geometry
circumcircle
analytic geometry
graphing lines
slope
parallelogram
geometric transformation

Let $ AD$ is centroid of $ ABC$ triangle. Let $ (d)$ is the perpendicular line with $ AD$. Let $ M$ is a point on $ (d)$. Let $ E, F$ are midpoints of $ MB, MC$ respectively. The line through point $ E$ and perpendicular with $ (d)$ meet $ AB$ at $ P$. The line through point $ F$ and perpendicular with $ (d)$ meet $ AC$ at $ Q$. Let $ (d')$ is a line through point $ M$ and perpendicular with $ PQ$. Prove $ (d')$ always pass a fixed point.

To: Thanhnam2902

Herebelow an easy approach to this nice problem:
Take B' and C' on (d) as the orthogonal projections of B and C respectively onto (d).
When M comes to B', P goes to B, and Q to A, while when it comes to C', P goes to A and Q to C, therefore the fixed point must be the common point of the perpendiculars from B' onto AB and from C' onto AC respectively, let it be F, and it's fixed.
We have now to show that, for any position of M on (d), the perpendicular from M to PQ, got as required, passes through F.
But there is a well-known theorem stating that: [b]The perpendiculars from the vertices of a triangle onto the sides of another triangle concur iff the perpendiculars from the vertices of the latter onto the sides of the first one concur[/b]. (Carnot??).
For our case we take the [quote]triangle[/quote] B'MC' and the triangle APQ. Since the perpendiculars from A, P and Q onto (d) are parallel (we assume they concur somewhere), we get that the perpendiculars from B' to AP. C' to AC and M to PQ will be concurrent as well and we are done.

We can see from here that when (d) is moving, F describes a straight line, parallel to AD.

Best regards,
sunken rock

A blind man sees better the details

[quote="sunken rock"] [b]The perpendiculars from the vertices of a triangle onto the sides of another triangle concur iff the perpendiculars from the vertices of the latter onto the sides of the first one concur[/b]. (Carnot??).
[/quote]
Two triangles satisfying this property are called [b]orthologic[/b].
For more information, see here
http://www.mathlinks.ro/Forum/viewtopic.php?t=4337

My solution:
[img]http://img397.imageshack.us/img397/1431/pham20van20quy2020thpt2ss3.jpg[/img]
We will prove that $ d';d_1;d_2$ are concurrent   [b](*)[/b]
We have $ B'A\equal{}C'A;B'P\equal{}MP;C'Q\equal{}MQ$ so
$ B'A^2\minus{}B'P^2\plus{}MP^2\minus{}MQ^2\plus{}C'Q^2\minus{}C'A^2\equal{}0$
By Carnot's theorem with triangle $ APQ$ we have $ d';d_1;d_2$ are concurrent
Done!

[b]Problem stated by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=37447]mr.danh:[/url]

Let ABC be a triangle. The line $ d\parallel BC$. A point M moves on $ d$. The perpendicular line from the midpoints of MB,MC to BC meets AB,AC at D,E respectively. Prove that the perpendicular line from M to DE passes through a fixed point when M moves on $ d$.

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=37447]mr.danh:[/url]

My opinion:
D,E lie on a perp bisector of M'B,M'C where M' is the orthogonal projection of M on BC.
N is the point symmetric to M' w.r.t DE. Easy to show that E lie on the circumcircle (ABC), which easily gives us M'N passes through the fixed point F', which is the intersection of altitude AH and the circumcircle (ABC). Hence, if F is the point satisfying $ \vec{F'F} \equal{} \vec{M'M}$, F is the fixed point that the perp line from M to DE passes through.
The problem was proven.

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=38081]Nemion:[/url]

What popped first to my mind was an analytic solution:
$ A \equal{} (0,a)$, $ B \equal{} (b,0)$ $ C \equal{} (c,0)$ $ M \equal{} (t,m)$ with $ m$ fixed. midpoints $ P$, $ Q$ of $ BM$, $ CM$ are: $ P \equal{} (\frac {t \plus{} b}{2},\frac {m}{2})$ and $ Q \equal{} (\frac {t \plus{} c}{2},\frac {m}{2})$ equation of $ AB$: $ y \equal{} \frac {a}{ \minus{} b}x \plus{} a$ eq $ AC$: $ y \equal{} \frac {a}{ \minus{} c} \plus{} a$ then $ E \equal{} (\frac {t \plus{} b}{2},(\frac {a}{ \minus{} b})(\frac {t \plus{} b}{2}) \plus{} a))$ and $ F \equal{} (\frac {t \plus{} c}{2},(\frac {a}{ \minus{} c})(\frac {t \plus{} c}{2}) \plus{} a))$ slope of $ EF$ (Reducing terms and making the computations) is: $ \frac {at}{bc}$ then the slope of the perpendicular to $ EF$ trough $ M$ is: $ \minus{} \frac {bc}{at}$ then the equation of the perpendicular to $ EF$ trough $ M$ is: $ y \equal{} \minus{} \frac {bc}{at}(x \minus{} t) \plus{} m$ the intersection of this line with the $ y$ axis is: $ y \equal{} m \plus{} \frac {bc}{a}$ (point: $ (0,\frac {bc}{a} \plus{} m)$) but this last point doesn't depend on $ t$ (but on $ m$) therefore it is fixed for $ m$ fixed.

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=5895]yetti:[/url]

As long as $ d \parallel BC,$ the line $ DE$ does not depend on $ d.$ Assume the proposition is true for the line $ BC$ itself. Let $ M' \in BC$ be the foot of normal from $ M \in d,$ let $ p \parallel p'$ be the perpendiculars from $ M, M'$ to $ DE,$ let $ F'$ be the fixed point of $ p'$ and let a parallel to $ MM'$ through $ F'$ cut $ p$ at $ F.$ $ MM'F'F$ is a parallelogram, $ FF' \equal{} MM'$ and $ (FF' \parallel MM') \perp BC.$ Since $ F' \in p'$ is fixed, $ F \in p$ is also fixed. Therefore, it is sufficient to prove the proposition for $ BC$ itself, hence assume $ M \in BC.$

Let $ MD, ME$ cut perpendiculars to $ BC$ at $ B, C$ at points $ U, V.$ $ D, E$ are midpoints of $ MU, MV,$ $ DE \parallel UV$ is a midline of the $ \triangle MUV.$ Since $ D, E$ are on perpendicular bisectors of $ MB, MC,$ $ \angle UMB \equal{} \angle B,$ $ \angle VMC \equal{} \angle C.$ Parallels to $ UM, VM$ through $ C, B,$ respectively, intersect at the reflection $ A'$ of $ A$ in the perpendicular bisector of $ BC.$ Define 3 points on the line at infinity: $ X^* \equiv MU \cap CA',$ $ Y^* \equiv MV \cap BA',$ $ Z^* \equiv BU \cap CV.$ Consider the line $ MBC$ and the line at infinity $ X^*Y^*Z^*.$ By Pappus theorem, the intersections $ U \equiv MX^* \cap BZ^*,$ $ V \equiv MY^* \cap CZ^*,$ $ A' \equiv CX^* \cap BY^*$ are collinear.

Let $ W$ be the foot of a normal from $ M$ to $ UV,$ i.e., $ MW \perp DE$ is the line supposedly passing through a fixed point. $ MUWB$ is cyclic because of the right angles $ \angle MBU, \angle MWU,$ hence $ \angle A'WB \equal{} \angle UMB \equal{} \angle B.$ Since $ A', B$ are on the circumcircle of the $ \triangle ABC$ and $ \angle A'CB \equal{} \angle ABC \equal{} \angle B,$ it follows that $ W \in (O).$ Let $ MW$ cut $ (O)$ again at $ F.$ Since the angle $ \angle A'WF$ is right, $ A'F$ is a diameter of $ (O)$ from a fixed point $ A',$ so that $ F$ is also fixed. Obviously, the angle $ \angle A'AF$ is also right and $ F$ is the other intersection of the A-altitude of the $ \triangle ABC$ with it circumcircle $ (O).$

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=21968]vittasko:[/url]

The [b][size=100]orthopole[/size][/b] $ K$ of the triangle $ \bigtriangleup ADE,$ with respect to the sideline $ BC,$ lies on the midperpendicular of the segment $ HA',$ where $ H$ is the orthocenter of $ \bigtriangleup ABC$ and $ A'$ is the orthogonal projection of $ A$ on $ BC$ $ ($ $ AA'$ = $ A$-altitude of $ \bigtriangleup$ $ ABC$ $ ).$

The line segment $ HM',$ where $ M'$ is the orthogonal projection of $ M$ on $ BC,$ always passes through the point $ K.$

So, the line through the point $ M$ and perpendicular to $ DE$ as the problem states, as parallel to $ DK,$ intersects the line segment $ AA'$ at one point so be it $ T,$ such that $ A'T \equal{} HA' \minus{} MM'$ $ ($ in my drawing, the point $ M$ is inwardly to $ \bigtriangleup ABC$ and $ MM' < HA'$ $ )$ and so, it is a fixed point.

I will check my notes and if I am not mistaken, I will post here next time the details about the proof of this result.

Kostas Vittas.

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=21968]vittasko:[/url]

Let $ P,\ Q$ be, the midpoints of the segments $ MB,\ MC$ respectively and we denote as $ D'\ M'\ E',$ the orthogonal projections of $ D,\ M,\ E$ respectively, on $ BC.$

It is well known (and easy to prove by [b][size=100]Carnot theorem[/size][/b] ), that the lines through the points $ D',\ A',\ E',$ where $ A'$ is the orthogonal projection of $ A$ on $ BC$ $ ($ $ AA',$ is the $ A$-altitude of $ \bigtriangleup ABC$ $ ),$ are concurrent at one point so be it $ K,$ as the [b][size=100]orthopole[/size][/b] of the triangle $ \bigtriangleup ADE,$ with respect to the sideline $ BC$ of $ \bigtriangleup ABC.$

Because of now, $ M'D' \equal{} D'B$ and $ M'E' \equal{} E'C$ and $ D'K\parallel BH$ and $ E'K\parallel CH,$ we conclude that the points $ M',\ K,\ H,$ are collinear and $ M'K \equal{} KH$ $ ,(1)$

It is easy to show that the line through the point $ M$ and perpendicular to $ DE,$ as parallel to $ A'K,$ intersects the line segment $ AA',$ at one point so be it $ T,$ such that $ A'T \equal{} MZ,$ where $ Z\equiv A'K\cap M'M,$ as the fourth vertex of the rectangle $ HA'M'Z.$

It is clear that $ T$ is a fixed point on the line segment $ AA',$ because of $ A'T \equal{} MZ \equal{} M'Z \minus{} M'M \equal{} HA' \minus{} M'M$ $ ,(2)$

Hence, the line through the point $ M$ and perpendicular to $ DE$ as the problem states, passes through the fixed point $ T$ and the proof is completed.

Kostas Vittas.

PS. It is easy to show that the line through the point $ M'$ and perpendicular to $ DE,$ passes through the point $ H',$ as the intersection point of the circumcircle of $ \bigtriangleup ABC,$ from its altitude $ AA'.$

[url=http://www.mathlinks.ro/viewtopic.php?p=1238906#1238906]An extension of this problem[/url]

Move $M$ linearly on $d$, in terms of the variable $t$ denoting time. Observe that $P, Q$ also move linearly on $AB, AC$ respectively. As a result of the above, for a particular point $S$ in the plane, the function $f(t) = MP^2 - MQ^2 + SQ^2 - SP^2$ is a quadratic in $t.$ Hence, if it is zero for $3$ points, it is identically $0$ and we'd be done.

We therefore just need to show that the lines $d'$ pass three a fixed point for some three choices of $M$. We will take the feet of the altitude of $B, D, C$ respectively to $d$ as our three choices. It's easy to check that these three choices of $M$ indeed lead to an intersection point for $S$.

$\square$


AMC AIME geometry USA(J)MO USAMO MATHCOUNTS HMMT

What is the best way to prepare for AIME? And which books should I use? Also does any one know when AOPS Intermediate Geometry comes out or other books on intermediate geo? :?:

[quote="PowerOfPi (on July 22 09)"]
When do you expect Intermediate Geometry and Intermediate Number Thoery to come out?[/quote]
[quote="rrusczyk (on July 23 09)"]
A long, long time from now.[/quote]

Hopefully answers your third question :)

[quote="Jongao"]Two books that are good for the AIME are Volume 2 and AcoPS (Art and Craft of Problem Solving).  AIME Class and AIME Seminar would also help.  But the best way to prepare for the AIME is undoubtedly doing AIME Problems (many, many of them).  Trust me: doing AIME problems took my score from a 2 to a 6. :)[/quote]

Hmm. doing AIME problems brought my score from an 8 to a 9...LOL

AoPS V2 is definitely good, but I think the AIME class is better. For me, talking to people is better than reading a book. And less boring.

For me, to get good at AIMEs, do Olympiad training. Most of the concepts used in AIMEs are just simplified versions of the stuff in Olympiads. My AIME score hovered around a 5-7. Then I did Olympiad training (not even that intensely) and my average went up 2 points. Then when I came back from Red MOP, I just looked at an AIME and solved the first 10 questions in an hour.

Olympiad training, in my opinion, is probably the easiest way to get better at AIME. In addition, if you end up making USAMO, you'll be... a lot better prepared.

[quote="Ubemaya"]Doing a lot of Mathcounts helped me greatly for both AIME and USAMO practice. I highly recommend doing some Mathcounts or Mathcounts-esque tests every once and a while (every other day or so if pretty good, but if you can't, at least twice a week is a good goal to shoot for).[/quote]

Note-This is a joke, I talked to him on AIM.

USAMO Prep helps you on aime more then AIME prep.

The thing is unfortunately, AIME is very computationally heavy, so Mathcounts probably IS the best way to train for AIME.

Well still...at least for an unexperienced person like me, AIME problems after #5 stop being trivial, and require some decent problem solving, so doing and learning olympiad math will help a lot, I believe. Haven't actually tried this myself though...

But yes, many problems on AIME require a lot of computation. Mathcounts would invariably help with this, and also would help speed up your first few problems. AMC 10 without the answer choices would probably work too.

Hmm, so basically, to prepare for AIME (in order of priority):

1. Do and Learn Olympiad Math
2. Do a lot of AIME problems, but on and off with this.
3. Do some Mathcounts/AMC10/AMC12 tests without the answer choices.
4. Stop failing

[quote="Arachnotron"]I suggest you do the opposite: master the basics, then start tackling progressively more difficult stuff. Odds are, if you're a problem solving newbie, Olympiad stuff will only scare you away.[/quote]

I'm not a "problem solving newbie", because I can get 144 on AMC 10 and 3-5 on AIME, Olympiad may be okay after some of Vol.2 because I've already mastered the Intro Books.

HMMT, ARML, and Mandelbrot problems can also help with AIME, too.

The easier Olympiad problems can prove quite useful as well, such as JBMO (Jr. Balkan Math Olympiad), CMO (Canadian Olympiad), and possibly the very very old IMO #1's (they're trivial).

Additionally, you should try doing some of the problems in the High School Intermediate Topics and Pre-Olympiad forums; those can help you master almost all the levels of AIME problems.

BMO (old ones) aren't much different from JBMO's I zink...


Let $0 \leq x,y,z,t \leq 1$.Find max
$M=\frac{\frac{y+t}{x+z}+\frac{z+t}{t+x}}{\frac{y+z}{x+t}+\frac{z+x}{t+y}}$[/COLOR]

$x=\alpha ,y=-\alpha ,z=\alpha$

${\frac{\frac{t+y}{x+z}+\frac{t+z}{t+x}}{\frac{x+z}{t+y}+\frac{y+z}{t+x}}=\frac{1}{4}\left(\frac{t^{2}}{\alpha^{2}}-1\right)}$

${\lim_{\alpha \to 0}\, \frac{1}{4}\left(\frac{t^{2}}{\alpha^{2}}-1\right)=\infty }$

is it rigth?

Yes, although it looks to me like there are probably several typos in the question.


Hello, does someone knows what age structure diagrams with a broad base suggests?, i have this options and little confuse ...
a population will continue to grow for some time
b enviromental resistance is not at work
c the biotic potential must be larger than usual
d the individuals must be larger than usual
e all of the above is true

thanks any opinions will be appreciate,

A broad base indicates a "baby boom", meaning the population is growing.  I'm pretty sure that age structure diagrams have nothing to do with the size of the organisms, so I'm gonna say B.


geometry
3D geometry
prism
combinatorics proposed
combinatorics

We say that a prism is [i]binary[/i] if there exists a labelling of the vertices of the prism with integers from the set $\{-1,1\}$ such that the product of the numbers assigned to the vertices of each face (base or lateral face) is equal to $-1$.

a) Prove that any [i]binary[/i] prism has the number of total vertices divisible by 8;

b) Prove that any prism with 2000 vertices is [i]binary[/i].

(a) Denote the prism by the vertices $A_1A_2...A_NB_1B_2...B_N$, so that the $A$'s denote one base, the $B$'s the other, and for every $i$ $A_i$ is joined to $B_i$ by an edge. Let  $a_i$ and $b_i$, where $a_i, b_i \in \{1, -1\}$ be the labels at these vertices. If the prism is to be binary, we are to have

$a_1a_2b_1b_2 = -1$;
$a_2a_3b_2b_3 = -1$;
....
$a_{N}a_1b_{N}b_1 = -1$.

Multiplying all these equations, we get

$(a_1...a_Nb_1...b_N)^2 = (-1)^N$.

Since the LHS is obviously 1, $N$ must then be even. Let $N = 2k$. Then multiplying all the equations

$a_1a_2b_1b_2 = -1$;
$a_3a_4b_3b_4 = -1$;
...
$a_{2k-1}a_{2k}b_{2k-1}b_{2k} = -1$

we get

$(a_1...a_{2k})(b_1...b_{2k}) = (-1)^k$.

The LHS = $(-1)(-1) = 1$, so $k$ must be even. $N$ is then divisible by 4. So the total number of vertices, $2N$, is divisible by 8.

(b) Given any prism $A_1A_2...A_NB_1B_2...B_N$ where the number of vertices is divisible by 8, that is, where $4|N$, the labelling

$a_1 = -1, a_2 = a_3 = ... = a_N = 1$,

$a_1 = 1, b_2 = b_4 = ... = b_N = 1, b_3 = b_5 = ... = b_{N-1} = -1$

shows that the prism is binary. In particular, this applies to $N = 1000$.


inequalities
geometry
circumcircle
geometry proposed

This might be trivial, but anyways...

Let $O$ be the circumcenter of acute $\triangle ABC$, and let $AO$, $BO$, and $CO$ meet their opposite sides at $A'$, $B'$, and $C'$, respectively.  Prove that

\[\frac{1}{A'O}+\frac{1}{B'O}+\frac{1}{C'O} \geq 3\left(\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\right)\]

multiply by R on both sides, then with areas it reduces to prove that $\sum{\frac{x+y}{z}} \geq \sum{\frac{x+y}{x+y+z}}$ where $x=(BOC), y=(AOC), z=(AOB)$. And the above inequality it's equivalent to $\sum{\frac{x+y}{z}} \geq \frac 32$ wich is known as Nesbitt's inequality, and is well known (add 3 on both sides and then use AM-HM).


advanced fields
advanced fields unsolved

Let $ (X,\tau)$ be a Baire space and $ (D_n)_n$ a sequence of open dense sets.

Prove that $ \displaystyle \bigcap_{n\geq 1} D_n$ is a Baire space with the induced topology.


trigonometry quadratics algebra polynomial

$n$ dots are drawn on the circumference of a circle. By joining all the dots to one another by straight lines, the maximum number of regions that can be formed is counted. For example, when $n=4$, the maximum number of regions is $8$. So, what is the maximum number of regions that can be formed when $n=7$

[hide="My Solution"]When $n=2$, the number of regions ($r$) is also $2$--the line splits the circle in half. When $n=3$, $r=4$ (an inscribed triangle and the three regions around it); when $n=4$, $r=8$ (an inscribed convex quadrilateral split into four regions by its diagonals plus the four regions around it). This seems consistent with the equation $r=2^{n-1}$. Therefore, when $n=7$, $r$ should equal $2^{7-1}=2^{6}=64$.[/hide]

[quote="KrazyNerd"]This seems consistent with the equation $r=2^{n-1}$.[/quote]

It's also consistent with the equation \[r = 4(2\sqrt{2}+2\sqrt{3}+\sqrt{6})\sin^{2}(\pi/n)+\frac{4\sin (\pi/n)}{2+\sqrt{2}-2\sqrt{3}}+\frac{32+10\sqrt{2}-28\sqrt{3}}{2+\sqrt{2}-2\sqrt{3}}.\]  Come up with a better (and more correct) solution.

Wow. I-I-I'm sorry. I diagrammed the problem, tried it up through $n=5$, and it seemed to work. :maybe: I hardly understand the equation you made, so maybe I don't belong in this forum, but I gave the problem my best try and put my solution and reason up. I don't think an honest try deserves a rating of spam, as I wasn't being off-topic or mean, but I'm sorry my solution was incorrect.

How did you come up with your equation, anyway? Since mine is wrong, I'd like to know why.

What he is saying is that just because a pattern is true for the first 5 terms doesn't mean it's true for the rest. He just came up with a really complicated equation that comes up with those same terms to show you the formula is not neccessarily 2^(n-1) and I do believe it's not.  :roll:

Sigh. I had a feeling it was too easy for intermediate. And for once I thought I'd solved something... :blush:

Anyway, I was just a little upset that I received such a low post rating for something that was simply a wrong answer, not spam or a flame. I'll keep looking for a better answer.

[quote="JBL"][quote="KrazyNerd"]This seems consistent with the equation $r=2^{n-1}$.[/quote]

It's also consistent with the equation
\[r = 4(2\sqrt{2}+2\sqrt{3}+\sqrt{6})\sin^{2}(\pi/n)+\frac{4\sin (\pi/n)}{2+\sqrt{2}-2\sqrt{3}}+\frac{32+10\sqrt{2}-28\sqrt{3}}{2+\sqrt{2}-2\sqrt{3}}. \]
Come up with a better (and more correct) solution.[/quote]

CAN YOU PLEASE EXPLAIN HOW YOU CAME UP WITH THIS?

I didn't mean to be harsh: as rofler said, I was merely pointing out that matching a pattern is a good way to guess what an answer could be, but it doesn't make a solution all by itself.  The guessing stage is how you come up with an answer -- after that, you still need a justification stage (a proof, for example).

As for how I came up with my answer:  you matched a pattern based on three values, $n = 2, 3, 4$.  I happen to know that $\sin \pi/2$, $\sin \pi/3$ and $\sin \pi/4$ are all relatively nice values.  So, I took the three points $\{(\sin \pi/2, 2), (\sin \pi/3, 4), (\sin \pi/4, 8)\}$ and found a quadratic polynomial $f(x) = ax^{2}+bx+c$ which passed through those three points.  (I could have done this by hand -- it happens that I have computer algebra software which can do those calculations more quickly than I can, so I used it instead.)

Also, about post ratings: usually, post ratings change over time.  I notice that the rating on your post has improved somewhat, for example.  If a single post is judged too harshly, I'm not sure it's worth getting upset over.  (At this point, people seem to like your post nearly as much as they like mine :).)

Now, some actual comments on the problem: I don't like this problem very much, because if you've seen things like it before, you know what to expect (for example, the fact that the sequence went up by powers of 2 initially is something I knew was coming).  Then the process of actually showing what the answer is can be quite messy.  An easier example is the following question: $n$ lines are drawn in the plane, so that no two are parallel and no three are concurrent.  Into how many pieces (including some unbounded ones) do they divide the plane?  (The first few values are 1, 2, 4, for $n = 0, 1, 2$.)  The next question of this sort is identical but with planes in space.  And the given problem is of a very similar sort.  (A less-obvious but more-easy problem that really belongs in the same sequence as the two I gave: $n$ points are drawn on a line.  Into how many pieces do they divide the line?)

Since you haven't seen these problems before, I suggest you play around with them on your own for a while.  The one hint I'll give you is, "Think inductively."


function
integration
geometry
perimeter
inequalities
IMC
college contests

Let $f,g:[a,b]\to [0,\infty)$ be two continuous and non-decreasing functions such that each $x\in [a,b]$ we have
\[ \int^x_a \sqrt { f(t) }\ dt \leq \int^x_a \sqrt { g(t) }\ dt  \ \  \textrm{and}\ \int^b_a \sqrt {f(t)}\ dt = \int^b_a \sqrt { g(t)}\ dt. \]

Prove that
\[ \int^b_a \sqrt { 1+ f(t) }\ dt \geq \int^b_a \sqrt { 1 + g(t) }\ dt. \]

Let's change the notations a bit:

$\sqrt f\to f,\sqrt g\to g$. We thus have $F(x)\le G(x),\ \forall x\in [a,b],F(b)=G(b)$, where $F(x)=\int^x_a f(t)dt,G(x)=\int^x_a f(t)dt$. We want to show that $\int^b_a \sqrt {1+[F'(t)]^2}dt\ge \int^b_a \sqrt{1+[G'(t)]^2}dt$, but these are the lengths of the graphs of $F,G$ between $a,b$. Additionally, we know that $F,G$ are increasing (actually, non-decreasing) and convex, s.t. $F$ is always below $G$ and that they share the chord $((a,F(a)=G(a)=0),(b,F(b)=G(b)))$, so we may approximate them with polygons which have the same properties. From here it becomes a simple geometry question:

Two polygons share a side, they're both convex and the other sides of the first one, say $P_1$ (corresponding to the graph of $G$), are inside the other one, say $P_2$ (corresponding to the graph of $F$). We have to show that the perimeter of $P_2\ge $the perimeter of $P_1$.

With the change of notations of Grobber, let's prove that if $f$ is a $C^1$ convex function on $[a,b]$, and $g$ is a $C^1$ function on $[a,b]$ such that $g(a)=f(a)$ and $g(b)=f(b)$ and $g(x) \leq f(x)$ for all $x \in [a,b]$ then $ \int_a^b \sqrt {1+ (f'(t))^2 } dt \leq \int_a^b \sqrt {1+ (g'(t))^2 } dt$.

Note that $h(t)= \sqrt {1+t^2} $ is strictly convex on $R$, so that $h(t) \geq h(s) + (t-s)h'(t)$ for all $s,t$.
It follows that :
$ \int_a^b \sqrt {1+ (g'(t))^2 } dt \geq \int_a^b \sqrt {1+ (f'(t))^2 } dt + \int_a^b [g'(t) - f'(t)]h'(f'(t)) dt.$
Thus it suffices to prove that :  $ \int_a^b [g'(t) - f'(t)]h'(f'(t)) dt \geq 0.$

But $h$ and $f$ are convex so that $h'$ and $f'$ are nondecreasing and then $h'of'$ is nondecreasing. It is known that (I do not know the english name for it. In France, we call it 'theorme de la moyenne') there exists $r \in [a,b]$ such that :
$ \int_a^b [g'(t) - f'(t)]h'(f'(t)) dt = h'(f'(r)) \cdot  \int_a^r [g'(t) - f'(t)]dt + h'(f'(b)) \cdot  \int_r^b [g'(t) - f'(t)]dt = [f(r) - g(r)][h'(f'(b)) - h'(f'(a))]$
where we have used that $^f(a)=g(a)$ and $f(b)=g(b)$.

The desired inequality now follows easily from the facts that $f \geq g$ and $h'of'$ is nondecreasing.

Note that it is not difficult to prove that the equality occurs if and only if $f=g$.

Hmmm... I think I interchanged $f$ and $g$ from Grobber's original statement.

Pierre.

At Oral Examination Ecole Normale Suprieure (ULM) 2002 there was a similar
problem
$f:[a;b] ->R$ concave function, continous, piecewise $C^1$.
Let $\gamma_1$ the parametric curve $x->(x;f(x))$ and
$\gamma_2$  continous arc, picewise $C^1$ with the same extremity of
$\gamma_1$ above $\gamma_1$.
Prove that the lenght of $\gamma_2$ is greater or equal to the lenght of
$\gamma_1$

[quote="pbornsztein"]It is known that (I do not know the english name for it. In France, we call it 'theorme de la moyenne') there exists $r \in [a,b]$ such that :
$ \int_a^b [g'(t) - f'(t)]h'(f'(t)) dt = h'(f'(r)) \cdot  \int_a^r [g'(t) - f'(t)]dt + h'(f'(b)) \cdot  \int_r^b [g'(t) - f'(t)]dt = [f(r) - g(r)][h'(f'(b)) - h'(f'(a))]$
[/quote]
In China, we call it 'The second mean-value theorem for integral'..
And the first mean-value theorem is
\[\int_a^n f(x)g(x) dx = \eta \int_a^n g(x) dx\]
where $g(x)$ does not change its sign on $[a,b]$.

There is a typo in your equation, it should be
\[ \int_a^b [g'(t) - f'(t)]h'(f'(t)) dt = h'(f'(a)) \cdot  \int_a^r [g'(t) - f'(t)]dt + h'(f'(b)) \cdot  \int_r^b [g'(t) - f'(t)]dt\]

[quote="grobber"]This is nice, but I think the inequality sign in the statement we have to prove should be reversed[/quote] the inequality sign is good. :)

Of course it is, and what I wrote after that sentence shows that I realized that the sign was Ok :). I wanted to write that, realized I was making a mistake, and then left the sentence there by accident. I'll edit it right away.

[quote="Valentin Vornicu"]Let $ f,g: [a,b]\to [0,\infty)$ be two continuous and non-decreasing functions such that each $ x\in [a,b]$ we have
\[ \int^{x}_{a}\sqrt{ f(t) }\ dt \leq \int^{x}_{a}\sqrt{ g(t) }\ dt \ \ \textrm{and}\ \int^{b}_{a}\sqrt{f(t)}\ dt = \int^{b}_{a}\sqrt{ g(t)}\ dt. \]
Prove that
\[ \int^{b}_{a}\sqrt{ 1+f(t) }\ dt \geq \int^{b}_{a}\sqrt{ 1+g(t) }\ dt. \]
[/quote]

We see that $ \int^{x}_{a}\sqrt{ f(t) }\ dt \leq \int^{x}_{a}\sqrt{ g(t) }\ dt$ can be changed to $ \int^{b}_{x}\sqrt{ f(t) }\ dt \geq \int^{b}_{x}\sqrt{ g(t) }\ dt$. Here is a same problem.

Let $ f: [0,1]\to [0,\infty)$ be a continuous function such that each $ x\in [a,b]$ we have
\[ \int_{x}^{1}{f\left( t \right)dt}\geq \int_{x}^{1}{tdt}\quad ,\quad \forall x \in \left[{0,1}\right]. \]
Then
\[ \int_{0}^{1}{f^{\alpha+1}\left( x \right)dx}\geq \int_{0}^{1}{x^\alpha f\left( x \right)dx}. \]
and
\[ \int_{0}^{1}{f^{\alpha+1}\left( x \right)dx}\geq \int_{0}^{1}{x f^\alpha \left( x \right)dx}\]
for every positive real number $ \alpha > 0$.


[quote="ShoeFactory"]How many 7-letter permutations are there of the letters in the word Okanagan?

6 letter?[/quote]
[hide="1"]
There are 8 letters in Okanagan. There are 2 repeats, 'a' and 'n'. Take out an 'a' and you get 1260. Take out an 'n' and you get 840. Take out any other letter and you get 420. So it's $420(3)+840+1260=\boxed{3360}$.[/hide]

[hide="2"]Take out 'a' and another letter (not 'n') and you get 180. Take out 'n' and another letter (not 'a') and you get 120. Take out two letters (not 'a' or 'n') and you get 60. Take out 'a' and 'n' and you get 360. Take out two 'n's and you get 120. Take out two 'a's and you get 360. So it is $360+120+360+60(3)+120(3)+180(3)=\boxed{1920}$. [/hide]

[quote="surge"]Why doesn't $\frac{P(8,6)}{2!3!}$ work for the second one?[/quote]

Thats exactly what i was thinking but also if you think about it, if you take 8P1/2!/3!, it doesnt make any sense at all.

Well, if you are only taking six letters, you don't necessarily have all 3 a's or both n's in the permutation.  Same is true for problem 1.  That's why you have to do the casework like ch1n353ch3s54a1l demonstrated...

Well for the first one...

$\frac{P(8.7)}{2!3!}$ works... maybe because dividing by the extra "1" or "1!" doesn't change the answer (for each time one of the "n"'s or "a's" are missing)?

I think so.  Because I'm pretty sure that's the only one it works for...


[quote="learningfish"]At a dog show 6 dogs ran off and began fighting.  The owners of each grabbed a lead.
What is the probability that 5 owners, and only 5, had the right dog?

30/42?[/quote]
If 5 owners got the right dog, doesn't that mean that the 6th no matter what got the right dog?

[quote="learningfish"]Maybe I am supposed to use a probability tree?[/quote]
You don't have to, but your question states that only 5 have to get their own dogs. But if 5 get their dogs then the 6th person automatically gets their dog because there's only one left. And this doesn't follow the question. So I'm confused.

yeah, unless there was a situation like the sixth was a stray or something then Walk Around The River is correct.

[quote]The owners of each grabbed a lead.[/quote]You can't grab a lead, you grab a leash.

[quote="xpmath"]I think it is 0.[/quote]
That's already been confirmed. Brainteasers are fun!

[quote="alanchou"][hide] If five people get the right dog, then the last person has to get their dog as well. But this means that it's impossible for exactly 5 people to get their own dog, so the probability is [b]0[/b]. This type of problem was on a State Countdown before.[/hide][/quote]
This is actually sort of a logic problem... I forget the notation.

(Not that I can post it anyways, LaTeX doesn't incorporate all logic symbols)

[quote="Walk Around The River"][quote="learningfish"]At a dog show 6 dogs ran off and began fighting.  The owners of each grabbed a lead.
What is the probability that 5 owners, and only 5, had the right dog?

30/42?[/quote]

this is more of a brainteaser...

[hide="the answer"]Its 0.  if 5 get the leash correct, the 6th must as well.  so its impossible for only 5 to get the correct leash.[/hide][/quote]

I agree with Walk Around the River.

[hide]It can't be "only" five, cuz then the sixth one has to grab the right one too, unless he/she doesn't care about the dog and decides to be amused by watching the dogs. Hey, what a question! If only 5, and only five owners grab dogs, not including the sixth owner, what will be the possibility of grabbing the correct dog?[/hide]

[quote="dragonx"]this is a trick question
[hide] if u have the 5th one then obviously the 6th one will also have his dog correct. so theres 0% chance[/hide][/quote]

Please don't revive old topics, Reds.


function
inequalities
number theory proposed
number theory

[size=150]P[/size]roblem.  so let n be a natural number greater one. Show that if  a, b, c are also interegers,  such that two conditions are true: $1 \leq a <b <c$ and a+b+c = 3n, ; then we have $abc+b \leq n^{3}$.  Is possible equation in last formula...?!

note that for numbers $x_{1}<...<x_{n}$ where $x_{1}+...+x_{n}=c$ where $c$ is a constant number,then the product of these numbers is maximum when their difference is minimum,i.e the function $f(x_{1},...,x_{n})$ is maximum when $g(x_{1},...,x_{n})=\sum_{1\leq i<j\leq n}(x_{j}-x_{i})$ is minimum...(its easy to show...)
[hide]
assume that $x>y$ then in inequality $\frac{x+y}{2}\geq \sqrt{xy}$ if we put $x-1$ instead of $x$ and $y+1$ instead of $y$ then the $LHS$ dont change and the right hand side becomes $\sqrt{xy+x-y-1}$ but from $x>y$ we get that $x-y-1>0$ so $xy+x-y-1\geq xy$ so the RHS becomes greater than before...[/hide]
now for $x,y,z$ denote $f(x,y,z)=xyz+y$ and $x+y+z=3n$ and $1\leq x<y<z$ hence we get that $y<n$ now as showed above we get that $f(n-1,n,n+1)\geq f(x,y,z)$ so we get that $f(n-1,n,n+1)\geq f(a,b,c)$ so we get that $(n-1)n(n+1)+n\geq abc+b$
which gives us:
$n^{3}\geq abc+b$ so if we put $a=n-a,b=n,c=n+1$ and we are done...


A verse can climb Everest, but a poem cannot.
A dart can climb Everest, but an arrow cannot.
A stag can climb Everest, but a lamb cannot.
A grader can climb Everest, but a judge cannot.
Fear can climb Everest, but love cannot.

And of course,

Everest can climb Everest, but a mountain cannot.

Question: What can climb Everest?

This reminds me of the "Ms. M" game we did in art class a while ago.

[quote="mgao"]types and examples of things?
for example mt everest is an example of a mountain.[/quote]

A stag is not an example of a lamb.

And neither is fear an example of love.

[hide]Words typed with only the left-hand keys.[/hide]

It's clearly not an accurate rule if it says that a sherpa can't climb Everest.

[quote="Arrange your tan"]cf249,

I don't know what the solution is, but your idea

doesn't pan out fully because the letters

"e" and "a" (for example) are on left-hand keys.[/quote]

I don't get what you want to point to, his rule at least applies (and is probably the intended one).

[hide]
"Cannot be typed only on left hand keys" does not equal "Can be typed only on right hand keys"[/hide]

By the way, that's correct.

How is your left hand related to Everest?
It does not make sense at all :huh:

Type the word "everest".

Notice what your right hand is doing.

What he means is that the concept of "left hand" and "everest" are totally unrelated.

It doesn't have to make sense, modeler, it's a riddle.

I retract my earlier post.

(It's too late to delete it.)

[quote="randomdragoon"]I also could have named the riddle "What do stewardesses like" but I think that would be a lot more obvious.

Desegregated stewardesses can climb Everest.[/quote]

Okay. So you can climb Everest using your left hand only, but left hand cannot climb Everest, right? :lol:

LOL no
the words which can be typed using the left hand solely, on the keyboard,... are the ones that can scale Everest
:)

Yes, but one can not type "left hand" with the left hand only.


abstract algebra
superior algebra
superior algebra solved

this one is quite easy. let G be a group and Z(G) denote its centre. if $G/ Z(G)$ is cyclic, then prove that G is abelian

well i just found that this has been asked before or even then its simple. anyway sorry, for not checking before . moderator please shift it to the solved problems section

This is quite classical ;-)

Take a generator $g Z(G)$ of $G/Z(G)$. For all $x,y \in$ there are integers $i,j$ such that $x Z(G) = (g Z(G))^i = g^i Z(G)$ and $y Z(G) = (g Z(G))^j = g^j Z(G)$. Then $x g^{-i} , y g^{-j} \in Z(G)$, hence
\[ x y = x g^{-i} g^i y g^{-j} g^j = x g^{-i} g^i g^j y g^{-j} = x g^{-i} g^j g^i y g^{-j} = y g^{-j} g^j x g^{-i} g^i = yx \]

By the way, this you should keep this theorem in mind when dealing with group theory. Sometimes, it's very useful.

More generally it is true that if the quozient $G/H$ is cyclic and $H\leq Z(G)$, them G is abelian


inequalities
geometry
circumcircle
trigonometry
incenter
trig identities
Law of Sines

$ABC$ is a triangle. The internal bisector of angle $A$ meets the circumcircle again at $P$. $Q$ and $R$ are similarly defined with $B$ and $C$ respectively.

Prove that $AP+BQ+CR > AB+BC+CA$.

I will prove that $\left|AP\right| > \frac{\left|AB\right|+\left|AC\right|}{2}$. Summing up for the other sides will then give the desired result.

If $\left|AB\right|=\left|AC\right|$, this claim is trivially true because then $\left|AP\right| = 2R$

If $\left|AB\right| \neq \left|AC\right|$:
Using the extended law of sines, we can show that $2\left|AP\right| = 2R\left(\sin{(B+\frac{A}{2})}+\sin{(C+\frac{A}{2})}\right)$.
Therefor, the inequality I intend to prove is equivalent to \[\sin{(B+\frac{A}{2})}+\sin{(C+\frac{A}{2})}> \sin{B}+\sin{C}\iff 2\cos{\frac{B-C}{2}> 2 \sin{\frac{B+C}{2}}\cos{\frac{B-C}{2}}\iff 1 > \sin{\frac{B+C}{2}}}\]
And there is no doubt about this last inequality.

[color=indigo]Let $I$ be the incenter of triangle $ABC.$

Easy to show that: $PI=PB=PC,\,QI=QC=QA,\,RI=RA=RB$

We have: $2IP=PB+PC>BC$ and similarly $2IQ>CA,\,2IR>AB$

Also $BI+CI>BC,\,CI+AI>CA,\,AI+BI>AB$

Adding all these inequalities above, we have: $AP+BQ+CR>AB+BC+CA$[/color]


How about having a folder in which one can discuss recent books read and tips of books to read
create book lists and so on.
(Books mainly in math)

Just some thoughts  :)

This forum http://www.mathlinks.ro/Forum/index.php?f=319 is about what you ask.


AMC
AIME
probability
analytic geometry
graphing lines
slope
ARML

I'll post the first problem.
I assume everyone knows how a marathon works.

I'll start with two problems. Pick one, and answer it. Then post two of your own. (or one, whatever, I'm not going to complain =0)

Here's the first.

[size=200]Problem $ \sqrt {1}$[/size]

[size=100]How many different strings of ones and zeros are there of length 2007 such that no two ones are adjacent?
[/size]
[size=200]Problem 2[/size]
2008 non-zero integers between one and nine are generated, find the probability that their product is evenly divisible by 10.

man i suck at counting so much....

I don't get the second problem. we would get probabilities with infinites, and that would be bad. is there some limit on the chosen integers?

Oh shoot. Let me reword. I forgot to do some editing last night :/

I have 2 nice ways to do #1.

[hide="First way"]
This requires [i]a lot[/i] of casework. There's no way you can do this by hand.

We can have 0, 1, 2, ..., up to 1004 1s. For each number of 1s, the number of ways that no 1s are together is a bijection to a much easier problem: between each pair of adjacent 1s (not really adjacent, but as in there are no 1s in between them). So for example, for 6 1s, take out 5 0s in between them. Now to convert it back, for n 1s, take a string of 2007 - (n - 1) numbers and choose n of them to be 1s, and then add 0s in between every pair of 1s. So the sum would be
\[\sum_{i=0}^{1004}\binom{2008-i}{i}\]
[/hide]

[hide="Second Way (much better)"]
We define a recursion for this method.

Let's take a look at a specific case. Say, three digits total. There are 5 ways:
000
100
010
001
101

To get to the next case, add a 0 to the end of each string which ends with a 1, and add a 1 and a 0 (for 2 separate strings) to the end of each string which ends with a 0. So for four digits, we have:
0000
0001
1000
1001
0100
0101
0010
1010

Another way to put it would be that for $n$ digits, we add a 0 to the end of all $n-1$ digit strings, and add a 01 to the end of all $n-2$ digit strings. So if $S_{n}$ is the number of possible $n$-digit strings, we have the relationship $S_{n}= S_{(}n-1)+S_{(}n-2)$. Since $S_{0}= 1$ and $S_{1}= 2$, we have the Fibonacci sequence, except the index is shifted over twice, so $S_{n}= F_{(}n+2)$. Now just plug into Binets formula for $2009$ and we have the answer!
[/hide]

Okay, Number 2...

[hide]
This is just PIE and complementary counting. First, let's count the ways that it's NOT divisible by 10:
First we count the number of ways that no five is used: $8^{2}008$.
Next we count the number of ways that no evens are used: $5^{2}008$.
Now we count the overlap between the two groups, which is where neither evens nor five is used: $4^{2}008$.

So there are $8^{2008}+5^{2008}-4^{2008}$ ways that the product is NOT divisible by 10. Now we just subtract that from the total number of ways to choose 2008 integers, which is $9^{2008}$, to get the final answer of $9^{2008}+4^{2008}-8^{2008}-5^{2008}$.
[/hide]

@DFT: You should specify something for the way you want your answer to be presented. For example, "The answer is in the form of $a^{n}\pm b^{n}\pm c^{n}...$. Find $n+a+b+...$

[hide="One more thing..."]
I suck at math.
[/hide]

The binet's formula solution was correct.

However, for the secondary problem, I wanted the probability, not the amount of numbers (trivial final step however.) Post another problem?

Oops, that was my bad. I'm too lazy to edit it...

Here's an easy one:

[u][size=200]Number 3.[/size][/u]
Let $a$ be the sum of the digits of $4444^{4444}$. Let $b$ be the sum of the digits of $a$. Find the sum of the digits of $b$.

[hide="3"]
Number of digits of $4444^{4444}$ is $4444\log4444<4444\cdot4=17776.$
So $a$ is less than $9\cdot17776=159984,$ so at max is has 6 digits.
Then $b$ is less than $9\cdot6=54.$
So sum of digits of $b$ is less than $13.$
But the sum of digits of a number is equal to the number mod9.
Using Euler's totient theorem,
$4444^{4444}=7^{4444}=7^{4}=7\mod9.$
But since the sum of digits is less than $13,$ it has to be $7.$(it can't be 16 or greater)
[/hide]

[color=red]
[size=200]PROBLEM 4 AND CHEESEBURGERS[/size][/color]
Let $w_{1}, w_{2}, \dots, w_{n}$ be complex numbers.  A line $L$ in the complex plane is called a mean line for the points $w_{1}, w_{2}, \dots, w_{n}$ if $L$ contains points (complex numbers) $z_{1}, z_{2}, \dots, z_{n}$ such that
\[\sum_{k = 1}^{n}(z_{k}-w_{k}) = 0. \]
For the numbers $w_{1}= 32+170i$, $w_{2}=-7+64i$, $w_{3}=-9+200i$, $w_{4}= 1+27i$, and $w_{5}=-14+43i$, there is a unique mean line with $y$-intercept 3.  Find the slope of this mean line.

Correct! Post a problem.

BTW, go compare your answers with mine on the practice ARML problems (FYI, they're not a real ARML).

EDIT: Somehow I missed the fact that you posted a problem...

EDIT again: I've seen that problem before.

EDIT again: For some reason the edits don't show up in the post. (It doesn't say how many times it has been edited).

[hide="#4"]
Let us call the mean line $y = mx+3$. Now pick 5 points on it: $a_{k}+(ma_{k}+3)i$ for $k = 1, 2, 3, 4, 5$.

Now add up $\sum_{k=1}^{5}(z_{i}-w_{i})$ to get $\sum_{k=1}^{5}a_{k}+mi\sum_{k=1}^{5}a_{k}+15i-(3+504i)$, which is equal to $0$. Make the substitution $\sum_{k=1}^{5}a_{k}= n$. Now we have $n-3+mni-489i = 0$. Setting reals and imaginaries equal, we have the system $n-3 = 0$ and $mn = 489$, so $n = 3$ and $m = 163$.

\[\boxed{163}\]
[/hide]

lol i forgot the problem so i edited it a while later
and yes, it doesnt show the edit, if you edited before someone else posted.
your turn

Okay.
[u]Problem 5[/u]
There are 3 distinguishable hockey pucks, non-collinear, on a rink. A hockey player shoots one puck through the other two. He does this 25 times. Can he return the pucks to their starting position? (Give a proof for why or why not).

[hide]
me: alright
me: so everytime you shoot a puck
me: the side of BC that A is on changes
me: if it was on the right
me: then it goes to the left
me: vice versa
me: (from B's perspective)
Ubemaya: that's not very clear
Ubemaya: maybe you should phrase it this way
Ubemaya: if you rotate the field so that B is straight above C, A changes sides every time
Ubemaya: anyway
Ubemaya: that's correct
me: same thing
Ubemaya: except
Ubemaya: an easier way to prove it is to consider the order of the pucks. Call ABC "good" and ACB "bad", every time the pucks go from good to bad
Ubemaya: well that's pretty much the same, just easier to see
me: eh
me: ok
me: im too lazy to write up the solution
Ubemaya: so type your solution
Ubemaya: 'give a problem
Ubemaya: OMG
me: ill just post this conversation
[/hide]

[size=200]PROBLEM 6[/size]
(Engel)
There are $13$ white, $15$ black, $17$ red chips on a table. In one step, you may choose 2 chips of different colors and replace each one by a chip of the third color. Can all chips become of the same color?

[hide="6"]
There are $45$ chips total. It is easy to see that the difference between the number of chips of each pair of colors is invariant modulo 3. Since the difference start off at 2, 2, 4, the differences can never attain 0, 0, 45.
[/hide]

I'm now going to post a series of problems leading up to a final problem. You guys can still post whatever you want.

Problem 7
On a chessboard, a queen starts in the lower left hand corner. Players take turns moving the queen up, right, or diagonally up-right. Whoever cannot move on his turn wins. Assuming the players use their best possible strategies, does the first player or the second player win?

By the way, we should start citing where our problems come from. This one is from Mathematical Circles (Russian Experience).

EDIT: Sorry, I wasn't keeping track of the problem numbers :oops: .

Doesn't every position have a legal move from it?

Sorry about that, it's been edited (x3 :lol:).

let BC and AO intersect at P

chase chase chase those angles!

[hide="proof:"]
draw a circle, pick an arbitrary point A outside of the circle.
draw the indicated segments. omg AZC looks like a right angle.
q.e.d (c?) :P
[/hide]

lol wow this took me way too long but I had to resort to some trig :(!

[hide]
Let $\angle ADC = \alpha$ and $\angle ZDC = \beta$. Since $AC$ is tangent to $\omega$, $\angle ACZ = \angle ZDC = \beta$.

Let the radius of $\omega$ be $R$. Then by the Law of Sines, $ZC = 2R \sin{\beta}$.

Since $\angle AOC = 2\angle ADC = 2\alpha$, we find that $AC = R \tan{2\alpha}$.

It remains to show that \[\frac{ZC}{AC} = \cos{\beta}
\Leftrightarrow 2\frac{\sin{\beta}}{\tan{2\alpha}} = \cos{\beta} \Leftrightarrow 2\tan{\beta} = \tan{2\alpha}.\]

But looking at triangles $DYX$ and $DBX$, this is equivalent to $\frac{2(YX)}{XD} = \frac{BX}{XD}$ which is true since $Y$ is the midpoint of $BX$.

[/hide]

[quote="AceOfDiamonds"][quote="pythag011"]Find the set of numbers k such that there exist infinitely many distinct positive integers x, y such that (x+i)(y+i) is a square for i = 1, 2, ..., k.

Wait what since when was this thread Tough AMC/Medium AIME? I remember it saying Tough AIME/Medium USAMO.

[hide="A RELATED PROBLEM"]Let a be a positive integer. p_i denotes the ith prime. Is the lim inf of $\log_{p_i}(\text{ord}_{p_i^2}(a)) > 1?$[/hide][/quote]
wha'ts ord :([/quote]

http://en.wikipedia.org/wiki/Order_(number_theory)

wait rofl I created that page

bump

guys you should do my problems :(

i have a new [size=50]conjecture[/size]problem for everyone!

k is a positive integer

there is an infinite graph where each finite subgraph can be colored with k colors

prove or disprove the statement that then it always possible to color the infinite graph with k colors

[quote="pythag011"]Bump. Why does nobody ever do my problems :([/quote]

Maybe someone would do your problems if they weren't always impossible. This marathon needs some non-impossible problems, so here's one: $x$, $y$, and $z$ are positive reals such that $x > y > z$. Prove that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \neq \frac{y}{x} + \frac{z}{y} + \frac{x}{z}$

These are tough AMC/Medium AIME problems. That means NO PROOFS.

[quote=v_Enhance][quote="vinayak-kumar"]That means NO PROOFS.[/quote] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3306342#p3306342]What's wrong with proofs[/url]? Ignoring the fact that you're bumping a thread that has been dead for three years, anyways. :wink:

Since I'm here I guess I might as well post an easier problem, though.

A positive integer $n$ is called [i]mythical[/i] if every divisor of $n$ is two less than a prime.  Find the unique mythical number with the largest number of divisors.[/quote]

Note that a mythical integer cannot be even, since $2$ is not $2$ less than a prime. Assume $n$ is a mythical integer with a prime power $p^k$ as its divisor, such that $p$ is not $3$ and $k$ is larger than $1.$ Then, $p^2$ is a divisor of $n$, so $p^2=q-2$ for some prime $q.$ But, $p^2$ would be $1 \pmod 3,$ so $q$ would be $0 \pmod 3,$ a contradiction to it being a prime (since $q>3$). For the same reason, all prime factors must be either $3$ or $2 \pmod 3.$ But if $p$ and $q$ are both prime factors, both being $2 \pmod 3,$ $pq$ gives a contradiction. Thus, $n$ must be of the form $3^k$ or $p\cdot 3^k$ with $p$ as $2 \pmod 3.$ $k=4$ works, but $k=5$ doesn't. If $n$ is $p\cdot 3^4,$ we must have $p+2, 3p+2, 9p+2, 27p+2, 81p+2$ all be prime. $p=5$ does not work. For other primes $p,$ by mod $5$ reasons, would not work either. Thus, the candidates are $3^4$ and $p\cdot 3^3.$ Indeed, $p\cdot 3^3$ gives more divisors, and since $p=5$ works, and we are given there is only one such mythical integer with maximal number of divisors, we conclude the answer is $135.$


anyone have a spare lj code? I'll make ya 2 blinkies and a icon...

Invite code required to get an LJ. Unfortunately I don't.

well use xanga or some other blog site if all else fails

livejournal was nice to me though, so use that preferably. but if youre trying to get many friends to use it, use xanga.

lj has more features and schtuff

[quote="Tare"]Why would they adopt such system?[/quote]

According to the LiveJournal website,

[quote="LiveJournal"]As a means to keep the community tight, control growth, and prevent abuse, joining LiveJournal now requires either buying an account or getting invited by an existing user. The point of this is to make LiveJournal a better place for the people that use it, at the added cost of a little inconvenience for new users.[/quote]

The way I see it, you have to either know somebody or be willing to spend a few bucks to start a paid account (and then let it expire).  This way, not anybody can just join.

That said, I happen to have an extra code from working in support.  Anyone interested should PM me, and I'll consider it and (probably) give you it.

*sigh*..i've been trying to get LJ forever...


function
real analysis
real analysis unsolved

c = 1449.2 + 4.6t - 0.055t^2 + 0.00029t^3 - 35(1.34 - 0.01t)+0.016z
derive the algebraic expression for dc/c as a function of dt/t and dz/z
HELP
so i differentiate all the t terms w.r.t. t * dt, add the z term differentiated w.r.t. z *dz
and divide by c?
if so it looks a bit complicated and long

It surely does, but then, the form of $c(t,z)$ is already complicated and long, and the answer will be even more so. The problem looks strange; are you sure you understood its statement correctly?

the answer is:
dc/c=0.00087t-.11+4.95/t dt + 0.016/z dz
just incase anyone wondered. i worked it through and its correct. it came from an expression for the speed of sound in water, that's why its complicated

To be honest, I don't understand how you got this particular answer :?. On the other hand, I did not quite understand the formulation of the problem either: what is $\frac{dt}{t}$ for example? The only meaning I could assign to the whole thing was that the problem asked to find the relative rate of change of $c$ given the relative rates of change of $t$ and $z$. But then the answer would be completely different: among ohter things it would contain the entire expression for $c$ in the denominator. Can you explain what was really meant and how you obtained your answer?


inequalities
trigonometry
inequalities proposed

$ x^2 \plus{} y^2 \equal{} 14x \plus{} 6y \plus{} 6$

Find max $ 3x \plus{} 4y$

x,y are reals

$ (x\minus{}7)^2 \plus{} (y\minus{}3)^2 \equal{} 8^2$ Hence we can write $ x \equal{} 7 \plus{} 8 \cos \theta$ and $ y \equal{} 3\plus{}8 \sin \theta$

$ \therefore \ 3x\plus{}4y \equal{} 33 \plus{} 8 (4\sin \theta \plus{} 3 \cos \theta) \le 33\plus{}40 \equal{} 73$

$ x^2\plus{}y^2\equal{}14x\plus{}6y\plus{}6 \iff (x\minus{}7)^2 \plus{} (y\minus{}3)^2 \equal{} 8^2$
よって, 中心が (7, 3) で半径が 8 の円.
これに $ 3x\plus{}4y\equal{}a$ が共有点を持てばよいので, 点と直線の距離の公式より,
\[ \frac{\left| 3 \cdot 7 \plus{} 4 \cdot 3 \minus{} a \right|}{\sqrt{3^2\plus{}4^2}} \equal{} \frac{\left| 33\minus{}a \right|}{5} \equal{} 8\]
この方程式を解いて, $ a\equal{}\minus{}7, 73$.

よって最大値は $ 73$.

let $ a \equal{} \frac {x \minus{} 7}{\sqrt {64}}, b \equal{} \frac {y \minus{} 3}{\sqrt {64}}$ so $ a^2 \plus{} b^2 \equal{} 1$ so its known that $ 3a \plus{} 4b\le5$ so
$ \frac {3x \minus{} 21 \plus{} 4y \minus{} 12}{\sqrt {64}}\le5$ or $ 3x \plus{} 4y\le 5\sqrt {64} \plus{} 33\equal{}73$

But dima, max value of $ (3x\plus{}4y)$ was asked.... :maybe:
Anyway, it gives that $ 71.0788655 \leq (3x\plus{}4y) \leq 73$. I wonder why the range of $ (3x\plus{}4y)$ is so small....

Sorry I make a typo, I edit my post. Sunkern_sunflora: $ \minus{} 5\cdot8 \plus{} 33 \equal{} \minus{} 7\le3x \plus{} 4y\le5\cdot8 \plus{} 33 \equal{} 73$

[quote="dima ukraine"]Sorry I make a typo, I edit my post. Sunkern_sunflora: $ \minus{} 5\sqrt {58} \plus{} 33\le3x \plus{} 4y\le5\sqrt {58} \plus{} 33$[/quote]
Try $ x\equal{}11.8, y\equal{}9.4$

hello, it must be $ \minus{}7\le3x\plus{}4y\le73$.
Sonnhard.

Sorry Dr Sonhard I dont see 6 in condition I reedit it :blush:  :)

Another way Lagrange multipliers, for $ F(x,y) \equal{} 3x \plus{} 4y$ if $ (x \minus{} 7)^2 \plus{} (y \minus{} 3)^2 \equal{} 64$ Then we can say $ a \equal{} x \minus{} 7,b \equal{} y \minus{} 3$ we will decide system
$ a^2 \plus{} b^2 \equal{} 64$
$ 3 \minus{} 2La \equal{} 0$
$ 4 \minus{} 2Lb \equal{} 0$ then $ x\pm\frac {24}{5}$ and $ y\pm\frac {32}{5}$ so $ maxF(x,y) \equal{} F(\frac {24}{5},\frac {32}{5}) \plus{} 3\cdot7 \plus{} 4\cdot3 \equal{} 73$

$ (x\minus{}7)^2\plus{}(y\minus{}3)^2\equal{}64$

And we need to find the max. of $ 3(x\minus{}7)\plus{}4(y\minus{}3)\plus{}33 \le \sqrt{(3^2\plus{}4^2)((x\minus{}7)^2\plus{}(y\minus{}3)^2)}\plus{}33\equal{}73$.

Equality when $ 3/(x\minus{}7)\equal{}4/(y\minus{}3)$ <--> $ (y\minus{}3)\equal{}4(x\minus{}7)/3$ which you can use to find such a pair (x,y) by plugging into the original equation.

[quote="B_Gasperov"]$ x^2 + y^2 = 14x + 6y + 6$

Find max $ 3x + 4y$

x,y are reals[/quote]

Yes, the answer is $ 73$, but no one didn't point out the equality case.

Anyway, my solution is here.

$ \{4(x - 7) + 4(y - 3)\}^2 + \{4(x - 7) - 3(y - 3)\}^2 = 25\{(x - 7)^2 + (y - 3)^2\} = 40^2$,

$ 3x + 4y = 3(x - 7) + 4(y - 3) + 33\leq 40 + 33 = \boxed{77}$ for $ x = \frac {59}{5},\ y = \frac {47}{5}$.

Japanese highschool students usually solve in such way as Kouichi Nakagawa's solution.

[quote="Dr Sonnhard Graubner"]hello, it must be $ - 7\le3x + 4y\le73$.
Sonnhard.[/quote]

That's right! :)

The equality case is yielded just as easily by the parametric representation. $ \sin \theta \equal{} \frac{4}{5}$ and $ \cos \theta \equal{} \frac{3}{5}$ gives $ x \equal{} \frac{59}{5}$ and $ y \equal{} \frac{47}{5}$

Indian students would handle it this way :lol:


$\log_{x}3^{x}= x \log_{x}3 = x \frac{\ln 3}{\ln x}$.

Thus, the derivative is then $\ln 3 \frac{\ln x-1}{(\ln x)^{2}}$.

Also, this problem does not belong in this section.

[quote="bos1234"]differentiate $log_{x}3^{x}$[/quote]

[hide="Alternate solution"]Let $y = \log_{x}3^{x}$. Then:

$x^{y}= 3^{x}$

$\ln (x^{y}) = \ln(3^{x})$

$y \ln(x) = x \ln(3)$

Differentiating both sides,

$y \cdot \frac{1}{x}+\frac{dy}{dx}\ln(x) = \ln 3$

Thus,

$\frac{dy}{dx}= \frac{\ln 3-\frac{y}{x}}{\ln(x)}$
$= \frac{\ln 3-\frac{\log_{x}3^{x}}{x}}{\ln(x)}$
$= \frac{x \ln 3-\log_{x}3^{x}}{x \ln(x)}$
[/hide]

vishalarul, your answer is also correct. :)


group theory
combinatorics theorems
combinatorics

is anyone know about hook length formula?
please explain specifically.

There is a nice article by Novelli, Pak and Stoyanovskii at http://www.emis.de/journals/DMTCS/volumes/abstracts/pdfpapers/dm010104.pdf

It's a formula for counting the number of standard Young tableaux of a given shape.  You can read about it in many texts on enumerative or algebraic combinatorics, e.g., Stanley's [u]Enumerative Combinatorics, Volume 2[/u] or Sagan's [u]The Symmetric Group[/u].

can anyone explain how i can get hook length at each cells?(young tableau)

and is this right?  ->   the number of standard tableau=(number of cells)!/each cells' hook length

It would probably serve you well to look at one of the source materials we have suggested (or even Wikipedia) if what you're interested in is straightforward definitions.  The hook length of a cell is (in English notation) the total number of cells directly below (the leg) or directly to the right (the arm) of the cell, counting the cell itself.  Your statement of the hook length formula is correct assuming you meant for the denominator of the right-hand side to be the product of the hook lengths.


limit
probability
logarithms
probability and stats

1. Let $ \{X_n,n\geq 1\}$ be a sequence of independent r.v, If for some $ p\in (1,2]$, $ \sum_n n^{\minus{}p} \mathbb{E}|X_n|^p<\infty$, prove  that \[ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j\equal{}1}^n(X_j\minus{}\mathbb{E}X_j)\equal{}0\quad a.s.\]

2. Let $ \{X_n,n\geq 1\}$ be a sequence of independent and identically distributed r.v., prove $ \limsup_{n\rightarrow\infty}\frac{|X_n|}{n}<\infty\quad a.s.$, implies $ \sum_n\mathbb{P}(|X_n|\geq n)<\infty$.

3. Let $ \{A_n, n\geq 1\}$ be a sequence of independent event. If $ \sum_n\mathbb{P}(A_n)\equal{}\infty$, prove
\[ \lim_n\frac{\sum_{m\equal{}1}^n 1_{A_m}}{\sum_{m\equal{}1}^n \mathbb{P}(A_m)}\equal{}1\quad a.s.\]

Thank you very much!!!

[quote="siiz"][quote="hubanya"]
2. Let $ \{X_n,n\geq 1\}$ be a sequence of independent and identically distributed r.v., prove $ \limsup_{n\rightarrow\infty}\frac {|X_n|}{n} < \infty\quad a.s.$, implies $ \sum_n\mathbb{P}(|X_n|\geq n) < \infty$.
[/quote]
This problem can be solved using 0-1 Law and Borel-Cantelli lemma.
[quote="hubanya"]
3. Let $ \{A_n, n\geq 1\}$ be a sequence of independent event. If $ \sum_n\mathbb{P}(A_n) \equal{} \infty$, prove
\[ \lim_n\frac {\sum_{m \equal{} 1}^n 1_{A_m}}{\sum_{m \equal{} 1}^n \mathbb{P}(A_m)} \equal{} 1\quad a.s.
\]
[/quote]
See Williams, Probability with Martingales, theorem 12.15.[/quote]

Thank u very much
I have solved the 1st and the 2nd question. To the second one, I used Borel-Cantelli lemma. To the first one, I used Kronecker's Lemma.

Thanks for your reference to the book.

[quote="hubanya"]1. Let $ \{X_n,n\geq 1\}$ be a sequence of independent r.v, If for some $ p\in (1,2]$, $ \sum_n n^{ \minus{} p} \mathbb{E}|X_n|^p < \infty$, prove  that
\[ \lim_{n\rightarrow\infty}\frac {1}{n}\sum_{j \equal{} 1}^n(X_j \minus{} \mathbb{E}X_j) \equal{} 0\quad a.s.
\]
[/quote]
Because the above result holds for any $ p > 1$ (and smaller than 2), I wonder if the following propositions are correct?
First, let denote $ \log^ \plus{} (x) \equal{} \log(|x| \vee 1)$.
Prop. 1) Let $ \{X_n,n\geq 1\}$ be a sequence of independent r.v, if  $ \sum_{n > 1} (n\log n)^{ \minus{} 1} \mathbb{E}(|X_n|\log^ \plus{} (X_n)) < \infty$, prove  that
\[ \lim_{n\rightarrow\infty}\frac {1}{n}\sum_{j \equal{} 1}^n(X_j \minus{} \mathbb{E}X_j) \equal{} 0\quad a.s.
\]
Prop. 2) Let $ \{X_n,n\geq 1\}$ be a sequence of independent r.v, if  $ \sum_{n > 1} (n(\log n)^2)^{ \minus{} 1} \mathbb{E}(|X_n|(\log^ \plus{} (X_n))^2) < \infty$, prove  that
\[ \lim_{n\rightarrow\infty}\frac {1}{n}\sum_{j \equal{} 1}^n(X_j \minus{} \mathbb{E}X_j) \equal{} 0\quad a.s.
\]
Any comment?


Let S be a set of n points in the plane such that the distance between any two points in S is at least one. Show that there are at most 3n pairs of points at distance exactly one.

This can be easily sharpened to $3n-6$. It is an old result from Erdös, using planar graphs.
You may find a solution, in french as exercise #14 in
http://www.animath.fr/cours/graphes.pdf

Pierre.


This was posted in another topic. Let's bring it out here where it can be seen.

[quote="Moubinool"]another question
$x_{0}>0$

$x_{n+1}=\frac{x_{n}}{1+n^{2}x_{n}}$

Find an equivalent of $x_{n}$[/quote]

[quote="mao_zai"]i hope i did not read wrongly.

but isnt $x_{n}=\frac{x_{0}}{1+(\sum_{i=1}^{n-1}i^{2}) x_{0}}$?[/quote]You are indeed correct, mao_zai!
We could certainly write the closed form expression for $\sum_{i=1}^{n-1}i^{2}$, but that would just make the final answer look very ugly.


AMC
AIME
USA(J)MO
USAMO
geometry
AMC 10
videos

I started practicing for AIME a lot recently, and it started off really well, I was able to do 5 or 6 more questions than I used to be able to do, but about a week later, after strenuous practicing, my head started to hurt, and sometimes I couldn't even solve some problems I was able to solve before...It doesn't seem like I'm getting any better either, only getting worse, is this because I'm working too hard? (4-5 hours/day doesn't seem like a long time...)

Four to five hours a day seems like a long time to me  :huh:  Take a break for awhile.

4-5 hours doesn't seem like a long time to do math, but it sounds like a REALLY long time to do AIME problems. You have plenty of time to prove yourself on the AIME, and if you're as interested as you sound, you'll do fine. Most importantly, only do math when you want to, don't try to make a schedule for yourself, and if you get tired or your head starts to hurt, STOP.

On down days, I suggest you just think about harder problems off and on throughout the day. You will find that working a few problems without a time limit is much less strenuous than working lots of problems with a strict time limit.

Don't overwork yourself and think that you need to qualify for USAMO. Just spend time doing the problems if you're having fun. Last summer, when I was just a newbie at competitive math, I think I made my biggest improvements ever just because I didn't feel too much pressure.... I just did problems because they were fun. But as the school year went by (although I liked (and still like) math), I felt more and more that if I didn't qualify for USAMO, I'd be a failure. So I spent at least 6-8 hours a day the week before the AIME (along with school) doing old AIMEs and reviewing. Then the day before the AIME, I got sick, and when I took the AIME, I had a $103^{\circ}$ fever, later going up to $105^{\circ}$ a few hours after the test. Needless to say, I didn't qualify for USAMO. In retrospect, I'd have done a lot better if I didn't overworked myself the week before the AIME and gotten myself sick (and the entire school year in general).

Do as much as you want to do...... challenge yourself, but not to a level in which you're not having fun and are stressing yourself. At that point, you're going to hurt, not help, yourself. The situation you're in right now can certainly be explained by overwork.

Well, I took just 4 AIME practice tests and (barely) qualified for USAMO  :D

But seriously, at this point, I wouldn't recommend practicing that much each day. First of all, as others said, use resources that actually teach you things, not just old contests that give you practice. Doing AIME Problems Series is one of the natural ideas if you're preparing for the AIME. As well, you should want to read and fully comprehend AoPS Volume 2 (and AoPS 1 if you feel a need to) among other books, as the others mentioned. As you learn more and get better, just do a practice AIME now and then. When I get bored, I sometimes go and take a practice AIME (but I don't do one or two every day!).

[quote="pkerichang"]I'll say start to get away from the practice problems...Read books that actually TEACHES you stuffs (Geometry Revisited, Introduction to Diophatine Equations, etc.) instead of pure problems (like old AIMES and AMC problems) and do the problems following each chapter of the teaching books.  This way you actually learn new things and new tools to help you solve problems.

Then after a while you can get back and do practice problems (or else you'll forget what you learned) then you should find improvement (my score went from 7-8 to 12 on AIME...)[/quote]
Agreed - concept books teach you in a more complete and systematic way than just doing problems.  However, if you're working with a guided problem-solving book (for example, ACoPS), then working problems can still really increase your knowledge of mathematical concepts because the books introduce you to new concepts and how to apply them to solve difficult problems.  So, really a combination of both is preferrable.  It's probably less helpful to just sit down and do 10 old AIMES - if you don't know the concepts underpinning the problems, you might after reading a solution understand how to solve that one problem, but most likely you won't be able to readily apply that knowledge to solving other problems, which is the whole point of preparation, isn't it?

You are not getting worse at all.  You just need to give yourself more breaks.  You should try to reduce your time.  Then, do the problems that you can do, and if you can't do it, just skip it and go to the next one.  If you still can't do any, just take a break, and give yourself a rest.

I hope I helped.

You can't expect to see much improvement in just one week. Good performance is a result of months of preparation leading up to the test. Don't worry too much about scores or time limits now- just focus on solving the problems and understanding the underlying 'big ideas.'

http://en.wikipedia.org/wiki/Diminishing_returns

Basically, repeated practices may result in declining of your score. So take a break. After all, USAMO is a big thing but you don't have to kill your summer vacation doing it. Play sport, get together with friends, and do whatever you want to do other than math.

That really isn't what's meant by diminishing returns, although I guess it could be taken away.

What is meant is that, for example, it's much easier to improve from scoring 2 to 5 on the AIME than it is to improve from 11 to 14.

[quote="mathgeniuse^ln(x)"]Then, do the problems that you can do, and if you can't do it, just skip it and go to the next one.[/quote]

How can you ever improve if you only do math problems that you can already do (or am I misunderstanding you)?

I guess he means that you shouldn't get bogged down on one problem. You should certainly attempt problems above your comfort level.

Thanks for all the advice. You guys are absolutely right, I should spend more time with other stuff rather than AIME problems. I do have the AoPS books. I feel much better now knowing that this was due to my working too hard (for my own limits), and not because my problem solving skills have diminished.   :)  Thanks a lot, you guys are great!  :D  :D

[quote="1234567890"] So I spent at least 6-8 hours a day the week before the AIME (along with school) doing old AIMEs and reviewing.[/quote]

wow, how do you have so much time?  6 hours plus school plus homework. I'm only doing 3 hours of studying n this is during the summer.  i feel so lazy :)

i can do most of the problems on the AMC10, but I always make a lot of stupid mistakes.  advice anyone?

[quote="myc"]i can do most of the problems on the AMC10, but I always make a lot of stupid mistakes.  advice anyone?[/quote]

Organize your work so that you can check your answers just by looking over your work, rather than having to solve the problem again.  For example, on the AIME I like to give myself a separate sheet of paper for each problem.

Also, leave yourself some time at the end of the test to check your work.  Remember that fixing an error on an AMC problem gives you more points than solving a problem you would otherwise leave blank.

In fact,human's brain is more complex than that we image.
So,I think you will get better on a lucky day.

Obviously, doing problems isn't going to teach you very much, if the solution to the problem involves concepts unknown to you.

So yeah, as someone else said, read books teaching actual concepts, not just problems.

Huh?? I'd think it would be opposite...how is it going to help to do problems that you already know how to solve?

[quote="Iversonfan2007"][quote="jclarkemathL314159"]Obviously, doing problems isn't going to teach you very much, if the solution to the problem involves concepts unknown to you.
[/quote]

uh, you learn those unknown concepts through doing those kinds of problems...[/quote]

Well...if you don't even know factoring, solving $xy-x-y+1 = 4$ will be VERY hard...(I know this is a lame example)

The AoPS intermediate classes are not only good for the concepts, they teach you how to think, so I would recommend that.

Okay.

I hope it is possible, is the deadline already expired?

No it's not (well for counting/probability and number theory it's almost over, so I'll assume you mean the upcoming ones).

[quote="Iversonfan2007"][quote="sen"]you cant get worse at math in a week.

do some english essays or something. :D[/quote]

no no the best way to get ur mind working again is by playing video games excessively and eating ice cream for one week straight[/quote]

Agreed.  :D Taking a break once in a while is a good idea, but perhaps one day should suffice? For AIME preparation, I have found that reading AoPS Volumes 1 and 2 helps immensely with concepts. After this, just go crazy by doing AIME problems (in the contests sections of this site (go Mr. Vornicu!)). These problems will help you think. Trust me, there is almost no other way to get better (I've tried all of them  :) ). Well, apart from taking AIME Problem Series A, there isn't a better way to get better at doing AIME problems.

[quote="Iversonfan2007"][quote="sen"]you cant get worse at math in a week.

do some english essays or something. :D[/quote]

no no the best way to get ur mind working again is by playing video games excessively and eating ice cream for one week straight[/quote]
 :rotfl: I don't know if it's just me, but my brain feels completely empty everytime I play games, and I play games a lot  :lol:

do like 2 hours and take a break for 2 hours then do 2 hours again.

I've been doing much less math than usual in the past few months, but I'll chime in with knowledge from another field. To most effectively build muscle, the key is to make sure it is well-worked, but not overworked. The latest fad is heavy exercise 3 times a week of compound exercises for most effective gaining, and no more than 5 times a week (split, so at least 48 hours between working the same muscle, better 72). I think this could apply to math. Maybe you should try algebra and related fields on mondays and fridays and geometry and related fields on Wednesdays and Sundays?  :D

Also, good nutrition is key. I'm currently most of the way into a (amateur) bodybuilder's diet, and I can't express how much better I feel. It's like I'm on a sugar high 24/7 in terms of thinking speed, except I'm not bouncing off the walls, and there's never any crash. It could also be that I'm now in good shape.

Look for a short essay on these topics in Serious Discussions in the next few months.


can anyone tell me why r there 2 AMCs?
the first one for our school is mandatory, but the second one seems optional

There are two AMC dates because of the fact that many schools are on vacation or otherwise closed during one of the dates. Also, some(potentially top) students tend to get sick or be absent on one of the test dates. Its not as likely to lose these students with two separate test dates.

As for being mandatory or not - thats your school's policy. For most of us, we have to beg the school to let us take it, not the other way around.


vector
geometry
rhombus
angle bisector
geometry unsolved

Here is the problem
Can anybody use complex number or vector to solve this?
 :)

i can't solve it by complex number or vector  :blush:
but i have an easy solution . . . do you want that?

1.
BF/CJ=BK/KC=AB/AC
AB/BD=BF/(DC-BF-CJ)  =>  BD=AB*CD/BF-AB-AC
Analogusly, CE=AC*BE/FC-AC-AB

2.
DC/BF=BP/BF=BC/FX
CF/BE=CF/CQ=YF/BC
(AB/AC)*(CD/BF)*(CF/BE)= BC/FX * YF/BC * AB/AC =1  =>  BD=CE

3.
RMSN is a rhombus  =>  RS⊥MN
Notice MR//AB  NR//AC  and  RS is angle bisector of ∠BAC   =>  AF//RS

Done!


function
topology
algebra
domain
limit
real analysis
real analysis unsolved

"A $C^1$ function $f: \mathbb R^n\to\mathbb R^n$ which increases distances is a diffeomorphism of $\mathbb R^n$ onto itself, and a continuous function $f: \mathbb R^n\to\mathbb R^n$ which increases distances is a homeomorphism of $\mathbb R^n$ onto itself."....

I don't really know what you meant with that Laplacian there (first of all, $\nabla$ usually denotes the gradient of a scalar field, not the Laplacian, and second of all, I would expect someone to say that $\Delta f>0$ for scalar fields $f$, not for $f: \mathbb R^n\to\mathbb R^n$ :)). Anyway, here's what I know for sure holds:

A $C^1$ function $f: \mathbb R^n\to\mathbb R^n$ which increases distances is a diffeomorphism of $\mathbb R^n$ onto itself, and a continuous function $f: \mathbb R^n\to\mathbb R^n$ which increases distances is a homeomorphism of $\mathbb R^n$ onto itself.

Sorry if I misunderstood or if I said anything stupid.

No.Actually I said a stupid thing.I had in my mind the same problems you mentioned,but I posted also some scraps of the demonstration I remembered then.There was nothing with the laplacien(it is called like this in my math course) in the statement of the problems.Only in their solutions.
Sorry for this.

Well for the second part (the one where the functions are continuous), you could do this (I've used this argument before, for a problem posted by harazi):

Clearly, $f$ is one-to-one. If we prove that it's also onto, we're done, since it would then be a continuous bijection whose inverse decreases distances, and is thus continuous as well. On the one hand, $f: \mathbb R^n\to\mathbb R^n$ is a continuous injection, so it's an open map, by Brouwer's Domain Invariance Theorem (this is a very strong and quite difficult to prove theorem, by the way :)), which means that $f(\mathbb R^n)$ is a (non-empty) open subset of $\mathbb R^n$. On the other hand, $f$ is a closed map. Indeed, if $f(x_n)\to y,\ (f(x_n))_n$ is a Cauchy sequence, which means that $(x_n)_n$ is also Cauchy (because the inverse of $f$ decreases distances), so it's convergent to some $x$, and we get $f(x_n)\to f(x)\Rightarrow f(x)=y$. This means that $f(\mathbb R^n)$ is also closed, so it's a non-empty clopen subset of the connected space $\mathbb R^n$; it must thus be the entire $\mathbb R^n$.

For the first part:

Suppose we can find $x\in\mathbb R^n$ for which $Df(x)$ is singular, and take some unit-length $v\in\mathbb R^n$ with $Df(x)v=0$. We then have $0=Df(x)v=\frac{\partial f}{\partial v}=\lim_{t\to 0}\frac{f(x+tv)-f(x)}t$, which means that for some small enough $t$, the condition $\|f(x+tv)-f(x)\|\ge\|x+tv-x\|=|t|$ is violated, and we have a contradiction. This shows that $Df$ is non-singular, i.e. $f$ is an injective local diffeomorphism. By an argument similar to the one used above (actually, we can even avoid using Brouwer's Theorem now, since $f$ is C^1, and we have other options), we find $f: \mathbb R^n\to\mathbb R^n$ to be onto as well, and the inverse is $C^1$ by the Inverse Function Theorem.


By a variant of the [url=http://mathworld.wolfram.com/ChristmasStockingTheorem.html]Hockey-Stick/Christmas Stocking Theorem[/url], it goes that:
$ \binom{n}{k}\plus{}\binom{n}{k\plus{}1}\equal{}\binom{n\plus{}1}{k\plus{}1}$

Now all we need is to apply back to the problem, keeping in mind our ultimate goal is 19C9.

[size=134][b]1) Solving for 18C8:[/b][/size]

$ \binom{17}{7}\plus{}\binom{17}{7\plus{}1}\equal{}\binom{17\plus{}1}{7\plus{}1}$

$ 19448\plus{}24310\equal{}\binom{18}{8}$

$ 43758\equal{}\binom{18}{8}$

[size=134][b]2) Solving for 18C9:[/b][/size]

$ \binom{17}{8}\plus{}\binom{17}{8\plus{}1}\equal{}\binom{17\plus{}1}{8\plus{}1}$

$ 24310\plus{}24310\equal{}\binom{18}{9}$

$ 48620\equal{}\binom{18}{9}$

[size=134][b]3) Solving for 19C9:[/b][/size]

$ \binom{18}{8}\plus{}\binom{18}{9}\equal{}\binom{19}{9}$

$ 43758\plus{}48620\equal{} \binom{19}{9}$

$ \boxed{92378}\equal{} \binom{19}{9}$

This one I used pascal's identity twice and it led me to the correct answer in a roundabout  manner.

[quote=Possible][quote=eduD_looC]Me. I mean, is there any other way?[/quote]

Yes, See Post #2[/quote]

I mean, you could use a calc and stuff.

[quote=Aryamans][quote=Possible][quote=Eat314]Do not use calculators on Alcumus unless it says you can.[/quote][/quote]

OK... but still, it is easy to just bash it out.

$$\frac {19!}{9!10!} \implies \frac {19*\cancel {18}\cancel {2}*17*\cancel {16}\cancel{2}*\cancel {15}\cancel {3}*\cancel {14}\cancel{2}*13*\cancel {12}2*11}{\cancel {9}*\cancel {8}*\cancel{7}*\cancel {6}*\cancel{5}*\cancel{4}\cancel {2}*\cancel{3}*\cancel{2}}$$[/quote]

I think you have some latex errors. Where did you get 182 and 162 and 153 and 142 and 122?

It means \cancel {18} \cancel {2} and same for the rest of them. I canceled it with the 9, and I had 2 left.


algebra
polynomial
modular arithmetic
number theory proposed
number theory

Find a polynomial $ P(n)$,with integer coefficients,such that $ P(n)\plus{}4^n$ is divisible by 27 for any positive integer $ n$.

[quote="SUPERMAN2"]Find a polynomial $ P(n)$,with integer coefficients,such that $ P(n) \plus{} 4^n$ is divisible by 27 for any positive integer $ n$.[/quote]

Nice problem

[hide="My solution"]
We need to have $ P(0)\equal{}\minus{}1 \pmod {27}$ and $ P(x\plus{}1)\equal{}4P(x) \pmod {27}$ $ \forall x\in \mathbb N$

Trying $ P(x)\equal{}ax\minus{}1$ immediatly lead to an impossibility : $ 4P(x)\minus{}P(x\plus{}1)\equal{}3ax\minus{}a\minus{}3$ which cant be null for any $ x$ mod 27.
Trying then $ P(x)\equal{}ax^2\plus{}bx\minus{}1$ implies $ 4P(x)\minus{}P(x\plus{}1)\equal{}4ax^2\plus{}4bx\minus{}4\minus{}ax^2\minus{}2ax\minus{}a\minus{}bx\minus{}b\plus{}1$
So $ 4P(x)\minus{}P(x\plus{}1)\equal{}3ax^2\plus{}(3b\minus{}2a)x\minus{}a\minus{}b\minus{}3$

We can then take $ a\equal{}9u$, then $ b$ such that $ 3b\minus{}18u\equal{}27v$, so $ b\equal{}6u\plus{}9v$ and $ a\plus{}b\equal{}\minus{}3\pmod {27}$ so $ 15u\plus{}9v\equal{}\minus{}3\pmod {27}$ or again $ 5u\plus{}3v\equal{}\minus{}1\pmod 9$

So, we can take for example $ u\equal{}v\equal{}1$ which gives $ P(x)\equal{}9x^2\plus{}15x\minus{}1$

We have $ P(0)\equal{}\minus{}1$ and $ 4P(x)\minus{}P(x\plus{}1)\equal{}27x^2\plus{}27x\minus{}27$

And so $ 9n^2\plus{}15n\minus{}1\plus{}4^n\equal{}0\pmod {27}$ $ \forall n\geq 0$
[/hide]

Indeed, nice problem.

My solution:
[hide]
By Newton's Binomial formula, $4^n=(3+1)^n=\sum_{k=0}^n{\binom{n}{k}3^k}\equiv 1+\binom{n}{1}3+\binom{n}{2}3^2 \pmod{27}=1+3n+\frac{n(n-1)}{2}3^2$.

Notice how the division by $2$ is the only problem. We multiply by $28$, because it solves this problem and preserves the equality mod $27$.

Therefore, $p(x)=-28(1+3n+\frac{n(n-1)}{2}3^2)$ is a polinomial which works.

[/hide]


grr can u simplify $5\sqrt{-1}$ any further?

if ur wondering why it is a negative 1 it's because there was a question involving imagenary numbers

 :wallbash:

I think only of writeing i instead of square root of -1.

[quote="b-flat"][quote="7h3.D3m0n.117"]ok nvm how bout instead of $i$ we use $5\sqrt{2}$?[/quote]

What do you mean?  $\imath\neq5\sqrt2$...

Also, please try not to use chatspeak in your posts.

Thanks.[/quote]
I think they mean that they changed the problem.

To answer your question, there is no way to simplify the expression further without losing accuracy. If you had something like 5 root 8, you could (since 8 is divisible by 4, a perfect square)

Yeah that's the simplest form you can get.

For a positive integer $p$, you can simplify $\sqrt{p}$ only if $p$ is a multiple of a perfect square.

Ok sorry for using "chatspeak" as you call it. I was on a chatroom earlier and that kinda threw me off track.

So basically

$5\sqrt{2}$ is the simplest form.


modular arithmetic
function
floor function
number theory
binomial coefficients

oh... Wait, I messed up, your right bengal.

Oh yeah, that's true, it should be $ \frac{2007!}{1916! 91!*91}$

Shouldn't it be 0 because at some point you are going to hit a multiple of 91?

expression is $ \frac{2007 \ \times \ 2006 \ \times \ ... \ 1917}{91 !}$

and the ONLY multiple of 91 in between 2007 and 1917 is 2002

But how does that help in finding remainder ???  :(

If you just want the answer then:

[hide="Answer"]
$ \binom{2007}{91}\equiv \boxed{5} \pmod{91}$

I got it from [url=http://www60.wolframalpha.com/input/?i=%282007+choose+91%29+mod+91]here[/url]
[/hide]

I actually doubt it helps. If you use the Legendre function, you'll see that all powers of 91 cancel out.

[quote="FantasyLover"]Find the remainder when $ \binom{2007}{91}$ was divided by 91.[/quote]

fanfan you can do better than this:

here is another way, using mods, to prove it is not 0:

in 2007 nCr 91 we just have to find the multiples of 7 and 13 in 2007!, 1961!, and 91!. if the number of 7s and 13s in 2007! is greater than the number of 7s and 13s in 1961! and 91! combined the remainder is 0

let us test this:

in 2007 there are

$ \lfloor\frac {2007}{7}\rfloor \plus{} \lfloor\frac {2007}{49}\rfloor \plus{} .... \equal{} 331$
$ \lfloor\frac {1961}{7}\rfloor \plus{} \lfloor\frac {1961}{49}\rfloor \plus{} ... \plus{} \lfloor\frac {91}{7}\rfloor \plus{} \lfloor\frac {91}{49}\rfloor \plus{} ... \equal{} 339$

so the answer is not 0

This problem seems to be causing a fair amount of havoc  :D Can someone come up with a method other than brute force and explain to us how  :( ?

How about [url=http://en.wikipedia.org/wiki/Wilson%27s_Theorem]Wilson's theorem[/url] over and over again (since you know it's nonzero)?

The most straightforward solution that I see is to use Pascal's Identity to show that:

 $ \binom{2007}{91} \equiv \binom{2002}{91} \pmod{91}$

You can then widdle this down to some modular congruences with workable numbers by using the fact:

 $ \binom{pa}{pb} \equiv \binom{a}{b} \pmod{p}$

I'll post a more complete solution later.

[hide]I believe this problem can be solved with a combination of [url=http://www.math.hmc.edu/funfacts/ffiles/30002.4-5.shtml]Lucas' Theorem[/url] and the Chinese Remainder Theorem. Find $ \binom{2007}{91} \bmod {7}$ and $ \binom{2007}{91} \bmod {13}$ with Lucas' Theorem, then use the C.R.T. to find the value $ \mod{91}$.

Lucas' Theorem is one of those very cool, and sometimes handy, theorems that is not widely known. I found it when trying to solve problem number 20 from the [url=http://bamm.usfca.edu/toi2001.pdf]2001 BAMM Competition[/url].[/hide]

Thanks to gauss202, we have:

[hide="Full solution"]$ 2007\equal{}5\cdot 7^3\plus{}5\cdot 7^2\plus{}6\cdot 7\plus{}5$

$ 91\equal{}1\cdot 7^2\plus{}6\cdot 7\plus{}0$

By Lucas' Theorem, $ \binom{2007}{91}\equiv \binom{5}{0}\binom{5}{1}\binom{6}{6}\binom{5}{0}\equiv 5\bmod 7$.

By the same method, we find $ \binom{2007}{91}\equiv 5\bmod 13$.

From the first, $ \binom{2007}{91}\equal{}7k\plus{}5\equiv 5\bmod 13\implies 7k\equiv 0\bmod 13\implies k\equal{}13j$.

Therefore, $ \binom{2007}{91}\equal{}91j\plus{}\boxed{5}$.[/hide]

Lucas' theorem is pretty great :D. I'll be sure to keep it in the back of my mind for whenever I'm dealing with binomial coefficients in modular arithmetic.


:( Using the numbers 10, 1, 5, 4, 25 and 2, I need to make 983. You can use the numbers only once each but you can use these: add, subtract, divide and multiply as many times as you like.

Is this allowed?

$ 4(25*10\minus{}5)\plus{}2\plus{}1 \equal{} 983$


Does anyone know anything about how the spelling check in google or Microsoft softwares is implemented?
Spelling check means the correction of the wrong user input.

Can anyone give me some related materials or websites?

Thank you very much.

I think it works by checking a word with a dictionary.  If the word isn't in the dictionary, it looks for the word with the closest spelling, that is, the fewest differences, possibly in pronunciation.  On google, I think it just looks for rearrangements of the letters that are commonly searched for.


limit
logarithms
real analysis
real analysis unsolved

[b]Let the reccurent sequence [/b]$a_1=a>0,\ a_{n+1}=\frac {na_n}{n+a_n},\ n\in N$. [b]Prove that :[/b]

$1.\ \lim_{n\to\infty} a_n=0\ ;\ \ \ 2.\ \lim_{n\to\infty} \sum\limits_{k=1}^n a^2 _k=\infty\ ;\ \ \ 3.\ \lim_{n\to\infty} a^2_n\cdot \ln n=\frac 12 .$


solve for x:
$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$

[hide]

x= :sqrt: (4+y)

y= :sqrt: (4-x)

x^4-8x^2+x+12=0

y^4-8y^2-y+12=0

so if a is a solution to x^4-8x^2+x+12=0, then -a is a solution to y^4-8y^2-y+12=0.

setting y to -x...

x= :sqrt: (4-x)

x^2+x-4=0 are 2 solutions to x^4-8x^2+x+12=0

we can then divide/factor this into

(x^2+x-4)(x^2-x-3)=0

and the 2 positive solutions are  (:sqrt: 17-1)/2 and  (:sqrt: 13+1)/2, however only one of these is larger than 2.

Note: u can just start off with x= :sqrt: (4+ :sqrt: (4-x)) and get x^4-8x^2+x+12=0, but i couldn't factor that by myself heh

[/hide]

[quote="GeoMar"]x= :sqrt: (4+y)
y= :sqrt: (4-x)
[/quote]

who says you can do this substitution?

Working out you get taht $x-4=\sqrt{4-x}$ so are you just "assuming" that relation is correct? Moreover, you continue your build on it. It may be correct but please explain how you did this :)

Hi,

I guess this is similar to GeoMar's method:

[hide]In place of x in the RHS, we can substitute for the expression for x again and so on.

So, what we get is x =  :sqrt: 4 +  :sqrt: 4 -  :sqrt: 4 +  :sqrt: 4 - .....

implies x =  :sqrt: 4 +  :sqrt: 4 - x

which implies that x^4 - 8x^2 + x +12 = 0. I couldn't find a root for this.

[/hide]

But, I think there should be a simpler method.

*********************************************************************

Sambit

No, also wrong... basicly what you're say is that when f(f(x))=x then f(f(f(...f(x)...)))=x

A simple counterexample is f(x)=-x where f(f(x))=x but f(x)=-x again

Ok, that doesn't mean what you said is necessarily wrong, it simply means you'll gonna have to find a better argument... or show that there is a reason why it is valid here ;)

Hi Peter,

I don't get it why you think the method is flawed or at least lacks explanation.
Both GeoMar and my attempts allude to the fact that
if f(f(x)) = x, then f(f(f(f(f(f(....)))))) = x, where f(f(...)) is iterated, and hence we can iterate it infinite number of times. It just follows by substituting the definition of x=f(f(x)) in place of the argument x in the equation, and continuing infinitely.

Fo your example, let f(x)= -x; so f(f(x)) = x; and hence you can see that f(f(..)) iterated even number of times will still give x.

Kindly comment on this.

Thanks,
Sambit

you just said the magic word there : "[b]even[/b] number of times" ;)

[i]Edited from a Sony Ericsson K700i [/i]

I'll give a guess, although I dont ahve much of an idea
I solved for
$x=\sqrt{4+\sqrt{4-x}}$
so substituting x back into itself we get the original problem:
$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$
i guess we are saying that if x=f(x) then x=f(f(x)) and x = f(f(f..(x)))

yeah... but: how do you [b]prove[/b] that the original has no solutions besides that one?

I mean, if I'm getting back to my example: it's not because f(f(x))=x that necessarily f(x)=x, there's also another solution f(x)=-x

Can you give me some hint how you know that the solution you found (and it's correct) is the only one? :)


algebra
function
domain
inequalities unsolved
inequalities

Let $u = 3{x^{2}}+2{y^{2}}$ such that $x+y = 2$ where $x$ and $y$ are positive real numbers
.Determine the maximum value of $u$.

Hope it's good;)

As we have $x+y=2$, we get $x^{2}+y^{2}+2xy=4$, $3x^{2}+2y^{2}=12-y^{2}-6xy$, so we want to find a minimum of $y^{2}+6xy$ and it's assumed for $y\to 0$ (y^2 is geting smaller, and for constant sum $x+y=2$ the minimum is when $y\to 0$)

Substituting  $y = x-2$ iin the given expression we get-
           $u = 5{x^{2}}-8x+8$
Clearly $u$ is increasing throughout its domain and since $x >0 x+y = 2$ we come to the conclusion that $u$ has its domain as $[0,2]$ .Hence u attains its maximum value at $x = 2$ and the maximum value is equal to $12$

Your answer is incorrect.What do you mean by maximum value? :maybe:

I think he is correct .The function $f(x)=5x^{2}-8x+8$ is monotonically increasing in $\left[ 0,\frac{4}{5}\right]$ so
$f(0)\geq f(x)$ so every $x\in\left[ 0,\frac{4}{5}\right]$
and it is monotonically increasing at $\left[ \frac{4}{5},\ 2\right]$ so $f(x)\leq f(2)$ for each $x\in \left[ \frac{4}{5},\ 2\right]$

So its local maximum attaint at 0 and 2 and f(2)>f(0) so f(2) is the maximum . So $f(x)\leq f(2)=12$  fo every $x\in\left[ 0,\ 2\right]$

Aren't $x,\ y$ positive numbers?
$x$ should have a range of $0<x<2?$
I can't understand what you said.

As I understood from post of ashwinkrishnan we meant non negative numbers .


geometry
rectangle
search
AMC 10
angle bisector

I don't understand how to do these following AMC 10 questions from last year's test(2009 B)

1. Rectangle ABCD has AB=8 and BC=6. Point M is the midpoint of diagonal AC and E is on AB with ME perpendicular to AC. What is the area of triangle AME?

2.Triangle ABC has right angle at B, AB=1 and BC=2. The bisector of <BAC meets BC at D. What is BD?

For #2, you need to know the Angle Bisector Theorem.

Basically first find the hypotenuse of ABC, which is $ \sqrt{1^2\plus{}2^2}\equal{}\sqrt{5}$. You need to find $ BD$, and since $ BC\equal{}2$, $ DC\equal{}2\minus{}BC$. Apply the angle bisector theorem and you get:

$ \frac{1}{BD}\equal{}\frac{\sqrt{5}}{2\minus{}BC}$
$ 2\minus{}BD\equal{}\sqrt{5}BD$
$ (1\plus{}\sqrt{5})BD\equal{}2$
$ BD\equal{}\frac{2}{1\plus{}\sqrt{5}}\equal{}\frac{2\sqrt{5}\minus{}2}{4}\equal{}\boxed{\frac{\sqrt{5}\minus{}1}{2}}$

To learn more about the Angle Bisector Theorem, see [url=http://www.artofproblemsolving.com/Wiki/index.php/Angle_bisector_theorem]here[/url].

[hide="hint #1"]Use similar triangles[/hide]

[hide="hint #1 - 2"] Triangle AME is similar to triangle ABC. [/hide]

There are posts for these problems, so you should search for them. Here are the links:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1420316#1420316]AMC 10 2009B Problem 18[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1420460#1420460]AMC 10 2009B Problem 20[/url]


function trigonometry geometry conics algebra polynomial quadratics

Hi everyone!
I'm a new member in this forum. :)  I am not good at math like all of u here. That's why I wanna improve my Math skill by asking u guys for some experiences. I'm now learing at highschool and I want to ask you what should I learn first and next......
Plz give me your exp. Thank you a lot!!!
I want to learn from all of you!!! :lol:

Welcome!

What exactly do you want us to give? How to improve?

First off, this depends on your skill level. What level problems can you solve?

I would get a textbook that matches up with your skill level (check the AoPS Bookstore) and study up on the problems.

I hope you like this site!

(this thread should be moved) :)

I'm making a set of notes for high school math. It's basically all the theory you'll encounter in your last years of high school math and it includes plenty of proofs and derivations. If you want, I could send it to you when I'm done or if you're in a rush, I can send you what I've done so far.

Thank you a lot [b]Buzzer11[/b] and [b]Prime factorization[/b] for your help!
I'd like to see your work some days. I have looked for this one for a long time but today is the 1st time I heard about this. Thank you too much [b]Prime factorization[/b]!!!

What I suggest you to do is to first order AoPS books 1 (the Basics) and 2 (and Beyond). Read through all of the chapters in AoPS book 1 very very carefully and absorb in all the information, understand how they derive various formulas, and independently work through every single problem that you can possibly do by yourself. DO NOT SIMPLY MEMORIZE ALL OF THE FORMULAS, CLEVER MANIPULATIONS, ETC. If you spend 2+ hours (or just keep hacking at a problem until you've run out of ideas on how to solve the problem), then get help. If you feel as if you easily grasped all of the concepts and easily solved almost all of the problems by yourself in AoPS book 1, then move on to AoPS book 2. However, if you constantly struggled with a certain area/areas of math while completing AoPS book 1, I would suggest ordering an AoPS Introduction to [insert whatever subject(s) you struggled with] and read the book(s)/work through all the problems in the book(s). Then, go back to AoPS book 1 and re-work all of the problems of the certain subject(s) that you struggled with. After this, move on to AoPS book 2. The AoPS books are great ways to both improve your math skills and get ready for math contests. Hope this helps!

Thank you LOKESH!!!
But I have a big problem when I am following your help. I'm just a student so how can I order this book? I don't have money and I have never ordered online before. Hix hix :(
Poor me!!! :huh:
Anyway thank you so much LOKESH!

Ok. Here's the attachments of my work in pdf format. The topics I've included so far are the study of functions, trigonometry, geometry and the conic section.

I have to warn you that it is about 200 pages long. Most notably, the "functions" document is 63 pages long and you might want to either skim through it or skip it entirely. Also remember that I haven't completed the notes so there might be a few mistakes in the derivations here and there. Finally, this is mostly theory so no practice questions are included.

Now, the upside :D I do cover some topics in more detail than what you may learn in school, for example, trig inverse formulas, the appolonius circle constant,  the broken chord theorem and some study on the fractional and integer part functions. Also, I try to assume as few things as possible as "by definition", so you will see proofs of things as simple as the ASA, SSS, SAS postulates.

If the attachments look like word documents, right click "open with" and choose adobe pdf (whatever version number you have).

Any critique is welcome, just keep in mind that I'm not trying to get this stuff published, it's more of an appendix of formulas with their derivations/proofs included.

very nice prime factorisation. Some of the proofs I have not seen so i learnt something with your documents. I like the geometry ones. Maybe after you finish high school or not busy you could make one (hard geometry problems and how to solve them)

good work once again!!!!!

Thank you bos1234. Any other critiques? I noticed that the geometry pakcage seems to be by far the most popular one. Is there a reason for that?

Also, does anyone have requests for a certain high school topic? I'm trying to figure out what to include next, though I'm pretty intruiged to do something related to polynomial theorems.

My beta version of my math program (ti-83, 84 basic) is almost done.
It's original focus was attempting to mimic those great quadratic programs... but then I realized: [i]why stop there?[/i]

Not only does it do the quad stuff, it also...;

can take a difference or sum in squares then simplifies it under a radical.
this is basically an uber sweet pythagorean theorem without a menu, it does all the square factoring for you, which is nice.

speaking of radicals n stuff, it also does....

heron's formula (find area or a side)
law of sines     (find a side or angle)
law of cosines  (find a side or angle)

it can solve trig function, but it takes a minute.
(it's real fun to proof trig ID's with this lol)

factors quadratics, foils quadratics,
ANY POSSIBLE ROOT is disp'd perfectly.

shows ALL TRIG IDS (readable, in the standarm forms.)...
[hide="THERE IS ONLY ONE PROBLEM"]
I can't spot radical fractions. ex: $ \frac{\sqrt{3}}{2}$[/hide]
Anyways, this program is great for troubleshooting any type of proof really, pythagorean theorem, quadratics, and half of trig is just bursting at your fingertips....

Takers?

you might get some help lin the computer science and informatics forum.


Here we can discuss books(please not LoTR). I would be discussing old ones.

Has anyone read Tristram Shandy? I have volume II, and I am thinking of reading it!

nope haven't read that series/book.

good books/series
~maximum ride
~twilight
~harry potter
~uglies,pretties,specials,extras

plus lots more that i can't think of right now

I recommend Flatland and any mythology. Although Flatland was a bit dense and Victorian era, I felt having annotations really helped.

Didn't really like Extras that much. Sorta ruined the beauty of the Tally Youngblood trilogy....


okay, I was told from someone it is possible to make 1=2.

Here is what i did.

[code]
(.5-1)=(-.5)
(.5-1)squared=(-.5)squared
(.5-1)squared=.25
square root(.5-1)squared=square root(.25)
(.5-1)=.5
    +1   +1
.5=1.5
+.5  +.5
1=2
[/code]

SUPPOSEDLY I did something wrong. If so, what did I do and how do i fix this?

there are [i]two[/i] square roots for every positive real number.
example,

$\sqrt{9}=3,-3$

oh yeahhhhhhhhhhh

Thank you, so it's all just an illusion that makes it LOOK like as if it's true?

I could trick a lot of people with that.

I've seen this so many times, but once in a while, its good to see it again. :)

Isn't the problem line 5, on the left side?
.5-1=-.5, not .5
I don't get how to make 1=2, can someone explain to me how it works or doesn't. I assume that it doesn't work.  :?


function
geometry
geometric transformation
reflection
calculus
calculus computations

Hi I was wondering how to go about finding the equation of a graph that results from the reflection of the function y = e^x about the line y = 4. I wasn't sure where to put the 4 - either in the exponent area or not.. thanks so much!!

Draw the picture. What happens to a point $ (a,b)$ under this reflection?


Pat and Chris ran a race where the finish line is 100 meters away (from the start).  Pat made it to the 80 meter mark at the same time that Chris crossed the finish line.

They will run a race again, but this time Chris will be backed up
25 meters behind the starting line, while Pat begins at the starting line.  Each person will still have the same goal of the 100 meter finish line.

If each person keeps the same respective speed each had in the prior race, what will be the outcome of this race?

[hide]If Chris’s speed is $ v$, Pat’s speed is $ 0.8v$. So Pat can run 100 metres in a time of $ \frac{100}{0.8v}$ while Chris can run 125 metres in a time of $ \frac{125}{v}$. Since $ \frac{100}{0.8v}\equal{}\frac{125}{v}$, the result of the second race will be a dead heat. :)[/hide]

Chris will win because although they cross the line at just about the same time, Chris is the faster runner and will win... but by about a millimeter!

Thanks,

but I'll go with Jane_Bennet's answer and support work instead.


abstract algebra
algebra
function
domain
irrational number
superior algebra
superior algebra theorems

Someone told me that, if $R$ is a domain with $K$ its quotient field, then $R'$ a ring such that $R \subseteq R' \subseteq K$ is always noetherian. Is this true? It does not sound true to me!

No, it is not true (even if you require $R$ to be Noetherian ;-). A simple example is as follows: take a field $k$ and some irrational number $\alpha>0$. Then the valuation ring in $k(X,Y)$ for the valuation satisfying $v(X)=1$, $v(Y)=\alpha$, and $v|_{k^\times}=0$ is not discrete, hence not Noetherian, but it contains $k[X,Y]$. Cf. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=151759]this topic[/url]. owk

Thanks for the example. The statement in the other post also seems very strong. How would one prove that?