This is an all-out competition I'm asking for, here. We will have multiple coaches, multiple contestents, and a whole bunch of mathcounts problems.How about it?

P.S. This is not the competition topic, I may not make that topic.

Here's howwe are gonna sign up:

contestents
regular moderators(coaches)
head coach

The contestents solve the problems, the regular moderators score them, the head coach gets all of the scores and places nice and easy, and then we continue until a winner is determined.

ok?

contest[b]a[/b]nt
I'll coach. Maybe make some problems.

How about sign-up for AoPS Kings in the SC forum?  It's already extremely big.  And the problems are like amazing.  Also, this wouldn't be a good idea right now until a few more weeks since so many competitions are going on right now.

[quote="daermon"]if you do make a big competition, consider planning it now but not starting it until next summer.  I think I speak for a lot of us when I say that I am extremely busy will school  :wink:[/quote]
I agree with daermon. I think a LOT of people will sign up before next summer.

I'll be a contestant...  All the 8th graders won't get any use out of this if you wait until the summer...

how about you start this later, like after aops kings but before mathcounts starts

I'll be a contestant...i hope..well i'll participate if it is at a convenient time. :lol:

What do coaches do? Perhaps we should have people in 9th grade coach, so they can feel...included. :( :D

more rounds=more opportunities for people to forget+less time to do each problem

I'll be a contestant or whatever you're calling it.

nutz_for2.718281828, could you post the competiton topic :?:

[quote="Klebian"]I believe nutz has left aops[/quote]
What do you mean?

yeah. what do you mean??? I'm on almost daily!

Now, here are the conditions of the topic:

There will be three rounds, 15-20 problems in each round, and about 5 to 7 days in each round? If at least three people agree with that, thenI will post the competition and advertise the competition on my signature.

Unfortunately, I may not be able to take part of this contest, so I will not be able to supervise it unless there aren't a few head coaches.

[quote="SplashD"][quote="bpms"]Perhaps more time... Like 2 weeks...[/quote]

then people will procrstinate, and forget. i like the 5-7 days[/quote]
But some people will be busy...
Are we supposed to time ourselves (honor system) or can we take as long as we want?

ok. I'll pm Treething, and ask him if he really wants to be a coach, and I'll make the competition after he responds positively and if.

eh i'll only be coach if there's gonna be more than one co-head coach or something


Euler
geometry
trapezoid
circumcircle
cyclic quadrilateral
geometry solved

Let $E$ be the intersection of the diagonals in cyclic quadrilateral $ABCD$, and let $F$ and $G$ be the midpoints of sides $AB$ and $CD$, respectively. Prove that the three lines through $G,F,H$ perpendicular to $AC,BD,AD$, respectively, intersect at one point.

[quote="warut_suk"]Let $ E$ be the intersection of the diagonals in cyclic quadrilateral $ ABCD$, and let $ F$ and $ G$ be the midpoints of sides $ AB$ and $ CD$, respectively. Prove that the three lines through $ G,F,H$ perpendicular to $ AC,BD,AD$, respectively, intersect at one point.[/quote]

What is H?

Anyway, if the problem is to prove that the perpendiculars from the midpoints of the segments AB, BC, CD, DA, AC, BD to the lines CD, DA, AB, BC, BD, AC concur at one point, then see Theorem 6 in the note "The Euler point of a cyclic quadrilateral" [url=http://www.cip.ifi.lmu.de/~grinberg/]on my website[/url].

[b]EDIT:[/b] Ah, I see you mean E instead of H. So the question is:

[color=blue][b]Problem.[/b] Let E be the point of intersection of the diagonals AC and BD of a cyclic quadrilateral ABCD, and let F and G be the midpoints of the sides AB and CD, respectively. Prove that the perpendiculars to the lines AC, BD, AD through the points G, F, E concur at one point.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°.

Let R be the orthocenter of triangle EAD. Then, $ AR\perp DE$, what rewrites as $ AR\perp BD$, and $ DR\perp AE$, what rewrites as $ DR\perp AC$, and finally $ ER\perp AD$, what shows that the point R lies on the perpendicular to the line AD through the point E.

Now, let the perpendiculars to the lines AC and BD through the points C and B intersect at a point Q. Since < QBE = 90° and < QCE = 90°, the points B and C lie on the circle with diameter QE. Thus, < QEC = < QBC, so that < QEC = < QBE - < CBE = 90° - < CBE = 90° - < CBD. Since the quadrilateral ABCD is cyclic, < CBD = < CAD, so this becomes < QEC = 90° - < CAD. In other words, < (QE; AC) = 90° - < (AC; AD), and thus < (QE; AC) + < (AC; AD) = 90°, so that < (QE; AD) = 90°, and thus $ QE\perp AD$. Hence, the point Q lies on the perpendicular to the line AD through the point E.

Now, let P be the midpoint of the segment RQ. Since the points R and Q lie on the perpendicular to the line AD through the point E, so does the midpoint P of the segment RQ. Now, the quadrilateral ARQB is a trapezoid, with its bases AR and BQ both being perpendicular to the line BD; hence, the line joining the midpoints F and P of its legs AB and RQ must also be perpendicular to the line BD. In other words, the point P lies on the perpendicular to the line BD through the point F. Similarly, the same point P must lie on the perpendicular to the line AC through the point G.

Thus, the perpendiculars to the lines AC, BD, AD through the points G, F, E concur at one point, namely at the point P. And the problem is solved.

[b]EDIT EDIT:[/b] The problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=83358 .

[b]PS.[/b] Please specify the source of your problems when posting them.

  Darij

$\boxed {A\ happy\ new\ year\ to\ the\ all\ mathlinks-users\ !}$
(now, here are five hours until the beginning of the new year).

A nice problem and a nice Darij's solution !

[hide="Here is a solution using the Carnot-lemma !."]
I note the circumcircle $C(O,R)$ of the cyclical convex quadrilateral $ABCD$,
the intersection $E\in AC\cap BD$ ($p\equiv EA\cdot EC=EB\cdot ED$),
the middlepoints $F$, $G$ of the sides $[AB]$, $[CD]$ (respectively) and the projections
$X$, $Y$, $Z$ of the points $E$, $F$, $G$ on the lines $AD$, $DE$, $EA$ (respectively).

I shall apply the [u]Carnot's lemma[/u] in the triangle $ADE$ for the points
$X\in AD$, $Y\in DE$, $Z\in EA$: " The perpendicular-lines from the points
$X\in AD$, $Y\in DE$, $Z\in EA$ on the lines $AD$, $DE$, $EA$ (respectively)
are concurrently iff $\Omega\equiv (XA^2-XD^2)+(YD^2-YE^2)+(ZE^2-ZA^2)=0\ \".$
Indeed, $EX\perp AD$, $FY\perp DE$, $GZ\perp EA$ $\Longrightarrow$
$\Omega =(EA^2-ED^2)+(FD^2-FE^2)+(GE^2-GA^2)\Longleftrightarrow$
$4\Omega =4EA^2-4ED^2+2(BD^2+AD^2)-AB^2-2(EA^2+EB^2)+$
$+AB^2+2(EC^2+ED^2)-CD^2-2(AC^2+AD^2)+CD^2\Longleftrightarrow$
$2\Omega=EA^2-ED^2+BD^2-EB^2+EC^2-AC^2\Longleftrightarrow$
$2\Omega=(EA+EC)^2-2EA\cdot EC-(EB+ED)^2+2EB\cdot ED+BD^2-AC^2\Longleftrightarrow$
$2\Omega=AC^2-2p-BD^2+2p+BD^2-AC^2\Longleftrightarrow \Omega=0.$
[/hide]
[hide="Here is a little generalization of the proposed problem."]
[b]The first (little) generalization.[/b] $\"$ Let $C(O,R)$ be the circumcircle of the convex quadrilateral $ABCD$, the intersection $E\in AC\cap BD$ and two points $F\in (AB)$, $G\in (CD)$. Then the perpendiculars to the lines $AD$, $BD$, $AC$ through the points $E$, $F$, $G$ (respectively) are concurrently $\Longleftrightarrow\frac{FA}{FB}=\frac{GD}{GC}\ \vee \ AD=2R.\ \"$
[u]A sophistical enunciation.[/u] If exists two points $F\in [AB]$, $G\in [CD]$ so that $\frac{FA}{FB}\ne \frac{GD}{GC}$ and the perpendiculars to the lines  $AD$, $BD$, $AC$ through the points $E$, $F$, $G$ (respectively) are concurrently, then $AD=2R.$[/hide]
[hide="Here is a great generalization of the proposed problem."]
[b]The second (great) generalization.[/b] $\"$ Let $ABCD$ be a convex quadrilateral, $E\in AC\cap BD$ and two points: $F\in AB$, $\overline {AF}=x\cdot \overline {AB}$; $G\in DC$, $\overline {DG}=y\cdot \overline {DC}$. Prove that the perpendiculars to the lines $AD$, $BD$, $AC$ through the points $E$, $F$, $G$ (respectively) are concurrently if and only if there is the following relation: $(x-y)\cdot \left(EA^2+ED^2-AD^2\right)=2(y\cdot EA\cdot EC-x\cdot EB\cdot ED).\ (1)\"$

[u]Remark.[/u] If the quadrilateral $ABCD$ is cyclically, then $EA\cdot EC=EB\cdot ED$.
I note the middlepoint $M$ of the side $[AD]$. Then the relation $(1)$ becomes
$(m-n)(EA^2+ED^2-AD^2+2p)=0\Longleftrightarrow$ $m=n\ \vee\ ME^2+MO^2=EO^2\Longleftrightarrow$
$m=n\ \vee\ M\equiv O$, i.e. the first generalization.[/hide]
[u]Remark.[/u] The two generalizations are proved with the same [u]Carnot's lemma[/u] and the [u]Stewart's theorem.[/u]


calculus
integration
geometry
calculus computations

Find the minimum area bounded by the graphs of $ y\equal{}x^2$ and $ y\equal{}kx(x^2\minus{}k)\ (k>0)$.

hello kunny, solving the equation $ x^2\equal{}k\cdot x(x^2\minus{}k)$ with respect to $ x$ for $ k>0$ we get
 $ x_1\equal{}\minus{}\frac{1}{2}\frac{\minus{}1\plus{}\sqrt{1\plus{}4k^3}}{k}$, $ x_2\equal{}0$, $ x_3\equal{}\frac{1}{2}\frac{1\plus{}\sqrt{1\plus{}4k^3}}{k}$ so we get the searched area as
 $ \displaystyle{A\equal{}\int_{x_1}^{0}kx(x^2\minus{}k)\minus{}x^2\,dx\plus{}\int_{0}^{x_2}x^2\minus{}kx(x^2\minus{}k)\,dx\equal{}\frac{1}{12}\frac{1\plus{}6k^3\plus{}6k^6}{k^3}\geq\frac{1}{12}(2\plus{}\sqrt{6})\sqrt{6}}$.
The equal sign holds if $ k\equal{}\frac{1}{6}6^{\frac{5}{6}}$.
Sonnhard.


geometry rectangle calculus conics parabola algebra polynomial

A rectangle is bounded by the x and y axis and the line $y=\frac{6-x}{2}$. Find the dimensions that maximize the area of the rectangle.

(NOTE: I don't think you can assume the sides are parallel to the x and y axis; if you can make that assumption, please explain? Also, the solution may involve calculus, I don't know. Help please.)

From your description, one of the points of the rectangle is clearly the origin, and the opposite vertex is on the line $y = \frac{6-x}{2}$

The area of the rectangle is $X \cdot y$, or $X \cdot \frac{6-x}{2}$
Which works out to be $A = \frac{-x^{2}+6x}{2}$
We want to find the values for which this parabola is maximised.
From algebra, we know the maximum valu, when a parabola is in standard form, is $\frac{-b}{2a}$. We put out area formula in standard form:

EDIT: More to be posted later. Ran out of time during lunch.

You don't need calculus for this question, if you can assume that the rectangle is parallel to the x and y axis.

[hide]If you let b=x then h=(6-x)/2. Thus the area will be A=x(6-x)/2=6x-x^2/2. Complete the square and you get A=1\2(x-3)^2+9/2 so the maximum possible area is 9/2 (when x=3) [/hide]

As for if the rectangle isn't parallel, that gets complicated really fast, and I don't think it will increase the area  because there are just too many areas left out.

neither of you are necessarily correct; the description does not necessarily imply that a vertex is on the origin, and "just too many areas left out" does not cut it. Consider the below case (this is a possible case, by no means the one that maximizes area).

You are right, though, the problem becomes quite simple if you assume that the sides are parallel to the x and y axis, but you cannot simply make that assumption. If indeed that assumption leads to the correct dimensions, then prove why.

Thinking rationally, we can see that it makes the most sense that the area would be maximised when the sides are parallel to the axes. Try and construct a rectangle with more area, under the given restraint. You'll find it's impossible.

However, for a challenge, consider the case where the sides *are not* allowed to be paralell to the axes. Obviously, one of the sides has to be collinear with the line $y=\frac{6-x}{2}$, and the other two points are on the axes.
Another good challenge.

You can do this with a differential equation, but you can also do it with simple algebra, so you really don't need calc. (2nd degree polynomials are quite nice.)

The area of the triangle it has to fit in is 9, and so far the greatest area we've come up with is 9/2, exactly 1/2 of that area. That is assuming the sides are parallel to the x and y axis. Can we show that the rectangle can never take up more than half, somehow?

I don't really know how but I just had an idea.

[quote="bthings_2000"][quote]Thinking rationally, we can see that it makes the most sense that the area would be maximised when the sides are parallel to the axes. Try and construct a rectangle with more area, under the given restraint. You'll find it's impossible. [/quote]

Unfortunately, asserting that something is true does not constitute a proof, or any kind of adequate justification. Nor does "find a counterexample, I dare you." You must give some kind of mathematical reason. Yes, once you make the assumption you're making, you can use second-degree polynomials easily. But we cannot so easily make an unsupported assumption.

I'm open to ideas.[/quote]

hehe. I agree completely. I don't have a proof, and I'm not trying to provoke anything. I'm making an assumption, and this may or may not be correct. The problem isn't entirely clear, and I'll try to find a proof. I tend to lack rigor when I post, by I have things laid out nicely in my mind. Now if only I could get the two together...  :lol:
Not trying to provoke any flames. You are 100% absolutely correct.

At what level this problem should be solved? If we're dealing with a rectangle in a general position, we'll unavoidably face a two-variable constrained maximization problem, which is, even in the simplest of its forms, probably well above Intermediate.


This is a very nice inequality. Here's a very short solution, though not easy and not completely mine:
[hide="Solution"]
We replace $ l_a,l_b,l_c$ by any three cevians. Without Loss of Generality, let $ a\le b\le c$. It is well-known that each median is smaller than half of the sum of the adjacent sides, hence each median is smaller than the largest side - $ c$. Then $ \dfrac{l_a}{m_a}\plus{}\dfrac{l_b}{m_b}\plus{}\dfrac{l_c}{m_c}\ge\dfrac{l_a\plus{}l_b\plus{}l_c}c>1$, because $ l_a\plus{}l_b>c$
As far as I remember, the book said that the proposer (and the authors of the book) didn't know a complete solution using the usual formulas for the length of the bisector / median. The computations are indeed very heavy.
[/hide]


algebra
system of equations
algebra unsolved

Find all $ (y,z) \in \mathbb{R}$ such that, they satisfies the system of equations:
\[ z\plus{}y^3\equal{}2
\\y\plus{}y^2z\equal{}2
\]

[quote="Moonmathpi496"]Find all $ (y,z) \in \mathbb{R}$ such that, they satisfies the system of equations:
\[ z \plus{} y^3 \equal{} 2 \\
y \plus{} y^2z \equal{} 2
\]
[/quote]
\[ z \plus{} y^3 \equal{} y \plus{} y^2z
\]

\[ \rightarrow (y^2 \minus{} 1)(z \minus{} y) \equal{} 0
\]


inequalities
combinatorics proposed
combinatorics

Let $ M(x_1,\dots,x_n) \equal{} \prod_{i\equal{}1}^{n}x_i ^{e_i}$ be a monomial. Let $ f(x) \in \mathbb{F} [x_1,\dots,x_n]$ be a polynomial. We say that $ M$ is not dominated in $ f$ if $ M$ is one of the monomials in $ f$ and in every other monomial $ M' \equal{} \prod_{i\equal{}1}^{n}x_i ^{d_i}$ appearing in $ f$ we must have $ d_i < e_i$ for some $ i$.

Let $ S_1, \dots, S_n \subseteq \mathbb{F}$ be sets of size $ |S_i| > e_i$. Prove that if the monomial $ M(x_1,\dots,x_n) \equal{} \prod_{i\equal{}1}^{n}x_i ^{e_i}$ is not dominated in $ f$ then there is some point $ \alpha \in S_1 \times S_2 \times \cdots S_n$ such that $ f(\alpha) \ne 0$.

Managed to use a variant of Nullstellenstaz to reduce the theorem to the following claim:

Let $ S_1 , ... , S_n$ be subsets of the field $ \mathbb{F}$, such that $ |S_i| > e_i \ge 0$ for all i. We define $ g_i \equal{} \prod_{\alpha \in S_i} (x_i \minus{} \alpha)$ for all $ i$. Let $ h_1, \dots, h_n \in \mathbb{F} [x_1,\dots,x_n]$ and define $ f \equal{} \sum_{i \equal{} 1}^{n}h_i g_i$. Suppose that $ deg(h_i) \le deg(f) \minus{} deg(g_i)$ for all $ i$. Then either the monomial $ \prod_{i \equal{} 1}^{n}x_i ^{e_i}$ doesn't appear in $ f$ (meaning that its coefficient in $ f$ is zero) or there exists a monomial $ \prod_{i \equal{} 1}^{n}x_i ^{d_i}$ in $ f$ such that $ d_i \ge e_i$ for all $ i$ and there's a strict inequality at least for one $ i$.

Anyone got any idea how to prove this?


I've seen some people use applets for shapes. I wish to do this, too. I don't understand the Java tutorial. Since Asymptote is extremely hard for me, and I hope this is much easier, can someone tell me how to draw shapes (I really need to explain a solution for someone)?

This is very urgent, so please help me!
I can't make any diagrams with Asymptote and thus need this.

And please tell me to how to use this; not tell me Asymptote is easy, etc. other stuff that I don't want to hear.

Thanks a bunch.


geometry
geometric transformation
rotation

How many ways can 8 people sit around a round table?
What if 2 people have to sit adjacent to each other?
What if 2 people hate each other and can't sit next to one another?

#1: 8*7*6*5*4*3*2*1=8!. But we must divide by 8 to eliminate any rotations (assuming that those are not allowed). So 7!=5040.

For the second one, treat those two people as one. Then 7*6*5*4*3*2*1=7!, but we divide by 7, so we get 6!, but we must multiply by 2, because those two people may switch seats and still be next to each other. So 2(6!)=1440.


I assume we can neglect Buoyant forces and viscous forces??

And does the boy just drop the stick just like that???

this is supposed to be an optics problem...neglect every possible force on the rod even gravity
the boy aims at the stone(that he sees)

Oh~ Great, I was solving it as a rotmech problem. :D:D

Back to square one then.

I really don't understand how this could be an optics problem. Could you please re-phrase your question. What do you mean, hit the bottom?

And here I was solving it using rotational mechanics. :rotfl:

why can't this be one..,.
the boy looks at the balll at an angle 45 degrees and the place where he sees the stone(apparent to him) he throws the stick since this isn't the place where the stone lies it hits the bottom some distance away from the stone.u have to find this distance

i think a diagram would make things much clearer.
i am not able to see how ONLY the angle at which he holds the stick can tell us how far the stick will be from the ball. there must be some velocity with which he throws the stick  :maybe:

the stick will land somewhere on the lake bed
assume that it has sufficent speed to move and that it travels straight....in other words assume that to be a light travelling at 45 degrees so no external effetcs whatsoever

is refractive index of air-water given??  :maybe:

aah, then the sum is 'killed' :D
it's simple geometry after that, draw a normal.
angle of incidence $ i$ = 45
$ \equal{}> sinr \equal{} \sqrt 2/3
\equal{}>tanr \equal{} \sqrt {2/7}
\equal{}> x \equal{} 32.tanr \equal{} 32\sqrt {2/7}$

so, distance between the apparent stone and the real stone
= 32 - x

which, unfortunately i am not jobless to simplify at the moment :D
is it right? :)

what?!  :o
if my diagram is right, then i'm quite sure this is right. unless i've made a calculation mistake.
and also, there are no forces acting on the rod like u stated earlier, so the stick HAS to go in a straight line..

EDIT: is my diagram right? :maybe:

ur diagram isn't perfect but then telling u that is like telling u the answer.....i thought u got it...shud i give u a hint?

hey i'm absolutely confident i'm right unless i've understood the sum wroing ar you've understood my soln wrong  :|
anyway, i redrew my diagram... check it out now and tell me..

sorry u r wrong....
the answer is $ 12 cm$

or shud we wait for others with a nice hint
[hide="hint"]
the stone as observed by the person will have apparent position of the ball in [b] both[/b] vertical and horizontal directions[/hide]

oh shoot i didnt think of that ! :blush:  anyway, ill try it tomorrow.. am off for the night  :roll:

Maybe we can assume the rod as a vector for sake of problem solving

$ \frac{L}{\sqrt{2} } \hat{i} \plus{} \frac{L}{\sqrt{2} } \hat{j}$

yes u can but i don't see any gr8 help being there.....

pardesi cant u try that urself? Only ur gf posted that  :D   :P   :rotfl:

can u give the soln pardesi  :maybe:
plz


What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$?

[hide="Solution"]
Take the reciprocal of the whole inequality, remembering to flip the signs. \[ \frac{15}{8} > \frac{n+k}{n} = 1+\frac{k}{n} > \frac{13}{7},  \] or \[ \frac{7}{8} > \frac{k}{n} > \frac{6}{7}.  \] Consider one side at a time. From the left side, we get $7n>8k$, or $49n>56k$. From the right side, we get $7k>6n$, or $56k> 48n$. So we need to find the largest integer $n$ such that there is only $k$ that satisfies \[ 48n < 56k < 49n.  \] In other words, there is only one multiple of 56 in the interval $(48n,49n)$. It is easy to see that for $n=56\cdot 2 = 112$, there will be 3 multiples of 56; however, two of them will be on the endpoints of the interval, so there will only be 1 such $k$. But for $n>112$, there will be at least 2 $k$ inside the interval, so $n_{\max}=\boxed {112}$.
[/hide]

[hide]By cross-multiplying and simplifying, we see that $\frac67n<k<\frac78n$, or $48n<56k<49n.$ We want to find the largest $n$ such that there is only one integer $k$ in the interval. Notice that when $n=56\cdot 2=112$, there are two multiples on the endpoints of the interval ($96\cdot 56$ and $98\cdot 56$), but only one $k$ in the interval. ($k=97$) Therefore, the largest possible value of $n$ is $n=112$. [/hide]

Wait, is the reason you can cross multiply in this situation because you know for a fact that both k and n are positive (or is it some other reason)

[hide="Alternate Solution"]
Rearranging, we get $\frac{6}{7} < \frac{k}{n} < \frac{7}{8}$. Note that if there is a unique value of $k$ such that this inequality is satisfied for every $n$, then we must have the inequality chain

$\frac{k - 1}{n}\le  \frac{48}{56} < \frac{k}{n} < \frac{49}{56} \le \frac{k + 1}{n}$.

Checking each endpoint chain, we have $56(k - 1) \le 48n\implies 56k - 56 \le 48n$. The other endpoint yields $49n \le 56(k + 1)\implies 49n \le 56k + 56$. Combining the inequalities, we have $56k - 56 + n \le 49n \le 56k + 56\implies n \le 2(56) = \boxed{112}$.
[/hide]

[hide=Solution]Flip the fractions and subtract one from all sides to yield $$\frac{7}{8}>\frac{k}{n}>\frac{6}{7}.$$  Multiply both sides by $56n$ to get $$49n>56k>48n.$$ This is equivalent to find the largest value of $n$ such that there is only one multiple of 56 within the open interval between $48n$ and $49n$.  If $n=112,$ then $98>k>96$ and $k=97$ is the unique value. For $n\geq 113,$ there is at least $(49\cdot 113-48\cdot 113)-1=112$ possible numbers for $k$ and there is one $k$ every 56 numbers. Hence, there must be at least two values of $k$ that work.  So, the largest value of $n$ is $\boxed{112}$.[/hide]


search
geometry
perimeter
geometric transformation

On a bright summer day, life is normal in the village. Until around noon, when a rumbling sound is heard, and the ground begins to shake! Cracks are seen on the ground, and chunks of earth begin falling into the ground. Soon, entire buildings are consumed and eventually the entire village falls into the ground. You black out, and by the time you awake it is the middle of the night. The full moon shines brightly directly above the hole the town fell through. You are the only survivors.

Application: pm me if you wish to play, I'll wait until I feel I have enough people then if you want to join I'll let you in by different means than falling with the village.

Current players:

Mathgeek_88: Yuki. A bookbinder's apprentice, knowledgable about the mythology and history of the land.

Hedgehog12: Wolfgang Heart. A hard working Blacksmith. In his free time he sleeps a lot.

Joe10112: Joe. A horseback rider that is specialized with a pole arms, has a pole axe but his horse was killed by the fall.

eggylv999: Bob. A lumberjack. Not very successful and not at all bright.

Aleph0: Aleph. A traveling merchant that had the misfortune to be in town on the day of the earthquake.

1=2: Harka. A peasant with a sword. Harka and his sword are like a donkey and a spinning wheel; nobody knows how he got it and danged if he knows how to use it.

kdange: Celdis Dendramor. A retired knight of 40 years of age. Has a sword and knows how to use it, but is weakened by age and wounds that never fully healed. Lactose intolerant.

I ran out of edits, so the character list continues here. (thanks for the help, could you [me's no being, just an old force remaining in that thread, left behind by ancient mages] tell me who you are so I can request your help in the future?):

And again, a still unknown force comes to help the mighty necromancer whowland. Nobody has a clue what happened, but obviously, whowland could edit his post again. After that, the unknown force vanishes into the mist, only to be seen if needed again.

yeaarrrrrgghhhh take out that lactose intolerant thing, what if i need to drink milk to pass or something? and seriously, wounds that never fully healed, how bout just "not in his prime"

The mysterious Moderators helped the dead Necromancer. The mysterious Moderators are almost all powerful, but the Council of Admins are the most powerful forces in the Universe.

kdange: PM me about this kind of thing, you're cluttering the thread.

Ok, we've got all the descriptions, so lets start.
Yuki: You pull your way out of the pile of books that fell on top of you. The wood building you were in was horribly mangled in the fall, so you manage to get out of it through a shattered wall. You stand amongst the debris and see a relatively intact island of ground and decide to walk over to it. When you arrive at the "island" you realize someone is already there. He is sprawled on the ground and at first you think he's dead, but upon closer inspection he is still breathing. You shake him out of his slumber. You recognize him as the town's blacksmith.

Wolfgang: You awake to find Yuki, the local bookbinder's apprentice, shaking you. She stops as you awake and asks you how you're doing. You find that you have suffered no injuries and feel lucky that you weren't in the forge during the earthquake as the fire and weapons would have surely killed you. You stand up and try to spot any survivors, and see two people you don't recognize walking in your direction, one with a large pack on his back and another with a pole arm. You beckon them over.

Aleph: When you awake, you find yourself lying on your back with your bag underneath you. You are immensly thankful that you were currently selling cloth and not cutlery. You stand up and see that your guard, Joe, is still alive. You pull him out of the saddle of his dead horse and wake him.

Joe: You finally open your eyes to see your employer, a trader named Aleph, shaking you. You thank him, and then you see your beloved horse and for a few moments mourn his passing. Aleph kindly suggests that you should get going and try to find a way out of here. You pick up your pole axe and begin to follow him. After a little stumbling around in the rubble you hear a man beckonng you over and you and Aleph go to join him.

Harka: You pull yourself from the ground with the help of your sword, using it like a crutch. You survey your suroundings and see a man next you. You decide to wake him. He introduces himself as Bob, and asks what happened. You explain as well as you can, but before you can really finish you hear a man beckoning someone. You decide you should go check it out.

Bob: Some guy just woke you up. You ask him what happened, he says something about earthquakes, but you don't undertand a word he says. He seems nice though. You hear some person yelling. The guy decides to go check it out and you follow him.

Celdis: You wake up to the sound of a man yelling, you pull yourself up with a piece of earth that landed dangerously close to your head. You check that your sword is still at your waist and start moving toward the yelling.

All: You have all gathered on the "island". You begin to discuss what to do next.

You can talk as much as you want during a round, but you may only take one action that requires an outcome. PLEASE don't determine outcomes more important than walking by yourself, that's my job. New rounds start whenever I post.

I think we should all check for any injuries first. any minor problems can get easily infected because of the dust and dirt. If everyone's okay, I vote we check out this place. We need food, and shelter. Maybe we can split up into groups...? and try to start building a shelter and gather firewood? We can also forage for food, or hunt, since some have swords.

Celdis says "[color=darkred]Ughhh, i hope Baxter (his dog) is ok... hi i'm Celdis. who are all of you?[/color]"

Hi Celdis, I'm Yuki. I'm an apprentice bookbinder here. Hmm...I don't think I've seen you around...

i guard the city from external dangers, however i didn't see this coming... who built this village anyway? its on top of a fault line!!!

Hmmm...thats a good question....Oh God! I must check to see if my books are alright! My poor books! I've been working on one for a while!

hmmmm i will let us use my horse as food or clothes.

thanks horsey!

Now why did this have to happen when I was here?...Oh well. Any injuries?

I'm feeling fine, a bit bruised on my legs. All we need to worry about is cuts and burns. I think my books are fine, some old ones have collapsed.
Thanks Joe for your horse! Well maybe we should consider skinning it and smoking it? Idk its your horse you can cook it.

Yawn Free weapons for everybody!!!!!!!! Yawn

snore

As you stand around discussing what you're going to do next, beams of light begin to shine through the whole. You realize that day has come, and with this new light you get your first glimpse of your surroundings. Shockingly, you appear to be in what used to be a massive dome. You can see the exquisite masonry that has held up this structure for who knows how long. The wreckage of the entire village fits easily into this huge dome, which is easily a half mile in diameter. On the outside edge of the dome, you see 4 massive wooden doors. Aleph looks at his compass and determines that they are due east, west, and north of you, respectively.

Don't we have a psychic bond??? We had similar dreams...

Good idea... but not an active one. You'll see. Anyways, sorry for the short break there, I had a procedure in the hospital and I couldn't get their wireless to work.

Now, you continue north. Aard'lak is able to walk, albeit at a kind of slow pace, so you just go on normally. to the north of the shrine is another tunnel, and beyond that is a door leading to a large circular room. In this room there is a massive iron gate covered in shallow channels arranged in intricate designs.

Spiderboy, can I take control of Aard'lak for a moment?

omg eggy ur alive too i thought u were dead!!! =)

sure! take control of me!

as long as no exorcist stuff happens. like, if my head starts spinning around, get outta there!

Sorry I haven't posted, but during this time I have thought more about what will happen so the plot will be better and make more sense.

Back in the game, Aard'lak walks up to the gate and begins to examine the patterns. He seems to find what he was looking for and beckons Harka to come over.

Do the rest of us just stay where we are?

Also, I think we should make a map. It could be useful.

You're all just standing around, maybe looking at carvings on the walls.
Aleph, being a seasoned traveler, knows that maps can be crucial to survival. Thus, he whips out parchment, his quill, and some ink and begins making an approximate map of the area, trying to make it as small as possible so he can add more as you explore.

[code]
 /\
 ||
 ||
 N

    C
    |
    |
    B
    |
    |
    A
[/code]

A=where we entered
B=shrine, where we found Aard'lak
C=where we are now (gate room)
|=passage

That's the map so far. 4 months and we've gone about 200 yards (if the second tunnel is 100 yards, like the first one). :(

Is the gate on the north, east, or west side of the room?

(Harka) Aard'lak motions for you to put your hands on some runes inscribed in the left door, and then puts his hands on some runes on the right door. You glance over at him and he seems to be in deep concentration so you close your eyes and try to concentrate. You feel a presence in the back of your mind. It's weak, but it feels warm and bright. You try and concentrate on it and it seems to grow bigger, flowing through your mind and body and drowning your thoughts.

The others look on in amazement as the runes begin to glow. Aard'lak's are glowing a pale yellow and Harka's a bright red. The color begins to flow from the runes and along the lines, filling the room with color. Eventually the lights move toward each other  between the doors and weave a complex pattern. When the lines meet in the center of the glowing symbol, there is a large clanking sound and the doors begin to slowly move inward and the color begins to fade.

(Harka) The presence begins to ebb, and you begin to feel weak. When it is gone you are dazed and kneeling on the ground, catching your breath.

Why don't we go through the doors? And are the doors on the north, east, south, or west side of the room?

The door is to the North, and when you go through you are shocked by what you see. Through the doorway is a massive dome (shocking, huh?). the door leads to a ledge about five feet across that runs around the dome about halfway up. In front of you is a large stone statue of a knight in ornate armor and holding an impressive great sword, and behind him is an impressive cityscape. The dome is about 6 miles in diameter and filled with stone buildings, and in the center is a massive ad hoc looking complex. It seems to have been built by connecting all the buildings in the center of the city and constructing a tall in the middle using the rest of the complex as a foundation. The spire is currently about 600 feet tall, but it still seems to be under construction. there are several spiral staircases that go from the ledge to the city, one in each of the compass directions, including one directly in front of you.

[code]
 /\
 ||
 ||
 N

    D
    |
    |
    C
    |
    |
    B
    |
    |
 ?-A-?
    |
    ?
[/code]

A=where we entered
B=shrine, where we found Aard'lak
C=gate room
D=dome
|=passage

sorry just got back from a vacation. and i thought this was dead. but yay


$ f(x)\equal{}\frac{1}{1\plus{}25x^{2}}$ and $ g_{n}$ is a polynomial of degree $ \le n$ such that $ g_{n}\left(\minus{}1\plus{}\frac{2(i\minus{}1)}{n}\right)\equal{}f\left(\minus{}1\plus{}\frac{2(i\minus{}1)}{n}\right)$ for $ i\equal{}1,2,\dots,n\plus{}1$

show that \[ \max_{\minus{}1\le x\le 1}\left|f(x)\minus{}g_{n}(x)\right|\to\infty\]

I don't think this belongs in HSB  :D . Mods, can you move this to a calculus forum?


Alcumus AMC AIME geometry LaTeX USA(J)MO USAMO

I realized that the AMC 10 is approaching quickly and I want to make AIME. My scores on practice tests are usually 102-114 and I want to make the AIME. I have the following books: Intro Geo (about half done), Intro NT (not much done), Intermediate Algebra (just started), or I could use Alcumus (for C&P). Which should I focus on the most before AMC 10 if I just want to make AIME, nothing more (and definitely nothing less!)?

I would get volume 1 & do a bunch of practice tests.

If you can't get volume 1 and can only use the resources you've listed, I would focus on Intro to Geo and Alcumus. Number theory on the AMC 10 is pretty trivial and algebra doesn't go higher than algebra 1. I would also recommend practice tests.

Finish Volume 1, take practice tests, and do the first 5 AIME problems of random AIMEs. That should get you about ready for AMC10.

In general, do practice tests. Also work through Intro to Geometry. Mostly, just be really familiar with working with sectors of circles and spotting similar triangles. C&P in the AMC 10 is pretty easy.

most of the AMC 10 problems are algebra/very simple geometry.

study algebra, geometry, and take past tests. taking past tests are the most helpful.

I would say do 15-25 of AMC 12s, and do 1-5 of AIMEs.

(That plus doing AMC 10 practice tests, obviously.)

[quote="myyellowducky82"]What sections (if any) should I study out of the Precalc book?
I own all of the books, and I know there's no way that I'm going to be able to work through each book completely.[/quote]

Precalculus book won't be much of a help.

OK - what to do for prep? First of all, relax - when I'm typing this, it's the first day in 2010, so relax and spend sometime with family to have a math-free moment for at least couple of hours.

Secondly, once you are relaxed enough, I highly recommend you to start taking practice tests and REVIEW after each one. REVIEW each one, including the very easy ones you got right. If there's something you don't understand, then go back to the books for those concepts and review them. But, your ultimate focus should be on the tests themselves (you have like a month left). If you have done all the practice tests, then review GEOMETRY unless you are good at it.

Now, having said that, I am going to follow up by some other things I picked up for studying for more standardized tests (I've not been very active here because I'm in college. I was here before $ \text{\LaTeX}$ was around, so I may be able to tell you something useful...or not  :D ).

There are some things that you want to do when you take practice tests.

1. [u]Be Confident[/u]. If you have been like most students at AoPS, you have spent weeks and if not, months, preparing for AMC by reading Intro books and/or AoPS Volume I/II. You know all these stuffs - you just have to tell yourself that you can do it.
2. [u]Keep your energy fresh for each problem[/u]. Students tend to get burned out quickly in AMC's, especially at lower levels like AMC 10 and 12. If you were to solve problems backwards, there may be higher chances of getting harder questions right because your brain is that much "fresher" (I'm assuming here that you know the concepts to the harder questions). You can maintain this high level of energy if you stay calm and take [b]one problem by one[/b]. If needed, take like 3 seconds break before you attack last 5 questions on the test because for most, not having that 3 seconds will not make much difference. It's all about how you approach the test.
3. [u]Practice is practice - don't burn out[/u]. I recall reading back several years ago - okay, quite sometime ago - that a student barely made the cut for AIME from AMC 12, somehow made the floor value for USAMO (this was before USAJMO and all that stuffs that I didn't even bother to read yet), and got like one problem right on USAMO. My point here is, the real tests are unpredictable, and so don't be down because of mediocre scores on the practice tests. You never know how the real one's going to be - and always keep that in your head.

Cheers to 2010 again!

[quote="giratina150"]I am preparing for the AMC 10 and hope to reach at least a score of 120.  I've already done all the practice tests and am doing Vol. 1 of AoPS.  What else can I do to raise my score from about 90-110 to over 120? :([/quote]
Practice AMC 12 and beginning AIME? You can also take the AHSMEs.

i agree with AdithyaGanesh about working on a specific range of problems
so if you can do the first 10 pretty easily,
try only working on
#15-25(as he mentioned)  but dont give up easily on any problems you cant get and dont necessarily time ur self either
also, try to determine which subject is your weakest by solving lots of problems
vol.1 is pretty helpful but again , you have to really sit and think about each of the questions you find difficult, or you wont improve
the main thing is, dont freak out
you have a month and if you practice solving LOTS of problems that are harder than an average amc 10 problem. you will improve greatly
or atleast enough to reach ur goal

Depends on what you expect to get :lol:
150:finish volume 1&2, a bunch of past AMC12 test(do them first then do AMC10,you will find AMC10 much easier than before), AIME ramdomly pick 1-10
120:volume1(plus volume2 would be awesome),a bunch AMC10&12(I'd suggest AMC10 last 10 problems and AMC12 last 15)
Both of them can make you into AIME!
Work hard man!


Let a,b,c,d be positive integers; a>b>c>d and ac+bd=(b+d+a-c)(b+d-a+c)
S=ab+cd
Prove that S is not a prime.
(I'm having a solution but it's too long and complicated )

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119217#119217]I don't think an IMO #6 is appropriate for this forum[/url].


Could you please pm me in how to get into MI (private) forums or you could just post here. Thank you ahead of time.  :oops:

Go to the usergroups page and select join Michigan.  Your membership will be pending I believe.  Also, you should probably pm the Mod 1234567890 in order to be accepted.

If 1234567890 is unresponsive, maybe you should try pming Allen (kyyuan), since he's still in high school and is more likely to check the boards these days.

SMF, I think I added you. Can you doublecheck?

Um...no. I can't seem to access it... :(


Do Protons Decay? If so what is their Half Life?

Protons do Decay. Which means we have to find the half life. The answer is hidden behind the formula
                         N=N0^e^-frequency t
No matter what type of particles  are emmited, all  nuclear decays follows this law of particle decays.
The rapidity of decay of a particular radioactive sample is usually measured by the Half Life, T1/2, defined as the time interval  in which the number of parent nuclei at the beginning  of the interval as reduced by a factor of one-half. The half life is readily obtained in terms of  frequency as

                                           T1/2=Ln2/frequency=0.693/frequency

/=means divided

and so the Half Life of the proton is 10^36

Uhh,
I don't think anyone's observed proton decay yet, though.  Aren't there detector experiments still going on that're trying to look for this?


trigonometry
modular arithmetic
real analysis
real analysis unsolved

Let $ x_0,x_1,\cdots$ be a sequence of reals such that
1) $ x_n>1\forall n\equal{}0,1,...$ and
2)$ \frac{1}{2}\left(x_{n\plus{}1}\minus{}\frac{1}{x_{n\plus{}1}}\right)\equal{}\frac{x_n\plus{}1}{x_n\minus{}1}\forall n\equal{}0,1,\cdots$.

Is it convergence?

Let $ x_0, x_1, x_2 \ldots$ be a sequence of reals such that
[b]1)[/b]  $ x_n > 1 ~~\forall ~~ n \in \mathbb{N} \cup \{ 0 \}$ and
[b]2)[/b]  $ \frac {1}{2}\left(x_{n \plus{} 1} \minus{} \frac {1}{x_{n \plus{} 1}}\right) \equal{} \frac {x_n \plus{} 1}{x_n \minus{} 1}~~\forall ~~ n \in \mathbb{N} \cup \{ 0 \}$.

Using $ 2)$:

$ \frac {1}{2}\left(x_{n \plus{} 1} \minus{} \frac {1}{x_{n \plus{} 1}}\right) \equal{} \frac {x_n \plus{} 1}{x_n \minus{} 1}$
$ \iff \frac{1}{2} \left( x_{n\plus{}1} \minus{} \frac{1}{x_{n\plus{}1}} \right) \equal{} 1 \plus{} \frac{2}{x_n \minus{}1}$
$ \iff x_{n\plus{}1} \minus{} \frac{1}{x_{n\plus{}1}} \equal{} 2 \plus{} \frac{4}{x_n \minus{}1}$
$ \iff (x_{n\plus{}1})^2 \minus{} 1 \equal{} 2x_{n\plus{}1} \plus{} \frac{4x_{n\plus{}1}}{x_n \minus{}1}$
$ \iff x_n \cdot (x_{n\plus{}1})^2 \minus{} (x_{n\plus{}1})^2 \minus{} x_n \plus{} 1 \equal{} 2x_n \cdot x_{n\plus{}1} \minus{} 2x_{n\plus{}1} \plus{} 4x_{n\plus{}1}$
$ \iff x_n [ (x_{n\plus{}1})^2 \minus{}1 \minus{} 2x_{n\plus{}1} ] \equal{} (x_{n\plus{}1})^2 \minus{} 1 \plus{} 2x_{n\plus{}1}$
$ \iff x_n \equal{} \frac{(x_{n\plus{}1} \plus{}1)^2 \minus{} 2}{(x_{n\plus{}1} \minus{}1)^2 \minus{}2}$

I'm not quite sure where to go from there, maybe it will help someone else though.

[hide]
Let $ x_n\equal{}\tan \theta_n$ where $ \theta_n \in \left( \frac{\pi}{4},\frac{\pi}{2}\right)$ for all $ n$.

Then the left hand side is $ \frac{x_{n\plus{}1}^2\minus{}1}{2x_{n\plus{}1}}\equal{}\minus{}\cot 2\theta_{n\plus{}1}$ and the right hand side is $ \minus{}\tan \left( \theta_n \plus{}\frac{\pi}{4} \right)$. So we get
\[ \frac{\pi}{2}\minus{}2\theta_{n\plus{}1}\equiv \theta_n \plus{}\frac{\pi}{4} \pmod{\pi}\]
Now the right hand side is between $ \frac{\pi}{2}$ and $ \frac{3\pi}{4}$, so we actually get
\[ \frac{3\pi}{2}\minus{}2\theta_{n\plus{}1}\equal{}\theta_n \plus{}\frac{\pi}{4}\]
\[ \Rightarrow \minus{}2\left( \theta_{n\plus{}1}\minus{}\frac{5\pi}{12}\right) \equal{}\theta_n\minus{}\frac{5\pi}{12}\]
From this recurrence, it follows that
\[ (\minus{}2)^n\left( \theta_n \minus{}\frac{5\pi}{12} \right) \equal{}\theta_0\minus{}\frac{5\pi}{12}\]
\[ x_n\equal{}\tan \left( \frac{5\pi}{12}\plus{}\left( \minus{}\frac{1}{2}\right)^n \left( \arctan x_0\minus{}\frac{5\pi}{12}\right) \right)\]
The sequence works because it is always the tangent of something strictly between $ \frac{\pi}{4}$ and $ \frac{\pi}{2}$. So the sequence always converges to $ \tan \frac{5\pi}{12}\equal{}2\plus{}\sqrt{3}$.
[/hide]


email
factorial
MATHCOUNTS
floor function
ceiling function
function
Harvard

 :welcome: Hey guys, this is a good challenge problem that took me about 1 and a half hours in total (non consecutively) to do. I hope you guys enjoy this little challenge. ( don't push yourself too hard just to beat me; i'm supposedly a super genius. not trying to be conceited or anything, but i just don't want your parents to sue me when you drop dead from overexertion. oh, and you most likely won't be able to get it if you didn't get to alg. 1~2 yet. this problem IS solvable and there's no tricks. im a 12 year old and i got it. you can get it too. oh and 1 more thing, if you get this, i suggest you don't post it and just email me that you got it or something. we want the others to work too! :D )

Using only two 2s, write an equation that equals 5 using any symbols. You may use concatenations, decimal points, sq. roots, multiplication, factorials, powers, etc etc.

FAQ Questions

You can't use ANY NUMBER besides the TWO 2s!!! not even 0!

1 big hint!!!!!u need a negative sign!!!!!!!!!!!

watever, just figure it out. GOOD LUCK! :winner_first:  :starwars:  :ddr:  :trampoline:

[quote="Treething"]*drops dead from overexertion*

What's a concantation?[/quote]

a concatenation is putting two numbers together to make another.

for example, from 2 and 2,  u can make 22.

HOPE u understood. :idea:

[quote="hwenterprise"]Wow, this problem is HARD!!!

I see why it took you 1.5 hrs.

Also, are you sure this is MATHCOUNTS????

PS: What are concatenations?????[/quote]

This is a brain teaser mathcounts problem. and yes, it is hard.
look above for concatenations.

[quote="236factorial"]okay.... that won't help at all... because you just used all the two's up there.  :?[/quote]

obviously,...hope u figure it out!!! :D

AAAAAHHHHHHHHHHHHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Sorry to SPAM, but that's just pure incredible.

Plus, this topic should be moved.

[quote="chess64"][hide="my answer"]
$\lfloor \sqrt{\sqrt{(\lceil 2\sqrt{2} \rceil !)!}} \rfloor$
[/hide]

Took me exactly 11 minutes. :)[/quote]

cool u must hav had luck with u, but congratulate urself
most ppl take weeks, o well depends on ur age.
email me with wat u got; dont worry i will just correct it.

email:rainochaos@yahoo.com  :clap2:

[quote="chess64"][hide="my answer"]
$\lfloor \sqrt{\sqrt{(\lceil 2\sqrt{2} \rceil !)!}} \rfloor$
[/hide]

Took me exactly 11 minutes. :)[/quote]

oops, thats not really correct, try again

[quote="236factorial"]I knew there was that floor function in there!!![/quote]

u thought wrong.

I don't see why you need me to email it, but whatever.

I'm only 13, entering 9th grade.

can we define a new function? as in: a @ b = a + b + 1

in which case 2@2 = 5!

if not then oh well....

how about just:
[hide="this"]
$\lceil\sqrt{22}\rceil$
[/hide]

but chess64's solution works too...

[quote="flamesofdestiny"][quote="ToBe"]haha im starting to laugh at this as well.

Flamesofdestiny, could you explain how that answer is "technically incorrect'?[/quote]
it means that the solution does not solve the problem legitimately. :rotfl:[/quote]

Why is it not legit?  And why don't mine work either?

$log_a(a^2a^2a)$
or
$log_a(a^{2^2}a)$
or
$!2 + 2\textdollar$

This thread has turned from a serious problem to a flame war to a complete joke  :rotfl:  :rotfl:  :rotfl:

[quote="flamesofdestiny"]THERE'S 1 answer to this prob, so GET IT!!![/quote]This earlier statement has been refuted many times over  :D

[quote="AntonioMainenti"]I tried to post another reply but apparently I never clicked "Submit." Basically, I think my other solution works (even though it technically might not  ;)), and I have another:

$5=2.2$

It's in base $\frac{2}{3}$ by the way.  :D[/quote]

In base 2/3, there's a 3  ;)  :rotfl:

I like Hamster's last solution the best, the one with the logs.

work in the ring $\mathbb{Z}/1\mathbb{Z}$ so that $2+2=5$  :):):):):)

Hmm...

$ \sqrt{.2^{-2}} = \sqrt{\frac{1}{0.04}} = \sqrt{25} = 5 $

If not writing the 0 before a decimal point is legal (which, of course, is :D)

OMG this so funny :rotfl:
Any way about flamesofdestiny being a genuis....genuises (spell?) don't say other people that they're stupid from what ive heard.

I used to think I was pretty smart(NOT any where CLOSE to a genuis), but ever since I've found this website, I've learned I wasn't quite as smart as I thought I was, and that has made me a lot smarter. IF you know what I mean..
Flamesofdestiny you should know that the world doesn't revolve aroud you (or me for that matter).(It revolves around the sun :lol: )

:rotfl: . That moderatorship part got me.

He's gonna win the Darwin award some day. Pure genius.

Many people think they are geniuses before AoPS gets to them.  :P

This is one hilarious thread. I'm regretting moving it to The Round Table, which is for *serious* things, instead of Games and Fun Factory :rotfl:  :rotfl:

no, don't regret it; in games and fun factory they'd have locked it long before it reached its full potential

"full potential" ??? :rotfl: :rotfl: this is getting funnier and funnier... thanks flamesofdestiny! :)

hay guys i got tested at an iq center my score was liek 93 bur im still higher than george bush

we should make a spam forum  :D

yeah...i seem to have a lot of potential spam topics lol :roll:  :rotfl:

[quote="4everwise"]Well, I used to consider myself pretty smart and after I found AoPS, I realized I wasn't all that exceptional. On the bright side, AoPS has motivated me to beome smarter by studying.  :)

And regarding the "Spam Forum," it would be really popular.[/quote]

As soon as I logged on to this forum, I felt unexpectadly stupid, since I joined at the time of the start of national preparation (and states). That time, I was pretty inexperienced with math problem solving.


inequalities
logarithms
inequalities unsolved

[b]Let a, b, c be three real numbers with  0 < a, b, c < 1

and a + b + c = 2.

Prove that[/b]

$ \frac{abc}{(1\minus{}a)(1\minus{}b)(1\minus{}c)} \geq 8$

[quote="Nora.91"][b]Let a, b, c be three real numbers with  0 < a, b, c < 1

and a + b + c = 2.

Prove that[/b]

$ \frac {abc}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)} \geq 8$[/quote]

$ \frac {abc}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}\equal{}\frac{8abc}{(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}\geq 8$

esay

Let, $ f(x) = \log \left(\frac {x}{1 - x}\right)$.
So, $ f(x)$ is concave function.
Thus,
\begin{eqnarray*} \frac {\log\left( \frac {a}{1 - a} \right) + \log\left( \frac {b}{1 - b} \right) + \log\left( \frac {c}{1 - c} \right)}{3} \ge \log\left( \frac {\frac {a + b + c}{3}}{1 - \frac {a + b + c}{3}} \right) & \iff & \log\left( \frac {abc}{(1 - a)(1 - b)(1 - c)} \right) \ge 3\log\left(\frac {\frac {2}{3}}{1 - \frac {2}{3}}\right) \\
& \iff & \log\left( \frac {abc}{(1 - a)(1 - b)(1 - c)} \right) \ge \log \left(2^3\right) \\
& \iff & \frac {abc}{(1 - a)(1 - b)(1 - c)} \ge 8 \end{eqnarray*}

Setting $ a\equal{}1\minus{}x\ ,\ b\equal{}1\minus{}y\ ,\ c\equal{}1\minus{}z$ the inequality is equivalent with :

$ (x\plus{}y)(y\plus{}z)(z\plus{}x)\ge 8xyz$ , trivial .

[quote="Nora.91"]Thanks for the replies but i did not understand this one line in fesha's solution ....

[quote="fesha3t"]
$ \frac {abc}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)} \equal{} \frac {8abc}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}\geq 8$

esay[/quote]

How do you get the last part       [b]... > 8[/b]

am i missing something?

[hide="Pardon me"]
i am just a beginner at Inequalities  :(  [/hide][/quote]
[hide]take $ a\plus{}b\minus{}c\equal{}x$ and cyclic[/hide]


probability
MATHCOUNTS
algebra
binomial theorem
\/closed

We'll be offering our Introduction to Counting course in January.  Classes will be once a week and will cover such counting courses as combinations, permutations, probability, combinatorial identities, the binomial theorem, and more.  The class will cover nearly all aspects of counting needed for success on MATHCOUNTS, and AMC-8/10/12.

If you are considering taking the course, please post your time/day of week preferences here (most notably the days you can't make it), along with any questions you have about the course.

Tuesday would be just excellent.
Sundays arent too bad also.

Saturdays, Thursday, Fridays, Mondays are the ones that I may not be able to make.
As far as time goes, 7:30-9 EST or a little bit later would be fine.

Thanks!


$ \because a^2\plus{}2b^2\plus{}3\equal{}(a^2\plus{}b^2)\plus{}(b^2\plus{}1)\plus{}2 \geq 2(ab\plus{}b\plus{}1)$
$ \therefore LHS \leq \sum_{cyc}\frac{1}{2(ab\plus{}b\plus{}1)} \equal{} \frac{1}{2}$

It seems that he derived it from the fact that:

$ \sum_{cyc} \frac{1}{ab\plus{}b\plus{}1} \equal{} \sum_{cyc} \frac{b}{ab\plus{}b\plus{}1} \equal{} \sum_{cyc} \frac{ab}{ab\plus{}b\plus{}1}$

[quote="wxydx00"][quote="apollo"]$ \therefore LHS \leq \sum_{cyc}\frac {1}{2(ab \plus{} b \plus{} 1)} \equal{} \frac {1}{2}$[/quote]
how do you know it?[/quote]
because $ abc \equal{} 1$
$ \rightarrow a\equal{}\frac{x}{y},b\equal{}\frac{y}{z},c\equal{}\frac{z}{x}$
therefore $ \sum{\frac{1}{ab\plus{}b\plus{}1}}\equal{}\sum{\frac{z}{x\plus{}y\plus{}z}}\equal{}1$  :lol:


Let $P_{1}P_{2}\dots P_{2n}$ be a convex polygon with an even number of corners. Prove that there exists a diagonal $P_{i}P_{j}$ which is not parallel to any side of the polygon.

Suppose that for every $i,j$ , $P_{ij}$ is parallel to one of the sides of a polygon. We have $(2n-3) n$ diagonals, and only $2n$ sides, which means that there is a side which is parallel to at least $n-1$ diagonals, thus these $n-1$ digonals are pairwise parallel, which means that they cannot share a common end. So these diagonals give us $2(n-1)$ different enpoints. Together with our chosen side, this gives us $2n$ different endpionts (obvoiusly these diagonals cannot share a common endpoint with our side), which are all vertex of our polygon. Thus one of our diagonals must be a side of the polygon ( impossible because of assumption about convexity)

The same problem is [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=diagonal+parallel&t=85912]here[/url].


superior algebra
superior algebra unsolved

$G$ and $G'$ are two additive groups. $f$ is an epimorphism from $G$ to $G'$. $g$ is a homomorphism from $G'$ to $G$. $f\circ g$ is the identity mapping in $G'$.

Show that $G=\mbox{im }g\oplus\ker f$.

(Note: In my book, An additive group is commutative)

Now my question is : Why do we need $f$ to be an EPIMORPHISM? The condition "$f\circ g$ is the identity mapping" is very strong.

I think $G$ is isomorphic to the semi-direct product of those two groups. Is this what you were referring to? :?

I think it should be that too... (I still don't know how to prove it)
btw, why do you say "semi-direct product", not "direct product"?

Because I'm not sure it's always direct. I the setting is the same as the following: given $f$ s.t. $f\circ f=f$, show that $G=\ker f\oplus\mbox{Im} f$. I don't think this is always true. We would still need the fact that $\mbox{Im} f$ is normal in $G$, and I'm definitely not sure about that. I could be wrong, of course, but it looks to me like any [url=http://planetmath.org/encyclopedia/SemidirectProductOfGroups.html]semi-direct product[/url] works, if we take $f$ to be the projection of an element onto its component coming from the group which is not normal (the group $Q$ in that link).

"Semi-direct product" is not mentioned in my book  :D
So it must be direct product.  :P

I know how to do now... And I make a mistake... I didn't understand the term correctly; so I didn't use the proper word in the description of the problem. Sorry for that.

Now re-read the problem and you will understand what I put wrongly.


A convex polygon with $ n$ sides has 14 diagonals. How many diagonals does an $ (n\plus{}1)$-sided convex polygon have?

the formula to find the number of diagonals in a regular polygon which has n sides is [b]n(n-3)/2. [/b]
 in this case there are 14 diagonals, therefore we have to make the equation n(n-3)/2=14.
                                                                    Solving the equation by multiplying by 2 gives us...
                                                         n(n-3)=28  n=7 using quadratic equations.
                                                n+1 sides= 8 sides. Plugging 8 sides in the diagonal formula gives us...
                                               8(5)/2=20 diagonals.
                                              And that is the answer to the simple problem!  :roll:  :)


I wanted to buy the AoPS books from the website, although the problem is that I am residing in a country, which is not listed in the 'shipping information' section. Do you have any idea where else I can get them from? I already tried amazon and ebay, although they weren't available.

They ship the books internationally from this site.  Send an email to orders at artofproblemsolving dot com (where "at"=@ and "dot"=.)  and they should give you more information.

Yes, but only to the countries listed. My billing adress is in a country, which is listed, although I live in a different place right now. :?

Where's this list of countries?  I couldn't find anything of the sort.

Anyway, did you email them like I said?  I'm 99% certain they'll be able to arrange something with you.

I am not sure why you don't see the list of countries. Maybe because you are recognized, as from the US. Anyway, I emailed and I got a reply, saying that something would be likely arranged. thanks

I have a question: Does anyone how long it taks to ship 3 books nationally / internationally? And what is the cost foe shipping nationally / internationally ?
Thanks in advance.

International takes about 2-3 weeks and is about $\$$25. National is cheaper and takes much less time to deliver (like 2-3 days for Priority), however I don't know how much it is (I think one of the options is $\$$2.50 to $\$$4 for larger orders) . ;)

It depends on the country.  Typically, priority international shipping takes 1-2 weeks and economy international shipping takes about a month (though it can be less depending on the country).  It is difficult to estimate international shipping costs because the rates are different to every country.

Within the US, priority shipping takes 2-5 business days and economy (media mail) shipping takes 2-3 weeks.  Priority shipping is the larger of $8 or 15% of the order; media shipping is
$3 for small orders and $4.50 for larger orders.

If you let us know which books and what country, we can provide exact shipping costs.

[quote="atm"]Oh thank you.
I decided to buy The Art Of Problem Solving Volume 1 solution and Volume 2 with both text and solutions. I'm in California. I[b]f I need them by 06/18 (I'll go to Vietnam on 06/19 night[/b], and I need those books with me), is it late right now to order them? (reply me according to the day that you reply). 06/19 is the lastest day (of course it has to be to me before 4 PM.) [b]How many days exactly does it take to be delivered in my case?[/b] And how much do I have to pay in total, including tax, shipping fee (the books are 52 bucks)? And do you have any [b]express shipping system[/b] that will take less time to deliver? If so, what is the cost and time it takes?
Thanks in advance.[/quote]

If you order today by priority mail, you should be ok because we're in CA, too.  It's not clear to me which books you want because the prize you cite does not match the books you mention above.  For AoPS vol1 + vol2 + solutions, the price is $\$$73, the fast shipping is $\$$10.95, and CA tax is 7.75% (of the $\$$73)

[quote="atm"]No, I only buy Volume 1 solution, I got the text book already. :)
(Volume 1 solution + Volume 2 text +Volume 2 solution = 52 bucks)
[b]So how many days exactly would it take and how much would it cost in my case?[/b]
I dont know if I can buy them today or not, since I have some personal problems that may take 1-2 days to solve before buying the books :-)
Thanks for reply though. :)[/quote]

Tax is the rate I quoted above; shipping would be $\$8$.  If you order in 2 days, you'll definitely be taking a chance.  I can't say how long it would take to get there; I would guess 2-4 days for CA, but wouldn't bet my life on it.


Note the logical reasoning used below and determine the value of x. Justify !

[img]http://img18.imageshack.us/img18/7865/calss.jpg[/img]

(My answer = 9)

It looks like for the first triangle,
4 + 2 - 8 + 5 + 9 = 12

And for the second,
4 + 7 - 17 + 6 + 10 = 10

And for the third,
4 + 5 - 13 + 12 + 8 = 16

I think you add up the 4 numbers on the sides of the triangles and subtract the number on the top, and the result is the same as the number on the bottom.

So I also get x=9.

Thank you very much ;)
I think this is the correct logic :)


geometry
3D geometry
trigonometry
ratio
rectangle
conics
ellipse

Put solutions for the USC test here.  [url=http://www.math.sc.edu/contest/problems.html]Here's[/url] the link to the problems.

[hide="16"] $(e)$ Using the difference of two squares, the sum evaluates to:$-(3+7+11+.....+3995)+1999^{2}=1999^{2}-(\frac{999(3998)}{2})=1999000$[/hide]
[hide="17"]$(a)$ Let the sum we are tring to find be $S$. Notice that if we multiply both sides by $10!$, we get:

$\frac{10!}{1!9!}+\frac{10!}{3!7!}+\frac{10!}{5!5!}+\frac{10!}{7!3!}+\frac{10!}{1!9!}=10!S$

Notice the LHS evaluates to:

$\binom{10}{1}+\binom{10}{3}+\binom{10}{5}+\binom{10}{7}+\binom{10}{9}=\frac{2^{10}}{2}=2^{9}=10!S$. Therefore $(a,b)=(9,10)$. [/hide]

[hide="18"] $(e)$ Let us assume $x>0$ and $y>0$, therefore we have $2x+y=10$ and $x=12$, which gives us that $y=-14$, which is a contradiction. Let us now assume that $x>0$ and $y<0$, from which we get  $x=\frac{32}{5}$ and $y=-\frac{14}{5}$ The answer follows. [/hide]
[hide="19"]
$(a)$ The roots of $x^{2}-x-1$ (let us call them $p$ and $q$) must satisfy $ax^{3}+bx^{2}+1$. We also know that $pq=-1$.

We have:

$ap^{3}+bp^{2}=-1$, multiplying by $q^{3}$ gives us $-a-bq=-q^{3}$

Similarily:

$aq^{3}+bq^{2}=-1 \Longrightarrow-a-bp=-p^{3}$

Solving this system of equations and getting rid of the $a$ term gives us $b=\frac{q^{3}-p^{3}}{q-p}$,from which we use difference of two cubes and using $q+p=1$, we simplify our answers to $b=-2$[/hide]

[hide="20"] $(c)$ Let $\sin{15}=x$ and $\cos{15}=y$

We have $x^{4}+y^{4}=(x^{2}+y^{2})^{2}-2(xy)^{2}=1-2(xy)^{2}$. Using our trig addition formulas ($\sin$ and $\cos(45-30)$) we simplify the sum to $1-\frac{1}{8}=\frac{7}{8}$[/hide]

[hide="21"] $(d)$ This one you just have to reason, $d$ is the correct answer because not all angles must have rational measures, therefore cannot be truly trisected. [/hide]

[hide="22"] $(d)$ The cube can be expanded as:

$x^{3}+3x^{2}(iy)+3x(iy)^{2}+(iy)^{3}$. Working with the real parts, we have that $x^{3}-3xy^{2}=-74$. The solution $(1,5)$ works[/hide]

[hide="23"]$(b)$ Let $x=-3$, we can easily find out the answer.[/hide]

[hide="24"] $(b)$ WLOG, let $\triangle{ABC}$ be iscoceles with $AC=CB=1$. Notice that $\triangle{ABC}\sim \triangle{FCG}\sim \triangle{DCE}$. Notice that the side ratio of similar triangles is the square root of the area ratios. From there, we see that $CD=\sqrt{1/3}$ and $FC=\sqrt{2/3}$. From here, we just do a little bit of arithmetic and rationalizing square roots and the answer follows.[/hide]

[hide="25"] $(c)$ Using the chicken mcnugget theorem, the largest integer that cannot be in the form $7x+4y$ is $(28-(7+4)=17$. Therefore, the scores that can't be attained are $1,2,3,5,6,9,10,13,17$.[/hide]

[hide="26"] $(c)$ WLOG, let the cylinder be a can of chicken soup with the label covering the entire outer layer (apart from the bases). Peeling the label off, we see a rectange with height $16$ and base $3$. The squirrel travels $4 \cdot$ the hypotenuse of the right triangle with base $3$ and height $4$. The answer is $20$.
[/hide]

[hide="27"] $(d)$ We can perform long division on $\frac{N^{2}+2N+1}{N+23}$ and we get our remainder to be $\frac{1+21 \cdot 23}{N+23}$, the largest integer $N$ therefore is $461$.[/hide]

[hide="30"]$(b)$ Notice that the given equation is equal to $x+\frac{1}{x}=-1$. Let this be $S_{1}$. Also notice that $(x+\frac{1}{x})(x^{n}+\frac{1}{x^{n}})=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}$. Or in other words, $S_{n+1}=S_{1}\cdot S_{n}-S_{n-1}$.
Therefore we have $S_{1}=-1, S_{2}=-1, S_{3}=2$ and from there, it repeats in cycles of three. Therefore, our sum is $9(1+1+4)=54$
[/hide]

posting the rest later...

Shobhit mentioned he didn't know how to do 28, so I'll do that.

[hide="28"](c) Since $F_{1}F_{2}$ is a diameter of a circle, and P is on the same circle, $F_{1}F_{2}P$ is a right triangle, and we know its area is 26, so $F_{1}P*F_{2}P=52$. We are also given that the major axis is 15, so by the definition of an ellipse, $F_{1}P+F_{2}P=15$. Then by the Pythagorean Theorem, the quantity we want is $\sqrt{F_{1}P^{2}+F_{2}P^{2}}=\sqrt{(F_{1}P+F_{2}P)^{2}-2(F_{1}P*F_{2}P)}=11$.[/hide]


vector
analytic geometry
LaTeX
linear algebra

S={ x e R3 / x1 + 2x2 - x3 = 0}. Find a base B from R3 that contains a base of S and a  base of S orthogonal, and  that the vector (0,5,-2) hast coordinates (0,1,1) in the base B.

I have come till here:

S=<(1,0,1)(0,1,2)>
S orthogonal=(-1,-2,1)

Putting these two together i get a Base, not B, lets call it Z.

Z={(1,0,1)(0,1,2)(-1,-2,1)}
Coordinates of (0,5,-2) in Z are (-2,1,-2).

How can i go on? Please help.

Are you sure this is intermediate??? May be I misundertood!!!

i think that this is just analytic geo, but it is hard to read because it is not in $\LaTeX$

$\LaTeX$ed by me. I'll have an attempt in latexing it correctly.

He wrote:

$S=\{x\in \mathbb{R}^3|x1+2x2-x3 = 0\}$. Find a base $B$ from $\mathbb{R}^3$ that contains a base of $S$ and a base of $S$ orthogonal, and that the vector $(0,5,-2)$ has coordinates $(0,1,1)$ in the base $B$.

I have come till here:

$S\leq (1,0,1)(0,1,2)$>
S orthogonal=(-1,-2,1)

Putting these two together i get a Base, not $B$, lets call it $Z$.

$Z=\{(1,0,1),(0,1,2),(-1,-2,1)\}$
Coordinates of (0,5,-2) in $Z$ are (-2,1,-2).

How can i go on? Please help.

me: Is that what you meant?

EDIT: I don't think this should be in the intermediate section.  I believe it involves 3D vectors/space and therefore advacned calculus/mathematics.

In fact this should go to the linear algebra section.


How do you prove that the determinant is distributive?

|AB| = |A||B|

?

"Distributive" is the wrong word for that property, since it has nothing to do with addition. The determinant is a homomorphism from the multiplicative group of invertible matrices into the multiplicative group of nonzero scalars.

It's probably best to build the proof up piece by piece, looking at elementary row operations.

There are three types of elementary row operation: interchange two rows, or multiply a row by a nonzero scalar, or add a scalar multiple of one row to another row. An [i]elementary matrix[/i] is the matrix that results from performing an elementary row operation on the identity matrix.

$ E_1\equal{}\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix},\ E_2\equal{}\begin{bmatrix}1&0&0\\0&2&0\\0&0&1\end{bmatrix}$ and $ E_3\equal{}\begin{bmatrix}1&0&0\\0&1&0\\3&0&0\end{bmatrix}$

are examples of the three types of elementary matrix. We can realize any elementary row operation as multiplication on the left by an elementary matrix. (If we multiply on the right by such a matrix, we perform an elementary column operation.)

Each kind of elementary row operation changes the determinant in a predictable way. This is a theorem with three parts, one for each type. After we've finished proving it, we encapsulate the result as this: if $ E$ is an elementary matrix, then $ \det(EA)\equal{}\det(E)\cdot\det(A).$

If $ A$ is elementary, we can completely factor $ A$ as a product of elementary matrices. (This is the Gaussian elimination algorithm.) $ \det(A)$ is then just the product of the determinants of those elementary matrices. If $ A$ and $ B$ are both invertible, then $ AB$ can be factored into elementary matrices just by concatenating the factors of $ A$ and of $ B.$ It follows that $ \det(AB)\equal{}\det(A)\cdot\det(B).$

We need a separate argument to deal with the cases when either $ A$ or $ B$ is singular, but that isn't too hard.


I really want to do well on the AIME next year and am willing to put in a lot of work, but I don't really know how to prepare.  Should I just work a lot of practice problems?  What books should I read to help me?  How did you people who go on to the USAMO and IMO get to where you are?

Read AoPS I and II and work on past AMCs and AIMEs-- which is what I did, but other people here could probably give you better advice. I only made AIME.  :(

Btw, this question is asked so many times. Didn't someone make a thread about this? You could look there. If not, I am sure there are tons of posts on this subject. Use the "Search" button at the top to get some more information.

This is a very frequent question.  Searching AoPS is a good idea.  In general doing as much amc/aime is good.  However, once you do an amc you should look it over and learn as much stuff that you didn't know before you took it as possible.  Just try learning concepts and then learn how to apply them.  I haven't takent the AoPS classes but I know that they will be good when I take some this summer.  Most importantly you have to work a lot and work hard.


geometry
circumcircle
function
similar triangles
geometry proposed

Given are $\triangle ABC$ with circumradius $R$ and three medians $m_{a},m_{b},m_{c}$ prove that
$\triangle ABC$ is equaliteral triangle $\Leftrightarrow$
$(1) m_{a}+m_{b}+m_{c}=\frac{\sqrt{3}}{2}(a+b+c)$
or $(2)\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}=3\sqrt{3}R$

I think it is wrong. (1) is equivalent to $a^{2}, b^{2}, c^{2}$ forms an arithmetic progression.

Both (1) and (2) must be wrong. As I told you, Gemath, there are triangles arbitrarily close to equilateral (i. e. their angles lie in an arbitrarily small neighbourhood of 60°) which satisfy $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}>\frac{\sqrt3}{2}\left(a+b+c\right)$ and there are triangles arbitrarily close to equilateral (i. e. their angles lie in an arbitrarily small neighbourhood of 60°) which satisfy $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}<\frac{\sqrt3}{2}\left(a+b+c\right)$. And since the set of non-equilateral triangles (parametrized by their angles, e. g. similar triangles are considered to be the same) is connected, and the function $\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)/\left(\frac{\sqrt3}{2}\left(a+b+c\right)\right)$ as a function of the angles of a triangle is continuous, there must be non-equilateral triangles satisfying $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}=\frac{\sqrt3}{2}\left(a+b+c\right)$. Similarly for (2). But Treegoner, I don't think that (1) is equivalent to the existence of a 60° angle either... am I doing something wrong?

  dg

ok I see Darij, it is my only conjecture  :blush: , but as you say if we have a function $f(a,b,c),a=b=c\Rightarrow f=0$ but $f$ never always $\ge 0$ or  $\le 0$ then we never have $f(a,b,c)=0\Rightarrow a=b=c$  really is it true?
and I only assert that $(1)$ have the nice equivalent
$\Leftrightarrow (m_{a}-\frac{\sqrt{3}}{2}a)(m_{b}-\frac{\sqrt{3}}{2}b)(m_{c}-\frac{\sqrt{3}}{2}c)=0$

[quote="gemath"]ok I see Darij, it is my only conjecture  :blush: , but as you say if we have a function $f(a,b,c),a=b=c\Rightarrow f=0$ but $f$ never always $\ge 0$ or  $\le 0$ then we never have $f(a,b,c)=0\Rightarrow a=b=c$  really is it true?[/quote]

If f is continuous and homogeneous in a, b, c, then I think yes... Can't be sure of pathological cases however.

  darij


function
complex analysis
complex analysis unsolved

f(z) = $ \sqrt {z \minus{} 1} \sqrt [3] {z \minus{} i}$

1)Find the branch points and their respective orders
2)It is not possible using a single branch cut that connects to branch points to define the branches of f(z). Why is this so?
3)At a point $ z \in C$ , how many values f have?

I'd write this as:

$ g(z)\equal{}\sqrt[6]{(z\minus{}1)^3(z\minus{}i)^2}$

guess I had some problems earlier convincing myself this was ok.  So this is a root-6 object with of course its real and imaginary (Riemann) surfaces folded over itself 6 times giving rise to the six roots of its argument. You could also interpret the function:

$ f(z)\equal{}\sqrt{z\minus{}1}\sqrt[3]{(z\minus{}i)}$

as just the multiplication two multifunctions:  one has two values, the other has three values.  So . . . for each of the two values of the first, I can multiply those by each of the three values of the second.  Six.  There you go.

The branch points are the values of the root arguments which equal zero or 1 and i above. The branch cuts depend on how you wish to define them.  Usually they are based on the principal branch cut of the root object which lies along the negative real axis.  To determine these cuts for $ g(z)$ above, I set $ (z\minus{}1)^3(z\minus{}i)^2\equal{}w$ and solve for $ z$ in terms of $ w$.  This gives me five root objects.  I then evaluate each of these for negative values of $ w$ (remember, I want the expression inside the root object to be along the negative real axis or $ w<0$).  I can do this with the following Mathematica code where I split up the calculation into 2 parts to get a better plot:

[code]Clear[z, w]
g[z_] := (z - 1)^3*(z - I)^2;
branchcuts = Solve[g[z] == w, z];

tlist1 = Table[N[Evaluate[({Re[#1], Im[#1]} & ) /@
             (z /. branchcuts)]], {w, -50, -2, 0.1}];
bplot1 = ListPlot[tlist1, PlotRange -> {{-3, 3}, {-3, 3}},
       AspectRatio -> 1];
tlist2 = Table[N[Evaluate[({Re[#1], Im[#1]} & ) /@
             (z /. branchcuts)]], {w, -2, 0, 0.001}];
bplot2 = ListPlot[tlist2, PlotRange -> {{-3, 3}, {-3, 3}},
       AspectRatio -> 1];

Show[{bplot1, bplot2}]
[/code]

[img]http://img156.imageshack.us/img156/2203/cutsforroot6us6.jpg[/img]

I can't answer the second part of the question.


Hi

We just had a contest and I'm trying to figure out the answers to some of the questions that I didn't answer. I would appreciate if you could help me with any of those.

This one looks easy, but I can't find the solution.

1) If p,q and r are positive intigers and (p+1/(q+1/r))=25/19, the the value of q is...

a)1 b)2  c)3  d)4 e)5

2) Quicy and Celine have to move 16 small boxes and 10 large boxes. Quincy can move small box in 3 min.'s, large box in 5 min.'s. Celine can move small box in 2 min.'s and large box in 6 min.'s.They start moving the boxes at 9:00 a.m. The earliest time they can finish moving all boxes is...

a)9:41   b)9:42       c)9:43             d)9:44     e)9:45

3) Two players play a game where you have to take out n number of toothpicks. Person A goes frist. He can remove 1,3 or 4. The person who removes the last one wins. Which should be the  vales of n so that B can win no matter how many toothpicks person A picks...

a)31      b) 32          c)33          d)34       e)35

1.

$p+\frac{1}{q+1/r}=p+\frac{r}{qr+1}=\frac{p(qr+1)+r}{qr+1}=\frac{25}{19}$

So we have

$p(qr+1)+r=25s$ and $qr+1=19s$ for some $s>0$

Substituting we have $p\cdot19s+r=25s$, so $p$ must be $1$ (if it were greater, $r$ would be negative). That yields $r=25s-19s=6s$.

So since $qr+1=19s$ and $r=6s$ we have $q=\frac{19s-1}{6s}$. For which $s$ does this give us an integer $q$? Well, we need that $19s\equiv1\;(\mathrm{mod}\;6s)\Rightarrow s\equiv1\;(\mathrm{mod}\;6s)$. This is only true for $s=1$. Inserting that gives us $q=\frac{19-1}{6}=3$.

[hide="Easier one for #1"]From ${25\over 19}=p+\frac{1}{q+{1\over r}}$ it follows that $p$ is the largest integer not greater than ${25\over 19}$, since the second term on the RHS is less than one. Therefore $p=1$. Subtract $1$ from the both sides and take the reciprocal:

${19\over 6}=q+{1\over r}$

Similarly, $q$ is the largest integer not greater than ${19\over 6}$, hence $q=3.$ Subtract $3$ from the both sides: ${1\over 6}={1\over r}$, hence $r=6$.[/hide]


quadratics
function
linear algebra
linear algebra unsolved

I have a problem which is elementary but looks difficult to solve:

Minimize $f(x)=x^{T}Q_{2}x+2x^{T}Q_{1}+Q_{0}$ where $Q_{2}\in{\mathbb R}^{n\times n}$ is singular.

Of course when $Q_{2}$ is non-singular, the solution is easy and well known as $x_{*}=-{Q_{2}}^{-1}Q_{1}$ and $f(x_{*})=Q_{0}-{Q_{1}}^{T}{Q_{2}}^{-1}Q_{1}$ by Lagrange Multiplier Method.  :?:

Watch out for those easy and well-known things: the minimum needs not exist even with a nonsingular $Q_{2}$ (consider $Q_{2}=-I_{n}$).

If the quadratic form is positive semidefinite, then you can reduce it to $y_{1}^{2}+\dots y_{m}^{2}$ by a change of variables. If the linear term depends on $y_{j}$ for some $j>m$, then the minimum does not exist. Otherwise, disregard $y_{m+1},\dots y_{n}$ and get back to the nonsingular case.

Yes, I have already realized. The function $f(x)=x^{T}Q_{2}x+2x^{T}Q_{1}+Q_{0}$ should be separated into two parts, one quadratic and one linear where the linear part has no optimum values. Hence $x=[{\bar x},{\tilde x}]$ where ${\bar x}$ is the minimum of the quadratic part and ${\tilde x}$ corresponds to the linear part. Thanks anyway :)


The diagram shows block 1 of mass 0.200kgg sliding to the right over a frictionless elevated surface at a speed of 8.00m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in simple harmonic motion with a period of 0.140s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h = 4.90m. What is the value of d?

Please help, I'm so confused! Thanks

Multistep problem.  We know that $ T\equal{}2\pi\sqrt{m/k}$ so then you can easily solve for the mass of the second block.  Now energy is conserved in the system so we need conservation of energy.  The total energy is initially all in the form of kinetic energy so use $ 1/2mv^{2}$ for the first block.  Momentum is conserved as well and initially, all momentum is in block 1.

Now you will have to recover the exit velocity of block 1, after which, the problem is easy.

[quote="blueeverest"]i don't think conservation of momentum is used numerically here except in principle.

After finding m2, use conservation of energy to say m1*v1^2 = (m1+m2)*v2^2, where v1 is the initial velocity given, v2 is to be found. This v2 is the max possible velocity and occurs at the mean position i.e. uncompressed spring after a time of T/2 i.e. half time period (i.e. after the spring gets compressed and comes back again to mean position). its interesting to note that after this point, m1 and m2 will lose contact since m2 will be traveling slightly slowly than m1.

Since its frictionless, v2 will remain same till the drop. then its routine projectile calculations.

I hope this helps.[/quote]

I am editing my own post here. I completely ignored the elastic collision aspect here. JRav is right. I think this amounts to setting up equations to solve for v. The masses dont behave in the way I explained previously. Sorry.


function
calculus
integration
geometry
combinatorial geometry
real analysis
real analysis unsolved

Suppose that $f$ is a nonnegative increasing function defined on $[0,1]$. Prove that one can find a nonnegative convex function $g$ such that $g(x)$ is smaller than $f(x)$ for all $x$ and the integral of $g$ between 0 and 1 is at least half of the integral of $f$ between the same numbers.

I'll just outline what I did, because I find it a bit difficult to write down everything. Hopefully, it will turn out to be correct :).

The perfect candidate would be the part $B_f$ of the border of the closed convex hull $K_f$ of $\Gamma_f=\{(x,f(x))\ |\ x\in I=[0,1]\}$ that lies below $\Gamma_f$ ($\Gamma_f$ is just the graph of $f$). What I tried to do was find a function $g\le f$ (not the $g$ that harazi mentioned; I'm using different notation) of the form $\sum_{j=1}^na_j\chi_{I_j}$ (where $I_j$ are disjoint intervals closed to the left and open to the right whih cover $I$, and $\chi_A$ represents the characteristic function of $A$) which approximates $f$ uniformly, arbitrarily well, except for a set of small measure. Then $B_g$ would lie below $B_f$, and it's clear to the naked eye that the area $S(B_g)$ under $B_g$ is $\ge\frac 12\int_Ig\ \mbox dt$. Since by this procedure $\int g$ can be made arbitrarily close to $\int f$ and $S(B_f)\ge S(B_g)$, the conclusion follows.

In order to find such a $g$, fix some $\varepsilon>0$. The set of points $x\in I$ where $f$ has variation $>\varepsilon$ is discrete, and hence finite. Choose a small interval around each such point (these will be some of our $I_j$'s). Now, if $\delta>0$ is small enough, and $|x-y|<\delta$ with both $x,y$ lying outside these first few intervals, we'll have $|f(x)-f(y)|<\varepsilon$ (this shouldn't be hard to prove), and all that's left is to partition the set difference between $I$ and the first intervals we've chosen into finitely many intervals $[u_k,v_k)$ of length $<\delta$, and then take for each interval $[u_i,b_i),\ a_i=f(u_i)$.

Sorry for making it sound so vague..


A large cube is constructed from individual unit cubes and then
dipped into paint. Thereafter, it is disassembled into the
original unit cubes; 486 of these have exactly one face painted.
How many unit cubes were used to construct the large cube?

Let the side length of the cube be $ x$.

Then the total number of cubes with one sided painted is

$ 6(x\minus{}2)^2\equal{}486$
$ (x\minus{}2)^2\equal{}81$
$ x\minus{}2\equal{}9$
$ x\equal{}11$

The total number of unit cubes used to construct the large cube is $ 11^3\equal{}\boxed{1331}$ cubes.


Where can i find more collection "Getting Started" level problems? All the contest problems are quite hard except for the AMC 10's.

You could try the easier problems of the AMC12.
http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm

all the olympiad problems AoPS has:
http://www.artofproblemsolving.com/Forum/resources.php

Try those maybe there's some easy problems.


hello, here is a few matrix computations, can some one please explain to me how to do them?

1. Show that the matrix $ \begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}$ is not diagonalisable

2. Let $ A\equal{}  \begin{pmatrix}
\minus{}2 & 1 \\
\minus{}4 & 2 \\
\end{pmatrix}$ find an invertible matrix $ P$ such that $ P^{\minus{}1}AP\equal{}  \begin{pmatrix}
0 & 1 \\
0 & 0 \\
\end{pmatrix}$

3. Find the Jordan canonical form of $ \begin{pmatrix}
0 & 1 \\
\minus{}4 & \minus{}4 \\
\end{pmatrix}$

1. Both eigenvalues of A are equal to 1. Solving the system (A-I)x = 0 we see that the eigenspace of $ \lambda \equal{} 1$ is spanned by the vector (1,0). Since the dimension of this space is only one, A is not diagonalisable.

2. We have $ \det(A) \equal{} 0$ and $ tr(A) \equal{} 0$, so both eigenvalues of A equal zero, and their eigenspace is spanned by (1,2). Now let P be a 2x2 matrix whose first column is (1,2) and the second column is (0,1). Then

$ AP \equal{} \begin{bmatrix} \minus{} 2 & 1 \\
\minus{} 4 & 2 \\
\end{bmatrix} \begin{bmatrix} 1 & 0 \\
2 & 1 \\
\end{bmatrix} \equal{} \begin{bmatrix} 0 & 1 \\
0 & 2 \\
\end{bmatrix} \equal{} \begin{bmatrix} 1 & 0 \\
2 & 1 \\
\end{bmatrix} \begin{bmatrix} 0 & 1 \\
0 & 0 \\
\end{bmatrix} \equal{} P \begin{bmatrix} 0 & 1 \\
0 & 0 \\
\end{bmatrix}$

and so

$ P^{ \minus{} 1}AP \equal{} \begin{bmatrix} 0 & 1 \\
0 & 0 \\
\end{bmatrix}$.

Observe that the second column of P can be found by solving $ A \begin{bmatrix} x\\ y \\ \end{bmatrix} \equal{} \begin{bmatrix} 1\\ 2 \\ \end{bmatrix}$.

3. Both eigenvalues are -2, and their eigenspace is spanned by (1,-2). The Jordan canonical form is therefore

$ J \equal{} \begin{bmatrix} \minus{} 2 & 1 \\
0 & \minus{} 2 \\
\end{bmatrix}$,

and we have

$ \begin{bmatrix} 0 & 1 \\
\minus{} 4 & \minus{} 4 \\
\end{bmatrix} \equal{} P J P^{ \minus{} 1}$

with, for example,

$ P \equal{} \begin{bmatrix} 1 & 0 \\
\minus{} 2 & 1 \\
\end{bmatrix}$.


Let $ a_1<a_2<a_3...<a_n<...$ be postive integers such that $ a_{2n}\equal{}a_n\plus{}n$ for $ n\equal{} 1,2,3,...$ It is known that if $ a_n$ is a prime number, then $ n$ is a prime number. Find, with proof, $ a_{1993}$.

Set $ a_1 \equal{} k$. Prove by induction that $ a_{2^n} \equal{} 2^n\minus{}1\plus{}k$. Indeed $ a_{2^{n\plus{}1}} \equal{} 2^n\minus{}1\plus{}k\plus{}2^n \equal{} 2^{n\plus{}1}\minus{}1\plus{}k$. And since $ a_1 \equal{} 2^0\minus{}1\plus{}k$ it holds.
$ a_{2^{n\plus{}1}} \minus{} a_{2^n} \equal{} 2^n$ and since $ a_1 < a_2 < a_3 < ...$ we have that $ a_n \equal{} n\minus{}1\plus{}k$ Which means that $ a_{1993} \equal{} 1992\plus{}k$ or $ \boxed{a_{1993} \equal{} 1992\plus{}a_1}$

but you also have to prove that $ a_1\equal{}1$

then $ n \equal{} a_{2n} \minus{} a_n \equal{} \sum_{j \equal{} n}^{2n \minus{} 1}a_{j \plus{} 1} \minus{} a_j \ge \sum_{j \equal{} n}^{2n \minus{} 1}1 \equal{} n$
so for $ j \equal{} n...2n \minus{} 1: \ a_{j \plus{} 1} \minus{}a_j\equal{}  1$
then $ \forall n\in\mathbb{N} \minus{} \{0\}: \ a_{n \plus{} 1} \equal{} a_n \plus{} 1$
so $ a_{1993} \equal{} a_1 \plus{} 1992$
and we know that $ a_n \equal{} a_{1} \plus{} n \minus{} 1$ prime ==> $ n$ prime is gives necceasy that $ a_1 \equal{} 1$
then $ a_{1993} \equal{} 1993$

[quote="(^_^)"]but you also have to prove that $ a_1 \equal{} 1$[/quote]
Forgot the sentence with prime numbers.. :s


function
Asymptote
limit
calculus
calculus computations

For f(x) = (2x + 5) / |3x - 4|, use graphs and tables to find the limit as x approaches infinity of f(x) and the limit as x approaches negative infinity of f(x)... Also identify any horizontal asymptotes...

When i graph the function in my graphing calculator, it looks like y will never reach a specific value as x approaches pos or neg infinity.. i went to table and checked at x value or 20000000 and -200000000 and i get 2/3 and -2/3 respectively

is this correct?
limit as x approaches pos inf = 2/3
limit as x approaches neg inf = -2/3

I'm not sure about the horizontal asymptotes?
would they be y = 2/3 and y = -2/3? I think? or is it something in between that?

Thanks for the input.

If we have $\lim_{x\rightarrow\infty}\frac{2x+5}{|3x-4|}$, then $x>0$, thus
$\\\lim_{x\rightarrow\infty}\frac{2x+5}{|3x-4|}\\ =\lim_{x\rightarrow\infty}\frac{2x+5}{3x-4}=\frac{a}{b}=\frac{2}{3}$
for $a$ from $2x$ and $b$ from $3x$, since the degree of both functions, $2x+5$ and $3x-4$, are the same.
Now, if we have$\lim_{x\rightarrow -\infty}\frac{2x+5}{|3x-4|}$, then $x<0$, thus
$\\\lim_{x\rightarrow -\infty}\frac{2x+5}{|3x-4|}\\ =\lim_{x\rightarrow -\infty}\frac{2x+5}{-3x+4}\\=-\frac{2}{3}$

Usually for such things, you can easily check on google.
Yes, your answer is correct, and the horizontal aymptotes are $y=2/3$ and $y=-2/3$.

Masoud Zargar


The next [url=http://imo2006.dmfa.si/]IMO 2006[/url] takes place from 6th July until 18th July in Slovenia. The contest questions are to be answered on July 12 (Wednesday) and July 13 (Thursday).

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=366[/img]

:lol: Good luck to all countries and their contestants :!:

[quote="zhaobin"][quote="esti"] Good luck to all countries and their contestants [/quote]
this may means good luck to no one :P  :lol:[/quote]
If all particpants would have 42 points, it would be better, if all participants would have 0 points.

[quote="zhaobin"]Yes,you are right,I just kidding.And will you go to?pavel kozlov(a member of Russian team on the IMO 2005) :)[/quote]
I know that now Pavel studing at an institut so........,
I saw that you studing at an institut too([quote="zhaobin"]I'm a university man in a not very famous university.I am interest in maths.I believe one coin have two sides.[/quote] :D ),but I am interesting did you be in any IMO?,if yes,what a medal did you take :blush: .

I haven't go to any Imo,and cmo,either :blush:,so no medal  :lol:
I know you will go to IMO 2006.And if you get a medal(I think you must will),can you give half of it to me :lol:

[quote="zhaobin"]I haven't go to any Imo,and cmo,either :blush:,so no medal  :lol:
I know you will go to IMO 2006.[/quote]
Oh,I know that in China that is most difficult to do :( ,but don't wory [b]you are wery good mathematic[/b] :lol:  ;) .
[quote="zhaobin"]And if you get a medal(I think you must will),can you give half of it to me :lol:[/quote]
If that will be again Bronze medal I will send you it [b]entirely [/b] :lol:.

[quote="Tiks"][quote="zhaobin"]I haven't go to any Imo,and cmo,either :blush:,so no medal  :lol:
I know you will go to IMO 2006.[/quote]
Oh,I know that in China that is most difficult to do :( ,but don't wory [b]you are wery good mathematic[/b] :lol:  ;) .
[/quote]
Thanks,but I think I should still work harder and harder :) And you have to.
[quote="Tiks"]
[quote="zhaobin"]And if you get a medal(I think you must will),can you give half of it to me :lol:[/quote]
If that will be again Bronze medal I will send you it [b]entirely [/b] :lol:.[/quote]
Wow,how many IMO's have you been?
I hope that will not be Bronze again.Then you can also send the Bronze to me :lol:

[quote="zhaobin"][quote="Tiks"][quote="zhaobin"]I haven't go to any Imo,and cmo,either :blush:,so no medal  :lol:
I know you will go to IMO 2006.[/quote]
Oh,I know that in China that is most difficult to do :( ,but don't wory [b]you are wery good mathematic[/b]   .
[/quote]
Thanks,but I think I should still work harder and harder , And you have to.
[quote="Tiks"]
[quote="zhaobin"]And if you get a medal(I think you must will),can you give half of it to me [/quote]
If that will be again Bronze medal I will send you it [b]entirely [/b] .[/quote]
Wow,how many IMO's have you been?
I hope that will not be Bronze again.Then you can also send the Bronze to me :lol:[/quote]
I think all of as still must work hard,if hi want get to the good rezults :)
I have been in tree IMO's :blush: ,and I have one honor(Japan) and two Bronze medals(Greese and Mexico) :( ,I hope that will be not bronze too :oops: ,(and not honor :lol: ).
Anyways I work hard for it ;) .

[quote="zhaobin"]You are very good you can go to IMO four times at least.Congratilation.Rememeber to tell me if the result of IMO 2006 is coming out.I hope it is good new. :)[/quote]
Thank you Zhaobin,but in Armenian that is not difficult to do :( :) ,and mybe I will go also to Vietnam(IMO 2007) :blush: ,of cours I will say you results,when I will know   ;) .

[quote="nonIMO"]Unfortunately, no link to the Google Maps provided on the website. But [url=http://maps.google.com/maps?f=q&hl=en&q=Ljubljana&ll=46.042549,14.489561&spn=0.001735,0.005364&t=k&om=1]we can find it ourselves[/url], don't we?  :?  :D[/quote]

Jajajaaj lol nice!

I must admit, that this website is much better, than any previous one. In particular, I like the way they presented the results. And photos of all team members... It's absolutely fantastic!

[quote="zhaobin"]You are very good you can go to IMO four times at least.Congratilation.Rememeber to tell me if the result of IMO 2006 is coming out.I hope it is good new. :)[/quote]
Dear $zhaobin$,I would like to keep my promes,and  tell you that fortunately this year my result was better than last tree,and I got silver medal with 24 point,I think it is a good result for a student from Armenia,becous laste silver medal we had in ... IMO 2000,so we were waiting for this medal 6 year :wink: .


[color=blue]¿Alguien conoce algunos métodos para aproximar el valor de $ \pi$?
Se ve interesante como tema de investigación para ratos libres.
Tengo en mente uno por cálculo de límites, tomando un polígono regular de muchos lados, aunque aún no he probado.

¿Existe alguno mejor?

Algo debe ocurrírseme para revitalizar una de las comunidades antaño más importantes de Mathlinks.[/color]

encontré un método muy interesante en un libro:

$ \pi \equal{} 4(\frac11 \minus{} \frac13 \plus{} \frac15 \minus{} \frac17 \plus{} ...)$

[quote="R.G.A.M."][color=blue]Se ve realmente muy interesante, especialmente por lo simple que aparenta ser.

¿Alguien tiene pruebas acerca de su precisión?[/color][/quote]La simpleza de este método está basada en la serie de Maclaurin (serie de Taylor centrada en 0) de la función arcotangente.

Esta igualdad dice que la serie del lado derecho converge a $ \pi$. Sin embargo, si por "precisión" entendemos "rapidez de convergencia", entonces los resultados no son muy buenos. Es decir, esta serie converge muy lentamente.

Existen muchos procedimientos para aproximar $ \pi$, y muchos de ellos son aún más rápidos.

Hay muchas identidades que envuelven a $ \pi.$

En cuanto a la serie, la escribimos como $ \sum\limits_{k\,\equal{}\,0}^{\infty }{\frac{(\minus{}1)^{k}}{2k\plus{}1}},$ y notando que $ \frac1{2k\plus{}1}\equal{}\int_0^1x^{2k}\,dx,$ la serie se reduce a resolver $ \int_0^1\frac1{1\plus{}x^2}\,dx$ que da el valor esperado.

(¿Está bien mi español?)

[quote="|-.-|"](¿Está bien mi español?)[/quote]
Está muy bien el español, mejor que el de la mayoría de nosotros... :lol:

Como mostró |-.-|, la serie $ \sum\limits_{k\, \equal{} \,0}^{\infty }{\frac {( \minus{} 1)^{k}}{2k \plus{} 1}},$ es equivalente al método del polígono del que hablaba [b]R.G.A.M.[/b] ya que es medir la longitud de la función $ f(x)\equal{}\sqrt{1\minus{}x^2}$.

Por otro lado tengo dos aproximaciones más:

$ \frac{\pi}{2}\equal{} \frac{2\cdot 2}{1\cdot 3} \cdot \frac{4\cdot 4}{3\cdot 5} \cdot \frac{6\cdot 6}{5\cdot 7}$...

$ \frac{\pi^2}{6}\equal{} \frac{1}{1} \plus{} \frac{1}{4} \plus{} \frac{1}{9}$...


geometry
circumcircle
perpendicular bisector

Let $ABC$ be an isosceles triangle $(BA=BC)$  . $(O,R)$ is the circumcircle of $\triangle ABC$ . It's known that : There exists a point $D$ inside $(O)$ such that $\triangle BCD$ is an equilateral triangle . $AD$ intersects $(O)$ at $E$ . Prove that : $DE =R$

Since $BA \equal{} BD \equal{} BC,$ it follows that $\triangle BAD$ is isosceles with apex $ B.$  Thus, $\angle BDA \equal{} \angle BAE \equal{} \pi \minus{} \angle BCE$ $\Longrightarrow$ $BCED$ is a kite $\Longrightarrow$ $BE$ is the perpendicular bisector of $DC.$ Consequently, $\angle COE \equal{} 2\angle CBE \equal{} 60^{\circ}$ $\Longrightarrow$ $\triangle OCE$ is equilateral with side lenght $R$ $\Longrightarrow$ $DE \equal{} EC \equal{} EO \equal{} R.$


geometry
3D geometry
analytic geometry
quadratics

Imagine a room, a cube, 10x10x10 metres. Inside the cube there is a tiny object, and I want to know precisely where that object is, in X-Y-Z coordinates. I have reference points in the middle of three faces (in the X, Y and Z planes, ie one each in the centre of two adjacent walls, and one in the centre of the floor), located at 0,5,5; 5,0,5; and 5,5,0 and I measure the distances from these points to the object (for example: 2.2m, 3.4m, 6.1m).

How do I convert the three lengths measured to X-Y-Z coordinates?

This is driving me crazy, and it DOES have a very real application.

Wibs

Sorry, here it is again, with attachment for clarity.

I say go with the tried and true method of setting up equations for the spheres.

If the three distances are $ d_{1}, d_{2}, d_{3}$, then
\[ (x\minus{}5)^{2}\plus{}(y\minus{}5)^{2}\plus{}z^{2}\equal{}d_{1}^{2}\]

\[ (x\minus{}5)^{2}\plus{}y^{2}\plus{}(z\minus{}5)^{2}\equal{}d_{2}^{2}\]

\[ x^{2}\plus{}(y\minus{}5)^{2}\plus{}(z\minus{}5)^{2}\equal{}d_{3}^{2}\]
Subtracting the third equation from the first and dividing by $ 10$ gives you $ x\minus{}y$.  You similarly subtract the third equation from the second and divide by $ 10$ to get $ x\minus{}y$.  Now express $ y, z$ in terms of $ x$, put it into one of these equations, and solve.

And, well...this doesn't belong in Olympiad Geometry, if someone wants to move it.

Let $ a \equal{} y\plus{}z$, $ b \equal{} z\plus{}x$, and $ c \equal{} x\plus{}y$; $ \alpha \equal{} x^{2}\plus{}y^{2}\plus{}z^{2}$.

After a scaling, the system is equivalent to $ \alpha\plus{}a \equal{} k_{1}$, $ \alpha\plus{}b \equal{} k_{2}$, and $ \alpha\plus{}c \equal{} k_{3}$.

This has a solution in quadratics after using $ \alpha \equal{} (s\minus{}a)^{2}\plus{}(s\minus{}b)^{2}\plus{}(s\minus{}c)^{2}$.