geometry
geometric transformation
rotation
superior algebra
superior algebra solved

Let $G$ be the affine group of a comutative field $k$ (that is, the group of mappings from $g$ from $k$ into $k$ which can be written in the form: $g(x)=ax+b$ where a=a(g) is nonzero), then for each triple $(f,g,h)$ of elements of $G$ such that $j=a(fgh)$ is not the identity and such that $fg$, $gh$, $hf$ are not translations, the following two assertions are equivalent.
a) $f^3g^3h^3=1$ (the identity mapping)
b) $j^3=1$ and $\alpha+j\beta+j^2\gamma=0$, where $\alpha$ is the unique fixed point of $fg$, $\beta$ that of the $gh$, $\gamma$ that of $hf$.

From this lemma follows Morley theorem which states that three intersection points of the trisectors of a triangle form an equilateral triangle.

Well, this is Alain Connes' proof for Morley's theorem. :)

With these keywords, you should be able to find it on the web, else, I'll post here my own proof of this result.

More exactly,

http://www.cut-the-knot.org/ctk/MorleysRedux.shtml

Napoleon theorem:

Given a triange $ABC$ and equilateral triangles $ABM$, $BCN$, $ACP$  erected on sides of $ABC$, Prove that $MNP$ is equilateral.

The problem is from some point of view similar with morley's theorem. I wonder if the same method, (the same lemma) doesn't work for this problem (for well chosen $g_i$).

I thought about choosing $g_i$ rotation of angle $\frac{2\pi } 3$ centers about vertex of $ABC$. Points $M,N,P$ are the fixed point of $g_1g_2$ and $g_2g_3$ and $g_3g_1$. And a relation like that in the point b) of the lemma would be helfull. The problem is that $g_1g_2g_3$ is a translation.

Can this solution be fixed? I see two ways to fix it :

1) to extend the lemma
2) to choose another tranformation which have $M,N,P$ as fixed points.

At a first look I do not expect to find $g_i$ which are not translations.

Thank you in advance,

Cristi


If a,b,c be Positive real number with:Sum(a.^2)=3,Then Prove That:
       (a/b)+(b/c)+(c/a)>=(9/a+b+c)
                                                                       Enjoy it!

Question $\sum a^{2}=3$ Prove $\sum_{cyclic}\frac{a}{b}\geq \frac{9}{a+b+c}$

Proof
$LHS=\sum_{cyclic}\frac{a^{2}}{ab}$
Use Cauchy, we have
$LHS\geq \frac{(a+b+c)^{2}}{ab+bc+cd}$
Let $x=\sum a$; $\sum ab=\frac{x^{2}-3}{2}$
Then it's to prove $\frac{2x^{2}}{x^{2}-3}\geq \frac{9}{x}$
$\iff 2x^{3}-9x^{2}+27\geq 0$
$\iff (x-3)^{2}(2x+3)\geq 0$ :)


equation
integer equation
Germany
TST
Team Selection Test

Let the positive integers $x_1$, $x_2$, $...$, $x_{100}$ satisfy the equation
\[\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+...+\frac{1}{\sqrt{x_{100}}}=20.\]
Show that at least two of these integers are equal to each other.

Let $ xn \geq ... \geq x2 \geq x1 $
We have $ 20 \geq 100/ \sqrt  x1 $
So $ 5 \geq \sqrt x1 $
Again $ 19 \geq 99/ \sqrt x2 => 5 \geq \sqrt x2 $
We can prove that $ 6 \geq  \sqrt x7$ too.
So we have seven integer numbers which less than 6.
That means at least two of these integers are equal to each other. :)

A slight different answer : It is easy to prove that $ \frac 1 {\sqrt n} < 2 \sqrt n - 2 \sqrt{n-1}$, from which we deduce the well-known inequality $ \sum^n_{k = 1} \frac 1 {\sqrt n} < 2 \sqrt n - 1$.
Now suppose that the $x_i^$'s are pairwise distinct, then
$ 20 = \sum \frac 1 {\sqrt {x_i}} \leq \sum^{100}_{k=1} \frac 1 {\sqrt k} <2 \sqrt{100} - 1 = 19$. A contradiction.

Pierre.

This was used in the Mathematical Olympiad Correspondence Program in Canada some years ago.


limit
inequalities
algebra
binomial theorem
real analysis
real analysis unsolved

Expand out the product:

$(n+a_1)\cdots(n+a_p)=n^p+(a_1+\cdots+a_p)n^{p-1}+O(n^{p-2})$

$=n^p\left(1+(a_1+\cdots+a_p)n^{-1}+O(n^{-2})\right)$

Take the $p$th root and use the binomial theorem:

$\left[(n+a_1)\cdots(n+a_p)\right]^{\frac1p}=n\left(1+\frac1p\cdot\frac{a_1 +\cdots+a_p}n+O(n^{-2})\right)$

$=n+\frac{a_1+\cdots+a_p}p+O(n^{-1})$

Subtract $n$ and take the limit to get the answer of

$\frac{a_1+a_2+\cdots+a_p}p$

Note that the AM-GM inequality says that we must be less than or equal to this number. As $n\to\infty,$ we approach the bound given by AM-GM as the limit. This makes sense, since the condition for equality in the AM-GM inequality is that the terms be equal, and (relatively), as $n\to\infty$ the terms get more and more alike.

My solution is different . I used Bernuli's ineq :
    $ (1+\frac{a_1}{n})(1+\frac{a_2}{n})...(1+\frac{a_p}{n}) \geq 1+\frac{a_1+a_2+..+a_p}{n} $
   and the lemma : $ \lim_{x\to 0}\frac{(1+x)^m-1}{x} = m $

$\displaystyle \lim_{x\rightarrow \infty}\frac{f(x)}{x^p}=1 \Longrightarrow  \lim_{x\rightarrow \infty}(\sqrt[p]{f(x)} -x)  = \lim_{x\rightarrow \infty}\frac{f(x)-x^p}{px^{p-1}}$
 :cool:


Find all polynomials $ P(x,y)\in R[x,y]$ such that for all $ x,y\in R$,

$ p(x\plus{}y,x\minus{}y)\equal{}2p(x,y)$


Prove that
\[ \prod_{n=2}^{\infty}\left(1-\frac{1}{n^2}\right)=\frac{1}{2} \]

Masoud Zargar

[hide]The product is \[ \left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{3^2}\right) \left(1-\frac{1}{4^2}\right) \cdots,  \] or \[ \left( \frac{2^2-1}{2^2} \right)\left( \frac{3^2-1}{3^2} \right) \cdots,  \] and factoring the numerators, we get \[ \frac{(1)(3)\cdot(2)(4)\cdot(3)(5)\cdots}{2^23^24^2\cdots} , \] and lots of terms cancel. We're left with \[ \frac{1}{2},  \] as desired.[/hide]

it is just (n-1)(n+1)/n^2, and it cancels

Yes.  I just factored and created two different product functions.  Then they could be simplified and multiplied together.  Then you get $\frac{1}{2}+\frac{1}{\infty}=\frac{1}{2}$.  I'll be back with a complete solution. :)

Masoud Zargar


search
geometry
circumcircle
rectangle
geometry proposed

Let $ABCD$ be a convex cyclic quadrilateral. I note $AB=a$, $BC=b$, $CD=c$, $DA=d$. Prove that for any point $P$ exists the folowing relation:

$\frac{bc\cdot PA^2+ad\cdot PC^2}{bc+ad}=\frac{cd\cdot PB^2+ab\cdot PD^2}{cd+ab}$.

[u]Remark.[/u] Search several particular cases ($P\in \{A,B,C,D,O,E,\ldots\}$), where the point $O$ is the circumcenter and  $E\in AC\cap BD$.

After two years this topic is  neatly. It is one from the first my topics on this side. "Kill" it, please !

We use vectors.  WOLOG, assume that points $ A$, $ B$, $ C$, $ D$ lie on a circle centred at the origin with radius $ R$.  Then
\begin{align*}
\frac{bc \cdot PA^2 + ad \cdot PC^2}{bc + ad} &= \frac{bc |\vec{P} - \vec{A}|^2 + ad |\vec{P} - \vec{C}|^2}{bc + ad} \\
&= \frac{bc (|\vec{P}|^2 - 2 \vec{A} \cdot \vec{P} + |\vec{A}|^2) + ad (|\vec{P}|^2 - 2 \vec{C} \cdot \vec{P} + |\vec{C}|^2)}{bc + ad} \\
&= \frac{(bc + ad) |\vec{P}|^2 - 2 (bc \vec{A} + ad \vec{C}) \cdot \vec{P} + (bc + ad) R^2}{bc + ad} \\
&= |\vec{P}|^2 - \frac{2 (bc \vec{A} + ad \vec{C})}{bc + ad} \cdot \vec{P} + R^2.
\end{align*}
Similarly,
\[ \frac{cd \cdot PB^2 + ab \cdot PD^2}{cd + ab} = |\vec{P}|^2 - \frac{2 (cd \vec{B} + ab \vec{D})}{cd + ab} \cdot \vec{P} + R^2.\]
Hence, it suffices to show that
\[ \frac{bc \vec{A} + ad \vec{C}}{bc + ad} = \frac{cd \vec{B} + ab \vec{D}}{cd + ab}.\]

Let $ X$ be the intersection of $ AC$ and $ BD$.  Then by similarity of triangles $ XAB$ and $ XDC$, and $ XBC$ and $ XAD$, $ XA: XC = ad: bc$ and $ XB: XD = ab: cd$.  It follows that
\[ \vec{X} = \frac{bc \vec{A} + ad \vec{C}}{bc + ad} = \frac{cd \vec{B} + ab \vec{D}}{cd + ab}.\]

If ABCD is a rectangle then we have
   PA^2 + PC^2 = PB^2 + PD^2

   For non-rectangle it has to be somewhat adjusted,
   so I guess with some help from Ptolemy, the factors
   in Virgil's  enunciation do this adjustment job
   cyclicaly and rather nicely.

   T.Y.
   M.T.

[quote="Virgil Nicula"][color=darkred] Let $ ABCD$ be a convex [b]cyclic[/b] quadrilateral. I denote $ \{\begin{array}{c} AB = a\ ,\ BC = b \\
\ CD = c\ ,\ DA = d\end{array}$ .

Prove that $ (\forall )\ P$ we have $ \frac {bc\cdot PA^2 + ad\cdot PC^2}{bc + ad} = \frac {cd\cdot PB^2 + ab\cdot PD^2}{cd + ab}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] For $ E\in AC\cap BD$ denote $ \{\begin{array}{c} AC = e\ ,\ BD = f \\
\ AE = x\ ,\ BE = y \\
\ CE = z\ ,\ DE = t\end{array}$ . Prove easily that

$ \frac {x}{ad} = \frac {y}{ab} = \frac {z}{bc} = \frac {t}{cd} = \frac {e}{ad + bc} = \frac {f}{ab + cd} = \sqrt {\frac { - p_w(E)}{abcd}} = \lambda\ \ (*)\ .$

where $ p_w(E)$ is the power of the point $ E$ w.r.t. the circumcircle $ w=\mathcal C(O,R)$ of $ ABCD$ .

Apply the [b]Stewart's relation[/b] to the cevian $ PE$ in the triangles $ APC$ and $ BPD$ :

$ \{\begin{array}{c} z\cdot PA^2 + x\cdot PC^2 = (x + z)\cdot PE^2 + xz(x + z) \\
 \\
t\cdot PB^2 + y\cdot PD^2 = (y + t)\cdot PE^2 + yt(y + t)\end{array}$ . Using $ (*)$ obtain :

$ \{\begin{array}{c} bc\cdot PA^2 + ad\cdot PC^2 = (bc + ad)\cdot PE^2 + abcd(bc + ad)\cdot\lambda ^2 \\
 \\
cd\cdot PB^2 + ab\cdot PD^2 = (ab + cd)\cdot PE^2 + abcd(ab + cd)\cdot\lambda ^2\end{array}$ $ \implies$

$ \boxed {\frac {bc\cdot PA^2 + ad\cdot PC^2}{bc + ad} = \frac {cd\cdot PB^2 + ab\cdot PD^2}{cd + ab}} =$

$ PE^2 + abcd\cdot\lambda ^2 = PE^2 - p_w(E) = PE^2 - OE^2 + R^2\ .$

[b][u]Remark.[/u][/b] $ \{\begin{array}{ccc} P: = A & \implies & e^2 = \frac {ad + bc}{cd + ab}\cdot (ac + bd) \\
 \\
P: = B & \implies & f^2 = \frac {ab + cd}{ad + bc}\cdot (ac + bd)\end{array}\|$ $ \implies$

the [b]Ptolemaeus' relations[/b] $ \{\begin{array}{cc} 1\blacktriangleright & \boxed {ef = ac + bd}\ . \\
 \\
2\blacktriangleright & \boxed {\frac ef = \frac {ad + bc}{ab + cd}}\ .\end{array}$[/color]


When do we say that two measures are [b]measure theoretically equivalent[/b]?

Can someone please give the definition. I am still not getting it!

Measures $ \mu$ and $ \nu$ are called equivalent if they have the same null sets.

Read more on [url=http://en.wikipedia.org/wiki/Equivalence_(measure_theory)]wiki[/url].

(Note: googling on "measure equivalence" is your friend.)

Or is this not what you meant?

Two meausres on a set $ S$ are equivalent if they both give rise to the same open sets that is iff a set $ A$ in $ S$ is open under measure $ \mu$ then it should be open under measure $ \nu$ and vice-versa,then $ \mu \equiv \nu$

What do you mean by open under a measure?

Pardesi appears to be confusing a "metric" with a "measure."  The condition listed is then the condition for two metrics to be (topologically) equivalent.


geometry
MATHCOUNTS
ratio
inradius
trapezoid
trigonometry
similar triangles

http://mcis.jsu.edu/mathcontest/geo02.pdf

$42,43,44,46,48,50$

I only got 39/50.

I think I messed up on some similar triangles stuff again. I couldn't find elegant solutions for these.

I HATE similar triangle stuff that I can't see. Still getting used to it. iI guess it's part of the learning process.

25,29,31,39,40 are CORRECTED  :lol:

I'll do more later, too tired.
[hide="29"]We have similiar triangles which are triangles AED to triangle ACB.So $\frac{3}{15}$=$\frac{2}{x}$. Solving for x which is the side length of CB is 10.[/hide]

lol...thanks for posting these tests.

25.

[hide]$7^{3}=343$.

there are $5\times 3=15$ holes drilled through so that makes $15\times 7=105$ units lost.

however, there are 9 places where the holes intersect 3 times so add $9\times 2=18$.  $343-105+18=256$

$\boxed{D}$[/hide]

31.

[hide]Drawing AC and BC, we see that it is just composed of two 1/6 circles minus $[ABC]$.

$2\cdot\frac{4^{2}\pi}{6}-\frac{4^{2}\sqrt{3}}{4}=\frac{16\pi-12\sqrt{3}}{3}$

$\boxed{A}$.

BTW. if you dont already know, the area of an equilateral triangle is $\frac{s^{2}\sqrt{3}}{4}$ where $s$ is the side of the triangle.  i dont like memorizing formulas a lot but this is one you should definetly remember (especially if ur in mathcounts)[/hide]

post more later.  gonna watch the apple :D
HAPPY NEW YEAR!
actually i think it already is new year for the east coast.  oh well, $1\frac{5}{12}$ more hours to go.

HAPPY NEW YEAR TO YOU TOO!  :lol:

Well, I figured out why I missed 29. I thought they were asking for length of a hypotenuse so I just didn't see the answer.

31, I think I didn't mean to post that number. oh well.

Well I still got a good bit more left.

Any more solutions?  :lol:

25 is done

[hide="39"]since the ratio of the the areas is the square of the ratio of the sides, we have that it is $\sqrt{9}=3$. so since the side length is 1, the base of the smaller triangle is 1/3. the altitudes have this same ratio too, so we have x+3x=1$\rightarrow$ x=1/4. the areas of the triangles are (1/3*1/4)/2=1/24 and (1*3/4)/2=3/8 so 3/8+1/24=5/12, E [/hide]

[hide="47"]this is simply the diagonal of a 4 by 5 rectangle. $\sqrt{41}$[/hide]

[/hide]

40.

[hide]We know $OF=4$ because it is one half of $AC$.  $FE=\sqrt{6^{2}-4^{2}}=2\sqrt{5}$.  $\triangle AFE~\triangle OAE$.

$\frac{4}{2\sqrt{5}}=\frac{OA}{6}\Rightarrow AO=\frac{12\sqrt{5}}{5}$.

$\triangle AOE~\triangle DPE$.  The ratio of the sides is 6 to 1 so $DP$ is one sixth of $AO$.  The difference of the radii is just five sixths of $AO$ which is $2\sqrt{5}$.

$\boxed{D}$[/hide]

note: I was purposefully brief so you have to work at the problem yourself.

42 [hide="hint"]is a nice problem...show that $\triangle EAD\sim\triangle \triangle ECA$, and others....you should be able to find AC...[/hide]

43 [hide]find the length of their common external tangent, and use the similarity to find the lengths coming from vertex A, then use the pythagorean theorem to find the tangent from C to the large circle, then use semiperimeter *inradius =area[/hide]

44 [hide]let AF=x, and FB=1, then CD=x+1, and use the trapezoid formula, and solve for x[/hide]

46 [hide]find a relation between the distance between centers, the radii, and the length of the common external tangent...[/hide]

48 [hide]if x is a side of the octagon, then $x=2\sin \frac{45}{2}$, , then the area is $8*\frac{x^{2}}{4}$...use double angle formulas...[/hide]

50 [hide]suppose it took 5 seconds to get from A to E, then the horizontal velocity is 3ft/s, and the vertical velocity is 4ft/s, to hit AD, it must travel 20ft horizontal...20/3 sec, that means teh vertical distance traveled was 80/3, but we have to account for bouncing, we can subtract 10s...to get 20/3 and that going up subtract 5, then going down, 5/3, so the answer is 5-5/3=10/3[/hide]

assuming that the ball doesn't lose speed, [hide="50"] We can use similar triangles over an over again until the distance of the ball to C is 1. Then, by similar triangles, when the ball hits BC, it is 5/3 away from C. When the ball hits AB again, it is 2 units away from B. We bounce it off of CDand back to AB, and then it is 2 units away from A. When the ball hits AD, it is 10/3 away from A.

C[/hide]


Just wondering, about how many people participate in USAMTS?  Does anyone have a rough figure or something?

If you look at the overall statistics, there's about 500 participants this year..

What percentage of them make it to AIME?

500 is really a lot of people.

My impression of the leaderboard right now is that a ton of people have over a 43, which is the minimum right now to meet the 68 cutoff, so I'm guessing 15-20%??


Let $ f$ a continous function in $ R$
$ k$ a positive real number

and:  $ |f(x) \minus{} f(y)|\geq k|x \minus{} y|$

Prove that f is a bijection.

Injective is clear
But surjective....

[hide="sketch of a solution"]
I don't have a lot of time at the moment, but if you divide each side by |x-y| then you get $ \left|\frac {f(x) \minus{} f(y)}{x \minus{} y}\right| < k$.  Using this you can show that the function is always increasing or the function is always decreasing.  Suppose that $ f$ isn't bijective.  In the case for always increasing consider the set $ S$ = {all points that aren't in the image of f}.  Since $ f$ is increasing, continuous, if any real is in $ S$ then all smaller reals are also in $ S$.  Since $ f$ isn't bijective, $ S$ is non-empty, so $ S$ has a supremum, $ a$.  Then since $ f$ is increasing there has to be some real number $ r < a$ such that $ r$ is in the image of $ f$ (I'm not quite sure how to make this rigorous).  but because $ r < a$, $ r \in S$ and because $ r$ is in the image of $ f$, $ r \notin S$, which is a contradiction.  The case for decreasing is similar.

I think that's one approach to take, but it would take some work to make it rigorous... maybe?

[/hide]


Let $a;b$ be real numbers satisfying:$a^{3}+2b^{2}-4b+3=0$ và $a^{2}+a^{2}b^{2}-2b=0$
Caculate the value of: $A=a^{2}+b^{2}$

Locked, see http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 .


Let q a positive rational numer.Prove that there exists an infinity of natural numbers n such that [ nq] and n are coprimes.

The following proof shows that all we need to assume is that q is not an integer. Let P_1=2 p_2=3 and so on the sequence of prime. Suppose the conclusion is false. Then we can find k such that for any n>k we have p_n| [qp_n]. The sequence [qp_n]/p_n tends to q and it's a sequence of integers. So it is stationary and its limit must also be an integer. Therefore q must be an integer.

assume that q=a/b such that b>1 and (a,b)=1 so we have x,y s.t:ax-by=1 so [(bm+x)a/b]=[am+y+1/b]=am+y .note that (a,y)=1 so as a result of Drichlit theorem am+y for infinitly many m is equal prime number
so we prove that for infinitly many n ,[nq] is prime number, I think that we are done! am I wrong :D


algebra
polynomial
superior algebra
superior algebra unsolved

Let f(x) in k[x] be an irreducible polynomial with coefficients in a field k.  Suppose that f(x) has a root of multiplicity greater than one in some extension of k.  We want to prove char(k)= p for some prime p and f(x)=g(x^p) for some g(x) in k[x].

We want to find all the values of a in Z for which the polynomial f(x)=x^5-ax-1 is not irreducible in Z[x].  For each such a find the corresponding factorization of f(x) into a product of irreducible polynomials in Z[x]

At first, note that having a multiple root at $x=a$ means that $f(a) = f'(a) =0$. So when $f$ is irreducible, so when $f$ is the minimal polynomial of $a$, we can only have $f'(a)=0$ when $char(k) \neq 0$, since otherwise $f'$ would be nontrivial for sure and would have a lower degree (and characteristics are prime when not $0$).
Similar, when you look at the derivation $f'(x)$, it has to be the zero-polynomial (otherwise it would again have lower degree than the minimal polynomial, contradiction).
But when the coefficient $a_n$ of $x^n$ is not zero, then by $(x^n)' = nx^n$ we need that $n=0$ (seen in the field, not in integers), thus $p|n$. This gives the second part of the first question.


Find all integer solutions to $2(x^5 +y^5 +1)=5xy(x^2 +y^2 +1)$.

I have an ugly way to solve this problem.But I believe it is not the most beutiful and simple one.I am busy these days,Maybe I can post the complete solution some days.But now only an hint:
Notice that the degree of LHS and RHS is 5 and 4,so we may use some inequality:
Case 1.$x<0,y<0$,then  $LHS<0,RHS>0$
Case 2.$x\ge 3,y\ge3$,$LHS\ge RHS$
Case 3.$x>0,y<0$.first we get $x+y\le-1$,and we can deduce that there must exist $A$ such that when $x\ge A$ $LHS<RHS$( I didn't compute carefully,maybe$A=4$ can work.
The other case is easy.(we only need to solve a few equation with one variable.)
I am sorry for the not complete solution :oops:

[quote="Hawk Tiger"]I have an ugly way to solve this problem.But I believe it is not the most beutiful and simple one.I am busy these days,Maybe I can post the complete solution some days.But now only an hint:
Notice that the degree of LHS and RHS is 5 and 4,so we may use some inequality:
Case 1.$ x < 0,y < 0$,then  $ LHS < 0,RHS > 0$
Case 2.$ x\ge 3,y\ge3$,$ LHS\ge RHS$
Case 3.$ x > 0,y < 0$.first we get $ x \plus{} y\le \minus{} 1$,and we can deduce that there must exist $ A$ such that when $ x\ge A$ $ LHS < RHS$( I didn't compute carefully,maybe$ A \equal{} 4$ can work.
The other case is easy.(we only need to solve a few equation with one variable.)
I am sorry for the not complete solution :oops:[/quote]
Your solution is not full.
Try again with result:
If $ x\plus{}y\plus{}z\equal{}0$ then $ 2(x^5\plus{}y^5\plus{}z^5)\equal{}5xyz(x^2\plus{}y^2\plus{}z^2)$
Imply that it contain term $ x\plus{}y\plus{}1$ .

Solution was posted here:
http://www.mathlinks.ro/viewtopic.php?t=166171

$ 5|x^5 + y^5 +1 $ which is impossible because $x^5$ mod 5 can be just 1 or 0


Now that its April 3, I thought a thread for state results would be good. Feel free to list score, rank, school, name, comments ect.

I think this is one of the lowest scoring competitions ever in Texas. 32 was the cut off for the top 10 and the best score was a 46 by Mark Zhang.

[quote="oneal3691"]I got 4th at chapter and got 2nd at state.
I live in Huntsville, AL so i competed in the Alabama state competiton.
I originally got 8th at state, and 8th in countdown, but after the competition, I was looking at the scoreshhet and realized that they had switched my scores with the girl who got 2nd place(also from my school) who was busy celebrating her trip to nationals. So unfortunately for her, we showed the problem to the state coordinator, and they finally decided to throw out the countdown round completely, and base the results on the written test, so I got second, and I'm going to nationals. I'm in the 8th grade.[/quote]
Weird. We had a similar problem where for countdown, two people had the same name and the wrong one went up. I think the name was like Xing-Xao. Anyway, its been a month and I havent gotten my scores back  :(  :(  :(  :(  :( . But I know my scores were bad. I got a 4 :( :( :( :( :( :( :( :( :( :( :(  on target and <24 on sprint :( :( :( :( :( :( . Thats a total of <32 :( :( :( :( :( . Maybe its because I prepared for like 1 hour total and only did school competitions and never got better than 40 and only had 5 (really) practices this school year. O well, at least I got 4th written and 1st countdown :first: official! :first:  :first: . See you at nationals!!

The cut for the countdown was a 34
hehe....
I'm from Fort Settlement
We got first at state and our individuals scores were 46, 45, 45, and 43 but the people going to nats got 46, 45, 45, and 45. We got a perfect score on the team round.  :)

[quote="tarquin"]I was crying after the sprint round because I thought I had lost the chance for nationals, probably barely above 20. Then after that it makes sense that I did so bad at target getting only 5 right. Cut off was a 34 so I probably made a 33 or a 32. When I found out the cutoff was so low, I was very, very, mad. Of course Fort Settlement owned the competition anyway... that's the best team I've ever seen at state...[/quote]Did you make Nationals?

O so thats a "no". At least theres next year if your in 6th/7th grade

[quote="Magixmaniak"]yes i did[/quote]Score, state?

[quote="tarquin"]Well the cutoff for [b]top ten[/b] was a 34. I got caught in some huge tie and made 15th... which is kind of sad...[/quote]Wow, that must have been a really hard test.  I remember two years ago I got a 37 or 38 and ended up 31st.  But those nationals scores...46,46,45,45...wow...

texas, and i got fourth, so whatever score corresponds...

It was more difficult then most. But I swear I should have gotten like a 43 or something in that range... The scores were so messed up...!

[quote="Secret Asian"][quote="mathgeek2006"]I'm not sure what my score was, (need to ask my coach still) but I know I got 5th individual and 2nd Countdown (Utah)--I think I got like a 32 or something...I'm pretty good at Countdown--If only I had a chance to get in there at nats.... :([/quote]

Um..32??  I tied for first in Utah and I had [hide][hide][hide]30:([/hide][/hide][/hide]
You were 5th place.  So you couldn't have gotten over my score.[/quote]

[hide="top secret information"][hide]It was a J-O-K-E
I wanted to see if people actually believed it....[/hide][/hide]

In Nevada, I came in 6th in chapter with a score of 31 (21/5). :?  However, I came in first at state with a 34 (24/5). :)

Massachusetts, my state got a 44, 43, 41, and 39 I think,

I came in second

Well unlike 2003, where you had Adam Hesterberg who was 5th as a seventh grader in 2002, or 2004, where you had Greg Gauthier who was 6th written as a seventh grader in 2003, there aren't any real big favorites. Sergei and David slipped into the countdown, but they got beat by several people at states. This year will be more open, and we'll just see who's better prepared and luckier.

Well, Sergei's slow with countdown, as I've beat him twice this year, and on chapter we both got that poorly worded #29 on sprint, problem wrong. On state his mistakes were just two stupid addition mistakes.

Don't worry about it.  Although I usually don't make too many stupid mistakes on test day, there's a lot of luck involved in MC.  You know, two years ago, I was realistically aiming at top 20 at nats, but of course I died on test day and got 50th or something.

[quote]nat mc
PostPosted: Mon Apr 11, 2005 4:27 am    Post subject:
Mark Zhang will be owned, he won't win[/quote]

Mark never gets owned except once by dennis (chapter)

Phelpedo, you did the same thing wrong as I did. The roles of the actors/actresses DID matter, so you wouldn't choose, you would permutate, or whatever the word is. Hope that helps.

[quote="Phelpedo"]I think that I got number 29 wrong but not because of the wording. I did 5c2= 10, 6c2 = 15, and then there are 7 left, 7c2 = 21. Then I multiplied these together, but got it wrong. Can someone suggest a reason why this approach is wrong?[/quote]

sirorange is right, since roles are different, order does matter, so you have to add in duplicates, (if man 1 and man 2 were chosen, you could choose man 2, and man 1 and that would be different), so you could just use the counting principle  5 men, 6 women, 2 male roles, 2 female roles, 2 interchangeable roles... 5 men for first role, 4 for second, 6 for first women role, 5 for second and (5+6-4  roles that have been cast) gives you 7 for the first interchangeable role, and 6 for the second interchangeable role...so basically 5*4*6*5*7*6=25200 ways to cast

Yeah.... I got that one wrong too, I thought the roles weren't distinct, so I didn't multiply by that extra 8.

9 people got 46s at state? I hate to think that there are liars on AoPS...


Given a line segment where you know both endpoints, and it is not vertical or horizontal, how would you find the midpoint?
Thanks,

Say the endpoints were $ (a,b)$ and $ (x,y)$.  The midpoint would be $ \left(\frac{a\plus{}x}{2},\frac{b\plus{}y}{2}\right)$.  Try proving it.

Thanks,
Hmm...thats odd, its like averaging the x and y values.

You can always draw a horizontal and a vertical leg, forcing this line segment to be the hypotenuse of a right triangle. The midpoint in the x direction would be the average of the x values (think of it as a number line ranging from 3 to 7). This also applies for the y direction.

[quote="eli140"]
Hmm...thats odd, its like averaging the x and y values.[/quote]

Exactly. If you think about it, that's really what a midpoint is. Directly between the line $ x \equal{} x_1$ and $ x \equal{} x_2$, as well as $ y \equal{} y_1$ and $ y \equal{} y_2$.

izzy is correct, midpoint mainly means like the middle of the point or the point where the distance from the other points is the same so averaging out the x's and the y's coordinates would help u find the midpoint


This is kinda related to the forum...I'm trying to make an avatar, but the program only uses .png, and I want to convert to .jpg to save size. I can't open it in paint to convert it and the program will only let you save as .png (fireworks). Does anyone know of a free program that will do this for me?

If you have Fireworks, you can use the 'Export' function to create a gif or jpg or whatever.  I think it's in the File menu -> go to Export Wizard (or something like that) and follow the directions.


The tallest inhabited building in the world is the Sears Tower in Chicago.  If the observation tower is 1450 feet above ground level, determine HOW FAR a person standing in the observation tower can see (with the help of a telescope) keeping in mind that the Earth's radius is 3960 miles.

HINT GIVEN:

1 mile = 5280 feet.

please don't ask for answers to your homework assigments on this board, however you are allowed to ask for tips...

[hide="tip"]theroritacally, if her looked up, then he could see for millions of miles.  Is the question "what is the furthest thing away he can see on the earth"? then, I would recomend using triangles.[/hide]

How do you know this is homework?

-especially, since [i]most [/i]schools are out already.

[hide="More Hints"]
The farthest you can see is the line tangent to the Earth's surface.

Let triangle $ABC$ be the triangle formed by joining the top of the Sear's tower $A$, the farthest you can see $B$, and the center of the Earth $C$. Then, angle $ABC$ is a right angle.

Using the values you know, solve for the side $\overline{AB}$. Depending on how precise you want your answer, you can solve for the triangle formed by the base of the Sear's Tower and $\overline{AB}$, though the difference between the two is probably negligible.
[/hide]

Here's a related [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1314609914&t=13321]problem[/url] that I posted several years ago.

Whenever I post questions here, I seek tips only and nothing more.

Thanks

Can this question be solved using the Pythagorean Theorem?

Yes, if you look at my previous explanation, it is evident that $\overline{AB}^{2}+\overline{BC}^{2}= \overline{AC}^{2}$ Plug in the values you know, and you have the answer.

AB = Distance from person to farthest away he can see (What you are solving for)
BC = Distance from farthest away person can see to center of the Earth (Earth's radius)
AC = Distance form person to center of the Earth (Earth's radius + Sear's Tower)

Similar problem here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152767

[quote="sharkman"]Gyan,

You link really helped me to fully grasped this question.

I know how to use a^2 + b^2 = (a + b)^2

Thanks[/quote]

wtf that's not true

[quote]I know how to use a^2 + b^2 = (a + b)^2 [/quote]

$a^{2}+b^{2}= (a+b)^{2}-2ab$. Just in case.  :D

I was looking at what I'd ever posted in the high school program, and came across this topic from many years ago. I decided to revive it with two comments.

1. The building formerly known as the Sears Tower is now officially called the Willis Tower. (I last lived in Chicago more than 35 years ago, but even so I have a sneaking suspicion that quite a few current residents of Chicago don't use the new name for the building.)

2. But at the same time, the Willis/Sears Tower is now nowhere close to being the tallest inhabited building in the world. So perhaps this problem should be re-posed, but with the building being the Burj Khalifa in Dubai. According to what I could look up:

Height of observation deck: 555 m (1821 feet)
Highest occupied floor: 584.5 m (1918 feet)
Top of spire/antenna: 829.8 m (2722 feet)


Take a parallelogram as in the figure. Pick any point in the interior and draw lines from that point to each of the four corners. Color opposite triangles red and blue (as in the figure). Use some basic geometry to show that no matter where the point is placed the total area of the two red triangles will equal the total area of the two blue triangles.

You can show that the sum of the areas all equal to half of the paralellogram.

yes(it's enough to choose  the point on the one of vertices) ,and i think it's very simple but nice problem !

Or if you want, draw a horizontal line through the point in question.

Same stands for a square. And we can use that property to prove the law of cosines.

i think the proof is obvious !
i have posted a same post with some differences for regular triangle i think.

I construct altitudes of the triangles through the point......But i think yours is better. :D

i think that proof works for every n-regular.


Let $S \in \mathbb{Q}$ such that:
[list]1. $\frac{1}{2} \in S$
2. If $x \in S$, then both $\frac{1}{x+1}$ and $\frac{x}{x+1} \in S$[/list]

Prove that if $k \in \mathbb{Q}^{(0,1)} $ then $k \in S$

[quote="Myth"]BTW. kueh! Do you have pdf?[/quote]

There is are pdfs of all the past british mo 1 and 2 on the official website: http://www.bmoc.maths.org/home/bmo.shtml

[quote="zhaobin"][quote="kueh"][quote="zhaobin"]I think it can be done by induction. :)[/quote]

Induction on?[/quote]
I have done it for some time.now let me raise my momery.
for a rational number $\frac{m}{n}\in(0,1),m<n,n\geq2$
for $2,3,\cdots,n-1$ is right.
than let $x=\frac{n-m}{m}\in S$ or$x=\frac{m}{n-m}\in S$
we can  $\frac{m}{n}\in S$[/quote]

oh that solution is nicer than mine. but mine was crazy - i showed for any p/q, i can do a list of functions on 1/2 as shown such that the result is p/q.


Fie $ n\in N$ si $ z\in C\minus{}R$ astfel incat $ z^n$ si $ (z\plus{}1)^n$ sunt numere reale.Determinati $ z$.


check this out

[url]http://www.1800gotjunk.com/genie[/url]

at first, i was amazed but then i realized the "trick"
how long did it take you to realize what they were doing?

Doesn't it say to only use a two-digit number?

But yea, it took me only a couple of seconds to get what they were doing.

[quote="mbabbitt2003"]I know how to make them WRONG.
Think of the same number over and over again.[/quote]

nope that's not it. unless you mess up in your arithmetic :rotfl: they'll be correct.

[quote]Think of the number 100.[/quote]

nope. must be a 2 digit number.

ah, i get it.  notice how some of the junk items are repeated in a predictable way on that grid...

there are certain "junk numbers" and those are the only ones you can choose. So, they keep the same junk on these junk numbers. pretty clever.

[quote="randomdragoon"]100 is two digits in base 11...[/quote]

No, 91 is two digits in base 11.  100 is always three digits (100 in base 11 is the same as 121 in base 10).

Ohhhhhh. Now I get it.
[hide]For those of you who don't, try subtracting the units digit from the two digit number first.[/hide]

Fine. $100_{10}$ is two digits in base 11...


linear algebra
matrix
induction
vector
linear algebra unsolved

Prove that $I_2$ can't be written as a finte sum of nilpotent matrix. :roll:

[quote="alekk"]trace ...[/quote]

 How by passing to "trace" directly?

THe trace of a nilpotent matrix is 0, the sum of the traces is the trace of the sum and the trace of the identity is not 0. This is what Alekk said. Nice problem: what matrices are linear combinations of nilpotent matrices? ;)

[quote="harazi"]Nice problem: what matrices are linear combinations of nilpotent matrices?[/quote]

The matrices with trace zero, right? :)

Guys, the question was for Cezar. ;)  :P

[quote="perfect_radio"]all the matrices with zero trace? if so, why?[/quote]

you break a matrix into two parts. One is the upper triangular part, one is the lower triangular part.

Not quite- you need another part for the diagonal. The matrix $N_d=\begin{bmatrix}1\\1\\ \vdots\\1\end{bmatrix}\cdot[d_1,d_2,\dots,d_n]$ where $d_j$ are the diagonal element works well here- then subtract off $N_d$ and take the upper and lower triangular parts of what remains.

to repare your proof you just have to proove that if a matrix has zero trace, it is similar to a matrix whose elements on the diagonal are null. (induction works very well ..)

when you say similar. are talking about $A=P^{-1} B P$. If so, how does that help?

In today's local olympiad, I got the following problem:

[i]4. Let $A \in \mathcal M _2 \left( \mathbb C \right)$. Prove that $\textrm{tr} \left( A \right) =0$ iff $A$ can be written as the sum of two nilpotent matrices.[/i]

My solution was about two pages long, considered many cases...

If alekk's assertion is true, then the problem would become easy. But I don't know how to prove that  :blush:

In you case when the dimension is just $2$ you can take a pair of linearly independent vectors $e_1,e_2=Ae_1$  (otherwise the matrix $A$ would be $%Error. "lamdaI" is a bad command.
_n$ which is easilty to deal out). In this base $e_1,e_2$ you will have $0$  in the position $(1,1)$ meaning that the element in $(2,2)$ is also zero (similarity by changing the base doesn't change the trace of $A$).


So, besides Miles, who else made All State Orchestra/Band this year?  One of my friends who somehow managed to procure a list of qualifiers told me just a few minutes ago that I made it, so I is happy. :)   Anyone else want to share in the smiles?

And plz don't ask me what chair I have.  Let's just say I'm somewhere in the middle of the 2nd violin section. :ninja:

Congrats to Miles for getting a ridiculous chair in cello!

YES!  After I bombed the first sight-reading by switching tempos halfway with no dynamic changes, I got 35th chair!

Though it's rather disappointing, since I got 39th, then 35th chair the last two years.  If only they could form an arithmetic sequence...

EDIT:  I can't believe I forgot Lillian!  Congrats!

Sitan, you have four more years to improve, and it only gets easier.  Stop getting emo. :help:

[quote="Math Geek"]There's a 35th chair?????????????????????????????????????????????????????????????????????????????????????????????????????

o.O

Is that considering the number of people in both violin sections in front of you or only in your section?

Dang...  I got:

[hide][hide][hide]
[size=9][color=white]18th chair[/color][/size][/hide][/hide][/hide]

Have fun clicking on all of those hide tags. :D[/quote]

*Pops in to forum*
Quoting is much easier. This also reminds me of those people who kept doing this in the Mathcounts Problem Series last year. Bah I used to play the violin, but my current school has no orchestra and I didn't really feel like playing anymore because of that. Sorry for spamming. *Leaves*

Thanks, guys.  Anyway, there's little shame in getting a bad seat at all-state.  In tenth grade I was last chair until some people dropped out, and last year I was 14th.  And before that I didn't make it.  Two years ago, this year's first chair cello was my replacement as last chair.  But now, 5.  Yay!  In conclusion, you violins should be thankful for the fact that you have 44 spots in each orchestra.

And yeah, two ARML people in the same section is pretty cool.

And it does [i]not[/i] get easier.  It's true that 10th and 12th are easier respectively than 9th and 11th, but 11/12 is a lot harder than 9/10, in my opinion.

By the way, is anybody else doing all-state chorus?

[quote="Boy Soprano II"]In conclusion, you violins should be thankful for the fact that you have 44 spots in each orchestra.

And it does [i]not[/i] get easier.  It's true that 10th and 12th are easier respectively than 9th and 11th, but 11/12 is a lot harder than 9/10, in my opinion.
[/quote]

Yes, cello is probably the most competitive instrument.

AHEM!  It is much easier to get into All-state as a junior/senior for one reason.  In 9-10 orchestra, nearly 200? violins audition.  In 11-12 orchestra, it's about 90.  So applying our mathematics understanding, you have nearly a flip of a coin chance to get in.  Sure the excerpts are harder, but you are challenged to play to that level.

All-state selection is not uniformly random.  It is true that a lot of good seniors feel that they are too busy doing college auditions, but I think it's also true that people who weren't very close to making it before tend to give up in junior and senior years.  Furthermore, the first audition is harder (more difficult material, but you need to play just as well), so the weakest of the ninety people is stronger than the weakest of the 200 people.

By the way, thanks for your remark about cellos.  It made me feel happy. :D

Due to Georgia Tech's wonderful scheduling of president's scholars finalist interviews, I can't make All-state.

Therefore, you must take me off the list of GA ARML people who made All-state.

Due to GMTA's delightful scheduling of the county piano competition on the same day as All-State, I can't perform, either.

But Jerry, technically you still made All-State, so stay on the list. :)

No.  I refuse.  Reason:

I like balance in life.  Now of the four people, two are confirmed to not go.  If we let $ A$ be the set of people from GA ARML who made all-state this year, $ B$ contain all the subsets of $ A$ of people who can go, and $ C$ contain all the subsets of $ A$ of people who cannot go, then $ B$ and $ C$ have the same cardinality if and only if I consider myself (and Sitan) off the list.

BTW, does anyone know where I can enlighten myself in set notation?

I suspect that all-state is more fun than a county piano competition.  For one thing, playing with other people is more fun than playing alone.  For another thing, you can play piano any time you want to, but you can't play in an orchestra any time you want to.  In particular, you can't play "Jupiter" any time you want to.  Furthermore, all-state lasts several days, which I imagine your piano competition doesn't.

Also, if you skip all-state, you can't do it next year, so you shouldn't do that unless you're a senior.  So unless county piano leads to something that's worth giving up two all-states, I recommend all-state, unless you like piano a lot more than violin.  But if you don't go, you will make another alternate happy.  I'm guessing that you've never gone before from the way you're choosing a county piano competition over this.

[hide="Set notation"]
Okay, $ A\in B$ means "$ A$ is an element of $ B$".  $ A \notin B$ is the negation of this.

$ A \subset B$ means "$ A$ is a subset of $ B$".  Some (weird) people like to use this notation to mean [i]proper[/i] inclusion, but usually this doesn't even matter.  If you want to be absolutely clear, then either specifify which convention you're using, or use $ \subseteq$ to denote inclusion with the possibility of equality, and $ \subsetneq$ to denote proper inclusion (i.e., "$ A \subsetneq B$" means that $ A$ is a subset of $ B$ and $ A$ is not equal to $ B$).

The notation "$ A \cup B$" means the union of $ A$ and $ B$, whereas "$ A \cap B$" means the intersection of $ A$ and $ B$.

The symbol $ \varnothing$ denotes the empty set.  Sometimes, people use $ \emptyset$, but I think the former symbol is preferred because it looks nicer and it won't be confused with 0.  Do not use phi ($ \phi$); this is typographically incorrect.

There are a few different notations for "the power set of $ A$".  The most popular one nowadays (I think) is $ \mathcal{P}(A)$.  People who like to use Fraktur (i.e., old German) fonts often write $ \mathfrak{P}(A)$.  Fraktur used to be much more popular than it is now.  There's also the notation $ 2^A$, but this is pretty obscure, and I think mostly used for pedagogical purposes anyway.

We denote the Cartesian product of $ A$ and $ B$ by $ A\times B$.  (This is the set of all ordered pairs $ (a,b)$ such that $ a\in A$ and $ b \in B$.)

Then there is set specification notation.  This looks something like this :
\[ \{ a \in A \vert P(a) \} , \]
where $ P(a)$ is some proposition involving $ a$ (e.g., "$ a \subset A$", "$ a \equal{} \varnothing$").  Sometimes a colon ( : ) is used instead of a vertical line.  It means, "the set of all elements $ a$ of $ A$ for which $ P(a)$ holds".  I personally don't like this notation very much.  Although it is concise, it is difficult for humans to read, and it lends itself to abuse.  For instance, you may become tempted to write things like
\[ \{ a \vert a \neq \varnothing \}, \]
which is not in fact a set.  That is why it is important to specify that $ a$ is an element of some larger set $ A$.  Furthermore, I think it's just kind of ugly.

In the old days, there were a lot of notations for complements.  But nowadays, we know that there's no such thing as a set of all sets, so "complement" is subject to whatever large set you're using, and I don't think people care about them that much any more, except maybe in topology and things like that.  I think notations for "the complement of $ A$" (with respect to a set $ U$) include $ \overline{A}$, $ A^C$, $ A'$, $ A^U$ (I think), and $ \complement A$ (maybe even $ \neg A$).

There are a few symbols for the cardinality (size) of a set.  The most popular one is $ \lvert A \rvert$.  There is also $ \# A$ (which I dislike, though not vehemently) and $ \text{card}\,(A)$.

Then there is set subtraction.  The subtraction of $ B$ from $ A$ is usually denoted $ A \setminus B$ or $ A \minus{} B$.  This is the set of elements of $ A$ that are not elements of $ B$.

I think that's about everything.  There are some other, more obscure notations (e.g., something for the disjoint union of two sets), but usually authors will tell you if they're going to use them, and they aren't really even standard, I think.
[/hide]

Um, no for the first time in my life I'd have to disagree with you.  In piano, I've won state two years in a row and placed 2nd two more years in a row (the latter two years I placed 2nd cuz MathCounts conflicted with the date and I could not practice for a whole week before the state competition).  The only reason I chose the county piano competition was that I enjoy piano more than violin and that, not to intentionally be immodest (though that's probably not the case), I will most definitely do as well if not better than I did last year, which was considerably well given the circumstances explained above.

Secondly, I asked my orchestra director, and he said that not participating in the performance in advance would only have the repercussions of my having to participate in the 1st round again next year.  So it turns out I'm only giving up one year of All-State.

Thirdly, you're right that I can play piano any time I want, but it's obvious that I can't participate in a statewide competition any time I want to.

And lastly, I understand that All-State is a wonderful experience and that I will be missing out on a lot, but if I have to choose between something at which I am adequate and something at which I am good, I will definitely choose the latter.  Of course, I'm not saying that I'm not interested in the violin.  Quite to the contrary, I highly enjoy playing the instrument.  But I also highly enjoy playing the piano, and I've been playing the piano for almost a decade more than I have been playing the violin.

Oh, and I've been preparing for this competition for almost half a year more than I have for All-State, so I think it's sensible to continue with the piano competition instead.  It's not like I'm giving up on All-State forever.  I will still have a chance next year.

PS: Jerry, your justification is interesting. :lol:

PPS: Miles, my judgment is oftentimes clouded, so do not blame me overmuch.  I'm just making what I think is the best possible choice in my situation. :maybe:

For people who didn't want to click on so many hide tags, Sitan got 18th chair.  :lol:

Thanks for the tips Miles!

Sitan, stringed instruments beat piano.  Always.  Even though there are many opportunities for scholarships via piano.  Always. :cursing:  :furious:  :bomb:

False.  Piano pwns stringed instruments in over 9000 ways.

Not many scholarships for country singers either :(.


Hello, I am studying for my SATs and I've come across a function problem that goes like this:

"The figure above shows the graph of the functions f and g. The function f is defined by f(x) = x^3 - 4x.  The function g is defined by g(x) = f(x+h) + k, where h and k are constants.  What is the value of hk?"

It shows the graph of f(x) and g(x) with points on f(x) - (-1,3), (0,0), (1, -3), (2,0) and g(x) - (2,1), (4, -5)

possible answers:

a) -6
b) -3
c) -2
d) 3
e) 6

I know that the answer is 6, but I'm trying to figure out how to solve these types of problems.

The graphs are exactly the same except that the points on g(x) are down 3, and 2 over on the left.

hope this makes sense, any help would be appreciated

Think about it.

If you add some number, say 2, to the argument of f(x) (argument is what is inside the parethases), to make f(x+2), then you need to put in a number thats 2 less to get out the same y value. The end result is that your graph shifts 2 to the left.

Similarly, f(x-1) would require you to put in 1 more, and the graph would be 1 to the right.

That takes care of the "h"

About the "k"? If you add k to the value of function f, your function is going to shift up by k. So if k is, say, -3, then -3 would be added to each y value, and the graph would appear to shift down 3 units.

In this case, your graph is 2 to the left, so h is 2, and your graph is down 3, so k is -3.

hk = -6


vector
analytic geometry
linear algebra
matrix
linear algebra unsolved

Let 2 bases in $R^{n}$ and k smaller than n. Take k vectors from the first base. Prove that we can exchange them with k convenable vectors from the second base such that in the end we have two new bases. It's a very nice problem, I think.

Assume WLOG that the first basis is the standard basis $(e_{i})_{i}$, and the chosen vectors are $e_{1},\ldots,e_{k}$. Also arrange the elements of the second basis, regarded as column vectors formed by their coordinates wrt $(e_{i})_{i}$, into a matrix (which will be, of course, invertible).

The problem is now reduced to the following:

Let $A$ be a non-singular $n\times n$ matrix (over some field; $\mathbb R$ isn't important here). Also take $k<n$. Then we can find a subset $J$ of $I=\overline{1,n}$ with $|J|=k$ such that both the minor of $A$ formed by the first $k$ rows and the columns with index in $J$ and the minor formed by the last $n-k$ rows and columns with index in $I\setminus J$ are non-zero.

This follows immediately from the fact that the determinant of $A$ is a linear combination (with signs $\pm 1$) of the products of the form $\Delta_{k}\cdot\Delta_{n-k}$, where $\Delta_{k}$ runs through all the minors formed by the first $k$ rows of $A$ and some $k$ columns, and $\Delta_{n-k}$ is the minor left by eliminating the rows and columns used in $\Delta_{k}$.

Exactly! I found this problem very funny and nice.


trigonometry
calculus
calculus computations

If $ F\in C^{4}[a,b] $ and $F(x)=F(a+b-x)\; ,\; \forall x \in [a,b] , $ prove that   for all $  t\in [a,b] $  there exists at least a point  $ \xi ,\xi=\xi(t,F) \in [a,b], $ such that
\[ \begin{array}{|c|}
\hline
\\
\displaystyle F(t)=\displaystyle F(a)+\frac{(t-a)(b-t)}{b-a}F^{\prime}(a)+(t-a)^2(b-t)^2\frac{F^{(4)}(\xi)}{4!}\\
\\
\hline
\end{array}\; .
\]
[b]Particular case :[/b] $[a,b]=[0,1]\; ,\; F(t)=\sin{\pi t}\; . $

The curve has an symetry axes with $x=\frac{a+b}{2}$

(1) $F(a)=F(b),\; F'(a)=-F'(b), \; F^{''}(a)=F^{''}(b) ,\; F^{'''}(a)=-F^{'''}(b), \; F^{'''}(\frac{a+b}{2})=0$

$g(t)=F(t)-F(a)+\frac{(t-a)(t-b)}{b-a}F'(a)-(t-a)^{2}(t-b)^{2}K$ where $K$ is a real

Apply four times Rolle theorem with (1) to get there exist w such that $F^{4}(w)-24K=0$


conics
ellipse
analytic geometry
trigonometry
vector
calculus
calculus computations

We know that the eq of the circle $x^{2}+y^{2}=r^{2}$ can be represented as $x=rcos \theta$ and $y=rsin \theta$. But how is about the eclipse:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .Can we change it into polar coordinate?
  Please help me! :)

Kalle - that's a parametric representation of the ellipse, not a polar representation.

Gibberergy - that's not really the representation of a circle.

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}$

isn't the equations of a curve so much as it is the formulas for converting between rectangular and polar coordinates. Note also that you're abusing the letter $r,$ trying to make it both the polar variable and the (constant) radius of a particular circle.

The polar coordinate equation of a circle centered at the origin of radius $R$ is simply $r=R.$

As for the ellipse: the "nice" representation of an ellipse in polar coordinates doesn't have the center at the center of the coordinate system but rather one of the two foci. The equation is of the form:

$r=\frac{a}{1-e\cos\theta}$

As long as $0<e<1,$ this gives an ellipse. You should be able to tell what happens for other possible values of $e.$

I didn't know it.
I study physic and derivate the formular for the particle moving around the circle (uniform motion). It is a fairly easy problem. But I ask my teacher that what happen if the particle moving around the eclipse. What is its velocity? what is its acceleration? What 's the direction of these vectors? My teacher doesn't give me the answer. So I decided to find it.
   I get stuck. How can we find the vector $r(t)$ to represent the eclipse. Please help me! :maybe:

That's "e[b]l[/b]lipse", not "e[b]c[/b]lipse."

Thank Kent,

   I just got the idea when I walk home. And I have proved that the direction of acceleration vector will point toward the center of the ellipse ( the origin of  the ellipse) ( I take the rectangular coordinator system, using the equation $x=acos \theta$, $y=bsin \theta$ to derivate the formular). Is it a correct answer? :maybe:


Is it possible to arrange the numbers 1,2,...,12 on a circle such that the absolute value of the difference between any two adjacent numbers is either 3,4, or 5?

No.  I think it's impossible.  I will try contradiction.

1, 2, 3, 10, 11, and 12 only have 3 possibilities each for adjacent numbers, namely {4,5,6} {5,6,7} {6,7,8} {5,6,7} {6,7,8} and {7,8,9} respectively.

2 and 10 have two adjacent numbers each.  Although, they may share at most 1.  The only possibilities for adjacent numbers to 2 or 10 are {5, 6, 7}.  So, 2 and 10 must be separated by exactly one number.  Similarly, 3 and 11 must be separated by exactly one number.  From this, we know two strings of 5 numbers with no order, {2,5,6,7,10} and {3,6,7,8,11}.  The intersection of these two strings is {6,7} which cannot be adjacent because they only differ by 1.


A person walks along a straight sidewalk at a constant speed of 6 feet/second. A spotlight is trained on the person from across the street, which is 20 feet wide.
How fast is the spotlight turning (in radians/second) when the distance between the person and the spotlight is 31 feet?

Thanks for any help i have no problem finding dy/dt or dx/dt in any of these situations but i am having trouble with finding the rate at which theta changes

Draw the picture. How are the angle and the length you care about related? Find an equation which is always true (not just at some instant) and differentiate it.


In a group of 12 people, among every 9 people one can find 5 people, any two of whom know each other. Show that there exist 6 people in the group, any two of whom know each other.

We invert the problem so we have to show:
In a graph with $12$ vertices from every $9$ vertices we can select a $\overline{K_5}$. Prove that we can select a $\overline{K_6}$.
If there is no odd cycle, the graph's bypartite and one piece contains at least $6$ vertices forming a $\overline{K_6}$ and we are done.
If there are odd cycles, let $C$ be the smallest odd cycle. Clearly that we do not have edges in the interior of $C$ from its minimality.
Now we distinguish some cases:
$C$ is a triangle:
From the remaining $9$ vertices, we can select a $\overline{K_5}$. Let $A$ be the set of it's vertices and let $B=V \setminus C \setminus A$, $|B|=4$. Now take $B,C$ and $4$ vertices from $A$. We find a $\overline{K_5}$ which must have at least two vertices from $B$ meaning there are two vertices from $B$ not connected.
Let them be $X,Y$.
Clearly we may find $U\neq V \in A$ with $U$ connected to $X$, $V$ connected to $Y$, otherwise add $X,Y$ to $A$ and drop the vertices from $A$ not connected to them to yield a $\overline{K_6}$).
Now take any other vertices $M \in A, N \in B$.
Take $C,X,Y,U,V,M,N$ - $9$ vertices. They contain a $\overline{K_5}$. Since from $C$, $X,U$, $Y,V$ we can select at most one vertex, we deduce that $M,N$ belong to it. Hence every vertex from $B\setminus \{X,Y\}$ is connected to every vertex from $A \setminus \{U,V\}$.
Now if $X$  belongs to the $\overline{K_5}$ then $X,N$ are not connected and the same rationament as above for not connected verices $X,N$ would produce a contradiction for $Y,V$ unless $N$ is connected to $V$ and then $Y$ belongs to $\overline{K_5}$ and so $X,Y,N$ is a triangle.
Clearly we may find $X',Y',N'$ from $A$ with $X,X'$, $Y,Y'$, $N,N'$ connected from Marriage Lemma (or else we could add $X,Y,N$ to $A$ and drop their neighbours from $A$ to get a $\overline{K_6})$.
Now from $C,X,X',Y,Y',N,N'$ we can't select a $\overline{K_5}$.Contradiction.
Hence $X$ doesn't belong to the $\overline{K_5}$ and $Y$ analogously. This means $U,V$ do and hence $N$ is not connected to any of $A$ thus $A \bigcup \{N\}$ form a $\overline{K_6}$, finishing this case

$C$ is a pentagon:
Then no other vertex is connected to $3$ vertices from $C$ otherwise we could find a triangle. Let $A=V\setminus C$ with $|A|=7$. If $A$ doesn't contain an edge we are done. Else, $MN$ be an edge. Choose $C,M,N$ and $U,V$ arbitrary vertices from $A\setminus \{M,N\]$. Since we have a $\overline{K_5}$ but we can select at most two vertices from $C$ and one from $M,N$ it means $U,V$ are not connected and belong to $\overline{K_5}$. It means $A \setminus \{M,N\}$ is a $\overline{K_5}$.Now we must have $M,N$ connected to one of $A\setminus \{M,N\}$ or else we could add them. They cannot form a triangle hence $M,X$, $N,Y$ are connected for some $X\neq Y$.
Now taking $C,M,N,X,Y$ would produce a contradiction.

$C$ is a heptagon, let $A=V \setminus C$ with $|A|=5$. Taking $C$ and two vertices form $A$ since we can select at most $3$ vertices from $C$ in the $\overline{K_5}$ we obtain that the other two vertices belong to the $\overline{K_5}$. Thus $A$ is a $\overline{K_5}$. Now every vertex from $A$ cannot be connected to more that $2$ vertices from $C$ from the minimality of $C$. Hence this vertex $X$  has degree at most $2$ thus there are at least $9$ vertices not connected to it. They contain a $\overline{K_5}$ which with $X$ forms a $\overline{K_6}$ as desired.

$C$ is a nonagon. This case is impossible because $C$ cannot contain a $\overline{K_5}$.

$C$ is ??$11$gon.

Label its vertices clockwise $1,2,3,4,5,6,7,8,9,10,11$.
The vertex $X$ cannot be connected to vertices from $1,3,5,7,9$ or $2,4,6,8,10$ from the minimality of $C$ hence one of $X,1,3,5,7,9$ or $X,2,4,6,8,10$ is a $\overline{K_6}$, as desired.
QED

[Edit: a small flaw noticed above has been corrected.]

[quote="iura"]We invert the problem so we have to show:
In a graph with $12$ vertices from every $9$ vertices we can select a $\overline{K_5}$. Prove that we can select a $\overline{K_6}$.
If there is no odd cycle, the graph's bypartite and one piece contains at least $6$ vertices forming a $K_6$ and we are done.

QED[/quote]

I'm sorry about the question that may be stupid but I wonder how a bypartite graph can contain $K_6$ as it's well known that $K_6$ contain a triangle which is an odd cycle, thus the graph ain't bypartite???


How would you maximize the following equation with $ P$ as a function of $ t_k$?

$ P\equal{}\displaystyle\sum_{j\equal{}1}^n (1\minus{}(1\minus{}s)^{t_j}*Q_j)$

under the following conditions:

$ \displaystyle\sum_{j\equal{}1}^n (t_j) \equal{} m$ and $ \displaystyle\sum_{j\equal{}1}^n (Q_j) \equal{} 1$

In addition, all of the $ Q_j$ are real numbers between 0 and 1.  All of the $ t_j$ are positive real numbers (not necessarily less than 1).  The $ n$, $ m$, and $ s$ are generalized constants with $ n$ and $ m$ positive integers and $ s$ is a real number between 0 and 1.

Is is possible to solve this without using Lagrange?

Let $ a\equal{}1\minus{}s$. By AM-GM, we have
\[ P\equal{}n\minus{}\sum a^{t_j}Q_j\le n\left(1\minus{}\sqrt[n]{a^m\prod_j Q_j}\right)\]
Sometimes that is all that you need: if we can make all $ a^{t_j}Q_j$ equal to each other, the equality will hold.

In order to understand what happens in other cases, let's assume that the sequence $ Q_j$ is non-increasing. Looking at two consequtive terms $ a^{t_j}Q_j\plus{}a^{t_{j\plus{}1}}Q_{j\plus{}1}$ and fixing all variables except $ t_j,t_{j\plus{}1}$, we see that at the point of maximum first, $ t_j\ge t_{j\plus{}1}$ (otherwise, swapping them results in larger $ P$) and, second, either $ a^{t_j}Q_j\equal{}a^{t_{j\plus{}1}}Q_{j\plus{}1}$, or $ t_{j\plus{}1}\equal{}0$ and $ a^{t_j}Q_j\ge Q_{j\plus{}1}$. Thus, in general, we have to make as many first terms in the sum equal as possible by distributing the corresponding $ t_j$ appropriately and put $ t_j\equal{}0$ for all other $ j$.


algebra
polynomial
abstract algebra
number theory solved
number theory

hey
this one is from polish selection tests
determine natural numbers a,b,c such that a^2 + 1  and b^2 + 1  are primes.and
 (a^2 + 1)(b^2 + 1) = c^2 + 1.
and here's one from romania 1994.
find the common root of the polynomials
x^(m+1) - x^n  + 1 = 0
x^(n +1) - x^m  + 1 = 0
thanx

Well, let's denote a ^2+1=p and b ^2+1=q, therefore c ^2+1=pq.

If p=q, then a=b and c ^2+1=(a ^2+1) ^2 , so c ^2 =a ^2 (a ^2+2), so a ^2+2 must be a square, and that's impossible.

So let's take p>q, therefore c ^2 +1<p ^2 , so c \leq p. Also, a \leq p.
but p|a ^2 +1, p|c ^2 +1, so p|a ^2 -c ^2  and p|(c-a)(c+a). Because p is prime, we have two cases:

p|c-a, but because c and a are natural, and lesser than p, that must mean that c=a, so q=1 . But 1 is not a prime.

p|c+a, but 0<c+a<2p, so c+a=p, so (p-a) ^2 +1=pq,  p ^2 -2pa+a ^2 +1=pq, therefore p ^2 -2pa+p=pq, so p+1-2a=q.

If p=2, then a=1, q=1, but 1 isn't prime.

If p>2, then p+1-2a is even so q is even so q=2, therefore b=1, so 2(a ^2 +1)=c ^2 +1=(p-a) ^2 +1=(a ^2 -a+1) ^2 +1. If a=1, that's impossible.If a=2 that's possible, and from here we obtain the solution (a,b,c)=(2,1,3) (and also (1,2,3), because we supposed that p>q).
If a>2 then a ^2 -a+1=a(a-1)+1>=2a+1, so (a ^2 -a+1) ^2 +1>=(2a+1) ^2 +1 > 2(a ^2 +1).

Therefore (1,2,3) and (2,1,3) are the only solutions.

The 2nd problem:

we presume that: m>n. Let x be their common solution of the two equations. Then x verifies x^n(x+1)(x^(m-n)-1)x=0 and x=-1 is not a solution. Let x be so x^(m-n)=1. We have x<> \pm 1, x from C\R, x satisfies: x^(m-1)=1, x^n*(x-1)=-1. We apply the module and get |x|=1 and |x-1|=1 => x cos(pi/3)+isin(pi/3) or x=cos(5pi/3)+i sin(5pi/3), which
means x=w or x=w (w has a line on it) , where w^2-w+1=0, w is the 6th order root of unity.
Taking modulo 6 of m and n we get that the only good cases are when m==1(mod 6) and n==1(mod6). So the common roots are given by:
1) the eqs x^2-x+1=0 for m==n==1(mod 6).
2) in other cases it doesn't exist!!!

cheers mate!


I come across one question and I do it in two ways , both ways "sounds" logical but there is something wrong in it ... here it goes

Question : There are 5 couples , 4 person are chosen among them with the condition the must be at least one male and one female . In how many ways can this be done ?

[b]First Method[/b] :

All ways - 4 male - 4 female = $10C4 - 5C4 - 5C4 = 200 $

[b]Second Method[/b] :

First we chose 1 male and 1 female which is $5C1\times 5C1$ then for the remaining two vacancies , we can choose it randomly from the remaining 8 person which is $8C2$ . So total ways = $5C1\times 5C1\times 8C2 = 700 $

Well , second method seems to flow logically but we know it is not true . Where actually goes wrong ?  :?

to me the second methodmakes more sense; on the other hand:
10C4=210.. ;)

its 12:50am, i'm too sleepy to think, so i'm wrong

[quote="a_vakilian"]your second solution is wrong because you count cases more than one.
i.e: $f_1 m_1 f_2 m_2$ and $f_1 m_2  f_2 m_1$ and $f_2 m_1 f_1 m_2$ and $f_2 m_2 f_1 m_1$ so you count this case $4$ times!!! ;)[/quote]

Hmm .... I guess that is why the second method gives us such big number  :rotfl:  .
Anyway , could there be any way to get the same answer by using the concept like second method ? Like , could we divide it by some number(or combination) to cancel off the repeated part ? :?

[quote="a_vakilian"]yes it is possible. you should divide the problem to $3$ case.when there is $2 woman ,2 man$ and $1 woman , 3 man$ and $3 woman,  1 man$ . ;)[/quote]

Well , what Im trying to ask is whether we can use the concept in second method , which is like setting a male and female first then the rest can be chosen at random . I know that we can divide it into 3 cases like you mention to solve it but what i want to know is whether the second method can be improve or not to make it right ?

I think there is not such a solution that you want. or it is not worth knowing :?  :oops:


geometry
3D geometry
sphere
calculus
integration
HMMT
LaTeX

A sphere with a radius of 5 is cut by a plane that is 4 away from the center of the sphere.  What is the volume of the larger of the two resulting solids?

[hide]I know that you could use calculus (take the equation for a circle, and use the cross sectional method and find the area beneath the graph from -5 to 4).

$\int_{-5}^{4}{\pi(\sqrt{25-x^{2}})^{2}dx}$
$\pi\int_{-5}^{4}{(25-x^{2})dx}$
$\pi((25*4-\frac{4^{3}}{3})-(25*-5-\frac{(-5)^{3}}{3}))$
$162\pi$

And you could imagine a hole with a radius of 3 being driven through the sphere.  Then if you compare each cross section of that $h$ away from the center with each cross section of a sphere with a radius of 4, $h$ away from the center you notice they are equal.  Then you add in the cyclinder that would fill up most of the hole and a sphere that is chopped off by two planes.

So end the end you would end up with $162\pi$ like before.[/hide]

Even if these answers aren't correct I think the process is correct.  I just want to find a solution without using calculus or using a crazy idea.

I don't believe a non-calculus formula exists for a sectoroid (a sector revolved about its axis) or a segmentoid although I may be mistaken

even if an antiderivative existed to the integral for the volume of a segmentoid, it would most definitely not be very nice. Calculus is the best method here

There is a non calculus way to find the volume of a sectoroid.

http://web.mit.edu/hmmt/www/datafiles/problems/2003/pgeom03.pdf

Problem #6 is not quite a sectoriod, but you can just add the cylinder in.  The answer is in the solutions:

http://web.mit.edu/hmmt/www/datafiles/solutions/2003/sgeom03.pdf

And I am pretty sure I have posted both the calculus way and the noncalculus (from the links) way above.  But I am guessing there really is not a way to find the volume without calculus or a great imagination so my question is answered.

The solution is perfectly reasonable if you know the geometric derivation of the formula for a sphere.

Yeah, I can't think of a non-calculus solution....

Hint: Cavalieri's principle

Use google to find out what it is.

The solution Altheman is hinting at (which is essentially what is used in the HMMT solutions linked to above) begins like this:

Sit a cone of radius $r$ and height $r$ down next to a hemisphere of radius $r$ so that the vertex of the cone is in the same plane as the base of the hemisphere and the plane of the base of the cone is tangent to the hemisphere.  (If the hemisphere is sitting on a table base-down, the cone is balanced on the same table on its tip.)  Now consider the combined area of a cross-section of the two shapes at height $h \leq r$ above the table.  This is a nice way to get the volume of a sphere.  It can also be used to get a nice formula for the "cap" of a sphere, as in the question here.

Side note: this requires you to know the volume of a cone, but this can also be derived using a Cavalieri argument.

o.O I am trying to force people to be independent and use google. This skill is very valuable.

Well...I suppose if you were being all rigirous and stuff, you would need to derive the volume of a cone by putting it in correspondence with a pyriamid. But obviously surface area/volume of a cone fall within most first year geometry classes.

[quote="Altheman"]I am trying to force people to be independent and use google.[/quote]  The effect you are having is of being cryptic and ensuring that most of your audience won't ever find out what you're talking about.

[quote]But obviously surface area/volume of a cone fall within most first year geometry classes.[/quote] But the volume is usually asserted, not proved.

[quote="JBL"][quote="Altheman"]I am trying to force people to be independent and use google.[/quote]  The effect you are having is of being cryptic and ensuring that most of your audience won't ever find out what you're talking about.
[/quote]

[hide="@JBL"]Okay. Lets consider an analogous situation. We are given a triangle concurrency problem. I suggest mass points [lets say in the HSB forum]. Then I should expect to get lots of stupid replies asking: "what are mass points?" "where can I learn more about mass points?" These are stupid questions because you can easiliy find the answer by a quick google search.  I know you have this ability. ex. consider your response to mathfanatic in that global warming thread about following fallacies created by those against global warming. Anyway, this situation is similar. You can easily type: "cavaleri principle sphere" into google and obtain the link: [url]http://www.walter-fendt.de/m14e/volsphere.htm[/url]. This takes a few seconds and you get a diagram which I find much more helpful than any solution provided in this thread. While you do not get a complete solution in this page, it is enough of a hint to finish the problem.

Furthermore, I think this type of hint is nicer than a full solution. This way the solver can still get a sense of accomplishment from solving the problem. I find that a fundamental part of my enjoyment of problem solving is when I come up with the solution on my own. By offering hints, I let the solver get started on the problem. The full solution should be left to those who have to struggle to solve the problem.

If I may predict your response, I think that you will think that this is a lot of ---- and I am just being lazy. I am not sure what position you think you are in so you can interpret my actions. I am in a better position than you are to I know what I am thinking.[/hide]

[quote="JBL"][quote]But obviously surface area/volume of a cone fall within most first year geometry classes.[/quote] But the volume is usually asserted, not proved.[/quote]

Really? I am suprised about that. In freshman year geometry, [i]before I was interested in math[/i] [i.e. so I took not particular action to figure this out], I was aware of this proof:
i) V=1/3 b*h where the base is a triangle; this comes from decomposing a parallelpiped [or w/e you call it]
ii) V=1/3 b*h where the base is a polygon; decompose the polygon into triangles and apply i)
iii) V=1/3 b*h where the base is a circle; a circle is a polygon with an infinite number of sides; apply ii)

Perhaps I had a better geometry class than I thought; I thought that this was the standard "non-calculus proof."

I suppose that is not 100% rigirous since we are dealing with limits. But that is being pretty picky.

[hide="@Altheman"]You think saying, "use mass points" in the HSB forum is educational -- I think it is pedagogically unsound.  Simply telling people to google things will lose most of your audience immediately -- they won't bother.  Of those who do google it, they won't know what to google for, and a substantial portion of them will fail to find an appropriate introductory source.  Far, far better (in my opinion) is to explain how to do something, and (if it's not obvious) show why it works.  (Cavalieri's Principle is something where the "why" is really calculus, but the result itself is very intuitively clear.)  In general, the general result is something best understood [i]after[/i] particular examples have been dealth with, and knowing the name (or the formula, or whatever) but not understanding it is worthless.  Teachers should do more than simply hand their students a textbook for each question, and this is all the more true when the textbook in question is the entire internet.

You're probably right that I've misinterpreted your intent in the past, so I retroactively apologize.  (That does not apply to my complaints about LaTeX non-usage, though  :wink: )[/hide]

[quote="JBL"]i) V=1/3 b*h where the base is a triangle; this comes from decomposing a parallelpiped [[i]maybe prism? -- JBL[/i]]
ii) V=1/3 b*h where the base is a polygon; decompose the polygon into triangles and apply i)
iii) V=1/3 b*h where the base is a circle; a circle is a polygon with an infinite number of sides; apply ii)

Perhaps I had a better geometry class than I thought; I thought that this was the standard "non-calculus proof."
I suppose that is not 100% rigirous since we are dealing with limits. But that is being pretty picky.[/quote]  It could just be that my memory of geometry class is very rusty :).  I'm not quite sure what you mean by (i), but it sounds plausible.  I don't like (iii) very much, because I think Cavalieri's is a nicer way to go from a single pointy thing to all pointy things (in other words, there's no particular reason to go via (ii) -- one can spit out (ii) and (iii) together without ever using the words "infinitely many").  A lot has to do with how well something is explained, I guess.

Well...I learn by reading the material first, then asking questions. And the lack of LaTex can be attributed to my procrastinating from school work...but w/e my posts have generally improved now that school has ended.

Anyway, Here is euclid's proof for $\frac{1}{3}bh$: http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII7.html

I think that it is reasonable to consider a circle as a limiting form of a polygon with a large number of sides. If you are particularly picky, you can consider the inscribed and circumscribed regular polygons with n sides. Then you can apply the method of exhaustion [originally an idea from archimedes]. To show that the circle is inbetween these two volumes. This is quite rigirous actually. In the development of integral theory by Tom Apostol in "Calculus," this is one of the 4 fundamental axioms that he takes for area. The analogue to volume is just as acceptable; if one object contains the other, then they have equal volume. Regardless, these are ideas that my dad would understand [and he does psychology]. I think they are sufficiently rigirous for say, the ancient greek geometers.


MATHCOUNTS
geometry
AMC
AIME
AMC 12
AMC 10

Are they helpful? I am deciding on getting them and I want your thoughts on the books. What are their levels and what contests do they prepare you for? Are they worth the money? Thanks.

They are very very very helpful! Vol 1 covers Mathcounts/AMC stuff, Vol 2 is more AIME/AMC/ARML stuff.

Vol 1 is recommended for middle school , Vol 2 is recommended for high school

Just make sure that if you get them, you do a lot of the problems and actually learn math rather than being a passive reader.

[quote="isabella2296"]They definitely are very helpful, at least in my opinion and what seems to be the majority of AoPSers. I haven't really started Volume 2 yet, but Volume 1 is extremely helpful for MathCounts and the AMCs, particularly the geometry section.[/quote] did you make aime isabella

[quote="Ktk"]If I was aiming for AIME via AMC12 (failed this year by a few mistakes, motivated to try AoPS now) should I just skip the first volume?[/quote]
AoPS Vol.1  doesn't go into details about the stuff that's on the AMC 12 and not AMC 10, but it covers basics for the AMCs and just knowing all of that stuff really will help you more than knowing more advanced stuff from Vol. 2. So you should get Vol. 1.


Recently, I've found that when I try to use the "hide" command, instead of applying the hide format, it will instead display the code.  Was this an intentional modification, or a bug?

Did you remember to close with (/hide)? (replace () with [])

[hide="Hiding"]...[/hide] works fine for me.

Oops...

At least one time it was because I used the wrong slash. :oops:

[hide="This is a test."] :maybe: [/hide]

Yep, that must have been it, because the above hide works just fine.


Prove that the elements of the set $ A\equal{}\{ (x,y)|x \in \mathbb{Z},y \in \mathbb{Z}, 2xy\minus{}6x\plus{}y\equal{}0 \}$ are on the graphic of the function $ f: \mathbb{R} \rightarrow \mathbb{R}, f(2x\minus{}3)\equal{}8x\minus{}7.$

[quote="moldovan"]Prove that the elements of the set $ A \equal{} \{ (x,y)|x \in \mathbb{Z},y \in \mathbb{Z}, 2xy \minus{} 6x \plus{} y \equal{} 0 \}$ are on the graphic of the function $ f: \mathbb{R} \rightarrow \mathbb{R}, f(2x \minus{} 3) \equal{} 8x \minus{} 7.$[/quote]

Unfortunately, this is wrong : $ (\minus{}1,6)\in A$ but is not on the required graphic since $ f(\minus{}1)\equal{}1\neq 6$ and $ f(6)\equal{}29\neq \minus{}1$)


[quote="solafidefarms"]Since this year seems to be easier than usual, what effect will that have on the cutoff? how far up will the cutoff move? Also, how is the cutoff determined? Is it the first score for which >x people got that score or above for some x?[/quote]
I think that there are an x number of people who qualify for the AIME. Anyways, if the third round is uber hard than the qualifying score might be as low as 60-63.

You start off with an index of 100 if you qualify through USAMTS, right?

also, people that qualify for the AIME through the USAMTS can qualify for the USAMO by both their index and the floor score, right?

I think the index only matters in USAMO qualification is you are in 11th or 12th grade.  Otherwise you go by floor which means if you score at least the score the lowest AIME score that the people that qualified by the index, you are in.  So I don't think the USAMO index really does matter for us...

But I am interested in how qualifications to AIME is decided.  It would feel good if I knew that even if I bombed the AMCs, I would still be in the AIME.


linear algebra
matrix
search
linear algebra unsolved

DO any of u know what a "[b][color=black][size=150]DIRAC MATRIX[/size][/color][/b]" is?

I have two words for you: [b]GOOGLE SEARCH[/b].

THanks for that!!!!!

@carcul- ia almost forgot that


There are n markers, each with one side white and the other side black,
aligned in a row so that their white sides are up. In each step, if possible, we
choose a marker with the white side up (but not one of the outermost markers),
remove it and reverse the closest marker to the left and the closest marker to the
right of it. Prove that one can achieve the state with only two markers remaining
if and only if n - 1 is not divisible by 3.

merged with http://www.mathlinks.ro/viewtopic.php?t=90046 .

  darij


find all pairs of natural numbers $ (n,m)$ such that  $ n!\plus{}1\equal{}m^2$

$ (4,5),(5,11),(7,71)$ is this all ?

This a very well known problem ( I think it is Brocard's problem) but unfortunately it hasn't been solved yet. I t seem that those are the only solutions but it is not known.
If someone can solve it I don't think he will post it here without publishing it :P   :rotfl:
But I recommend you to try it, it's very nice

Daniel


calculus
derivative
real analysis
real analysis unsolved

Show that $x^{n}+nx-1 =0 \quad n\geq 1$, has exactly one real root $a_{n}$. Does the series
\[\sum_{n}a_{n}\]
converge ?

(I'm not sure about the problem. Think it should be "one real [i]positive[/i] root")

It has at least one positive root by the Intermediate Value Theorem; its value at zero is negative, and it goes to $\infty$ at $\infty$. If $n$ is even, it has a negative root as well, so it should specify "positive real root".
For uniqueness, show that the derivative is positive for positive $x$; Rolle's theorem finishes the proof.

The series diverges: $\frac1n<a_{n}<\frac1{n-1}$ for $n>1$.

[quote="jmerry"]

The series diverges: $\frac1n<a_{n}<\frac1{n-1}$ for $n>1$.[/quote]

Thanks for your reply, but how did you get the bounds for a_n ?

Looking again, they're not quite right- I was rushed.

Let $f_{n}(x)=x^{n}+nx-1$. $f_{n}(\frac1n)=(\frac1n)^{n}>0$
$f_{n}(\frac1{n+1}=\frac1{(n+1)^{n}}+\frac{n}{n+1}-1< \frac1{n+1}+\frac{n}{n+1}-1=0$

By IVT, there is a zero of $f$ between $\frac1n$ and $\frac1{n+1}$.