analytic geometry
parameterization
linear algebra

Hey, I am in desperate need of help in lin alg! Could you please when responding give a detailed explaintation so that I can understand whats going on. Thanks, anywhoo here is my question.
Let P1 = P1(2, -1, -2) and P2 = P2(1, -2, 0). Find the cooridnates of the point P:

a) 1/5 the way from P1 to P2

b) 1/4 the way from P2 to P1

Thanks !!

~Geah

Try the formula $ P_t = (1 - t)P_0 + tP_1$ for $ 0\le t\le 1.$ This is the convex combinations of the two points, or a parameterization of the line segment connecting them. Note what you get if $ t = 0$ and what you get when $ t = 1.$ When $ t = \frac12,$ you get the midpoint. You should be able to reason out how to get some other fraction - in particular, the coefficient should be larger in front of the point you're closer to.


calculus
derivative
function
continued fraction
real analysis
real analysis unsolved

How do you find the derivative of the [url=http://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function]Question Mark Function[/url]?

According to the article,

[quote="wikipedia"]The question mark function is a strictly increasing and continuous, but not absolutely continuous function. The derivative vanishes on the rational numbers; however, since the rationals are a set of measure zero, this vanishing of the derivative at the rationals is not in contradiction with the non-absolute continuity of the function. It does not have a well-defined derivative, in the classical sense, on the irrationals; however, there are several constructions for a measure that, when integrated, yields the question mark function. One such construction is obtained by measuring the density of the Farey numbers on the real number line. The question mark measure is the prototypical example of what are sometimes referred to as multi-fractal measures.[/quote]

[quote]since the rationals are a set of measure zero, this vanishing of the derivative at the rationals is not in contradiction with the non-absolute continuity of the function.[/quote]
There would be no contradiction even if the zero set of derivative had positive measure: consider the Cantor staircase.

Oh, I didn't know that had a name.
The quote there is misleading; the derivative is defined almost everywhere, but I don't know any specific values except at the rationals. The computation relies on properties continued fractions: as one term $ q_k$ of the continued fraction goes to $ \infty$, our code for it looks like $ a\plus{}b\cdot 2^{\minus{}q_k}$ and the fraction looks like $ \frac{c\plus{}dq_k}{e\plus{}fq_k}$

A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=36145]previous post[/url] with this function.


Lagrangia urged me to note a few things about Canada, so I will do so.
Canada, as you know, is a huge place.  We have cities like every country does, and we also have beautiful beautiful places (maybe Rockies, or even more up north :D)  Canada hosted IMO a few years back, in Toronto which is a huge city.  For math, we do not get a lot of training, but there are some very smart ppl which keeps up Canada to about top 10-20 in IMO every year (no I am not one of them  :D).

I think Canada is an awsome place to live in  :D  :D  :D

i think that to. my godfather lives in canada for 10 years now, in Vancouver. I hope to get there and visit the place. From he tells me (he is Romanian, stayed a couple of years in Germany, a couple of years in Israel and some in France, and then went to US/Canada, and decided to remain in Canada) he says it's the best place to live :)

Canada is incredibly boring! (meh that exclamation mark proves I wasn't born in Canada)

I live in a rich, suburb of Toronto (150k people, 40 or so miles away from city centre)


AMC
USA(J)MO
USAMO
USAJMO
Harvard
college
AMC 10

In discussions with some of my underclassmen as to whether to take the AMC-10 or AMC-12 (or both) this year, I asked them the question: "Why do you, personally, want to make USAMO"?

I'd be interested in hearing peoples' responses from here.

If you mean just qualifying for USAMO in general, here are my reasons (even though I am far from actually making it):
 - looks good on a college application
 - I have to do lots of (entertaining) problems to train for it
 - it's fun. Just having a goal and finally reaching it is very satisfying.
 - I have nothing better to do

EDIT: Agree with thunder365.

Why [i]wouldn't[/i] you want to make usamo

Hm this seems rather amusing.

-because I like math (Who doesn't?)
-I'm bored at school with math classes, and want some harder problems to do
-sometimes normal math contest problems can be (*davidyang voice*) trivial (ohnoez pythag alert! AH! :o ) so interesting problems would be better
-USAMO helps develop brain capacity, more like "thinking outside the box"
-Math contests are cool in general
-Ahh, proofs. They help you develop mathematical reasoning and solution writing skills.
-Most importantly: You get to make MOP by doing well on USAMO. By going to MOP, you get several weeks of freedom, fun, and a good social experience.
-And, most most SUPERMOST importantly...if you win USAMO, and pwn TST, your big adventure is the IMO!!!! (Also, a great foreign travel experience too :) )

So yeah...USAMO has its reasons, a whole variety!

[quote="abacadaea"]Do you mean why usamo rather than usajmo, or just why we like qualifying for the usamo[/quote]

Both.

-Because I get out of 2 or 3 classes at the end of the day those 2 days
-Because I can brag to others which, even though they don't care too much, gives me some sense of satisfaction
-Because it's there
-Because hard math problems is all life is (I don't have a girlfriend yet...  :blush: )
-Because it gives me something to talk about on AOPS.  If I didn't take the USAMO, I couldn't answer the problems.
-Because it looks good on college applications
-Because I get a sense of satisfaction in answering a problem or 2 on USAMO.
To sum all these up:
-Because it's better than the alternative.

Because it's better than not making USAMO.

Hmm.. I really like an0maly's answer.  :D

[quote="buzzer11"]Do you think many people will be taking the usajmo this year?[/quote]
It's a fixed number, around 230.

well i guess i mean, will many people take the usajmo path this year.

Do you mean the USAJMO path to red MOP? That's what I'm trying to do this year.

[quote="tenniskidperson3"]-Because hard math problems is all life is (I don't have a girlfriend yet...  :blush: )
[/quote]

math problems > relationships!

Last shot at making MOP for me. Course I want to make USAMO.

[quote="Lazarus"]math problems > relationships!
[/quote]

interesting coming from you  :wink:

[quote="diophantient"]It's a well kept secret among mathematicians, but it turns out that qualifying for the USAMO helps you get girls.[/quote]

 :rotfl:

Source? Who said that? O__O

well they've done lots of studies, up at Harvard and Yale.

It's true, all else equal. "Intelligence is sexy," says Natural Selection.

On the other hand, if you spend all your extra time on math and less time on your physique or on developing social skills, then the law of diminishing marginal returns kicks in.

[b]Answer to Original Question:[/b] I did it for the experience, and to see if I could make it. I didn't particularly care about doing well on it though -- in my senior year, I handed in my answer to the first problem after 20 minutes and left to go to NYC for Moody's Mega Math Challenge (which I would highly recommend to anyone here).

Of course, I tried the problems on my own for "fun" when I got back.

USAMO = awesomeness. Awesomeness leads to good things... :lol:

Because I like competing.

[quote="diophantient"]It's a well kept secret among mathematicians, but it turns out that qualifying for the USAMO helps you get girls.[/quote]

Empirically proven.

[quote="1=2"][quote="diophantient"]It's a well kept secret among mathematicians, but it turns out that qualifying for the USAMO helps you get girls.[/quote]
This person is male.

Why would you want to make USAMO?
1. Because it looks/is awesome.
2. Because it leads to all sections of MOP, if you're good at it.
3. Because exposing yourself to USAMO problems is the first step to conquering USAMO problems, which is the ultimate (or penultimate) step to black MOP, and ultimately, the IMO.
4. Because saying "I made the USA Math Olympiad" sounds more impressive than saying "I made the American Invitational Mathematics Exam".
5. Boy/girlfriends. Especially those who are smart, like you.
6. College apps.

Why wouldn't you want to make USAMO?
1. Lame reasons.[/quote]

ok people. Telling random girls that you made USAMO will NOT increase your chances with them. Trust me. The average girl doesn't even know what USAMO is. However, if you were to tell a mathematically inclined girl...

anyways... It seems that the discussion has sort of gravitated toward "getting girls" so this post will bring this discussion back towards the real topic:

another reason to make USAMO: Making USAMO gives you a chance to make MOP. Making MOP makes you smarter. END RESULT: (unless you're like stupid) You make MOP again next year and (likely but not for sure) a higher level of MOP then last year.

Please forgive this return to the girl topic but...
In all seriousness, girls appreciate intelligence. Contrary to popular belief.
USAMO = intelligence.

[quote="diophantient"]It's a well kept secret among mathematicians, but it turns out that qualifying for the USAMO helps you get girls.[/quote]
i would hope not if you're a girl

Anyways, which situation seems more likely:
1.
Boy: Hey I made the USAMO
Girl: ZOMG LETS HAVE KIDS
2.
Boy: I made USAMO
Girl: Ummmm cool?

this topic is win

I wanna make USAMO (not this year though) so my hard work will have actually paid off.
It will also look good on my college resume.
USAMO also puts you in the Top 500 or so mathematicians in the United States :D
And is it really true that USAMO gets you girls?

@ thunder, wtf?

[quote="buzzer11"]
USAMO = intelligence.[/quote]

Is this equality symmetric, reflexive, and transitive?

[quote="the cliu"]I wanna make USAMO (not this year though) so my hard work will have actually paid off.
It will also look good on my college resume.
USAMO also puts you in the Top 300 or so mathematicians in the United States :D
And is it really true that USAMO gets you girls?

@ thunder, wtf?[/quote]

I believe it's top 500. :lol:

Top 270 if you take out the USAJMO people.

I would put it much higher because I think at least a few dozen good mathematicians don't do the AMCs, and don't forget to include high school or lower  :)

[quote="the cliu"]
USAMO also puts you in the Top 500 or so mathematicians in the United States
[/quote]

That is plausibly top ten million worldwide  :wink:


Learning conics for the first time...

Find the equation of the ellipse containing points $ (6, \minus{} 1)$, $ ( \minus{} 4, \minus{} 5)$, $ (6, \minus{} 5)$, and $ ( \minus{} 12, \minus{} 3)$.

What I've noticed:
[hide]The first three points mentioned make a right triangle. I hypothesized that the midpoint the hypotenuse of this right triangle (or any right triangle inscribed in an ellipse) must be the center of the ellipse (I believe this is so, but I could not prove it except by imagining it in my head). This does indeed place the center of the ellipse at $ (1, \minus{} 3)$, which would make $ ( \minus{} 12, \minus{} 3)$ a vertex which makes me think I'm on the right track, but I can't figure out where to go from there.[/hide]

hello, we have the equation of the given ellipse as
$ \frac{(x\minus{}x_0)^2}{a^2}\plus{}\frac{(y\minus{}y_0)^2}{b^2}\equal{}1$, inserting the cordinates of all given points in this equation we get an equation system, solving this we get $ a\equal{}13,b\equal{}\frac{13}{6},x_0\equal{}1,y_0\equal{}\minus{}3$.
Sonnhard.

Right -- assuming that the axes are parallel to the x and y axes, and assuming your observations to be true (which is reasonable [i]if you graph all four points and see where the ellipse should be[/i]), you then:

1. Suppose that $ (x_0, y_0)\equal{}(1,\minus{}3)$, the center.
2. Suppose that semi-axis in the x direction has length equal to the distance from center $ (1,\minus{}3)$ to vertex $ (\minus{}12,\minus{}3)$, that is $ a\equal{}13$.

Then, plug in any one of the points to solve for $ b$.

The thing is, your first hypothesis is far from true. Consider a right triangle with right-angle vertex at an endpoint of the minor axis, and symmetric over the minor axis (ie, a 45-45-90 triangle). So long as the ellipse is not also a circle, the resulting right triangle contains the center inside of it. There are many, many other such right triangles.


Let $ a,b$ be two positive integers ($ a\neq b$) such that $ a^2\plus{}ab\plus{}b^2\ |\ ab(a\plus{}b)$ . Prove that :

$ |a\minus{}b|> \sqrt[3]{ab}$

[hide="Solution"]Let $ gcd(a,b) \equal{} k$. Then $ a \equal{} km$, $ b \equal{} kn$, $ gcd(m,n) \equal{} 1$. It simplifies to $ m^2 \plus{} mn \plus{} n^2|mnk(m \plus{} n)$. But since $ (m,n) \equal{} 1$, $ m$ and $ m^2 \plus{} mn \plus{} n^2$ don't have any common factors (any common factor would have to divide $ n^2$). So $ m^2 \plus{} mn \plus{} n^2|k(m \plus{} n)$. But any common divisor of $ m^2 \plus{} mn \plus{} n^2$ and $ m \plus{} n$ divides $ (m \plus{} n)^2 \minus{} (m^2 \plus{} mn \plus{} n^2) \equal{} mn$, so they're also relatively prime. So $ m^2 \plus{} mn \plus{} n^2|k$. Substituting back, $ a^2 \plus{} ab \plus{} b^2|k^3$, so $ k^3 \ge a^2 \plus{} ab \plus{} b^2$, so $ k \ge \sqrt [3] {a^2 \plus{} ab \plus{} b^2}$. But $ gcd(a,b) \equal{} k|a \minus{} b$, so we have $ |a \minus{} b| \ge k \ge \sqrt [3] {a^2 \plus{} ab \plus{} b^2}$[/hide]
So we can tighten it to $ |a \minus{} b| \ge \sqrt [3] {a^2 \plus{} ab \plus{} b^2}$

Problem source : Russia 2001

[hide="Solution"]
Let $ \gcd(a,b) \equal{} g$;  $ a \equal{} gx$;  $ b \equal{} gy$;   $ \gcd(x,y) \equal{} 1$.

$ \frac{ab(a\plus{}b)}{a^2\plus{}ab\plus{}b^2} \equal{} \frac{gxy(x\plus{}y)}{x^2\plus{}xy\plus{}y^2}$

$ \gcd(x^2\plus{}xy\plus{}y^2,x) \equal{} \gcd(y^2,x)  \equal{} 1$. Similarly $ \gcd(x^2\plus{}xy\plus{}y^2,y) \equal{} 1$.

$ \gcd(x\plus{}y,x^2\plus{}xy\plus{}y^2) \equal{} \gcd(x\plus{}y,x(x\plus{}y)\plus{}y) \equal{} \gcd(x\plus{}y,y) \equal{} 1$.

$ \implies x^2\plus{}xy\plus{}y^2|g \implies x^2\plus{}xy\plus{}y^2 \le g$.

$ {|a\minus{}b|}^3 \equal{} g^3 \cdot {|x\minus{}y|}^3 \equal{} g^2 \cdot {|x\minus{}y|}^3 \cdot g   \ge  g^2 \cdot 1 \cdot (x^2\plus{}xy\plus{}y^2) > g^2xy \equal{}ab$

$ \implies |a\minus{}b|>\sqrt[3]{ab}$[/hide]


ratio
modular arithmetic
function
calculus
calculus computations

We know that $\{n\mod \pi, n \in N\}$ is dense in $(0,\pi)$ but how to prove that?
That is long long ago and I can't recall the proof. Please anyone shed me a light. A link would be nice.

TIA

Actually, if $a,b$ are incommensurable (their ratio is irrational), then $na\pmod b$ is dense in $(0,b)$.

Construct a circle with length $b$ and start putting points $A_n$ on it in a clockwise direction, for example, s.t. the arc $A_nA_{n+1}$ has length $a$ when measured in a clockwise direction(you may assume $a<b$). It's not ahrd from here. First show that there are infinitely many points; from here we can derive the fact that there is a $T=T(\varepsilon)$ s.t. $A_nA_{n+T}<\varepsilon,\ \forall n$ (in fact, all such arcs have the same length), and the conclusion easily follows: given an arc on our circle, if you find $T$ s.t. the corresponding $\varepsilon<$the length of the arc, one of the points $A_1,A_{1+T},A_{1+2T},\ldots$ must fall inside the arc.

it can be done in a more technical manner, but this really shows what's going on, I think.

[quote="grobber"]Construct a circle with length $b$ and start putting points $A_n$ on it.[/quote]
This is the point of the proof. Thank you.

[quote="Myth"]Indeed, this statement is trivial.
The more general statement is that $\{an\pmod{b}\}$ is evenly distributed on $(0,b)$.[/quote]
If a and b are incommensurable, we can prove that $f(n) =  an\mod b$ is not a period function of $n$. I will use this result and get back to you on the "Tough Series" problem


Is a line a convex function? Is a line a concave function?

Hmm... Wikipedia seems to give both definitions (one as a convex function and one as a strictly convex function)...

My question then is this (in addition to the first question): Would one be able to use a line in Jensen's inequality? (i.e. would a line satisfy the prerequisites of the jensen's inequalty?)

Thanks so much!

Line is the equality case in jensen's inequality. I think.  :lol:

A line function is of the form $ f(x) \equal{} ax \plus{} b$. Therefore, by Jensen's

$ w_1(ax_1 \plus{} b) \plus{} w_2(ax_2 \plus{} b) \plus{} \ldots \plus{} w_n(ax_n \plus{} b) \ge a(w_1x_1 \plus{} w_2x_2 \plus{} \ldots \plus{} w_nx_n) \plus{} b$

Equality holds because we have $ \sum{w_i} \equal{} 1$.

TachyonPulse, so you're saying that a line is the equality case of jensen's, right?

Thanks!

Yeah. Sorry or the confusion; I was going to have another line using an equal sign, but I forgot.

Thanks so much!

P.S. speaking of forgetting, you forgot the "f" in "for" :)


Given triangle ABC and D and E on AB and AC, respectively. Let M be the midpoint of BE and N be the midpoint of CD. Prove that the area of triangle AMN is 1/4 of that of quadrilateral BCED.

(Do not use coordinates.)

[hide="Well, you could use mass points or area ratios..."]Extend $AM$ and $AN$ to meet $BC$ at $P$ and $Q$.

Let $K=[ABC]$

$[AMN]=(\frac{AM}{AP})(\frac{AN}{AQ})[APQ]=(\frac{AC+AE}{AC+AE+CE})(\frac{AB+AD}{AB+AD+BD})([APC]+[AQB]-K)=(\frac{AC+AE}{2AC})(\frac{AB+AD}{2AB})(\frac{AC}{AC+AE}K+\frac{AB}{AB+AD}K-K)=(\frac{AC+AE}{2AC})(\frac{AB+AD}{2AB})(\frac{AC\cdot AB-AE\cdot AD}{(AC+AE)(AB+AD)})K=\frac{AC\cdot AB-AE\cdot AD}{4\cdot AB\cdot AC}K$

$[BCED]=K-[ADE]=K-(\frac{AD}{AB})(\frac{AE}{AC})K=\frac{AC\cdot AB-AE\cdot AD}{AB\cdot AC}K$[/hide]


Let $p(n)$ be the number of partition of n without any restriction.the prove that $p(5m+4)\equiv 0 (mod 5),p(7m+5)\equiv 0 (mod 7)$
nice :)

The problem is really nice. :clap: Well,I know that Ramanujan is the great genius from India,I am curious about him,it is said that he is one of the greatest mysteries of the math history.

Dear sam-n,where did you find the nice problem?

"Aspect of combinatorics and combinatorial Number theory" by S.D.Adhikari.

Dear sam-n,do you have the solution of the problem?

[quote="coffeechao"]sorry, I am not clear in the definition of $p(n)$ here, in particular, "the number of partition of n"
Could you give some examples? Cheers.[/quote]
ok,since 4=1+1+1+1=1+1+2=1+3=2+2,i.e 4 can be parted in 5 ways,thus p(4)=5.
Aother example,5=1+1+1+1+1=1+1+1+2=1+1+3=1+4=2+3=2+2+1,thus p(5)=7.
Hope you get it.

An extension of this question can be found at planetmaths.com I think.

Bomb


linear algebra
matrix
vector
abstract algebra
graph theory
combinatorics proposed
combinatorics

We are given two sets of people A and B.
Prove that there either exist a subset C of A (with at least 1 element) or a subset D of B such that : The number of people in D ( C) that  each person in A ( B ) knows has the same parity (i.e all even or all odd).

I think I have a short solution for this problem, but  I want some elegant solutions.

I found this very old post by Viet Bao, IMO 2003 perfect scorer, in the pre-olympiad forum: it hasn't received any replies yet :) Anyone?

An algebraic solution:

We construct a bipartite graph with bipartition $(A,B)$. We label the vertices of $A\ 1,2,\ldots,m=|A|$, and the vertices of $B\ m+1,\ldots,m+n$, where $ n=|B|$. Now we consider the adjacency matrix of the graph as the matrix of an endomorphism $T$ of $V=\mathbb Z_2^{m+n}$. Let $V_A$ be the subspace of $V$ consisting of vectors which have $0$ on the last $n$ positions, and $V_B$ the subspace of $V$ consisting of vectors which have $0$ on the first $m$ positions.

We have $T(V_A)\subseteq V_B,\ T(V_B)\subseteq V_A$. Let $T_A: V_A\to V_B,\ T_B: V_B\to V_A$ be the restrictions of $T$. There are two cases: either one of $T_A,T_B$ is not injective, in which case, if, say, $T_A$ is not injective, then we can take a non-null vector in its kernel and regard it as the characteristic vector of $C$, or both $T_A,T_B$ are injective. In this case, $V_A,V_B$ have the same dimension, and both $T_A,T_B$ are actually isomorphisms. This means that $T_Ax=v$, where $v$ has $1$ on its last $n$ positions and $0$ on the first $m$, has a solution $x\in V_A$, and, again, this $x$ can be regarded as the characteristic vector of $C$.


Sequance of polynomial  $P_{n}(x)$ is defined by  $P_{0}(x)=0,P_{1}(x)=x$ and

$P_{n}(x)=xP_{n-1}(x)+(1-x)P_{n-2}(x)$ for $n\geq 2$. Find $P_{n}(x)$.

Don't get angry over nothing. The proof is a trivial induction.
You want to know how I got this solution?

Well, the roots of $Y^{2}-xY+(x-1)\in\mathbb{Q}(x)[Y]$ are $Y_{1}=1$ and $Y_{2}=x-1$.
Then a standard trick about [url=http://en.wikipedia.org/wiki/Linear_recursive_sequence#Linear_homogeneous_recurrence_relations_with_constant_coefficients]linear recurrence[/url] gives some $\alpha(x),\beta(x)\in\mathbb{Q}(x)$
with $P_{n}(x)=\alpha(x)+\beta(x)(x-1)^{n}$ for all $n\in\mathbb{N}_{0}$.
Then $n=0,1$ finds $\alpha(x)$ and $\beta(x)$.

I feared you wouldn't understand all this, so I left that out,
and simply presented the result, which again can be checked trivially by induction.


Let $a,b,c> 1$ three distinct  natural  numbers , prove that

$(1+\frac{1}{a})(2+\frac{1}{b})(3+\frac{1}{c})\leq\frac{91}{8}$

Sorry see the edited version of the problem now and try to solve it. We have $a,b,c>1$

$S=(1+\frac{1}{a})(2+\frac{1}{b})(3+\frac{1}{c})\leq\frac{91}{8}$
If $a<b<c \Rightarrow S\le\frac{91}{8}$
If $a<c<b \Rightarrow S\le\frac{45}{4}$
If $b<a<c \Rightarrow S\le\frac{65}{6}$
If $b<c<a \Rightarrow S\le\frac{59}{8}$
If $c<b<a \Rightarrow S\le\frac{245}{24}$
If $c<a<b \Rightarrow S\le\frac{21}{2}$


inequalities
trigonometry
geometry
circumcircle
geometry proposed

There are three distinct points $A,B,C$ in the circle with  radius [b]1 [/b].

(1) If $AB^2+BC^2+CA^2>8$,prove that $\triangle{ABC}$ is acute triangle.

(2) Prove that $AB^2+BC^2+CA^2\leqq9$. When does equality hold?

[b](1)[/b] Suppose $ABC$ isn't an acute triangle, $AC$ is the largest side, i.e. $\angle ABC\geq \frac{\pi}{2}$. Then we conclude $AC^2\geq AB^2+BC^2$ (from cosine theorem). It follows that $AC^2>4$ -- contradiction, since $AC$ is a chord of unit circle.

[b](2)[/b] Due to Lagrange formula $AB^2+BC^2+CA^2=3I_M$, where $I_M$ is a moment of inertia of mass points $(1)A$, $(1)B$ and $(1)C$ regarding of their center of mass $M$. On the other hand, due to Steiner's teorem $I_M+3MO^2=I_O$, where $O$ is a circumcenter, so $3I_M\leq 3I_O=3(1+1+1)=9$. Equality holds iff $MO=0$, i.e. $ABC$ is an equilateral triangle.


[quote="SUPERMAN2"]Let $ I$ be the incenter of triangle $ ABC$ and $ A_1,B_1,C_1$ the incenters of triangles $ IBC,ICA,IAB$ ,respectively.Prove that $ AA_1,BB_1,CC_1$ are concurrent.[/quote]
[b]Generalized problem :[/b]
See on the figure: $ (AF, AE); (BF,BD); (CD, CE)$ are isogonal lines. Prove that $ AD, BE, CF$ are cocurrent.
We can prove this problem easily by using Tri-Ceva theorem.
Another concurrent: $ AX, BY, CZ$

[quote="jayme"]Dear Mathlinkers,
this is an application of the isogonal theorem.
The point of concurs is the first de Villiers's point (X1127).
Sincerely
Jean-Louis[/quote]
Tell me please dear [b][size=100]Jean Louis[/size][/b], have you in mid some reference about the [b][size=100]Isogonal theorem[/size][/b] you said, or you mean the generalization of the [b][size=100]Jacobi theorem[/size][/b] I have called as [b][size=100]Isogonic theorem[/size][/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=154396]Here[/url] ?

Thank you in advance, Kostas Vittas.

[quote="livetolove212"]...[b]Generalized problem :[/b]
See on the figure: $ (AF, AE); (BF,BD); (CD, CE)$ are isogonal lines. Prove that $ AD, BE, CF$ are cocurrent.
[b]Lemma:[/b] Let $ ABC$ be a triangle and $ P$ be a point distinct from $ A,B,C$. Then the isogonal of the lines $ AP, BP, CP$ wrt $ \angle BAC, \angle ABC, \angle ACB$ are concurrent.
We can prove this lemma easily by using Tri-Ceva theorem.
Return to problem.
Let $ X \equal{} BF\cap CE, Y \equal{} AF\cap CD, Z \equal{} BD\cap AE$ then using the lemma above we get $ AX, AD$ are isogonal lines wrt $ \angle BAC$, similarly to $ BY, BE$ and $ CZ, CF. (1)$
To prove $ AD, BE, CF$ are concurrent we need to prove $ AX, BY, CZ$ are concurrent...[/quote]
Another interesting and well known result in this general configuration, is as follows :

[b][size=100][color=DarkBlue]Let be the points $ P\equiv AD\cap BE\cap CF$ and $ Q\equiv AX\cap BY\cap CZ.$ Then, the line segments $ DX,\ EY,\ FZ$ are concurrent at one point so be it $ R,$ which lies on the line segment $ PQ.$[/color][/size][/b]

Kostas Vittas.

Dera Kostas and Mathlinkers,
in my terminologie, Jacobi's theorem is also the isogonal theorem ... Perhaps I make some confusion.
I know that J.J. A. Mathieu has worked on this notion, but I don't know the exact difference between the works of the two geometers.
Thanks in advance for your advices.
Sincerely
Jean-Louis


Let $ P(n)$ the sum of the digits of the number pairs n.For example, $ P(1234) \equal{} 2 \plus{}4 \equal{} 6$. What is the value of $ P(1) \plus{} P(2) \plus{} P(3) \plus{} ... \plus{} P(100)$?

A) 200
B) 360
C) 400
D) 900
E) 2250

Hi luiseduardo.

I don't understand: could you post the problem in the original Portuguese, so that I can try to help? Obrigado.

OK,

Seja $ P(n)$, a soma dos algarismos pares do número  . Por exemplo, $ P(1234) \equal{} 2\plus{}4 \equal{} 6$ Qual o valor de $ P(1)\plus{}P(2)\plus{}P(3)\plus{}...\plus{}P(100)$ ?

(Braziliam Olympic - 2008)

$ P(1) \plus{} \cdots \plus{} P(9) \equal{}
P(10) \plus{} \cdots \plus{} P(19) \equal{}
P(30) \plus{} \cdots \plus{} P(39) \equal{} \ldots \equal{}$
$ P(90) \plus{} \cdots \plus{} P(99) \equal{} 0 \plus{} 2 \plus{} 4 \plus{} 6 \plus{} 8 \equal{} 20$.

$ P(20) \plus{} \cdots \plus{} P(29) \equal{} 2\cdot 10 \plus{} (0 \plus{} 2 \plus{} 4 \plus{} 6 \plus{} 8) \equal{} 40$.
$ P(40) \plus{} \cdots \plus{} P(49) \equal{} 4\cdot 10 \plus{} (0 \plus{} 2 \plus{} 4 \plus{} 6 \plus{} 8) \equal{} 60$.
$ P(60) \plus{} \cdots \plus{} P(69) \equal{} 6\cdot 10 \plus{} (0 \plus{} 2 \plus{} 4 \plus{} 6 \plus{} 8) \equal{} 80$.
$ P(80) \plus{} \cdots \plus{} P(89) \equal{} 8\cdot 10 \plus{} (0 \plus{} 2 \plus{} 4 \plus{} 6 \plus{} 8) \equal{} 100$.

$ P(100) \equal{} 0$.

Therefore the result is C) $ 400$.

Ah, "the sum of the even digits of the number". Sorry Luiseduardo, I should have realised.


[quote="Ravi B"]I noticed that when posting a new topic, about half of the forums have a field for entering a source, and half don't.  For example, I was going to post a new topic in a local forum (NYC) just now, but I noticed there was no source field.  I think there must have been a source field in that forum about 2 weeks ago, because I posted a topic with a source then.

What's the policy on source fields?  Thanks.[/quote]Well that idea is that in lower-level forums and forums such as this one (no math involved) there is no source field needed, we took it off (at input). If you want to be able to input source fields in a particular forum (or in all for that matter) just let us know and we'll make it available (it's per user based).

[quote="Valentin Vornicu"] Well that idea is that in lower-level forums and forums such as this one (no math involved) there is no source field needed, we took it off (at input). If you want to be able to input source fields in a particular forum (or in all for that matter) just let us know and we'll make it available (it's per user based).[/quote]

What do you mean by lower level forums? Does that mean the middle school forums? I don't see a source box there.

[quote="236factorial"]Does that mean the middle school forums?[/quote]Yes.

Are you supposed to write where you got a posted problem? ie: from an actual test or one you made up??


calculus
derivative
rotation
trigonometry
probability
geometry
3D geometry

I decided to start a jokes thread. I need suggestions as to how to do this. Here is the rules I have thought of:

[b][u]RULES[/u][/b]

JOKES MUST BE RELATED TO MATH!!

1. The jokes can be made up or from anywhere.

2. Repeats must be caught before another joke is posted, or else we just go on. For it to be a repeat, it has to be the exact same word for word. If the repeat is caught, the joke earns 0 points.

3. To post a joke, you must rate the post before yours on a scale of 1-10. This becomes the points for that user.

4. With every joke, there should be a hide tag that contains a scoreboard. To make it easy, just copy and paste it from the last post and add points as necessary.

5. I can't keep a scoreboard myself, so for this to work, every player must follow the rules and post a scoreboard or they will lose half their points.

6. No signups necessary.

7. The first user to 250 points wins, then a new round will start.

I will start.

Teacher: "Who can tell me what 7 times 6 is?"
Student: "It's 42!"
Teacher: "Very good! - And who can tell me what 6 times 7 is?"
Same student: "It's 24!"

[hide="Scoreboard"]
PowerOfPi - [whatever the next person rates my joke]
[/hide]

Here's my joke :(you must know differentiation)
This is by far the best joke i have come across

$ e^x$ and $ x^2$ were walking down the road when they saw a differential operator coming from the other side. $ e^x$ told $ x^2$. My dear friend, if you are differentiated thrice, you shalt die whereas I on the other hand can survive the onslaught of the operator forever, so run away. I shall take care of him! $ x^2$ escapes.

$ e^x$ walks over to the differential operator and says, "Hi, I'm $ e^x$". With a grin, the differential operator replies, "Hi. I am $ (d/dy)$".

Hope you got it!!

[hide="scoreboard"]PowerOfPi - 6
_MM_ - [ next person gives][/hide]

@_MM_: I liked that one.

I just made this up.

[hide="Question"]
What do you call $ 16\pi$?
[/hide]
[hide="Answer"]
Two octupi (plural of octupus)  :rotfl:  :rotfl:  :rotfl:
[/hide]

[hide="Scoreboard"]
PowerOfPi - 6
_MM_ - 8
[/hide]

Is everyone just going to ignore this? (Say yes if you are, PLEASE REPLY)

It's kind of hard to think of a math joke.

Teacher: What is 2k + k?
Student: 3000!
 :rotfl:

scores:
PowerOfPi - 11
_MM_ - 8

I actually got this from a laffy taffy tag.

Why was the 8 afraid of the 7?

Because 7 8 9!   :rotfl:   :|

It was actually "why was 6 afraid of 7?".  :P

A Statistical Department is hiring mathematicians. Three recent graduates are invited for an interview: one has a degree in pure mathematics, another one in applied math, and the third one obtained his B.Sc. in statistics.
All three are asked the same question: "What is one third plus two thirds?"
The pure mathematician: "It's one."
The applied mathematician takes out his pocket calculator, punches in the numbers, and replies: "It's 0.999999999."
The statistician: "What do you want it to be?"

[hide="Scoreboard"]
PowerOfPi - 11
_MM_ - 8
abcak - 7
[/hide]

(I assume Jonathan's post was voided as he didn't actually follow the rules of the game.)

"The number you have dialed is imaginary. Please, rotate your phone by 90 degrees and try again..."

[hide="Scoreboard"]
PowerOfPi - 11
_MM_ - 8
isabella2296 - 8
abcak - 7
[/hide]

@Izzy: Yes, Jonathan's post does not get any points (Sorry Jonathon!)

Q: Why do you rarely find mathematicians spending time at the beach?
A: Because they have sine and cosine to get a tan and don't need the sun!

[hide="scoreboard"]
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abcak - 7
ytrewq - 8
[/hide]

A stats professor plans to travel to a conference by plane. When he passes the security check, they discover a bomb in his carry-on-baggage. Of course, he is hauled off immediately for interrogation.
"I don't understand it!" the interrogating officer exclaims. "You're an accomplished professional, a caring family man, a pillar of your parish - and now you want to destroy that all by blowing up an airplane!"
"Sorry", the professor interrupts him. "I had never intended to blow up the plane."
"So, for what reason else did you try to bring a bomb on board?!"
"Let me explain. Statistics shows that the probability of a bomb being on an airplane is 1/1000. That's quite high if you think about it - so high that I wouldn't have any peace of mind on a flight."
"And what does this have to do with you bringing a bomb on board of a plane?"
"You see, since the probability of one bomb being on my plane is 1/1000, the chance that there are two bombs is 1/1000000. If I already bring one, the chance of another bomb being around is actually 1/1000000, and I am much safer..."

[hide="scoreboard"]
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_MM_ - 8
isabella2296 - 8
ytrewq - 8
abcak - 7
[/hide]

How do you make 7 even?

Take away the s.

[hide="scoreboard"]
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abcak - 7
[/hide]

Two men are sitting in the basket of a balloon. For hours, they have been drifting through a thick layer of clouds, and they have lost orientation completely. Suddenly, the clouds part, and the two men see the top of a mountain with a man standing on it.
"Hey! Can you tell us where we are?!"
The man doesn't reply. The minutes pass as the balloon drifts past the mountain. When the balloon is about to be swallowed again by the clouds, the man on the mountain shouts: "You're in a balloon!"
"That must have been a mathematician."
"Why?"
"He thought long and thoroughly about what to say. What he eventually said was irrefutably correct. And it was of no use whatsoever..."  :P

[hide="scoreboard"]
PowerOfPi - 24
ytrewq - 15
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abcak - 7
[/hide]

A mathematician is flying non-stop from Edmonton to Frankfurt with AirTransat. The scheduled flying time is nine hours.
Some time after taking off, the pilot announces that one engine had to be turned off due to mechanical failure: "Don't worry - we're safe. The only noticeable effect this will have for us is that our total flying time will be ten hours instead of nine."
A few hours into the flight, the pilot informs the passengers that another engine had to be turned off due to mechanical failure: "But don't worry - we're still safe. Only our flying time will go up to twelve hours."
Some time later, a third engine fails and has to be turned off. But the pilot reassures the passengers: "Don't worry - even with one engine, we're still perfectly safe. It just means that it will take sixteen hours total for this plane to arrive in Frankfurt."
The mathematician remarks to his fellow passengers: "If the last engine breaks down, too, then we'll be in the air for twenty-four hours altogether!"

[hide="Scoreboard"]

PowerOfPi - 24
ytrewq - 21
_MM_ - 8
isabella2296 - 8
abcak - 7
[/hide]

[quote="PowerOfPi"]Q: What do you call a teapot of boiling water on top of mount everest?

A: A HIGH-POT-IN-USE[/quote]

I'm sorry, I'm ashamed of myself.

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ytrewq - 97
sdkudrgn88 - 50
1=2 - 41
_MM_ - 29
isabella2296 - 15
james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

A computer software engineer went into a software engineering interview. He was asked what 2 to the power of 16 was. He promptly said "0", and was then told that 32-bit integers were being used instead of 16-bit integers. He then gave the correct answer, 65536.

A mathematician went into the same interview, and was asked what 2 to the power of 16 was. He promptly said, "65,536". He was then told that 16-bit integers were being used instead of 32-bit integers, and then asked, "what's a bit?"

A physicist went into the same interview, and was asked what 2 to the power of 16 was. He promptly said, "Well, I can't give you the exact number, but I can give you an analogy of its magnitude..."

A statistician went into the same interview, and was asked what 2 to the power of 16 was. She told them, "Well, I can give you a quick estimate that it's around 65,000 give or take a myriad or two."

Another computer software engineer went into the same interview, and was asked what 2 to the power of 32 was. He responded, "zero", to which he was told that 16-bit integers were being used instead of 32-but integers. He said, "Oh, then in that case, it's 4,294,967,296."

And a bonus:

Bill Clinton went to the same interview, and was asked what 2 to the power of 16 was. He replied, "Well, depends on your definition of 'was', doesn't it."

It is only two weeks into the term that, in a calculus class, a student raises his hand and asks: "Will we ever need this stuff in real life?"
The professor gently smiles at him and says: "Of course not - if your real life will consist of flipping hamburgers at MacDonald's!"

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PowerOfPi - 112
ytrewq - 97
sdkudrgn88 - 56
1=2 - 41
_MM_ - 29
isabella2296 - 15
james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

(NOTE: Adapted from some Caltech story that I came across at the XKCD forums.)

There was a calculus professor at a university who was known for giving extremely hard tests. They were so hard, and he marked them so strictly, that merely passing the course could merit somebody for a Fields award.

One day, a few weeks before the final exam, the faculty in the math department decided to play a prank on this professor, and plant a student in the test room who would steal a copy and give it to the teachers.

The day of the exam, one of the teachers got the professor to leave the room for a moment just as he was about to start handing the tests out, while the planted student got up and took a copy and left out the back door. This planted student then handed the problems to the teachers, who photocopied the test and then started to work on one problem each.

Two hours later, the teachers got together again, and told the planted student to write the notes in his own handwriting. The teachers had double-checked their work to make sure that they didn't make that many errors, and the student got to copying.

Another hour later, it was time for the tests to be collected. The planted student had all his work done.

The next day, the calculus professor was absolutely jubilant. [i]There is a genius in my class,[/i] he said, [i]an absolute genius! He got 50% on one of my tests...[/i]

In case I get knocked off for plagiarism, here's another one:

You're facing North. You walk straight for two miles, but you realize that you've gone one mile north and then one mile south. Where are you?
Well, wherever it is, it must not be Euclidean space...

[hide="Scoreboard"]PowerOfPi - 112
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sdkudrgn88 - 56
1=2 - 41
_MM_ - 29
isabella2296 - 15
james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5[/hide]

Two men are having a good time in a bar. Outside, there's a terrible thunderstorm. Finally, one of the men thinks that it's time to leave. Since he has drunk a lot, he decides to walk home.
"But aren't you afraid of being struck by lightning?" his friend asks.
"Not at all. Statistics shows that, in this part of the country, one person per year gets struck by lightning - and that one person died in the hospital three weeks ago."

[hide="Scoreboard"]PowerOfPi - 112
ytrewq - 103
sdkudrgn88 - 63
1=2 - 41
_MM_ - 29
isabella2296 - 15
james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5[/hide]

Another mathematicians answering machine (cheesy):

"Please leave your name and number -- But first, a short algebra quiz: How much is 5Q + 5Q? (Pause while caller thinks) 10Q You're welcome!"

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PowerOfPi - 112
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sdkudrgn88 - 63
1=2 - 41
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isabella2296 - 15
james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

A mathematical biologist spends his vacation hiking in the Scottish highlands. One day, he encounters a shepherd with a large herd of sheep. One of these cuddly, woolly animals would make a great pet, he thinks...
"How much for one of your sheep?" he asks the shepherd.
"They aren't for sale", the shepherd replies.
The math biologist ponders for a moment and then says: "I will give you the precise number of sheep in your herd without counting. If I'm right, don't you think that I deserve one of them as a reward?"
The shepherd nods.
The math biologist says: "387".
The shepherd is silent for a while and then says: "You're right. I hate to loose any of my sheep, but I promised: One of them is yours. Have your pick!"
The math biologist grabs one of the animals, puts it on his shoulders, and is about to march on, when the shepherd says: "Wait! I will tell you what your profession is, and if I'm right I'll get the animal back."
"That's fair enough."
"You must be a mathematical biologist."
The man is stunned. "You're right. But how could you know?"
"That's easy: You gave me the precise number of sheep without counting - and then you picked my dog..."

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PowerOfPi - 115
ytrewq - 110
sdkudrgn88 - 63
1=2 - 41
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isabella2296 - 15
james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

Philosophy is a game with objectives and no rules.
Mathematics is a game with rules and no objectives.

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james4l - 14
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abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

A mathematician and a Wall street broker went to races. The broker suggested to bet $ \$10,000$ on a horse. The mathematician was skeptical, saying that he wanted first to understand the rules, to look on horses, etc. The broker whispered that he knew a secret algorithm for the success, but he could not convince the mathematician.
"You are too theoretical," he said and bet on a horse. Surely, that horse came first bringing him a lot of money. Triumphantly, he exclaimed:
"I told you, I knew the secret!"
"What is your secret?" the mathematician asked.
"It is rather easy. I have two kids, three and five year old. I sum up their ages and I bet on number nine."
"But, three and five is eight," the mathematician protested.
"I told you, you are too theoretical!" the broker replied, "Haven't I just shown experimentally, that my calculation is correct! $ 3\plus{}5\equal{}9$!"

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sdkudrgn88 - 63
1=2 - 41
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james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

What is the volume of a pizza, radius z and depth a?

Pizza of course!  :P

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ytrewq - 124
PowerOfPi - 122
sdkudrgn88 - 63
1=2 - 41
_MM_ - 29
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james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

[quote="PowerOfPi"]What is the volume of a pizza, radius z and depth a?

Pizza of course!  :P [/quote]

Well, if the pizza is a cone, then it's a third of a pizza, but whatever...

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james4l - 14
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erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

"The integral of three x squared?"
Said you as if you didn't care.
Then you said to me,
"It's x cubed you see,"
I told you "plus C" 'cause I care.

"The derivative of two x plus three?"
Said I with a hint of mad glee.
I thought I'd derived,
The answer of five,
But I fell in my own trap, poor me.

Why do mathematicians like national parks?

Because they have natural logs.

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PowerOfPi - 126
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sdkudrgn88 - 71
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james4l - 14
bluecarneal - 13
SnowHawk - 8
abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

[quote="PowerOfPi"]Why do mathematicians like national parks?

Because they have natural logs.[/quote]
Eheheh...

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PowerOfPi - 129
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sdkudrgn88 - 71
1=2 - 41
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james4l - 14
bluecarneal - 13
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abcak - 7
erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]

If God created the integers, then Satan created number theory.

$ \frac{12\plus{}144\plus{}20\plus{}3\times\sqrt{4}}{7}\plus{}5\times11\equal{}9^2\plus{}0$

A dozen, a gross, and a score
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is nine squared and not a bit more

dozen=12
gross=144
score=20
Look it up on wikipedia.

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abcak - 7
erw314 - 7
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ak909090-5
[/hide]

*bump* This thread is getting dusty.

A computer science teacher has a policy that if anybody in his class gets exactly 65, he will bump the mark up to a 72. When asked about why this happens, he said, "It's to upgrade them from C to C++." (For Americans, change the marks to 75 and 82 respectively. [i]Then[/i] you'll get it.)

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james4l - 14
bluecarneal - 13
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erw314 - 7
algebra2000 - 7
ak909090-5
[/hide]


algebra
polynomial
quadratics
pigeonhole principle

Let $a,b,c$ be distinct integers. Can the polynomial $(x-a)(x-b)(x-c)-1$ be factored into the product of two polynomials with integer coefficients?

[hide]
No. It's equivalent to finding an integer $x_{1}$ that satisfies the equation $(x_{1}-a)(x_{1}-b)(x_{1}-c)=1$, but since all factors must be integers, all must either be either $-1$ or $1$ and thus at least two of $a,b,c$ must be equal. (We know that, if factorable over the reals, the cubic would have at least one integer root $x_{1}$ since the monic linear factor would force an integer root.)  [/hide]

[quote="Altheman"]quadratic polynomials with integer coefficients don't have to have integer roots[/quote]
But the linear polynomial will. That's what I was talking about. I'll edit.  :)

Assume we can.  Let $P(x)=(x-a)(x-b)(x-c)-1=Q(x)R(x)$.  Plugging in $a,b,c$ we have $P(a)=-1=Q(a)R(a)$, $P(b)=-1=Q(b)R(b)$, and $P(c)=-1=Q(c)R(c)$  Since $Q(x)$ and $R(x)$ have integer coefficients and $a,b,c$ are integers $Q(a,b,c)$ and $R(a,b,c)$ are integers.  So $Q(a)=\pm 1$ then $R(a)=\mp 1$.  The same is true for $b$ and $c$.  By the pigeonhole principle (or common sense) we see that there will be two elements of the set$\{a,b,c\}$, lets call them $i$ and $j$ such that $Q(i)=Q(j)=\pm 1$ and therefore $R(i)=R(j)=\mp 1$.   Therefore the polynomial $Q(x) \mp 1$ and $R(x) \pm 1$ will both have at least $2$ roots, and we can say $deg Q, deg R \geq 2$.  However this would imply $deg P \geq 4$ and since $P(x)=(x-a)(x-b)(x-c)-1=Q(x)R(x)$, $deg P=3$ and we have a contradiction.


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241246

Not fixed for me.

Please describe exactly what appears to be wrong. It's working fine for me.

If I recall correctly, you use Safari. If that's correct, make sure you clear your cache as Safari likes to cache things like crazy, making changes not appear.

It appears to be working now.  However, at the time I posted, it wasn't.  It was the same problem that ernie said.

I use firefox, but sometimes use safari.  This time I was using firefox.


In the Course Information at My Classes section, it says that the time for the class is 7:30PM ET but I just got an email saying that it was on 7:00PM ET. Which is the right time?

I sent out an email earlier today to the entire class explaining that the classroom is opening half an hour ahead of time in order to let new AoPS students learn about the virtual classroom on the first day and make sure they have no problems with the classroom features.

ok. Thanks... but I didn't seem to get that email with the explanation... it only said

Greetings,

The next Art of Problem Solving Intermediate Algebra class is on Tuesday at 7:00
PM ET (4:00 PM PT).

See you there!

Sincerely,

Art of Problem Solving Instructors

I'm not trying to be picky or anything, but I'm just concerned about future emails...
 :oops:

It was a complete different email that I sent out today.


trigonometry
geometry
geometric transformation
reflection

A man directs a beam of sound of frequency $12000$ Hz , at an approaching motor car .He listen to the reflected sound that emitted . What is the apparent frequency heard by the man if the car is travelling at $10$m/s (taking velocity of sound = $330$ m/s )

I know the formula

$f'=\frac{v\pm v_{o}}{v\pm v_{s}}f$ of doppler effect .

In my opinion $v_{o}=0$ and $\pm v_{s}=-10$ . Somehow the answer suggested is using $\pm v_{o}=+10$ and $\pm v_{s}=-10$ . I am not pretty agree with it , so I need opinion from all of you .Thanks

shyong,

Let's see if we can derive the Doppler effect and then maybe it will become clear what values to use.

First, let us call the speed of sound, $c.$

The apparent [u]velocity[/u] of sound is then $\vec{c'}=\vec{c}-\vec{v}_{o}.$

We know then that

$\vec{c'}=f'\lambda'\hat{u_{c'}}=\vec{c}-\vec{v}_{o}$

$f' = \frac{\vec{c}\cdot\hat{u_{c'}}-\vec{v}_{o}\cdot\hat{u_{c'}}}{\lambda'}$

$\vec{c}\cdot\hat{u_{c'}}= c$

$\vec{v}_{o}\cdot\hat{u_{c'}}= v_{o}\cos \theta,$ where $\theta =\begin{cases}0\\\pi\end{cases}$

$\therefore f' =\frac{c \mp v_{o}}{\lambda'}$

If the source of sound is moving then the wavelength is compressed if the source is moving  in the same direction as the wavefront (use minus), or the wavelength is stretched if the source is moving in the opposite direction of the wavefront (use plus).

By similar reasoning as before:

$\lambda' =\frac{c \mp v_{s}}{f}$

$\therefore f' =\frac{c \mp v_{o}}{c \mp v_{s}}f.$

In the problem the source of sound is its reflection from the approaching car, therefore one chooses $+v_{o}= 10$, the wavefront is approaching the man $\theta =\pi$, and $-v_{s}=-10$, the car (source) is moving in the same direction as the wavefront.


For an integer $n \geq  0$ , a permutation $(a_0 ,a_1 ,a_2 ,... , a_n )$ of $(0,1,2,... ,n)$ is said to be a  square permutation if $k + a_k$  is a square for each k , $0 \leq  k \leq  n$ . Show that for every positive integer n there exists a square permutation of $(0,1,2,... ,n)$


For all integers $ n$ greater than $ 1$, define $ a_n \equal{} \frac {1}{\log_n 2002}$. Let $ b \equal{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5$ and $ c \equal{} a_{10} \plus{} a_{11} \plus{} a_{12} \plus{} a_{13} \plus{} a_{14}$. Then $ b \minus{} c$ equals

$ \textbf{(A)}\ \minus{} 2 \qquad \textbf{(B)}\ \minus{} 1 \qquad \textbf{(C)}\ \frac {1}{2002} \qquad \textbf{(D)}\ \frac {1}{1001} \qquad \textbf{(E)}\ \frac {1}{2}$

Hmm wow again if you knew that 2002=11*13*14 its a lot easier.

Pretty straightforward for #22

Notice that it just is (log n)/(log 2002).

So b=log(2*3*4*5)/(log 2002)-using the fact that log(a)+log(b)=log(ab)
and c=log(10*11*12*13*14)/(log2002)

so b-c=log(2*3*4*5/10*11*12*13*14)/(log 2002) using the fact that log(a)-log(b)=log(a/b)

So canceling its just log(1/2002)/(log2002)=-1, B


[img]http://www.mathhelpforum.com/math-help/attachments/number-theory/10534d1237150030-10-2006-n-2-n-0-n-10-2006-p1.jpg[/img]

is it no such number existed?
cos the minimium n is 10^2006?

please help, thanks.

[quote="Rust"]There are 4 solutions, $ n \equal{} 0,1$ and $ n \equal{} n_1 \equal{} 5^{2^{204}}\mod 10^{2006}, n \equal{} 10^{2006} \plus{} 1 \minus{} n_1$.[/quote]

m|n means n=mq for some integer q
$ n^{2}\minus{}n$ should be greater or equal to $ 10^{2006}$

isn't it?


advanced fields
advanced fields theorems

What do you think about [url=http://www.math.purdue.edu/~branges/riemann-hecke.pdf]this[/url] paper. It is poorly written but Yashowanto Ghosh says he is fairly certain that this approach is valid. I found this at NewScientist 22nd March 2008.

DeBranges comes up with a proof of Riemann Hypothesis about twice a year. I know a few people who honestly read the previous versions and detected the mistakes. It has always been some stupid little thing hidden somewhere in the middle of otherwise quite an ingenious argument. So I'm rather sceptical about the current version as well. I will be happy to be proved wrong, of course :).


trigonometry
calculus
calculus computations

For $ x\in\mathbb{R}$ let \[ f(b) \equal{} \max(|\sin{x} \plus{}\frac{2}{3\plus{}\sin{x}} \plus{} b|)\] Find the minimum of $ f(b)$ over $ b\in\mathbb{R}$.

[hide="solution"] Compute $ \max_x \left \{\sin x \plus{} \frac {2}{3 \plus{} \sin x} \plus{} b \right \} \equal{} \frac {3}{2} \plus{} b$ and $ \min_x \left \{\sin x \plus{} \frac {2}{3 \plus{} \sin x} \plus{} b \right \} \equal{} b$ by differentiating to find the extrema are at $ \pm \frac {\pi}{2}$.  Taking absolute values we therefore conclude
\[ f(b) \equal{} \max \left \{\frac {3}{2} \plus{} b, \minus{} b \right \}
\]
from which it is easy to see that $ \min_b \{ f(b) \} \equal{} \frac {3}{4}$. [/hide]


I have 7 books I want to arrange on a shelf. Two of them are math books, and one is a physics book. How many ways are there for me to arrange the books if I want to put the math books next to each other, and put both of them to the left of the physics book?

Can I suggest that "left" is clarified to not mean "immediate left"?

What's wrong with this method?

If we consider the physics book to be in the 3rd place from the left, then there's only one possible seating for the math book (considering that as a single entity first); then if it's in the 4th place, there will be 2 seatings for the math books (1,2 and 2,3). Continuing the pattern we find the total number of arrangements to be $ 5!*2!(1\plus{}2\plus{}3\plus{}4\plus{}5)\equal{}120*2*15 \equal{} 3600$.

Alternatively, one might say that the number of ways to choose spots for the math textbooks and physics textbook is $ \binom {6}{2}$. $ 2!$ ways to arrange the math textbooks, and $ 4!$ to arrange the remaining books. Thus: $ \binom {6}{2} 2! 4! \equal{} 720$

[quote="ilikecake"]There's something wrong or unclarified I found here- the math textbooks were not said to have distinguishability, so, they could be counted as the same since the unclarified usage was only "math books." Thus another answer could be if you didn't account for switching the math books around thus giving [b]360[/b].[/quote]
The math books are distinguishable

[hide="Alcumus Solution"]We can arrange the books on the shelf, pretending that the two math books are actually one book because they must be next to each other. In exactly half of these arrangements, the math books will be to the left of the physics book, giving $ 6!/2=360 $ arrangements. However, in each of these arrangements there are two ways to arrange the math books, so the total number of arrangements is $ 360\cdot2=\boxed{720} $.
[/hide]

If we consider the 2 math books as 1 book, then there are technically 6 books, so there are $6!$ ways to arrange them. In 3/6 of the different ways of ordering the subjects, the math books are to the left of the physics book, so we divide $6!$ by 2 to get $360.$ Then we multiply by $2!$ because there are $2!$ ways to arrange the 2 math books, giving us an answer of $720.$

[quote=MathArt4]
[hide=What's wrong with this?]
There are $5$ ways for the physics book. There are $2$ ways to arrange the math books. For the rest of the books there are $24$ ways.
$$5*2*24=\boxed{240}$$[/quote]

You did not assume the math books were together or to the left of the physics books.

[hide=Solution]My solution: You want to put the 2 math books together. There are 6 places you can do this. Next to find the number of ways, we have 2!*5!=240. 240*6=1440. You want the math books to be on the left of the physics book, so you divide 1440 by 2 to get $720$ as our final answer.[/hide]

[quote=kr1234]I thought that the math books had to be to the immediate left of the physics book - I wish this problem was more clear.[/quote]

You can notify the aops staff about this. I wish it were more clear as well.

[quote=kr1234]I thought that the math books had to be to the immediate left of the physics book - I wish this problem was more clear.[/quote]

The question says they have to be to the left, but it doesn't say they have to be to the immediate left, and so you would be safe to assume they could be anywhere to the left. However, you would not be safe to assume they must be placed to the immediate left, because it does not say that specifically in the problem. You can ask the AoPS staff to help with this, but I hope this helped for the future! :)

It also doesn't clarify that the books are indistinct. They seem like they should have been...

[quote=Williamz1483][u]DO NOT[/u] POST THE ALCUMUS SOLUTION!!![/quote]

You can if you specify you did. However, it is not recommended.

[hide=Solution]We know that by symmetry, half of the time the math books will be to the left of the physics book and half of the time they will be to the right of the physics book. So, we have 7!/2 = 2520 ways. But, the math books must be together, which is like 6 books, so it is 6!/2 = 360. However, we know that we can order the books 2 ways, so the answer is 720.[/hide]

The answer you have seems to assume the math books must be next to the physics book, not just somewhere to the left of it. It is still not right. You want a block of math-math-physics and there are 2! ways to arrange these books into a block. Counting the block as a book, you now have five books, so there are 5! ways to arrange the books, for a total of 2!⋅5!=240 ways to arrange them.

If you don't require that the math books and physics book be next to each other, there are 2! ways to arrange the two math books into a pair. There are then 6! ways to arrange the pair and the five other books. Half of these have the math books somewhere to the left of the physics book, so there are 2!⋅6!⋅12=720 ways to arrange the books now.

[quote=mathapple101]I thought it said "immediate left." I got 240.[/quote]

Yeah that's what I thought at first

Did anyone bash casweork and get 720 only to see urself being dumb after reading the solution? lol me

[hide]
MM _ _ _ _ _, 5x4!, 5 ways for the physics book, and 4! ways for the other books.
Notice that each time MM goes to the left, the places for the physics book get less by 1, so we have 4!(5+4+3+2+1) = 360, and then since all books are distinguishable, even the math ones, multiply it by 2 for arranging the math books, to get 720
[/hide]

Tip: Consider the maths books as one unit to start with, as they must be together. Remember to account for the arrangements within the two books though!
[hide=Solution]First, consider the two math books as one unit, since they must be next to each other. Then, there are 6 units to arrange: the math books unit, the physics book, and the other $4$ books. Since the math books unit must be to the left of the physics book, there are $3$ possible positions for the math books unit: it can be in the first, second, or third position from the left. For each of these positions, there are $5!$ ways to arrange the other $5$ units. So far, we have $3×5!$ arrangements.
Next, consider that within the math books unit, there are $2!$ ways to arrange the two math books. So, the total number of arrangements is $3\times5!\times2!=\boxed{720}$. [/hide]


Does anybody have any Text about Vibration for free???;)  ;)

You can get this book [b]Vibrations and Waves[/b]
[url=http://www.lightandmatter.com/bk3a.pdf]Page 1 to 50[/url]
[url=http://www.lightandmatter.com/bk3b.pdf]Page 51 to end[/url]


The question is in the attachement, please!

[hide]
Supposed that the inverse is : $ {C}^{\minus{}1}$ which : $ {X}_{11}$ is the block in the top left, $ {X}_{12}$ is the block in the top right, $ {X}_{21}$ is the block in the bottom left and $ {X}_{22}$ is the block in the bottom right, then we have the equation :
$ A{X}_{11}\equal{}I$, then $ {X}_{11}\equal{}{A}^{\minus{}1}$, $ {X}_{11}\plus{}B{X}_{21}\equal{}0$ then $ {X}_{21}\equal{}\minus{}{(AB)}^{\minus{}1}$, $ A{X}_{12}\equal{}0$ then $ {X}_{12}\equal{}0$ and $ {X}_{12}\plus{}B{X}_{22}\equal{}I$, then $ {X}_{22}\equal{}{B}^{\minus{}1}$, then we can construct $ {C}^{\minus{}1}$ as the inverse.
[/hide]

AkiM, couldn't you have written that, instead of attaching a word file? Here's the problem:

Let $ A$ and $ B$ be $ n \times n$ matrices such that they both have inverses $ A^{\minus{}1}$ and $ B^{\minus{}1}$, respectively. Then show that

$ \left [ \begin{array}{cc} A & 0 \\
I_n & B \end{array} \right]$

also has inverse and express it in terms of $ A^{\minus{}1}$, $ B^{\minus{}1}$, and $ 0$.

Thank you, Carcul and Qruni! Had no time to do that.


50 samurai live in a castle across a road from 50 ninjas.

Samurai rules:
All the samurais meet every day for lunch - this is the only time they see each other.
If you discover that you have a dot on your forehead, kill yourself that night alone.
You can think as logically as the best mathematicians.

Ninja rules:
You are at war with the samurai.
If you commit a crime to the samurai, you must tell them somehow.
You cannot tell a lie.

One fateful night, a ninja crosses the road to the samurai castle.  He places dots on some amount of samurais' foreheads.  Because he has to tell, he writes on the samurais' lunch table "I placed dots on anywhere from 1 to 30 samurais."  The next day at lunch, the samurai see the note and see each other.

17 days later, all the dotted samurai are dead.  No samurai looked in a mirror, talked to another samurai, or drew a smiley face with a dot on its forehead and pointed at a samurai sitting next to him.  How many samurai did the ninja dot?

The solution to this problem is in "Solution to 50 samurai"

Hint: [hide]Think of small examples.[/hide]

DO NOT POST ANSWERS HERE!  DELETE THIS AS QUICKLY AS POSSIBLE!!
Post your answers on "open this quickly please"

do the samurai know when there are no dots left?


i tink we need to revive this forum...

so how did everyone do? i felt the AIME II was mildly harder than AIME I. i got a 10, so i don't tink i'm gonna make the USAMO... :(

how did everyone else do?

10 will definitely make it unless you got < 110 on the AMC12, in which case you need to worry a little.

my AMC12 score was monsterously bad...so i do need to worry a little...


geometry
geometric transformation
reflection
conics
ellipse
perpendicular bisector
advanced fields

Find out the locus of the points from which we can draw perpendicular tangents to an ellipse.

I'll use the following 2 lemmas:

Lemma 1

The locus of the reflections of F in a variable tangent is a circle with center F' and radius 2a (standard notations).

Proof: Assume th tangent touches the ellipse at T. The tangent is the exterior bisector of FTF' (well-known optical property of the ellipse), so the reflection of line TF in the tangent is the same as the line TF'. Let f be th reflection of F in the tangent. Then T, f anf F' are collinear and F'f=F'T+Tf=F'T+TF=2a and we're done.

Lemma 2 (I'm a bit lazy to prove it here; try it yourself, it's not hard)

The tangents through M touch the ellipse at T and T'. Then the angles <TMF and <T'MF' are equal.

To our problem:

Assume M is on the required locus. Let f be the reflection of F in the tangent through M touching the ellipse at T. From the conditon <T'MT=90 and lemma 2 we easily see that <F'Mf=90. At the same time, Mf=MF because MT is the perpendicular bisector of Ff. All of this translates to MF2+MF'2=F'f2=4a2 (by applying Pythagoras in the right triangle MF'f and by using F'f=2a and Mf=MF), which shows that M is on a circle centered at the center of the ellipse, of radius.. well, you calculate it by using the median theorem in the triangle MFF' :).

Lets go for a shorter solution!  :blush:

Because the 2 tangents are perpendicular then they will have m and -1/m as angular  coeficients. So we write their eqs(we shall use standard notations):
y=mx+sqrt(a^2m+b^2)
y=-1/m+sqrt(a^2/m^2+b^2)

so eliminating m we get that x^2+y^2=a^2+b^2, so the locus is Monge's circle. I hope I am right! :maybe:

cheers! have fun! :D :D :D

I'm not sure but I think this locus is named Monge's circle


n(n+1)(2n+1)/6

Then, for n=50 the answer is 42925.[/code]

[quote="Ceva Theorem"]What is the sum of the following sequence:?

$1^2 + 2^2 + 3^2 + 4^2 +.........+50^2$[/quote]

[hide]
The answer is:
$\frac{n(n+1)(2n+1)}{6}$

So, just substitute $n=50$, and you should get the answer.

Therefore, the answer is $\boxed{42925}$.[/hide]

[quote="hoshattack"]how did you get the n(n+1)(2n+1)/6 ?[/quote]

Formula for sum of squares.

What ignite means is the sum of consecutive perfect squares starting with 1. ;). For the sum of 1^3+2^3+3^3...+(x-1)^3+x^3=(1+2+3+...+(x-1)+x)^2. So 1^3 +2^3+3^3=(1+2+3)^2=36.

[quote="pgpatel"]What ignite means is the sum of consecutive perfect squares starting with 1. ;).[/quote]

Well couldn't you apply it to any consecutive sequence of squares with a little extra work?

Question: Find the sum of $10^2+11^2+12^2+...+50^2$

$\frac{(50)(51)(101)} {6}=42925=\text {Sum of first 50 squares}$

$\frac{(9)(10)(19)} {6}=285=\text {Sum of first 9 squares}$

Subtract: $42925-285=\boxed{42640}$

You can do any arithmetic sequence of squares (the number squared)

because is you do something like

3^2+6^2+9^2+12^2+15^2

that is equal to (substitute the 3 instead of a 1)

(3+6+9+12+15)(33)/3

if you want something like 2^2+5^2+8^2+11^2

then you do

9-4-5
36-25=11
81-64=17

and you have an arithmetic sequence to just subtract (or add in other cases)

If anybody finds an easier way to do that, I want to know too

You can pretty much do the sum of any squares (3^2 + 6^2 + ...) using n*(2n+1)*(n+1)/6... just write in summation notation and expand.

The above's short form would be (3x)^2, for example, and expanding yields 9x^2.

what about 2^2+5^2+8^2+11^2...98^2 for example?

*ehehheehee is not happy about (3x-1)^2=9x^2-6x+1*

Well, that can be written as:

$\displaystyle\sum_{i=1}^{k}(3i-1)^2$, right?

Expanding yields:

$\displaystyle\sum_{i=1}^{k}9i^2-6i+1$, and break this up, and use individual sums.

$9\displaystyle\sum_{i=1}^{k}i^2 - 6\displaystyle\sum_{i=1}^{k}i + \displaystyle\sum_{i=1}^{k}1$

$\frac{9i(i+1)(2i+1)}{6} - \frac{6i(i+1)}{2} + i$. You can just simplify the rest out  ;)

Note here that we have used

$\displaystyle\sum_{i=1}^{k}i = \frac{i(i+1)}{2}$

and

$\displaystyle\sum_{i=1}^{k}i^2 = \frac{i(i+1)(2i+1)}{6}$.

sorry, I don't know calculus or algebra 2...I guess my crude arithmetic sequence thing is going to be the best thing I can do...

Well, if you have AoPS 1 or AoPS 2, try to read up on the "sequences and series" areas. Especially volume 2 (it might be a bit intimidating first, but you'll get used to it), because it has really good tips on how to solve those types of problems. And especially read the sections on summation notation.


calculus
integration
trigonometry
limit
calculus computations

last [b][color=red][u]tough[/u][/color][/b] problem for today...  :D

show

$\int_{0}^{\infty}\;\left\{ \;\left( \frac{\sin{\theta}}{\theta}\right) \cdot \left( \frac{\sin{\frac{\theta}{3}}}{\frac{\theta}{3}}\right) \cdot \left( \frac{\sin{\frac{\theta}{5}}}{\frac{\theta}{5}}\right) \;\right\}\;\;d\theta\;\;=\;\;\boxed{\frac{\pi}{2}}$

can't we use complex variables to solve the problem ?  :D

wouldn't something like $\boxed{15\;\int_{\mathbb{C}}\;\left\{ \frac{\mathbb{I}m (e^{\imath z}) \cdot\mathbb{I}m (e^{\frac{1}{3}\imath z}) \cdot\mathbb{I}m (e^{\frac{1}{5}\imath z})}{z^{3}}\right\}\;dz}$ work?

[quote="Timestopper_STG"]Since $\int_{0}^\infty\frac{\sin\theta}{\theta}d\theta=\int_{0}^\infty\frac{\sin\theta}{\theta}\left(\frac{\sin\frac{\theta}{3}}{\frac{\theta}{3}}\right)d\theta=\int_{0}^\infty\frac{\sin\theta}{\theta}\left(\frac{\sin\frac{\theta}{3}}{\frac{\theta}{3}}\right)\left(\frac{\sin\frac{\theta}{5}}{\frac{\theta}{5}}\right)d\theta=\frac{\pi}{2}$
So what about $\int_{0}^\infty{\Pi_{k=0}^{n}\left(\frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)d\theta}$  :D[/quote]

you mean what happens if and when $n\;=\;7$ ?  :rotfl:

:blush: ...I mean for any non-negative integer n... :blush:

the thing is, as borwein and borwein have shown,

$\int_{0}^{\infty}\;\prod_{k=0}^{1}\left( \frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)\;d\theta\;=\;\boxed{\frac{\pi}{2}}$

$\int_{0}^{\infty}\;\prod_{k=0}^{2}\left( \frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)\;d\theta\;=\;\boxed{\frac{\pi}{2}}$

$\cdots$
$\cdots$
$\cdots$

$\int_{0}^{\infty}\;\prod_{k=0}^{6}\left( \frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)\;d\theta\;=\;\boxed{\frac{\pi}{2}}$

but

$\int_{0}^{\infty}\;\prod_{k=0}^{7}\left( \frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)\;d\theta\;<\;\boxed{\frac{\pi}{2}}$

and onwards.... (which i what i meant earlier). they also showed that

$\lim_{N\to\infty}\int_{0}^{\infty}\;\prod_{k=0}^{N}\left( \frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)\;d\theta\;=\;\boxed{\alpha\cdot\frac{\pi}{2}}$

$(where\;\;\;\alpha\to0\;\;\;as\;\;\;N\to\infty)$

perhaps it is a [u]strict[/u] [b]monotonic[/b] convergence, i don't know yet for sure ?  :maybe:

i suppose one interesting thing to study would be to observe the "general" behavior of

$\int_{0}^{\infty}\;\prod_{k=0}^{N}\left( \frac{\sin\frac{\theta}{(2k+1)}}{\frac{\theta}{(2k+1)}}\right)^{M}\;d\theta\;\;\;\quad (where\;\;\;\{M,N\}\in\mathbb{N}^{+})$

and essentially see what else happens as $\lim_{M\to\infty}$, and also as $\lim_{N\to\infty}$ for one particular $M$! :D

i find didilica's approach quite interesting.  :ddr:


For n natural n>6 prove that are a,b natural a,b>1 ,gcd(a,b)=1 and a+b=n.

Pick $a$ a prime $<n, n-a \neq 1$ that does not divide $n$.  For $n > 6$ this is always possible.  Then $gcd(a, n-a) = 1$.


geometry
analytic geometry
graphing lines
slope

[b]Find where the line $ y = \frac{3}{2}x - 7 $ cut the axes:[/b]

When the line cute the y-axis we have $x = 0$, then $y = \frac{3}{2}0 - 7 \quad \therefore \quad y = -7$.
It's similar for x-axis. We have $y = 0$, then $\frac{3}{2}x - 7 = 0 \quad \therefore \quad x = \frac{14}{3}$

[b]Does $(-2, -2)$ lie on the line with equation $3x + 4y = -14$ ? [/b]

$(a,b)$ lies on the line $\Leftrightarrow x = a \wedge y = b$ satisfies line equation.

Then check if $x = -2 \wedge y = -2$ satisfies $3x + 4y = -14$.

$3(-2) + 4(-2) = - 6 - 8 = -14$

That means the pois lies on the line.

[quote="Ash_"]hello, we recently got given this assignment to do. im usually pretty good with maths but for this assignment i've been struggling. i've found this forum as a result of an internet search.

i have a few questions that i can't find the formula for, and don't understand.

such as this one.

'Find where the following lines cut the axes

y=(3/2)X - 7

'

i have found x and y by using.

X = (3/2) - 7 (please correct me if im wrong)

and Y = (3/2)x1.5 - 7 (again please correct me if im wrong, i got the 1.5 from X in the previous answer.

----

this question aswell.

'Does (-2, -2) lie on the line with equation 3x + 4y = -14 ?

i have no clue what to do here.

---
theres also

'Find the equation of the line:
 through P(-1, 5) which is parallel to the line with equation 3x + 4y = -8
 and through Q(-4, -1) which is perpendicular to the line with equation y = 3x +5.

i remember someything about a parallel line being inverse opposite of something.

---

thanks for any help.[/quote]

This should be in the Basics Forum...

[hide]
Where the line hits the x-axis, the y variable is equal to 0. When the line hits the y-axis, the x variable is equal to 0.

X-Axis hit:
$0=(3/2)x-7$
$\frac{3x}{2}=7$
$3x=14$
$x=\frac{14}{3}$

Y-Axis hit:
$y=0-7$
$y=-7$
[/hide]

[hide]
For this problem, just plug in the point into the equation:

$3x+4y=-14$
$3(-2)+4(-2)=-14$
$-6-8=-14$
$-14=-14$

Since this fact is true, the point IS on the line...
[/hide]

[hide]
$3x+4y=-8$
$y=-\frac{3}{4}x-2$
Plug in the point to find the y-intercept
$5=\frac{3}{4}+?$
$?=\frac{17}{4}$
The equation of the line is $y=-\frac{3}{4}x+\frac{17}{4}$

$y=3x+5$
Perpendicular lines have negative reciprocal slopes of each other.
$y=-\frac{1}{3}x+?$
Plugging in the point...
$-1=\frac{4}{3}+?$
$?=-\frac{7}{3}$
The equation of the line is $y=-\frac{1}{3}-\frac{7}{3}$
[/hide]


inequalities
logarithms
number theory
real analysis
real analysis unsolved

Let $H_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n}$. Show that, for each positive integer $n$,

        \[\sum_{d/n}d\leq H_{n}+2\exp(H_{n})\log(H_{n}).\]

Now cut the 2 in the right hand side and prove it!  :wink:

In "A beautiful mind" John Nash is quoted as saying
[quote]people have a mental picture that this is a difficult problem. Maybe that's the stumbling block. Maybe, if people didn't realize that the problem was "hard", they could solve it.[/quote]
(in defense of putting an equivalent of FLT on a final exam.)

But seriously, I get stuck after writing $n=\prod_{i}p_{i}^{r_{i}}$ ($p_{i}$ prime) and $\sum_{d|n}d=n\prod_{i}(1+p_{i}^{-1}+p_{i}^{-2}+\dots+p_{i}^{-r_{i}})\le nH_{n}$...

Mlok, this is not such an easy problem as you think. Like Harazi said, if you cut $2$ we obtain an inequality equivalent to Riemann hyphotesys.
I will post soon some ideas of mine on this topic concerning the inequality Harazi was talking about. :wink:

Indeed, this is not an easy problem and the one without $2$ is intractable. Cezar, are you planning a proof of RH?  :wink:  I can prove, using Merten's theorem, that for all $\epsilon>0$ there exists an explicitly calculable $C$ such that $\sigma(n)\leq (1+\epsilon)e^{H_{n}}\ln H_{n}$; but that's not nice, so that I'm interested in a nice approach for the inequality with $2$, that is the original post by Cezar.

How easy is to prove a weaker one  :wink: :
\[\sum_{d/n}d\leq c \exp(H_{n}) H_{n}. \]
And if you prove this one:
\[\sum_{d/n}d\leq H_{n}+\exp(H_{n})\log(H_{n}). \]
anyway don't do like Perelman.   :P
BTW Lagarias seems to be one of the living math gods.

Harazi, can you shows us how you prove that using Mertens theorem? :)
I think I have two solutions for Lagarias problems: one of them is mine and the other oen is due to Walter Janous. :wink:

After the fantastic course you have in analytic number theory, I will not even think to post the way in which you can deduce those results from Merten's theorem: that's a piece of cake! And by the way, I don't think it is very useful to say you have a solution and someone else has (of course!) if you don't post it...

Can anyone post a solution to this anyway?  :(


Find all solution of the equation $ a^{n}\plus{}b^{n}\equal{}c^{n}$ where $ n\geq3$

[quote="Elio (n)"]Find all solution of the equation $ a^{n} \plus{} b^{n} \equal{} c^{n}$ where $ n\geq3$[/quote]
$ \{\left(a,b,\left(a^n\plus{}b^n\right)^{\frac{1}{n}}\right)|a>0,b>0\}.$ :lol:

[quote="mathson"]but if you mean $ n \in \mathit{Z}$
there are no solution[/quote]
What is your proof?  :lol:

I've got a very nice proof; however, it does not fit into this webpage.

Refering to Fermat's Last theorem is true that there are no integers that sodisfy the equation.

 But we have the solutions       1)$ a \equal{} 0 ;b \equal{} 1 ; c \equal{} 1$
                                             2)$ a \equal{} 1 ; b \equal{} 0 ; c \equal{} 0$
                                             3)$ a \equal{} 0 ; b \equal{} 0 ; c \equal{} 0$
                                                                                Isn't it?

@AlbanianEagle That is no other proof of this exept ANDREW WILES's proof. Correct me if I am wrong.

If you have your own proof you are goning to win the ABEL prize  this year   :lol:

Elio, bertram was refering to what Fermat said in his book,

Fermat:  I have discovered a truly marvellous proof of this, which this margin is too narrow to contain.

, a little subtle though lol.

[quote="BOGTRO"][quote="Elio (n)"]Find all solution of the equation $ a^{n} \plus{} b^{n} \equal{} c^{n}$ where $ n\geq3$[/quote]

There are no positive integer solutions.

If you don't understand why, then... uh...

[url=Fermat's Last Theorem]http://www.fermatproof.com/[/url][/quote]

lol

[quote="Sage!"]Ufortunatelly this isn't a correct proof.[/quote]

Unfortunately this man is correct!

So did I miss that elementary proof was found?

[quote="Elio (n)"]Find all solution of the equation $ a^{n} \plus{} b^{n} \equal{} c^{n}$ where $ n\geq3$[/quote]

Well, if you notice, there is [b]no[/b] indication that $ a$, $ b$, or $ c$ have to be integers. Thus there are infinitely many solutions.


Bueno Leonardo,

El post era más como un proyecto personal mío al que queria invitar a los demas a participar que una propuesta para hacer mas grande el foro. He hablado con Valentín, y el dice que mientras lo que se pida sea razonable, pues que el no tiene problema.

Ahora, como vos decís, es necesario tener más participación de la gente, me parece que quienes mas participamos somos vos, yo y Pascual.

Por cierto, que división tenías en mente cuando escribiste el post anterior?

[quote="manuel"]Buena idea David vamos a dividirlo...[/quote]

Hey hey, ¿Alguien leyó el primer post? Yo lo que estoy proponiendo es [b]discutir un temario de olimpiadas[/b], que abarque todos los niveles, desde las más simples olimpiadas, hasta la IMO.

La idea de [i]dividir[/i] el foro es en realidad de Leonardo. Yo la única modificacion que he propuesto es la de hacer un foro para ayudar en español a extrangeros, por que muchos mensajes son precisamente para eso, y me parece que deberian ir aparte.

Yo no tengo ningún problema en proponer una división del foro a los administradores, sin embargo no algo a tontas y a locas, debería ser algo planeado y con algún propósito.

Por favor, ¿A alguien le interesa lo del temario?

Hola  a todos¡¡¡ djimenez: yo opino que tu idea acerca del temario es maravillosa me encantaría trabajar contigo y prepararlo. en términos de la dificultad de los problema yo opino que por cada a pesar q  determinar la dificultad de un problema es dificil yo creo que esto se debe a que un problema tenga mas de una solucion supongamos una larga y complicada y otra corta e ingeniosa sin embargo la solucion corta e ingeniosa generalmente requiere de herramientas mui complicadas y dificiles de dominar y y la larga requiere de mucha paciencia  no mucho pensamiento y por lo tanto un problema en terminos de sus soluciones diferentes no varía tanto en terminos de su dificultad. ahora yo creo mas que la dificultad de un problema depende mas de quien lo contempla. pero yo opino que uno al resolver un problema sabe para que nivel seria facil el problema y para cual no?? es decir yo me acuerdo cuando apenas iniciaba que veia problemas muy dificiles q ahora ya no. y yo puedo identificar por alumno quien se le puede hacer facil y a quien no. yo opino que para empezar seria util que cada quien explique su nivel. problemas clasicos que ha resuelto. herramientas que utiliza etc. y con esta informacion seria mas facil dividir los problemas etc. que opinan???

David, no entiendo bien lo del temario, me gustaria me lo explciara, porque estoy interesado :)

Por lo que veo, hay un gran malentendido, pues hay unos hablando de dividir el foro, otros de un temario que no se que es.... espero alguien aclare todo...

Y obviamente culquiera sea la resolucion final, pues pienso ayudar con mi material....

Suerte a todos!

Jeje, al parecer yo tampoco tengo claro que es un temario, porque pense que se referian a dividir el foro, de ahi saque mi idea  ;). Si alguien puede aclarar seria excelente.

Bueno, explico nuevamente. Mi idea es hacer un listado lo mas comprensivo posible de los temas y los resultados que un muchacho debe aprender (fuera de, digamos, lo que se ve en un colegio regular hasta por alli del noveno grado) para tener alguna oportunidad en la IMO. Tengo una version preliminar, en Ingles y medio incompleta. Si estan deacuerdo podriamos empezar a discutirlo.

Aaaaaaaah, ya, bueno, eso suele ayudar mucho para la gente que esta comenzando porque uno no sabe que estudiar, pero tiene los animos. Si puedo ayudar en algo, dejenme saber.

[quote="Pascual2005"]listo, me gustaria conocerlo David![/quote]

Bueno, dado que es muy largo voy a hacer distintos post para cada topico, asi queda un poco mas clara la discucion. Pero sera en la noche por que tengo que salir y no he terminado la edicion del documento.

Bueno señores, revisen los post que voy a hacer en cada uno de los temas en especifico. Porfa, que lo de combi se quede en el topic de combi, lo de geo (tal vez se mezcle un poco con trigo) pero que no se salga de alli, y asi sucesivamente.

Espero sus criticas!!


calculus
geometry
trigonometry
calculus computations

An area of farmland along a stone wall is to be fenced. (No fence is needed along the stone wall.) There are 600m of fencing available. What is the maximum rectangular area that can be fenced?

A conical tank has a radius of 5 ft and a height of 10 ft. Water runs into the tank at the constant rate of 2 cubic feet per minute. How fast is the water level rising when the water is 6 ft deeps Round your answer to the nearest hundredth.

Two drivers begin at the same location and drive in perpendicular directions. Driver A travels due north at a rate of 30 miles per hour; Driver B travels due west at a rate of 50 miles per hour. At what rate, in miles per hour, is the distance between the two drivers changing 4 hours into the hike?

An area of farmland along a stone wall is to be fenced. (No fence is needed along the stone wall.) There are 600m of fencing available. What is the maximum rectangular area that can be fenced?

 _____x_____| Wall
|
| y
|__________|

A =xy ; 2x+y=600

y=600-2x

A=x(600-2x) = 600x-2x^2

A'(x) = 600-4x=0 when x= 150

Max area  = 150(600-300) = 45,000. Correct?

A conical tank has a radius of 5 ft and a height of 10 ft. Water runs into the tank at the constant rate of 2 cubic feet per minute. How fast is the water level rising when the water is 6 ft deeps Round your answer to the nearest hundredth.

V=(1/3)x(pi)x(r^2)x(h) ;  r/h = 5/10=1/2 , r = h/2

V=(1/3)x(pi)x[(h/2)^2]x(h) =  (pi/12)x(h^3)

.... dh/dt = 0.07 ft./min.?

Two drivers begin at the same location and drive in perpendicular directions. Driver A travels due north at a rate of 30 miles per hour; Driver B travels due west at a rate of 50 miles per hour. At what rate, in miles per hour, is the distance between the two drivers changing 4 hours into the hike?

s^2=x^2+y^2

(2s)x(ds/dt) = (2x)x(dx/dt) + (2y)x(dy/dt)

(ds/dt) = (1/2s)[(2x)x(dx/dt) + (2y)x(dy/dt)]

ds/dt= 58.3 mph?

Those all seem to be correct.

If you plan to post here again, it would be very helpful if you'd learn a little bit about $ \text{\LaTeX}.$ You can find tutorials on the site, and if you look as someone else's posts, you can hold your mouse over the mathematical notation and see the code they used to produce that.

One comment on the last one: the elapsed time (4 hours) turns out not to matter. If the two cars start together and move apart on perpendicular axes at speeds $ a$ and $ b$ respectively, then at all times they will be increasing the distance between them at a rate of $ \sqrt{a^2+b^2},$ which in this case is $ \sqrt{50^2+30^2}=10\sqrt{34}\approxeq 58.3.$

What if, on the last problem, Driver B was driving due west while Driver A was driving $ 30^{\circ}$ east of north?

Then I'd use s^2=(a^2+b^2 - 2ab cos 120) , right?

Let $ t$ be the elapsed time.

There is a triangle formed with sides $ 30t$, $ 50t$, and let $ c$ denote the length of the last side.
By cosine law:
$ c^2 \equal{} (30t)^2 \plus{} (50t)^2 \minus{} 2(50t)(30t)\cos 120$
$ c^2 \equal{} 900t^2 \plus{} 2500t^2 \plus{} 1500t^2 \equal{} 4900t^2$
$ c \equal{} 70t$
$ \frac {dc}{dt} \equal{} 70 mph$

Unless I did something wrong?

[i]Edit: @Alpine, yes :).[/i]


can you guys hlep me do this problem.

If log10 5 is that log 10/log 5 or log 5/log 10.

Thanks it will be on the core 40 for indiana  so I want to make sure

$\log_{a}{b}= \frac{\log{b}}{\log{a}}$

well the $b$ on the left side is higher up than $a$ is, so $b$ goes up on top of the fraction :D

OF COURSE I AM!!!!!!

*HYPERVENTILATION**spazzzzzzzz*

usaha, maybe i AM high.. jk

jason, another post was made about this:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=744010335&t=136826[/url]

wikipedia: [url]http://en.wikipedia.org/wiki/Multinomial_theorem[/url]

[quote="jasonwingofcarmelin"]shut up

I have another question

(a+b)^10 when you expand using biomial theorm, what aoub t(a+b+c)^10, is there a trinomial theorm?[/quote]

why would you care?  Itgs not going to be covered for school so its not that imiportnat for the core 40 test.  If you do want to learn for fun, undefined has some good websites for you.

Whats 4096-2048+1024..if its geometric and has 16 terms?

[quote="undefined117"]for geometric series,

$S_{n}= a_{1}\cdot\frac{1-r^{n}}{1-r}$

in this case, $a_{1}= 4096$, $r =-\frac{1}{2}$, and $n = 16$.

so $S_{n}= 4096\cdot\frac{1-(\frac{1}{2})^{16}}{1+\frac{1}{2}}= \boxed{\frac{21845}{8}}= 2730.625$

or you can just type in $\sum_{k=0}^{15}{(4096\cdot(-\frac{1}{2})^{k})}$ into a calculator :D[/quote]

Hmm this was a core 40 quetion today..I got that answer too, and I think the right answer is going to be in common fraction, but I dont think they care if you put it as decimal or mixed number


The official results for the 2005 Collaborative Problem-Solving Contest were:

Place	Score	School
1	158	Vestavia Hills High School
2	143	Science Academy of South Texas
3	138	Palo Alto High School
4	134	Honourable Vincent Massey Secondary School
5	131	Skyview High School
6	130	Bainbridge High School
7	129	Racine Case High School
8	125	Freedom High School
9	122	Summerville High School
10	121	Mississippi School for Mathematics & Science
11	118	Huntsville High School
12	118	A&M Consolidated High School
13	117	The Webb Schools
14	109	Homestead High School
15	105	Rockdale Magnet School
16	103	Thomas Jefferson High School
17	102	John F. Kennedy High School
18	96	Hopewell High School
19	96	The Peddie School
20	92	Willowbrook High School
21	91	Ladue Horton Watkins High School
22	88	Sycamore High School
23	87	Grissom High School
24	86	Lakeside School
25	84	Texas Academy of Mathematics & Science
26	84	Wyoming Seminary

Results of our competitions are always distributed by e-mail to all participating schools and posted on our website.

>158 is a very very very very very incredible score....

     Yes, it is!  Vestavia Hills has a very good team, with a long history, a very experienced coach, and a period every day for practice.  They were first in several of our contests this year, but were second a time or two, I believe.  Schools with new teams need only work hard and wait, however... Schools such as 5th-place Skyview have only had math teams for a few years.


Let $ ABCD$ be a convex quadrilateral, Prove that $ AB^2\plus{}BC^2\plus{}CD^2\plus{}DA^2 \ge AC^2\plus{}BD^2$

There is a formula, due to Newton, if I am not mistaken, stating:

[i]Sum of squares of sides of a quadrilateral equals sum of squares of main diagonals plus 4 times the square of the segment joining the midpoints of these diagonals[/i]

Hence the inequality is trivial.

Best regards
sunken rock

Let $ X,Y,Z,W$ be the midpoints of $ AB,BC,CD,DA$ and $ M,N$ the midpoints of $AC,BD.$ It is easy to se that $XYZW,$ $ XMZN,$ $WMYN$ are parallelograms.

$ 2(XN^2\plus{}NZ^2)\equal{}XZ^2\plus{}MN^2 \ , \ (WM^2\plus{}MY^2)\equal{}WY^2\plus{}MN^2$

$ 2(XN^2\plus{}NZ^2\plus{}WM^2\plus{}MY^2)\equal{}XZ^2\plus{}2MN^2\plus{}WY^2$

$ XZ^2\plus{}WY^2\equal{}2(WX^2\plus{}WY^2)$

$\Longrightarrow \  XN^2\plus{}NZ^2\plus{}WM^2\plus{}MY^2\equal{}WX^2\plus{}WY^2\plus{}MN^2$.

Substituting $ XN\equal{}\frac{_1}{^2}BC \ , \ NZ\equal{}\frac{_1}{^2}DA \ , \ WM\equal{}\frac{_1}{^2}AB \ , \ MY\equal{}\frac{_1}{^2}CD$

$\Longrightarrow \  AB^2\plus{}BC^2\plus{}CD^2\plus{}DA^2\equal{}AC^2\plus{}BD^2\plus{}4MN^2$

Prove that: $ AB\plus{}BC\plus{}CD\plus{}DA \ge AC\plus{}BD\plus{}2MN$.Try it :wink:


A chess grandmaster practices at least once a day, to keep in shape. However, our grandmaster does not practice more than ten times a week, to ensure that he won't get too tired. Prove that if he plays long enough, there will be a sequence of consecutive days where he played 21 games.

I like this problem

[hide]Let R = 10, the maximum number of practises in a week. Let A = 21.

Let N a number of weeks during which he practises (at most R*N times). It's what we are looking for.

Let p(n) the cumulative number of practises after the first day :

1 <= p(1) < p(2) < ... < p(7*N) <= R*N

it means he practised p(2)-p(1) >= 1 times the second day, and so on.

Then A+1 <= p(1)+A < ... < A+p(7*N) <= R*N + A.

Then we get 2*7*N numbers between 1 and R*N+A

So as soon as we have 2*7*N > R*N+A, by pigeon holing we get (at least) two equals numbers.

Since the first 7*N and the last 7*N are different, there exist m and n such as p(n) = p(m) + A.

so condition is N > A / (14-R)

Here N is 6 weeks.

Hope it's clear and correct.[/hide]


Given n>1 and real s>0,find the maximum of$ {x_1}{x_2}\plus{}{x_2}{x_3}\plus{}...\plus{}{x_{n\minus{}1}}{x_n}$for non-negative reals $ {x_i}$such that$ {x_1}\plus{}{x_2}\plus{}...\plus{}{x_n}\equal{}s$

$ x_1x_2\plus{}\cdots\plus{}x_{n\minus{}1}x_n\le\frac12(x_1^2\plus{}2x_2^2\plus{}2x_3^2\plus{}\cdots\plus{}2x_{n\minus{}1}^2\plus{}x_n^2)\le x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2\le(x_1\plus{}x_2\plus{}\cdots\plus{}x_n)^2\equal{}s^2$.

Equality holds for example when $ x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}x_6\equal{}\frac{s}2$ and $ x_i\equal{}0$ for $ i\ne2,3,4,5,6$.

The small cases $ n\le 6$ can be checked easily, I think.

[quote="Johan Gunardi"]$ x_1x_2 \plus{} \cdots \plus{} x_{n \minus{} 1}x_n\le\frac12(x_1^2 \plus{} 2x_2^2 \plus{} 2x_3^2 \plus{} \cdots \plus{} 2x_{n \minus{} 1}^2 \plus{} x_n^2)\le x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2\le(x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n)^2 \equal{} s^2$.

Equality holds for example when $ x_2 \equal{} x_3 \equal{} x_4 \equal{} x_5 \equal{} x_6 \equal{} \frac {s}2$ and $ x_i \equal{} 0$ for $ i\ne2,3,4,5,6$.

The small cases $ n\le 6$ can be checked easily, I think.[/quote]
Hello,I think it has some thing wrong with your reply~~
becasue $ {x_1}\plus{}{x_2}\plus{}...\plus{}{x_n}\equal{}s$

Sorry, I was wrong. I'll try again.

Let $ x_i\equal{}\frac{s}{n}\minus{}k$ and $ x_{i\plus{}1}\equal{}\frac{s}{n}\plus{}k$ for $ i\equal{}2,3,\ldots,n\minus{}2$.

Note that $ x_{i\minus{}1}\left(\frac{s}{n}\minus{}k\right)\plus{}\left(\frac{s}n\minus{}k\right)\left(\frac{s}n\plus{}k\right)\le x_{i\minus{}1}\frac{s}{n}\plus{}\frac{s}{n}\cdot\frac{s}{n}$. So the maximum value is attained when $ x_2\equal{}x_3\equal{}x_4\equal{}\ldots\equal{}x_{n\minus{}1}\equal{}\frac{s}{n}$. The rest is easy.

Answer: $ (n\minus{}1)\left(\frac{s}{n}\right)^2$

I hope I didn't make stupid mistake again.

[quote="xiaohui"]Given n>1 and real s>0,find the maximum of$ {x_1}{x_2} \plus{} {x_2}{x_3} \plus{} ... \plus{} {x_{n \minus{} 1}}{x_n}$for non-negative reals $ {x_i}$such that$ {x_1} \plus{} {x_2} \plus{} ... \plus{} {x_n} \equal{} s$[/quote]

$ {x_1}{x_2} \plus{} {x_2}{x_3} \plus{} ... \plus{} {x_{n \minus{} 1}}{x_n}\le ({x_1}\plus{}{x_3}\plus{}...\plus{}{x_{2k\plus{}1}}\plus{}...)({x_2}\plus{}...{x_{2k}}\plus{}...)\equal{}uv\le \frac{(u\plus{}v)^2}{4}\equal{}\frac{s^2}{4}$

Equlty holds for example for $ {x_1}\equal{}\frac{s}{2},{x_2}\equal{}\frac{s}{2},{x_i}\equal{}0$


let $ p_k$ be the $ k^{th}$ prime number. Show that $ p_1p_2p_3...p_n \plus{} 1$ cannot be the perfect square of an integer.

This question doesn't belong here.  It's also been posted several times before.

can you send me a link to where it has been posted?

This is the kind of problem that is annoyingly difficult to search for, but if you don't know the solution, here's a [hide="hint"] This problem has almost nothing to do with the primes. [/hide]

[hide="Solution"]
Let $ a^2 \equal{} p_1p_2p_3 \cdots p_n \plus{} 1$
$ (a \minus{} 1)(a \plus{} 1) \equal{} p_1p_2p_3 \cdots p_n$
Either both of $ a\minus{}1, \ a\plus{}1$ are even, or neither are, therefore, since the first prime is $ 2$, numbers of this form must have a $ 2^2$ or $ 2^0$ in it. By the construction, this is not the case, so contradiction... Assuming I interpreted the problem correctly.
[/hide]

oh... lol... I solved it... I am pretty dumb... I can't believe I didn't see that! Thanks for the little hint! Thanks for a nice solution!


function
calculus
derivative
geometry
3D geometry
quadratics
symmetry

Given $x_1, x_2, x_3 ..., x_n$ are sets of random numbers selected from an interval with length of 1. Let $x={\frac 1 n}\displaystyle\sum_{j=1}^n x_j, y={\frac 1 n}\displaystyle\sum_{j=1}^n x^2_j$. Find the maximum value of $y-x^2$

It has two separate cases when n is even and when n is odd

my solution
[hide] First Case: when $n$ is even
we let $n=2k$, thus from the question we get,
 $\frac{2k(x_1^2+x_2^2+x_3^2+...+x_{2k}^2)-(x_1+x_2....+x_{2k})^2}{(2k)^2}$
we can easily see that it would be equal to
$\frac{(2k-1)(x_1^2+x_2^2+x_3^2+...+x_{2k}^2)-2(x_1x_2+x_1x_3...+x_{2k-1}x_{2k})}{4k^2}$
We can see that ${(x_1x_2+x_1x_3...+x_{2k-1}x_{2k})}$ contain each $x_n$   $2(2k-1)$times
The top part of the fraction would equal to $(x_1-x_2)^2+(x_1-x_3)^2+....+(x_1-x_{2k})^2+(x_2-x_3)^2+....(x_{2k-1}-x_{2k})^2$ since we wanted to find the maximum, we use the greatest difference between  the numbers; thus we set first $k$ terms be the largest number in the interval and second $k$ terms be the smallest number in the interval;
In    $(x_1-x_2)^2+(x_1-x_3)^2+....+(x_1-x_{2k})^2+(x_2-x_3)^2+....(x_{2k-1}-x_{2k})^2$

when $x_l$ is chosen from first $k$ term, and $x_m$ is chosen from second $k$ terms. the term $(x_l-x_m)^2$  would be $1$, otherwise it would be $0$. Thus there is $k$ choices for $x_l$, and $k$ choices for ${x_m}$, so the sum of th top will be $k^2$
Therefore when $n$ is an even number, the maximum will be $\frac{k^2}{4k^2}$=$\frac{1}{4}$
Second Case: When $n$ is odd
we let $n=2k+1$
we can let either $(2k+1)$th term equal $1$ or $0$,
if $(2k+1)$th term is $0$ there is $k$ choices for $x_l$, and $(k+1)$ choices for $x_m$, the sum would be $k(k+1)$.
if $(2k+1)$th term is $1$ there is $(k+1)$ choices for $x_l$, and $k$ choices for $x_m$, the sum would be $(k+1)k$, thus either value of $(2k+1)$th term would yield the same sum for the top.
thus the value of the expression will be
$\frac{(k+1)k}{(2k+1)^2}$ Since $n=2k+1$,$k$=$\frac{n-1}{2}$, subtitute into the expression thus we get the maximum value for $n$ when $n$ is odd is $\frac{n^2-1}{4n^2}$
[/hide]

Consider $f(x)=y-x^2$. Consider this a function of one of the variables, while the others are fixed. The second derivative of $f(x_i)$ is $f''(x)=\frac{2}{n}$, and since $n$ is positive, the second derivative is positive everywhere on the interval. Thus, the function is convex. This process can be repeated for all variables, and thus the function is convex in all variables. As a result, it obtains its maximum at the vertices of the $n$-space cube, and is less elsewhere.

The problem is that we can't just check all the vertices for each $n$, cause the operation is approximately $O(2^N)$.

Full solution::

[hide]

All sums in this proof will be symmetric. For example, if we have three variables, this sum would be :

First of all, let's plug our expressions in, and rearrange a bit. We get the equivalent expression

Now, consider the function of the variable , while all the others are held fixed. The second derivative is

Since , then the second derivative is positive, and thus the function is convex in . Since the function is homogenously symmetric in all its variables, this process is equivalent for all , and thus the function is convex in all variables. As such, it attains it's maximum when each variable is either  or  (the endpoints).

Now, suppose  of the variables are equal to  and  are equal to . Then it is clear to show that , and that . Plugging this straight into our expression, we get

Where  and  must be integers. Our final expression is a quadratic in .

Suppose  is even, so that . Then our expression becomes

Since  is an integer, we can let , and the square term becomes , and we're left with a maximum of . Since this is the vertex, the function is less everywhere else. The function achieves this maximum in any permutation of  1's and  0's.

Now, suppose  is an odd integer, so that . Our expression becomes

Since  must be an integer, we must choose the nearest integer. Due to symmetry, both  and  minimize this. Plugging this in and simplifying the entire expression, we get

Then, plugging , we get the minimum to be . The function achieves this maximum for any permutation of  1's and  0's, or vice versa.

Thus, our maximum for all values of  is

[/hide]


The domain of $f(x)$ is $R$. If the maximum value of $f(2006x+2006)$ is 2006, then find the maximum value of $f(x^{2005}+1)$.

I hope I'm not doing anything wrong...

The functions $g_1(x)=x^{2005}$ and $h_1(x)=2006x$ are bijective and range over the reals.

Thus, the functions $g_2(x)=x^{2005}+1$ and $h_2(x)=2006x+2006$ are bijective, mapping from the reals to the reals.

Hence, if $y\in\Re$ such that $y=x^{2005}+1$, then there exists some $t\in\Re$ such that $2006t+2006=y$.

Similarly, if $t\in\Re$ such that $t=2006x+2006$, then there exists some $y\in\Re$ such that $y^{2005}+1=t$.

Since $f(2006t+2006)\le2006$, then $f(y)=f(x^{2005}+1)\le2006$.

Since $f(2006t+2006)=2006$ for some $t$, then $f(y)=f(x^{2005}+1)=2006$ for some $y$.

Therefore, the maximum value is $2006$.


A medieval weapon in the shape shown consists of 4 arcs which are semicircles of radius 6 inches. Point $ X$ is the centroid of the weapon. How many square inches are in the area of the cross-section of the weapon shown?

[asy]pair X; label("X",X,S);
for(int i = 0; i < 4; ++i){
draw(Arc(dir(90*i),1,-90*(i+1),-90*(i-1)));
}[/asy]

We can rearrange this, utilizing congruency, to a square with side length $ 12$, so the area is $ 12^2 \equal{} \boxed{144}$.