Έστω  $ AA_1$  η διχοτόμος οξυγωνίο τργιγώνου $ ABC$, και έστω $ M$ το μέσον του $ AA_1$. Παίρνουμε σημείο  $ P$ στην  $ BM$ έτσι ώστε  $ \angle APC \equal{} 90^{\circ}$ και ένα σημείο $ Q$ στην $ CM$ τετοιο ώστε  $ \angle AQB \equal{} 90^{\circ}$.
Νδο τα σημεία $ A_1$, $ P$, $ M$ and $ Q$ είναι ομοκυκλικά .

Περιμένω τις λύσεις σας  :)

[quote="silouan"]Έστω  $ AA_1$  η διχοτόμος οξυγωνίου τργιγώνου $ ABC$, και έστω $ M$ το μέσον του $ AA_1$. Παίρνουμε σημείο  $ P$ στην  $ BM$ έτσι ώστε  $ \angle APC = 90^{\circ}$ και ένα σημείο $ Q$ στην $ CM$ τετοιο ώστε  $ \angle AQB = 90^{\circ}$.
Νδο τα σημεία $ A_1$, $ P$, $ M$ and $ Q$ είναι ομοκυκλικά .

Περιμένω τις λύσεις σας  :)[/quote]

[color=red][b]ΛΥΣΗ[/b][/color]

Εύκολα προκύπτει ότι το ίχνος $ D$ του ύψους $ h_a$ είναι το δεύτερο σημείο τομής των κύκλων $ AQB$ και $ APC$. Επίσης έχουμε ότι $ MD = MA = MA_{1}$ από το θεώρημα της διαμέσου σε ορθογώνιο τρίγωνο. Είναι ακόμη γνωστό ότι $ \angle{DAA_{1} = \theta = \frac {|\angle{B} - \angle{C}|}{2}}$ ως γωνία μεταξύ ύψους και διχοτόμου.

[b]ΛΗΜΜΑ 1.[/b]Οι περιγεγραμμένοι κύκλοι των τριγώνων $ AQC$,$ QCA_1$,$ BPA_1$ και $ APB$ εφάπτονται της πλευράς $ AA_1$.

[i]Απόδειξη.[/i]  Ας θεωρήσουμε ένα σημείο $ Q'\in CM$ τέτοιο ώστε ο περιγεγραμμένος κύκλος του τριγώνου $ AQ'C$ να εφάπτεται της πλευράς $ AA_1$. Σημειώστε ότι είναι σχετικά απλό να αποδείξει κανείς πως υπάρχει τέτοιο σημείο πάνω στη $ CM$ και επιπλέον είναι μοναδικό.
Έχουμε $ MQ'\cdot MC = MA^2 = MD^2$. Αυτό δείχνει ότι ο περιγεγραμμένος κύκλος του τριγώνου $ Q'CD$ εφάπτεται της $ MD$. Χρησιμοποιώντας το θεώρημα που συνδεέι την υπο χορδής και εφαπτομένης γωνία με την εγγεγραμμένη που βαίνει στο ίδιο τόξο, λαμβάνουμε τις ακόλουθες γωνιακές σχέσεις:

$ \angle{MQ'A} = \frac {\angle{A}}{2}$

$ \angle{MQ'D} = \angle{MDA_{1}} = \angle{MA_{1}D = 90 - \theta}$
Συνεπώς $ \angle{B} + \angle{AQ'D} = 180$, δηλαδή το $ Q'$ ανήκει στον κύκλο $ AQB$ και επομένως $ Q' = Q$.
Όμοια αποδεικνύουμε το ζητούμενο και για το σημείο $ P$.

Tώρα, γνωρίζοντας ότι $ \angle{PBA_{1}} = \angle{PA_{1}A}$ και $ \angle{MCA_{1}} = \angle{MA_{1}C}$ έχουμε τις εξής γωνιακές σχέσεις που αποδεικνύουν ότι τα τετράπλευρα $ MPDA_{1}$ και $ MDA_{1}Q$ είναι εγγράψιμα:

$ \angle {MPA_{1}} = \angle{PBA_1} + \angle{PA_{1}B} = \angle{MA_{1}D} = \angle{MDA_1}$ και ομοίως $ \angle{MQD} = \angle{MA_{1}D}$.

Συμπερασματικά, τα σημεία $ M,P,D,A_{1},Q$ είναι ομοκυκλικά.

[i]Παρατήρηση.[/i] Αν σταθεροποιήσουμε τις κορυφές $ B$ και $ C$ και επιτρέψουμε στην κορυφή $ A$ να κινείται ώστε η γωνία $ \angle{BAC}$ να παραμένει σταθερή τότε ο γεωμετρικός τόπος του περικέντρου των προαναφερθέντων ομοκυκλικών σημείων είναι κύκλος κέντρου $ T$, όπου το $ T$ είναι το μέσο του ελάσσονος τόξου $ \overarc{BC}$.

Νίκος Ράπανος

Ωραία λύση Νίκο!

Μια άλλη προσέγγιση θα ήταν αφού αποδείξουμε ότι $ MQ\cdot MC\equal{}MA_1^2$ να πάρουμε αντιστροφή με κέντρο $ M$ κι ακτίνα $ MA_1$. Ετσι τα $ P,Q$ θα πήγαιναν στα $ B,C$ αντίστοιχα και αφού το $ A_1$ θα έμενε στη θέση του, θα είχαμε $ P',A_1,Q'$ συνευθειακά κι άρα $ M, P, A_1, Q$ ομοκυκλικά :wink:


algebra
polynomial
Rational Root Theorem

Well, one of the factors is x.

Another one may be found by rational root theorem.

well, by the RRT, you can find that 1 is a root, but I dont know how to solve anymore for that third degree polynomial-try graphing it(on a calculator) and finding the zeros..

I graphed it and the roots were appaent. It factors as [hide]$(x+5)x(x-1)(x-3)(x-7)$[/hide]

But yeah, if you can't use a graphing calc., go RRT.

Not sure if this is the most efficient way, but here goes.
First we know that $0$ is a root, since $x$ divides the polynomial. We now know one root out of five.
A good next step is to see what $x=1$ yields, since it is easy to evaluate. In this case, it works as well. Two roots out of five.
We divide the polynomial by $x$ and $(x-1)$ which gives us $x^{3}-5x^{2}-29x+105$.
We use the Rational Root Theorem to find that our next root must divide $105$. We try $-5$, and it works. We can continue guessing according to RRT, or we can simply just...
Divide the polynomial we got by $(x+5)$, and we get $x^{2}-10x+21$. The remaining roots are $3$ and $7$.

Of course, we could've just started using RRT after dividing by x. This is just how I did it.

o wow..hmm im stupid, when I tested out 7, it didnt work.. and i was using a calculator


geometry function algebra domain perimeter inequalities

A wire 10 meters long is to be cut into two pieces.  One piece will be shaped as an equilateral triangle, and the other piece will be shaped like a circle.

(a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle.

(b) What is the domain of A?

[quote="The QuattoMaster 6000"]I am not sure if this is correct:
[hide]
a. Since the side length of the equilateral triangle is x, we have that the perimeter of the triangle is 3x.  Therefore, it takes 3x meters of wire to make the equilateral triangle.  The area of the equilateral triangle thus formed is $\frac{x^{2}\sqrt{3}}{4}$.  Now, there is 10-3x meters of wire left for the circle.  Say that r is the radius of the circle.  $2r\pi=10-3x$, so $r=\frac{10-3x}{2\pi}$.  This implies that the area of the circle formed is $r^{2}\pi=\frac{100+9x^{2}-60x}{4\pi}$.  Therefore, the total area of the triangle and circle is $\frac{x^{2}\pi\sqrt{3}+100+9x^{2}-60x}{4\pi}$.
b. Since 3x cannot exceed 10 and 3x cannot be less zero, we have the domain of x is $0\geq(x)\geq\frac{10}{3}$, so the domain of A is $\frac{100}{4\pi}\geq(A)\geq\frac{25\sqrt{3}}{9}$.  [/hide][/quote]

[hide="Pretty much, but..."]
You've got a mistake in part b. First of all, $3x$ and $10-3x$ cannot be $0$ either because you must make both a circle and a triangle. Also, you meant $\leq$ where you put $\geq$. So it should be:

$0 < x < \frac{10}{3}$

Also note that the domain of $A(x)$ is the possible values of $x$, not the possible values of $A$. So you computed (with your reversed inequalities) the range of $A$, not its domain. The interval I just described, $(0 , \frac{10}{3})$ is the domain of $A$.[/hide]

:blush: Oh, yes.  You are right, sorry about that.

Great math notes for me to review and study.

Thanks!


function algebra polynomial inequalities LaTeX quadratics AMC

Given the proportion $ \frac{x\plus{}y}{3}\equal{}\frac{y\plus{}z}{4}\equal{}\frac{x\plus{}z}{5}$, determine the value of $ \frac{xy\plus{}yz\plus{}xz}{x^2\plus{}y^2\plus{}z^2}$

Note that the thing you want to compute is homogeneous in $ x, y, z$, so you can just set $ x \plus{} y \equal{} 3, y \plus{} z \equal{} 4, x \plus{} z \equal{} 5$.

A function $ f(x, y, z)$ is homogeneous of degree $ k$ if $ f(tx, ty, tz) \equal{} t^k f(x, y, z)$.  The usual example is [url=http://en.wikipedia.org/wiki/Homogeneous_polynomial]homogeneous polynomials[/url]; here we have a quotient of two homogeneous polynomials of the same degree.

If you don't like buzzwords, let $ \frac {x \plus{} y}{3} \equal{} ... \equal{} d$ and set $ a \equal{} \frac {x}{d}, b \equal{} \frac {y}{d}, c \equal{} \frac {z}{d}$.  Then $ \frac {xy \plus{} yz \plus{} zx}{x^2 \plus{} y^2 \plus{} z^2} \equal{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$.

so $ (x\plus{}y)(y\plus{}z)\plus{}(y\plus{}z)(z\plus{}x)\plus{}(z\plus{}x)(x\plus{}y) \equal{} xy\plus{}zx\plus{}yz\plus{}y^2 \plus{}yz\plus{}yx\plus{}xz\plus{}z^2 \plus{} zx\plus{}zy\plus{}yx\plus{}x^2 \equal{} 3xy\plus{}3yz\plus{}3xz\plus{}y^2\plus{}x^2\plus{}z^2 \equal{} p$... perhaps not so useful.... yet.

$ (x\plus{}y)^2\plus{}(x\plus{}z)^2\plus{}(y\plus{}z)^2 \equal{} 2x^2\plus{}2y^2\plus{}2z^2 \plus{} 2xy\plus{}2yz\plus{}2xz \equal{} q$

Now $ q\minus{}2p \equal{} \minus{}4(xy\plus{}xz\plus{}yz) \implies \frac{2p\minus{}q}{4} \equal{} xy\plus{}yz\plus{}xz$ and $ 2p\minus{}3q \equal{} \minus{}4(x^2\plus{}y^2\plus{}z^2) \implies \frac{3q\minus{}2p}{4} \equal{} x^2\plus{}y^2\plus{}z^2$.

Now it follows for t0r's remarks that we can let $ x\plus{}y\equal{}3$, $ y\plus{}z\equal{}4$, and $ x\plus{}z\equal{}5$.  Then we get $ p\equal{}3*4\plus{}4*5\plus{}5*3 \equal{} 12\plus{}20\plus{}15 \equal{} 47$ and $ q\equal{}3^2\plus{}4^2\plus{}5^2 \equal{} 9\plus{}16\plus{}25 \equal{} 50$.  Then we have $ xy\plus{}yz\plus{}xz \equal{} (2p\minus{}q)/4 \equal{} (2*47\minus{}50)/4\equal{}11$ and $ x^2\plus{}y^2\plus{}z^2 \equal{} \frac{3*50\minus{}2*47}{4} \equal{} 14$, so our answer is $ \frac{11}{14}$.

comments:
1) pretty much I just found the two symmetric polynomials that I could knowing that I could set $ x\plus{}y$ and such to whatever I like.  The only polynomials with a degree of 2 were $ \sum_{cyc} (x\plus{}y)(y\plus{}z)$ and $ \sum_{cyc} (x\plus{}y)^2$, and then I saw they were both linear combinations of the two I wanted to find.
2) by rearrangment inequality, it is clear our answer will be less than 1.

Ooh. Hmm. You guys make it sound so easy. And that is beastly LaTeX people. Wow... Thanks!  :D

[quote="jonathanchou711"]Given the proportion $ \frac {x \plus{} y}{3} \equal{} \frac {y \plus{} z}{4} \equal{} \frac {x \plus{} z}{5}$, determine the value of $ \frac {xy \plus{} yz \plus{} xz}{x^2 \plus{} y^2 \plus{} z^2}$[/quote]

This is easy for Japanese high school students.

Let $ \frac {x \plus{} y}{3} \equal{} \frac {y \plus{} z}{4} \equal{} \frac {x \plus{} z}{5}\equal{}k$, we have $ x\plus{}y\equal{}3k,\ y\plus{}z\equal{}4k,\ z\plus{}x\equal{}5k$.
Adding these gives $ 2(x\plus{}y\plus{}z)\equal{}12k\Longleftrightarrow x\plus{}y\plus{}z\equal{}6k\Longrightarrow x\equal{}(x\plus{}y\plus{}z)\minus{}(y\plus{}z)\equal{}2k$, yielding $ y\equal{}k,\ z\equal{}3k$.
Substitue these for the given expression, we have $ \frac{xy\plus{}yz\plus{}zx}{x^2\plus{}y^2\plus{}z^2}\equal{}\frac{(2*1\plus{}1*3\plus{}3*2)k^2}{(2^2\plus{}1^2\plus{}3^2)k^2}\equal{}\boxed{\frac{11}{14}}$

[quote="t0rajir0u"][hide="Or..."] $ x \equal{} 2, y \equal{} 1, z \equal{} 3$. [/hide][/quote]

(EDIT: seems kunny beat me to it.)
Of course, I missed the simplest way.  note we could also let $ d$ be the common value, then solve the system in terms of $ d$ to get $ x\equal{}2d$, $ y\equal{}d$ and $ z\equal{}3d$.  The $ d$'s would cancel out in the end.

by the way... I did not know if my method would work out at the beginning, I just suspected.  Strongly.  The fundamental theorem of symmetric polynomials says there is a way for any symmetric polynomial to be expressed in terms of elementary symmetric polynomials (for example, for three variables $ x\plus{}y\plus{}z$, $ xy\plus{}yz\plus{}xz$, and $ xyz$ are the elementary symmetric polynomials).  So it seems it could be possible to write non-elementary polynomials in terms of other polynomials.

Practice 1

Find the value of $ p$ such that $ \frac{a}{b\plus{}c}\equal{}\frac{b}{c\plus{}a}\equal{}\frac{c}{a\plus{}b}\equal{}p$.

1985 Tama Art University entrance exam/Architecture

[quote="kunny"][quote="jonathanchou711"]Given the proportion $ \frac {x \plus{} y}{3} \equal{} \frac {y \plus{} z}{4} \equal{} \frac {x \plus{} z}{5}$, determine the value of $ \frac {xy \plus{} yz \plus{} xz}{x^2 \plus{} y^2 \plus{} z^2}$[/quote]

This is easy for Japanese high school students.

Let $ \frac {x \plus{} y}{3} \equal{} \frac {y \plus{} z}{4} \equal{} \frac {x \plus{} z}{5} \equal{} k$, we have $ x \plus{} y \equal{} 3k,\ y \plus{} z \equal{} 4k,\ z \plus{} x \equal{} 5k$.
Adding these gives $ 2(x \plus{} y \plus{} z) \equal{} 12k\Longleftrightarrow x \plus{} y \plus{} z \equal{} 6k\Longrightarrow x \equal{} (x \plus{} y \plus{} z) \minus{} (y \plus{} z) \equal{} 2k$, yielding $ y \equal{} k,\ z \equal{} 3k$.
Substitue these for the given expression, we have $ \frac {xy \plus{} yz \plus{} zx}{x^2 \plus{} y^2 \plus{} z^2} \equal{} \frac {(2*1 \plus{} 1*3 \plus{} 3*2)k^2}{(2^2 \plus{} 1^2 \plus{} 3^2)k^2} \equal{} \boxed{\frac {11}{14}}$[/quote]

Ok you know what? Don't be bragging about your "super awesome skillz"! I'm only in 7th grade! And I have to do this crap! K? You probably only finished pre-algebra or something in 7th grade!  :mad:  Don't give yourself compliments! The real skilled waits for the compliments and earns them!  :mad:

BTW: Thnx for the practice anyways...

My solution is not a super awsome skills, just put the common value of three fractional expressions as $ k$.

What's pre- algebra?

t0rajir0u and facis made the solution more difficult, don't they?

[quote="kunny"]t0rajir0u and facis made the solution more difficult, don't they?[/quote]
$ x \equal{} 1, y \equal{} 2, z \equal{} 3$ is difficult?  All I claimed is that the answer is independent of the choice of $ x, y, z$, and all that's left to do is justify that.

Students less than junior highschool students usually don't notice $ homogeneous$.
By the way, what's level of 7 th grade?

[quote="kunny"]Practice 1

Find the value of $ p$ such that $ \frac {a}{b + c} = \frac {b}{c + a} = \frac {c}{a + b} = p$.

1985 Tama Art University entrance exam/Architecture[/quote]

Cross multiplying, we clearly have
\begin{align*}
a(a+c) &= b(b+c) \\
a(a+b) &= c(b+c).
\end{align*}
Dividing the two, we have
\begin{align*}
\frac{a+c}{a+b} &= 1\\
a+c &= b+c \\
b&=c.
\end{align*}
So it follows that $ a^2 + ab - 2b^2 = 0$ so $ a=b$ or $ a=-2b$. Thus $ p$ is equal to $ \frac12$ or $ -1$.

Kunny: Normal 7th grade differs from school to school, but the stuff taught in public schools is usually quadratic equations, maybe a little bit of Pythagorean theorem. (That's why I spend so much time on AoPS :P .)

Uh... 7th grade? 7th grade should be Pre-Algebra. You don't know what Pre-Algebra is??? Uh... http://en.wikipedia.org/wiki/Pre-algebra  It's wiki, and it's not that reliable, but it's good enough. But I'm in Geometry. That's 2 levels up. But what I posted wasn't even Geometry. That was like Algebra 2! I'm in Kumon: http://www.kumon.org And that was hard for me. Have you never been a 7th grader before?

[quote="ThinkFlow"][quote="kunny"]Practice 1

Find the value of $ p$ such that $ \frac {a}{b + c} = \frac {b}{c + a} = \frac {c}{a + b} = p$.

1985 Tama Art University entrance exam/Architecture[/quote]

Cross multiplying, we clearly have
\begin{align*} a(a + c) & = b(b + c) \\
a(a + b) & = c(b + c). \end{align*}
Dividing the two, we have
\begin{align*} \frac {a + c}{a + b} & = 1 \\
a + c & = b + c \\
b & = c. \end{align*}
So it follows that $ a^2 + ab - 2b^2 = 0$ so $ a = b$ or $ a = - 2b$. Thus $ p$ is equal to $ \frac12$ or $ - 1$.

Kunny: Normal 7th grade differs from school to school, but the stuff taught in public schools is usually quadratic equations, maybe a little bit of Pythagorean theorem. (That's why I spend so much time on AoPS :P .)[/quote]

Thank you for the explanation.

Alternative Solution:

$ \frac {a}{b + c} = \frac {b}{c + a} = \frac {c}{a + b} = p\Longrightarrow a = p(b + c),\ b = p(c + a),\ c = p(a + b)$

Adding both sides gives $ a + b + c = 2p(a + b + c)$.

Case 1. $ a + b + c\neq 0$, we have $ 1 = 2p$, yielding $ p = \boxed{\frac {1}{2}}$.

Case 2. $ a + b + c\neq 0$, we have $ p = \frac {c}{a + b} = \frac {c}{ - c} = \boxed{ - 1}$.

Genellary it is holds that $ \boxed{\frac {a}{b} = \frac {c}{d} = \frac {e}{f} = \frac {pa + qc + re}{pb + qd + rf}}$, which is called "componend" by Mathematics Japanese English dictionary.

Use the fact, for $ a + b + c\neq 0$, we have  $ \frac {a}{b + c} = \frac {b}{c + a} = \frac {c}{a + b} = \frac {a + b + c}{(b + c) + (c + a) + (a + b)} = \frac {1}{2}$

[quote="jonathanchou711"]I'm in Geometry. That's 2 levels up. But what I posted wasn't even Geometry. That was like Algebra 2! I'm in Kumon: http://www.kumon.org And that was hard for me. Have you never been a 7th grader before?[/quote]
You're being [b]extremely[/b] rude.  It should be obvious that kunny is from Japan and if you aren't aware of how much more difficult the math curriculum is in most countries, including Japan, than it is here, you're in for a surprise.  And I don't know why you would expect either that middle-school math levels line up within the country or that they would line up [b]outside[/b] the country.

[quote="jonathanchou711"]You probably only finished pre-algebra or something in 7th grade!  :mad:[/quote]
And not only is this ad hominem likely to be totally incorrect, it reveals a lot more about your character than about kunny's.

[quote="jonathanchou711"]Uh... 7th grade? 7th grade should be Pre-Algebra. You don't know what Pre-Algebra is??? Uh... http://en.wikipedia.org/wiki/Pre-algebra  It's wiki, and it's not that reliable, but it's good enough. [/quote]

Where do you live? The system of education has different style by country.

[quote="jonathanchou711"]

I'm in Kumon: http://www.kumon.org And that was hard for me. Have you never been a 7th grader before?[/quote]

I know Kumon well, in my opinion, Kumon is effective for rapid calculation, but it is just repeating skills for solving, but I don't like the style.

I think that my ideal mathematics education is "Guess and Checking", say, U.S.A.

Practice 2

Let $ a,\ b,\ c$ be non-zero real numbers such that $ \frac{\minus{}a\plus{}b\plus{}c}{a}\equal{}\frac{a\minus{}b\plus{}c}{b}\equal{}\frac{a\plus{}b\minus{}c}{c}$.

Find the value of $ \frac{(b\plus{}c)(c\plus{}a)(a\plus{}b)}{abc}$.

1985 Tokyo Women University entrance exam/Mathematics

Using the componend, we find that the three fractions equal one. Thus, we find
\begin{align*}
b+c &= 2a \\
a+c &= 2b \\
a+b &= 2c
\end{align*}
if we set the fractions equal to one and cross-multiply. Solving, we find that $ a=b=c$, so the desired value is $ 2^3 = 8$. Thanks for the practice. :)

You have almost solved it, but not perfect.

[hide="Solution"]
Let $ k\equal{}\frac{\minus{}a\plus{}b\plus{}c}a\equal{}\frac{a\minus{}b\plus{}c}b\equal{}\frac{a\plus{}b\minus{}c}c$, so that $ b\plus{}c\equal{}a(k\plus{}1)$, $ a\plus{}c\equal{}b(k\plus{}1)$, and $ a\plus{}b\equal{}c(k\plus{}1)$, so adding yields $ 2(a\plus{}b\plus{}c)\equal{}(a\plus{}b\plus{}c)(k\plus{}1)\implies(a\plus{}b\plus{}c)(k\minus{}1)\equal{}0$. If $ a\plus{}b\plus{}c\equal{}0$, then substitute $ b\plus{}c\equal{}\minus{}a$, $ c\plus{}a\equal{}\minus{}b$, and $ a\plus{}b\equal{}\minus{}c$ to get $ \frac{\minus{}a\cdot\minus{}b\cdot \minus{}c}{abc}\equal{}\minus{}1$. If $ a\plus{}b\plus{}c\ne0$, then $ k\equal{}1$, so multiplying the second group of equations yields $ \frac{(b\plus{}c)(c\plus{}a)(a\plus{}b)}{abc}\equal{}\frac{abc(k\plus{}1)^3}{abc}\equal{}2^3\equal{}8$.
[/hide]

Different Solution:

I see, I forgot the case when $ a+b+c=0$. If $ a+b+c=0$, then $ a+b=-c$ so
\begin{align*}
\frac{(c+b)(c+a)(a+b)}{abc} &= \frac{-c(c^2 + (a+bc + ab))}{abc} \\
&= \frac{c^2-c^2 + ab}{-ab} \\
&= -1,
\end{align*}
and our values are $ 8$ and $ -1$.

[quote="math154"]
Let $ k = \frac { - a + b + c}a = \frac {a - b + c}b = \frac {a + b - c}c$, so that $ b + c = a(k + 1)$, $ a + c = b(k + 1)$, and $ a + b = c(k + 1)$, so adding yields $ 2(a + b + c) = (a + b + c)(k + 1)\implies(a + b + c)(k - 1) = 0$. If $ a + b + c = 0$, then substitute $ b + c = - a$, $ c + a = - b$, and $ a + b = - c$ to get $ \frac { - a\cdot - b\cdot - c}{abc} = - 1$. If $ a + b + c\ne0$, then $ k = 1$, so multiplying the second group of equations yields $ \frac {(b + c)(c + a)(a + b)}{abc} = \frac {abc(k + 1)^3}{abc} = 2^3 = 8$.[/quote]

My solution is same as yours, but you need to examine the case $ a + b + c = 0$.

[quote="ThinkFlow"]Different Solution:

I see, I forgot the case when $ a + b + c = 0$. If $ a + b + c = 0$, then $ a + b = - c$ so
\begin{align*} \frac {(c + b)(c + a)(a + b)}{abc} & = \frac { - c(c^2 + (a + bc + ab))}{abc} \\
& = \frac {c^2 - c^2 + ab}{ - ab} \\
& = - 1, \end{align*}
and our values are $ 8$ and $ - 1$.[/quote]

Now We have just finished Practice 2 :lol: .

[hide="Or..."] [b]Lemma:[/b]  $ \frac {a}{b} \equal{} \frac {c}{d} \Rightarrow \frac {a}{b} \equal{} \frac {a \plus{} c}{b \plus{} d}$.

Summing pairwise, the original three fractions are equal to $ \frac {2a}{b \plus{} c} \equal{} \frac {2b}{c \plus{} a} \equal{} \frac {2c}{a \plus{} b}$ and summing all three, the original three fractions are equal to $ 1$ (after the provision about the $ a \plus{} b \plus{} c \equal{} 0$ case, where we can write

$ \frac{(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \equal{} \frac{\minus{}abc}{abc} \equal{} \minus{}1$

and in the $ a \plus{} b \plus{} c \neq 0$ case none of $ a \plus{} b, b \plus{} c, c \plus{} a$ are equal to zero either.) [/hide]

[quote="kunny"]Practice 2

Let $ a,\ b,\ c$ be non-zero real numbers such that $ \frac { \minus{} a \plus{} b \plus{} c}{a} \equal{} \frac {a \minus{} b \plus{} c}{b} \equal{} \frac {a \plus{} b \minus{} c}{c}$.

Find the value of $ \frac {(b \plus{} c)(c \plus{} a)(a \plus{} b)}{abc}$.

1985 Tokyo Women University entrance exam/Mathematics[/quote]

Add two to each of those fractions to get $ \frac{a\plus{}b\plus{}c}{a}\equal{}\frac{a\plus{}b\plus{}c}{b}\equal{}\frac{a\plus{}b\plus{}c}{c}$.

If $ a\plus{}b\plus{}c\equal{}0$, $ b\plus{}c\equal{}\minus{}a$ a.s.o. and the fraction is $ \minus{}1$. If $ a\plus{}b\plus{}c\ne 0$, then $ a\equal{}b\equal{}c\ne 0$ and we get $ 8$.

@altheman: Dang almost exactly what I did, except I didn't immediately realize I should add 2 so I went through a little work.

@kunny:
Sorry to ruin the math but...

[quote="kunny"]I think that my ideal mathematics education is "Guess and Checking", say, U.S.A.[/quote]

Was this a joke or were you being serious?  I'm not totally sure what you mean (I want to know, whatever it turns out to be)[/quote]

[hide][quote="jonathanchou711"]Given the proportion $ \frac {x \plus{} y}{3} \equal{} \frac {y \plus{} z}{4} \equal{} \frac {x \plus{} z}{5}$, determine the value of $ \frac {xy \plus{} yz \plus{} xz}{x^2 \plus{} y^2 \plus{} z^2}$[/quote]
This seems quite straightforward if you note the homogenuity (hinted by t0rajir0u)

$ x \plus{} y \equal{} 3; y \plus{} z \equal{} 4; x \plus{} z \equal{} 5 \implies x \plus{} y \plus{} z \equal{} (3 \plus{} 4 \plus{} 5)/2 \equal{} 6$
$ \implies(x \plus{} y \plus{} z)^2 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx \equal{} 36$

$ x^2 \plus{} 2xy \plus{} y^2 \equal{} 9;y^2 \plus{} 2yz \plus{} z^2 \equal{} 16;x^2 \plus{} 2xz \plus{} z^2 \equal{} 25$

$ \implies 2(x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx) \equal{} 9 \plus{} 16 \plus{} 25 \equal{} 50$

So we have

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx \equal{} 36 ...eqn(1)$

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx \equal{} 25 ...eqn(2)$

$ eqn(1) \minus{} eqn(2)$ yields $ xy \plus{} yz \plus{} zx \equal{} 11$

$ 2*eqn(2) \minus{} eqn(1)$ yields $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 14$.

so the desired answer is $ 11/14$

... by the way, don't judge your math ability to regular school curriculums... they aren't great tests for your creativity and problem solving abilities... (not to brag, but guess what... I self-studied up to diff eqns before 9th grade but sometimes I cannot even solve an AIME or AMC problem  :blush: )[/hide]

[quote="kunny"]Practice 2

Let $ a,\ b,\ c$ be non-zero real numbers such that $ \frac { - a + b + c}{a} = \frac {a - b + c}{b} = \frac {a + b - c}{c}$.

Find the value of $ \frac {(b + c)(c + a)(a + b)}{abc}$.

1985 Tokyo Women University entrance exam/Mathematics[/quote]

Here is my solution:

$ \frac { - a + b + c}{a} = - 1 + \frac {b + c}{a}$, we have

$ \frac { - a + b + c}{a} = \frac {a - b + c}{b} = \frac {a + b - c}{c}$

$ \Longleftrightarrow \frac {b + c}{a} = \frac {c + a}{b} = \frac {a + b}{c} = k\Longrightarrow b + c = ak, \c + a = bk,\ a + b = ck$.

Adding these both sides gives $ 2(a + b + c) = k(a + b + c)\ \cdots [*]$

$ \therefore \frac { - a + b + c}{a} = \frac {a - b + c}{b} = \frac {a + b - c}{c} = \frac {ak\ \cdot bk\cdot ck}{abc} = k^3$.

Thus, from $ [*]$, the answer is for $ a + b + c\neq 0$, yielding $ = 2^3 = \boxed{8}$, for $ a + b + c = 0$, yielding  $ k = \frac { - c}{c} = - 1\Longrightarrow ( - 1)^3 = \boxed{ - 1}$.

Of course, I have the solution using the componend.

[quote="kunny"][quote="math154"]
Let $ k \equal{} \frac { \minus{} a \plus{} b \plus{} c}a \equal{} \frac {a \minus{} b \plus{} c}b \equal{} \frac {a \plus{} b \minus{} c}c$, so that $ b \plus{} c \equal{} a(k \plus{} 1)$, $ a \plus{} c \equal{} b(k \plus{} 1)$, and $ a \plus{} b \equal{} c(k \plus{} 1)$, so adding yields $ 2(a \plus{} b \plus{} c) \equal{} (a \plus{} b \plus{} c)(k \plus{} 1)\implies(a \plus{} b \plus{} c)(k \minus{} 1) \equal{} 0$. If $ a \plus{} b \plus{} c \equal{} 0$, then substitute $ b \plus{} c \equal{} \minus{} a$, $ c \plus{} a \equal{} \minus{} b$, and $ a \plus{} b \equal{} \minus{} c$ to get $ \frac { \minus{} a\cdot \minus{} b\cdot \minus{} c}{abc} \equal{} \minus{} 1$. If $ a \plus{} b \plus{} c\ne0$, then $ k \equal{} 1$, so multiplying the second group of equations yields $ \frac {(b \plus{} c)(c \plus{} a)(a \plus{} b)}{abc} \equal{} \frac {abc(k \plus{} 1)^3}{abc} \equal{} 2^3 \equal{} 8$.[/quote]

My solution is same as yours, but you need to examine the case $ a \plus{} b \plus{} c \equal{} 0$.[/quote]

I think I did examine that case...

[quote="ThinkFlow"][quote="kunny"]Practice 1

Find the value of $ p$ such that $ \frac {a}{b + c} = \frac {b}{c + a} = \frac {c}{a + b} = p$.

1985 Tama Art University entrance exam/Architecture[/quote]

Cross multiplying, we clearly have
\begin{align*} a(a + c) & = b(b + c) \\
a(a + b) & = c(b + c). \end{align*}
Dividing the two, we have
\begin{align*} \frac {a + c}{a + b} & = 1 \\
\end{align*}
[/quote]Shouldn't the RHS be $ b/c$ instead of $ 1$ (of course $ b = c$ but we do not know that yet at this point of the solution...)  :huh:

[quote="kunny"]Practice 3

Let $ a,\ b,\ c$ be non-zero real numbers. Find the values of $ k,\ l$ such that $ \frac {a^2 \plus{} b^2}{c^2} \equal{} \frac {b^2 \plus{} c^2}{a^2} \equal{} \frac {c^2 \plus{} a^2}{b^2} \equal{} k$,
$ \frac {a^3 \plus{} b^3}{c^3} \equal{} \frac {b^3 \plus{} c^3}{a^3} \equal{} \frac {c^3 \plus{} a^3}{b^3} \equal{} l$.

2002 Tokai University entrance exam/Humanities[/quote]
We have $ \frac {a^2 \plus{} b^2 \plus{} b^2 \plus{} c^2 \plus{} c^2 \plus{} a^2}{c^2 \plus{} a^2 \plus{} b^2} \equal{} 2(a^2 \plus{} b^2 \plus{} c^2)/(a^2 \plus{} b^2 \plus{} c^2)$

If $ a^2 \plus{} b^2 \plus{} c^2\neq 0$ the answer is $ 2$.

If $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$, then $ b^2 \plus{} c^2 \equal{} \minus{} a^2$ etc

So we are left with $ \minus{} c^2/c^2 \minus{} a^2/a^2 \minus{} b^2/b^2 \equal{} \minus{} 3$ (of course $ a,b,c\neq0$)

The second problem can be solved similarly.

EDIT: Yeah, forgot to mention that -3 won't work...


Does anybody know how did they do at Northwest chapter in CT?
It was supposed to be last Chapter competition in CT, right?

I'm from Litchfield County! But I wasn't in Mathcounts that year, sorry!


group theory
abstract algebra
vector
search
superior algebra
superior algebra unsolved

Since I am taking a course in group & rings this semester I am revising the beginner's course in algebra that I took 1.5 years ago since I've forgotten all of it. I have a problem from a past exam which is probably very simple to you. I found it moderately difficult, but there is a detail in my solution that I feel uneasy about, even though it's correct.

"Prove that every element $ z$ in a finite field $ F$ can be expressed $ z \equal{} a^2 \plus{} b^2$ with $ a,b \in F$."

If the characteristic of $ F$ is two then every element is a square so we are done.

If the characteristic is larger, then let the order of $ F$ be $ r$. $ r$ is odd. We consider how many squares there are in $ F$. Squaring $ 0$ gives $ 0$ and after that we choose a remaining element and it's (distinct) additive inverse; these give the same square. Then we keep choosing two elements at a time like that (each time we get a new square because $ x^2 \equal{} c$ has at most two solutions), and so we realize there are $ \frac{r\plus{}1}{2}$ squares in $ F$.

Now comes the part where I am "uneasy": What feels natural to me is to prove that this now implies that when going through $ a^2 \plus{} b^2$ for all $ a, b \in F$ we must get $ r$ distinct elements. However, the only argument I can find is this: Choose a $ z \in F$. Varying $ a$ in $ F$, $ a^2$ assumes $ \frac{r\plus{}1}{2}$ different values, and varying $ b$, $ z \minus{} b^2$ also assumes $ \frac{r\plus{}1}{2}$ different values. Since there cannot be $ r\plus{}1$ distinct elements in $ F$ we get a solution to $ z \equal{} a^2 \plus{} b^2$.

How can I do the last bit more directly, i.e. directly showing that $ a^2\plus{}b^2$ gives $ r$ distinct elements? Maybe a weird question that does not have too much to do with mathematics, but I feel the current solution is too tricky  :P

This is the only way I know how to prove the theorem.

I believe that the solution you proposed is essentially the most direct and simple way to prove this (at least that I am aware of). You can shorten your argument slightly by looking at cosets, I think. But I'm not sure.

There is one cool proof:

Let $ F^\plus{}\equal{}F$ and $ F^*\equal{}F\backslash \{0\}$ be the additive and multiplicative subgroups of $ F$, respectively.

We consider the set $ S$ of that nonzero elements that are a sum of two squares.
If $ x\equal{}a^2\plus{}b^2, y\equal{}c^2\plus{}d^2 \in S$, then so is $ xy\equal{}(a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\minus{}bd)^2\plus{}(ad\plus{}bc)^2$. Thus we have a subset of a finite group $ F^*$ that is closed under multiplication, so $ S$ is a subgroup of $ F^*$.
As $ S$ contains all squares, it has index $ \leq 2$ in $ F^*$.
If index $ \equal{}1$, then we are done.
Otherwise, index $ \equal{}2$.
Then every sum of squares is a square, thus $ S \cup \{0\}$ is closed under addition. But now we got a subgroup of $ F^\plus{}$ having $ \frac{|F|\minus{}1}{2}\plus{}1 > \frac{|F|}{2}$ elements, so this set must be the full $ F^\plus{}$. Again, we are done.

[quote="ZetaX"]
As $ S$ contains all squares, it has index $ \leq 2$ in $ F^*$.
[/quote]
Why is this? Sorry, I don't get it. Index is the number of cosets, right?

The squares $ T$ have index $ 2$ (or $ 1$, but then $ |F|$ is a power of two; but we never need to make this case work, every step works independent from this). We have $ T \subset S \subset F^*$, thus the index of $ S$ is at most that of $ T$, being $ 2$ (every coset $ xS$ of $ S$ contains a coset $ xT$ of $ T$).
To see that $ T$ has at most index $ 2$:
The squaring map $ f(x)\equal{}x^2$ has kernel $ \mathrm{ker}(f) \equal{} \{ \pm 1\}$. As $ |T| \equal{} |\mathrm{im}(f)| \equal{} \frac{|F^*|}{\mathrm{ker}(f)} \geq \frac{|F^*|}2$, we get what we wanted to show.

uhm.. why should this $ S\cup\{0\}$ be a subgroup of $ F^\plus{}$?

anyway, i think there's a linear way to get to the conclusion: $ S\cup\{0\}$ is a $ \mathbb{F}_p$-vector subspace of $ F$, since every element in $ \mathbb{F}_p$ is a sum of two squares (pigeonhole).
but it can't be contained in any proper subspace, because of its cardinality, so we are done.

(and this would help in case of two-squares-like formulas, if there's any for higher powers)

[quote="ma_go"]uhm.. why should this $ S\cup\{0\}$ be a subgroup of $ F^ \plus{}$?[/quote]
As the sentence before that part says: the sum of two squares is a square.

And my goal was to give a proof based purely on the groups.

sorry, you're right :) i didn't quite follow... and, by the way, nice solution ;)

do you know any higher-degree identity along the line of the two- and four-squares ones? i'm too lazy to search for it on wikipedia ;)


algebra
difference of squares
special factorizations

Ok, Ive been given this algebraic fraction to simplify

x^2 - 9
______
x^2+x-6

I think I need to do this but Im not sure what else to do.

x^2 - 9
______
(x-2)(x+3)

If im right what can I do next, if not where have I gone wrong, please don't only post the answer. Thankyou

[hide=""Difference of squares--so CLICK ME if you still want to know""]$a^2-b^2=(a-b)(a+b)$[/hide]

Yes -- you've factored the bottom, now you need to factor the top.

Just wondering, isn't this should be on Getting Started Forum? ;)

Should it be in getting started?  Probably.  But as a rule of thumb I generally don't move topics down the ladder.

A *very good* rule of thumb.  Occasionally easier than expected level is much better than what happens when posts move down regularly - lots of stuff that's too hard.

well don't bother about where it should be now.  Once you factor the top using a difference of squares problem, cross out the common factors in the top and bottom.  note 9=3^2.


[b][u]The author of this posting is: Megus.[/u][/b]
_____________________________________________________________

Given is function $f(x) = \frac{1}{x+1}$ where $x > 0$.  For all natural $n$ we know that:

$g_n(x) = x+f(x)+f(f(x))+\ldots+f(f((\cdots f(x)))$ ($n$ times iterated)

Prove:
(a) function $g_n$ is strictly increasing
(b) $g_n(1)=\frac{F_1}{F_2}+\frac{F_2}{F_3}+\cdots +\frac{F_{n+1}}{F_{n+2}}$ where $F_0 = F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ for $n \geq 1$.
_________________
Przemyslaw Chojecki

[b][u]The author of this posting is: Myth.[/u][/b]
_____________________________________________________________

For (a).
It is easy to show that for positive increasing function $h(x)$:

1) ${h(x)+f(h(x)}$ is increasing
2) $f(f(h(x)))$ is increasing

_________________
Amicus Plato, sed magis amica est veritas


Four basketball players, whose heights in inches are 60, 62, 64 and 70, are standing on the court. Their teammate Natalie, who is 74 inches tall, walks out onto the court, and the mean height increases by $ m$ inches while the median height increases by $ n$ inches. What is the positive difference between $ m$ and $ n$ ?

To elaborate:

Original heights: ${60}, {62}, {64}, {70}$ <---Median is $\frac{62+64}{2}$, or $63$

Mean is $\frac{60+62+64+70}{4}$, or $64$.

After the new height is added: Median=middle number--> now $64$

Mean: $\frac{60+62+64+70+74}{5}={66}$

Median increases by 1.
Mean increases by 2.

${2}-{1}={1}$


[quote="Alisher"]a,b,c>o  prove that
\[ \frac{1}{{5(a^{2}+b^{2})-ab}}+\frac{1}{{5(a^{2}+c^{2})-ac}}+\frac{1}{{5(b^{2}+c^{2})-bc}}\ge \frac{1}{{a^{2}+b^{2}+c^{2}}}\]
[/quote]
Easy or hard?
I think, it's transparent. SOS kills it easy enough.   :wink:

$ \frac{1}{{5(a^{2}+b^{2})-ab}}+\frac{1}{{5(a^{2}+c^{2})-ac}}+\frac{1}{{5(b^{2}+c^{2})-bc}}\geq\frac{(1+1+1)^{2}}{{10(a^{2}+b^{2}+c^{2})-(ab+bc+ca)}}$

I used here CBS:$ \frac{x^{2}}{a}+\frac{y^{2}}{b}+\frac{z^{2}}{c}\geq\frac{(x+y+z)^{2}}{a+b+c}$
for positive numbers.

Unfortunately we cannot use now:$ a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ in the initial inequality.
It's no good using this method  :blush:

[quote="Alisher"]Let $ a,b,c>0,$  prove that

$ \frac {1}{5b^2\minus{} bc\plus{}5c^2} \plus{} \frac {1}{5c^2\minus{}ca\plus{}5a^2} \plus{} \frac {1}{5a^2\minus{}ab\plus{}5b^2}\ge \frac {1}{a^2 \plus{} b^2\plus{} c^2}.$[/quote]$ \sum{\frac {1}{5b^2\minus{} bc\plus{}5c^2}}\minus{}\frac {1}{a^2 \plus{} b^2\plus{} c^2}$

$ \equal{}\sum{\frac{(b\minus{}c)^2\left[(b\plus{}c\minus{}a)^2\left(2a^2\plus{}40b^2\plus{}23bc\plus{}40c^2\right)\plus{}31a^2bc\plus{}10\left(b^4\plus{}c^4\right)\plus{}57bc\left(b^2\plus{}c^2\right)\right]}{4\left(a^2\plus{}b^2\plus{}c^2\right)\left(5b^2\minus{}bc\plus{}5c^2\right)\left(5c^2\minus{}ca\plus{}5a^2\right)\left(5a^2\minus{}ab\plus{}5b^2\right)}}$

$ \geq0.$

By the way, this inequality holds for [b]real[/b] $ a,b,c$ with $ 9bc\plus{}a^2>0,9ca\plus{}b^2>0,9ab\plus{}c^2>0.$

Ji Chen, did you use computer to find this hard formula?

Thank you very much. :)

Yes, manlio. Another formula relate to Vasc's inequality

$ \frac {7}{4b^2 \minus{} bc \plus{} 4c^2} \plus{} \frac {7}{4c^2 \minus{} ca \plus{} 4a^2} \plus{} \frac {7}{4a^2 \minus{} ab \plus{} 4b^2}\ge \frac {9}{a^2 \plus{} b^2 \plus{} c^2}$

see : http://www.mathlinks.ro/Forum/viewtopic.php?t=185227


A woman stands on a scale in a moving elevator.  Her mass is 60.0kg, and the combined mass of the elevator and the scale is an additional 815kg.  Starting from rest, the elevator accelerates upward.  During the acceleration, there is tension of 9410N in the hoisting cable.  What does the scale read during the acceleration?

[color=blue]Denote woman's mass $m$ and the mass of the elevator and the scale $M$.
$T_0$ is the tension of the hoisting cable and $T$ is the scale read during the acceleration.
$a$ is the acceleration of the system
We have these equations
$\large T_0-(m+M)g=(m+M)a$
$T-mg=ma$
From these two equations, you can get $T=mg+\frac{m}{m+M}[T_0-(m+M)g]$[/color]


To fulfil the demand of light speed :D , i ll post some problems that is too hard for me.

1. 10 sqs. each with sides of 1in, arranged in a plain to form a figure so that no sq has all four edges on the perimeter of the figure, no 2 sq. overlap, and where two sq share a point they share edge....(thats the original problem i m serious)

2.A model train is made by connecting cars with possible lengths of 2in and 5 in, what is the greatest unachieveable length?(easy!)

1.[hide]Don't know lol[/hide]

2.[hide]using the formula ab-a-b, you get
2*5-2-5
10-7
=3[/hide]
3. Thanks for fulfilling my demands lol  :D

i dun even understant wats N.O1 asking for

Answer is

[hide]18[/hide]

here is a tricky one

The purity of gold is measured in karats. 24 karat gold is puregold and k karat gold is k/24 by mass. one piece of 16-karat gold jewelery has a mass of 30 grams . Another piece of 10-karat gold has a mass of 40 gram. Find the numbe of grams in the positive difference between the amount of pure gold in each piece of jewelry. express answer as a mixed number

answer:

[hide]3(1/3)[/hide]

[hide="question 1"] This is a problem? They don't even ask a question! Where did you get this from?[/hide]

[hide="question above"]
16/24*30=2/3*30=20
10/24*40=5/12*40=50/3
20-50/3=60/3-50/3=10/3=3 1/3
>>3 1/3<<[/hide]
Checks his answer...YES!
BTW, kidwithshirt, the way you express the answer to some say multiply. better remove the parenthesis and add a space instead.


geometry
incenter
trapezoid
geometric transformation
reflection
symmetry
circumcircle

Dear Mathlinkers,
let ABCD a square, M a point on segment CD, I, X, Y the incenters wrt MAB, MAD, MBC,
P, Q the meetpoints of the parallel to AD passing through X with BI, AC,
S, R the meetpoints of the parallel to AD passing through Y with AI, BD.
Prove that the center of the circle passing through X, Y, Q, R lies on PS.
Sincerely
Jean-Louis

a is the square side, E intersection of its diagonals. x,y are inradii of triangles MAD, MBC. MD = m, MC = n.

$ x \equal{} \frac {a \plus{} m \minus{} \sqrt {a^2 \plus{} m^2}}{2} \Longrightarrow m \equal{} \frac {2x(a \minus{} x)}{a \minus{} 2x},\ n \equal{} \frac {2y(a \minus{} y)}{a \minus{} 2y}$

$ a \equal{} m \plus{} n \equal{} \frac {2x(a \minus{} x)}{a \minus{} 2x} \plus{} \frac {2y(a \minus{} y)}{a \minus{} 2y}.$

$ \triangle BPX \sim \triangle BYC \Longrightarrow PX \equal{} \frac {YC \cdot BX}{BC} \equal{} \frac {2y(a \minus{} x)}{a}$
$ \triangle ASY \sim \triangle AXD \Longrightarrow SY \equal{} \frac {XD \cdot AY}{AD} \equal{} \frac {2x(a \minus{} y)}{a}$

K, L are midpoints of XQ, YR. G is midpoint of KL. PS || QR cuts KL at O. WLOG, m < n => x < y.

$ KO \equal{} \frac {2 \cdot KL \cdot KP}{XP \minus{} YS} \equal{} \frac {(a \minus{} x \minus{} y)}{y \minus{} x} \cdot \left(\frac {2y(a \minus{} x)}{a} \minus{} \frac {a \minus{} 2x}{2}\right) \equal{}$

$ \equal{} \frac {(a \minus{} x \minus{} y)}{y \minus{} x} \cdot \left(\frac {a(x \plus{} y) \minus{} 2xy}{a} \minus{} \frac {a \minus{} 2y}{2}\right) \equal{}$

$ \equal{} \frac {1}{y \minus{} x} \cdot \left(\frac {x(a \minus{} x)(a \minus{} 2y) \plus{} y(a \minus{} y)(a \minus{} 2x)}{a} \minus{} \frac {(a \minus{} x \minus{} y)(a \minus{} 2y)}{2} \right) \equal{}$

$ \equal{} \frac {1}{y \minus{} x} \cdot \left(\frac {(a \minus{} 2x)(a \minus{} 2y)}{2} \minus{} \frac {(a \minus{} x \minus{} y)(a \minus{} 2y)}{2} \right) \equal{}$

$ \equal{} \frac {1}{y \minus{} x} \cdot \frac {(y \minus{} x)(a \minus{} 2y)}{2} \equal{} \frac {a \minus{} 2y}{2} \equal{} LY \equal{} LR \equal{} LE$

E is also diagonal intersection of isosceles trapezoid XQRY with perpendicular diagonals => E is its anticenter. Reflection O of E in its centroid G is its circumcenter.
________________________________

EDIT:

Perpendicular to $ CD$ through $ M$ cuts symmetry axis $ KL$ of trapezoid $ XYRQ$ at $ N.$ $ QR$ cuts $ AB$ at $ T.$ $ \triangle MQR$ is isosceles with $ MQ \equal{} MR$ $ \Longleftrightarrow$ perpendicular bisector of $ QR$ goes through $ M$ $ \Longleftrightarrow$ $ \angle QTA \equal{} \angle OMN$ $ \Longleftrightarrow$

$ \frac {y \minus{} x}{a \minus{} x \minus{} y} \equal{} \frac {KO \minus{} KN}{MN} \equal{} \frac {(a \minus{} 2y) \minus{} 2(m \minus{} x)}{a} \equal{} \frac {(a \minus{} 2y)(a \minus{} 2x) \minus{} 2ax}{a(a \minus{} 2x)}$

$ \Longleftrightarrow$ (multiplying this through and rearranging)

$ a(a \minus{} 2x)(y \minus{} x) \plus{} 2ax(a \minus{} x \minus{} y) \plus{} (x \plus{} y)(a \minus{} 2x)(a \minus{} 2y) \equal{} a(a \minus{} 2x)(a \minus{} 2y)\ \Longleftrightarrow$

$ 2[a(x \plus{} y) \minus{} 2xy] (a \minus{} x \minus{} y) \equal{} a(a \minus{} 2x)(a \minus{} 2y)\ \Longleftrightarrow$

$ 2[x(a \minus{} x)(a \minus{} 2y) \plus{} y(a \minus{} y)(a \minus{} 2x)] \equal{} a(a \minus{} 2x)(a \minus{} 2y),$

which is an obvious identity, in view of $ a \equal{} m \plus{} n \equal{} ...$ (see above).

Dear Yetti and Mathlinkers,
it seems that we can say more about this figure :
for example,
that the center of the circle passing through X, Y, Q, R is the orthopole of the triangle PXY wrt AB.
Sincerely
Jean-Louis

[quote="jayme"]... the center of the circle passing through X, Y, Q, R is the orthopole of the triangle PXY wrt AB.[/quote]
You mean orthopole of the $ \triangle MXY,$ I trust. This is easy to prove and on top of it, it gives synthetic proof of the $ \triangle MQR$ being isosceles.

Let the incircles $ (X), (Y)$ touch $ CD$ at $ U, V,$ let $ M', U', V' \in AB$ be feet of perpendiculars from $ M, U, V$ to $ AB.$ Otherwise, keeping the same notation as above: $ E$ is diagonal intersection of the square $ ABCD$ and of the isosceles trapezoid $ XYRQ.$ $ K, L$ are midpoints of $ XQ, YR.$ The right $ \triangle MUX, \triangle UKE$ with corresponding sides $ UM \perp KU, UX \perp KE$ are similar by SAS, because $ \frac {UM}{UX} \equal{} \frac {a}{a \minus{} 2x} \equal{} \frac {KU}{KE}$ $ \Longrightarrow$ $ MX \perp UE$ $ \Longrightarrow$ reflection of $ UE$ in the trapezoid centroid $ G$ (common midpoint of $ KL, OE$) goes through $ V'$ and through the trapezoid circumcenter $ O$ and $ (V'O \parallel UE) \perp MX.$ Similarly, $ (U'O \parallel VE) \perp MY.$ Then $ O$ is orthopole of $ AB$ WRT $ \triangle MXY$ $ \Longrightarrow$ perpendicular to $ XY$ from $ M'$ goes through $ O$ $ \Longrightarrow$ $ M'O$ is perpendicular bisector of $ XY$ $ \Longrightarrow$ $ \triangle M'XY$ is isosceles with $ M'X \equal{} M'Y.$ The same holds for $ \triangle MQR,$ who is reflection of $ \triangle M'XY$ in the trapezoid symmetry axis $ KL.$

_________________________________________

EDIT: Using the two problems about inradii (see [url]http://www.mathlinks.ro/viewtopic.php?t=282316[/url]), the above metric proof of the trapezoid $ XQRY$ circumcenter on $ PS$ can be re-written as follows:

As before, $ a$ is the square side, $ E$ intersection of its diagonals. $ x, y, r$ are inradii of triangles $ \triangle MAD, \triangle MBC, \triangle MAB.$ $ K, L$ are midpoints of $ XQ, YR.$ $ G$ is midpoint of $ KL.$ $ PS \parallel QR$ cuts $ KL$ at $ O.$

$ \left(1 \minus{} \frac {2x}{a}\right)\left(1 \minus{} \frac {2y}{a}\right) \equal{} 1 \minus{} \frac {2r}{a}\ \Longrightarrow\ r \equal{} x \plus{} y \minus{} \frac {2xy}{a},$

$ \frac {1}{r} \equal{} \frac {1}{a \minus{} 2x} \plus{} \frac {1}{a \minus{} 2y} \equal{} \frac {2\ KL}{XQ \cdot YR}.$

[quote="yetti"]$ \triangle BPX \sim \triangle BYC \Longrightarrow PX \equal{} \frac {YC \cdot BX}{BC} \equal{} \frac {2y(a \minus{} x)}{a}$
$ \triangle ASY \sim \triangle AXD \Longrightarrow SY \equal{} \frac {XD \cdot AY}{AD} \equal{} \frac {2x(a \minus{} y)}{a}$.[/quote]

$ \Longrightarrow PQ \equal{} SR \equal{} a \minus{} \left(2x \plus{} 2y \minus{} \frac {2xy}{a}\right) \equal{} a \minus{} (x \plus{} y \plus{} r)$

$ KO \equal{} \frac {2\ KL \cdot KP}{XP \minus{} YS} \equal{} \frac {KL \cdot (XQ \minus{} 2\ PQ)}{XQ \minus{} YR} \equal{} \frac {KL\ (2r \minus{} YR)}{XQ \minus{} YR} \equal{}$

$ \equal{} \frac {YR\ (XQ \minus{} KL)}{XQ \minus{} YR} \equal{} \frac {_1}{^2} YR \equal{} LR \equal{} LY \equal{} LE$

[quote="yetti"]$ E$ is also diagonal intersection of isosceles trapezoid $ XQRY$ with perpendicular diagonals $ \Longrightarrow$ $ E$ is its anticenter. Reflection $ O$ of $ E$ in its centroid $ G$ is its circumcenter.[/quote]

Dear Mathlinkers,
for a proof see:
http://perso.orange.fr/jl.ayme  vol.5,  le résultat de Larrosa Canestro, p.
Sincerely
Jean-Louis


[i]Moderator's note:  This was split from the discussion about who should win the American presidential race because it was silly and irrelevant. --JBL
[/i]

CJH, why must you always stir up politics.
You even have taiwan politics in your sig

That started with TokenAdult posting pro-Chen Sui Bian comments in some Taiwan article. (and not telling the dark side of it like what Lee Teng Hui did). In this Democratic World we like to express our own opinions. That is why we have nice discussions in these forums and carefully written facts of our views.

Plus my skateboard guy looks cool. I'm going to have it for a while.

[quote="tokenadult"][quote="churchilljrhigh"]That started with TokenAdult posting pro-Chen Sui Bian comments in some Taiwan article. (and not telling the dark side of it like what Lee Teng Hui did).[/quote]

Well, that's not an accurate statement, as anyone who looks at the board can verify. Moreover, there isn't anyone I have mentioned whose name is Chen Sui Bian.

Taking care to read carefully and avoid careless mistakes is helpful in math contests and helpful in discussions of public policy.[/quote]

Well I can verify that you had many pro pan-green statements and made a statement about neck to neck which was the Taiwan Election 2000 involving Pan Green candidate Chen Sui Bian,

[quote="tokenadult"]I'm not a citizen of that country, so I don't get a vote over there, but I would like to see the incumbent (the "pan green" candidate) get reelected. From what I hear about polling data (which tends to be skewed somewhat by who takes a particular poll), the race is neck-and-neck, with a statistically insignificant lead enjoyed by the challenger, the KMT-PFP coalition ("pan blue") candidate.[/quote]

As you can see Tokenadult publicly stated his views on which side he was on and gave alot of posts supporitng his cause (just visit the thread). Looking at the statement your statement is not very accurate either, thus undermining the credibility of your argument.

Finally the whole Taiwan thread was started by you talking about the Great Wall of People in Taiwan. This was organized by Chen Sui Bian and the Pan-Green Coalition. Although you haven't mentioned his name you have mentioned his party countless times and have firmly established your views. I notice that you are an very educated person on the case of Taiwan and I am also too. It was good to have these political discussions.

That signature is so stupid churchilljrhigh.

but Chen was stupid enough so anything could happen (remember the time when he revealed a letter about China trying to compromise the political issue?)

joel: why didn't you just delete it as a whole.  I don't foresee anything fruitful coming out of this discussion.

In general, I'd rather not delete anything with actual content.  If no one says anything, it will eventually die a lonely death at the bottom of the forum.

[quote="theone853"]CJH, why must you always stir up politics.
You even have taiwan politics in your sig[/quote]

[i]exactly.[/i] ur location too.

i agree with cats delete this i bet im not the only person that's offended

I can't believe everybody is making a big fuss about being Taiwanese or Chinese. I was at the National Geography Bee last year and I met a  guy who was from Taiwan, but he accepted that he was chinese when he was talking to other people and never made refutes that he was Taiwanese. I mean all these "Radical Taiwanis" are forgetting their culture and ancestors because of some stupid political division that was started by two crazy dictators who killed many. Do we math students have to care about this that was started in the past and elsewhere and won't effect us because most of us on this forum are in U.S?  [b](Fact 98% of Taiwanese are Han Chinese)[/b] Boy these radical pan green party need to get less liberal and have some conservatism!

More than 330,000 votes were declared invalid in the presidential election

There is a Vote-Rigging and a Assasination Attempt Questioning which gives basis to my signature.

BTW The U.S. is the greatest country in the world so what they say (like if they wanted to invade Iraq they have the right to) is true so we have to listen them.

Propositions from two presidents

President Clinton pledged "three nos" - no to Taiwan independence, no to two Chinas, and no to Taiwan joining international organisations that need statehood for membership.

The administration of George W Bush has remained careful not to encourage Taiwan's President Chen Shui-bian, warning him against any actions which risked altering the status quo.

Just wondering: Why is the ratio of discussion between Taiwanese politics and U.S. presidental election like 5:1? :lol:

the chen sui bien guy is kinda illogical.
china has already pledged to keep taiwan under its control and will use military force if neccesary.
thats why the usa doesnt support taiwan rifting from china.
has the election ended yet?

Should Singapore be under the control of the P.R.C. because of the ethnicity of its citizens? Should Tibet NOT be under the control of the P.R.C. because of the ethnicity of its citizens? How about Eastern Turkestan?

[quote="fubu"][quote="Scrambled"]
has the election ended yet?[/quote]

I think the electron ended, Chen won. But they think there are about erm. 300,000 or something number like that invalid votes and Chen only won by like 30,000.  Some numbers like that.  So they're doing a recount.[/quote]

Yeah, that's what some Houston Chinese newspapers are saying. The difference was too little.

I think we should do what the Taiwanese wish to do. Perhaps have some sort of vote (sponsored by the UN or some independent third party) asking the Taiwanese themselves what their opinion is, namely, to truly grant themselves independent nation status or to be a part of China. Should they choose to be their own nation succession would occur and I guess it'd get a little more complicated. Personally I recognize Taiwan as independent from China, I think the olympiads do so as well, though I am unsure of this.


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If tanx*tany= (a-b)/(a+b).
show that
(a-bcos2x)*(a-bcos2y) is independent of x and y.

[quote="Flame"][hide]
$ \tan x \cdot \tan y = \frac {a - b}{a + b}$

$ \frac {\sin x \cdot \sin y}{\cos x \cdot \cos y} = \frac {\frac {1}{2}(\cos (x - y) - \cos (x + y))}{\frac {1}{2}(\cos (x - y) + \cos (x + y))}$

We can assume:

$ a = \cos (x - y)$
$ b = \cos (x + y)$

$ (a - b\cos 2x)(a - b \cos 2y) = a^2 - ab(\cos 2x + \cos 2y) + b^2(\cos 2x \cdot \cos 2y)$

$ \cos 2x + \cos 2y = 2 \cos \frac {2x + 2y}{2} \cos \frac {2x - 2y}{2} = 2ab$

$ \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align*} \cos 2x \cdot \cos 2y & = \frac {1}{2}(\cos 2(x - y) + \cos 2(x + y)) \\
& = \frac {1}{2}(\cos^2 (x - y) - \sin^2 (x - y) + \cos^2(x + y) - \sin^2 (x + y)) \\
& = \frac {1}{2}(2 \cos^2 (x - y) - 1 + 2 \cos^2(x + y) - 1) \\
& = \frac {1}{2}(2a^2 - 1 + 2b^2 - 1) \\
& = a^2 + b^2 - 1 \end{align*} \end{document}$

So the beginning expression is independent of $ x$ and $ y$ but this is somewhat awkward to say. It would be more appropriate to say that it depends only on $ a$ and $ b$.

But there is something that should be added to this solution. We actually should have assumed that

$ \cos(x - y) = ka$
$ \cos(x + y) = kb$

where $ k$ can be any real number. So:

$ \cos 2x + \cos 2y = 2k^2ab$

$ \cos 2x \cdot \cos 2y = k^2(a^2 + b^2) - 1$

So when we put it in the expression we get that it equals to

$ (a^2 - b^2)(1 - k^2b^2)$

where $ k$ is "stabilizing" factor such that $ ka$ and $ kb$ are in interval $ [ - 1,1]$. That way we can input any real $ a$ and $ b$ to get valid solution. But then again, it doesn't depend on $ a$ and $ b$ only. Could someone help me with this?[/hide][/quote]

Hmm try this for root of (a-b)/(a+b).
Maybe then we can work it out.

Where did you get this problem? Probably my approach is wrong.

[quote="Flame"]But then again, it doesn't depend on $ a$ and $ b$ only. Could someone help me with this?[/quote]

That's fine; it's still independent of $ x,y$. (as the original expression is homogeneous in $ a,b$, the ending expression can be scaled to reach many different values)


function
quadratics
calculus
derivative
algebra
domain
analytic geometry

Find all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that:
$f(x-f(y))=f(x)+f(f(y))-axf(y)-bf(y)-c$ $\forall x,y\in \mathbb{R}$
(a,b,c are given real numbers)

may i ask what level of math this is from?

it is the generalisation of a number 6 problem in IMO, it is thus difficult. The solution uses elementary mathematics (maybe one can call this technical ,though). Looking at your signature I think you should try some easier ones :roll:
Ps: next time use PM-s please

[hide]
[b]As Albanian Eagle correctly pointed out, it was not mentioned that the function is continuous and differentiable. Redone with finite differences.[/b]

---------------

[i]Not too sure about this, but... here's a swing... :maybe: [/i]

Let $y$ be any real constant, and $f(y)$ therefore be any constant in the range of $f(x)$.

$f(x+1-f(y))=f(x+1)+f(f(y))-axf(y)-af(y)-bf(y)-c$ and $f(x-f(y))=f(x)+f(f(y))-axf(y)-af(y)-bf(y)-c$ → $(f(x+1-f(y))-f(x-f(y)))=(f(x+1)-f(x))-af(y)$

Let us use $f_{1}(x)=f(x+1)-f(x)$ represent the first row of finite differences:
$f_{1}(x-f(y))=f_{1}(x)-af(y)$→$f_{1}(x)-f_{1}(x-f(y))=af(y)$

Over a distance of $f(y)$ in the $x$ direction, $f_{1}(x)$ changes by $af(y)$. Second level of finite differences $f_{2}(x)$ is proportional to change in $x$ by proportion constant $a$ → $f_{2}(x)=a$.

$f_{1}(x)=ax+k+p(x)$ where $k$ is some unknown constant and $p(x)$ is some periodic function with period that makes some number of complete cycles over an $x$ distance of $f(y)$.
$f(y)$ can be any number in the range of $f$.
Let $m$ be a rational value in the range of $f$
Let $n$ be a trascendental value on that range
$p(x)$ must make some integer number of complete cycles across the $x$ interval $[x,x+m]$ and across the $x$ interval $[x,x+n]$.
This is impossible unless $p(x)$ is constant.
Therefore, $p(x)$ is a constant, and $f_{1}(x)=ax+k+p(x)=ax+b$ for all x.
It follows that $f(x)$ is a quadratic with an $x^{2}$ coefficient of $\frac{a}{2}$

$f(x)=\frac{a}{2}x^{2}+Bx+C$
Use $x=0$ and $z=f(y)$ → $f(-z)=f(0)+f(z)-bz-c$

$f(z)-f(-z)=bz+c-f(0)$ → $2Bz=bz+c-C$ → $(2B-b)z=c-C$
$z$ is a variable and $c-C$ is not, so equality occurs only when both sides are 0.
$2B-b=0$ → $B=\frac{b}{2}$
$c-C=0$ → $C=c$.

The final equation is now:
$f(x)=\frac{a}{2}x^{2}+\frac{b}{2}x+c$
[/hide]

no one mentioned conditions of differentiability or even continuity... so your solution is not acceptable but you did find the answer :P

:oops_sign: Good point.

You can arrive at the same conclusion using finite differences instead of derivatives, and finite differences don't require those conditions, right?

Hmm....
[hide]$f(x+1-f(y))=f(x+1)+f(f(y))-axf(y)-af(y)-bf(y)-c$
$f(x-f(y))=f(x)+f(f(y))-axf(y)-af(y)-bf(y)-c$

Subtracting,
$(f(x+1-f(y))-f(x-f(y)))=(f(x+1)-f(x))-af(y)$

Let us use $f_{1}(x)=f(x+1)-f(x)$ represent the first row of finite differences:
$f_{1}(x-f(y))=f_{1}(x)-af(y)$

$f_{1}(x)-f_{1}(x-f(y))=af(y)$

So, the over a distance of $f(y)$ in the $x$ direction, $f_{1}(x)$ changes $af(y)$. Therefore, the second level of finite differences $f_{2}(x)$ is proportional to change in $x$ by a proportion constant of $a$, and so $f_{2}(x)=a$.

It follows that $f(x)$ is a quadratic with an $x^{2}$ coefficient of $\frac{a}{2}$[/hide]
Does that work better?

I'm trying to avoid using $x==f(x)$ to avoid limiting the domain to the range.

here is some things that are true :    :lol:
(let f(y0)=0 for some y0. then f(x) = f(x) + f(0) -c
so f(0) = c if we know that f=0 for some y.)

Now let x= f(y) then f(0) = 2f(f(y)) -af(y)^2 -bf(y) - c

so f(f(y)) = a/2 f(y)^2 + b/2 f(y) + [c+f(0)]/2

so f(z) = a/2 z^2 + b/2 z + k for all z in range of f.
Now let x -> f(y)/2 - b/2a
then we see that f(-b/2a + z) = f(-b/2a -z ) for all z in range of f.

Also we can easily see that f(-z) = a/2 z^2 - b/2 z + f(0) for all z in range of f.

I am not sure that u actually have the right idea here, because using your finite difference ideas you won't be able to prove that f(x) is a quadratic for all x. rather u can only show it for all points in range of f(x).

Finite differences = slope of secant lines.

What it does is show that $f_{1}(x)=ax+k+p(x)$ where $k$ is some unknown constant and $p(x)$ is some periodic function with period that makes some number of complete cycles over an $x$ distance of $f(y)$ . However, $f(y)$ can be any number in the range of $f$. Let $m$ be a rational value in the range of $f$ and $n$ be a trascendental value on that range. Thus, $p(x)$ must make some integer number of complete cycles across the $x$ interval $[x,x+m]$ and across the $x$ interval $[x,x+n]$. This is impossible unless $p(x)$ is constant. Therefore, $p(x)$ is a constant, and $f_{1}(x)=ax+k+p(x)=ax+b$ for all x.

Sorry but you can't conclude that p(x)=0 so try some other argument :(

I did this in about 5 minutes over a coffee, knowing it to be rubbish but I was astounded when this completely daft approach coincided with the vastly superior solution given about.

let x = f(y).
Then f(x) = [f(0) + ax^2 + bx + c]/2 for all x in the range of f.
Assume continuity from range to the rest of the domain.
x = 0 lies in the domain hence c = f(0).
Hence f(x) = [ax^2 + bx] + c.
Direct substition verifies the solution.

This makes me suspect the prescence of some principle to some extent justifying these gung-ho shortcuts.   Does anyone know of any general results governing the connections between the behaviour of a function evaluated at all points in its range and the extension of the function to the superset consisting of the whole domain.  What restrictions need to be placed on:  If for all y = f(x), f(y) = g(y), then f(x) = g(x) for all x in the domain?  It fails for constant polynomials but otherwise it is true for all polynomials defined on the whole of R. (A polynomial is completely determined by any segment of itself consisting of more than one point, and the range of a polynomial always contains more than one point unless the polynomial is constant).

Is there any criterion for a functional equation to have a polynomial solution?

If I understood correctly the post above :
Yeah it's quite easier in this case to guess the answer to the problem. But as you see there is still no solution.

As for your questions, well they are a bit general :P


$ 1)$.Find all primes $ p$ and $ q$ such that $ p \plus{} q \equal{} (p \minus{} q)^3$ (easy)

$ 2)$.Let $ a,b$ be dinstict positive integers such that $ ab(a \plus{} b)$ is divisible by $ a^2 \plus{} ab \plus{} b^2$.Prove that $ |a \minus{} b| > \sqrt [3]{ab}$ (normal)

[hide]Ωραια προβληματακια..

για το δευτερο, παρε  $ gcd(a,b) \equal{} d$
Αρα $ a \equal{} d*a_1, b \equal{} d*a_2$ με $ gcd(a_1,a_2) \equal{} 1$
 και μετα εκμεταλλευσου οτι $ a_1a_2 < a_1^2 \plus{} a_1a_2 \plus{} a_2^2 | d$ για να σχηματισεις την ζητουμενη ανισοτητα.

για το πρωτο, ειναι $ p \equal{} \frac {(p \minus{} q)( (p \minus{} q)^2 \plus{} 1)}{2}$ και $ q \equal{} \frac {(p \minus{} q)( (p \minus{} q)^2 \minus{} 1)}{2}$
και επειδη p,q πρωτοι $ p \minus{} q \equal{} 2$[/hide]

Για το πρώτο μία διαφορετική λύση:
Έστω $ p\minus{}q\equal{}x$
Τότε
$ p\plus{}q\equal{}x^3$
$ p\minus{}q\equal{}x$
Με πρόσθεση κατά μέλη
$ x|2p$,but $ x \not| p$,so $ x|2 \Leftrightarrow x\in \{1,2\}$
Aλλά $ p\minus{}q\equal{}1$ ικανοποιείται μόνο για $ p\equal{}3$ και $ q\equal{}2$ which is impossible


ARML AMC AIME MATHCOUNTS Harvard college MIT

Is ARML comparable to AIME? It has some same concepts, the AIME is harder but ARML lots less time.

So it seems that doing AIME won't help much with ARML because the formats are drastically different. What about MC National Target?

I don't see how AIME won't help with ARML, because practicing AIMEs helps with your problem solving, and problem solving is used in ARML (obviously). It might not be super effective because of drastically different formats, but having good problem solving skills will help on any math competition (obviously again). I agree that the ARML is not comparable to AIME though.

MC National Target is nowhere near as hard as ARML. I'd say that only 2-3 problems each year in the individual part of ARML are easy enough to be in mathcounts, and even the hardest mathcounts problems would probably only be about average difficulty in ARML (although I don't remember much of the mathcounts tests I practiced, so I could be wrong).

Seriously?  A mere six threads down in this forum is a 24-post thread titled "ARML and AIME," subtitled "whats easier," and you decided to create a thread about how the difficulty of AIME and ARML compare?

JBL makes a good point: look around a little before posting duplicate topics.

However, let me throw in my two cents. The very fact that people frequently compare AIME and ARML difficulty levels suggests that there are plenty of people who lean one way or the other. I've had this very discussion with fellow AIME and ARML problem-writers, and the lack of a clear consensus leads me to believe that the problem-by-problem difficulty levels are very comparable.

Maybe the easy problems in ARML are easier than the easy problems in AIME; maybe #15 on the AIME is harder than I-10 at ARML, but those are outlier problems. By and large, both exams cover the same topics, so to any given student it'll come down to taste. If you thrive in one exam setting you'll find that exam easier than the other, simple as that.

I you handed the competitors all 10 individual problems at once and gave them 50 minutes to do the lot, allowing for highly variable per-problem time allocations, then the number of people getting 2009 #10 would have been more than 8.

Sorry about the repeated thread:blush:
Got to be more careful later.

What I meant was I need to get used to ARML format sort of, since the AIME has a drastically different format, I think I should just do ARML problems.

How does this compare to MC national target?

IMHO, ARML and AIME are very comparable in difficulty, but they have very different styles of problems.  ARML problems tend to have a solution that's a little more straightforward (assuming you can find it), whereas AIME problems often require a bit more slogging in the solution.

For problems very similar in style to ARML, try Harvard-MIT or Stanford/Rice Math Tournament problems.  I can't speak for HMMT, but when I was writing for SMT, we tried to write very ARML-like problems.

[quote="AwesomeToad"]How does this compare to MC national target?[/quote]

Although the format of targets is more similar to that of ARMLs, the questions are much too easy to be of any good practice.  You even have a calculator on targets, so there is a lot more room for calculator-bashing in problems.
I tend to think that ARML problems aren't so much about speed as they are about insight.  If you get used to the kinds of ideas that ARML problems use often, you will most likely be able to answer a set of 2 questions in well under the 10 min time allotment. (Of course, I-10 last year was just mean  :P )

[quote="cyberspace"](Of course, I-10 last year was just mean  :P )[/quote]
About 1900 students (I can't be more exact than that since not all teams were full teams) participated in ARML. Here are how many got each individual question:
\[ \begin{tabular}{rr}1. &1672\\
2. &1256\\
3. &1011\\
4. & 470\\
5.  &749\\
6.  &263\\
7.  &541\\
8.  &284\\
9. &1172\\
10.  & 8\end{tabular}\]
But, as I said above, more time would have made a considerable difference with that problem.


In the grid below, place one digit in each of 24 non-shaded squares such that 14 numbers occurs below either vertically or horizontally in adjacent squares. What is the number A,BCD,EFG where each letter denotes the digit in the so lettered square?

[img]7507[/img]

The answer key says that it is 2228657 but I got 8222657 :maybe:

66236 has to go into the 5-digit rectangle and use logic for everything else.


floor function
advanced fields
advanced fields unsolved

Let $ X$ and $ Y$ be two disjoint finite set, $ \mathcal{S} \equal{} \{A\subseteq X\cup Y| A\cap X\neq\emptyset\}$.

Let $ \mathcal{M}$ be a subset of $ \mathcal{S}$ satisfying:

1) there exists $ A\in\mathcal{M}$ such that $ A\subseteq X$
2) if $ A, B\in\mathcal{M}$, neither $ A\subseteq B$ nor $ B\subseteq A$.

Then what's the maximal cardinality of $ \mathcal{M}$ given $ |X| \equal{} n$, $ |Y| \equal{} m$.

I know if $ |X| \equal{} n$, then any subset $ \mathcal{M}$ of $ 2^X$ satisfying 2) must satisfying $ |\mathcal{M}|\leq C(n,[n/2])$. But how to solve this new problem or estimate a tighter upper bound than just $ C(n\plus{}m, [(n\plus{}m)/2]$.

Any hints are welcomed. Thanks!

Well, if $ n>m$, the bound stated by you is correct, since any set in the Sperner antichain, having cardinality $ \lfloor (n\plus{}m)/2 \rfloor$ has nonempty intersection with $ X$, and there exists one included in $ X$.

If $ n \leq m\minus{}2$, the Sperner antichain will both contain sets with empty intersection with $ X$, and none included in $ X$. The issue of calculating the exact length of a maximal antichain might be arduous. There may be some considerations in an excellent book of Anderson, Combinatorics of sets, dedicated to antichains in sets environment.

[quote="mavropnevma"]Well, if $ n > m$, the bound stated by you is correct, since any set in the Sperner antichain, having cardinality $ \lfloor (n \plus{} m)/2 \rfloor$ has nonempty intersection with $ X$, and there exists one included in $ X$.

If $ n \leq m \minus{} 2$, the Sperner antichain will both contain sets with empty intersection with $ X$, and none included in $ X$. The issue of calculating the exact length of a maximal antichain might be arduous. There may be some considerations in an excellent book of Anderson, Combinatorics of sets, dedicated to antichains in sets environment.[/quote]

Thanks. Thanks for your recoomendation of that book. I think I have to spend a lot of time to work on it.

Can we prove $ C(n\plus{}m, [(n\plus{}m)/2])\minus{}C(m, [(n\plus{}m)/2])\plus{}1$ is a upper bound?
(let $ C(m, [(n\plus{}m)/2])\equal{}0$ if $ m<[(n\plus{}m)/2]$). I guess the first term is the maximal lenth of antichain in $ X\cup Y$, the second term is just to remove the set with $ [(n\plus{}m)/2]$ element but included in $ Y$?

I can't prove it.....


calculus
integration
limit
calculus computations

For positive integers $n,$ let

$S_{n}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots\cdots+\frac{1}{\sqrt{n}},\ T_{n}=\frac{1}{\sqrt{1+\frac{1}{2}}}+\frac{1}{\sqrt{2+\frac{1}{2}}}+\cdots\cdots+\frac{1}{\sqrt{n+\frac{1}{2}}}.$

Find $\lim_{n\to\infty}\frac{T_{n}}{S_{n}}.$

Use Cesaro Stolts lemma.
We have that
$\lim\frac{T_{n}}{S_{n}}=\lim\frac{T_{n+1}-T_{n}}{S_{n+1}-S_{n}}=\lim\frac{\sqrt{n+1}}{\sqrt{n+3/2}}=1$.

Your answer is correct, but I would like to see a basic solution.

I do not know what you mean by a  more basic solution, however another approach of this problem, without making use of Cesaro-Stolz lemma, would be this:

Let $x_{n}=1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}-2\sqrt{n}$,

and

$y_{n}=\frac{1}{\sqrt{1+\frac{1}{2}}}+\cdots+\frac{1}{\sqrt{n+\frac{1}{2}}}-2\sqrt{n+\frac{1}{2}}$.
Then it is not hard to prove that these two sequences are convergent, say $x_{n}\rightarrow a$ and $y_{n}\rightarrow b$.
Then we see that $S_{n}=x_{n}+2\sqrt{n}$ and $T_{n}=y_{n}+2\sqrt{n+\frac{1}{2}}$.

Thus $\frac{T_{n}}{S_{n}}=\frac{y_{n}+2\sqrt{n+\frac{1}{2}}}{x_{n}+2\sqrt{n}}=\frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n}}\frac{\frac{y_{n}}{2\sqrt{n+\frac{1}{2}}}+1}{\frac{x_{n}}{2\sqrt{n}}+1}\rightarrow 1$

[b]Is it a "basic solution" ?[/b]

[b]Lemma.[/b] $|\alpha |\le 1$ $\implies$ $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f(\frac{k+\alpha }{n})=\int_{0}^{1}f(x)\ dx\ .$

$\{\begin{array}{c}f(x)=\frac{1}{\sqrt x}\\\\ \{t,\alpha ,\beta \}\subset [-1,1]\\\\ S_{n}(t)=\sum_{k=1}^{n}f(k+t)\\\\ \end{array}$ $\implies$ $\{\begin{array}{c}\int_{0}^{1}f(x)\ dx=2\\\\I_{n}\equiv\frac{1}{n}\cdot\sum^{n}_{k=1}f(\frac{k+\alpha}{n})\rightarrow 2\\\\ J_{n}\equiv\frac{1}{n}\cdot\sum^{n}_{k=1}f(\frac{k+\beta}{n})\rightarrow 2\end{array}$ $\implies$ $\frac{S_{n}(\alpha )}{S_{n}(\beta )}=\frac{I_{n}}{J_{n}}\rightarrow 1\ .$


I am still trying to grasp what this means. :(
I mean, what kind of a perverted joke is this huh? :mad:
I doubt I'll be able to sleep tonight. The mental peace seems to be in no mood to return till I get a reply to the PM I just sent to him. :(

Who's perfect628?

And why is he dead?  :maybe:

http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=21169

wait is this a sick joke or did an AoPSer really die? how d'you know?

How would we know if he was dead? He had no way to tell us he was dead.  :|

Last post by perfect628:

Yesterday, at 1:37 pm

If true, how did you find out so fast?

[quote="mustafa"]The topic for mafia game 23 says Mafia Game 23-Day 5 Perfect628 died

 :roll:[/quote]

yeah i think zeb misenterpreted that    :D

I [i]hope[/i] Zeb misinterpreted it.
And if that is the case, perfect628 won't be happy to find this topic out.
But why hasn't he posted yet after his last post about six hours before Zeb's post?

perfect628 was in the USAMTS math jam last night. Not dead, just was killed in mafia.

...

That's reassuring.
What sort of an irresponsibel idiot are you zeb huh? :mad:
Don't you see how serious this can get?

Yeah, making jokes about this is no laughing matter. I'll give zeb another chance to explain himself before this topic is locked...

[quote="zeb"]Is there gonna be a funeral or something?

I have to feel sorry for him, even though I don't know him very well - the poor guy died before he could even finish a game of Mafia! (I can only hope his replacement wins it for him.)[/quote]

I am not dead. I died in mafia #23. And what are you talking about, I was not in the USAMTS math jam.

[quote="perfect628"]I am not dead. I died in mafia #23. And what are you talking about, I was not in the USAMTS math jam.[/quote]

proof enough  :D

zeb, explain yourself please

[quote="perfectnumber628"][quote="besttate"]perfect628 was in the USAMTS math jam last night. [/quote]
nope, that was me.  I am not perfect628.[/quote]
 :rotfl:

Wow, this is getting confusing... :rotfl:

[quote="xxazurewrathxx"]Roflmao. I thought someone actually made a thread to talk about a person's death. That would be sick.[/quote]

There was a thread about Anna Nicole Smith's death :roll:

I meant about how we have perfect628 and perfectnumber628.  I had a feeling that zeb misinterpreted the topic title from the Mafia game, saying that "perfect628 died".  He's just out of that game; he's not really dead.

Or this whole thread was just a joke, and people didn't want to point it out just in case perfect628 really had died.

But now that we know he didn't, I can say this:

If you die in Mafia, you die in real life!  *Sinister grin*

imo, it's idiotic that this topic has gotten 2 pages of posts...

Okay cheers everyone! This topic has got a happy ending, and that's all that matters. But zeb should be more careful in future. :D

And if I was dead, I would hope that the topic wouldn't go in GFF...


$ \frac {9x^2}{({1 \minus{} \sqrt {1 \plus{} 3x}})^2}$ is less than $ 3x \plus{} 5$

find the range of real values of  $ x$

[quote="123456789"][hide]$ \minus{} \frac {1}{3}\leq x < \frac {5}{12}, x\neq0$[/hide][/quote]

can you show the steps involved?


Find all primes $ p$ such that $ \frac{p\plus{}1}{2}$ and $ \frac{p^2\plus{}1}{2}$ are both squares.

[hide]Let $ \frac {p \plus{} 1}{2} \equal{} m^2, \frac {p^2 \plus{} 1}{2} \equal{} n^2.$ The second equation implies that $ p^2 \equal{} 2n^2 \minus{} 1 > n^2$, so $ p > n$, and similarly $ p > m$. Subtract the two equations to obtain $ p(p \minus{} 1) \equal{} 2(n^2 \minus{} m^2) \equal{} 2(n \minus{} m)(n \plus{} m).$ $ p$ is an odd prime, so either $ p|(n \minus{} m)$ or $ p|(n \plus{} m).$ If $ p|(n \minus{} m)$, since $ p < n$ we get that $ m \equal{} n$, but then $ p \equal{} 0$ or $ 1$, a contradiction, so $ p|(n \plus{} m).$ In addition, since $ n, m < p$ then $ n \plus{} m < 2p$ and thus $ n \plus{} m \equal{} p$. Substituting this into the equation yields $ m \equal{} \frac {p \plus{} 1}{4}$, thus $ \frac {p \plus{} 1}{2} \equal{} (\frac {p \plus{} 1}{4})^2$ with solution $ p \equal{} 7.$[/hide]


Let's work in $\mathbb{R}^{2006}$. Find the biggest positive integer $n$ for which there exist linear subspaces $V_1,V_2,\,\cdots,V_n$ of $\mathbb{R}^{2006}$ such that $\dim V_k = k$ for all $k$ and such that $V_{i + j} = V_i + V_j$ for all $i, j$ satisfying $i + j \leq n$.

Sorry, I was confused, this is rubbish :D


[quote="NkMAsTeR"]Prove that
$\sqrt[n]{\frac{a}{b+c}} +\sqrt[n]{\frac{b}{c+a}} +\sqrt[n]{\frac{c}{a+b}} > 2 \geq \frac{n \sqrt[n]{n-1}}{n-1}$
where n is an integer and not less than 2 and a,b,c >0[/quote]
I think i prove that when a,b,c is sides of a triangle the ineq holds:
$\sqrt[n]{\frac{a}{b+c}} +\sqrt[n]{\frac{b}{c+a}} +\sqrt[n]{\frac{c}{a+b}}\geq \sqrt[2]{\frac{a^2}{a(b+c)}} +\sqrt[2]{\frac{b^2}{b(c+a)}} +\sqrt[2]{\frac{c^2}{c(a+b)}}\geq \sum \sqrt \frac{a^2}{(\frac{a+b+c}{2})^2}=2$

[quote="Hawk Tiger"][quote="NkMAsTeR"]Prove that
$\sqrt[n]{\frac{a}{b+c}} +\sqrt[n]{\frac{b}{c+a}} +\sqrt[n]{\frac{c}{a+b}} > 2 \geq \frac{n \sqrt[n]{n-1}}{n-1}$
where n is an integer and not less than 2 and a,b,c >0[/quote]
I think i prove that when a,b,c is sides of a triangle the ineq holds:
[b]$\sqrt[n]{\frac{a}{b+c}} +\sqrt[n]{\frac{b}{c+a}} +\sqrt[n]{\frac{c}{a+b}}=\sqrt[2]{\frac{a^2}{a(b+c)}} +\sqrt[2]{\frac{b^2}{b(c+a)}} +\sqrt[2]{\frac{c^2}{c(a+b)}}[/b: 6bf2465f5f]\geq \sum \sqrt \frac{a^2}{(\frac{a+b+c}{2})^2}=2$[/quote]
Can you explain this point ? But I think it true where a,b,c be positive reals.

sorry the first $=$should be $\geq$
and i have only found a ugly solution when $a\geq b+c$
but i don't want to post it

[quote="Frey"][quote="NkMAsTeR"]Prove that
$\sqrt[n]{\frac{a}{b+c}} +\sqrt[n]{\frac{b}{c+a}} +\sqrt[n]{\frac{c}{a+b}} > 2 \geq \frac{n \sqrt[n]{n-1}}{n-1}$
where n is an integer and not less than 2 and a,b,c >0[/quote]

*This is inequality true:
                $\sqrt[n]{\frac{a}{b+c}} +\sqrt[n]{\frac{b}{c+a}} +\sqrt[n]{\frac{c}{a+b}} \geq \frac{n \sqrt[n]{n-1}}{n-1}>2$[/quote]
I'm sorry but I think my Inequalitity is true! You can try with some values of n
Ex:
$A= \frac{n \sqrt[n]{n-1}}{n-1} If n=2 then A=2 If n=3 then A \approx 1.89 ..$

We have $\frac{a}{b+c} \geq \frac{4a^2}{(a+b+c)^2}$ in fact we have:
$a(a+b+c)^2 \geq 4a^2(b+c)$ and so $a(a-b-c)^2 \geq 0$ that it's true. So we have:

$\sum \sqrt[n] {\frac{a}{b+c}} \geq \sum \sqrt{\frac{a}{b+c}} > \sum \frac{2a}{a+b+c} =2$

For the second part we have:

$\left ( 1+ \frac{n-2}n \right) ^n \geq 1+ \frac{n-2}n * n = n-1$ for bernoulli

Taking $n$-th root we have:

$\frac{2n-2}n \geq \sqrt[n]{n-1}$ and thus : $2 \geq \frac{n \sqrt [n] {n-1} } {n-1}$

[quote="Simo_the_Wolf"]
$\sum \sqrt[n] {\frac{a}{b+c}} \geq \sum \sqrt{\frac{a}{b+c}}$[/quote]

Why is this true ? :?:

[quote="silouan"]The above holds if $b+c>a$ and I think for $a>b+c$  it is trivial ;)[/quote]

if $a\geq b\geq c$ and $b+c\geq a$ then the inequality is true .

But if $a>b+c$ , when $a\rightarrow\infty$ and fix $b+c=1$ then it becomes

$\sqrt[n]{a}>\sqrt{a}$

which is not true  :?:


probability
expected value
combinatorics unsolved
combinatorics

Suppose a computer can generate a whole number from 1 to k in one instance randomly.

What is the expected number of instances, n required for each given k such that a complete set of 1 to k has been produced? Ie, the first time where you get all of the numbers from 1 to k.

How do I calculate Prob(set first completed in exactly nth instance) for all n?

Thanks all!

the expected number is $\sum_{i=1}^k \frac{k}{i}$

only have the proof in german. dont know the specific termology in english.
could somebody else prove it plz?

just keep computing the conditional expectations of the waiting time for arrival of the next new symbol given that we have already gotten $i$ of the symbols from $\{1,2,\ldots,k\}$ and finally add them all up.


Could someone help me by telling me where I can find the 2006 AIME II solutions?
Thanks.  :D

Click the number next to a particular problem. That will bring you to a thread. Then scroll down to find the solution to the problem (someone should have posted it).


we may post here some information about who is going at the imc as it is done elsewhere in the forum it may be also important for some to know how to get to macedonia (i've heard for example that you cant go there from bulgaria by train)
generally feel free to submit any information

I am going to drive a Romanian Team to the IMC this year. Probably Horia is going to join me with his car, because we are about 8 people wanting to come (amfulger, claude, kappa, phoria, ana caraiani, tanase raluca, a friend of mine and me).

Are you going to come qw?

btw, the train information is acurate. Also you can't travel by air from Romania to Skopje direclty. You have to take the plane to Milano (or something) and then back to Skopje ... which is kind of weird ...

well if all goes ok i'll come our team will probably consist of 4 people but we are struggling to find a means on transport.it's up to the university's authorities but i have completely no idea what will they do. it may finish with hiking :)

our univ. decide not to pay for our being there, so it's totaly a private thingy :) (except for Ana Caraiani, for which Princeton is paying) we did not find any other (safe) way to get than, besided driving a car, so we'll just do that. :)

oh, and i suppose everybody knows that we have high competition fee this year as far as i remember it was 225 euro for the "poorer countries"

Hi!
I'm going to IMC too. :)

[quote="Valentin Vornicu"]the fee we (the romanians) are requied to meet is smaller than that ... so I guess we are the poor-est country :)[/quote]
I don't think so because we also got a discount  :)

cool, so we can make this topic like the one for IMO "who's coming to" :)

[quote="qw"]maja do you know how to reach macedonia from your location are there any buses or trains? i would be grateful for any information[/quote]

I think there are regular lines (both bus and train lines) from Serbia to Macedonia. In next few days, Im going to look for more information about this and then Ill tell you what I found out.

...

OK, here is what I found out.
From Novi Sad to Skopje, one can go only by bus which departs at 6:10 a.m. and 10:20 a.m. every day. The trip takes about 12 hours (at least I was told so). The ticket costs about 15 EURO and it should be booked in advanced (3-5 days).
If you would go from Belgrade, then there is an additional option to go by train (which I dont prefer because trains are very dirty here). Anyway, it departs at 7:00, 13:50, 17:45, 23:20 every day and the ticket costs about 25 EURO, which is kind of strange because train transportation is usually cheaper here.
Also I believe that the daily number of buses which goes from Belgrade to Skopje is probably greater that two, but I havent checked that.

Maja


trigonometry
quadratics
algebra
geometry proposed
geometry

Fin the angles of $ \triangle ABC$ if $ a^2\equal{}R^2\plus{}b^2\plus{}c^2$(usual notations)

I don't understand your solution.:(...the idea to compose this pb was from this relation:$ \cos AsinBsinC\equal{}\minus{}\frac{1}{8}$, which is equivalwnt with the ip.
But $ sinBsinC\equal{}\frac{1}{2}\left(\cos(B\minus{}C) \plus{}cosA \right)$ and we will obtain a quadratic equation in $ cosA$, with $ \Delta\equal{}16(\cos^2(B\minus{}C)\minus{}1)\leq 0$, so $ B\equal{}C$.The only solution is $ A\equal{}120,B\equal{}C\equal{}30$


function
real analysis
real analysis unsolved

all Lipschitz functions are absolutely continuous? Do you look at the length of the interval? And choose $ \delta$ based on that?

State the definition of absolute continuity. Write the Lipschitz condition right below it. Stare at them for a minute. Then tell what exactly your difficulty is.

Suppose $ f: X \to Y$ is a function. Let $ k$ be a positive integer. If $ d_{Y}(f(a), f(b)) \leq k d_{X}(a,b)$ for all $ a,b \in X$ then $ f$ is Lipschitz.

A function $ I \to X$ is absolutely continuous on $ I$ if for every positive number $ \varepsilon$, there is a positive number $ \delta$ such that whenever a sequence of pairwise disjoint sub-intervals $ [x_{k}, y_{k}]$ of $ I$ satisfies $ \sum_{k} |y_{k}\minus{}x_{k}| < \delta$  then $ \sum_{k} d(f(y_{k}), f(x_{k})) < \varepsilon$.

Since $ I \subseteq \mathbb{R}$ it has a finite diameter. Call this $ p$. Let $ \varepsilon > 0$. Choose $ \delta \equal{} \frac{\varepsilon}{kp}$. Then:

$ \sum_{k} |y_{k}\minus{}x_{k}| < \delta \implies  \sum_{k} d(f(y_{k}), f(x_{k})) \leq kd_{X}(a,b) < \frac{\varepsilon}{p}$. Hence $ f$ is absolutely continuous.

Is this correct?

you are using the same k :)
$ \sum_{k} |y_{k}\minus{}x_{k}| < \delta \implies \sum_{k} d(f(y_{k}), f(x_{k})) \leq  \sum_{k} Kd(y_{k}, x_{k}) \equal{} K \sum_{k} d(y_{k}, x_{k})$

then choose $ \delta\equal{}\epsilon/K$

I don't know why you have to assume I to be an interval?

[quote="1234567a"]you are using the same k :)
$ \sum_{k} |y_{k} \minus{} x_{k}| < \delta \implies \sum_{k} d(f(y_{k}), f(x_{k})) \leq \sum_{k} Kd(y_{k}, x_{k}) \equal{} K \sum_{k} d(y_{k}, x_{k})$

then choose $ \delta \equal{} \epsilon/K$

I don't know why you have to assume I to be an interval?[/quote]

I think you are proving uniform continuity. I am trying to prove absolute continuity.

No, 1234567a is proving exactly what is needed: absolute continuity (though, of course, uniform continuity is proved in the same way and even with the same $ \delta$ in the end).


I've just installed asymptote on my new Windows Vista (premium 64bit) machine having used it in the past on my old one. I've got ghostscript and GSView installed and checked that they are working by opening a .ps file. Next I installed asymptote thought I'd check it was OK by trying to produce a simple output in interactive mode:

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

but I just get the told:

Cannot write to out.eps

Any ideas?

Vista puts restrictions on the Program Files directory which some but not all programs can deal with. In particular it won't let programs save files there and expect them to save them elsewhere. As a workaround (there are probably better ways) re-install Asymptote to another directory outside Program Files.

Thanks a lot. I just saved it directly onto my C: drive. Also, I had to change the default path in asymptote to look for gsview64.exe rather than gsview32.exe.

Now to configure with TeXnicCenter (fingers crossed).

I've just tried to configure aysmptote to work with TeXnicCenter but I'm having a slight difficulty:

I set up asymptote by going to Tools-> Customise->Tools then creating "Asyptote" before entering the following info into the three fields:

"C:\Asymptote\asy.exe"

-batchView -tex "pdflatex" -align C -f "pdf" %tc

%dc

which is the same setup as on my other machine (except, of course, the path).

When I try to build a document involving the asymptote environment it gets stuck straight away in the package-loading stage and says:

[code]! LaTeX Error: File 'asymptote.sty' not found.[/code]

Incidentally, when I execute the same command on my XP machine, the corresponding line in the dialogue box is:

 [code](C:\Program Files\MiKTeX 2.5\tex\latex\Asymptote\ayymptote.sty")[/code]

whence it proceeds to build the document with no problems whatsoever.

Now I suspect this is related to not having asymptote installed in Program Files but I'm out of ideas. I've tried copying the folder "Asymptote" which contains asymptote.sty and two other .sty files from the XP machine into various folders within TeXnicCenter without joy.

You've forgotten to do the bit that tells LaTeX all about asymptote - you have to put some asymptote files where LaTeX can find them. You'll find the details at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]Using Asymptote in LaTeX[/url].


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

nice one:
assume that $G$ is p-subgroup of $S_n$ .and $n<p^2$.prove that $G \cong \mathbb{Z}_p\oplus \dots \oplus \mathbb{Z}_p$.

sam-n,this problem is very beautiful.Do you have a solution fo r this one? I tried something but it doesn't work.
 I'm just courious.

Cezar, don't be surprised you cannot kill the problem from the first attempt. Do not forget that it is posted by sam-n. I have rarely seen easy problems posted by him on the forum. ;)  Please, sam-n, don't post solution even if you have.

Suppose that $ |G| = p^{k} $. Then if we take some $ x \in G $,
then number of elements in each cycle in decomposition of $ x $
divides $ p^{k} $, therefore it is equal to $ p^{j} $ for some $ j
\geqslant 1 $. Now, because of $ n < p^{2} $, we have $ j=0,1 $.
So cycles of $ G $ consist only of $ 1 $ or $ p $ elements. Hence
$ x^{p} = 1 $ for every $ x \in G $. So if we will prove that $ G
$ is abelian, we will solve the problem, because every finite
abelian group, such that $ x^{p} = 1 $ for every $ x $ in it must
be $ \mathbb{Z}_{p} \oplus \ldots \oplus \mathbb{Z}_{p} $.
 Let us prove by induction on $ n $, that $ G $ is abelian.
If we have $ 1 \leqslant n < p $, then $ |G| = p^{k} | n! $, so $
k=0 $ and $ G $ is a trivial group. If $ n = p $ and $ G $ is
nontrivial, every nontrivial element of $ G $ consists of a single
cycle of length $ p $. If we take some nontrivial element $ a \in
G $, then for every $ x \in G $ if we have that $ x(1) = i $, then
for some integer $ m $, we have $ a^{m}(1) = i $, so $ a^{-m}x (1)
= 1 $, therefore $ a^{-m}x $ is a trivial element, so $ x = a^{m}
$. So we see that in this case $ G $ is isomorphic to $
\mathbb{Z}_{p} $. Now suppose that $ n > p $. It is well known
that every $ p $ - group has nontrivial center. Take some element
$ a $ , which lies in the center of $ G $. Then it has a
decomposition to nonintersecting cycles, which consist of $ 1 $ or
$ p $ elements. Denote by $ C_{1},C_{2}, \ldots , C_{m} $ the
cycles which consist of $ p $ elements ( by $ C_{j} $ I mean the
set of elements in corresponding cycle). We have that $ m
\geqslant 1 $, because $ a $ is nontrivial. Now, for each $ x \in
G $, we have that $ xa = ax $, therefore $ x $ maps each cycle $
C_{i} $ to some cycle $ C_{j} $. Therefore every $ x \in G $ acts
on the set of cycles $ \{ C_{1}, C_{2} , \ldots , C_{m} \} $, so $
G $ acts on the set $ \{ C_{1}, C_{2} , \ldots , C_{m} \} $. Each
orbit of $ G $, acting on $  \{ C_{1}, C_{2} , \ldots , C_{m} \} $
consists of $ p^{j} $ elements, because it's order must divide $
|G| = p^{k} $. Therefore if we take the union of $ C_{i} $ lying
in this orbit, then it has $ p^{j+1} $ elements. But again, by $ n
< p^{2} $ we have that $ j=0 $, so this orbit consists of only $ 1
$ element. So we have that every element $ x \in G $ maps $ C_{j}
$ to itself for every $ j $. Take $ A = C_{1} $, $ B = \{
1,2,\ldots,n \} \backslash C_{1} $. Then $ |A| = p $, $ |B| = n-p
$, and $ G $ acts separately on $ A $ and on $ B $. If we denote
by $ H $ the group of permutations on $ A $, which comes from $ G
$, and by $ K $ the group of permutations on $ B $, which comes
from $ G $, then by induction hypothesis we have that $ H,K $ are
commutative, hence $ G $ is commutative.

I hope it is correct.


function
topology
real analysis
real analysis unsolved

Suppose that the function $ f: [a,b] \to [c,d]$ is absolutely continuous and bijection.
Is it true that $ f^{\minus{}1}: [c,d] \to [a,b]$ is absolutely continuous?
Can we weaker the assumption?,
i.e. what is necessary and sufficient for an absolutely continuous function to have inverse image is also absolutely continuous?

Here's a counterexample: let $ g$ be the Cantor devil's staircase function on $ [0,1]$ (increasing, not absolutely continuous) and let $ f^{\minus{}1}(x)\equal{}g(x)\plus{}x$. $ f$ then maps $ [0,2]$ to $ [0,1]$ bijectively, and is absolutely continuous since it satisfies a Lipschitz condition $ |f(x)\minus{}f(y)|\le |x\minus{}y|$.

The necessary and sufficient condition to be added is that $ f'\ne 0$ almost everywhere. (Which means either $ f'>0$ a.e. or $ f'<0$ a.e., since $ f$ is monotone.)

In higher dimensions this is a topic of current research. There the analogue of the class of absolutely continuous functions is the Sobolev space $ W^{1,1}$ of mappings $ f\colon \mathbb R^n\to\mathbb R^n$. In two dimensions the situation is now [url=http://adsabs.harvard.edu/abs/2006ArRMA.180...75H]clear[/url]: if such $ f$ is a homeomorphism and has positive Jacobian determinant a.e., then its inverse is also in $ W^{1,1}$. When $ n>2$, things get more complicated.


Hello

Is there people here which speaks spanish? (or at least write it)

Well, the "Spanish Communities" forum is reasonably active. It's too bad that AoPS/Mathlinks stopped posting flags beside posters' screen names, but I think most of the participants in that forum live in the Americas, from Mexico through Argentina and Chile, scattered across quite a few of the countries in between.

Okay

Well actually I posted this 'cause I want to see if there're some people (excluding Spanish-speaking countries) which at least writes in spanish :), not necessary to talk the language.

Yo se un poco de espanol. Pero es muy muy muy mal.

yo voy a estudiar espanol en mi collegio en la ciudad de los angeles.

Sí, habla español porque apprendo español en mi escuela en Canadá... Me gusta español mucho, pero mi español es muy mal :( ¡Viva Español!)

Yo hablo español, soy de San Luis Potosi, Mexico. Me llamo Ernesto H. Escobedo. No se cual es la comunidad para hispanoparlantes aqui en MathForums, o una comunidad para la olimpiada Mexicana de Matematicas

[b][color=red]Creo que la comunidad en Español se encuentra por acá:[/color][/b]

[url=http://www.artofproblemsolving.com/Forum/index.php?f=255]*****[/url]

[color=red][b]Hasta la vista, [i]baby[/i].[/b][/color]

¡Hola! Soy de El Salvador y hablo español. :D

¡Hola!

               Estoy bien. [color=cyan]:lol: [/color]

Hola, yo puedo hablar un poco espanol. Buena suerte en sus estudias de matematicas.

Hola Como Estas

I forgot most of what I learned....shoot....thats no good

I have soo much espanol tarea :(

I remember this though

Nos Vemos!![/b]

Yo soy de Puetro Rico y quiero un forum para Puerto Rico

Yo soy Louisville,  you hablo un poco espan^~ol y mucho americano.  Me gusta deportiste y matematicas.  Yo estudio para los exams mucho.  Yo (Aprendo pretéritos en español. ~ Utilicé a un traductor para las partes en el ()).

Hablo un poco espanol.

Soy de los Estados Unidos.

Tengo nada hablar (esta fraze tiene razon?  No se?).

The $\text{Spanish Communities}$ is not very active.

Las comunidades en español no son muy activas, creo que se debe a los moderadores del mismo, pues desde hace tiempo ya no se muestra alguna actividad. Espero que los participantes de foro que hablen o lean en el idioma español se manifieste.

You hablo y escribo español porque nací y crecí en Uruguay (South America). Allí estudié química en la universidad.
También hablo y escribo en inglés porque hace más de 30 años que vivo en EEUU (USA).

"Tengo nada hablar" no tiene sentido (means nothing). I don't know what was meant by that. If I knew what was meant, I could translate.

Absolutamente nada que decir, señorita???

Ignore the last word.

what the heck...
I only took half a year of Spanish, and I found spanish so uninteresting.
Hola
Bueno Dias----> good morning
Bueno Tardes----> good afternoon
Bueno Noches----> good night
Papel----> paper
Boligrafo---> Pen i think...
Marcadora----> Marker
Pizara----> board

[quote="carlosmath"][quote="3333"]All I can say is:

El coche es grande

El coche es pequenio

El coche es nuevo

El coche es viecho

La camionetta es grande

La camionetta es pequenia

La camionetta es nueva

La camionetta es viecha

Did I spell everything correctly?[/quote]

There are some mistakes

El coche es pequenio $\Rightarrow$ El coche es pequeño.

El coche es viecho $\Rightarrow$ El coche es viejo.

La camionetta es grande $\Rightarrow$ La camioneta es grande.

La camionetta es pequenia $\Rightarrow$ La camioneta es pequeña.

La camionetta es viecha $\Rightarrow$ La camioneta es vieja.

Saludos y qué bueno que estés aprendiendo es idioma español (spanish)[/quote]

Meh

[quote]Bueno Noches----> good night[/quote]
That is awesome because in Russian, night is ночь which is pronounced "noch."
Wouldn't it be buenas noches though instead of bueno?

[quote="yugrey"]I was trying to say that I have nothing to say.[/quote]
No tengo nade que decir. = I have nothing to say.
No tengo nade de que hablar. = I have nothing to talk about.
No sé que decir. = I don't know what to say.
(I used "alt 0233" for the é).


number theory
greatest common divisor
algebra proposed
algebra

it' anyone here give us a solution

it' so easy  :?

If one of the 18 cosecutive numbers is 999, we are done. If not, at least one of the 18 cosecutive numbers is divisible by 18. Hence the sum of its digits must be 9 or 18, and our proof is complete.

[quote="k2c901_1"]If one of the 18 cosecutive numbers is 27, we are done. If not, at least one of the 18 cosecutive numbers is divisible by 18. Hence the sum of its digits must be 9 or 18, and our proof is complete.[/quote]

Isnt the question asking for three digit consecutive number ? So i doubt 27 will be one of them .

Since we know that if a number is divisible by 3 , the sum of its digit must also be a multiple of three . So in any 18 consecutive number , there will always be at least one of it which can be divisible by 3 .Or stronger one is at least 6 number which can be divisible by its own sum of digit .

A correction has been posted. As for your arguement, I don't think it's correct, as having the sum of the digits of a certain number divisible by 3 does not guarantee that the it is divisible by the sum of its digits. Take, for example, 984.

Ya , i agree with you k2c901_1 and I think I have mess up with the question itself .My arguement can only take out the gcd of the number itself and its sum of digit.  :blush:  :blush: I first thought that we can see it through divisibility of 9 and then come to divibility of 3 . I guess divisibilty of 18 is more accurate here ....  :P .


modular arithmetic
number theory
prime factorization

[quote="Hikaru79"][quote="cincodemayo5590"][quote="N.T.TUAN"]Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.[/quote]

[hide="Would it make the most sense to..."]
prove that any number $A$ is divisible [i]not only[/i] by a number in the set $\{ 3,5,7,9 \}$? (That is, in the prime factorization of $A$, those numbers are not the only primes.) [/hide][/quote]

I don't think that would work; an even number would work by your condition, but not by his, since $2$ is a prime less than $11$, but it is not in that set.[/quote]

But $A$ can't be a multiple of $2$ with these digits of 1,3,7,9....

[quote="HTA"]if A is a prime number then A is divisible by itself (we easily have $A \geq 11$)
if A is a composit number , assume that A has no divisot greater than 11 , we obtain $A$ max $=3.7=21$ , but $11;13;17;19$ are all primes , so A min $=21$ , BUT $21$ IS NOT CONTRACDITION !![/quote]
It's not that easy, A is not necessarily squarefree.
[hide]
Clearly $A$ can't be divisible by 2 or 5, so suppose all of its prime factors are less than 11. Then $A$ is a power of 3 multiplied by a power of 7, that is, $3^{a}\cdot 7^{b}$. We now prove that the tens digit of $A$ is a problem (since $A>9$, it actually has a tens digit).

To see this, consider parity of $a$ and $b$. If the same, then $A\equiv 1\pmod{4}$ and $A\equiv \pm 1\pmod{5}$. If different, then $A\equiv 3\pmod{4}$ and $A\equiv \pm 2\pmod{5}$. (This is so because $A\equiv 3^{a+b}\pmod{4}$ and $A\equiv 3^{a-b}\pmod{5}$.) In particular, $A$ must be congruent to one of 1,3,7,9 modulo 20, making its tens digit even, a contradiction.
[/hide]

[quote="HTA"]if A is a prime number then A is divisible by itself (we easily have $A \geq 11$)
if A is a composit number , assume that A has no divisot greater than 11 , we obtain $A$ max $=3.7=21$ , but $11;13;17;19$ are all primes , so A min $=21$ , BUT $21$ IS NOT CONTRACDITION !![/quote]
oh , i'm sorry , i read the question wrong :D

[quote="N.T.TUAN"]Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.[/quote]
if we assume that A has no prime divisors greater than 11 then A=3^x*7^y (checking that other cant be divisors)

but A=10^a_1+3*10^a_1+7*10^a_1+9*10^a_4
a_i are distinct arbitary integers.
so,3^x*7^y=10^a_1+3*10^a_2+7*10^a_3+9*10^a_4

3^x*7(7^{y-1}-10^a_3)=10^a_1+3*10^a_2+9*10^a_4
taking mod 3 left hand side is 0 mod 3 while rhs=1 mod 3
nd its the contradiction we desired.

What about the possibility of repeated digits? Also your numbers allow for 0 to be among the digits.

Oups...I forgot to mention that a_i=(0,1,2,3) and distinct.

But do i need to prove it for repeated digits?
If so then it is really hard to seek for contradiction :(
bcz thr is 1 so u may take as many as u want.So i thought that probably reapeated digits are not allowed ;)
whatever, if repeated digits are allowed then i need to think again!