a + b + c = 1. The functions $ x^2$ and $ \frac{1}{x}$ are convex for positive $ x$.

By applying Jensen's twice we get

$ \frac{(\frac{a}{a \plus{} 2b})^2 \plus{} (\frac{b}{b \plus{} 2c})^2 \plus{} (\frac{c}{c \plus{} 2a})^2}{3}$
$ \geq \left(\frac{\frac{a}{a \plus{} 2b} \plus{} \frac{b}{b \plus{} 2c} \plus{} \frac{c}{c \plus{} 2a}}{3}\right)^2$
$ \geq \frac{1}{9}\left(\frac{1}{a(a \plus{} 2b) \plus{} b(b \plus{} 2c) \plus{} c(c \plus{} 2a)}\right)^2$
$ \equal{} \frac{1}{9}\left(\frac{1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ac}\right)$
$ \equal{} \frac{1}{9}\left(\frac{1}{(a \plus{} b \plus{} c)^2}\right)$
$ \equal{} \frac{1}{9}$

and done


hint για αρχη!
[hide]
Εστω $ k > 4$ ενας απο τους φυσικους που εχουμε επιλεξει για να μας δινουν αθροισμα $ n$.Αν τον αντικαταστησουμε με τους δυο φυσικους $ 2,k \minus{} 2$ δεν χαλαμε προφανως το  σταθερο αθροισμα αλλα μεγαλωνουμε το γινομενο! [/hide]

[hide]Αφου περασε μια μερα χωρις απαντηση δινω την στρατηγικη που σκεφτηκα για το παραπανω προβλημα.

Αρχικα οπως ειπα στην υποδειξη παρατηρουμε οτι με τον τυχαιο $ k > 4$ φυσικο που εχουμε επιλεξει οι δύο αριθμοί $ 2,k \minus{} 2$ μας δίνουν μεγαλυτερο γινόμενο.

Συνεπως η αναλυση καθε αριθμου ως αθροισμα καποιων $ 2$ μας μεγαλωνει το γινομενο. Αυτο μας οδηγει
στην εξης στρατηγικη,

Αν ο αριθμος ειναι αρτιος δηλαδη $ 2k$ τον γραφουμε σαν αθροισμα $ k$ ,$ 2$-αριων με γινομενο $ 2^k \geq 2k$

Aν ο αριθμος ειναι περιττος $ 2k \plus{} 1$ τον γραφουμε ως αθροισμα $ k \minus{} 1$, $ 2$-αριων και ενος $ 3$.

Τωρα ας δουμε αν επιδεχεται και αλλη βελτιωση το γινομενο. Εχουμε αποδειξει οτι το μεγιστο γινομενο θα περιεχει μονο $ 2,3$. Ομως $ 2 \plus{} 2 \plus{} 2 \equal{} 3 \plus{} 3$ και $ 3^2 > 2^3$. Συνεπως μολις φερουμε το γινομενο στην παραπανω μορφη καθε $ 3$ δυαρια τα μετατρεπουμε σε $ 2$ τριαρια ωστε τελικα το γινομενο να αποτελειται απο τριαρια και το πολυ $ 2$ δυαρια. Η τελευταια αναλυση μας δινει προφανως το μεγιστο.  :)

Για 100 δηλαδη μεγιστο γινομενο το $ 3^{32}2^2$ κοκ[/hide]

Tο εχουμε συζητήση παλαιότερα  Το είχα προτείνει σαν θεμα απο Ολυμπιαδα.. Δεν θυμαμαι το Post...

Πράγματι το είχατε προτείνει κ.Χρίστο. Βρίσκεται εδώ http://www.mathlinks.ro/viewtopic.php?search_id=36036783&t=214796

[quote="bilot4"]Μιας και ο Θαλής πλησιάζει και βλέπω ότι δεν υπάρχει και μεγάλη κίνηση στο forum λέω να βάλω ένα απλό προβληματάκι για προθέρμανση (μάλλον επιπέδου Ευκλείδη ή ίσως και 1ο Αρχιμήδη...) :maybe:

Εστω ότι κάποιοι φυσικοί αριθμοί έχουν άθροισμα 100. Ποιο είναι το μέγιστο γινόμενο που μπορεί να έχουν?

Υ.Γ. Να λυθεί το πρόβλημα και όταν οι αριθμοί έχουν άθροισμα $ n$[/quote]

Να βρεθεί επίσης το μέγιστο γινόμενο αν γνωρίζουμε ότι όλοι οι αριθμοί είναι διαφορετικοί μεταξύ τους.

[quote="Demetres"][quote="bilot4"]
Εστω ότι κάποιοι φυσικοί αριθμοί έχουν άθροισμα 100. Ποιο είναι το μέγιστο γινόμενο που μπορεί να έχουν?
[/quote]Να βρεθεί επίσης το μέγιστο γινόμενο αν γνωρίζουμε ότι όλοι οι αριθμοί είναι διαφορετικοί μεταξύ τους.[/quote]
Έστω $ a_{1}<a_{2}<...<a_{n}$ θετικοί ακέραιοι με $ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\equal{}100$ και ώστε το γινόμενο $ a_{1}a_{2}...a_{n}$ να είναι το μέγιστο δυνατό.

[u][b]Ισχυρισμός[/b][/u] Οι $ a_{1},a_{2},...,a_{n}$ είναι διαδοχικοί ακέραιοι εκτός το πολύ από μια εξαίρεση.
  [Αν υπάρχουν $ 1\leq i<j\leq n$ ώστε $ a_{i\plus{}1}\minus{}a_{i}\geq 2$ και $ a_{j\plus{}1}\minus{}a_{j}\geq 2$ τότε αλλάζουμε τους $ a_{i},a_{j}$ με τους $ a_{i}\plus{}1,a_{j}\minus{}1$. Τότε
$ (a_{i}\plus{}1)(a_{j}\minus{}1)>a_{i}a_{j},$ δηλαδή το άθροισμα παραμένει 100 ενώ το γινόμενο μεγαλώνει, άτοπο αφού υποθέσαμε ότι το γινόμενο είναι το μέγιστο δυνατό.]

[u][b]Περίπτωση 1η[/b][/u] Οι $ a_{1},a_{2},...,a_{n}$ είναι διαδοχικοί φυσικοί, έστω οι χ,χ+1,...,ν. Αν χ>4 τότε αν αλλάξουμε τον χ με τους 2,χ-2 το άθροισμα μένει 100 ενώ το γινόμενο μεγαλώνει αφού 2(χ-2)>χ. Δηλαδή $ x\leq 4$. Οι εξισώσεις χ+(χ+1)+...+ν=100 είναι αδύνατες για χ=1,2,3,4.

[b][u]Περίπτωση 2η[/u][/b] Οι $ a_{1},a_{2},...,a_{n}$ είναι διαδοχικοί φυσικοί εκτός από μία ακριβώς εξαίρεση. Τότε $ 2\leq a_{1}\leq 4$ (Αν $ a_{1}\equal{}1$ τότε αλλάζουμε τους $ a_{1},a_{n}$ με τον $ a_{n}\plus{}1$ οπότε το γινόμενο μεγαλώνει).
Έχουμε τις περιπτώσεις:
2+...+χ+(χ+2)+...+ν=100 που δίνει ν(ν+1)=204+2χ οπότε ν=14 και χ=3 και το γινόμενο είναι 14!/4.
3+...+χ+(χ+2)+...+ν=100 που δίνει ν(ν+1)=208+2χ οπότε ν=15 και χ=16 που απορρίπτεται
4+...+χ+(χ+2)+...+ν=100 που δίνει ν(ν+1)=214+2χ οπότε ν=15 και χ=13 και το γινόμενο είναι 15!/(6.14)
Είναι 14!/4>15!/(6.14)
Άρα το μέγιστο είναι  14!/4

Πολύ ωραία Παύλο, αυτήν την απάντηση βρήκα και εγώ. Γενικά αν εχουμε άθροισμα $ n$, $ k$ ελάχιστο ώστε $ 1 \plus{} 2 \plus{} \cdots \plus{} k > n$ και $ \ell \equal{} 1 \plus{} 2 \plus{} \cdots \plus{} k \minus{} n$ τότε (αν δεν έχω κάνει κάποιο λάθος) το μέγιστο γινόμενο (για $ n > 4$) ισούται με

- $ k!$ αν $ \ell \equal{} 1$
- $ (k\plus{}1)!/2k$ αν $ \ell \equal{} 2$
- $ k!/(\ell\minus{}1)$ αν $ \ell \geqslant 3$


geometry
3D geometry
graph theory
combinatorics unsolved
combinatorics

we have 4 cubes.every cubes has 6 faces that faces of every cube paint with these 4 colors as below
cube 1 :face 1=red;face 2=white;face 3=white;face 4=red;face 5=blue;face 6 =green
cube 2 :face 1=white;face 2=white;face 3=blue;face 4=red;face 5=blue;face 6 =green
cube 3 :face 1=red;face 2=white;face 3=green;face 4=blue;face 5=red;face 6 =blue
cube 4 :face 1=green;face 2=blue;face 3=white;face 4=red;face 5=green;face 6 =blue
we want to put cubes on together that we can see every color from one side mius up and down.
can we do this work or not?
how I can find a method to solve it with graph theory?

I don't know if it's the same colours, but a similar puzzle called "Instant Insanity" is solved using graph theory in the following article:
[url]http://www.geometer.org/mathcircles/graphprobs.pdf[/url]


one of my friends told me a geometry book which is written by parsalow, but i googled and turned nothing. he or I might be misspelled it, does anyone knows the name which can be misspelled as parsalow  :roll:
thanks..

I guess you mean Prasolov's book
 http://students.imsa.edu/%7Etliu/Math/planegeo.pdf

that is exactly what i am looking for :) thanks a lot.. :)