geometry
power of a point
radical axis
cyclic quadrilateral
geometry unsolved

Points $E,F$ are on the sides $AC,BC$ of triangle $ABC$ such that $AE=BF$. The circles passing through $A,C,F$ and through $B,C,E$ intersect again at point $D$. Prove that the line $CD$ bisects angle $ACB$.

[b]Lemma:[/b]
Given two circle $(O_{1})$ and $(O_{2})$ and $\Delta$ is the radical axis of them. $M$ is an arbitrary point in the plane, and $H$ is the point lies on $\Delta$ such that $MH \perp \Delta$. So we have $pow_{(O_{1})}M-pow_{(O_{2})}M = 2 \bar{MH}. \bar{O_{2}O_{1}}$
( It's very easy to prove it  :wink:  )
[b] Back to the problem : [/b]
Let $\Delta$ is the radical axis of the two circles $(ACF)$ and $(BCE)$. Of course $\Delta$ is joining in two points $C$ and $D$
Let $A_{1},B_{1}$ are the points lie in $\Delta$ such that $AA_{1}\perp \Delta$ and $BB_{1}\perp \Delta$.
Now use the lemma, we have: $\frac{AA_{1}}{BB_{1}}= \frac{|pow_{(ACF)}A-pow_{(BCE)}A|}{|pow_{(ACF)}B-pow_{(BCE)}B|}= \frac{|0-\bar{AE}.\bar{AC}|}{|\bar{BF}.\bar{BC}-0|}= \frac{AE.AC}{BF.BC}= \frac{AC}{BC}$
It means that $\Delta$ is the bisector of angle $\hat{ACB}$, which ends the proof.

This can be proven readily with Euclidean geometry; properties of the cyclic quadrilateral and then congruence.  Difficult to describe otherwise [and it's late here], but if someone can hint on how to send a graphic I'll do that ...if within the next day or so.  I will not be available for a couple of weeks after that.

When you reply a topic, look at the Options ( to be the left of the page) , click "Add an Attachment" to attach your figuer :wink:
If you don't have enough time I can creat a figure and attach it for you  :) . Yesterday while solved this problem I didn't creat it yet :D
I guess that your means is by the cyclic quadrilateral $BDEC$ and $ADFC$ we have $\hat{DBF}= \hat{DEA}$ and $\hat{DFB}= \hat{DAE}$ so with $BF$ = $AE$ we will have $\Delta BDF = \Delta EDA$. It means that $BD=ED$, so $CD$ is the bisector of angle $\hat{ACB}$

[quote="PDatK40SP"]I guess that your means is by the cyclic quadrilateral $BDEC$ and $ADFC$ we have $\hat{DBF}= \hat{DEA}$ and $\hat{DFB}= \hat{DAE}$ so with $BF$ = $AE$ we will have $\Delta BDF = \Delta EDA$. It means that $BD=ED$, so $CD$ is the bisector of angle $\hat{ACB}$[/quote]

Yes, the external angle of a cyclic quadrilateral being equal to the interior and opposite angle.

Thanks for the tip.  A lot on my plate these days, so I wasn't paying close attention.

I´ve been trying to prove the lemma that PDatK40SP uses but I can´t ( :blush:  it says too easy aaa!) Can someone show me the proof plz?

[quote="malinger"][color=darkred]Points $ E,F$ are on the sides $ AC,AB$ of $ \triangle ABC$ such that $ BF=CE$.
The circumcircles of $ \triangle ABE$ and $ \triangle ACF$ intersect again at point $ D$.
Prove that the line $ CD$ bisects angle $ \widehat{BAC}$.[/color][/quote]
[color=darkblue][b]Proof.[/b] The quadrilaterals $ ABDE$, $ ACDF$ are concyclically $ \implies$ $ \{\begin{array}{c}m(\widehat{BDE})=m(\widehat{CDF})=180^{\circ}-A\implies \widehat{BDF}\equiv\widehat{EDC}\\\\ \widehat{DFB}\equiv \widehat{DCE}\ ,\ \widehat{DBF}\equiv\widehat{DEC}\ ,\ BF=EC\end{array}\|$ $ \implies\over{A.S.A.}$

$ \triangle BDF\equiv\triangle EDC$ $ \implies$ $ [BDF]=[EDC]$ $ \implies$ $ BF\cdot\delta_{AB}(D)=CE\cdot\delta_{AC}(D)$ $ \implies$ $ \delta_{AB}(D)=\delta_{AC}(D)$,

i.e. the point $ D$ belongs to the bisector of the angle $ \widehat{BAC}$, where $ \delta_{XY}(P)$ is the distance from the point $ P$ to the line $ XY$.

[b]Remark.[/b] Denote $ L\in BE\cap CF$. Prove easily that the quadrilaterals $ BFLD$, $ CELD$ are cyclically, $ \{\begin{array}{c}DF=DC\\\ DE=DB\end{array}$

and $ MN\parallel AD$, where the points $ M$, $ N$ are the middle of the segments $ [BC]$, $ [EF]$.[/color]

Consider an inversion of center $ A$ and power $ k\in \mathbb{R_{+}}$.
Let $ B', C', E',F', D'$ be the inverses of points $ B,C,E,F,D$ respectively.
Because $ A,B,D,E$ are conciclic,  $ B', D', E'$ will be colinear.
Similarily, because $ A, F, D, C$ are conciclic, $ C', D', F'$ are colinear.
In triangle $ AF'C'$ we apply Menelaus` theorem for the transversal $ B'-D'-E'$: \[  \frac{AB'}{B'F'}\cdot\frac{F'D'}{D'C'}\cdot\frac{C'E'}{E'A}=1.\]But we easily see that $  \frac{AB'}{B'F'}=\frac{AF}{BF}$ and $ \frac{C'E'}{E'A}=\frac{CE}{AE},$ therefore, by multiplying the last two relations and keeping into acount the Menelaus relation, we obtain \[ \frac{F'D'}{D'C'}=\frac{AE}{AF}=\frac{AF'}{AE'}\](Because $ AE\cdot AE'=AF\cdot AF'=k$ and $ BF=CE$)

Thus $ AD'$ is the internal bisector of angle $ F'AE'$, but because $ D'\in (AD$, we obtain that AD is the internal bisector of $ BAC$, q.e.d.

[quote="malinger"]Points $ E,F$ are on the sides $ AC,BC$ of triangle $ ABC$ such that $ AE=BF$. The circles passing through $ A,C,F$ and through $ B,C,E$ intersect again at point $ D$. Prove that the line $ CD$ bisects angle $ ACB$.[/quote]
Let intersection of $ AR,BE$ be $ X$. By miquel theorem, $ E,X,D,A$ and $ F,X,D,B$ is cyclic.
By Law of sin, circumradi of $ XEA$,$ XFB$ is equal. So  $ DE=DB$(By Law of sin in triangle DXE,DXB) thereofore $ DC$ bisect angle $ C$


uhhhh go wiki them.
they have good detailed explanations

[quote="moogra"]is there actually a reason why the contests named their contests p versus np and  Navier-Stokes Equations, Poincare Conjecture, etc/[/quote]

There are only seven of them, they are the Milllenium Prize problems. Mathemathical problems, although two are borrowed from physics (YM and NS) and one from computer science (P vs NP), whose solution, it is assumed, will involve new key ideas for te development of mathematics.

Yang MIlls theories are quantum field theories based on non abelian groups like SU(N). A complete description of a QFT is given by its Lagrangian(*) and if for example you had a Lagrangian like L = a* · a + b* · b (* means complex conjugate), you can perform a transformation U belonging to the group SU(2), over the vector (a,b) leaving the Lagrangian invariant. This is what is meant when one says 'based on a non abelian group'. You see, physics at this level, is actually mathematics so it turna out that YM theories provide almost inm,ediately a very interesting tough mathematical problem which is one of hte MP problems and which I shall not explain but you should visit maybe the official site.

Also one of the MP problems seems already solved, it's the one concerning the Poincare conjecture. It was solved (at least the mathematical community seems convinced)by a russian mathematician and former imo perfect scorer, Grigori Perelman. For more, google or wiki all this stuff

Ursula

(*) Except for topological issues


What is the maximum number of consecutive positive integers
that can be added together before the sum exceeds 400?

We know the formula for sum of consecutive integers is $ \frac {n(n \plus{} 1)}{2}$, so we want that to be equal to $ 400$. Solving for $ n$, and rounding up, we get $ \boxed{28}$.

We want to round down. Our answer is therefore $ \boxed{27}$.

No, we do not want to round down, if we round down, then sum is not greater than $ 400$. Thus $ \boxed{28}$ is correct.

The question stipulates "before the sum exceeds $ 400$". So should it not be lower than $ 400$? Either that, or I'm missing something really obvious ...

Yes, it is rounded down since it is asking for maximum number of positive integers BEFORE it exceeds 400... not after it exceeds.

so our answer is,
$ \boxed{27}$ :lol:

I agree with KitKat.LoL,

the answer is $ \boxed{\boxed{27}}$

Dope, yes the answer is $ \boxed{27}$. I did not notice that it said "before the sum exceeds $ 400$". I thought is said "What is the minimum number of consecutive positive integers that can be added together before the sum exceeds $ 400$?"   I apologize for my stupidness..  :ewpu:


Zach and Mandy are traveling with their parents to Disneyland. They got bored and started looking at the cars, and they made these statistics: the probability that a car with an Ohio license in a time of 15 minutes is 1/6, the probability that a car with a Georgia license shows up in 10 minutes is 1/8, and the probability that a car with a California license plate in 5 minutes is 3/5. What is the probability that in 20 minutes, Zach and Mandy see a car with a Georgia license plate, a car with a California license plate, and do not see a car with an Ohio license plate in 30 minutes?

[hide]Hopefully what you mean is all of them have to be true. Let us consider what you were saying a while ago.
The probability of seeing a Cali license plate within the given time frame is $ \frac{609}{625}$, and the probability of not seeing a car with an ohio license plate in 30 min is $ (\frac{5}{6})^{2}$. Then, we consider the Geor license plate. The probability here would be $ (1-\frac{7}{8}^{2})$. That would be 15/64. Multiplying all of that together, we get the answer should be
$ \frac{609}{625)\times\frac{25}{36}\times\frac{15}{64}=\frac{208}{1280}$.
I'm new to this forum, and I hope I got this problem right.[/hide]

[quote="stupididiot1337"][hide]Hopefully what you mean is all of them have to be true. Let us consider what you were saying a while ago.
The probability of seeing a Cali license plate within the given time frame is $ \frac {609}{625}$, and the probability of not seeing a car with an ohio license plate in 30 min is $ (\frac {5}{6})^{2}$. Then, we consider the Geor license plate. The probability here would be $ (1 \minus{} \frac {7}{8}^{2})$. That would be 15/64. Multiplying all of that together, we get the answer should be
$ \frac {609}{625) \times \frac {25}{36} \times \frac15}{64} \equal{} \frac {208}{1280}$.
I'm new to this forum, and I hope I got this problem right.[/hide][/quote]
fixed, but i think you mean
$ \frac{15}{64}\frac{609}{625}\frac{25}{36}\equal{}\frac{203}{1280}$


[quote="DavinciCoder"]On question 2... according to the bottom of my solution, I only have 11 solutions, but the 12th is also in my work, I just forgot to type it.  How does the grading system handle mistakes like that?[/quote]

It depends on how well organized your solution is and how alert your grader is. We do not take points off for trivial errors, such as leaving a known answer out of the final list. We would take a point off for missing an answer completely. The question is, in a quick scan of your work, will the grader see that you did find the twelveth answer?

When I grade, I like to write a comment about how to correct the error in each erroneous solution, so I would be scanning for your mistake and would likely see you have the twelveth solution. Therefore, unless I am half-asleep at that moment (grading starts way too early for me), I would give you the full five points and write a snide remark about being more careful with your lists.

Erin Schram
USAMTS grader

[quote="ZennyK"]Last year I completely misinterpreted problem 3 on round 3 (the one with the weird graph and all the "pebbling number" stuff) and the grader gave me a four (SOMEhow)... I didn't get the correct answer, but I had a correct solution and answer within the confines of my interpretation. I would guess your solution is worth 3 points.[/quote]

And the moral of the story is, if I can't solve it, I'll just misinterpret into an easier problem :-)

[quote="Magnara"]
And the moral of the story is, if I can't solve it, I'll just misinterpret into an easier problem :-)[/quote]

Watch the debate tonight - you'll see you aren't the only who likes this approach!

[quote="rrusczyk"]Watch the debate tonight - you'll see you aren't the only who likes this approach![/quote]
Haha - 'tis sooth, Mr. Rusczyk. ;)

I believe the reason I still got credit for my solution was because I [u]didn't[/u] trivialize the problem in my misinterpretation. In fact, based on the sample solution, I think I made made the problem harder.

[quote="countersixte"]I made a similar error in my solution to problem 5.  However, instead of omitting something in my final list, I had a direct contradiction.  In the body of the proof, I said that BC = 2.  At the end of the proof when I listed the answers, I said BC = 4.  I think that since problem 5 has only three answers (the lengths of the sides), it would be harder to miss than an unlisted solution to problem 2.[/quote]

If you use BC = 2 in the calculations of your solution, we graders will count the BC = 4 at the end as a typographical error and ignore it. That is if the grader looks at the right part of your calculations, which he or she likely will, since I presume you got the right answer for AB and AC.

Erin Schram
USAMTS grader

[quote="PepeCamello"]What about this:

For problem three, I mistook the problem for saying that it had to be integer solutions. If I worked it out with a method that [i]would[/i] work had I not limited myself to integers, how many points do you think that deserves?

I got r+s+t+u=24 when (r,s,t,u) = (10,8,6,0)[/quote]

That one we graders will have to put our heads together to decide how to grade. It will be worth at least 4 points. We might decide it is an interesting enough variant to give it all 5 points.

Erin Schram
USAMTS grader

[quote="ZennyK"]Last year I completely misinterpreted problem 3 on round 3 (the one with the weird graph and all the "pebbling number" stuff) and the grader gave me a four (SOMEhow)... I didn't get the correct answer, but I had a correct solution and answer within the confines of my interpretation. I would guess your solution is worth 3 points.[/quote]

I was the head grader for that problem. We were giving 3 points to anyone who had the correct 13-pebble diagram that was the arrangement of the most pebbles that could not pebble a corner. The rest of your work must have been impressive enough to bump you up to 4 points, even if you were solving the wrong problem.

Erin Schram
USAMTS grader

[quote="thoobik"]I was foolish on #2 and didn't explicitly say that the $(\pm5,\pm6)$ solutions were in fact solutions.  However, I did say that those were the solutions to $x^2+y^2+11=0$ when I divided by $a-b$ in devolving the cubics so I'm hoping for some mercy.[/quote]

Maybe you will get a grader more merciful than me.

You posted your solution in another thread, so I peeked at it. You called the $x=\pm 6$ and $y=\pm 5$ "restrictions" rather than solutions. I would rule that you did not view them as solutions, and award you only 4 points for that problem.

Erin Schram
USAMTS grader

I reread my proof and have come to the same conclusion regarding my wording.  Thanks for the input.

Heh, how's this for a dumb mistake:  In Round 1 Problem 2, I simplified the equation to a=b and solved for (x+y)(x-y)=11 but I ignored the possibility of a=0 or b=0.  Thus, I only got 4/12 solutions.  How many points would be deducted for a mistake like this?

The annoying part about it is, when I first started messing with the problem (far before I typed up my answers) I noticed that a and b were zero, but completely forgot about that when typing up my solution.

Oh well, I guess I should be more careful when checking my solutions in the future. :'(

I have a friend, who, on the polygon problem with the aritmatic sequence of angles(4/1), he did everything perfectly, he got to the part of the (x+10)(x-12)=0.  This was all done really well, then he said "Therefore the answer must be a hexagon" and he got 3 points out of 5.  That is sort of the mother of all dumb mistakes, but I don't think he should have gotton a 3, if he had just left out the last part and not finished the problem I would think that would be enough for a 4.  What do you think?

Strictly speaking, he should get no points because the rubric only gives points if the correct answer is given.  I agree, though.  Holistically, if all the logic and reasoning was correct, I think he should only be deducted one point for not naming the polygon correctly.


function
floor function
arithmetic sequence
algebra unsolved
algebra

Given is the function $ f\equal{} \lfloor x^2 \rfloor \plus{} \{ x \}$ for all positive reals $ x$. ( $ \lfloor x \rfloor$ denotes the largest integer less than or equal $ x$ and $ \{ x \} \equal{} x \minus{} \lfloor x \rfloor$.)
Show that there exists an arithmetic sequence of different positive rational numbers, which all have the denominator $ 3$, if they are a reduced fraction, and don’t lie in the range of the function $ f$.

[quote="valerie"]Given is the function $ f \equal{} \lfloor x^2 \rfloor \plus{} \{ x \}$ for all positive reals $ x$. ( $ \lfloor x \rfloor$ denotes the largest integer less than or equal $ x$ and $ \{ x \} \equal{} x \minus{} \lfloor x \rfloor$.)
Show that there exists an arithmetic sequence of different positive rational numbers, which all have the denominator $ 3$, if they are a reduced fraction, and don’t lie in the range of the function $ f$.[/quote]
I suppose it should be  $ f(x) \equal{} \lfloor x^2 \rfloor \plus{} \{ x \}$.

Now if $ f(x) \equal{} \frac{n}{3}$ where $ n \in \mathbb{R}$. It is easily seen that $ 3x$ is an integer and, that $ 3x \equiv n \bmod 3$.

The first guess is to find a aritmethic sequence on where all numbers are on the form: $ m \plus{} \frac{1}{3}$.
So I'll check what values $ f(m \plus{} \frac{1}{3})$ can take.
$ f(m \plus{} \frac{1}{3}) \equal{} \lfloor m^2 \plus{} \frac{2}{3}m \plus{} \frac{1}{9} \rfloor \plus{} \frac{1}{3} \equal{} m^2 \plus{} \lfloor \frac{2}{3}m \plus{} \frac{1}{9} \rfloor \plus{} \frac{1}{3}$
Then $ f(m \plus{} \frac{1}{3}) \equal{} k \plus{} \frac{1}{3} \iff$
$ m^2 \plus{} \lfloor \frac{2}{3}m \plus{} \frac{1}{9} \rfloor \equal{} k$
Then I'll take three cases:
$ m \equal{} 3n$: $ m^2 \plus{} \lfloor \frac{2}{3}m \plus{} \frac{1}{9} \rfloor \equal{} 9n^2 \plus{} 2n \equal{} k \iff$
$ (3n\plus{}\frac{1}{3})^2 \minus{} \frac{1}{9} \equal{} k \iff$
$ (9n\plus{}1)^2 \equal{} 9k\plus{}1$. (So assume $ 9k\plus{}1$ isn't a square)
$ m \equal{} 3n\plus{}1$:  $ m^2 \plus{} \lfloor \frac{2}{3}m \plus{} \frac{1}{9} \rfloor \equal{} k \iff$
$ 9n^2\plus{}6n\plus{}1 \plus{} 2n \equal{} 9n^2 \plus{} 8n \plus{} 1 \equal{} k \iff$
$ (3n \plus{} \frac{4}{3})^2 \plus{} 1 \minus{} \frac{16}{9} \equal{} k \iff$
$ (9n \plus{} 4)^2 \equal{} 9k\plus{}7 \iff$ (So assume $ 9k\plus{}7$ isn't a square)
$ m \equal{} 3n\plus{}2$:  $ m^2 \plus{} \lfloor \frac{2}{3}m \plus{} \frac{1}{9} \rfloor \equal{} k \iff$
$ 9n^2\plus{}12n\plus{}4 \plus{} 2n\plus{}1 \equal{} 9n^2 \plus{} 14n \plus{} 5 \equal{} k \iff$
$ (3n\plus{}\frac{7}{3})^2 \minus{} \frac{49}{9} \plus{} 5 \equal{} k \iff$
$ (9n\plus{}7)^2 \equal{} 9k\plus{}4 \iff$ (So assume $ 9k\plus{}4$ isn't a square)

Now we just have to find cases where $ 9k\plus{}1, 9k\plus{}4, 9k\plus{}7$ isn't squares. If $ 9k\plus{}1 \equiv 7 \bmod 8$, then $ 9k\plus{}7 \equiv 5 \bmod 8$ thus none of $ 9k\plus{}1$ and $ 9k\plus{}7$ are squares. Then let $ 9k\plus{}4$ be such that $ 9k\plus{}4 \equiv 2 \bmod 5$ then $ 9k\plus{}4$ isn't a square.

$ 9k\plus{}1\equiv 7 \bmod 8 \iff k \equiv 6 \bmod 8$
$ 9k\plus{}4 \equiv 2 \bmod 5 \iff k \equiv 2 \bmod 5$

Then $ k \equiv 22 \bmod 40$. So the aritmethic progression $ 22 \plus{} \frac{1}{3}, 62 \plus{} \frac{1}{3}, 102 \plus{} \frac{1}{3}, ...$ works.

(It became a bit sketchy so please ask, if there's a bad explanation)


Let $ a,b,c>0$. Prove that $ \frac{1}{a^2\minus{}ab\plus{}b^2}\plus{}\frac{1}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{1}{c^2\minus{}ca\plus{}a^2}\ge \frac{3}{ab\plus{}bc\plus{}ca}$ .

[quote="Mateescu Constantin"]Let $ a,b,c > 0$. Prove that $ \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2}\ge \frac {3}{ab \plus{} bc \plus{} ca}$ .[/quote]
Denote $ f(a, b, c) \equal{} \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2}$
Assume $ c\geq b \geq a$
we will prove: $ f(a, b, c) \geq f(0, b, c) \geq \frac {3}{ab \plus{} bc \plus{} ca}$
That is very easy

[quote=caubetoanhoc94][quote="DangChienbn"][quote="Mateescu Constantin"]Let $ a,b,c > 0$. Prove that $ \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2}\ge \frac {3}{ab \plus{} bc \plus{} ca}$ .[/quote]
Denote $ f(a, b, c) \equal{} \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2} \minus{} \frac {3}{ab \plus{} bc \plus{} ca}$
Assume $ c\geq b \geq a$
we will prove: $ f(a, b, c) \geq f(0, b, c) \geq 0$
That is very easy[/quote]

Assume:$ a \ge b \ge c$.
We have:
$ LHS \ge \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{a^2\minus{}ab\plus{}b^2} \ge \frac{3}{ab} \ge RHS$
[hide] Sao phải khổ vậy trời [/hide][/quote]

and why $\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{a^2\minus{}ab\plus{}b^2} \ge \frac{3}{ab}$ ??

[quote=mudok][quote=Understandingmathematics]
and why $\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{a^2\minus{}ab\plus{}b^2} \ge \frac{3}{ab}$ ??[/quote]
By AM-GM:

$\frac{a^2-ab+b^2}{(ab)^2}+\frac{1}{a^2\minus{}ab\plus{}b^2} \ge 2\sqrt{\frac{a^2-ab+b^2}{(ab)^2}\cdot\frac{1}{a^2\minus{}ab\plus{}b^2}}=\frac{2}{ab}$

$\Rightarrow \frac{1}{a^2}\plus{}\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2\minus{}ab\plus{}b^2} \ge \frac{2}{ab}$

$\Rightarrow \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{a^2\minus{}ab\plus{}b^2} \ge \frac{3}{ab}$[/quote]

thanks


In the finger game Alice and Bob each hold out two hands with one finger per hand.  The moves are redistributions of fingers on your hands or tapping an opponent's hand with your own. The opponent has to hold out a number of fingers equal to the sum of the number of  extended fingers on his hand and your hand. If it is greater then five he puts the hand down.  The opponent loses if both hands are down.   Alice goes first. What are optimal stratagies for Alice and Bob?

lol i played this game with friends before, never really thought about the mathematical aspect of it.

Question: Can 0 and 3 be redistributed to 1 and 2, or can redistribution only occur on two hands that are in the game?

In any event, here's another version of the same game.

A and B each start with 1 finger out on each hand. A goes first and choses one of her hands and one of B's hands, adds the two together and replaces this with the hand she chose from B (same as before). A and B take turns in this manner. If the number of fingers that are supposed to be on a hand is greater than 5, take it mod 5 (ie, if 2 gets added to 4, it becomes 1 because 2+4=1 mod 5).
If one hand is even and the other hand is zero, a player may take his turn by splitting. In other words a hand can go from 4 and 0 to 2 and 2, or from 2 and 0 to 1 and 1.
The loser is the person who has both hands at 0.

Find a winning strategy for either A or B, or prove that neither A nor B can win assuming both play optimally.

I played this once in the car on the way to an away basketball game. I didn't notice the mathematical aspect either.

in gopherhole112's version of the game with the mod 5 there is a loophole when you play: hands with zero are out. This is how my sister and i played waiting in line at Busch Gardens. It's a great time passer.
When both players have one hand down and it is player one's turn
player one___player two
___2___________4____
___2___________1____
___3___________1____
___3___________4____
___2___________4____
___2___________1____ etc.
there may be others but we haven't stumbled upon them yet...
post them if you find them.   :D

I think this game is called chopsticks.

By the way, are you taking into account of splits? Like if you only have one hand out and it has 4 extended fingers, you can split it into 2 fingers on each hand.

we didn't play with splits because the rules we came across said that once a hand had 0 it was out.

[quote="Ignite168"]I've found tons and tons of infinite repeating loops in this game which is why I don't play it much anymore.[/quote]

My friends and I play it where you put the hand down only when there's 6 fingers on it. If it's 7 or more, you subtract 6 and that's how many fingers are on that hand. We also play splits with all numbers greater than one when you only have one hand up. That makes the game very strategic

chopsticks!!!
pretty fun, but it doesn't end if you know how to play it right (I do  :lol: )

:diablo:


conics
parabola
inequalities unsolved
inequalities

a, b, c, A, B, C are reals with a, A non-zero such that |ax2 + bx + c| ≤ |Ax2 + Bx + C| for all real x. Show that |b2 - 4ac| ≤ |B2 - 4AC|.

Well, there is one parabola inside another, and a line intersecting both...

Very nice problem. This problem was in IMO selection test of my country last year. It is also from Putnam. Actually I don't remember my solution of that time, but I found another one.Here is it.
We can assume that  $a, A>0$,because if it is not, we can write it it in the following way:
$|ax^{2}+bx+c|=|(-a)x^{2}-bx-c|$,and now  $-a>0$. Now I'll prove that $|A|\geq{|a|}$. We can rewrite  the given condition in this way: $|a+\frac{b}{x}+\frac{c}{x^{2}}|\leq{|A+\frac{B}{x}+\frac{C}{x^{2}}|}$, when $x\rightarrow{\infty}$ , LHS tends to $|a|$ and RHS tends to $|A|$, so $|A|\geq{|a|}$.
If RHS has two real roots, then it is obvious that the same roots has LHS.Using Viett's theorem we obtain that $(\frac{b}{a})^{2}-4\frac{c}{a}= (\frac{B}{A})^{2}-4\frac{C}{A}\Rightarrow((\frac{b}{a})^{2}-4\frac{c}{a})a^{2}\leq{((\frac{B}{A})^{2}-4\frac{C}{A})A^{2}}$ or $|b^{2}-4ac|\leq{|B^{2}-4AC|}$ (since $|A|\geq{|a|}$).
If RHS has one real root, then it has two real roots, but they are just equal, so the method above works.
If RHS has no real roots(LHS has no real roots too), then it can be easily proved that the minimum of LHS is less or equal to the minimum of RHS.But we know that $LHS_{min}=\frac{-b^{2}+4ac}{4a}$ and  $RHS_{min}=\frac{-B^{2}+4AC}{4A}$.So
$\frac{-b^{2}+4ac}{4a}\leq{\frac{-B^{2}+4AC}{4A}}\leq{\frac{-B^{2}+4AC}{4a}}, $thus $-b^{2}+4ac\leq{-B^{2}+4AC}$, this is what we need to prove, since $|b^{2}-4ac|=-b^{2}+4ac$, because it has no real roots.
QED
P.S. Hope you like my solution.


function
calculus
derivative
limit
calculus computations

The following limit represents the derivative of some function f at some number a. State f and a in each case.
lim(x approaches to 0) \frac{3^x-1}/{x}\.
I have no idea how to do this type of problem if I can't calculate the limit given(as above, without using L's Hospital rule). Can anybody give me a help?

The derivative of a function $ f$ at a point $ a$ is
\[ \lim_{x\to 0}\frac {f(a \plus{} x) \minus{} f(a)}{x}
\]
So if you can find an $ f$ and an $ a$ so that $ f(a) \equal{} 1$ and $ f(x \plus{} a) \equal{} 3^x$, you've answered the question. (try $ a\equal{}0$).

The content of this problem is not "calculate the limit". It's "know the definition of the derivative".

There are actually many correct answers; translating the function and point doesn't change the form of the limit in the derivative.


1. Suppose you deposit 5oo dollars  in a savings account that has an interest rate r, compounded annuall. At the end of three years, the value of account will be
500(1+r)^3. simplify 500(1+r)^3.

And

solve x^2+3=2x

Tip- (a+b)³ = a³ +3a²b+3ab²+b³ .

500(1+3r+3r²+r³)= 500+1500r+1500²+500r³
2.)

x²+3= 2x ←  NOT in standard form. Subtract 2x, subtract 3 from both sides.
=x²-2x= -3 ←  Perfect square method.
x²-2x+1²=-3+1²
=x²-2x+1=-2

Factor.

(x-1)²= -2
square root.
x-1= ±√2i

x= 1±√2i

Solution set is, {1+√2i, 1-√2i}

Correct me in I'm wrong please.  :maybe:

#2

$ x^2\minus{}2x\plus{}3\equal{}0$

Quadratic Formula $ \frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$

$ \frac{2\pm\sqrt{4\minus{}12}}{2}\equal{}1\pm\sqrt{2}i$


I need help....

Find the number of order triples (a,b,c) where a, b, and c are positive integers. A is a factor of B, A is a factor of C, and A + B + C = 100.

:huh:
Is this AIME?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=798316#798316

You can also do it by counting the possible values of b and c when a=1, 2, 4, 5, 10, and 20

no... I dun think so =) obviously not xD

Let $ b \equal{} ka,c \equal{} na$

Then... $ a(1 \plus{} k \plus{} n) \equal{} 100\rightarrow a|100$

Test... lol.

$ a \equal{} 1, b \equal{} 1\hdots98, c \equal{} 1\hdots98$ $ b$ determines $ c$ obviously =) xD.  So $ 98$ possible since the order actually matters (o.O).

$ a \equal{} 2,b \equal{} 2, 4, \hdots 96, c \equal{} 2, 4, \hdots 96$  $ 48$ possible i think.... or something of the sort

$ a \equal{} 4, b \equal{} 4,8,\hdots 92, c \equal{} 4, 8, \hdots 92$ ya blah blah etc etc. $ 23$ possible... something like this lol. might've messed up xD
AND SO ON
If you really wanted to, you can use balls and urns to find combinations... like... $ k \plus{} n \equal{} 99,49,$ etc.


A Martian day, called a sol, is 24 hours, 39 minutes long in Earth time. If it is noon at a particular location on Earth and "noon" at the same moment at a particular location on Mars, after how many sols willl it first be noon at these same locations at the same moment in time again?

[hide="ANSWER"]Answer: 480[/hide]

[hide="i think..."] wouldnt you have a sol is 39/3600ths of a day more than a day on earth. we want both of them to be at noon. so wouldnt we want to find the least possible value of x where x and 39x/3600 are integers. so i get 1200 for x so the next time its noon on both planets is in 13 days. [color=brown][b][u]lucky 13![/u] [/b] [/color][/hide]

[quote="abacadaea"] a sol is 39/3600ths of a day[/quote]

Um actually it would be $ \frac{39}{60\cdot 24}$.

[hide] I get the answer 13 as well, but wouldn't a sol be 39/1440ths of a day more then in earth instead?[/hide]
Oops, buzzer 11 beat me to it.


calculus
combinatorics proposed
combinatorics

Call a permutation  \pi on {1,2..,n} good iff | \pi (j)-j| is constant for all j ,1 \leq j \leq n .Determine the number of good permutation for n=1996

with n is positive number,define P_n={a in Z+/ EXIST permutation  \pi  of{1,2,..,n}satisfy| \pi (i)-i=a,i= 1..n}
we can show that :a in P_n <--->(2*a=n)or(a in P_(n-2*a)) (not so hard)
define k is the greatest nonegative  integer such that a in P_ (n-2*k*a)
--->2*a<= n-2*k*a
if 2a<n-2*a*k then ain P_n-2*(k+1)*a(contract with the way to choose k)
so 2*a=n-2*k*a--->(k+1)a=n/2
.............
finally,we calculus :there are 1+d(n/2) good permutation(d(n)is the number of divisor of n)
sorrry because my english is so poor


Find $k$ if $P,Q,R,$ and $S$ are points on the sides of quadrilateral $ABCD$ so that \[ \frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = k,  \] and the area of the quadrilateral $PQRS$ is exactly 52% of the area of the quadrilateral $ABCD$.

For picture, go [url=http://www.cms.math.ca/Competitions/IMTS/imts3.html]here[/url].

$k=\frac32$ or $\frac23$. It is the root of the equation: $\dfrac{48}{100}=\dfrac{2k}{(k+1)^2}$

[hide="Here is a solution."]$\frac{13}{25}=\frac{52}{100}=\frac{[PQRS]}{[ABCD]}=1-\sum \frac{[SAP]}{[ABD]}\cdot \frac{[ABD]}{[ABCD]}=$
$1-\frac{k}{(k+1)^2}\sum \frac{[ABD]}{[ABCD]}=1-\frac{2k}{(k+1)^2}\Longrightarrow$
$\frac{k}{(k+1)^2}=\frac{6}{25}\Longrightarrow \boxed {\ k\in \left\{\frac 23 ,\frac 32\right\}\ }\ .$[/hide][hide="Another two same problems."]
 $\blacktriangleright\ 1^{\circ}.$ Let $ABCD$ be a convex quadrilateral the points $\{M,N\}\subset (AB)$ and $\{P,Q\}\subset (CD)$ so that $N\in (MB)$ and $Q\in (PD)$.
Prove that: $\frac{AM}{CP}=\frac{MN}{PQ}=\frac{NB}{QD}\Longrightarrow [MNPQ]=\frac{MN}{AB}\cdot [ABCD]\ .$

$\blacktriangleright\ 2^{\circ}.$ Let $ABCD$ be a convex quadrilateral and given are the points
$M\in (AB)$, $N\in (CD)$ so that $\frac{MB}{AB}+\frac{NC}{CD}=1\ .$
Prove that $[MCD]+[NAB]=[ABCD]\ .$[/hide]


If ($a,b \in \mathbb{Z}$) $ab\equiv 0 (mod$ $a+b)$ then ${a^2b^2\over a+b}$ can be written if form $x^2y^3$ where $x,y \in \mathbb{Z}$.

Let  gcd(a,b)=d, a=x.d and b=y.d . We know gcd(x,y)=1

  We know  ab=0(mod a+b). So xd.yd=0(mod xd+yd)
  Then d(x+y) divides x.y.(d^2) . We know that gcd(x+y,xy)=1 (Because gcd(x,y)=1 )
  Then d(x+y) divides d^2 . Then x+y divides d . So there is a positive  integer k such that d=(x+y).k

 We get  (a^2)(b^2)/a+b = (xd)^2.(yd)^2/xd+yd = x^2.y^2.d^4/d(x+y)
 If put in there d=(x+y)k we get;

(a^2).(b^2)/a+b = (x^2).(y^2).(d^4)/d(x+y) = (x^2).(y^2).(d^3)/x+y=
=x^2.y^2.[(x+y)k]^3/x+y = (x^2).(y^2).[(x+y)^2].(k^3)=
=c^2.d^3 (where x.y.(x+y)=c and k=d)

Proof completed.

[/code]


[quote="hunter_cvp_ic"]$\text{A,C,B,D are 4 point on the line such that 3AC=3BD=BC. M,N.P are on the circle through B,C such that }\widehat{NBC}= 2 \widehat{NCB}, \widehat{MBC}=2\widehat{MCB}and \widehat{PAD}= 2 \widehat{PDA}. \text{Prove that MNP is regular triangle}$[/quote]

1-Are $A$, $C$, $B$, $D$ placed in that order??

2-You mean \emph{any} circle through $B$ and $C$, don't you?

3-Finally, is the last relation OK? Assuming 1, all points $M$, $N$ and $P$ seem to be closer to $C$ than to $B$, as they all belong to the circle, I don't see how they can form an equilateral triangle. This of course, might be wrong, is just what it seems at first sight.

And now somebody comes, solves the problem and I make a fool of myself

Daniel

$\text{sorry DHERNANDEZ}\\ \text{The order of A,B,C,D is A,C,D,B And for any circle through Band C}\\ \text{So you can find soln}\\ \text{Thanks}$

actually there are two points with the propetry$\angle PAD=2.\angle PDA$ which one do you mean?

and in my sketch, $\triangle MNP$ is not a regular triangle.

are you sure about the correctness of the problem.

oh ! Ithink it have unique P !! Can you show me two point P satisfy this condition  AMIR.S

[quote="Amir.S"][quote="hunter_cvp_ic"]$A,C,B,D$ are 4 points on the line such that $3AC=3BD=BC$. $M,N.P$ are on the circle through $B,C$ such that $\widehat{NBC}= 2 \widehat{NCB},\ \widehat{MBC}=2\widehat{MCB}$and $\widehat{PAD}= 2 \widehat{PDA}$.Prove that $MNP$ is regular triangle[/quote]

look there are two points like $X$,that $\angle XBC=2\angle XCB$ these points are $M$ and $N$ as the problem said.
what is the reason that don't exsits two $Y$ that $\angle YAD=2\angle YDA$ ?

PS:it is impossible to find $M$ and $N$ with compas and ruler.[/quote]
Oh !! I don't think so .On circle through B C is must be have two point X such $\widehat{XBC}=2\widehat{XCB}$ but the circle not through A,D and P on this circle ,this property not same!!!!! Are you agree! And i think if you find locus of X such $\widehat{XAD}=2\widehat{XDA}$ you can check there exist unique point P satisfy it!!!!!!!!!!!!!  :wink:

[quote="hunter_cvp_ic"]
Oh !! I don't think so .On circle through B C is must be have two point X such $\widehat{XBC}=2\widehat{XCB}$ but the circle not through A,D and P on this circle ,this property not same!!!!! Are you agree! And i think if you find locus of X such $\widehat{XAD}=2\widehat{XDA}$ you can check there exist unique point P satisfy it!!!!!!!!!!!!!  :wink:[/quote]

look:the locus of $X$ that $\widehat{XAD}=2\widehat{XDA}$ is a conic and this conic intersect the circle at two distinct point. :wink:

[quote][size=134][color=darkred][b][u]A probable enunciation.[/u][/b] Let $w$ be a circle and let $B,C$ be two points which belong to given circle. Denote the point $D\in (BC)$ so that $CD=2\cdot DB$ and the point $A$ so that $C\in (AD)$ and $AC=BD\ .$ Consider : two points $M\in w\ ,\ N\in w$ so that the line $AB$ separates these points and ${m(\widehat{BMC})> m(\widehat BNC})\ ;$ the point $P\in w$ so that the line $AB$ separates the points $P$ and $M\ .$ Suppose that $M(\widehat{CBM})=2\cdot m(\widehat{BCM})$ , $m(\widehat{CBN})=2\cdot m(\widehat{BCN})$ and $m(\widehat{PAD})=2\cdot m(\widehat{PDA})\ .$ Prove that $\triangle MPN$- regular.[/color][/size][/quote]

[size=150][color=darkblue][b]Hunter_cyp_ic[/b] , is it O. K. ?[/color][/size]

[quote="Virgil Nicula"][quote][size=134][color=darkred][b][u]A probable enunciation.[/u][/b] Let $w$ be a circle and let $B,C$ be two points which belong to given circle. Denote the point $D\in (BC)$ so that $CD=2\cdot DB$ and the point $A$ so that $C\in (AD)$ and $AC=BD\ .$ Consider : two points $M\in w\ ,\ N\in w$ so that the line $AB$ separates these points and ${m(\widehat{BMC})> m(\widehat BNC})\ ;$ the point $P\in w$ so that the line $AB$ separates the points $P$ and $M\ .$ Suppose that $M(\widehat{CBM})=2\cdot m(\widehat{BCM})$ , $m(\widehat{CBN})=2\cdot m(\widehat{BCN})$ and $m(\widehat{PAD})=2\cdot m(\widehat{PDA})\ .$ Prove that $\triangle MPN$- right.[/color][/size][/quote]

[size=150][color=darkblue][b]Hunter_cyp_ic[/b] , is it O. K. ?[/color][/size][/quote]
No !!! Prove that MNP is regular not right
  :) AMIR.S   !!! Can you post soln for locus of P????

Oh !! I have locus of P it through C and P so I think P is unique !!!!!????

If my result is true .So who can solve it

CA :( N ANYONE HELP ME
NOONE SOLVE MY PROBLEM

this problem is forgoten !! no body solve it?? :(

[quote="hunter_cvp_ic"]Prove that MNP is regular triangle[/quote]
What do you mean by "regular triangle" :?:

This pro in CRUX math so it mean ABC is regular triangle ( equilateral triangle) but in Crux ,it use calculate and it is not nice .I need some nice prove  :lol:


superior algebra
superior algebra unsolved

Supposed that gcd(p,q)=1, G is finite group and elements a,b belong to G such that o(a)=p,o(b)=q. If o(ab)=pq, then whether or not ab=ba?

If G satisfied the condition above for every element. Then what is the structure of G? Do the simple groups satisfy?


Dear community
Dear every body
I am working on calculus through history, and especially root number. A mathematician from Nishapour (al Haçan al Nishapouri) wrote some manuscript about nth-root of number.
So any information about him,his life,his work, is welcomed.
Younes from Algeria.

علیک السلام یونس
هل تستطيع أن تقول لي اسمه في العربية؟
امید

سلام عليكم و عيد مبارك
 . من اسمه ما لي سوى الحسن النيشابوري
سوف أحقق نسختين موجودتين في مكتبة أحمد الثالث بإستمبول تركيا
رقم النسختين هو 3149 ,3152
لا خبر لي بحياته ، أساتذته ...
شكرا

علیک سلام
أنا حاولت أیضا ولیکن أنا لا وجدت شيئا. سأحاول مرة أخرى


Let $ a,b \in \mathbb{Z}$ so that $ a^2 \not\equal{} 4b.$ Prove that the set $ M\equal{}\{ n \in \mathbb{Z}|\sqrt{(n^2\plus{}an\plus{}b)}\in \mathbb{N} \}$ is finite.

[hide]
Let $ K\equal{}n^{2}\plus{}an\plus{}b$.  If $ a$ is even, denote $ a'\equal{}a/2$.  If $ b>(a')^2$, then $ K>(n\plus{}a')^2$.  Note that $ (n\plus{}a'\plus{}1)^2\minus{}K\equal{}2n\plus{}(\text{some constant})$ and thus will become positive for sufficiently large $ n$, which results in $ K$ being caught between 2 perfect squares and thus $ K$ not a perfect square itself.  Similar processes can be used to prove in the subcase where $ b<(a')^2$ and when $ a$ is odd that $ K$ for sufficiently large $ n$ is between 2 squares.[/hide]


usually every year on mathcounts, they give a problem where it's like find the number of times 6 (randomly chosen) is used when all the numbers from 1-600 (randomly chosen) are written down. what's the method of solving these kinds of problems. i know organized casework is the most obvious answer, but it never works for me! :(
help?

How many times is 6 the last digit, middle digit, and first digit?

1-600.

Obviously for it to be the first digit, we can have $ 6, 60\minus{}69$ and $ 600$.

12

Middle digit (in case of three digit numbers): $ 5*1*10$

50

Last digit: $ 16,26...96$

10

and $ 5*10*1$ 50 again.

So er r... 122. or something

It pretty much is the only way to find them. So get used to it :)

I get like 121 for number of times 6 appears (it has to 1 mod 6 just think about it :) )

also there are 115 numbers that have a 6 in them but that wasnt the question (it said number of times 6 appears and some numbers use 6 twice)

_ _ _

if the first digit is 6 and we dont want to have anymore digits to be 6 we will have 1.

second digit is 6 no more others to be 6 we have 6*9=54

same if third digit is 6 thats 54....you can use 0 as first digit

if both the second and third digit is 6 we have 12 because it asked number of times 6 is used not how many numbers have a 6 so its 6*2=12.

54+54+12+1=$ 121$ times 6 appears

There should be only $ 115$ numbers that have a 6 in them

Mathcrazed; first youve counted something like 166 has having only 1 6.
ALso youve overcounted the number of terms from 16-96, thats only 9 not 10.  ALso by doing that your overcounting numbers like 66.

BTW theres a lot easier solution but I gotta go to sleep...
Its just 20*6+1.  Just think about why....its really easy to see.


geometry
topology
advanced fields
advanced fields open

It is a well-known fact (see e.g. Burago et al., A course in metric geometry) that a compact metric space cannot be mapped isometrically into a part of itself, namely, if  f:X->X' is a distance-preserving mapping, then X' = X.

I feel that a similar thing should be generalized for almost-isometries, i.e. such mapping that | d(x,y) - d(f(x),f(y)) | <= epsilon. My claim is that a 2epsilon ball around X' is X, i.e. that an almost isometry is "almost surjective".
I claim that the latter holds for a general compact metric space.

Does anybody know how to prove it?

Unfortunately, this is false. Let $a$ be a big number. Consider the interval $[0,a]$ with the metric $d(x,y)=\min(1,|x-y|)$. Then the mapping given by $f(x)=(1-\varepsilon) x$ is an $\varepsilon$-isometry but the distance from $a$ to the image is $1$ if $a>\dfrac{1}\varepsilon$.


[b][i]A Pentagon is inscribed in a Hexagon. If they share a common side of length $ 2$ inches, find for the area within the Hexagon but outside of the Pentagon.[/b][/i]

Well, the easiest way is to use the formulas: $ \frac{3(2)^2\sqrt3}2\minus{}\frac{(2)^2\sqrt{25\plus{}10\sqrt5}}4\equal{}\boxed{6\sqrt3\minus{}\sqrt{25\plus{}10\sqrt5}}$.

[quote="AIME15"]Are they regular?  Otherwise, the problem is unsolvable.[/quote]

Yes, they are regular.... :)

I got the solution $ 6\sqrt3\minus{}5\tan54^\circ$ which is equivalent to math154's solution but I don't know how to calculate $ \tan54^\circ$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1335229#1335229]With the Golden Triangle[/url] ($ 36^\circ \minus{} 72^\circ \minus{} 72^\circ$), you can find the trigonometric values of multiples of $ 18^\circ$.

[hide="Alternatively, an algebraic approach..."]
All primitive $ 5^{th}$ roots of unity satisfy $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0$. Dividing both sides by $ x^2$ yields

$ x^2 \plus{} x \plus{} 1 \plus{} \frac1x \plus{} \frac1{x^2} \equal{} 0$, and letting $ y \equal{} x \plus{} \frac1x$, this simplifies to $ y^2 \plus{} y \minus{} 1 \equal{} 0$, which has roots $ y \equal{} \frac { \minus{} 1\pm\sqrt5}2$.

Substituting this back in, we get $ x \plus{} \frac1x \equal{} \frac { \minus{} 1\pm\sqrt5}2\implies x^2 \plus{} \left(\frac {1\pm\sqrt5}2\right)x \plus{} 1 \equal{} 0$, and

$ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} \left(x^2 \plus{} \left(\frac {1 \minus{} \sqrt5}2\right)x \plus{} 1\right)\left(x^2 \plus{} \left(\frac {1 \plus{} \sqrt5}2\right)x \plus{} 1\right) \equal{} 0$. But the original expression is also

equivalent to $ \left(x \minus{} \text{cis}\left(\frac {2\pi}5\right)\right)\left(x \minus{} \text{cis}\left(\frac {4\pi}5\right)\right)\left(x \minus{} \text{cis}\left(\frac {6\pi}5\right)\right)\left(x \minus{} \text{cis}\left(\frac {8\pi}5\right)\right) \equal{} 0$! We can rewrite this as

$ \left(x^2 \minus{} \left(\text{cis}\left(\frac {2\pi}5\right) \plus{} \text{cis}\left(\frac {8\pi}5\right)\right)x \plus{} \text{cis}\left(\frac {2\pi}5\right)\text{cis}\left(\frac {8\pi}5\right)\right)\left(x^2 \minus{} \left(\text{cis}\left(\frac {4\pi}5\right) \plus{} \text{cis}\left(\frac {6\pi}5\right)\right)x \plus{} \text{cis}\left(\frac {4\pi}5\right)\text{cis}\left(\frac {6\pi}5\right)\right)$

$ \equal{} 0$, and since $ \text{cis}\left(\frac {2\pi}5\right) \equal{} \overline{\text{cis}\left(\frac {8\pi}5\right)}$ and $ \text{cis}\left(\frac {4\pi}5\right) \equal{} \overline{\text{cis}\left(\frac {6\pi}5\right)}$, the Fundamental Theorem of Algebra tells us that

$ x^2 \plus{} \left(\frac {1 \minus{} \sqrt5}2\right)x \plus{} 1 \equal{} x^2 \minus{} \left(\text{cis}\left(\frac {2\pi}5\right) \plus{} \text{cis}\left(\frac {8\pi}5\right)\right)x \plus{} \text{cis}\left(\frac {2\pi}5\right)\text{cis}\left(\frac {8\pi}5\right)$ and

$ x^2 \plus{} \left(\frac {1 \plus{} \sqrt5}2\right)x \plus{} 1 \equal{} x^2 \minus{} \left(\text{cis}\left(\frac {4\pi}5\right) \plus{} \text{cis}\left(\frac {6\pi}5\right)\right)x \plus{} \text{cis}\left(\frac {4\pi}5\right)\text{cis}\left(\frac {6\pi}5\right)$, so

$ \text{cis}\left(\frac {2\pi}5\right) \plus{} \text{cis}\left(\frac {8\pi}5\right) \equal{} 2\cos\left(\frac {2\pi}5\right) \equal{} \frac {\sqrt5 \minus{} 1}2$

$ \implies\boxed{\cos\left(\frac {2\pi}5\right) \equal{} \sin\left(\frac\pi{10}\right) \equal{} \frac {\sqrt5 \minus{} 1}4}$ and

$ \text{cis}\left(\frac {4\pi}5\right) \plus{} \text{cis}\left(\frac {6\pi}5\right) \equal{} 2\cos\left(\frac {4\pi}5\right) \equal{} \frac { \minus{} 1 \minus{} \sqrt5}2$

$ \implies\boxed{\cos\left(\frac {4\pi}5\right) \equal{} \frac { \minus{} 1 \minus{} \sqrt5}4}\implies\boxed{\sin\left(\frac {3\pi}{10}\right) \equal{} \frac {1 \plus{} \sqrt5}4}$.

[b]Note:[/b] We could have assumed instead that $ x^2 \plus{} \left(\frac {1 \plus{} \sqrt5}2\right)x \plus{} 1 \equal{} x^2 \minus{} \left(\text{cis}\left(\frac {2\pi}5\right) \plus{} \text{cis}\left(\frac {8\pi}5\right)\right)x \plus{} \text{cis}\left(\frac {2\pi}5\right)\text{cis}\left(\frac {8\pi}5\right)$. However, $ \cos\left(\frac {2\pi}5\right) > 0 > \cos\left(\frac {4\pi}5\right)$, so this is incorrect. (Why?)

And also, why doesn't \operatorname work anymore...???
[/hide]

So in this particular problem, we directly get from above that $ \sin54^\circ \equal{} \frac {1 \plus{} \sqrt5}4\implies\cos54^\circ \equal{} \frac {\sqrt {10 \minus{} 2\sqrt5}}4\implies\tan54^\circ \equal{} \frac {\frac {1 \plus{} \sqrt5}4}{\frac {\sqrt {10 \minus{} 2\sqrt5}}4}$

$ \equal{} \frac {1 \plus{} \sqrt5}{\sqrt {10 \minus{} 2\sqrt5}}\cdot\frac {\sqrt {10 \plus{} 2\sqrt5}}{\sqrt {10 \plus{} 2\sqrt5}} \equal{} \frac {2\sqrt {(3 \plus{} \sqrt5)(5 \plus{} \sqrt5)}}{4\sqrt5} \equal{} \frac {2\sqrt {(5 \plus{} 2\sqrt5)}}{2\sqrt5}$

$ \equal{} \frac {\sqrt {5 \plus{} 2\sqrt5}}{\sqrt5} \equal{} \sqrt {1 \plus{} \frac {2\sqrt5}5}\implies5\tan54^\circ \equal{} \sqrt {25 \plus{} 10\sqrt5}$.

Thank you. It must have taken you some time to write it all down :)

Whatever, I can use this for future reference. :P


Let $ X_1,X_2,...$ be independent random variables with the same distribution and let $ N$ be a random variable with value in $ \mathbb{N}$. and $ Y: \equal{} \sum_{i \equal{} 1}^nX_i$ ,Prove:
$ i)\ \ \ E(Y) \equal{} E(N).E(X)$
$ ii)\ \ \text{Var}(Y) \equal{} E(N)\text{Var}(X) \plus{} E(X)^2\text{Var}(N)$
$ iii)\ M_Y(t) \equal{} E(M_X(t)^N)$

for example:
\[ E[Y] \equal{} \sum_{n\equal{}0}^{\infty} E[\sum_1^N X_k|N\equal{}n] P[N\equal{}n]\]
and because $ N$ and $ (X_i)_i$ are independent, \[ E[\sum_1^N X_k|N\equal{}n] \equal{} n E[X] \] the  other equalities should be proved the same way.


A 20- pound flamingo stands 3.0 feet in height.  If the dimensions and densities of all flamingos are similar, what is the height, in feet, of a 100- pound flamingo? Express your answer as a decimal to the nearest tenth.

*The answer is supposedly 5.1, but could somebody verify this? I believe that it is 15 instead.

Well, since density is constant, an 100-pound flamingo has five times the volume of a 20-pound flamingo. Since the dimensions are in the same ratio, then length, width, and height must increase by the same factor $ x$, and $ x^3 \equal{} 5$, so $ x \equal{} \sqrt[3]{5}$. The cube root of 5 is approximately 1.71, so $ 3 \times 1.71 \approx \boxed{5.1}$.

all you have to do is use proportions:

20/3=100/x
a 20 pound flamingo is 3 ft. tall, so a 100 pound flamingo is how much?
100/20 = 5
therefore 3 * 5 = [b]15[/b]

Hmmmm.... It should be 15. Think of it as a proportion.

20:3 as 100:?

20/3= 6.666667

100/3= about 15

Therefore, I believe it should be 15.

Weird....... :|

btw, the reason why it isn't 15 is because of the fact that if the weight increases, the dimensions of the flamingo only increases to the third root of the weight increase.
An example of this would be, lets say, a square.
if the dimensions was 3x3, then the area would be 9.
However, if we increased the area to 81, then the dimensions would be 9x9.
Even though the area increased 9x, the actual dimensions only increased by the square root of 9.
As result, darkdieuguerre is correct, and the answer, 5.1 stands.


calculus
integration
floor function
calculus computations

$ \int_0^{100} e^{x\minus{}\lfloor x \rfloor} dx \equal{} 100\int_0^1 e^x dx \equal{} 100e\minus{}100$


A penny, a nickel and a dime are tossed simultaneously. What is the probability that exactly two of the coins show heads when they land? Express your answer as a common fraction.

penny and nickel, nickel and quarter, penny and qurter

answer: $ \frac{3}{8}$

There are $ 2^3 \equal{} 8$ ways to *organize* the coins, and there are 3C2 (3) ways to get heads.
So we have a $ \frac {3}{8}$ chance.

why is there 3*2 ways to get heads?

its 3C2, not 3P2.


pigeonhole principle
induction
graph theory

hello

please help by exaplaining this question in steps thank much

Show that in a simple graph with at least two vertices there must be two vertices that have the same degree. use the Pigeonhole principle

Perhaps

[hide="induction"]Let $G_{n}$ be a simple graph of $n$ vertices.

For $G_{2}$, we have two vertices with degree 1.

Assume that it is true for $G_{k}$.

In the graph $G_{k+1}$, we have a subgraph $\{v_{1},v_{2},\ldots,v_{k}\}$ and a new vertex $v_{k+1}$. Note that $v_{k+1}$ will introduce edges to $\{v_{1},v_{2},\ldots,v_{k}\}$ and changes some of the degrees of the vertices in $G_{k}$. If $\{v_{1},v_{2},\ldots,v_{k}\}$ has at least two vertices with same degree, we are done as guaranteed by the induction hypothesis.

EDIT PART: Suppose that $\{v_{1},v_{2},\ldots,v_{k}\}$ all have different degrees. Note that $\deg(v_{j})\le k$ for $j\in[1,k]$. There are at most $k$ vertices that $v_{k+1}$ can connect to, so $\deg(v_{k+1})\le k$. By pigeonhole, there must exist another vertex with the same degree as $v_{k+1}$ since $\{\deg(v_{1}),\deg(v_{2}),\ldots,\deg(v_{k})\}$ can be at most a set of $k$ consecutive positive integers and at least a vertex $v_{m}$ must have $\deg(v_{m})=\deg(v_{k+1})$.

[/hide]
I hope this is alright... I haven't done graph theory for a while.
EDIT: fixing second part of proof
EDIT2: removed the unnecessary paragraph

[hide="or we could just be like"]
$n$ vertices, degree ranges from $1$ to $n-1$.... PIGEONHOLED!!!
[/hide]

thanx 10000th .... id this final solution ot there any other way to explain this

thanx


[quote="math92"]I have flipped a coin 49 times, and each time it came up heads. The probability of this happening is 1/562,949,953,421,312. What is the probability that the 50th flip will turn up heads?[/quote]

[hide]the flips are all independent so .5[/hide]


Prove that the number 0.14916253649... is irational.

The number should mean that the infinite sequence of squares of natural numbers are written one next to the other. Sorry

Briefly, this number is irrational since it contains blocks of $ 0$'s of unbounded length (consider the squares of powers of $ 10$). A rational number cannot have that in its decimal expansion unless the expansion terminates ends with all $ 0$'s, which clearly isn't the case here.


[b]Problem.[/b][i]In how many ways can you select $7$ distinct integers from the set $\{ 3,4,5,\ldots, 97\}$ such that they differ by at least $3$?  Find the remainder modulo $1000$.[/i]

[hide=my attempt]
In the diagram below, let each $|$ represent one of the $7$ numbers chosen from the set, and the $*'s$ represent the consecutive integers between them.
\[* | * | * | * | * | * | * | *. \]

Each distribution of the numbers for the $*'s$ will correspond to one of the desired sets of $7$ numbers. Out of the total $95$ numbers, $7$ of them must be set aside for the numbers that form the desired set, and since each must be differ by at least $3$, the $6$ $*'s$ in the middle, excluding the far left and right ones, must receive at least $3$ numbers.

The total number of possible distributions is then
\[\binom{(95-7-6\times3)+(8-1)}{8-1}= \binom{77}{7}, \]
since after accounting for those restrictions, we are free to distribute the numbers however we want.

Er, I don't know how to evaluate $\binom{77}{7}$ mod 1000 w/o a calculator, or if that's even correct  :oops:

With your logic, try to solve the exact same problem when you have $7$ consecutive integers to choose from.

[hide]
Start with the set $\{ 1,2,3,\cdots ,83\}$ which has 83 elements, and pick 7 of them. Then, after each chosen integer, insert two integers before it, shifting it and the ones after it forward 2. The $n$th element, ordered from least to greatest, will increase by $2n$. (For example, if we start by choosing 1,3,6,10,15,21,28, we will change it to 3,7,12,18,25,33,42.)

This will send 7 integers in the set $\{ 1,2,3,\cdots ,83\}$ to 7 integers in the set $\{ 1,2,3,\cdots ,97\}$ such that each chosen integer has at least two spaces before it. This forces any two chosen integers to be separated by at least 3, and 1 and 2 cannot be selected, so this is equivalent to 7 integers in the set $\{ 3,4,5,\cdots ,97\}$ such that any two are separated by at least 3. Therefore, the total is $\binom{83}{7}$. (Anyone want to reduce this number modulo 1000 by hand?)
[/hide]

Yes, that is the answer I obtained.  I created a bijection between sets with a difference of at least one and the sets with a difference of at least 3. The modulo $1000$ was just there to give it a nice structure like AIME problems.  You can prime factorize $1000=2^{3}5^{3}$, use a congruence for binomial coefficients modulo a prime power (there is a stronger version of Lucas' Theorem), and then use the Chinese Remainder Theorem to obtain an answer of $628$.  You could also just expand it, which is probably easier for such numbers.  I did not bother doing the calculations by hand; I just used a calculator. :P

When you said each number is three apart, does that mean that there are at least three consecutive integers between them, like between $3$ and $7$, or that their difference is $3$, like for $3$ and $6$?. Because if it's the latter, then I subtracted and extra $6$ and otherwise would have gotten $\binom{83}{7}$ as well  :maybe: Or maybe that's just a coincidence  :lol:


People in the Bay Area forums (like me, and I remember Ubemaya getting loads of it a while ago) have been getting large amounts of random "1" and "2" ratings.

I think we should get rid of the ratings:
People use the "1" and "2" ratings to annoy others
Only some people actually give ratings based on post content.

EDIT: Mods feel free to move this if it should be in Communities Round Table

Although I agree with you in some respects, I have only one thing to say: If you don't like ratings, don't pay attention to them.

I once had a rating of 4.5, and then it dropped to 3.6 in a matter of days and has stayed there ever since.

Somebody apparently likes rating everything spam in the Bay Area forum, so I have stopped posting there completely. Other forums also have this problem. For example, the posts in one perfectly normal topic with a problem, several solutions, and a normal discussion in the Number Theory forum were all rated as spam. Some users enjoy abusing our rating system, and I'm afraid we can't stop them all. I thought Mr. Vornicu said something about "users abusing the rating system will have their rights disabled", but obviously this guy's rights have not been disabled.

About your idea of removing ratings, unfortunately you won't get anywhere with that. The best you'll accomplish is disable ratings in our forum (like the Test Forum, G&FF, etc.).

I don't know how many times this needs repeating, but the solution to "bad" (i.e. incorrect) ratings is to give "good" (i.e. correct) ratings, not to complain about how someone else is doing it wrong.

Well then, we'll have to work a little harder at that.   :lol:  Mods, lock this thread.

[quote="i_like_pie"]
Somebody apparently likes rating everything spam in the Bay Area forum, so I have stopped posting there completely. [/quote]
Which is a pity :(. You shouldn't take your ratings that seriously. I would say that well-known users like you should not be concerned with hexagons at all: their reputation (whether good or bad) is known to almost everybody and cannot be changed by placing a few (or even a multitude of) "spam" or "excellent" ratings on their posts. I am not visiting the Bay Area forum myself, but I'm still pretty sure some people there will miss you, so I encourage you to return and not to pay too much attention to a few jerks or to any particular single digit number.

Of course I understand that post ratings don't really mean anything and I don't care much about them, but nevertheless, there's no point in posting somewhere when you know you're almost guaranteed a spam rating. It's just silly how there absolutely [i]has[/i] to be at least one person who just [i]has[/i] to break the rules, and even worse that he hasn't been stopped.

Anyway, the BA forum is pretty much dead; the only semi-alive topic is the marathon, and even that isn't getting much attention.

[quote="i_like_pie"]About your idea of removing ratings, unfortunately you won't get anywhere with that. The best you'll accomplish is disable ratings in our forum (like the Test Forum, G&FF, etc.).[/quote]

Ratings in a forum which does not raise your post count doesn't affect your average.

[quote="i_like_pie"]
Somebody apparently likes rating everything spam in the Bay Area forum, so I have stopped posting there completely.[/quote]
The private Michigan forum has the same problem.

[quote="Temperal"][quote="i_like_pie"]About your idea of removing ratings, unfortunately you won't get anywhere with that. The best you'll accomplish is disable ratings in our forum (like the Test Forum, G&FF, etc.).[/quote]

Ratings in a forum which does not raise your post count doesn't affect your average.[/quote]
Did I say they did? :huh:

[quote="i_like_pie"]there absolutely [i]has[/i] to be at least one person who just [i]has[/i] to break the rules[/quote]
That's one of the fundamental laws of nature :D, probably even more fundamental than the physics laws.
[quote]
and even worse that he hasn't been stopped.
[/quote]
I agree with you here but the only ones who can do "by force" are admins and they seem to be overloaded with other stuff that, at least in their opinion, has higher priority. For better or for worse, there is no "forum police" here though Valentin tries to implement some "automatic censoring" now and then. The only way for a general user to stop this kind of abuse is the one mentioned by JBL: be an active rater yourself and rate the posts properly. I also agree that we may underestimate the negative impact of people's hurt feelings when their posts get inappropriate ratings and I myself am definitely guilty of not rating posts I read at all most of the time. But I strongly disagree with your statement that "there's no point in posting somewhere when you know you're almost guaranteed a spam rating". You post not to get ratings but to convey information, to ask for help, to challenge people, or, sometimes, just for the general fun of communicating with somebody else. None of those can be taken away from you by assigning less than 3 hexagons to your post.

[quote]Did I say they did? :huh:[/quote]

No, but you implied it by suggesting removing ratings in those fora would be helpful.

I think he was just implying that the ratings there shouldn't count, just like the ones in the other forums he mentioned don't count.

Oh, he meant the BA forum's ratings. Maybe the entire N/L C forum post ratings should be disabled.

[quote="fedja"]The only way for a general user to stop this kind of abuse is the one mentioned by JBL: be an active rater yourself and rate the posts properly.[/quote]
I constantly do this, but one person's efforts are not enough. :(

[quote="fedja"]I strongly disagree with your statement that "there's no point in posting somewhere when you know you're almost guaranteed a spam rating". You post not to get ratings but to convey information, to ask for help, to challenge people, or, sometimes, just for the general fun of communicating with somebody else. None of those can be taken away from you by assigning less than 3 hexagons to your post.[/quote]
Point taken, you're obviously correct. :)

[quote="Temperal"]Oh, he meant the BA forum's ratings.[/quote]
Yes; now I understand what the confusion was about.

[quote="Temperal"]Maybe the entire N/L C forum post ratings should be disabled.[/quote]
I thought about this, and considering that almost every single community forum has at most 2 active topics, I don't think many people would have a reason against this proposal. If some more people agree, we can push this idea until it becomes a reality. :)

is it really true that some people give others low ratings just because they dislike them? :maybe:

it really doesn't make sense nor is it fair for other people who don't deserve low ratings. :(  :huh:

You think too well of mankind :). It is worse than that: it seems that some people give others low ratings just for the fun of making other people upset.

Is it possible for a user to access the ratings log? That way, people can pinpoint specific spam-raters and report them to the admins.

[quote="gauss1181"]is it really true that some people give others low ratings just because they dislike them? :maybe: [/quote]

This is a maybe. If I see someone I dislike I'm not going to rate them spam if they're post is good. I'm simply not going to rate it. However, if a person I disliked spammed then I will poorly rate him without thinking about it.

This has to be a joke, right? Who rated my previous post < 3? So now even the topic that discusses bad ratings gets bad ratings? :maybe:

[quote="archimedes1"]Assume it's a joke. I would stop wasting energy on things like this.  This topic is probably loads of fun to read for the person(s) who rates all posts as spam.[/quote]

that's funny to all.

I think that someone is abusing the ratings as well. I thought everyone hated me with the 2.8 rating...

I think the posts that are the most easily targeted are posts that are not important enough to get a single rating from most people, in which case a single rating from an abuser makes the whole post rated as spam.

So a possible solution is for everyone to try to rate every unrated post that you read.


Euler
geometry
geometric transformation
reflection
geometry proposed

Let k be line with arbitrary direction, $ k_a, k_b, k_c$ its reflections in the triangle sides BC, CA, AB and $ p_a \parallel k_a,\ p_b \parallel k_b,\ p_c \parallel k_c$ their parallels through the triangle vertices A, B, C. Show that $ p_a, p_b, p_c$ are concurrent iff $ k \parallel OH$ (OH is the triangle Euler line).

Reflections $ k_a, k_b, k_c$ in BC, CA, AB of any line k through orthocenter H, including the Euler line OH, concur in the pole $ P \in (O)$ of the Simson line parallel to k. If $ k \equiv OH,$ $ k_a, k_b, k_c$ pass through reflections $ O_a, O_b, O_c$ of O in BC, CA, AB. in The 9-point circle N is the midpoint of $ AO_a, BO_b, CO_c \Longrightarrow$ parallels $ p_a \parallel k_a,\ p_b \parallel k_b,\ p_c \parallel k_c$ through A, B, C pass through reflection Q of P in N.


if you have a recursive formula like $2*a_n  + 5*n_{n+2}= 4*a_{n+1}$ with $a_0 = 3$
how would you find an explicit formula for this, i know there is a way, but i don't really understand any of the explanations
and also a more general procedure for finding the explicit formula for something like
$\sum_{i=0}^m k_i*a_{n+i} = 0$ where $a_n$ is the nth term of the sequence, $k_n$ is the constant that goes along with $a_n$ and $a_0 = r$

The initial condition $a_0=r$ does not suffice to define a unique sequence if $m>1$. Rather, you would need the first $m$ initial values. To solve, let $a_n=b^n$ and solve the resulting equation, obtaining solutions $b_1, b_2, \ldots b_m$. Then the general solution is $a_n=\sum_{i=1}^mc_ib_i^n$. Use the initial values to find all $c_i$.

I think what you're asking for is the Characteristic Equation.

You apply a z-transform and solve for the roots of the equation.

Then, plug these in and find the coefficients based on initial conditions.

For example, given the Fibonacci sequence, solve for the roots of $x^2-x-1=0$

and solve for A and B in $a_n=Ar_1^n+Br_2^n$

the thing is that i don't know what a characteristic equation, or any of that is, i would just like to know how step by step how to do it, if there is a website that shows it, then fine, but i cannot seem to find a good explaination

Here is an example. The Fibonacci sequence is generated by $f_0=1$, $f_1=1$, and $f_{n+2}=f_n+f_{n+1}$. To get the characteristic equation, we replace $f_{n+k}$ by $b^k$, giving $b^2=1+b$, or $b^2-b-1=0$. This quadratic equation has solution $b=\frac{1\pm\sqrt{5}}{2}$. These roots are commonly known as $\phi$ and $1-\phi$. We use the solutions of the characteristic equation to form a general solution to the recurrence relation: $f_n=c_1\phi^n+c_2(1-\phi)^n$ for some $c_1$ and $c_2$. But since we want a specific solution, we substitute the values of $f_0$ and $f_1$. Using $f_0$ gives us $1=c_1+c_2$, and using $f_1$ gives us $1=c_1\phi+c_2(1-\phi)$. Solving simultaneously, we find that $c_1=2-2\phi=1-\sqrt{5}$ and $c_2=2\phi-1=\sqrt{5}$. This gives us the final answer, $f_n=(1-\sqrt{5})\phi^n+\sqrt{5}(1-\phi)^n$.

[quote="Altheman"]alright thank you...that helps, but why does this work? i know nothing about characteristic formulas, my school's math program is pathetic[/quote]

It isn't something covered in almost any school's math program.

There's lots of information online about how to solve linear recurrence relations. One popular way is to use generating functions.

Here's a few pages describing another way to solve them.


At an immunization clinic, there is one scheduled appointment every five minutes starting at noon. Each patient arrives exactly at the scheduled time, spends 10 minutes in the lobby doing paperwork, then five minutes out of the lobby getting shots, and finally 20 minutes in the lobby to be sure there is no adverse reaction. How many patients are in the lobby at 12:34 p.m.?

Patients are at the clinic for $ 10 \plus{} 5 \plus{} 20 \equal{} 35$ minutes. Therefore, we care about all patients who arrive from 12:00 to 12:30, inclusive. This is 7 patients. At any given time after 12:10, one patient is getting shots, and thus out of the lobby. Thus the answer is $ \boxed{6}$ patients; the patient who arrived at 12:20 isn't in the lobby.


To make lemonade, I use a ratio of $ 7$ parts water to $ 1$ part lemon juice. If I want to make a gallon of lemonade, and there are four quarts in a gallon, how many quarts of water do I need?

We want $ 4*\frac {7}{8} \equal{} \frac {7}{2}$ quarts of water.


$x,y,z \in \mathbb{R}$ , and $x+y+z=6$ . find the maximum possible value of $m$ if $x^{2}+y^{2}+z^{2}\ge m$

Perpendiculjar to $x+y+z=6$ is line $x=y=z$.

more details please ... please be more explicit . anyone ?

I have a question: Shouldn't $x,y,z$ rather be positive reals, as, if they can be negative, the sum $x^{2}+y^{2}+z^{2}$ can diverge to infinity...

Rust i don't understand. why do you get maximum for x=y=z. i knew that, but i want proof, that's why i posted the problem in the first place

anyway  , i think that the problems asks for the minimum value of the expression . read carefully

If I understand correctly, the line $x=y=z$ is perpendicular to $P\equiv z=6-x-y$ so the minimal distance from the origin to $P$ is the distance from the origin to point $A(2;2;2)$, which is the intersection of the line  $x=y=z$ with $P$

For $x=n,y=0,z=6-n$ we have that $x+y+z=6$ and $x^{2}+y^{2}+z^{2}=2n^{2}+36-12n$ which can be made arbitraly large as $n$ can be arbitraly large.
On the other hand, by Cauchy's $(x^{2}+y^{2}+z^{2})(1+1+1)\geq(x+y+z)^{2}$ so $m\cdot 3\geq 36$ so $m\geq 12$ with equality when $x=y=z$.


I just started this in order to find out what other high school freshman got on the AMC

Which one should I choose if I got a 125 on the AMC 12?

Haha. Yeah. Some freshman are really smart...

I thought you were an 8th grader Jarey  :?

Well, I didn't take the AMC 12 when I was a freshman (only 8th grade and sophomore), so I guess I'll put my 8th grade score (140.5).  The AMC12B in 2002 was extremely easy.


\sqrt[x]{y} yields
[tex]
\displaystyle \sqrt[x]{y}
[/tex]

Of course, there is always [tex]y^{\frac{1}{x}}[/tex]

[quote="Rep123max"]Let me test it out:

[tex]\displaystyle\frac{a_1+a_2+a_3+...+a_n}{n}\geq \sqrt[3]{a_1a_2a_3...a_n}[/tex].[/quote]

Shouldn't

[tex]\displaystyle\frac{a_1+a_2+a_3+...+a_n}{n}\geq \sqrt[3]{a_1a_2a_3...a_n}[/tex]

be

[tex]\displaystyle\frac{a_1+a_2+a_3+...+a_n}{n}\geq \sqrt[n]{a_1a_2a_3...a_n}[/tex]

by AM-GM (I know this isn't the point of the topic)?

Yea but he is applying it to a certain problem where n=3.  This is a problem in another section though.

[quote="Rep123max"]Let me test it out:

[tex]\displaystyle\frac{a_1+a_2+a_3+...+a_n}{n}\geq \sqrt[n]{a_1a_2a_3...a_n}[/tex].[/quote]

You might also try \cdots instead of ... , as shown below:

[tex]\displaystyle\frac{a_1+a_2+a_3+\cdots+a_n}{n}\geq \sqrt[n]{a_1a_2a_3\cdots a_n}[/tex]

\ldots gives you dots along the bottom; \cdots gives you dots in the middle.

[tex](1 + 2 + \ldots +4) \cdots(7 + 8 + \ldots + 10)[/tex]

Also, \cdot will give you one single dot for multiplication.

[quote="confuted"][quote="JBL"]Also, \cdot will give you one single dot for multiplication.[/quote]
It's also quite nice for a dot product.[/quote]

You can also use \times for multiplication, though most people prefer a dot.  \times works for cross-products though.

[quote="Rep123max"]Let me test it out:

[tex]\displaystyle\frac{a_1+a_2+a_3+\cdot \cdot \cdot +a_n}{n}\geq \sqrt[n]{a_1a_2a_3\cdot \cdot \cdot a_n}[/tex].[/quote]
I think the little n is too low. LaTeX will let you move it using \leftroot and \uproot
[tex]\sqrt[\leftroot{-2}\uproot{3}n]{a_1a_2a_3\cdot \cdot \cdot a_n}[/tex] which looks a lot better don't you think?
[code]\leftroot(2) means 2 to the left
\leftroot(-2) means 2 to the right
\uproot(3) means 3 up
\downroot(-3) means 3 down[/code]


:w00t: The admins listened! Now there's somewhere for all of the astronomy topics to go, and for all of us astronomy/astrophysics people to discuss things. I'm excited... :laugh:

Oh, would it be possible for someone to move my dark matter thread and other random astronomy threads (i.e. Astronomy Olympiad) over here?

:lol:
It is quite exciting.  :)
Let the posting begin.
There is an Astronomy Olympiad?

There should be a wikipedia forum....
http://en.wikipedia.org/wiki/International_Astronomy_Olympiad

Hm, astronomy and engineering is a slightly odd match given that physics and astronomy would probably be better.  Oh well, all the more better for the forum (hopefully :P)!


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

I draw 2 axes, one horizontal and one vertical, and label the horizontal one 6:30, 7:00, and 7:30 for the newspaper, and the vertical one 7:00, 7:30, and 8:00. What I'm confused about is what to shade in for the region where he gets the newspaper before he leaves for work. Could you please explain this one to me?

There is a 1/2 chance that he leaves from 7:30-8:00. He then automatically gets the paper. There is also a 1/2 chance that the paper comes from 6:30-7:00. He then automatically gets the paper. So we have 1/2+1/2-1/4=3/4. Now we have 3/4+the probability that he leaves after the paper arrives when they are both leaving/arriving between 7:00-7:30. This is clearly 1/2 since each have an equal chance of occuring first. So we have 3/4+1/4*1/2=7/8. Was that of help?

Of course, if I were him, I would always leave at 7:30...

[quote="keta"]I draw 2 axes, one horizontal and one vertical, and label the horizontal one 6:30, 7:00, and 7:30 for the newspaper, and the vertical one 7:00, 7:30, and 8:00. What I'm confused about is what to shade in for the region where he gets the newspaper before he leaves for work. Could you please explain this one to me?[/quote]
You must shade the region where $y>x$ (or $y\geq x$, it doesn't matter), i.e. the region above the angle bisector of the first quadrant. This will cover 7/8 of the square of all possibilities.


induction
number theory proposed
number theory

Let $ d_{1}, ..., d_{n}$ be all positive divisors of an arbitary positive integer $ m$ ( 1 and $ n$ included). For each $ i=1, ..., n$ denote by $ a_{i}$ the number of positive divisors of $ d_{i}$. Prove that: $ a_{1}^{3}+...+a_{n}^{3}= (a_{1}+...+a_{n})^{2}$

First note that it is true for all powers of primes, for which it becomes $ \sum_{i=1}^{n}i^{3}=(\sum_{i=1}^{n}i)^{2}$. To complete the proof, we show that if it is true for $ a$ and $ b$ coprime, then it is true for $ ab$. Call the divisors of $ a$ $ a_{i}$, $ 1\le i\le m$ and the divisors of $ b$ $ b_{i}$, $ 1\le i\le n$. Then the divisors of $ ab$ are precisely the numbers $ a_{i}b_{j}$, $ 1\le i\le m$ and $ 1\le j\le n$. Hence we see that $ \sum_{(i,j)=(1,1)}^{(m,n)}(a_{i}b_{j})^{3}=(\sum_{i=1}^{m}a_{i}^{3})(\sum_{j=1}^{n}b_{j}^{3})$ =$ (\sum_{i=1}^{m}a_{i})^{2}(\sum_{j=1}^{n}b_{j})^{2}=(\sum_{(i,j)=(1,1)}^{(m,n)}a_{i}b_{j})^{2}$, as desired. Then we are done by induction on the number of prime divisors.


inequalities
search
inequalities solved
3-variable inequality
cauchy schwarz

Lets see :
$Case I$ one of the numbers is $0$ so
$a^2+b^2=1$ and
$a+b\leq\sqrt2$ This is easy to prove because $a^2+b^2=(a+b)^2-2ab=1$ and we have to prove that
$(a+b)^2\leq2$. $2ab$ takes it's maximum when $a=b$ and so on..(This is when we get the equality)
$Case II$ none of them is $0$ and WLOG $a \geq b \geq c$.
$abc=P$.
$a \rightarrow \frac{a}{\epsilon}$
$b \rightarrow b\epsilon$
$\epsilon\geq1$
$a\geq b\epsilon$.
it is easy to see that $a+b+c\geq\frac{1}{\epsilon}+b\epsilon+c$ so when all of the numbers are equal the $LHS$ takes its maximum now by simply checking we get that the inequality is true.

You didn't use in your solution the condition
$a^2+b^2+c^2=1$!!  :(

What do you mean??? I have just not written the whole solution....

[color=blue][b]Problem.[/b] For any three reals a, b, c satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, prove the inequality $ a \plus{} b \plus{} c \minus{} 2abc\leq\sqrt2$.[/color]

[i]Solution.[/i] The following beautiful solution is not mine - it is stolen from a paper by Harazi, Gil and Vasc (or, at least, a subset of them) called "The Cauchy-Schwarz inequality" (it is a 17-pages long paper containing some exercises, partially with solutions, probably meant for publication in some magazine):

Since $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ and $ a^2\geq 0$ yield $ b^2 \plus{} c^2\leq 1$, and since $ 2bc\leq b^2 \plus{} c^2$ by a well-known inequality, we have $ 2bc\leq 1$.

Now,

$ a \plus{} b \plus{} c \minus{} 2abc \equal{} a\cdot\left(1 \minus{} 2bc\right) \plus{} \left(b \plus{} c\right)\cdot 1$.

By the Cauchy-Schwarz inequality,

$ a\cdot\left(1 \minus{} 2bc\right) \plus{} \left(b \plus{} c\right)\cdot 1\leq\sqrt {\left(a^2 \plus{} \left(b \plus{} c\right)^2\right)\cdot\left(\left(1 \minus{} 2bc\right)^2 \plus{} 1^2\right)}$.

Hence, we have

$ a \plus{} b \plus{} c \minus{} 2abc\leq\sqrt {\left(a^2 \plus{} \left(b \plus{} c\right)^2\right)\cdot\left(\left(1 \minus{} 2bc\right)^2 \plus{} 1^2\right)}$
$ \equal{} \sqrt {\left(a^2 \plus{} \left(b^2 \plus{} c^2 \plus{} 2bc\right)\right)\cdot\left(2 \minus{} 4bc \plus{} 4b^2c^2\right)}$
$ \equal{} \sqrt {\left(\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 2bc\right)\cdot\left(2 \minus{} 4bc \plus{} 4b^2c^2\right)} \equal{} \sqrt {\left(1 \plus{} 2bc\right)\cdot\left(2 \minus{} 4bc \plus{} 4b^2c^2\right)}$

(since $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$). Thus, in order to establish $ a \plus{} b \plus{} c \minus{} 2abc\leq\sqrt2$, it remains to show that $ \sqrt {\left(1 \plus{} 2bc\right)\cdot\left(2 \minus{} 4bc \plus{} 4b^2c^2\right)}\leq\sqrt2$. This is equivalent to

$ \left(1 \plus{} 2bc\right)\cdot\left(2 \minus{} 4bc \plus{} 4b^2c^2\right)\leq 2$.

But this is easy, since

$ 2 \minus{} \left(1 \plus{} 2bc\right)\cdot\left(2 \minus{} 4bc \plus{} 4b^2c^2\right) \equal{} 4b^2c^2\left(1 \minus{} 2bc\right)$,

what is $ \geq 0$, since $ 2bc\leq 1$ yields $ 1 \minus{} 2bc\geq 0$. Thus, the problem is solved.

Note that the paper where I have found this solution contains the problem in a slightly different form:

[color=blue][b]Alternative problem.[/b] For any three reals x, y, z satisfying $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$, prove the inequality $ x \plus{} y \plus{} z\leq 2 \plus{} xyz$.[/color]

But of course, it's the same problem, since the inequality $ x \plus{} y \plus{} z\leq 2 \plus{} xyz$ rewrites as $ x \plus{} y \plus{} z \minus{} xyz\leq 2$, and the substitution $ a \equal{} \frac {x}{\sqrt2}$, $ b \equal{} \frac {y}{\sqrt2}$, $ c \equal{} \frac {z}{\sqrt2}$ brings this into the form $ \sqrt2\left(a \plus{} b \plus{} c\right) \minus{} 2\sqrt2abc\leq 2$, what is equivalent to $ a \plus{} b \plus{} c \minus{} 2abc\leq\sqrt2$.

  Darij

[quote="zhaobin"]let $x=\sqrt{2} a,y=\sqrt{2} b,z=\sqrt{2} c$
the inquality is equal to $x+y+z \leq xyz+2$
and I think it has be sloved in the forum.
you can search it :)  :)[/quote]

The proof for example can be found  in post 6 or 7 http://www.mathlinks.ro/Forum/viewtopic.php?t=27862

How is one supposed to come up with that kind of solution? What is the motivation behind all the random steps, esp. Cauchy?????

I like to believe that it's divine inspiration.

[quote="wisemaniac"]Let $ a$, $ b$, $ c$ be reals such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$.
Prove that $ a \plus{} b \plus{} c \minus{} 2abc\leq\sqrt2$.[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ where $ v^2\in\mathbb R$ and $ abc \equal{} w^3.$
Hence, $ a \plus{} b \plus{} c \minus{} 2abc\leq\sqrt2\Leftrightarrow$
$ \Leftrightarrow3u\sqrt {3(3u^2 \minus{} 2v^2)} \minus{} 2w^3\leq3\sqrt {6(3u^2 \minus{} 2v^2)^3}\Leftrightarrow f\left(w^3\right)\geq0,$ where
$ f\left(w^3\right) \equal{} 2w^3 \minus{} 3u\sqrt {3(3u^2 \minus{} 2v^2)} \plus{} 3\sqrt {6(3u^2 \minus{} 2v^2)^3}.$
But $ f$ is increasing function. Hence, $ f$ gets the minimal value, when $ w^3$ gets a minimal value, id est,
when two numbers from $ \{a,b,c\}$ are equal.
Thus, it remains to check the following inequality.
$ (a^2 \plus{} 2b^2)(a \plus{} 2b) \minus{} 2ab^2\leq(a^2 \plus{} 2b^2)\sqrt {2(a^2 \plus{} 2b^2)},$ which is easy for making.

[b](American Mathematical Monthly 2004)[/b]
[quote=wisemaniac]Let $a$, $b$, $c$ be reals such that $a^{2}+b^{2}+c^{2}=1$.
Prove that $a+b+c-2abc\leq\sqrt2$.[/quote]
[b]Proof:[/b]
$(a+b+c-2abc)^2=\left((a+b)\cdot 1+c(1-2ab)\right)^2$
$\leq\left((a+b)^2+c^2\right)\left(1+(1-2ab)^2\right)$
$=2-4a^2b^2(1-2ab)\leq 2.$
[b]Strengthening:
[/b]Let $a$, $b$, $c$ be reals such that $a^{2}+b^{2}+c^{2}=1$.Prove that $$a+b+c-2abc\leq \sqrt{2- 5a^2b^2c^2}.$$


Prove that there exists a natural number $k$ for which
$k\cdot2^n+1$ is composite for each natural number $n$.


geometry
perimeter
trigonometry
ratio
analytic geometry
graphing lines
slope

I plan to make a little contest for my school (or the middle schooled attached to my school, depending on difficulty).

Unlike a friend of mine who also did this, I plan to make my questions [u]original[/u].

Difficulty can be varied, I want to start on about amc10 question 10 level, and possibly move up to amc12 question 20 level.

Would anyone like to help, just post w/e questions that come to mind.

Note: Must be original.

EDIT: I plan on making the contest full-solution.
EDIT: If anyone wants to post a question, can you pm me that question with your full name so I can give credit. If you do not want to post your real name (hunted by FBI, etc.), tell me so and then tell me what you should be refered as.

Can you give a more detailed range of the problems you want??
Is trig, vector and conics allowed?

Well, my initial plan is to write a contest for the middle school, as my high school has too much good people for me to manage. (2 past imo contestants :? ), so I want to keep it simple.

Maybe on the last questions we might add a little basic trig on it, but for everything else, I'd stick to areas, basic slopes, and stuff that may be solved my a gifted middle-schooler.

Acrobat Distiller?

Here's a problem I thought up:
Find $\sqrt{18}+\sqrt{9}+\sqrt{4.5}+...$. Express your answer in simplest radical form. It's not the hardest problem in the world, but I bet some middle schoolers won't be able to get it. :D
I have another good ratios problem, and I'll post that soon.

Very good question. Difficulty might be a little hard, but acceptable for one of the later questions.
BTW, is the answer [hide]$6(\sqrt2+1)$[/hide]

Thats a geometric sequence. Each time you divide by $\sqrt2$...So use the formula and rationalize.

How about a question like this (I just made the numbers on this up, you might want to change the numbers for a cleaner answer):

Marc drove for 4 hours.
Marc traveled for 2 hours at an average of 60 mph. In the next hour, he caught traffic and his average was reduced to 45 mph. He tried to make up for time by averaging 72 mph in the last hour.

What was Marc's average speed?

Questions like that test to make sure they understand how to manipulate time and they are forced to either use variables or find the total distance....

To shinwoo:
It's an infinite geometric sequence that has a common ratio $\sqrt{\frac{1}{2}}$, which equals $\frac{\sqrt{2}}{2}$.

So plugging into the formula $\frac{a}{1-r}$:
$\frac{\sqrt{18}}{1-\frac{\sqrt{2}}{2}}=$
$\frac{\sqrt{18}(1+\frac{\sqrt{2}}{2})}{1-\frac{1}{2}}=$
$\frac{\sqrt{18}+\frac{\sqrt{36}}{2}}{\frac{1}{2}}=$
$2(\sqrt{18}+3)=$
$2(3\sqrt{2}+3)=$
$6(\sqrt{2}+1)$
Which is hiitsme's answer.

[quote="hiitsme"]This is a question to all people. What are some good free pdf creators? I use linux, so a linux native program would be good, but I can try to emulate windows programs.[/quote]

I actually have the entire Adobe Creative Suite...So if you just send me the file...I would be glad to convert it for you.

[quote]I actually have the entire Adobe Creative Suite...So if you just send me the file...I would be glad to convert it for you.[/quote]

Great. What file should I send it to you as? Or should I jst post the questions.

I actually have a program called lyx, which can create pdfs, but your's might be more professional. I'm not sure  :? .

I dont know how much different it will come out...I am not familiar with lyx.

Anyway post the code here, send me a file (windows compatable) by e-mail...I really don't care.  Do whatever you find easiest!
 :D

The First MINGMO
1. A circle with center $\displaystyle O$ has circumfrence $\displaystyle 4\sqrt{3}\pi$. A chord is drawn so that $\displaystyle \angle AOB = \frac{2\pi}{3}$. find the length of $\displaystyle AB$.
2. In a Tic-Tac-Toe game, how many possible arrangements are there? (Assuming X goes first.)
3.Given $\displaystyle \log_ab+\log_ac=0$ and $\displaystyle \log_ab \times \log_ca=2$. Find the real solution to $\displaystyle \frac12a + 5b$.
4. a) Prove is isoceles triangle $\displaystyle ABC$, where $\displaystyle AB$ and $\displaystyle AC$ are equal, that the angle bisector of $\displaystyle A$ is perpendicular to side $\displaystyle BC$.
    b) Prove that the perpendicular bisector of a chord passes the radius.
5. Find sum of the sequence $\displaystyle \sqrt{18} + \sqrt9 + \sqrt{4.5}........$ (Credit goes to ln(dx/dy) (I seriously need his real name) for this one)

Here you go, I don't know if it would be easy for you to do considering the latex, so if there is a better way to post it, please tell.

I wonder if I can post this contest somewhere in the middle school forums, and and admin can sticky it.

(I know this contest is very primitive and the problems suck, but if people like the idea, I would take longer to write more creative problems. But hey, at least they're original :) )

ooo..

I kinda think I know it now...

well, I'm not really good at dealing with infinite sums

A GOOD pdf generator is .[url=http://www.primopdf.com]Primo PDF[/url] If you can print it, you can turn it into a PDF.

Here's a problem (but its really easy compared to some of the others):

(sorry i dont know how to use LaTeX)

what is the cube root of the quantity:

4^5 + 4^5 + 4^5 + 4^5

By the way, i think that some of those problems might be a little too challenging for middle schoolers (6th, 7th, and 8th graders)

Umm.....
$\displaystyle 4(4^5)=4^6=(4^2)^3=16^3$ ?
Is that what you mean by cube root? I think it would be better phrased. "Evaluate the sum as a cube root" or something.

EDIT: Yes, I'm an idiot. You're right, I totally misunderstood the question. I thought [quote]cube root of the quantity[/quote] meant put the thing in third power, rather than taking the root. Sorry.  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

no, the answer should be 16:

cubert of 4^6= 4^2=16

I really don't like people doing problems with calculators, I'm not going to put incredibly long questions with 5 digit multiplications anyways. For later contests, I'm going to take time and write good questions that tests brainpower instead of who has the most expensive calculator.

But to give an answer though, I probably won't allow calculators, but if people seriously need to cheat (in cases someone can't add or something :( ), the questions might be structured so even if people do cheat, they probably won't get the answer, as the quetions will test more creativity.

Here it is...Sorry for the delay:

http://yashar.biz/mathcounts/The%20First%20MINGMO.pdf

Alright, thanks for the pdf, but you should give credit to John Napp on the last questions. (The infinite geometric sequence)

Yay. Thanks again.
I wonder if anyone has done the questions...

It seems like you're fine on questions for now, but if you're still looking for questions, here are some questions I made up and posted in the geometry class forum (these are all original btw):

1. In any triangle, ABC, let E be a point on the altitude AD (between A and D).
a) Prove that AC^2-CE^2=AB^2-EB^2
b) Show that AC^2-CE^2=AB^2-EB^2 also holds if D is between E and A.

2. In quadlirateral ABCD, angles B and D are right angles. Altitudes drawn from B to AC and from D to AC meet AC at points P and Q respectively. P and Q trisect the line segment AC. Find the perimeter of the quadlirateral if AB=5 and AD = 3.

3. In right triangle ABC, with right angle at B, an altitude is drawn from B to AC. B meets AC at point F. Let AF=d, and FC=e.
a) Find the area of ABC in terms of d, and e
b) Find tan A in terms of d, and e.
For those who don't know trig, tan A=BC/AB

I could probably dig up some more questions like this also (I've got plenty).
Also, I'm a bit lazy right now to right up solutions, but if anybody is having trouble with any of the questions I'll post the solutions ASAP.

Also, here's an algebra (sort of) one:

If p and q are positive integers such that p/q is in lowest terms, and:

p/q=1/1+1/2-2/3+1/4+1/5-2/6+1/7+1/8-2/9+...1/478+1/479-2/480,

prove that p is divisible by 641.

Now, if anyone has any questions to add, please send them to me via pm, as I plan to post this contest in the middle school forums. The poll was a dazzling sucess. 100% of 2 people voted yes  :lol:  :lol:  :lol:  :lol: !!!!!!! (One being me  :blush:)

It's a good idea...I'd reccomend you post that in the MathCounts section. :lol:

for number 1 do u mean the length of arc AB is 2pi/3? im confused sorry or do u mean the actual angle is 2pi/3 degrees?

do we pm answers to u or post them here?

1 is in radians. Equivalently I could put $%Error. "dosplaaystyle" is a bad command.
120^\circ$. Post them here if you wish, as this is the sample contest, I'll post another in the middle school shortly. How do people find the difficulty?

The slope of a line is 3.  We shift it up by 5.  Then it is reflected on the y axis.  We then change the slope to make it the multiple of five (...-15, -10, -5, 0, 5, 10, 15 ...) nearest to it.  We add 2 to the b value in y = mx + b.  Then we change m to the old m - 2.  Then we multiply the whole equation by 2 and set the y intercept to 0.  Write the equation of that line.

Answer :

[hide]
y = -7x
[/hide]


What is the sum of the prime numbers between 10 and 20?

The prime numbers between 10 and 20 are 11, 13, 17, and 19. $ 11 \plus{} 13 \plus{} 17 \plus{} 19 \equal{} 60$.

EDIT: Beaten.


integration
function
calculus
calculus computations

If $ f: [a,b]\rightarrow R$ be a continuous on $ [a,b]$ and differentiable on $ (a,b)$ then prove or disprove that there is some $ c\in(a,b)$ such that

$ \int_a^{b}f(x)dx\equal{}f(a)(b\minus{}a)\plus{}f'(c)\frac{(b\minus{}a)^2}{2}$

In such questions one is required to assume some function for which conditions for Rolle's or LMVT are satisfied.What I want to know is how does one find such functions

Do you want an obvious example? How about $ f(x) \equal{} 1$?

How do you find such functions? That's just a "nice enough" condition, so almost anything you would write down is one.


Is it valid in Olympiad questions, when using Pell equations to find the fundamental solution and then extrapolate from that, or what? Is it possible to prove that no other fundamental solutions exist

Yes, but I'm not sure whether this is a citable result.  What you want is the notion of a [url=http://mathworld.wolfram.com/FundamentalUnit.html]fundamental unit[/url].


this is the ying yang game. you "want" or "dont want" two things and the person below you has to compramise them. example:

i want chicken but
i want beef

person below me:
you get chicken-beef

i want sandwhiches but
i dont want bread

ill start us off

i want lollipops but
i dont want lollipops

Err: Invalid Argument

I want world peace
but I don't want the world to die.

Everyone in the world dies, but the world doesn't die.

I want 1=2
but I don't want 1=1

You get 0, but you don't get 1.

I want calculus 2 to be done
but I want my sanity intact as well.

I want posts that are easier to understand, but I don't want
stupid posts. (like this one)

you get posts that say "SMART SMART SMART SMART" (you know? like a sign that says "BAD" and a person saying "well. thats not a good sign")

anyway
I want my math homework to be finished, but in the way that I like it.

you finish your math homework, but the way you like it is for every problem to be wrong so you fail :o

i want it to be Wednesday tomorrow, but i don't want to skip tuesday (idk...)

You get to skip Tuesday, but nobody else skips Tuesday, so you get extremely confused.

I want pancakes for breakfast everyday, but I want them to taste really good.

Wait isn't this basically a corrupt-a-wish game???

No, but it just turns out that everyone corrupts the requests anyway.

You don't HAVE to corrupt it.

You get awesome, delectable, delicious, fantastic, tasteful pancakes everyday, but you eventually get tired of it.

I wish my FTW score was really high (even higher than mews), and my reactions would be faster.

You are on top of FTW scoreboard until AIME15 pwns u 35000 times in a row and steals ur rating.

I want cheese
I dont want chess

Your friends get you cheese for your birthday, and throw away your chess set. (You meant to say cheese, not chess right?)

I want to be taller.
I want some candy right now.

You have a whole bunch of candy piling on top of you, squashing yourself and condemning your height to stay at 11 inches for the rest of your life.

I want Antium but I don't want Rome

A rich guy buys you Antium, but not Rome.

(Stop corrupting the wishes!)

I want to go to CCC, but I don't want to not go to math club.

[quote="Wickedestjr"]You get to go to CCC only once, and you never get to go to any more math clubs any more again.

I want a larger variety of fruits in this world, but a smaller variety of vegetables in this world.[/quote]

Please continue off of this.

WG. Now the only vegetable is Sauerkraut and you must have three servings of it every day!

I wish for a brand new (#*^)^_%*#)^*)#(#

You get a (#*^)^_%*#)^*)#(#, but you never use it at all.

ERROR: NO WISH!

don't be getting ideas, TPBM

ERROR: NO WISH! :P

get ideas from TPAM and TPATPAM and TPATPBM

I want this spam chain to end
but i want it to never return

The spam chain is over and will never return until a day has passed.

I want no school,
but I want to be really smart.


The point of tangency of an inscribed circle to a right triangle divides the hypotenuse into two segments of length 7 and 8. What is the area of the triangle?

Let $ ABC$ be the right triangle with $ A=90$ and $ D,E,F$ the points of tangency with $ BC,AB,AC$.
Then we have $ AB=7+r,AC=8+r$
$ AB^{2}+AC^{2}=BC^{2}\Rightarrow (7+r)^{2}+(8+r)^{2}=225\Rightarrow r^{2}+15r-56=0$.
The area is $  \frac{AB\cdot AC}{2}=\frac{(7+r)(8+r)}{2}=\frac{r^{2}+15r+56}{2}=56$

where can you obtain past westpac comp. papers?

Let triangle ABC be right angled at C. Let the incircle touch the hypotenuse AB at T.
Given c = 15,
s-b = BT = 8,
s-a = AT = 7
from which we get by subtraction,
a-b = 1.
Now 225 = c^2
             = a^2 + b^2
             = (a-b)^2 + 2ab
             = 1 + 2ab,
So     ab = 112,
      area = (1/2)ab = 56

See these sites:

[url]http://www.amt.edu.au/wuamcqs04.pdf[/url]

[url]http://www.amt.edu.au/wuamcqi04.pdf[/url]

[url]http://www.amt.edu.au/wuamcqj04.pdf[/url]


Let $(x_n)$ and $(y_n)$ be decreasing sequences of positive real numbers. We know that for every $k \in Z^+$ we have $x_1x_2...x_k \geq y_1y_2...y_k$.
Prove that for every $k \in Z^+$ $x_1 + x_2 + ... + x_k \geq y_1 + y_2 + ... + y_k$.

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=36610
http://www.mathlinks.ro/Forum/viewtopic.php?t=29109

  darij


modular arithmetic
function
absolute value
number theory
totient function

I'm rather new to all this so can someone please post some properties of the modulus ( the one that has to do with divisibility, i.e.  a=n (mod m) etc, not the absolute value )
By which I mean is it associative, commutative and distributive under multiplication or addition, and other relations?

I would really appreciate your help   :D

If $a \equiv b\pmod{m}$, then for all positive integers $c$:

1. $a+c \equiv b+c \pmod{m}$
2. $a-c \equiv b-c \pmod{m}$
3. $ac \equiv bc \pmod{m}$
4. $a^c \equiv b^c \pmod{m}$
5. $(a+b) \pmod{m} \equiv a \pmod{m}+b \pmod{m}$
6. $ab \pmod{m} \equiv \left(a \pmod{m}\right)\left(b \pmod{m}\right)$

[quote="prototypex24"]Thankls for the reply, NoSoupForYou

However, I'm wondering if anyone can explain to me what are "congruences" in the context of modulo arithmetic? :?[/quote]

I think you mean $x \equiv y \mod {n}$ if $x-y$ is divisible by $n$. You read this $x \text { is congruent to } y \text { mod } n$. Like 3 is congruent to 1 mod 2.

$x \equiv y \pmod {n}$ means that the last digits of $x$ and $y$ are the same in base $n$.

Thanks for all your replies  :D

But I do have one question:
Is   -a(mod n) equivalent to (n-a)(mod n)?

Yep

You can probably convince yourself by using rule #5 above.

Thanks WindSlicer and everyone else who answered my questions.

Another Question :P
Does identity 5. and 6. in NoSoupForYou's reply hold when
a is not congruent to b modulo m ?

Yes. For example,

$36 \cdot 3 \mod {2} = (36 \mod {2})(3 \mod {2})=0$. There is no relationship, necessarily between $a$ and $b$.

I've seen a lot of people use modulo in questions like 'find the last three digits of a certain number'.

Can someone explain to me how this is done? I don't get it :?

Thank you.

The main idea is to find what that number is congruent to in mod 1000.  :)

Find the last 3 digits of $3^{1234}$.

[hide="Solution"]
First of all, we're trying to find $3^{1234} \mod {1000}$.

Sorry, I have to go. I'll finish later.
[/hide]

Euler's Theorem is also useful:
$a^{\phi (n)} = 1(mod n)$
where $(a,n)=1$

Also, what's $\phi (n)$ too? I'm not very good at this, so I hope you don't mind... :blush:

(a,n) stands for the greatest common divisor. When (a,n) = 1, it means that a and n are relativly prime.
$\phi(n)$ is the totient function, and I think it gives you the number of integers less then n that are relativly prime to it. I think. Maybe.

What's the difference between relatively prime and coprime?

http://mathworld.wolfram.com/Coprime.html

They're basically same as relatively prime.

Note: Totient function is not getting started. So, if this discussion goes on, I'll lock the topic (you'll see that some topics are locked and the reason is because they are going over the level of Getting Started).

Please discuss the advanced materials in Intermediate or Pre-Olympiad.

Thank you.

[quote="chess64"]Find the last 3 digits of $3^{1234}$.

[/quote]

Can someone solve this as an example? I just need to see how it's done ;)

how would you find the last three digits of 3^3^3^3 ... and higher powers?


Find $ \sum^{49}_{k\equal{}0} (\minus{}1)^k \binom{99}{2k}$ where $ \binom{n}{j} \equal{} \frac{n!}{j!(n\minus{}j)!}$.

$ \textbf{(A) } \minus{}2^{50} \qquad \textbf{(B) } \minus{}2^{49} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 2^{49} \qquad \textbf{(E) } 2^{50}$

[quote="bbgun34"]Another way to do it without complex numbers is just with the binomial theorem, i.e. (1+1)^99 and (1-1)^99.[/quote]

can you elaborate?


function
real analysis
real analysis unsolved

Let function f;g:[a;b]->R (a<b), are sequent function such that 0<f(x)<g(x)  with $ x\in[a;b]$.Prove that always exists a>0 such that $ (1 \plus{} a)f(x)\le$g(x)with $ x\in[a;b]$

Please, edit your post to make it comprehensible (and [color=red]always[/color] use the "Preview" button before clicking the "Submit" one).

What is the definition of sequent function? I guess, that you meant that there exists $ x$ for which $ (1\plus{}a)f(x)\le g(x)$, right?

you can see again
Let function f;g:[a;b]->R (a<b), are sequent function such that 0<f(x)<g(x) with $ x\in[a;b]$.Prove that always exists c>0 such that $ (1 \plus{} c)f(x)\le$g(x) with $ x\in[a;b]$

@leshik: I'm pretty sure that "sequent" really means "continuous" and the whole problem is just to show that if a continuous on a compact interval function $ \frac{f}{g}$ is strictly less than $ 1$ everywhere, then there exists $ c>0$ such that $ \frac fg\le\frac 1{1\plus{}c}$ everywhere ;).

Ok, if this is the case then ... just Weierstrass theorem :wink: