geometry
trigonometry
function
calculus
derivative
inequalities
geometry unsolved

Let $ AB$ be the diameter of a given circle with radius $ 1$ unit, and let $ P$ be a given point on $ AB$. A line through $ P$ meets the circle at points $ C$ and $ D$, so a convex quadrilateral $ ABCD$ is formed. Find the maximum possible area of the quadrilateral.

Let $ \alpha \equal{} \hat{(AB,CD)}$  and H be the projection of O on CD.

We have: $ S_{ACBD} \equal{} \frac{1}{2} AB.CD.\sin{\alpha} \equal{} CD \sin{\alpha}$ (1)

We need to find $ alpha$ in $ (0,\pi)$ such that it will minimize (1).

Now: $ OH \equal{} OP \sin{\alpha}$ then $ CD \equal{} 2 \sqrt{1 \minus{} OH \sin^2{\alpha}}$

Now just use analysis to find the mimimum of the function $ f(x) \equal{} \sqrt{1 \minus{} OH \sin^2{\alpha}}\sin{\alpha}$

[quote="mathangel"]Let $ \alpha \equal{} \hat{(AB,CD)}$  and H be the projection of O on CD.

We have: $ S_{ACBD} \equal{} \frac {1}{2} AB.CD.\sin{\alpha} \equal{} CD \sin{\alpha}$ (1)

We need to find $ alpha$ in $ (0,\pi)$ such that it will minimize (1).

Now: $ OH \equal{} OP \sin{\alpha}$ then $ CD \equal{} 2 \sqrt {1 \minus{} OH \sin^2{\alpha}}$

Now just use analysis to find the mimimum of the function $ f(x) \equal{} \sqrt {1 \minus{} OH \sin^2{\alpha}}\sin{\alpha}$[/quote]

 And what will you get after taking the derivative to 0?
 I wonder if it will be anything constructible.
 Would be nice to know if the derivative equation is reasonably doable, but
  I stop the moment I see a shadow of unnatural irrationality (other than root of
  N(atural)).

  The fact that this problem was given at some math competition hints that
  it is not a straightforward problem.  And if it were up to calculus, then it would not be
  in the geometry department.

  This is a purely geometric problem and has a fine point to it.

  Thank you.
  M.T.

This is a rough problem because there are two different cases that produce different equality cases...

WLOG $ P$ is on $ AO$. Let $ PO\equal{}p$. Let $ C'$ and $ D'$ be the projections of $ C$ and $ D$ onto $ AB$, respectively. Call $ CC'\equal{}c$ and $ DD'\equal{}d$. The area of the quadrilateral is $ c\plus{}d$. Let $ M$ be the midpoint of $ CD$ (so $ CD\perp OM$). Then it follows that if $ OM\equal{}x$, then $ \frac{d}{x}\equal{}\frac{PD}{p}$ and $ \frac{c}{x}\equal{}\frac{CP}{p}$. Hence $ c\plus{}d\equal{}\frac{x}{p}\cdot CD$. Note that $ CD\equal{}2\sqrt{1\minus{}x^2}$. Hence \[ [ABCD]\equal{}\frac{2}{p}\cdot x\cdot \sqrt{1\minus{}x^2}\] where $ x\le p$.

Case 1: $ p\ge \frac{1}{\sqrt{2}}$. Note \[ [ABCD]\equal{}\frac{2}{p}\cdot x\cdot \sqrt{1\minus{}x^2}\equal{}\frac{2}{p}\cdot \sqrt{x^2\cdot(1\minus{}x^2)}\le \frac{2}{p}\cdot \frac{(x^2)\plus{}(1\minus{}x^2)}{2}\equal{}\frac{1}{p}\] by the AM-GM inequality. Equality occurs when $ 1\minus{}x^2\equal{}x^2\implies x\equal{}\frac{1}{\sqrt{2}}\le p$.

Case 2: $ p<\frac{1}{\sqrt{2}}$. Note \[ [ABCD]\equal{} \frac{2}{p}\cdot x\cdot \sqrt{1\minus{}x^2}\le \frac{2}{p}\cdot p\cdot \sqrt{1\minus{}p^2}\iff \frac{4}{p^2}\cdot (p^2\minus{}x^2)(1\minus{}x^2\minus{}p^2)\ge 0\] but $ x^2\plus{}p^2\le p^2\plus{}p^2\equal{}2p^2<1$, so the inequality above is true. Equality occurs when $ x\equal{}p$. That is when $ CD\perp AB$.

[quote="rem"][color=darkred]Let $ AB$ be a fixed diameter of a given circle $ w = C(O,1)$ and let $ P$ be a fixed point on $ (AB)$. A mobile line through $ P$ meets the circle $ w$ at $ C$ and $ D$, so a convex quadrilateral $ ACBD$ is formed. Find the maximum possible area of this quadrilateral.[/color] [/quote]
[color=darkblue][b]Proof[/b] (similarly with [u]Altheman[/u]'s). Suppose w.l.o.g. that $ P\in (OB)$ . Denote $ \phi = m(\widehat {OPD})$ and the distance $ x$ of $ O$ to $ CD$ . Observe that $ CD = 2\sqrt {1 - x^2}$ and $ \sin\phi = \frac xp$ . Since $ S\equiv [ACBD] =$ $ \frac 12\cdot AB\cdot CD\cdot\sin\phi$ , then $ S$ is [b]max.[/b] $ \Longleftrightarrow$ $ x\cdot\sqrt {1 - x^2}$ is [b]max.[/b] $ \Longleftrightarrow$ $ x^2\cdot (1 - x^2)$ is [b]max.[/b], where $ x\in [0,p]$ . But $ x^2 + (1 - x^2) = 1$ ([b]constant[/b]). Observe that $ x^2 = 1 - x^2$ $ \Longleftrightarrow$ $ x = \frac {\sqrt 2}{2}$ . In conclusion, $ x_{\mathrm {max}} = \{\begin{array}{ccc} \frac {\sqrt 2}{2} & \mathrm {for} & \frac {\sqrt 2}{2}\le p \\
 \\
p & \mathrm {for} & p < \frac {\sqrt 2}{2}\end{array}$ and $ S_{\mathrm {max}} = \{\begin{array}{ccc} \frac 1p & \mathrm {for} & \frac {\sqrt 2}{2}\le p \\
 \\
2\sqrt {1-p^2} & \mathrm {for} & p < \frac {\sqrt 2}{2}\end{array}$ [/color]


function
topology
complex numbers
complex analysis

Let $ U$ be an open set (non-empty) of $ \mathbb{C}$ and let $ f$ be an analytic function on the set. Define $ g$ on $ \tilde  U \equal{} \{ z : \overline{z} \in U \}$ by $ g(z)\equal{}\overline{f(\overline{z})}$. Prove $ g$ is analytic.

To use Cauchy-Riemann equations, write $ f\equal{}u\plus{}iv$ where $ u,v$ are real functions of $ x,y$. Then $ g(x,y)\equal{}u(x,\minus{}y)\minus{}iv(x,\minus{}y)$. Chain rule finishes the problem.

Other approaches: 1. $ (g(z\plus{}h)\minus{}g(z))/h \equal{} \overline{ (f(\overline{z}\plus{}\overline{h})\minus{}f(z))/\overline{h})}$.

2. If $ f(z) \equal{} \sum c_n(z\minus{}a)^n$, then $ g(z)\equal{} \sum \overline{c_n}(z\minus{}\overline{a})^n$


J have taken this problem from  some book

it probably comes from bulgaria 95 but I'm not sure

Find all positive integers(x,y) satisfying :
x^2 + y^2 is divisble by  1995(x-y)

[hide="Hint"]$ 1995(x\minus{}y)|x^2 \plus{} y^2 \iff 1995(x\minus{}y)|x^2 \plus{} y^2 \minus{}2xy\plus{}2xy \iff 1995(x\minus{}y)|(x\minus{}y)^2\plus{}2xy$[/hide]

Could you complete your solution please?

I think that you are talking about the following problem:
Find all $ (x,y)$ such that $ \dfrac{x^{2} \plus{} y^{2}}{x \minus{} y}$ is an integer which is a divisor of $ 1995$.
[b]Solution:[/b]
[b]Lemma[/b]: If $ p$ is a prime number of the form $ 4k \plus{} 3$ and $ p | x^{2} \plus{} y^{2}$ then $ p | x$ and $ p | y$.
Apply this for $ 3$, $ 7$ and $ 19$ and you can see that your problem become the following:
Find all pairs $ (x,y)$ such that $ x^{2} \plus{} y^{2} \equal{} 5(x \minus{} y)$ but it implies $ (2x \minus{} 5)^{2} \plus{} (2y \plus{} 5)^{2} \equal{} 50$. Solve that and you get the pairs $ (3,1)$ and $ (2,1)$.
Then the solution of the original problem are the pairs of the form $ (3a,a), (2a,a), (a,2a), (a,3a)$ where $ a$ is a divisor of $ n \equal{} 3.7.19$.


Let a,b,c,d≥0 satisfying $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$
Prove that (1-a)(1-b)(1-c)(1-d)≥1

I think it's not right...
Try $ a\equal{}b\equal{}c\equal{}d\equal{}\frac{1}{2}$

Oh,I'm really sorry.I take this problem from my book and my book has something wrong.In fact, what we need to prove is (1-a)(1-b)(1-c)(1-d)≥abcd


Let $ n$ be a given positive integer .Find the largest real number $ t_1(n)$ and the least real number $ t_2(n)$ Such that for all real numbers $ x_1,x_2,..,x_n$ , holds the following inequality :
 $ t_1(n)\sum_{k \equal{} 1}^{n}{(x_1 \plus{} x_2 \plus{} ... \plus{} x_k)^2} \leq \sum_{k \equal{} 1}^{n}{x_k^2} \plus{} (\sum_{m \equal{} 1}^{n}{x_m})^2 \leq t_2(n)\sum_{k \equal{} 1}^{n}{(x_1 \plus{} x_2 \plus{} ... \plus{} x_k)^2}$
[hide="Answer of me"] $ t_1(n) \equal{} 4sin^{2}(\frac {\pi}{2n \plus{} 2})$,
$ t_2(n) \equal{} 4cos^{2}(\frac {\pi}{2n \plus{} 2})$ True or false ?  :P [/hide]


geometry
Asymptote
geometric transformation

Hi, I'd like to know if it's possible to draw loci using Asymptote. Actually I'm looking for a geometry program that can draw loci, just like Sketchpad, but I want to get images with a good quality, so Sketchpad is not a good choice... :maybe:

Any ideas or suggestions? Thanks in advance for your help. :lol:

What's wrong with Sketchpad's image quality?

Don't forget you can increase the number of samples in a locus by selecting it and pressing '+'; and you can "overdrive" loci to much higher resolution when printing by using Advanced Preferences | Locus/Plot Export Quality.

Good luck,
Nick Jackiw
Sketchpad Projects
KCP Technologies

[quote="tpulak"]I have a question to you: What's a loci?[/quote]

Loci is the plural of locus. In general terms, a locus is a set of points that satisfy certain conditions.

[quote="njackiw"]
Don't forget you can increase the number of samples in a locus by selecting it and pressing '+'; and you can "overdrive" loci to much higher resolution when printing by using Advanced Preferences | Locus/Plot Export Quality.
[/quote]

Thanks for the tip, Nick! :D Indeed, my loci look much better now!! But I have another question for you: Can I also increase the quality of lines and circles, too? For example, lines sometimes look "broken" and not perfectly straight... What would you recommend to me then?

Thank you!!

[quote="tpulak"]Thanks for telling me what a loci was, but I'm sorry, but I don't really know what to do. Is there an option for higher resolution? If there is, then make your sketch higher in resolution.This should make it more better quality. If still it isn't good quality, then another program might be a better choice.[/quote]

That's what I pointed out in my first post; I can get a good quality for the locus curve, but not for the rest of the sketch... :(

So, I insist on my original question: Does anybody know if it's possible to draw loci in Asymptote?? :maybe: Thanks for replying!

[quote="Jutaro"]

[quote="njackiw"]
Don't forget you can increase the number of samples in a locus by selecting it and pressing '+'; and you can "overdrive" loci to much higher resolution when printing by using Advanced Preferences | Locus/Plot Export Quality.
[/quote]

Thanks for the tip, Nick! :D Indeed, my loci look much better now!! But I have another question for you: Can I also increase the quality of lines and circles, too? For example, lines sometimes look "broken" and not perfectly straight... What would you recommend to me then?

Thank you!![/quote]

I'm not sure if you're talking about while printing or on-screen. When printing, Sketchpad attempts to use device resolution, so I doubt you're seeing jaggies on printout. If you seem them onscreen, be aware that older versions of Sketchpad (pre-4.05?) used to ship with anti-aliasing turned off for speed purposes. (GDI+, the anti-aliased Windows drawing toolkit that Sketchpad uses, is substantially slower than the old GDI toolkit.) If you prefer smooth lines to fast lines, change Advanced Preferences | Anti-Aliased Graphics from "Never" to "When Speed Permits" or to "Always."

Nick Jackiw
Sketchpad Projects
KCP Technologies

Hi,
[quote="Jutaro"]Does anybody know if it's possible to draw loci in Asymptote??[/quote]
It's of course possible to draw loci with Asymptote.
A static example:
[code]import geometry_dev;
size(12cm,0);

circle C=circle(origin,1);
draw(C);
point A=point(C,0), B, M;
guide locus;

real eps=0.001;
for (real a=eps; a < 360; a += 1) {
  B=angpoint(C,a);
  triangle t=triangle(origin,A,B);
  triangle intouch=intouch(t);
  M=intouch(t.AC);
  locus=locus..M;
  if(a-eps == 30 || a-eps == 90 || a-eps == 225 || a-eps == 345) {
    draw(t, dot);
    draw(intouch, dot);
    draw(incircle(t), 0.8*blue);
    label("$N$", Label("$M$", 0.8*red), "$P$", intouch);
    dot(M, 0.8*red);
  }
}

draw(locus..cycle, bp+0.8*red);
[/code]

And a movie:
[code]import geometry_dev;
import animate;

animation animation;

size(12cm,0);

circle C=circle(origin,1);
draw(C);
point A=point(C,0), B, M;
guide locus;

for (real a=0.001; a < 360; a += 5) {
  save();
  B=angpoint(C,a);
  triangle t=triangle(origin,A,B);
  draw(t);
  draw(incircle(t), bp+0.8*blue);
  triangle intouch=intouch(t);
  draw(intouch, dot);
  M=intouch(t.AC);
  label("$N$", Label("$M$", 0.8*red), "$P$", intouch);
  dot(M, 0.8*red);
  locus=locus..M;
  draw(locus, bp+0.8*red);
  animation.add();
  restore();
}

animation.movie();[/code]
Result [url=ftp://download.tuxfamily.org/piprim/temp/locus.gif]here[/url].
geometry_dev is the development version of geometry.asy; the documentation (in French; English translation in progress) can be found [url=http://piprim.tuxfamily.org/asymptote/travaux/modules/doc/geometry.pdf]here[/url].


integration
logarithms
algebra proposed
algebra


A set $F\subset\mathbb R$ is called locally symmetric with respect to a point $x\in\mathbb R$ if there exists $\delta>0$ such that $F\cap (x-\delta,x+\delta)$ is symmetric with respect to $x$. Suppose that $F$ is closed, $F\neq \mathbb R$ and $F$ is locally symmetric with respect to every $x\in\mathbb R$ (with $\delta$ depending on $x$, of course). Prove that $F$ is at most countable.

To my shame I don't know what happens if $F$ is not necessarily closed. The conjecture is that then either $F$ or its complement is at most countable.

Let $\tilde F$ be the set we get after taking the points which are not condensation points out of $F$ (by condensation point I mean point for which all neighbourhoods contain uncountably many points of $F$). It's easy to argue that the points we have taken out are countably many. $\tilde F$ is also closed (it's perfect, actually).

Let $x_0$ be a point not in $\tilde F$, and assume $(x_0,b)$ is the largest such interval which has void intersection with $\tilde F$. It means that $b\in\tilde F\subset F$, and any interval centered at $b$ has at most countably many points of $F$ to the left of $b$ but uncountably many (points of $F$) to the right of $b$. This means that there are no points of $\tilde F$ to the right of $x_0$. In exactly the same way we show that there are no points of $\tilde F$ to the left of $x_0$, so $\tilde F=\emptyset$. The conclusion follows.

Isn't there a possibility that the general problem ($F$ not necessarily closed) depends on the Continuum Hypothesis?


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 10
probability

wht does it mean, and whats the point of doing it.. :?:

oh, well whats the point of that, isnt it just better to guess, lol.

Wrong answer = 0 points,
Omit = 1.5 points
Correct = 6 points.

what do u personally think is better, omitting or guessing, on a question u have noo idea on??

Omit.  Statistically, it will get you more points than a guess.  On the other hand, if you can eliminate at least one choice, guess.

Well, if your guesses are completely random, then guessing gives you an expected gain of zero, so ideally guessing or omitting would not give you any advantage one way or the other.

EDIT: lol

No.  Since there are five choices on the AMC rather than only four, the expected value of points per problem if you guessed all of them would be 6/5=1.2 compared with an expected value of 1.5 for every question omitted.

[quote="hjoon0125"]personally, i think guessing is better[/quote]
[color=blue][b]I also have the same opinion with u. May my idea be wrong ? Is there any thought?[/b][/color] :maybe:  :roll:

[quote="worthawholebean"]You're wrong. Guessing will, on average, reduce your score.[/quote]
[color=blue][b]Oh, thanks ever so much for ur kind attention if u explain more clearly about the difference between the "omitting" and "guessing".
Best regards,[/b][/color]

I'm rather surprised that this discussion is occurring on AoPS, of all places...  :huh:

Guessing has an expected value of $ 1/5 * 6 \equal{} 1.2$ while omitting gets you $ 1.5$. Obviously, omitting is the better choice. If you omit everything, you receive $ 37.5$, if you guess on everything, your expected value is $ 30$.

although you could have a good day, or just be a good guesser, and get more points from guessing.

lets say for 10 questions, i guess 3-4 rite, then my score would be 18-24 for those ten questions..

but for all ten questions i omit, then my score would me 15...

well, omitting is like a risk-free way to get 1.5 points, and guessing is all vegas, :pilot:

[quote="Paiev"]Guessing has an expected value of ...[/quote]
That assumes that you are doing something truly random to choose your answer - flipping a coin, rolling dice, choosing numbers from a random digit table. (Yeah, there are ways to generate 1/5 probabilities even with coins.)

But it's more likely that you're not doing that. You're looking at the answers, and saying, "I don't really know, but I kind of like that one ..."

Here's my experience with the AMC's question writers: if you're there, then you've been suckered. You're being led down the path of error, and if you go that way, it's not going to turn out well.

[quote="benzi455"]Well, if your guesses are completely random, then guessing gives you an expected gain of zero, so ideally guessing or omitting would not give you any advantage one way or the other.[/quote]

Okay, okay, I'm sorry!  I made a math mistake on a math forum.  Sorry.  I was assuming that every test maker made it that way, like the SAT and stuff.

Of course, there are a couple of standard situations where guessing is generally considered a good thing.  One is if you have 13 other questions answered (and think they're all correct).  Then the 14th question puts you over the AIME threshold, assuming no lowered cutoff, so most people will pick one and guess.  The other is if you have solved 24 questions, and a guess would give you a perfect score.

i should make a poll on this topic :blush:

personally i think just get it right xD. but if your really unsure and come up with an answer that happens to be an answer choice, dont put it down necessarily especially if its in a later problem. on amc 10 i did that and got a 139.5(i failed badly >.<). They like to take common errors and make them into an answer. and i really hate that -.- at that point i realized i wasnt gonna make usamo in middle school xD lol.

Yes...someone I know ended up answering 16 and leaving 9 blank in the amc 12.  this gave him a maxscore of 107.5, but he answered 2 of the 16 wrong.  he told me later he should have at least guessed on another one for a similar reason that AIME15 mentioned (passing score for amc12 is 100).

[quote="AIME15"]However, sometimes it is profitable to guess.  Let's say you would have a 121.5 score if all the problems you answered are correct, and your goal is to make AIME (120).  It is profitable to guess on one of the problems you ommitted, because if you get it incorrect, you'll still be at 120.  If you get it correct, you have a "cushion" point (so if one of the others you thought correct is incorrect, it's still ok).[/quote]
is 120 the cutoff for amc10 EVERY YEAR? I thought that 100 was for cutoff for amc12 every year but i was totally WRONG...

AMC 12 is 100 or top 5%, whatever is lower.

AMC 10 is 120 or top 1%, whatever is lower.


[quote="anirudh"]Can someone give me step by step directions to install mysql successfully. I want to install it, so I ccan use phpbb2. thanks.  :)[/quote]

Google it.  Have you tried installing Apache also?  Also try simply following the instructions in the installer.  it usually works.

it doesn't. it says that the installer for phpbb cannot find the database.  ;)


irrational number
number theory
prime factorization

Solve the equation of  $ \sqrt{2} \plus{}  \sqrt{3} \equal{} \sqrt{m} \minus{} \sqrt{n}$  in integer $ m, n$

Squaring both sides,
$ 5 \plus{} 2\sqrt{6} \equal{} m\plus{}n \minus{} 2 \sqrt{mn}$
$ m\plus{}n\minus{}5 \equal{} 2 \sqrt{6} \plus{} 2\sqrt{mn}$
LHS is integer, but RHS is an irrational number.
Thus, there are no solution.

CMIIW

[quote="TFPNH"]Squaring both sides,
$ 5 \plus{} 2\sqrt {6} \equal{} m \plus{} n \minus{} 2 \sqrt {mn}$
$ m \plus{} n \minus{} 5 \equal{} 2 \sqrt {6} \plus{} 2\sqrt {mn}$
LHS is integer, but RHS is an irrational number.
Thus, there are no solution.

CMIIW[/quote]

why RHS is irrational number ?  :)
In my thought, you must prove that RHS is irrational number

[hide="Quick proof that RHS is irrational"]
$ (\sqrt 6 \plus{} \sqrt {mn})^2 \equal{} 6 \plus{} mn \plus{} 2\sqrt {6mn}$.

If $ \sqrt {6mn}$ is irrational, then we're done, otherwise assume $ mn \equal{} 6x^2$ for a positive integer $ x$. Then we have $ 6 \plus{} 6x^2 \plus{} 12x \equal{} 6x^2 \plus{} 12x \plus{} 6 \equal{} 6(x \plus{} 1)^2$, which is not a perfect square since it has an odd $ 2\cdot 3$ in the prime factorization, QED.[/hide]

${ \sqrt{6}+\sqrt{mn}=k}$, k is an integer:
$ 6+mn+2\sqrt{6mn}=k^2$
so
$ \sqrt{6mn}=t$
$ mn=6l^2$, l is integer
so
$ \sqrt{6}+\sqrt{mn}=\sqrt{6}(1+l)$ ,RHS is not integer
proved!


hello, what are the questions here?
Sonnhard.


How can I write the long division and the synthetic divsion ways to divide any two polynomials?
(ie $ (x^3\plus{}3x^2\plus{}3x\plus{}1)\div (x\plus{}1)$)

[i]I'll repeat the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1171968#1171968]Synthetic Division[/url] post here. The first example is normal polynomial division, the second is synthetic division, and the third shows the working. There's a great demo [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/polynom/polydemo.pdf]here[/url].[/i]

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=polynom]polynom package[/url]
There's lots of possibilities:
1. \polylongdiv{2x^4+21x^3+35x^2-37x+46}{2x+7}

[img]http://sixthform.info/gblog/pictures/ba339d02eae4e6d1df38ece70632349a.gif[/img]

2. \polyhornerscheme[x=-7/2]{2x^4+21x^3+35x^2-37x+46}

[img]http://sixthform.info/gblog/pictures/f51d74edc9ffce501292604822d2813b.gif[/img]

and even
3. \polyhornerscheme[x=-7/2,tutor=true,resultstyle=\color{blue},tutorlimit=10,stage=10]{2x^4+21x^3+35x^2-37x+46}

[img]http://sixthform.info/gblog/pictures/65babad67cdb0455f69fc3f293294b3d.gif[/img]

As usual for synthetic division, the quotient coefficients need to be divided by 2 in cases 2 and 3 above.

thank you for responding. But it still isn't working for me.

Am I only supposed to add \usepackage{polynom} to my preamble and then type \polylongdiv{X^3+X^2-1}{X-1} or is there something else I need to do?

Thank you
luminate

That's all that's needed. What error message does the log give? If necessary, download a more recent version of polynom.sty (attached as a zip file)

It works beautifully now but I was wondering it is possible to have the divisor be more than a binomial e.g. \polylongdiv{2y^6 + 3y^4 + 4y^2 + 5}{y^2-1}

Thank you

luminate

I don't think you can use y^2-1 directly though you could divide by y-1 then by y+1 which would give the same result. See the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=polynom]polynom documentation[/url] for full details.


well i'm taking them in order :)
first the statement
=========================
Find the minimum value of
| sin x + cos x + tan x + cot x + sec x + csc x |
for real numbers x.
=========================

for those of you which don't know sec x= 1/sinx  and csc x = 1/cosx. I had the following "brilliant" idea:

denote a=sin x + cos x. it's well known that a \in [-\sqrt 2 , \sqrt 2] (use the fact that sin2x+cos2 x = 1 and prove it! :) )

then 2 sin x cos x = a2 - 1  which means that the expression above is in fact

| 2/(a2 -1 ) + a + 2a/(a2-1) | = | (a2 - a +2 )/ (a-1) | = (a2-a+2)/ |a-1|. after some computations (we are trying to find the minimal value k so we study the case a<1 to get rid of the modulus) and we obtain that we should have
a2 +a(k-1)+2-k \geq 0 and Delta yields 0 when k=2 \sqrt 2 -1. Obiviously we have equality for a=1- \sqrt 2, and it is just the case that a is in the required range. It's easy to check that for a>1 we also have a standing equality, so the minimal value is k= 2\sqrt 2 - 1.

[quote="Valentin Vornicu"]well i'm taking them in order :)
for those of you which don't know sec x= 1/sinx  and csc x = 1/cosx.[/quote]

I think you mixed these up (or actually you got them right but the english language got them wrong), but that's the great thing about the comm. law of addition!

Yeah, he mixed them up. Happens to me too from time to time. But it doesn't really matter here since $\frac1{\sin x}+\frac1{\cos x}=\frac1{\cos x}+\frac1{\sin x}$.

Anyway you could also use deriv... *shuts hole*  :blush:


abstract algebra
algebra
polynomial
Ring Theory
superior algebra
superior algebra solved

Hello

it is known that in $\mathbb{C} [x_{1},x_{2},\ldots,x_{n}]$ every ideal is generated by a finite number of polynomails (noetherian property)

Now I was just wondering, if I define a minimal generating set as a (finite) set of generators of an ideal, such that removing one of them will make the set lose its generating property, do all minimal generating sets of the same ideal have the same cardinality?

The reason I care so much is that with nullstellensatz, you can link algebraic varieties with radical ideals.  That way I could define a dimension of a variety.

It's not quite that simple- look up Krull dimension if you want more details. The dimension is defined in terms of a chain of prime ideals, not a set of generators

It's possible to require "too many" generators- that's why the definition is what it is.

so i see, this krull dimension is interesting

but simply, the answer to my question is no, and that is exactly why krull dim exists??

and please, is there an easy way to convince me in the complex case of nonconstantness of generators??

Does "minimal" mean that no proper subset works? If so, note that the ideal $(y)$ in the polynomial ring over one variable is generated by the two elements $y^2+y,y^2-y$, but not either one alone.

A more subtle example, from a textbook exercise:
Let $Y$ be the curve parametrically defined by $x=t^3,y=t^4,z=t^5$ in affine three-space (over your favorite field). Let $A(Y)$ be the set of all polynomials which are uniformly zero on $Y$. Show that $A(Y)$ is a prime ideal of height (dimension) 2, which cannot be generated by two elements.


limit
integration
real analysis
calculus
calculus open

Prove that $ \lim_{n\to\infty} \left(\sum_{i\equal{}1}^n \frac{i^{n\minus{}i}i!}{n!}\minus{}\frac{5}{4}\sqrt{n}\right) \equal{} 0$.

[hide="Motivation"]It can be shown that in order to bubblesort a random array of $ n$ reals, one needs an average of $ n\minus{}\sum_{i\equal{}1}^n \frac{i^{n\minus{}i}i!}{n!}$ iterations.[/hide]

[hide="Work so far"]Using Stirlings Formula $ n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ (one would have to prove that the errors vanish as $ n\to\infty$...), we get for the sum
\[ \sum_{i\equal{}1}^n \frac{\sqrt{2\pi i}i^{n\minus{}i}i^ie^j}{\sqrt{2\pi n}n^ne^i} \equal{} \sum_{i\equal{}1}^n \sqrt{\frac{i}{n}}\cdot\left(\frac{i}{n}\right)^ne^{n\minus{}i}.
\]
Denoting $ t_i \equal{} \frac{i}{n}$, one gets $ n\sum_{i\equal{}1}^n t_i^{j\plus{}\frac{1}{2}}\left(e^{1\minus{}t_i}\right)^j$, and taking limits (again, this would have to be concluded more rigorously), one gets
\[ j\int_0^1 t^{1}{2}(te^{1\minus{}t})^j\mathrm dt
\]
Now I am kinda stuck, but I have the following ideas: First, one might try a power series expansion, but I think high powers of $ j$ would remain. Secondly, one might interpret it as a convolution of $ t^\frac{1}{2}$ and $ ((1\minus{}t)e^t)^j$, but I do not know whether there is a theory for convolutions over a finite interval. Lastly, one might take a decent CAD and try brute forcing it; unfortunately, I do not own one...[/hide]

The coefficient is $ \sqrt{\frac\pi 2}$, not $ \frac 54$ :roll:. Also, the asymptotics is not as sharp as you want (the difference does not really tend to $ 0$ though it stays bounded).

Anyway, try this. Rewrite your sum as $ \sum_{i\equal{}0}^n \frac{(n\minus{}i)^i}{(n\minus{}i\plus{}1)(n\minus{}i\plus{}2)\dots n}$ and notice that $ e^{\minus{}k/(n\minus{}i)}\le\frac {n\minus{}i}{n\minus{}i\plus{}k}\le e^{\minus{}k/n}$ for $ 1\le k\le i$. Thus we can squeeze our sum between $ \sum_{i\equal{}0}^n \exp\left\{\minus{}\frac {i(i\plus{}1)}{2(n\minus{}i)}\right\}$ and $ \sum_{i\equal{}0}^n \exp\left\{\minus{}\frac {i(i\plus{}1)}{2n}\right\}$. The second sum is very close to the middlepoint Riemann sum for $ \int_0^\infty e^{\minus{}t^2/2n}$ with step $ 1$, from where you can conclude that it is, indeed, $ \sqrt{\frac{\pi n}{2}}\plus{}o(1)$. The first sum can go down from that only by a constant (I leave the exact estimates to you) and does so. The truth is somewhere in between, so I'm pretty sure that you have the asymptotic formula $ \sqrt{\frac{\pi n}{2}}\minus{}C\plus{}o(1)$ with some $ C>0$ but you need to be a bit more accurate to show that.


[quote=Vasc]If $a,b,c$ are non-negative numbers, then
$\frac{b+c}{2a^2+bc}+\frac{c+a}{2b^2+ca}+\frac{a+b}{2c^2+ab} \geq \frac{6}{a+b+c}$.[/quote]
$\text{Prove that if} \; a,b,c\geq 0,\; \text{then}$
\[ \frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ca}+\frac{a+b}{c^2+ab}\geq\frac8{a+b+c}. \]
[url]https://artofproblemsolving.com/community/c6h72501p420082[/url]

[quote=Vasc]If $a,b,c$ are non-negative numbers, then
$\frac{b+c}{2a^2+bc}+\frac{c+a}{2b^2+ca}+\frac{a+b}{2c^2+ab} \geq \frac{6}{a+b+c}$.[/quote]
The stronger one
\begin{align*} \frac{b+c}{2a^2+bc} \ge \frac{2}{a+b+c}+ \frac{b+c-2a}{a^2+b^2+c^2} \end{align*}
also holds for all $a,b,c \ge 0$, there is a not hard proof.

[quote=KaiRain]
The stronger one
\begin{align*} \frac{b+c}{2a^2+bc} \ge \frac{2}{a+b+c}+ \frac{b+c-2a}{a^2+b^2+c^2} \end{align*}
also holds for all $a,b,c \ge 0$, there is a not hard proof.[/quote]
Let $S=b+c;P=bc \rightarrow P \in \left[0,\dfrac{S^2}{4}\right].$
Need to prove
$$f(P)=4{P}^{2}+ \left(8{a}^{2} -5{S}^{2}-aS \right) P+S \left( S+3a
 \right)  \left( a-S \right) ^{2}\geqslant 0$$
Since $$f''(P)=8>0 \Rightarrow f(P)\geqslant \min\left\{f(0),f\left(\dfrac{S^2}{4}\right)\right\}\geqslant 0$$
Also the SOS's proof
$$\sum\limits_{b,c}\left( a-b \right) ^{2} \left( a+b-c \right) ^{2} \left( b+2c
 \right)  +2 \left( b-c \right) ^{2} \left(b+c-a\right) ^{2}a
+4abc \left( 3{a}^{2}-3ab-3ac+{b}^{2}+bc+{c}^{2} \right) \geqslant 0$$
PS: I wish to know more about your SOS method, it's really nice!

[quote=KaiRain]
The stronger one
\begin{align*} \frac{b+c}{2a^2+bc} \ge \frac{2}{a+b+c}+ \frac{b+c-2a}{a^2+b^2+c^2} \end{align*}
also holds for all $a,b,c \ge 0$, there is a not hard proof.[/quote]
It was [url="https://artofproblemsolving.com/community/c6h2030698p14468307"]here[/url].


29. A carpenter has 40 boards and 600 nails. It takes 2 boards and 40 nails to make a bench, and it takes 5 boards and 50 nails to make a table. Benches sell for $ \$35$, while tables sell for $ \$80$. In order to maximize his earnings, the carpenter should make $ x$ tables and $ y$ benches. What is the value of $ x\plus{}y$?

Basically, if you buy all benches, you use all the nails, but not all the boards.
If you buy all tables, you use up all the boards, but not the nails.
Therefore, you must find a combination of benches and tables in order to use up all of the boards and nails.
10 benches, and 4 tables get you 40 boards and 600 nails.
Therefore, $ x \plus{} y \equal{} 10 \plus{} 4 \equal{} 14$

build on mathcrazed???

Do what peijin did =).

$ 2y\plus{}5x\equal{}40$
$ 40y\plus{}50x\equal{}600$
Subtract 10 times the first from the second to get $ 20y\equal{}200\rightarrow y\equal{}10$
$ x\equal{}4$. all done...
$ x\plus{}y\equal{}14$


geometry
circumcircle
projective geometry
geometry proposed

Let $ A,B,C,D$ are four points on circle.  Let $ P \equal{} AB\cap CD,Q \equal{} AC\cap BD$. Show that $ P$ lie on polar of $ Q$.
I showed by using Pascal thorem twice but I think there is more easy solution.

We denote as $ S,$ the orthogonal projection of $ P,$ on the line segment $ OQ,$ where $ O$ is the circumcenter of $ ABCD.$

In the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139564]Circumscriptible quadrilateral[/url] (post 4), has already been posted an elementary ( and easy I think ) proof of the result that the point $ S,$ is the harmonic conjugate of $ Q,$ with respect to the points $ X$ and $ Y,$ as the intersection points of $ (O),$ from the line segment $ OQ.$

Hence, the point $ P,$ lies on the polar of $ Q,$ with respect to the circle $ (O)$ and the proof is completed.

Kostas Vittas.

PS. The polar of $ P,$ with respect to the circle $ (O),$ is the line connecting the points of $ (O),$ in which the lines through $ P,$ tangent to it. This line passes through the point $ Q\equiv AC\cap BD,$ as well known (and may be proved elementary).


hello, do you mean $ \frac {6x^2 \plus{} 7x \minus{} 5}{2x \minus{} 1}$?
Sonnhard.

[quote="bybimbab"]3. Simplify
6x^2 +7x −5
-------------
2x −1

by dividing out (4)[/quote]
Just use the discriminant.....
$ \frac{(2x\minus{}1)(3x\plus{}5)}{(2x\minus{}1)}$


Try it!
(I don't know how to solve this board  :rotfl:)


Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $ \$1$ more than a pink pill, and Al’s pills cost a total of $ \$546$ for the two weeks. How much does one green pill cost?

$ \textbf{(A)}\ \$7 \qquad
\textbf{(B)}\ \$14 \qquad
\textbf{(C)}\ \$19 \qquad
\textbf{(D)}\ \$20 \qquad
\textbf{(E)}\ \$39$

[quote="Corrina Chang"][hide="My Solution"]
The cost of one day's pills is $ \frac {546}{14} \equal{} \$39$.  Let $ x$ denote the cost of one green pill.  Then $ x \plus{} (x \minus{} 1) \plus{} 39 \implies x \equal{} 20$, which corresponds to answer choice $ \textbf{(C)}$.
[/hide]

Hope I helped![/quote]

Corrina, where you say "$ x \plus{} (x \minus{} 1) \plus{} 39$" I think you mean to say Then $ x \plus{} (x \minus{} 1) \equal{} 39$

Actually, $ x \equal{} 20$ corresponds with answer choice $ \boxed{\text{D}}$, not $ \text{C}$.

Let the cost of pink pill be x
and the cost of green pill be x+1
Then (2x+1)*14=546.
2x+1=39   => x=19
then cost of pink pill is x+1 = $20
Option (C) is correct

This particular question has been given a wrong solution in the book Introduction to Algebra by Richard Rusczyk.
Page 72
Problem 3.15


algebra
function
domain
superior algebra
superior algebra unsolved

You were asked before what you mean by prime and you didn't answer. So don't expect any responses...

i think he means that spec(R) is irreducible

edit: apparently not, http://en.wikipedia.org/wiki/Prime_ring

[quote="wiosna"]
Show that prime and reduced ring R is a domain.[/quote]

If $ xy \equal{} 0,$ for some $ x,y \in R,$ and $ a \in R,$ then $ (yax)^2 \equal{} ya(xy)ax \equal{} 0.$ Thus $ yax \equal{} 0$ because $ R$ is reduced. So $ yRx \equal{} (0)$ and hence $ x \equal{} 0$ or $ y \equal{} 0,$ because $ R$ is prime.

(The converse is also true, i.e. every domain is prime and reduced.)


a,b,c \geq 0 . Prove  that  a^4+b^4+c^4 \leq 2(a 2 b 2 + b 2 c 2 + c 2 a 2 ) ===> a 2 +b 2 +c 2   \leq 2(ab+bc+ca)
Thanks

Well, the first condition says that (a+b+c)(a+b-c)(b+c-a)(c+a-b)>=0. Thus, (a+b-c)(b+c-a)(c+a-b)>=0. We cannot have two of the numbers a+b-c, b+c-a, c+a-b negative (easy to prove). Thus, a+b-c>=0, b+c-a>=0 and c+a-b>=0. Multiply these by c,a,b and add them up.


induction
function
modular arithmetic
geometry
3D geometry
quadratics

Prove that 1 + n(n+1)(2n+1)/6 is a non-square for any positive integer n.

If $ 1\plus{}n(n\plus{}1)(2n\plus{}1)/6 \equal{} a^2$

I don't know what to do with this... perhaps analysis in a mod (I don't know which).  It might also help to note that $ \sum_{i\equal{}1}^n i^2 \equal{} n(n\plus{}1)(2n\plus{}1)/6$.  perhaps some ideas about Pythagorean triples could help, but I can't find a way to apply any of this right now.

Would this be some sort of induction?
f(1) = 1 + (2 + 3 + 1) /6
      = 4 + 1
      = 5
not a square which is not a square
If you expand it out, the function becomes:
f(n) = 1 + (2n^3 + 3n^2 + n) / 6

If you expand and simplify
f(n+1) = (n + 1)^2 + f(n)
Now I'm stuck.

What do you mean by using pythagorean triples?

the problem with induction is that f(n) is not a perfect square is not a strong enough statement for $ (n \plus{} 1)^2 \plus{} f(n)$ not to be a perfect square.  Suppose $ f(n) \equal{} 2n \plus{} 3$; then clearly $ (n \plus{} 1)^2 \plus{} f(n) \equal{} n^2 \plus{} 4n \plus{} 4 \equal{} (n \plus{} 2)^2$.  Of course, though, this assumption that $ f(n) \equal{} 2n \plus{} 3$ is ridiculous - $ f(n)$ is a cubic!

Perhaps this would be better:
Prove that $ n(n \plus{} 1)(2n \plus{} 1)/6 \minus{} a^2 \equal{} 1$ has no solutions.
$ (2n^3 \plus{} 3n^2 \plus{} n)/6 \minus{} 1 \equal{} a^2$

let $ n \equal{} m \minus{} 1/2$

$ (2m^3 \minus{}m/2)/6 \minus{} 1 \equal{} a^2$
$ 1/3m^3 \minus{}1/12m \minus{} 1 \equal{} a^2$
$ 1/27m^3 \minus{}1/108m \minus{} 1/9 \equal{} (a/3)^2$

let $ a/3 \equal{} y$, $ m/3 \equal{} n/3 \minus{} 1/6 \equal{} x$.  then
$ x^3 \minus{}1/36x \minus{} 1/9 \equal{} y^2$

which is an elliptic curve.  We must show that for $ x > \minus{} 1/6$, there are no rational points of the form $ (n/3 \minus{} 1/6, a/3)$, where $ n$ and $ a$ are integers

I do not know how to do this but I think it is doable.

It seems like this problem hasen't been fully solved so here it goes:

For any integer $ x$, $ x^2 \equiv 0\pmod{4}$ or $ x^2 \equiv 1\pmod{4}$, which means that $ \dfrac{(n)(n\plus{}1)(2n\plus{}1)}{6} \equiv 3\pmod{4}$ or $ \dfrac{(n)(n\plus{}1)(2n\plus{}1)}{6} \equiv 0\pmod{4}$.

So we do case work:
Case 1:
$ n\equiv 0\pmod{4}$, since $ n>0$ we can use $ 4$ to work with.
$ \dfrac{(4)(5)(9)}{6} \equiv 2\cdot 3\cdot 5\pmod{4}$
$ \dfrac{(4)(5)(9)}{6} \equiv 2\pmod{4}$

Since $ 2\not\equiv 3$ or $ 0\pmod{4} \Rightarrow n\not\equiv 0\pmod{4}$

Case 2:
$ n\equiv 1\pmod{4}$
$ \dfrac{(1)(2)(3)}{6} \equiv 1\pmod{4}$

Since $ 1\not\equiv 3$ or $ 0\pmod{4} \Rightarrow n\not\equiv 1\pmod{4}$

Case 3:
$ n\equiv 2\pmod{4}$
$ \dfrac{(2)(3)(5)}{6} \equiv 5 \equiv 1\pmod{4}$
Since $ 1\not\equiv 3$ or $ 0\pmod{4} \Rightarrow n\not\equiv 2\pmod{4}$

Case 4:
$ n\equiv 3\pmod{4}$
$ \dfrac{(3)(4)(7)}{6} \equiv 14\equiv 2\pmod{4}$
Since $ 2\not\equiv 3$ or $ 0\pmod{4} \Rightarrow n\not\equiv 3\pmod{4}$

And that shows that for all positive values of $ n$, $ 1\plus{}\dfrac{(n)(n\plus{}1)(2n\plus{}1)}{6}$ is never a perfect square.

Nice solution, but is there any specific reason for using $ \mod 4$ ?

Mod 4 is nice to work with for perfect squares... As is mod 8.

For cubes: mod 9, mod 13(!). Of course these are only recommendations and sometimes problems need different mods.

[quote="tonypr"]$ n\equiv 0\pmod{4}$, since $ n > 0$ we can use $ 4$ to work with.[/quote]  Unfortunately, no: because you have division by 2, you need to consider the possibilities of $ n$ modulo 8 to get all possibilities for your expression modulo 4, and there's a difference bewteen $ n \equiv 0 \pmod 8$ and $ n \equiv 4 \pmod 8$.  For example, for $ n \equiv 0 \pmod 8$, we do have $ 1 \plus{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \equiv 1 \pmod 4$.  Indeed, it's clear that by setting $ n \equal{} 6m$ we can get $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1 \equiv 1 \pmod m$ for any $ m$, potentially giving us a perfect square, so a naive modular attack isn't going to work on this one.

But for modulos that are not 4, non-squares do not take on the form $ n\equiv 1\pmod{m}$ for any integer m. Sorry, I still dont see a problem with the above proof. Can you find a couterexample?

tonypr's proof is wrong -- that doesn't (necessarily)* mean the conclusion is.  I showed that for any $ m$, we will sometimes have that $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1 \equiv 1 \pmod m$.  Since 1 is a quadratic residue in every mod, this shows that $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1$ [i]can[/i] be a square modulo $ m$ for every $ m$.  Thus, no argument that begins, "Consider the given expression modulo $ m$," can possibly conclude that $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1$ is not a perfect square.  In particular, I showed that tonypr was wrong when he claimed $ n \equiv 0 \pmod 4$ implies $ \frac{n(n \plus{} 1)(2n \plus{} 1)}{6} \equiv 2 \pmod 4$.

* However, it turns out that the conclusion is also wrong, as there are at least two counter-examples (which I've just found with a computer search).  The smallest is $ 1 \plus{} \sum_{i \equal{} 1}^{47} i^2 \equal{} 189^2$.  The next is $ 1 \plus{} \sum_{i \equal{} 1}^{48} i^2 \equal{} 195^2$.  These are the only "small" examples.

[quote="JBL"]tonypr's proof is wrong -- that doesn't (necessarily)* mean the conclusion is.  I showed that for any $ m$, we will sometimes have that $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1 \equiv 1 \pmod m$.  Since 1 is a quadratic residue in every mod, this shows that $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1$ [i]can[/i] be a square modulo $ m$ for every $ m$.  Thus, no argument that begins, "Consider the given expression modulo $ m$," can possibly conclude that $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 1$ is not a perfect square.  In particular, I showed that tonypr was wrong when he claimed $ n \equiv 0 \pmod 4$ implies $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \equiv 2 \pmod 4$.

* However, it turns out that the conclusion is also wrong, as there are at least two counter-examples (which I've just found with a computer search).  The smallest is $ 1 \plus{} \sum_{i \equal{} 1}^{47} i^2 \equal{} 189^2$.  The next is $ 1 \plus{} \sum_{i \equal{} 1}^{48} i^2 \equal{} 195^2$.  These are the only "small" examples.[/quote]
Oh very good, thank you!


$PB=\sqrt{(x-8)^{2}+y^{2}}$ , $PA=\sqrt{x^{2}+(y-6)^{2}}$ ve $x^{2}+y^{2}=4$

bunları kullanarak $PB=\sqrt{68-16x}$ ve $PA=\sqrt{40-12y}$' yi elde ederiz .   $y$ yerine de $\sqrt{4-x^{2}}$ yazarsak $PA+PB$ sadece x cinsinden olur .x 'e göre türevini alıp sıfıra eşitlediğinizde sonuca gidersiniz


Heres a really hard "MATHCOUNTS" problem that was in a practice book...I cant find an easy way to do it

The units digit of a six-digit number is removed,leaving a 5 digit number.
The removed digit is then placed in the front of the 5 digit number.
The new number is 1/3 of the origional.
What is the sum of the digits of the original number?

if its too easy, ill move it

As a fact, I know the result is 142857, it's a magical rotating number when multiplied by 2, 3, 4, 5, and 6. I don't see an easy way to do this either.

[hide]Let the first five digit be $A$ , last digit be $B$ then we have

$\frac{1}{3}(10A+B)=10^5B+A$  :arrow: $7A=(3\times 10^5-1)B=299999B$

 :arrow: $A=42857B$ . Since $A$ has five digit , $B$ can only be $1$ or $2$ .

When $B=1$ , the number is $428571$ so sum=$\boxed{27}$

When $B=2$ ,the number is $857142$ which has same sum=$\boxed{27}$[/hide]

$M=10A + B$. Removing the last digit and placing in front, we have

$N=10^5B + A$.

$N=\frac{1}{3} M \Rightarrow 3(10^5B + A) = 10A+B$.

$(3 \cdot 10^5-1)B=7A \Rightarrow 42857B = A \rightarrow A = 42857, B = 1$

So $428571$

EDIT: Darn, shyong beat me. :mad:

EDIT 2: The expansion of $1/7$ shows up in the weirdest places :?


Let p^3-1 divisible p and p-1 divisible q. Prove that p = q^2+q+1

how can $ p^3\minus{}1$ be divisible by $ p$?

you must have meant $ q^3\minus{}1$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=320949

[quote="homang77"]Let $ p^3\minus{}1$ divisible $ p$ and $ p\minus{}1$ divisible $ q$. Prove that $ p \equal{} q^2\plus{}q\plus{}1$[/quote]

I presume you mean "Let $ q^3\minus{}1$ is divisible by $ p$ and $ p\minus{}1$ is divisible by $ q$. Prove that $ p \equal{} q^2\plus{}q\plus{}1$."

Let $ p\equal{}qk\plus{}1$. $ q^3\minus{}1\equal{}(q^2\plus{}q\plus{}1)(q\minus{}1)$, but $ p$ divides it and $ p\ge q\plus{}1> q$, so $ p$ divides $ q^2\plus{}q\plus{}1$, i.e. $ qk\plus{}1$ divides $ q^2\plus{}q\plus{}1$. Suppose $ qk\plus{}1\ne q^2\plus{}q\plus{}1$, we let that $ \ell(qk\plus{}1)\equal{}q^2\plus{}q\plus{}1$, where $ \ell>1$. By taking modulo $ q$, $ \ell\equiv 1\bmod{q}$, so $ \ell\ge q\plus{}1$. $ q^2\plus{}q\plus{}1\equal{}\ell(qk\plus{}1)\equal{}(q\plus{}1)(qk\plus{}1)\equal{}kq^2\plus{}(k\plus{}1)q\plus{}1>q^2\plus{}q\plus{}1$, leading to a contradiction. So $ p\equal{}q^2\plus{}q\plus{}1$.


Really need a solution to this question, been bugging me all day, hoping someone here can help me out. Can't figure out what the hell is going on with my answers as I hate these types of questions, so please, anyone have a crack. Cheers
------------------------------------

x+y+z = 4
x^2+y^2+z^2=10
x^3+y^3+z^3=22
x^4+y^4+z^4= ?

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

Hint: Every symmetric polynomial can be written in terms of the elementary symmetrical polynomials. For three variables they are $\sigma_{1}=x+y+z$, $\sigma_{2}=xy+yz+zx$ and $\sigma_{3}=xyz$.
Thus write them using only those terms and find out the values of $x+y+z$ (given), $xy+yz+zx$ and $xyz$ by this.
Then set it into the term for $x^{4}+y^{4}+z^{4}$.

[quote="ZetaX"]Hint: Every symmetric polynomial can be written in terms of the elementary symmetrical polynomials. For three variables they are $\sigma_{1}=x+y+z$, $\sigma_{2}=xy+yz+zx$ and $\sigma_{3}=xyz$.
[/quote]

ZetaX is right. And to create this formulas just compute $(a+b+c)^{i}$ for $i\in\{2,3,4\}$ and you will get what you want.

[quote="Rust"]Let $s_{i}=x^{i}+y^{i}+z^{i}$. Then $s_{4}=\frac{1}{6}s_{1}^{4}-s_{1}^{2}s_{2}+\frac{1}{2}s_{2}^{2}+\frac{4}{3}s_{1}s_{3}.$[/quote]

nice, but I bet you do not remember that.. the point is to remember so few things as possible..

There is a formula, decide yourself if you really want to remember ;)

$\sum_{i=1}^{k}x_{i}^{n}=n \cdot \sum_{l_{1}+2 \cdot l_{2}+...+k \cdot l_{k}=n}\frac{(-1)^{n+l_{1}+l_{2}+...+l_{k}}}{l_{1}+l_{2}+...+l_{k}}\cdot (l_{1},l_{2},...,l_{k})! \cdot (\sigma_{1}^{l_{1}}\cdot \sigma_{2}^{l_{2}}\cdot ... \cdot \sigma_{k}^{l_{k}})$

With $\sigma_{i}$ being again the elementary symmetrical polynomials (in $x_{1},x_{2},...,x_{k}$) and $(l_{1},l_{2},...,l_{k})!= \frac{(l_{1}+l_{2}+...+l_{k})!}{l_{1}! \cdot l_{2}! \cdot ... \cdot l_{k}!}$ being a multinomial coefficient.


limit
function
calculus
calculus computations

(1) A sequence $ \{a_n\}$ decreases monotonously and  coverges to the costant $ \alpha$. Under the condition,

supposed it is known that there exists $ \lim_{n\to\infty} \frac {1}{n}\sum_{k \equal{} 1}^n a_k$,

prove that $ \lim_{n\to\infty} \frac {1}{n}\sum_{k \equal{} 1}^n a_k\leq a_{N}$ for any positive integers $ N$.

(2) For a continuous function $ f(x)$, define a function $ g(x)$ by $ f(x \plus{} 1) \minus{} f(x) \equal{} g(x)$.

If $ g(x)$ decreases monotonously and $ \lim_{x\rightarrow \infty} g(x) \equal{} 0$, then find $ \lim_{x\rightarrow \infty} \frac {f(x)}{x}$.

Note that you may use the fact $ |f(x)|$ has maximum value in the arbiterary intervals.

Note: if $ \lim_{k\to\infty}a_k\equal{}\alpha,$ then $ \lim_{n\to\infty}\frac1n\sum_{k\equal{}1}^na_k\equal{}\alpha.$

You don't need monotonicity for that claim. For more, try searching on this site for "Cesaro" and seeing how many entries come up.

I would like to see a proof without using Cesaro, epsilon-delta for those who don't study them.

Did you say without using epsilon-delta arguments in the proof?

I wonder how the concept of limit is defined in Japan.  :huh:

One can prove (1) without showing the limit but the proof is quite trivial using the squeeze theorem. The interesting part is 2 and how it follows from 1.


1)Decidir si es posible ordenar los números enteros desde 1 hasta 2004 inclusive de manera tal que la suma de 10 números consecutivos sea siempre múltiplo de 10.
Si la respuesta es sí, decir cuál es ese orden. Si la respuesta es no, explicar el porqué.
2)Se tiene una bolsa con 111 bolitas de cuatro colores, verde, rojo, blanco o azul (cada bolita tiene un solo color). Si se sacan de la bolsa 100 bolitas, al azar, siempre hay entre ellas una de cada color. Determinar el número mínimo de bolitas que hay que sacar de la bolsa (sin espiar) para estar seguro de sacar al menos tres de tres colores diferentes.
3)Alan tiene que representar al 100 como suma de uno o más enteros positivos o bien todos iguales o bien tales que la diferencia entre el mayor y el menor sea 1. Determinar de cuántas maneras puede hacerlo.
ACLARACIÓN: Dos representaciones que usan los mismos números pero en otro orden se consideran una sola manera
4)Determinar para qué enteros n es posible ordenar los números enteros desde 1 hasta n inclusive de manera tal que el promedio de todo grupo de dos o más números consecutivos no sea nunca entero.

5)Alan y Lucía, que viven en la misma calle, salen en el mismo instante, cada uno de su casa hacia la del otro, caminando a velocidades constantes (pero no necesariamente iguales) hasta que se encuentran en un punto C. Si Alan hubiese salido 30 minutos antes, se habrían encontrado 2 km más cerca de la casa de Lucía que el punto C. Si, en cambio, Lucía hubiese salido 30 minutos antes, se habrían encontrado a cierta distancia x más cerca de la casa de Alan que el punto C. Calcular el valor de x.

6)En una gran ciudad, todas las calles son de doble mano, pero tienen solo dos direcciones posibles (Sur Norte o Este Oeste). Durante un paseo en auto por la ciudad, no se ha pasado dos veces por un mismo lugar y se regresó al punto de partida. Se realizaron 100 giros hacia la izquierda. Determinar cuántos giros a la derecha se deben haber hecho. (Dar todas las posibilidades.)

7)Se escriben los enteros positivos, uno a continuación del otro, sin separaciones, en una cinta infinita: 1234567891011121314.... A continuación la cinta se corta en tiras de siete dígitos cada una. Demostrar que cada entero de siete dígitos aparece en al menos una tira.

8)En el mismo plano coordenado se han graficado cuatro funciones de la forma y=x2+ax+b. Estos gráficos tienen en total exactamente cuatro puntos de intersección, y en cada punto de intersección se cortan exactamente dos de esos gráficos. Consideramos la coordenada x de cada punto de intersección. Demostrar que la suma del mayor y el menor es igual a la suma de los otros dos.

¿José, dónde encontraste estes problemas?

ellos son dificiles.

De una competencia argentina (1º y 2º etapa) que se desarrolló en 2005.......

aja .. eso es cierto .. ya los habia revisado

Pero no parecen muy dificiles ...

Saludos,

Carlos Bravo ;)
Lima - PERU

Vamos gente con ánimo!!!! Alguna sugerencia?????

Problema 1:

Dividamos el conjunto [1,2004] en clases residuales modulo 9

0,10,20,30,....,2000
1,11,21,31,....,2001
2,12,22,32,....,2002
3,13,23,33,....,2003
4,14,24,34,....,2004
5,15,25,35,....,2005
6,16,26,36,....,2006
7,17,27,37,....,2007
8,18,28,38,....,2008
9,19,29,39,....,2009

Para que un conjunto se pueda ordenar de tal manera que la suma de 10 elementos consecutivos sea multiplo de 10 este ha de tener el mismo numero de elementos en cada una de sus clases residuales modulo 10, en conjunto [1,2004] no cumple dicha condicion por lo tanto no puede ser ordenado de dicha manera.

Saludos

Me gustaria saber si mi solucion les parece correcta, ya que soy nuevo en esto. Saludos


Two members of an increasing arithmetic sequence are 1 and 27. The sequence has less than 40 terms and its median is 38. What is the value of the fourth term following 27.

Since $ 27\minus{}1\equal{}26\equal{}2\cdot13$, we see that the common difference of the sequence can be $ 1,2,13,26$.

It can't be $ 1$, since in order to have a median of $ 38$, it would have to be $ 1,2,3,\ldots,75$, which has $ 75$ terms.

It can be $ 2$: the sequence is $ 1,3,5,\ldots,37,39,\ldots,73,75$, with $ 38$ terms and median $ \frac{37\plus{}39}2\equal{}38$.

It can't be $ 13$, since that would result in $ 1,14,27,40,\ldots$, which clearly can't have a median of $ 38$.

It can't be $ 26$, because the sequence would be $ 1,27,53,\ldots$, which can't have a median of $ 38$ either.

Thus, the only possible sequence is $ 1,3,5,\ldots,25,27,29,31,33,\boxed{35},\ldots,73,75$.


I've just come to know that IIT-JEE 2010 is being prepared by IIT-Madras :o
Can anybody tell me anything about what kind of papers do they make?
I've heard that they are too good at chemistry and make good questions in it and also they don't leave the maths paper easy too.Is it true?
so overall the situation is dreadly as the subject I hate the most is expected to be the most difficult in IIT-JEE 2010. :(

If what you say is true, I'm in a mess as well :(
Chem SUCKS!

I have some suggestion for those guys preparing for IITJEE chemistry

1)DO NOT NEGLECT NCERT chemistry books : I know what iam saying may sound trivial but infact it is very essential towards ur jee preparation.
Most of the questions(esp INORGANIC), if u observe for past few years are from NCERT textbook - either the 11th standard book or 12th standard book.

For Inorganic chemistry,Use
a) The ncert chemistry 11th standard book vol 2 - I think it has Hydrogen,alkali,alkaline and early p block
b) The ncert chemistry 12th standard book vol 1 - It has the late p block,d and f block and coordination and stuff like that.

These two sources are according to current trend of jee chemistry and from what i asked other people are sufficient - so be sure to master these b4 going in for anything else.
I am a first hand example of the regrets one must face for neglecting the 12th standard chem textbook esp p-block : - I got screwed in jee chemistry and lost marks substantially in this area.

[b]@Ram:[/b] How did you manage to balance your preparation for IIT-JEE as well as the Olympiads? Could you give some advice regarding preparation for Maths Olympiad(like how to go about preparing each topic)? KK sir wants us to spend most of our time for JEE prep.(I'm a KK sir student, class XII) but I am interested mainly only in Maths (and Organic Chem. somewhat) Is it OK to spend more time on Olympiad Maths than JEE prep.?

@key..: Jee math is a good foundation for improving problem solving skills.
Attend sir's classes on math [b]properly[/b] and listen in class(that itself is good base for olympiad level math)

U can do olympiad math if u like provided u are [b]confident[/b] of tackling jee problems.However do not do this neglecting jee prep for physics and chem for iitjee is ur primary focus.(correct me if iam wrong)

just forget all such stuff..
i planned to do best in physics and i did worst in it....
i was hoping for 30% in math and got a lot more..was hoping for 80% in physics
and got a lot lot less..so all planning fails..just work hard..
be consistent and keep working
i regret giving up at the start of my 12th
got 1108 air
hope none of u end up like me..all the best

the piece of advise i can give u is , donot neglect NCERT chem  :o atleast for inorganic chemistry. do math well., do not screw up in chem as a lot of us did this yr. , i think the rest shud be fine  :D
cheers!!

Check out my guide at goiit. 100% success guaranteed. no hardwork , no intelligence. just circle bubbles ftw. :D

[hide]http://www.goiit.com/posts/list/community-shelf-my-jee-story-944601.htm[/hide]


Let $ S$ be an infinite set of real numbers and |$ x_1\plus{}x_2\plus{}...x_n$|<=$ 1$ for every finite subset {$ x_1,x_2...,x_n$} of $ S$. Prove that $ S$ is countable

Let $ S_k\equal{}\{x\in S: x>\frac1k\}.$ The claim is that $ S_k$ must be a finite set - in fact, the cardinality of $ S_k$ must be less than $ k,$ since the sum of any $ k$ elements of $ S_k$ would be greater than $ 1.$ Then $ S^\plus{}\equal{}\{x\in S: x>0\}\equal{}\bigcup_{k\equal{}1}^{\infty}S_k.$ As the union of countably many finite sets, $ S^\plus{}$ is countable.

Now look at $ T_k\equal{}\{x\in S: x<\minus{}\frac1k\}.$ Since the sum of any $ k$ elements of $ T_k$ would be less than $ \minus{}1,$ we must have that the cardinality of $ T_k$ is less than $ k.$ Then $ S^\minus{}\equal{}\{x\in S: x<0\}\equal{}\bigcup_{k\equal{}1}^{\infty}T_k$ is countable union of finite sets, hence countable.

Either $ S\equal{}S^\plus{}\cup S^\minus{}$ or $ S\equal{}S^\plus{}\cup S^\minus{}\cup\{0\}.$ Either way, $ S$ is a union of finitely many countable sets and hence countable.

Bolzano-Weierstrass is a slick way to prove this.

What if we specified that the finite subsets referenced in the problem also have minimum cardinality k?

[quote="t0rajir0u"]$ S$ can have no limit points except $ 0$.  Suppose otherwise; then there exists $ L \neq 0$ and an infinite number of elements $ x_k$ of $ S$ such that $ |x_k \minus{} L| < \frac {L}{2}$, whose sum is unbounded; contradiction.  (This is essentially a topological restatement of Kent Merryfield's proof - both are fundamentally based on the Archimedean property.)

It follows that every element of $ S$ is isolated with the possible exception of $ 0$, so it is possible to associate to every nonzero element $ s \in S$ a rational number $ r_s$ less than $ s$ with the property that $ (s, r_s)$ contains no element of $ S$.  Countability follows.

(Bolzano-Weierstrass appears only to show that $ S$ is either finite or has $ 0$ as a limit point, which is not actually necessary to prove the result, unless I'm misinterpreting what proof SouthStar is referring to.)[/quote]

That is what I had in mind; I didn't mean (s)he could just quote BW as an answer. Of course it's not necessary to prove that S has 0 as its sole limit point; that's why I said this is a "slick" proof. :)

[quote="RED-and-black"]What do you mean when you say "finite subsets referenced in the problem have minimum cardinality k". Can you also write shortly the proof with Bolzano-Weierstrass. Thanks in advance[/quote]

Instead of "for every finite subset {$ x_1,x_2...,x_n$} of $ S$", what if you had "for every finite subset {$ x_1,x_2...,x_n$} of $ S$ [i]such that n>=k"? where k is an arbitrary natural number?


algebra
polynomial
inequalities proposed
inequalities

Dear all

Can anyone solve the attahced proplem?

Thanks

Noting that p(t) is polynomial of degree one


Time limit of 3 minutes for these three problems:

1. Subtract and simplify the following expression to a single fraction.
$\frac{5}{2x-2}-\frac{2x-1}{2-2x}$

2. Working together, Kate and Jean can close the restaurant in  $\frac{2}{3}$ of an hour.Working alone, Jean can do the job in 1 hour less time than Kate, when she works  alone.  How long does it take Jean to do the job alone? Write your answer in  terms of hours.

3. If a, b, c, and d are all ≥ 0 and  $a+c=b$, $a+d=c$,$b-d=2$, and $b+c-d=3$, determine the sum of $(a + b + c + d)$.

[hide]1. Change sign of 2nd fraction and get $\frac{x+2}{x-1}$

2. $\frac1t+\frac1{t+1}=\frac1{\frac23}\Rightarrow t=1$

3. $a+c=b,a-c=-d\Rightarrow 2a=b-d=2\Rightarrow a=1$
$b+c-d=c+2=3\Rightarrow c=1$
$a+d=c\Rightarrow d=0$
$a+b+c+d=2a+2c=4$[/hide]

Time=2m18s

[hide]1) (x+2)/(x-1)

2) jean can close the restaurant in 1 hr

3) a=1
b=2
c=1
d=0

sum is 4[/hide]
just slightly under 3 mins.

is there a faster way to do problem 3?


Let $ a,b,c>0$ satisfy $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$≤$ 3$.
Prove that $ \sum_{cyc}\frac{a^2\plus{}1}{\sqrt{a^2\minus{}a\plus{}1}}$≥$ 6$

we have by Am-Gm,
$ \frac{a^2\plus{}1}{\sqrt{a^2\minus{}a\plus{}1}}\geq 2\sqrt{a}$ because $ \frac{a^2\plus{}1}{\sqrt{a^2\minus{}a\plus{}1}}\equal{}\frac{a}{\sqrt{a^2\minus{}a\plus{}1}}\plus{}\sqrt{a^2\minus{}a\plus{}1}$
then it's suffice to prove that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\geq 3$
which is true because from Holder $ 27\leq \left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2\leq 3(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2$

$ \frac {a^2 \plus{} 1}{\sqrt {a^2 \minus{} a \plus{} 1}} \plus{} \frac {1}{a} \equal{} \frac {a}{\sqrt {a^2 \minus{} a \plus{} 1}} \plus{} \sqrt {a^2 \minus{} a \plus{} 1} \plus{} \frac {1}{a}\geq3$ by AM-GM.
Thus, $ \sum_{cyc}\left(\frac {a^2 \plus{} 1}{\sqrt {a^2 \minus{} a \plus{} 1}} \minus{} 2\right)\geq3 \minus{} \frac {1}{a} \minus{} \frac {1}{b} \minus{} \frac {1}{c}\geq0.$
More way:
$ \frac {a^2 \plus{} 1}{\sqrt {a^2 \minus{} a \plus{} 1}} \plus{} \frac {1}{a}\equal{}\frac {a^2 \plus{} 1}{2\sqrt {a^2 \minus{} a \plus{} 1}} \plus{} \frac {a^2 \plus{} 1}{2\sqrt {a^2 \minus{} a \plus{} 1}}\plus{}\frac {1}{a}\geq$
$ \geq3\sqrt[3]{\frac{(a^2\plus{}1)^2}{4a(a^2\minus{}a\plus{}1)}}\geq3,$ where the last inequality is equivalent to $ (a\minus{}1)^4\geq0.$


[quote="qweretyq"]The polynomial x^6 - 6x^5 + 15x^4 - ax^3 + 15x^2 - bx + 1 = 0
has six positive roots. Compute a and b.[/quote]
[hide]
Coefficients are $\binom{6}{0}, \binom{6}{1}...\binom{6}{6}$
$a=20$ $b=6$[/hide]


geometry
circumcircle
parallelogram
Spiral Similarity
geometry solved
USA TST
Fixed point

[quote="orl"]Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG = \angle CAX$.[/quote]
An ugly solution  :) .
The first I will display two lemmas .
Lemma 1 : A,B is two fixed point on $ O_x,O_y$
Let M,N varies on $ O_x,O_y$ and satisfy :
$ \frac{OA}{OM}+\frac{OB}{ON}=1$
Then the line MN pass through a fixed point X such that $ OAXB$ is a  parallelogram.
Proof : Have some solutions for this problem . A simple way is draw $ A_t\parallel{}O_y$ ,it cuts $ PQ$ at the point X .
Apply Thales law we have : $ BX\parallel{}O_x$ ,therefore X is a fixed point .
Problem claim .
Lemma 2 : A,B is fix points on $ O_x,O_y$
Let M,N varies on $ O_x,O_y$ and satisfy $ \frac{OM}{OA}+\frac{ON}{OB}=1$ then the circles $ (OMN)$ pass through a fixed point T such that $ OT,OX$ conjugate with $ \angle{xOy}$
Proof : Invert pole O with $ k=OA.OB$ .Then we have :
$ A\to A',B\to B' ,M\to M',N\to N'$
Then $ (OMN)\to M'N'$
It becomes lemma 1 .
The lines $ M'N'$ pass through the  point X' which $ OA'X'B'$ is a parallelogram  .
Therefore $ (OMN)$ pass through X  that  is the imagine of X' .
Because $ OA=OA',OB=OB'$ so that $ OT,OX$ conjugate with $ \angle{xOy}$
Now apply to our problem .
Easy to check that $ \frac{AR}{AB}+\frac{AP}{AB}=1$ therefore the circle pass through fixed point X .
Morethan $ OX,OG$ conjugate with $ \angle{BAC}$ so we have : $ \angle{BAG}=\angle{CAX}$

Dear Tung, your proof is not ugly at all. It is really nice.  :clap:
In addtion, the fixed point X lies on segment HK where H is the orthocenter and K is the midpoint of median AM.

let $ \alpha \equal{} \angle BAG$, and let $ \frac {BP}{PC} \equal{} \frac {x}{y}$... let $ l$ be the line through $ A$ such that $ l$ and line $ AG$ are isogonal lines, and let it intersect the circumcircle of triangle $ AQR$ at $ T$...

from ptolemy's theorem we have that $ AQ\dfrac{\sin(A \minus{} \alpha)}{\sin A} \plus{} AR\dfrac{\sin\alpha}{\sin A} \equal{} AT$...

it's easy to obtain that $ \dfrac{\sin(A \minus{} \alpha)}{\sin A} \equal{} \dfrac{c}{2m_a}$ and $ AQ \equal{} \dfrac{bx}{x \plus{} y}$...

from these it's easy to conclude that $ \dfrac{bc}{2m_a} \equal{} AT$, which is fixed, and we're done  :D

[quote="orl"]Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG \equal{} \angle CAX$.[/quote]
[hide="Solution"]
We consider the point $ X$ so that $ \triangle XAC\sim \triangle XBA$.  From the lemma I proved here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1185976#1185976, we have that $ X$ lies on the $ A$-symmedian.  Now, since $ PQ\parallel AB$ and $ PR\parallel AC$, we have that $ \frac {CQ}{QA} \equal{} \frac {CP}{PB} \equal{} \frac {RA}{RB}$, so $ \frac {RA}{AB} \equal{} \frac {QC}{AC}$.  This means that since $ \triangle XAC\sim \triangle XBA$, we have that $ \triangle XRA\sim \triangle XQC$, so $ \angle XQC \equal{} \angle XRA$, which means that $ XRAQ$ is cyclic.  Then, all of the circumcircles of $ \triangle ARQ$ pass through the point, $ X$, which is fixed, and lies on the $ A$-symmedian, as desired. [/hide]

Just an outline.First,consider the midpoints of $AB$ and $AC$,$M$ and $N$ and let $X$ be the intersection of the circumcircle of $ABC$ with the symmedian of $ABC$.Now,it is easy to prove $XM/XN=AB/AC$ and now from this we just need to prove that for any two points $Q$ and $R$ with the point $P$ that $XMQ$ is similar to $XNR$,but this is an easy lenght chase,so we are finished.

We can rewrite the problem as following:

Let $ P_1, P_2, P_3 $ be three points on $ BC $ .
Let $ Q_1, Q_2, Q_3 $ be three points on $ AC $ satisfy $ P_1Q_1\parallel P_2Q_2\parallel P_3Q_3\parallel AB $ .
Let $ R_1, R_2, R_3 $ be three points on $ AB $ satisfy $ P_1R_1\parallel P_2R_2\parallel P_3R_3\parallel AC $ .

Then $ \odot (AQ_1R_1), \odot (AQ_2R_2), \odot (AQ_3R_3) $ are coaxial with common radical axis is A  -symmedian .

Proof:

Let $X \equiv  \odot (AQ_1R_1) \cap \odot (AQ_3R_3)  $ and $ Y \equiv AX \cap \odot (ABC) $ .
(i.e. $ X $ is the Miquel point which maps $ Q_1Q_3 $ to $ R_1R_3 $ )

Since $ \frac {Q_1Q_2}{Q_2Q_3}= \frac {P_1P_2}{P_2P_3}=\frac {R_1R_2}{R_2R_3} $ ,
so $ X $ is also the Miquel point maps $ Q_1Q_2 $ to $ R_1R_2 $ .
ie. $ X\equiv \odot (AQ_1R_1) \cap \odot (AQ_2R_2) \cap \odot (AQ_3R_3) $

Consider the case when $ P \equiv B $ and $ P \equiv C $ ,
then we get $ X $ is the Miquel point which map $ CA $ to $ AB $ .
i.e. $ X $ is a point satisfy $ \triangle XCA \sim \triangle XAB $

Since $ \angle XAC=\angle XBA=\angle YBC , \angle XCA=\angle XAB=\angle YCB $ ,
so we get $ \triangle XAC \sim \triangle XBA \sim \triangle YBC $ ,
hence $ AB\cdot CY=AX\cdot BC=AC\cdot BY $ .
i.e. $ ABYC $ is a harmonic quadrilateral and $ AX $ is A-symmedian

Q.E.D
____________________________________________________________
[b]Remark:[/b]

(1)
Since $ XY $ bisect $ \angle BXC $ ,
so from the property of harmonic quadrilateral we get $ X $ is the midpoint of $ AY $ .

(2)
This problem has something to do with Arzt parabola
Since $ QR $ is tangent to $ A $ -Arzt parabola when $ P $ varies on $ BC $ ,
so $ \odot (APQ) $ passes through the focus $ X $ of A -Arzt parabola .

Yes, it does Telv. Note that $QR$ is simply tangent to a fixed parabola (affine homography, so tangent at line at infinity). Hence, $APQ$ passes through a fixed point. Taking $P = B, C$, we have the point is the intersection of the circles through $B$ and $A$ tangent to $AC$ and $C$ and $A$ tangent to $AB$. We know this intersection lies on $A$-Symmedian.

Let the $A$-symmedian cut $\odot{AQR}$ at $X$ and $BC$ at $M'$ and $AM$ be the median.Then $\frac{RX}{XQ}=\frac{sin\angle{RAX}}{sin\angle{XAQ}}=\frac{sin\angle{MAC}}{sin\angle{MAB}}=\frac{AB}{AC}=\frac{BR}{AQ}$ where the last line follows from parallelity.This along with $\angle{BRX}=\angle{AQX} \implies \triangle{BRX} \sim \triangle{AQX} \implies \angle{AYB}=180-A$.
The equality may also be written as $\frac{RX}{XQ}=\frac{AR}{CQ}$ and this in combination with $\angle{ARX}=\angle{XQC}$ implies that $\triangle{ARX} \sim \triangle{CQX}$ so once again $\angle{AYC}=180-A$.Thus we get $\angle{BXC}=2A$ or $X$ is actually the point of intersection of $\odot{BOC}$ with the $A$-symmedian,where $O$ is the circumcenter of $\triangle{ABC}$.This is indeed fixed,as desired.

I've changed the notations a bit- $D$ is a variable point on $BC$ and $P, Q$ lie on $AB, AC$ respectively such that $DP \parallel AC, DQ \parallel AB$.

Let the tangents to $B$ and $C$ intersect at $U$. By letting $D$ be the midpoint of $BC$, we see that $X$ must be the other point of intersection of the $A$-symmedian and the circle with diameter $AO$, where $O$ is the circumcenter of $\triangle ABC$. Let $T$ be the said point. Note that $T$ also lies on the circumcircle of $\triangle BOC$ as $OT \perp TU$. Now,
\[\angle TCB = \angle TUB =  180^{\circ} - \angle ABU - \angle BAU \\ = 180^{\circ} - \angle BAC - \angle ABC  - \angle BAU = \angle ACB - \angle BAU,\]
so $\angle BAT = \angle TCA$. Similarly, $\angle TBA = \angle TAC,$ so $\triangle TBA \sim \triangle TAC \implies \dfrac{AB}{AC} = \dfrac{AT}{CT}$. Now, we have that
\[\dfrac{BD}{BC} = \dfrac{BP}{AB} \quad \dfrac{CD}{BC} = \dfrac{CQ}{AC} \\
\implies \dfrac{BP}{AB} + \dfrac{CQ}{AC}= 1 \implies \dfrac{AP}{CQ} = \dfrac{AB}{BC} = \dfrac{AT}{TC} \\
\implies \triangle APT \sim \triangle CTQ \implies \angle APT = \angle TQC = 180^{\circ} - \angle AQT,\]
so $T$ lies on $(APQ)$, as desired.

Here's a solution with a combination of synthetic geometry and complex numbers.

Note that $AQ \parallel PR$ and $AR \parallel PQ \implies AQPR$ is a parallelogram. Furthermore, from $AC \parallel RP$, it follows that \[\triangle BAC \sim \triangle BRP \implies \frac{AB}{AC} = \frac{RB}{RP} = \frac{RB}{AQ} = \frac{AB - AR}{AQ} \implies AB \cdot AQ + AC \cdot AR = AB \cdot AC.\] Now, let $A$ be the origin of the complex plane, and let us denote in lowercase letters the complex numbers corresponding to points. The above relation can then be rewritten as $|b||q| + |c||r| = |b||c|.$ But note that all three terms in this equation have equal arguments; hence, we may remove the magnitudes to obtain $bq + cr = bc.$ Now, we will prove that $x = \tfrac{bc}{b + c}$ is the fixed point. First, we show that $\angle GAB = \angle CAX.$ This is equivalent to $\tfrac{g}{b} / \tfrac{c}{x} \in \mathbb{R}.$ Note that $g = \tfrac{b + c}{3}$, so that \[\frac{g}{b} \bigg/ \frac{c}{x} = \frac{\frac{b + c}{3}}{b} \bigg/ \frac{c}{\frac{bc}{b + c}} = \frac{b + c}{3b} \bigg/ \frac{b + c}{b} \in \mathbb{R},\] as desired. Now, in order to prove that $A, Q, R, X$ are concyclic, we need only prove that $\tfrac{q}{r} / \tfrac{q - x}{r - x} \in \mathbb{R}.$ Using the condition $bq + cr = bc$, we find that \[\frac{q - x}{r - x} = \frac{q - \frac{bc}{b + c}}{r - \frac{bc}{b + c}} = \frac{(b + c)q - bc}{(b + c)r - bc} = \frac{cq + (bq - bc)}{br + (cr - bc)} = \frac{cq - cr}{br - bq} = -\frac{c}{b}.\] Then since $A, R, B$ and $A, Q, C$ are collinear, it follows that \[\frac{q}{r} \bigg/ \frac{q - x}{r - x} = -\frac{q / b}{r / c} \in\mathbb{R},\] as desired. $\square$

Let $D$ and $T$ be the intersections of the $A$-symmedian with $(ABC)$ and $(ARQ)$. I claim that $T$ is the midpoint of $AD$. Consider the inversion $\mathcal{I}(A,1)$ and denote by $X^\prime$ the inverse of $X$. Obviously, $A-D^\prime-T^\prime,\ R^\prime-T^\prime-Q^\prime$ and $B^\prime-D^\prime-C^\prime$ are collinear. Note that $\dfrac{D^\prime B^\prime}{D^\prime C^\prime}=\dfrac{\frac{DB}{AD\cdot AB} }{\frac{DC}{AD\cdot AC}}=\dfrac{DB}{DC}\cdot \dfrac{AC}{AB}=1$, so $D^\prime$ is the midpoint of $B^\prime C^\prime$. From $\dfrac{AR}{RB}=\dfrac{QC}{QA}$ it is easy to infer that $\dfrac{AB^\prime}{B^\prime R^\prime}=\dfrac{C^\prime Q^\prime}{C^\prime A}$. Let $S$ be a point on $Q^\prime R^\prime$ such that $B^\prime S\parallel AQ^\prime$. We have then $\dfrac{SR^\prime}{SQ^\prime}=\dfrac{B^\prime R^\prime}{B^\prime A}=\dfrac{C^\prime A}{C^\prime Q^\prime}$, so $SC^\prime \parallel AR^\prime$. Therefore, $AB^\prime SC^\prime$ is a parallelogram, which implies $A-D^\prime-S$ collinear, so $S=T^\prime$. This implies $$AT^\prime=2AD^\prime\Leftrightarrow AD=2AT\Leftrightarrow (ARQ)\ \mathrm{passes\ through\ the\ midpoint\ of}\ AD$$

My solution: :)

Let $\omega_1$ be a circle passes $A, B$ and tangents to $AC$. Let $\omega_2$ be a circle passes $A, C$ and tangents to $AB$. Let $X$ be intersection of $\omega_1$ and $\omega_2$, so $X$ is fixed. Since two triangle $ABX$ and $AXC$ are similar to each other, hence $\dfrac{AX}{BX}=\dfrac{AC}{AB}$. On the other hand, $\dfrac{AC}{AB}=\dfrac{PR}{BR}=\dfrac{AQ}{BR}$ so $\dfrac{AX}{BX}=\dfrac{AQ}{BR}$, also $\widehat{RBX}=\widehat{QAX}$ hence two triangle $BRX$ and $AQX$ are tangent to each other, therefore $\widehat{BRX}=\widehat{AQX}$, following that $A, Q, X, R$ is cyclic. Moreover, we have: $$\dfrac{\sin BAX}{\sin CAX}=\dfrac{\sin BAX}{\sin ABX}=\dfrac{BX}{AX}=\dfrac{AB}{AC}$$
So $AX$ is the symmedian line of triangle $ABC$, which means $\widehat{BAG}=\widehat{CAX}$.
We have done.

[hide=Barycentric solution]
We use Barycentric coordinates wrt $\triangle$ ABC. $BC = a, CA = b, AB = c, A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$.

Let point $P$ be $(0:d:a-d)$.

Since $ARPQ$ is parallelogram, by Parallelogram lemma, we have, $Q = (d:0:a-d)$ and $R = (a-d:d:0)$.

The general equation of circle is $a^2yz + b^2zx + c^2xy - (x+y+z)(px+qy+rz) = 0$.

Substitute coordinates of point $A, Q$ and $R$ into the equation, we obtain $p = 0, r = \frac{b^2d}{a}$ and $q = \frac{c^2(a-d)}{a}$ respectively.

So the equation of $(AQR)$ is $a^2yz + b^2zx + c^2xy - (x+y+z)( \frac{c^2(a-d)}{a} y+\frac{b^2d}{a} z) = 0$.

We can see that $d$ will be cancelled off if $c^2y = b^2z$, i.e., the coordinates of fixed point, $X$ has $\frac {y}{z} = \frac {b^2}{c^2}$.

Since the coordinates of the Symmedian point is $(a^2:b^2:c^2)$, so point $X$ lies on A-symmedian line.

Thus, we have $\angle BAG = \angle XAC$ as required. $\blacksquare$

[hide=Note]
If we substitute $ z = \frac{c^2}{b^2}y$ into the equation of $(AQR)$, we will get $\frac{x}{y} = \frac{b^2+c^2-a^2}{b^2}$.

Hence, the coordinates of the fixed point, $X$ is $(b^2+c^2-a^2:b^2:c^2)$. [/hide] [/hide]

[quote=Reynan]@fvlfollower why $d$ has to cancel out?[/quote]

$P= (0:d:a-d)$, so locus of point $P$ on side $BC$ depends on $d$ and this makes points $Q, R$, and the equation of $(AQR)$ contain $d$ as well.

$(AQR)$ passes through a fixed point implies that there is a point on $(AQR)$  which its coordinates do not contain $d$.

Hmm...Let $E$, $F$, and $M$ by the midpoints of $AB$, $AC$, and $BC$, respectively. Let the $A$-symmedian intersect the circumcircle again at $D$. Let $(AEF)$ intersect the $A$-symmedian at point $X$. It is easy to see that $X$ is the midpoint of $AD$ due to the homothety $AEF$ to $ABC$. It suffices to show that $ARXQ$ is cyclic. Also note that $ARPQ$ is a parallelogram.

[i]Lemma 1[/i]: $X$ is the center of the spiral similarity that takes $BA$ to $AC$
[i]Proof[/i]: It is easy to see that $\bigtriangleup ABD$ and $\bigtriangleup AMC$ are similar. Since $X$ and $F$ are midpoints of $AD$ and $AC$, respectively, $\bigtriangleup ABX \sim \bigtriangleup AMF$. Then, $\angle ABX = \angle AMF = \angle BAM = \angle CAX$. Similarly, $\angle ACX = \angle BAX$. This means $\bigtriangleup XBA \sim \bigtriangleup XAC$, and the lemma is proven.

[i]Lemma 2[/i]: $X$ is also the center of the spiral similarity that takes $ER$ to $FQ$
[i]Proof[/i]: It suffices to show $\frac{AR}{AB} = \frac{CQ}{CA}$. So, $\frac{AR}{AB} = \frac{QP}{AB} = \frac{CP}{CB}$. On the other hand, $\frac{CQ}{CA} = 1 - \frac{QA}{CA} = 1 - \frac{PR}{CA} = 1 - \frac{BP}{BC} = \frac{CP}{CB}$, and the lemma is proven.

Lemma 2 implies that $X$ is also the center of spiral similarity that takes $EF$ to $RQ$. Since $ER$ and $FQ$ intersect at $A$, $(AEF)$ and $(ARQ)$ concur at $X$. This implies $ARXQ$ is cyclic.

We use barycentric coordinates on $\triangle{ABC}$.
We let $BP=m$ and $PC=m$. Noting that $\triangle{BRP} \sim \triangle{BAC}$ and $\triangle{CQP} \sim \triangle{CAB}$. It follows that $P=(0:n:m)$, $Q=(bn:0:ab-bn)$ and $R=(cm:ac-cm:0)$.
We know the circle $(AQR)$ is of the form $$-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$$, for some real numbers $u$, $v$, and $w$. Plugging in $A$, $Q$ and $R$, we find that the equation is $$-a^2yz-b^2xz-c^2xy+\frac{1}{a}(c^2my+b^2nz)(x+y+z)$$.
The angle condition is easily translated to the fact that our magical point $X$, should it exist, lies on the $A$ symmedian of $\triangle{ABC}$, and as a result, we let $X=(t:b^2:c^2)$. Plugging $X$ into out circle formula and simplifying, we have $$-a^2b^2c^2-tb^2c^2-tb^2c^2+\frac{1}{a}(b^2c^2(m+n))(t+b^2+c^2)$$ At this point, the crucial observation is that $m+n=a$, which deletes $m$ and $n$ from the equation entirely, implying that $t$, and in turn $X$ is independent of our choice of $P$. $\square$

[hide=bary]
We use barycentrics wrt $\triangle ABC$. Let $P=(0,m,n)$ where $m+n=1$. Then $Q=(m,0,n)$ and $R=(n,m,0)$. A little bit of work shows that $(AQR)$ has equation
\[-a^2yz-b^2xz-c^2xy+(c^2ny+b^2mz)(x+y+z)=0.\]
We will show that the intersection of $(AQR)$ with the A-symmedian is independent of $m,n$. We know that the points on the A-symmedian are parameterized by $(t:b^2:c^2)$, so the intersection is given by
\[-a^2b^2c^2-b^2c^2t-c^2b^2t+(c^2nb^2+b^2mc^2)(t+b^2+c^2),\]
or
\[-a^2-2t+t+b^2+c^2=0\]
or $t=b^2+c^2-a^2$, which evidently is independent of $m,n$.
[/hide]

complex!

Let A=0, B=b, C=c. Let the intersection of the symmedian of ABC and (ABC) be D.
Since ABDC is a harmonic quad, BA * CD / BC *CA = -1. Therefore, (b-a)(c-d) / (b-d)(c-a) = -1. Simplyfying, D= 2bc/(b+c).

Let X the midpoint of AD. X=bc/ (b+c). Additionally, since we know CQ/QA= AR/RB, let, R= b(1-k) and Q=ck for some constant k. Now we wish to show if AQRX is cyclic and we know all of their coordinates which simplifies out nicely!

[quote=yunseo]complex!

Let $A=0$, $B=b$, $C=c$. Let the intersection of the symmedian of $ABC$ and $(ABC)$ be $D$.
Since $ABDC$ is a harmonic quad, $BA \cdot CD / BC \cdot CA = -1$. Therefore, $(b-a)(c-d) / (b-d)(c-a) = -1$. Simplyfying, $D= 2bc/(b+c)$.

Let $X$ [b]be[/b] the midpoint of $AD$. [b]We see that[/b] $X=bc/ (b+c)$. Additionally, since we know $CQ/QA= AR/RB$, let, $R= b(1-k)$ and $Q=ck$ for some constant k. Now we wish to show if $AQRX$ is cyclic and we know all of their coordinates which simplifies out nicely![/quote]

FTFY

Here is my solution for this problem
[b]Solution[/b]
Let $AD$ be $A$ - symmedian of $\triangle$ $ABC$ ($D$ $\in$ ($ABC$)); $X$ be second intersection of ($APQ$) with $AD$; $S$, $M$, $N$, $T$ be midpoint of $BC$, $AB$, $AC$, $AP$
It's easy to prove that: $\triangle$ $BDC$ $\sim$ $\triangle$ $RXQ$
But: $S$, $T$ are midpoint of $BC$, $RQ$, respectively then $\triangle$ $BDS$ $\sim$ $\triangle$ $RXT$
So: $\widehat{RXT}$ = $\widehat{BDS}$ = $\widehat{ABC}$ = $\widehat{AMN}$ or $X$ $\in$ ($TMR$)
Hence: $X$ is Miquel point of completed quadrilateral $ARTN.MQ$ or $X$ $\in$ ($AMN$)
Therefore: $X$ is midpoint of $AD$ or $X$ is fixed point which satisfies $\widehat{GAB}$ = $\widehat{CAX}$

[quote=orl]Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG = \angle CAX$.[/quote]

[b]solution(with amar_04 and RAMUGAUSS)[/b]:
I claim that $X$ is the midpoint of the $A$ symmedian chord in $\triangle ABC$. It is very well known that $X$ is the center of spiral similarity sending $BA \mapsto AC$. Now $$\frac{AR}{BR}=\frac{PC}{PB}=\frac{QC}{AQ}$$. Hence $X: R\mapsto Q \implies \triangle XRA \sim \triangle XQC \implies \angle XQC=\angle XRA \implies A,R,X,Q$ are conclyc

It suffices to show that there is a fixed point on the symmedian, or a point with coordinates $(t:b^2:c^2)$. Let $D = (0,m,n)$ where $m+n=1$. Note that $R = (n,m,0)$and $Q = (m,0,n)$. The equation of $(AQR)$ is
\[  -a^2 yz - b^2xz -c^2xy+(x+y+z)(vy+wz)=0 \]
where $v = c^2n$ and $w = b^2 m$.  If we plug in a point $X = (t:b^2:c^2)$, we obtain
\[ -a^2b^2c^2 - b^2c^2t - b^2c^2t + (b^2+c^2+t)(b^2c^2)(m+n) = 0 \]
\[ -a^2 - 2t + b^2 + c^2 + t = 0 \]
Therefore, $t = b^2 +c^2-a^2$, and there exists a fixed point on the symmedian, as desired.

It doesn't look like anyone posted a linearity solution yet, so here it is:

Define a function $F(P)=pow_{(ARQ)}P-pow_{(ABC)}P,$ which we know is linear.  Now let $D$ be the foot of the $A$-symmedian on $BC$, it suffices to show $F(D)$ is fixed. But $$F(D)= \frac{b^2}{b^2+c^2}F(B)+\frac{c^2}{b^2+c^2}F(C) = \frac{b^2}{b^2+c^2}(BR \cdot c)+\frac{c^2}{b^2+c^2}(CQ\cdot b )$$ and by similar triangles, $BR = \frac{c}{a} BP$, $CQ = \frac{b}{a}CP$, so hence $$F(D)= \frac{b^2c^2}{a(b^2+c^2)} (BP + CP) = \frac{b^2c^2}{(b^2+c^2)} $$ and we are done.

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As usual, set $A=(1:0:0), B=(0:1:0), C=(0:0:1)$, and let $P=(0:m:n)$, and $Q=(x:0:z)$
Now, $\overline{PQ}$ and $\overline{AB}$ intersect at the point at infinity along $\overline{AB}$.
One can easily check that the point at infinity along $\overline{AB}$ is just $(-t:t:0)$.
This means that $(x:0:z)$, $(0:m:n)$, and $(-t:t:0)$ are collinear.
Hence $\begin{vmatrix}
-t & t & 0 \\
0 & m & n \\
x & 0 & z \\
\end{vmatrix}=0$.$\implies Q=(m:0:n)$. Similarly, $R=(n:m:0)$.
Now, we can compute the equation of $(AQR)$.
It is just $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2ny+b^2mz}{m+n}\right)=0.$
Since $\angle BAG=\angle CAX$, it follows that $X$ lies on the $A-symmedian$, so $X=(k:b^2:c^2)$ for some $k$.
Putting this into the equation of $(AQR)$, we see that $k=b^2+c^2-a^2$.
Consequently, $X=(b^2+c^2-a^2:b^2:c^2)$, which does not depend on $m$ and $n$ and hence does not depend on $P$ and so we are done.

By the gliding principle, there is a fixed point $X$ lying on each of the circles $(AQR)$, which must lie on $(AAC)$ and $(AAB)$ but not be $A$, where $(AAC)$ is tangent to $AB$ and $(AAB)$ is tangent to $AC$. Then by $\sqrt{bc}$-inversion at $A$, $X$ must lie on the $A$-symmedian of $\triangle ABC$ implying the desired.

[hide=Solution]We use Barycentric Coordinates on $\triangle ABC.$ Let $P=(0,u,v)$ with $u+v=1,$ and notice that $Q=(u,0,v)$ and $R=(v,u,0).$ Since $\angle BAX=\angle GAC,$ $X$ lies on line $AK,$ where $K$ is the symmedian point. Hence, $X=(t:b^2:c^2),$ where $t\in\mathbb{R}.$ The equation of $(AQR)$ is then $$-a^2yz-b^2zx-c^2xy+(c^2vy+b^2uz)(x+y+z)=0.$$ Because $X$ lies on $(AQR),$ $$-a^2b^2c^2-b^2c^2t-c^2b^2t+(c^2vb^2+b^2uc^2)(t+b^2+c^2)=0,$$ which simplifies to $$-a^2-2t+(u+v)(t+b^2+c^2)=0,$$ or $t=-a^2+b^2+c^2.$ Thus, $X$ is independent of $P.$ $\square$[/hide]

Sol:- Inverting with radius √AB*AC and reflecting across A- angle bisector. We get the following problem.
P is an arbitrary point on (ABC) in arc BC not containing A. Let π be the circle through A and P tangent to AC & let ∆ be the circle through A and P tangent to AB. Let AB intersect π again at D & let AC intersect ∆ again at E. Prove that as P varies DE passes through a fixed point.

Proof:- Let A' be the reflection of A across midpoint of BC. We find that P is the center of spiral sim. Sending DA to AE and also sends B to C. So we get DB*CE=AB*AC and hence DBA' is similar to A'CB which implies that D-A'-E are collinear which regardless of any P. Hence we are done.

Dumpty point!
Let $S=PQ\cap (AQR)$ and $T=PR\cap (ARQ)$. Also, let $D=BS\cap (AQR)$.
By Pascal on $ARTDSQ$ we get $T-D-C$ collinear.
Now,
$$
\measuredangle DBA=\measuredangle SBA=\measuredangle DSP=\measuredangle DAQ=\measuredangle DAC
$$
$$
\measuredangle ACD=\measuredangle ACT=\measuredangle PTD=\measuredangle RAD=\measuredangle BAD
$$
Hence $D$ is the $A-Dumpty$ point. Since $D\in (AQR)$ por every point $P$, we've finished.

The spiral similiarity $\psi$ sending $B\to A$, $A\to C$ also sends $R \to Q$. This is because
$$\frac{BR}{RA}=\frac{BP}{PC}=\frac{AQ}{QC}$$

Now $\psi$ is centered at the Dumpty point $F$, so $AQRF$ is cyclic. Since $F$ is on the $A$-symmedian we are done. $\square$

Let $Q_a$ denote the $A$-Dumpty point. Because $Q_a$ lies on the $A$-Symmedian, we know $\angle BAG = \angle CAQ_a$, so it suffices to show $AQQ_aR$ is cyclic.

It's well-known that $Q_aBA \overset{+}{\sim} Q_aAC$ and $(Q_aBA)$ is tangent to $AC$. In addition, the parallel lines imply $$\frac{BR}{RA} = \frac{BP}{PC} = \frac{AQ}{QC}$$ so $Q_aBRA \overset{+}{\sim} Q_aAQC$ which means $Q_aBA \overset{+}{\sim} Q_aRQ \overset{+}{\sim} Q_aAC$. Thus, $$\angle QQ_aR = \angle AQ_aB = 180^{\circ} - \angle CAB = 180^{\circ} - \angle QAR$$ as required. $\blacksquare$

Let the $A-$symmedian intersects $BC$ at $J$ and $(AQR)$ at $K$. We want to show $K$ is the fixed point. It is equivalent to show $P(J, (AQR))$ is fixed. Consider $f:R^2\to R$ such that $$f(\bullet)=P(\bullet, (AQR))-P(\bullet, (ABC))$$ Or equivalently it is sufficient to prove $f(J)$ is fixed. We know that $J=(0:b^2:c^2)$. Let $W=b^2+c^2$. Now
\begin{align*}
f(J)&=\frac{c^2}{W}f(C)+\frac{b^2}{W}
f(B)\\
&=\frac{1}{W}\left(c^2\cdot CQ\cdot b+b^2\cdot BR\cdot c\right)\\
&=\frac{bc}{W}\left(CQ\cdot c+BR\cdot b\right)\\
&=\frac{b^2c^2}{W}
\end{align*}
Where $CQ\cdot c+BR\cdot b=bc$ follows after adding $\frac{CQ}{CA}=\frac{CP}{CB}, \frac{BR}{BA}=\frac{BP}{BC}$. So $f(J)$ is fixed and therefore $K$ is fixed. Hence $K\equiv X$.

Let $\Gamma$ be the circumcircle of $\triangle ABC.$ Then, let the $A$-symmedian of $\triangle ABC$ intersect $\Gamma$ in $X_1$. We claim that if $X$ is the midpoint of $AX_1$ then $AQXR$ is cyclic for any $P.$

Let $Q_1,R_1$ be the reflections of $A$ across $Q$ and $R$ respectively. It suffices to show $AQ_1X_1R_1$ cyclic. WLOG let $P$ be closer to $B$ than $C$ which will mean that $Q_1$ is on the interior of $AC$ and $R_1$ is on the exterior of ray $AB$. We claim that $X_1$ is the Miquel Point of quadrilateral $ABPQ_1.$

Since $ACX_1B$ is already cyclic, to prove the claim we need to show that $CQ_1PX_1$ is cyclic. We will do this by proving that $\angle CQ_1X_1=\angle CPX_1.$ Let $M$ be the midpoint of $BC$. Note that since $AX_1$ is the $A$-symmedian of $\triangle ABC$, $\angle CAM=\angle X_1AB.$ Furthermore, $\angle ACM=\angle AX_1B$ so $\triangle ACM\sim \triangle AX_1B.$ Thus, \[\frac{AX_1}{BX_1}=\frac{AC}{CM}=2\cdot\frac{AC}{BC}=2\cdot \frac{RP}{BP}=2\cdot\frac{AQ}{BP}=\frac{AQ_1}{BP}.\] Therefore we have $\triangle AQ_1X_1\sim \triangle BPX_1$ by $SAS$ similarity. We can therefore conclude that $\angle CQ_1X_1=\angle CPX_1$ finishing our proof that $X_1$ is the Miquel Point of quadrilateral $ABPQ_1.$

Since $X_1$ is the Miquel Point, $AQ_1X_1R_1$ is cyclic, as desired. Since we already established that $X_1$ is on the symmedian, $\angle CAX=\angle CAX_1=\angle BAG.$

We will use barycentric coordinates. Let $P=(0,p,1-p)$. Then, $$R=(1-p,p,0),Q=(p,0,1-p).$$ Let's find the equation for $(ARQ)$. Substituting $(1,0,0)$ gives us $w=0$, then using $(1-p,p,0)$ gives $v=c^2(1-p)$, and using $(p,0,1-p)$ gives $w=b^2p$. Therefore, the equation for $(ARQ)$ is $$(x+y+z)(c^2(1-p)y+b^2pz)=a^2yz+b^2xz+c^2xy.$$ Note that the "fixed point" is just a solution $(x,y,z)$ that works regardless of $p$. To find such a triple, we want $p$ to cancel, so we want $c^2y=b^2z$. This leads to the solution $$(x,y,z)=(-a^2+b^2+c^2,b^2,c^2),$$ which works regardless of $p$. Since the last two coordinates are $b^2$ and $c^2$, this lies on the A-symmedian, so we are done.

Here’s a pure angle chasing solution which only use the knowledge of dumpy point.

We claim that the desired fixed point is the dumpy point $X=D$ which lies on the symmedian as required. Let $T$ be the midpoint of $AP$. Let $AD$ intersects $(ABC)$ again at $S$.

By Miquel’s Theorem, it’s suffice to prove that $(BRD),(CQD)$ intersect on $BC$. We claim that the intersection point is $TD\cap BC=J$. We want $B,R,D,J$ to be concyclic.

Let’s delete point $Q$ and introduce point $I=AS\cap AP$.

We angle chase
$$\angle BDI=\angle DBA+\angle DAB=\angle DAC+\angle DAB=\angle BAC=\angle BRP=\angle BRI$$
so $B,R,D,I$ are cyclic. We now only need $B,D,I,J$ to be cyclic.

But first, note that $B,I,P,S$ are concyclic by Reim’s Theorem on $\overline{AIS},\overline{BPC}$.

Now let’s finish this problem by proving that $B,D,I,J$ are concyclic.
$$\angle IBJ=\angle IBP=\angle ISP=\angle ADT=\angle IDJ.$$
($DT\parallel SP$ because $DT$ is the $A$-midline of triangle $ASP$.)

Let $M,N$ be the midpoints of $AB,AC$. Notice by linear motion, we have $\frac{MR}{AB} = -\frac{NQ}{AC}$ while lengths are directed. Now take a homothety with scale factor $2$. Let $S$ be the intersection of the cirumcircle of $ABC$ and the symmedian point. I claim that this point, the images of $R,Q$ under the homothety, and $A$ are cyclic, which implies the problem statement. Let $R',Q'$ be the images of $R,Q$ under this homothety. This is equivalent to showing that $AS$ is the radical axis of the circumcircles of $(ABC), (AR'Q')$. Now consider the function taking points $X$ and returning the power of $X$ relative to circle $(AR'Q')$ minus the power of $X$ relative to circle $(ABC)$. It is well known that this function is linear. Then evaluated at $B$, the function gives $BR' \cdot BA - 0$, evaluated at $C$ this gives $CQ' \cdot CA - 0$. Let $AS$ meet $BC$ at $K$. By symmedian properties, $\frac{BK}{CK} = \frac{AB^2}{AC^2}$. Then the value of the function at $k$ is $\frac{AC^2 (BR')(AB) + AB^2 (AC)(CQ')}{AC^2 + AB^2} = 0$, thus $AK$ is the radical axis and since $AKS$ collinear we are done.

[hide=Solution]
Use barycentric coordinates with respect to triangle $ABC$. Let $P = (0, p, 1-p)$. Then, note that since $PQ$ is parallel to $AB$, we have that $$Q = P + (A-B)k = (0, p, 1-p) + (k, -k, 0) = (k, p-k, 1-p),$$ and since the $B$-component of $Q$ is $0$, $k = p$, so $$Q = (p, 0, 1-p).$$ We can similarly obtain that $$R = (1-p, p, 0).$$ Now, suppose that the circumcircle of $AQR$ had equation $$-a^2yz-b^2zx-c^2xy + (ux+vy+wz)(x+y+z) = 0.$$ Since $A$ lies on this circle, it follows that $u = 0$, so $(AQR)$ has equation $$-a^2yz-b^2zx-c^2xy + (vy+wz)(x+y+z) = 0.$$ Since $Q = (p,0,1-p)$ lies on this circle, we have that $$-b^2p(1-p) + (w(1-p)) = 0,$$ so since $p\neq 1$ (as $Q\neq A$ since $P$ is on segment $BC$, and the setup would make no sense if $P = B$ or $P=C$ because the circumcircle of $AQR$ would be ill-defined), we have $$w = b^2p.$$ Similarly, since $R = (1-p, p, 0)$ lies on this circle, we have $$-c^2(1-p)p + (vp) = 0,$$ so since $p\neq 0$ as $R\neq A$ since $P$ is on line segment $BC$, and $P\notin \{B,C\}$ as discussed earlier, we have that $$v = c^2(1-p).$$ Therefore, the equation of $(AQR)$ is $$-a^2yz-b^2zx-c^2xy + (c^2(1-p)y + b^2pz)(x+y+z) = 0.$$ Now, let $X = (b^2+c^2-a^2 : b^2 : c^2)$. Then, we have that
\begin{align*}
-a^2b^2c^2 - b^2c^2(b^2+c^2-a^2) - c^2(b^2+c^2-a^2)b^2 + (c^2(1-p)b^2+b^2pc^2)(2b^2+2c^2-a^2) \\ = -b^2c^2[a^2+2(b^2+c^2-a^2)] + b^2c^2(2b^2+2c^2-a^2) \\ = 0,
\end{align*}
so $X$ lies on $(AQR)$ regardless of the choices of $p$ (i.e. the choice of $P$). Also, since the ratio of the $B$ and $C$ components of $X$ is $\frac{b^2}{c^2}$, it follows that $X$ lies on the $A$-symmedian, so $\angle BAG = \angle CAX$, as desired. $\blacksquare$

We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle.

Let $P = (0,p,1-p)$. We can see that $Q = (p,0,1-p)$ and $R = (1-p,p,0)$.

Pluggin $A$ into the general circle we find that $u = 0$. Plugging in $Q$ we find that $w = b^2p$, and plugging in $R$ we find that $v = c^2(1-p)$. Thus we can see that the circle is given by $$-a^2yz -b^2xz - c^2xy + (c^2(1-p)y + b^2pz)(x+y+z) = 0$$. We find that the point $(-a^2+b^2+c^2:b^2:c^2)$ is a point which always satisfies this regardless of $p$. We can also see that it is parametrized by $(a^2:b^2:c^2)$ which is the isogonal conjugate of $G$. $\blacksquare$


The function $ P(x)\equal{}x^{2}\plus{}ax\plus{}b$ has roots $ c$ and $ d$, while the function $ q(x)\equal{}x^{2}\plus{}cx\plus{}d$ has roots $ a$ and $ b$. Find $ a\plus{}b\plus{}c\plus{}d$.

[hide="Solution (unsure)."]
$ P(x)$ can be written as $ (x\minus{}c)(x\minus{}d)$, so $ a \equal{} \minus{}(c\plus{}d)$ and $ b \equal{} cd$.  Thus, $ a\plus{}b\plus{}c\plus{}d$ $ \equal{}$ $ a\plus{}b\minus{}[\minus{}(c\plus{}d)]$ $ \equal{}$ $ b$.  Applying the same logic to $ q(x)$, $ d \equal{} ab$ and $ c \equal{} \minus{}(a\plus{}b)$ and the sum is equal to $ d$, so $ b \equal{} d$.  Because $ b \equal{} cd$, this means that $ c\equal{}1$; similarly, because $ ab\equal{}d$, then $ a\equal{}1$.  Plugging that into the original equation, we have that $ 1\equal{}\minus{}(1\plus{}d)$ and that $ b \equal{} d \equal{} \minus{}2$.  The sum is then $ 1\minus{}2\plus{}1\minus{}2$, or $ \boxed{\minus{}2}$.[/hide]

We use the Vieta's relation for roots and coefficients
c+d = -a ...   1
cd =b     ...     2

and
a+b=-c  ...  3
ab=d    ....    4

d- (-c) = ab - (a+b)
d+c  =  ab - (a+b)
-a = ab -a -b

ab - b =0.
b=0  or  a=1.

 case1          b=0 implies d =0 by ...4
from ...3
c =-a
therefore a+b+c+d  = a+0+-a+0
=0

case2
a=1.
from ...4   b=d
then from...2   c=1
put in ...1
d =-2
and b=-2 also
a+b+c+d =  1-2+1-2
=-2

ans are -2 or 0


Prove that $ \begin{cases} m,n \in Q ,  \  \   m \neq n\\
(\sqrt{m}\minus{}\sqrt{n}) \in Q \end{cases} \Rightarrow \sqrt{m}, \sqrt{n} \in Q$

[hide="solution"]First of all, we know that $ \mathbb{Q}$ is closed under division, multiplication, addition and subtraction, so that the final outcome of any two members of $ \mathbb{Q}$ under these operations, is also in $ \mathbb{Q}$.

Now if $ m$, $ n$ are in $ \mathbb{Q}$, then so is their difference $ m\minus{}n\equal{}(\sqrt{m}\plus{}\sqrt{n})(\sqrt{m}\minus{}\sqrt{n}).$  Thus $ \sqrt{m}\plus{}\sqrt{n}\in\mathbb{Q}$, so the sum of this and $ \sqrt{m}\minus{}\sqrt{n}$, or $ 2\sqrt{m}$, is in $ \mathbb{Q}$ and so is the difference of them, or $ 2\sqrt{n}$.  Dividing by 2, we get the desired result.[/hide]

[quote="tenniskidperson3"][hide="solution"]First of all, we know that $ \mathbb{Q}$ is closed under division, multiplication, addition and subtraction, so that the final outcome of any two members of $ \mathbb{Q}$ under these operations, is also in $ \mathbb{Q}$.

Now if $ m$, $ n$ are in $ \mathbb{Q}$, then so is their difference $ m \minus{} n \equal{} (\sqrt {m} \plus{} \sqrt {n})(\sqrt {m} \minus{} \sqrt {n}).$  Thus $ \sqrt {m} \plus{} \sqrt {n}\in\mathbb{Q}$, so the sum of this and $ \sqrt {m} \minus{} \sqrt {n}$, or $ 2\sqrt {m}$, is in $ \mathbb{Q}$ and so is the difference of them, or $ 2\sqrt {n}$.  Dividing by 2, we get the desired result.[/hide][/quote]

Correct. I like quick responses. :)


This is somewhat of a trick question :). We can prove that $|AB-CD|\ge|AC-BD|\ (*)$:

Let $P=AC\cap BD$, and assume WLOG that $DC>AB$ (when we have equality, it's clear). $PAB$ and $PDC$ are similar, so let $k=\frac{DC}{AB}>1$. We have $|AB-CD|=(k-1)AB$, and $|AC-BD|=|PA+PC-PB-PD|=|(PA-PD)-(PB-PC)|$, which is equal to $|(k-1)PA-(k-1)PB|=(k-1)|PA-PB|$, and $(*)$ is now reduced to $AB\ge|PA-PB|$ (true by the triangle inequality).

Suppose without loss of generality that $AB \ge CD$ and $AD \ge BC$.

We now consider two cases:

Case I: $AC \ge BD$

Since we have $AC \ge BD$, $AB \ge CD$, $AD \ge BC$,
$AC(AB+AD) \ge BD(BC+CD)$
$AB \cdot AC + AD \cdot AC + 2BD \cdot AC \ge BC \cdot BD + CD \cdot BD + 2AC \cdot BD$
$AB + AD + 2BD \ge BC + CD + 2AC$
$AB - CD + AD - BC \ge 2AC - 2BD$

Case II: $BD \ge AC$

We have $BD \ge AC$, $AB \ge CD$, $AD \ge BC$ and thus
$BD(AB+AD) \ge AC(BC+CD)$.
Using similar logic to Case I, we get a similar result:
$AB - CD + AD - BC \ge 2BD - 2AC$

So therefore $|AB - CD| + |AD - BC| \ge 2|BD - AC|$
QED.

[quote="MithsApprentice"]Let $ABCD$ be a cyclic quadrilateral. Prove that \[ |AB - CD| + |AD - BC| \geq 2|AC - BD|.  \][/quote]

This is an old, old post, anyway, I solved this problem recently and strangely enough my solution wasn't posted so here it goes.

1) Assume $AB \geq DC$ and $AD \geq BC$. Let $D'$· be the point in the circle with $BD' \parallel CD$ and $B'$ the point with $DB' \parallel BC$
2) Note $CD' = DB = CB'$, $DD' = BC$, $BB' = CD$
3) We have the two triangle inequalities
        $AD + D'C \geq DD' + AC$
        $AB + CB' \geq BB' + AC$
The result follows from summing the inequalities and using 2)

Daniel

sorry to bring this topic again to life, but i found an interesting "reference" previous to this olympiad... [url]http://www.imo.org.yu/othercomp/Rom/RomMO98.pdf[/url] check out the 9th-form problem 2... it's exactly what grobber said in his post... however both of them are not too difficult.

Let angles $\alpha,\beta,\gamma,\delta$ have sum $180^\circ$ so we want to show $|\sin\alpha-\sin\gamma|+|\sin\beta-\sin(\alpha+\beta+\gamma)|\ge 2|\sin(\alpha+\beta)-\sin(\beta+\gamma)|$. Equivalently, we want to show
\[|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|+|\cos(\tfrac{\alpha+2\beta+\gamma}{2})\sin(\tfrac{-\alpha-\gamma}{2})|\ge 2|\cos(\tfrac{\alpha+2\beta+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|.\]
With respect to $\beta$, note the difference between the two sides is minimized at the extrema of $|\cos(\tfrac{\alpha+2\beta+\gamma}{2})|$. Hence it suffices to argue
\[|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|\ge 0\]
and
\[|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|+|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{-\alpha-\gamma}{2})|\ge 2|
\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|.\]
The former is trivial. For the latter, it is equivalent to demonstrate
\[\sin(\tfrac{\alpha+\gamma}{2})\ge |\sin(\tfrac{\alpha-\gamma}{2})|.\]
This is trivial: WLOG $\alpha>\gamma$, then it follows from noting $0<\tfrac{\alpha-\gamma}{2}<\tfrac{\alpha+\gamma}{2}<\pi/2$.


What is the exponent of 10 when $ 4^{12}\times 5^{20}$ is written in
scientific notation?


function
algebra
domain
number theory
relatively prime
number theory proposed

Let k is an integer and S={-1,-2,-3,...}.Find all function $f:S \rightarrow Z$ satisfied the relation:
$\forall n \in S and \leq {-1}$ : $f(n)f(n+1)=(f(n)+n-k)^2$.

Looking through some of the older problems...

Let $g(n)=f(n-k)$. We have $g(n+1)g(n)=(g(n)+n)^2$, so $g(n+1)=g(n)+2n+\frac{n^2}{g(n)}$. We can extend the domain of g to all integers with this recursion; we are looking for functions $g$ such that $g(n) \in Z$   $\forall (n < -k)$.

Obviously, we must have $g(n) \mid n^2$ for all n so that $g(n+1)$ will be an integer. Let $-n$ be a prime greater than $k+2$; then there are 6 cases:
Case 1: $g(n) = n^2$. Then $g(n+1)= n^2+2n+1=(n+1)^2$. Obviously, this forces $g(m)=m^2$ for all $m>n$.
Case 2: $g(n) = n$. Then $g(n+1)= n+2n+n=4n$. Since $n$ and $n+1$ are relatively prime, $g(n+1) \nmid (n+1)^2$ and $g(n+2) \notin Z$, making this case impossible.
Case 3: $g(n) = 1$. Then $g(n+1)= 1+2n+n^2=(n+1)^2$ and $g(m)=m^2$ for all $m>n$.
Case 4: $g(n) = -n^2$. Then $g(n+1)=-1+2n-n^2=-(n-1)^2$. Since $(n-1)^2 > (n+1)^2$, $g(n+2) \notin Z$.
Case 5: $g(n) = -n$. Then $g(n+1) =-n+2n-n=0$ and $g(n+2)$ does not exist.
Case 6: $g(n) = -1$. Then $g(n+1)=-n^2+2n-1=-(n-1)^2$.  Since $(n-1)^2 > (n+1)^2$, $g(n+2) \notin Z$.

In the only two cases that allow $g$ to take integer values, $g(m)=m^2$ for all $m>-n$. Since there are infinitely many primes, we can choose $n$ arbitrarily large, and $g(n)=n^2$   $\forall n$.
Back to the original problem, the only solution is $f(n)=(k-n)^2$.