analytic geometry
graphing lines
slope
geometry
rotation

If x and y satisfy $x^2+y^2\le4$ and $x \ge-1$

What is the least value of $\frac{y+2\sqrt(3)}{x+2}$?

Suppose that
$\frac{y + 2\sqrt 3}{x + 2} = m$

Then
$y = (x + 2)m - 2\sqrt 3$

Now look, if $x \in \mathbb{R}$ then we have a straight-line passing through the point $(-2, -2\sqrt 3)$ and $m$ is the slope of the straight-line.

$x^2 + y^2 \leq 4$ is a circle centered on $(0,0)$ with radius $2$.

The least value of $m$ occurs when that straight-line is tangent to the circle. Note that one of the two tangent lines is invalid, because the circle isn't a complete circle since $x \geq -1$.

By plane geometry is possible to find the least value $m = \frac{\sqrt{3}}3$

[quote="F u r u y á"][OFF]
wow... essa foi pelo método ponciano hein???   :coolspeak:
[/OFF]

Can u explain me plz why "The least value of m occurs when that straight-line is tangent to the circle" ?[/quote]We want to find $x_0$ and $y_0$ satisying the conditions on $x$ and $y$ such that the line will passing through $(-2,-2\sqrt3)$ and $(x_0,y_0)$ will have the least slope. First let the line be horizontal, then rotate it counterclockwise until it is tangent to the circle. This is the line with the least slope.

ahhh now I got it. The position you said is the one which got the least slope, thus the least m value. thx dude


[hide="Hidden by mod"]some two of them leave the same remainder...subtract them the number is divisible by 987654321.clearly upto a certain point all are 0's from behind .
remove the 0's and u get the reqd. no.[/hide]

[hide="Hidden by mod"]I will prove a more general result. Let $ n$ be a positive integer which isn't divisible by $ 2$ and $ 5$. Then for any $ k$ there exists $ m$ such that $ n\mid 1 \plus{} 10^k \plus{} 10^{2k} \plus{} \dots \plus{} 10^{mk}$.

Proof: Let $ N_k(t) \equal{} 1 \plus{} 10^k \plus{} 10^{2k} \plus{} \dots \plus{} 10^{tk}$. Consider the $ n \plus{} 1$ numbers $ N_k(0),N_k(1),N_k(2),\dots,N_k(n)$. Two of these numbers must yield the same remainder upon division by $ n$. Thus there exists $ 0\le y < x\le n$ such that $ N_k(x)\equiv N_k(y)\pmod n\implies \sum_{i \equal{} 0}^x 10^{ik}\equiv \sum_{i \equal{} 0}^y10^{ik}\pmod n$ which implies $ 10^{(x \plus{} 1)k}\sum_{i \equal{} 0}^{y \minus{} x \minus{} 1}10^{ik}\equiv 0\pmod n$. But since $ (10,n) \equal{} 1$, we have $ n\mid \sum_{i \equal{} 0}^{y \minus{} x \minus{} 1}10^{ik} \equal{} N_k(y \minus{} x \minus{} 1)$. And we are done.[/hide]

[quote="hell_ever"]Show that there exists a number of the form $ 123456789123456789.......123456789$ that is divisible by $ 987654321$[/quote]
Where did you find it ?  :omighty:
because It was a contest problem in POSN(The contest in Thailand)
Thanks


function
real analysis
real analysis unsolved

Let $f : (a,b) \to \mathbb R$ be a function with the property that for all $x \in (a,b)$ there is a non-degenerated interval $[ a_x,b_x ]$ with $a < a_x \leq x \leq b_x < b$ such that $f$ is constant on $\left[ a_x,b_x \right]$.

(a) Prove that $\textrm{Im} \, f$ is finite or numerable.

(b) Find all continuous functions which have the property mentioned in the hypothesis.

For every value $a$ of $f$ there is a ratiaonal $q$ number s.t $f(q)=a$ because for $f(x)=a$ we can find a rational number in $[a_x,b_x]$ since in every interval lies a rational number. Hence we can define an injection from $Im(f)$ to $\mathbb{Q}$ and, so $Im(f)$ is numerable or finite.
For the second part we use the fact that the image of a continuous function must be an interval. And an interval is numerbale, only if it is degenerated, hebce $f$ is constant.

[quote="xirti"] Hence we can define an injection from $ Im(f)$ to $ \mathbb{Q}$ and, so $ Im(f)$ is numerable or finite. [/quote]
But $ \mathbb{Q}$ is infinite?? What means numerable by the way?

Numerable means countable I suppose.
A set $ X$ is countable if there exists a injection $ f : X \rightarrow \mathbb{N}$. ($ f$ being a bijection implies $ X$ to be countably infinite)

See [url]http://en.wikipedia.org/wiki/Countable_set[/url].


as always I'm having problems with my editor :alien: (I'm going to cry lol) , and I cannot edit this limit properly. I hope someone can solve it just with this typing ;(

lim [(x+a)^p/(x+b)^(p-1) -  (x+c)^q/(x+d)^(q-1)]  (x -> + infinite)

Well, the solution is this: [color=white](q-1)d+p*a-(p-1)*b-q*c[/color] but what do you do  to arrive there?

In $\LaTeX$:
Find
\[ \lim_{x \rightarrow \infty} \left( \frac{(x + a)^p}{(x + b)^{p - 1}} - \frac{(x + c)^q}{(x + d)^{q - 1}} \right). \]

This comes down to sorting out the 'top-heavy' fraction.

We want to choose K such that $(x+a)^p = (x+k)(x+b)^{p-1} + P(X)$ where P(X) is a polynomial of degree p-2, then $\lim_{x\to\infty}\frac{(x+a)^p}{(x+b)^{p-1}} = \lim_{x\to\infty}(x + k + \frac{P(X)}{(x+b)^{p-1}}) = x + k$

To do this we just have to match up coefficients:

$(x+a)^p = x^p + \binom{p}{1}ax^{p-1} + \binom{p}{2}a^2x^{p-2} + ....$
$(x+k)(x+b)^{p-1} = x^p + (k + b(p-1))x^{p-1} + ...$

So $k + (p-1)b = ap \to k = ap - (p-1)b$

So $\lim_{x\to\infty}\frac{(x+a)^p}{(x+b)^{p-1}} = x + ap - (p-1)b$

Apply this to the other bit to get the total limit as

$(x + ap - (p-1)b) - (x+cq - (p-1)d)$, then the two x cancel out and you are left with the constant terms.


AMC AIME trigonometry conics parabola function geometry

If $ f(x) \equal{} 3\sqrt{x}$, find all x for which $ f(x) \equal{} f^{\minus{}1}(x)$

so obviously I've found $ f^{\minus{}1}(x) \equal{} \frac{x^2}{9}$

setting the two equations equal to each other,

$ 3\sqrt{x} \equal{} \frac{x^2}{9}$

$ 3^3\sqrt{x} \equal{} x^2$

$ 3^6x \equal{} x^4$

$ 3^6 \equal{} x^3$

$ x \equal{} 3^2 \equal{} \boxed{9}$

but $ x \equal{} 0$ is obviously another solution, but how would I get this algebraically?

good thing the AIME doesn't rely too much on just algebra and stuff.. XD

[quote="undefined117"]If $ f(x) \equal{} 3\sqrt {x}$, find all x for which $ f(x) \equal{} f^{ \minus{} 1}(x)$

so obviously I've found $ f^{ \minus{} 1}(x) \equal{} \frac {x^2}{9}$

setting the two equations equal to each other,

$ 3\sqrt {x} \equal{} \frac {x^2}{9}$

$ 3^3\sqrt {x} \equal{} x^2$

$ 3^6x \equal{} x^4$

$ 3^6 \equal{} x^3$

$ x \equal{} 3^2 \equal{} \boxed{9}$

but $ x \equal{} 0$ is obviously another solution, but how would I get this algebraically?

good thing the AIME doesn't rely too much on just algebra and stuff.. XD[/quote]

Never cancel out the xs (unless you set them to 0 right away)
$ 3^6x \equal{} x^4$ implies that $ 3^6x \minus{} x^4 \equal{} 0$, then $ x(3^6 \minus{} x^3) \equal{} 0$ then x=0 and x=9 from setting both to 0.

You can cancel out the x's but you have to make sure they arent equal to 0 (or just set them equal to 0 at that step).

Check out this AIME problem (see what happens when you decide to randomly cancel out the xs)

Find the sum of the values of $ x$ such that $ \cos^3 3x \plus{} \cos^3 5x \equal{} 8 \cos^3 4x \cos^3 x$, where $ x$ is measured in degrees and $ 100 < x < 200$

UNDEFINED, the problem thread is dead already.

dudeee awesome. Thanks!

and yeah I'll try that AIME problem.. I still suck at trig though.

:D okay when I take a parabola, let's say like $ f(x) \equal{} x^2\plus{}8x\plus{}5$, and take its inverse, do I include both halves of the parabola, like $ f(x) \equal{} \pm\sqrt{x\plus{}11}\minus{}4$?

or just $ f(x) \equal{} \sqrt{x\plus{}11}\minus{}4$?

I knowww I'm getting so good at this stuff :rotfl:

It has to be plus/minus, otherwise your graph/function would not really be the inverse of the parabola.

[quote="undefined117"]hmm.. I was worried since that inverse doesn't pass the vertical-line test :huh:[/quote]

i guess you can just call it y=...

but then again i remember like learning in Dr.Ji's class that parabolas dont have an inverse because it isnt one-to-one or something like that.

not one to one.. I see. Thanks guys!

I should start reading my sister's algebra textbook -_-

[quote="the future"]hmm.. I was worried since that inverse doesn't pass the vertical-line test :huh:

i guess you can just call it y=...

but then again i remember like learning in Dr.Ji's class that parabolas dont have an inverse because it isnt one-to-one or something like that.[/quote]

parabolas don't have inverses cuz to graphically figure out the inverse of a graphed function, don't you need to reflect it over the line y = x. and if you reflected it, then you don't have a function by the vertical line test. i think........

[quote="undefined117"]I should start reading my sister's algebra textbook -_-.[/quote]

that is deeply disturbing, that you need to review alg. 1.

I also need to review geometry, so relatively speaking, not really (:

umm you can review geometry by doing that problem in the AIME probelm solving marathon  :D propoganda.

undefined, r u seriously going to exeter??

considering it, why? Do you think it's improbable at this stage (needing to review algebra/geo)? :lol:

propaganda! I'll _tryyy_


[quote="symmetry"]Statistics

Question :

The significance level of a hypothesis test is 0.33. What is the
probability of a Type I error?[/quote] The probability of making a Type I error is denoted by $\alpha$. The p-value denoted by $p$ is the significance level of the hypothesis test. If $p < \alpha$ then one is free to reject the null hypothesis.

That means in your case the probability of a Type I error, $\alpha > 0.33$

Okay....I will work with this information.

Thanks


algorithm
superior algebra
superior algebra solved

i)Prove that $S_n$ can be generated by $(1, 2),(1, 3),...,(1, n)$.
ii)Prove that $S_n$ can be generated by $(1, 2),(2, 3),...,(i ,i+1),...,(n-1, n)$.
iii)Prove that $S_n$ can be generated by $(1 ,2)$ and $(1, 2 ,..., n$.

[b]i)[/b] Use that $(1,k)(1,l)(1,k)=(k,l)$ to get all transpositions. But they generate the whole group, done.
[b]ii)[/b] Use that $(1,k)(k,k+1)(1,k)=(1,k+1)$ to reduce it inductively to case [b]i)[/b]
[b]iii)[/b] Use that $(1,2,3,...,n)(k,k+1)(1,2,3,...n)^{-1}= (k+1,k+2)$ with $(1,2,3,...n)^{-1}=(1,2,3,...,n)^{n-1}$ to reduce it to [b]ii)[/b]

I have a conjecture of that, that I sort of have proven
suppose I give you  a set of transpositions and I would ask you whether it genrerates the full group
Is this algorithm good ?  I put all elements in{1,2,......n} in a graph, and then I connect those that are together in one of the given transpositions
the generated group is the full group iff only the graph is connected? :|

Yes. If you have a path from $a$ to $b$, you can generate the transposition $(a,b)$ as in ZetaX's proof of (ii). If it's not connected, we can't exchange mebers of different connected components.

[quote="ZetaX"][b]i)[/b] Use that $(1,k)(1,l)(1,k)=(k,l)$ to get all transpositions. But they generate the whole group, done.
[b]ii)[/b] Use that $(1,k)(k,k+1)(1,k)=(1,k+1)$ to reduce it inductively to case [b]i)[/b]
[/quote]

this is helpful to me :) but is there any chance you or someone could go through parts i and ii in more detail?
thanks

$i)S_n$ consist of all permutations of the set $\{1,2,\cdots,n\}$. It follows from the theory of permutations that every permutation can be written as a product of transpositions. Now take any of those permutation and replace the element by three element where the first index is one. This can be done by the formula ZetaX gave.

For $ii)$ we know that elements of the form $(1,i)$ generates $S_n$. Can we make transpositions that changes $(i,i+1)$ to $(1,i+1)$? Then we know by $i)$ that our set generates $S_n$. Lets see. We have the element $(1,2)$. But $(1,3)=(1,2)(2,3)(1,2)$ so we have found also that we can express $(1,3)$ as the composition of our permutations. Again $(1,4)$ can be expressed as a composition of permutations of the form we were given. But hey! Continuing this way we can find all permutations of the form $(1,i)$ expressed as a composition of permutations of the elements we were given. But what did we prove in $i)$?


calculus
integration
logarithms
calculus computations

Evaluate $ \int{\frac{\sqrt{1\plus{}\ln{x}}}{x\ln{x}}}dx$.

I initially started with the substitution $ u\equal{}\sqrt{1\plus{}\ln{x}}$. Then $ x\equal{}e^{u^{2}\minus{}1}$ but then I'm at a loss for what to do next. Is this integral possible by rationalizing substitutions or is there something I'm just not seeing here?

$ \int\frac{\sqrt{1\plus{}\ln x}}{x\ln x}dx\equal{}\int\frac{u}{\left(u^{2}\minus{}1\right)e^{u^{2}\minus{}1}}d\left(e^{u^{2}\minus{}1}\right)\equal{}\int\frac{ue^{u^{2}\minus{}1}}{\left(u^{2}\minus{}1\right)e^{u^{2}\minus{}1}}2udu$
The rest is quite easy :)


The venusian prophet Zabruberson sent to his pupils a $ 10000$-letter word, each letter being $ A$ or $ E$: the [i]Zabrubic word[/i]. Their pupils consider then that for $ 1 \leq k \leq 10000$, each word comprised of $ k$ consecutive letters of the Zabrubic word is a [i]prophetic word[/i] of length $ k$. It is known that there are at most $ 7$ prophetic words of lenght $ 3$. Find the maximum number of prophetic words of length $ 10$.

Anyone have smarter ideas than the relatively brute-force method below?

Initially we have 8 possible missing triples, but obviously we can interchange E and A without changing the problem, so in fact we're dealing with 4 problems: where AAA is omitted, where AAE is omitted, where AEA is omitted or where EAA is omitted.  Also, the second and last of these are equivalent under reversal of the whole word.  So, lacking any better ideas, the thing to do now seems to be to enumerate the elements of $ \{A, E\}^{10}$ that do not contain AAA, enumerate those that do not contain AAE and enumerate those that do not contain AEA.  This is not hard to do recursively, say, and it gives that the maximum theoretical bound comes if the missing word was AAA, in which case there are at most 504 words of length 10.  But this can clearly be achieved, e.g., by placing each of the 504 words of length 10 in order, spaced by some extra Es (to make sure we don't accidentally create new AAAs -- note in particular that this trick wouldn't have worked if the length-three word that gave the largest number of length-ten words was AAE) -- this will only take 6000 or so characters, so we've got lots of space to do it in.


[quote="Singular"]Yes.  We prove the stronger:  $\sum_{sym} \frac{a^2}{a+b} \leq \sum_{sym} \frac{a}{2}$.

Multiply out.  Let $S_k$ be the $k$th powersum, and $q, r$ = $ab+bc+ca$, $abc$.

Then we have $S_1(S_1q - r) \geq qS_2 + q^2 - rS_1$.

Now, normalize $S_1 = 1$.  We have now $q-r = q - 2q^2 + q^2 - r$, equivalent to $q^2 \geq 0$ which is true.[/quote]

your "Stronger" inequality is false. By cauchy $2(a^2+b^2) \ge (a+b)^2$ so $\frac{a^2+b^2}{a+b} \ge \frac{a+b}{2}$ then you are wrong.

Oops, i missed 3 terms when I multiplied out. (i treated the sym as if it were cyc)

i can't prove this inequality with vasc's hint :(

can anyone help me?

Vasc's hint is just the original inequality with $ a \plus{} b \plus{} c$ added to both sides.

I guess in that context this inequality is a matching upper bound for the QM-AM inequality for three variables. AM-QM states that $ \sqrt{3(a^2\plus{}b^2\plus{}c^2)}\minus{}(a\plus{}b\plus{}c)\ge0$, that is, $ \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}} \minus{} \frac {a \plus{} b \plus{} c}{3}\ge0$, while Cezar's inequality rewrites as $ \frac {\sum\frac {(a \minus{} b)^2}{2\left(a \plus{} b\right)}}{3}\geq \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}} \minus{} \frac {a \plus{} b \plus{} c}{3}$.  I'm not sure if that's going down the right track though...

t's rather obvious that we can write vasc's inequality into
$ \sum (\frac {1}{k} \minus{} \frac {1}{2(b \plus{} c)}) (b \minus{} c)^2 \le 0$ where $ k \equal{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c)$, and it's obvious that $ k \ge 2(b \plus{} c)$ so $ \frac {1}{k} \le \frac {1}{2(b \plus{} c)}$ and the whole thing is proven.

Problem : Prove that :
  $ \sqrt{3(a^2+b^2+c^2)} \leq \frac{a^2+b^2}{a+b} +\frac{c^2+b^2}{c+b}+\frac{a^2+c^2}{a+c} $

Solution : Write the inequality as follows
$ \sqrt{3(a^2+b^2+c^2)} -(a+b+c) \le \sum{(\frac{b^2+c^2}{b+c}-\frac{b+c}{2})}$,
$ \frac{(a-b)^2+(b-c)^2+(c-a)^2}{\sqrt{3(a^2+b^2+c^2)} +a+b+c} \le \sum\frac{(b-c)^2}{2(b+c)}$.
It suffices to show that
$ \frac{(a-b)^2+(b-c)^2+(c-a)^2}{2(a+b+c)} \le \sum\frac{(b-c)^2}{2(b+c)}$.
The last inequality is equivalent to
$\sum{\frac{a}{b+c}(b-c)^2} \ge 0$,
which is clearly true.

I found Cezar's original inequality

$ \frac {b^2 + c^2}{b + c} + \frac {c^2 + a^2}{c + a} + \frac {a^2 + b^2}{a + b}\geq\sqrt {3(a^2 + b^2 + c^2)}$

rather hard to prove, but the stronger inequality from Nttu's post #14 is nice and easy. Let us give a non-Muirhead proof for this stronger inequality:

[color=blue][b]Theorem 1.[/b] For any three positive reals a, b, c, we have

$ \frac {c^2 + a^2}{c + a}\cdot\frac {a^2 + b^2}{a + b} + \frac {a^2 + b^2}{a + b}\cdot\frac {b^2 + c^2}{b + c} + \frac {b^2 + c^2}{b + c}\cdot\frac {c^2 + a^2}{c + a}\geq a^2 + b^2 + c^2$.[/color]

[i]Proof.[/i] Using the $ \sum$ notation for cyclic sums, we can rewrite this inequality as $ \sum\frac {c^2 + a^2}{c + a}\cdot\frac {a^2 + b^2}{a + b}\geq \sum a^2$, what is equivalent to $ \sum\frac {c^2 + a^2}{c + a}\cdot\frac {a^2 + b^2}{a + b} - \sum a^2\geq 0$.

Now,

$ \sum\frac {c^2 + a^2}{c + a}\cdot\frac {a^2 + b^2}{a + b} - \sum a^2 = \sum(\frac {c^2 + a^2}{c + a}\cdot\frac {a^2 + b^2}{a + b} - a^2)$
$ = \sum((a + \frac {c(c - a)}{c + a})\cdot(a + \frac {b(b - a)}{a + b}) - a^2)$
$ = \sum(\frac {c(c - a)}{c + a}a + \frac {b(b - a)}{a + b}a + \frac {c(c - a)}{c + a}\cdot\frac {b(b - a)}{a + b})$
$ = \sum\frac {c(c - a)}{c + a}a + \sum\frac {b(b - a)}{a + b}a + \sum\frac {c(c - a)}{c + a}\cdot\frac {b(b - a)}{a + b})$
$ = \sum\frac {ca(c - a)}{c + a} + \sum\frac { - ab(a - b)}{a + b} + \sum\frac {bc}{(c + a)(a + b)}(a - b)(a - c)$
$ = \underbrace{\sum\frac {ca(c - a)}{c + a} + \sum\frac { - ca(c - a)}{c + a}}_{ = 0} + \sum\frac {bc}{(c + a)(a + b)}(a - b)(a - c)$
$ = \sum\frac {bc}{(c + a)(a + b)}(a - b)(a - c)$.

So it remains to show that $ \sum\frac {bc}{(c + a)(a + b)}(a - b)(a - c)\geq 0$. This is equivalent to $ \sum\frac {b + c}{a}(a - b)(a - c)\geq 0$. But this follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=387639#p387639]the Vornicu-Schur inequality[/url] (in fact, if we WLOG assume that $ a\geq b\geq c$, then $ \frac {a + b}{c}\geq\frac {c + a}{b}$, since $ a + b\geq c + a$ and $ c\leq b$). Hence, Theorem 1 is proven.

  Darij

[quote=Nguyenhuyen_AG][quote=Cezar Lupu]If $ a$, $ b$, $ c$ are positive reals, then prove that
$ \sqrt {3\left(a^2 \plus{} b^2 \plus{} c^2\right)}\leq\frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}$.[/quote]
Stronger
\[\frac{b^2+c^2}{b+c} \geqslant \frac{3(b^2+c^2)-2bc+a(b+c)}{2\sqrt{3(a^2+b^2+c^2)}}.\][/quote]
It's a quadratic inequality of $a$.
It's enough to prove that:
$$(12(b^2+c^2)^2-(b+c)^4)(12(b^2+c^2)^3-(b+c)^2(3b^2+3c^2-2bc)^2)\geq(b+c)^6(3b^2+3c^2-2bc)^2,$$ which after substitution $b^2+c^2=2ubc$ gives
$$u^2(2u-1)(u-1)^2\geq0.$$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group and let $ G_n\equal{}$ {$ g^n|g\in G$}. Under what hypothesis about $ G$ can we show that $ G_n$ is a subgroup of $ G$?

commutativity. but can it be a subgroup without being commutative? example/proof?

what about $ G \equal{} Q_8$ (the quaternions) and $ n \equal{} 2$ ? or $ G\equal{}Q_8$ and $ n\equal{}3$ ?

[quote="question?"]ah ok, thxs. not familiar with the quaternions though.[/quote]

The quaternions $ Q_8\equal{}\{1,\minus{}1,i,\minus{}i,j,\minus{}j,k,\minus{}k\}$ are a [b]noncommutative[/b] group with $ 8$ elements.   If you take $ n \equal{} 2$, you got the subgroup $ \{1, \minus{} 1\}$ and if you take $ n \equal{} 3$ you obtain the whole group $ Q_8$. More information about the group in http://en.wikipedia.org/wiki/Quaternion

Regards

More generally, we have $ G_n\equal{}G$ whenever $ n$ is relatively prime to $ |G|$.

Also $ G_n$ is a normal subgroup of $ G\equal{}D_n$ , whatever $ n$ is.

Could we have a group which $ G_n$ is not subgroup ?

We need to think about nonableian group .

Does any of u know an example forall n ?


function
complex analysis
complex analysis unsolved

Let $ f : \mathbb C \to \mathbb C$ be a non-constant entire function such that $ f(z\plus{}i)\equal{}f(z)$ for all $ z \in \mathbb C$. Let $ U$ be an open subset of $ \mathbb C$, and let $ z_0 \in U$. Suppose $ g: U\backslash \left\{ {{z_0}} \right\} \to \mathbb{C}$ is an analytic function on $ U\backslash \left\{ {{z_0}} \right\}$. It is given that $ z_0$ is [u]not a removable singularity[/u] of $ g$. Is it true that $ f \circ g$ has an [u]essential singularity[/u] at $ z_0$?

Thank you  :maybe:.

[quote="1234567a"]how about $ f \equal{} e^{z2\pi}$ and $ U \equal{} \mathbb{C}\setminus \{0\}$ and $ g \equal{} \frac {1}{z}$ ?[/quote]

Cannot right? Since $ f \circ g(z) \equal{} e^{\frac{2\pi}{z}}$.


quadratics counting derangement search function blogs algebra

What are they? How do you prove them? What do you use them for in solving problems? How do you calculate them? If you didn't use them, how much extra work do you have to do? And are they used commonly (like quadratic formula, etc.)?

Sorry for so many questions (I actually have more).

See [url=http://www.lmgtfy.com/?q=derangement]here.[/url] Also, the [url=http://www.artofproblemsolving.com/Forum/search.php]search function[/url] will yield several results as well. I might also recommend my [url=http://www.lmgtfy.com/?q=encyclopedia+mathematica+derangement]blog[/url] [url=http://www.mathlinks.ro/Forum/weblog_entry.php?p=1253137&ppage=3]entry[/url] on the subject.


:D i'm in basic algebra  :rotfl:

my bad wrong section


AoPSwiki
articles
number theory
modular arithmetic

hello every one
i am new member here

i need some help

can any one explain to me the modulus\[\bmod \]

[url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]Try reading this from the AoPSWiki[/url].

[quote="Mr.Mathematics"]hello every one
i am new member here

i need some help

can any one explain to me the modulus
\[\bmod \]
[/quote]

hmm prehaps when you have finished with the Wiki article if there is a specific question that you would like us to answer?

[quote="ckck"][quote="7h3.D3m0n.117"][quote="Mr.Mathematics"]how can i use it in this question
\[\frac{{3^{2005}+7^{1426}}}{{16}}\]
[/quote]

Well i'm assuming that the question only asks the units digit? if not then you would have to solve the entire problem by hand  :D[/quote]
I think its asking for the remainder, because thats what most of these problems are like. But I may be wrong and its asking for the units digits, and then you don't have to use mods.[/quote]
well sir what would be the divisor hmmm

[quote="Mr.Mathematics"]
i want the division remainder
\[\frac{{3^{2005}+7^{1426}}}{{16}}\]
[/quote]
[hide="Method One"]
We can notice that $3^{4}= 81 =80+1 \equiv 1 \bmod 16$ and that $7^{2}=49=48+1 \equiv 1 \bmod 16$, so: \[3^{2005}+7^{1426}\equiv 3^{1}+7^{0}\equiv \boxed{4 \bmod 16}\]
[/hide]
[hide="Method Two"]
Euler's Totient Theorem (look it up): \[\varphi(16)=12 \implies 3^{12}\equiv 1 \bmod 16 \text{ and }7^{12}\equiv 1 \bmod 16 \] \[\implies 3^{2005}+7^{1426}\equiv 3^{1}+7^{10}\equiv 3+(7^{-1})^{2}\equiv 3+(7)^{2}\equiv \boxed{4 \bmod 16}\]
(Note that $7^{-1}\equiv 7 \bmod 16$) [/hide]

Could you explain method one a bit further? I'm not very comfortable with modular arithmetic but I've been starting to study it a bit more recently. Thank you!

bump... No one is kind enough to help me with this solution??? :(  :P

[quote="Mr.Mathematics"]about method one
why you took
\[3^{1}and7^{0}\]
[/quote]
It was simplified to that way. Since $3^{4}$ was taken out, and there were 501 of those that could go into 2005 and it left a remainder of 1. So there were only one more 3 left. Same goes for the 7.


vector
projective geometry
geometry solved
geometry

Vector solution is quite straightforward. If no simpler solution appears, I'll post it.

[quote="Fang-jh"]Let points A,B, and C be on line L, points D,E, F, and line L in a plane such that AD∥CF, AE∥BF,BD∥CE. Prove that the three points D,E, and F are collinear.[/quote]

Ok, judging from your source I assume you have read my opera omnia and particularly remember

[url=http://forumgeom.fau.edu/FG2004volume4/FG200406index.html]Atul Dixit and Darij Grinberg, [i]Orthopoles and the Pappus Theorem[/i], Forum Geometricorum, 4 (2004) p. 53-59[/url],

where I use the Pappus theorem - Theorem 1 in this paper - and prove a particular case of it (namely, the case when the points A', B', C' lie on the line at infinity). For your problem, apply this theorem in this particular case to the points A, B, C on one line and the points $A^{\prime}=BD\cap CE$, $B^{\prime}=CF\cap AD$, $C^{\prime}=AE\cap BF$ on the line at infinity, and you obtain that the points $BC^{\prime}\cap CB^{\prime}=F$, $CA^{\prime}\cap AC^{\prime}=E$, $AB^{\prime}\cap BA^{\prime}=D$ are collinear. It's that simple.

  Darij


The years from 1900 to 1999 are written down consecutively and then pluses and minuses are placed alternately between the digits as shown:
\[1+9-0+0-1+9-0+1-1+9-0+2-\dots-1+9-9+9=K\]
Compute the value of $K$.

[hide] First, take all of the digits that should be in the thousands place.  All of these are going to be 1.  However, the first 1 is a plus and the rest are minuses.  Thus, we have 99 -1s and 1 +1.  Our total for the thousands digits is -98.

Now, take the hundreds digits.  These will all be 9s.  Furthermore, they will all be +9.  Thus, we have that there are 100 9s, so 100(9)=900 is the sum of the hundreds digits.

Now, take the tens digit and ones digits together.  The tens will be minuses and the ones will be plusses.  Notice that there are going to be 10 1s, 10 2s, 10 3s, 10 4s... and 10 9s in the tens digit spot.  There are also going to be 10 1s, 10 2s, 10 3s, 10 4s... and 10 9s in the units digit spot.  Thus, the sum of the ten's and the sum of the ones digit will cancel.

Finally, we can conclude that the sum of the digits is $900-98=\boxed{802}$ [/hide]


geometry Euler incenter circumcircle vector trigonometry

1. What can you say about $\triangle ABC$ if the $\text{Euler line}$ is parallel to $BC$.

2. Let $P$ be an internal point if $\triangle ABC$ and let $L, M, N$ be the feet of the perpendiculars to the three sides. Define the point $P$ such that $\triangle LMN$ is equilateral.

Excuse me for being clueless, but what does the point $P$ have anything to do with $\triangle{LMN}$ :? , you just state that $P$ is an internal point, so say if we have an equilateral triangle, in which $\triangle{LMN}$ is obviously equilateral, $P$ can be any point wihin $\triangle{ABC}$ :|

[quote="pieterminate"]1. what was the euler line?  :oops:  :oops:

[hide="2."]
the incenter
[/hide][/quote]

Oh so $P$ was supposed to be defined for the intersection of the altitudes I see ;)

BTW, the euler line is the line which contains the circumcenter, orthocenter, and centroid on it

[hide="partial solution for 1"]
Set up a vector system with the circumcenter at the origin and WLOG let the circumradius be 1. (Scaling the figure won't affect whether the Euler line is parallel to $BC$.) Since $({\bf b}+{\bf c})\cdot ({\bf b}-{\bf c})=||{\bf b}||^2-||{\bf c}||^2=1^2-1^2=0$, ${\bf b}+{\bf c}$ and ${\bf b}-{\bf c}$ are orthogonal.

The Euler line contains the origin and the centroid, given by $\frac{{\bf a}+{\bf b}+{\bf c}}{3}$ so the line consists of scalar multiples of ${\bf a}+{\bf b}+{\bf c}$. Since this is parallel to ${\bf b}-{\bf c}$, ${\bf a}+{\bf b}+{\bf c}$ and ${\bf b}+{\bf c}$ must be orthogonal.

Setting dot product equal to 0, $({\bf a}+{\bf b}+{\bf c})\cdot ({\bf b}+{\bf c})=0$. This dot product equals ${\bf a}\cdot {\bf b}+{\bf a}\cdot {\bf c}+||{\bf b}||^2+2{\bf b}\cdot {\bf c}+||{\bf c}||^2$.

Now each individual dot product must be evaluated. Either $\angle BOC$ and $\angle A$ intercept the same arc or they intercept opposite arcs. If they are the same, then $\angle BOC=2\angle A$. If they are opposite, then $\angle BOC=360^\circ -2A$. Either way, ${\bf b}\cdot {\bf c}=\cos 2A$. The other dot products can be similarly found, so this becomes $2+\cos 2C+\cos 2B+2\cos 2A$.

Finally, when ${\bf b}+{\bf c}={\bf 0}$, then the orthogonal method does not work because ${\bf 0}$ is orthogonal with anything. This happens when ${\bf b}$ and ${\bf c}$ are opposite, so $\angle A=90^\circ$. But when $\angle A=90^\circ$, the Euler line clearly isn't parallel to $BC$ because it includes $A$ and a point on $BC$.

This means that the Euler line is parallel to $BC$ if and only if $2\cos 2A+\cos 2B+\cos 2C=-2$ and $\angle A\neq 90^\circ$. I'm not sure how to get this in a nicer form, but I do know that this should not include isosceles (except for equilateral) triangles.
[/hide]

$L, M, N$ are the feet of the perpendiculars from $P$.


[quote="Agr_94_Math"]If $ s(n)$ denotes the sum of the digits in $ n$ then find all natural numbers $ n$ for which
$ s(n) \equal{} s(2n) \equal{} s(3n) \equal{} ........ \equal{} s(n^2)$.[/quote]

answer is $ 1,10^k \minus{} 1\forall (k \in N)$ :wink:

Hello. Will someone please give me a proof?


calculus
integration
derivative
function
parameterization
calculus computations

I'm having difficulties in understanding some part of the proof in the following page:
http://en.wikipedia.org/wiki/Leibniz_integral_rule

First, how can the following be true:
$D_{x}I_{x}= I_{x}D_{x}$
If the differentiation is applied last, there'll never be a constant in the result, but if integration is applied last, there may be an integration constant.

Second, in that page, I didn't understand how they got steps (3) and (4). It says "by the fundamental theorem of calculus", but I didn't get it.

${\partial G \over \partial b}={\partial \over \partial b}\int_{a}^{b}f(t,x) \, dx = f(t,b) \qquad \qquad (3)$
${\partial G \over \partial a}={\partial \over \partial a}\int_{a}^{b}f(t,x) \, dx =-f(t,a) \qquad \qquad (4)$

I though, maybe
$I(t,x) = \int f(t,x) dx$
then
$G = I(t,a)-I(t,a)$
so
${\partial G \over \partial a}=-{\partial \over \partial a}I(t,a)$

How do we say ${\partial \over \partial a}I(t,a) = f(t,a)$? Yup, [i]this[/i] is the second question.

This method applies only to definite integrals. There is no "+C".

Those partial derivatives are exactly the FTC (derivative of an integral side). Look up a statement of the theorem.

A more general version of the Leibniz Integral Rule can be found [url=http://mathworld.wolfram.com/LeibnizIntegralRule.html]here[/url] (reference http://www.mathworld.com ), it allows for the bounds to be functions of the parameter as well as the integrand, that is it states that

${\partial \over \partial t}\int_{a(t)}^{b(t)}f(x,t) \, dx = \int_{a(t)}^{b(t)}{\partial f \over \partial t}\, dx+f(b(t),t){\partial b \over \partial t}-f(a(t),t){\partial a \over \partial t}$


limit
function
integration
calculus
real analysis
real analysis unsolved

[color=blue]

If the series $\sum_{n\geq 1} a_{n}$ with positive terms and decreasing general term $a_{n}$ is convergent, then:
$\lim_{n\to\infty} n\cdot a_{n}=0$. :)  ;)

[/color]

who is Olivier ? anyway it is very well known and already posted.

Well, I found this in a magazine without solution and I eanjoyed solving this problem. I didn't know that it has already been posted on this forum. :)

There is an almost instant proof using the condensation test: given our conditions, $\sum_n 2^na_{2^n}$ converges. The terms of a convergent series go to zero, so $a_k=o\left(\frac1k\right)$ if $k=2^n$. For other $m$, we have $a_m<a_k$, where $k$ is the largest power of 2 less than $m$; then $a_m=o\left(\frac1k\right)$; since $\frac km\ge\frac12$, $a(m)=o\left(\frac1m\right)$ and we are done.

[quote="grobber"]The continuous version has also been discussed several times:

If $f: [0,\infty)\to(0,\infty)$ is a non-increasing continuous function such that $\int_0^\infty f\mbox dt<\infty$, then $xf(x)\to 0$.[/quote]

Cauchy criterion for the integral and the non-increasing function
gives $\int_{x/2}^{x}f(t)dt\geq x/2f(x) \geq 0$

The above integral tends to 0 so x/2f(x) tends also 0

This method can be applied to series too.
$0<n x_{2n} \sum_{k=1}^{2n}x_k-\sum_{k=1}^n x_k \to 0$.
And $n x_{2n} \to 0 \Rightarrow 2n x^2n \to 0$.
It $0< (2n+1) x_{2n+1}< (2n+1) x_{2n}= x_n +2n x_n$ which both tend to 0. So $n x_n$ is convergent.


the perimeter of a triangle is 40. One side is cut by the bisector of the opposite angle into two segments having lengths 6 and10.. Find the twice the length of the shortest side of the triangle divided by 3.

[hide="Solution"]Let $ a$ and $ b$ be the other two sides of the triangle.

We know $ a+b+16=40$, or $ a+b=24$. Also, $ \frac{a}{6}=\frac{b}{10}$.

Substituting $ a=\frac{3b}{5}$ into the first known equation, we can solve for $ b$:
\begin{align*}\frac{3b}{5}+b&=24\\\\
3b+5b&=120\\\\
b&=15\end{align*}
The length of the the last side, $ b$, is $ 9$.

Finally, the shortest side of the triangle, $ 9$, divided by $ 3$ is $ \boxed{3}$.[/hide]


let $a,b,n>ab$ be nature numbers and $(a,b)=1$ prove that there are natural $x,y$ such that:$ax+by=n$

The coin exchange problem of Froebenius is solved here :

http://www.animath.fr/cours/arithm.pdf

Have a look at pages 30-32.

Pierre.


bn^b=t make b the subject of the formula

http://www.mathlinks.ro/Forum/viewtopic.php?t=90658
Why reposting¿ And especially: explain what the heck you are talking about there¿ (continue there please, not here)

Edit: and don't do like in http://www.mathlinks.ro/Forum/viewtopic.php?t=84038 again!


geometry
geometric transformation
rotation

Why the need of such definitions?

[quote="Wikipedia"]The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.
[/quote]
 :wow:

How else are you going to define time, when the period of Earth's rotation changes depending on various factors.  With the ability to measure things more exactly, it makes sense to define units in terms of physical constants that don't vary.  The meter is defined as a fraction of the distance light travels in a vacuum per second.


If $ f$ an analytic function so that $ f(0)\equal{}f'(0)\equal{}0$, then prove that $ |f''(0)| \leq 2$. Thanks in advance.

Obviously false. What did you really mean?

Maybe the missing hypothesis is that $ f$ maps the unit disk into the unit disk?

Yes, f maps unit disk into the unit disk.

OK, Qruni, we will all appreciate if you spend some more of your time verifying the conditions of the problems you post, so we'll not have spend ours trying to guess what was meant or asking for clarifications  :ninja:

As to the question itself, just divide $ f(z)$ by $ z^2$ and apply the maximum principle.


geometry
3D geometry
prism
calculus
integration
perimeter
ratio

I DON'T GET HOW TO DO THESE.

2.If the sum of the interior angles of a particular convex polygon is 9720°, how many sides does the polygon have?

3.The 442 students and 40 staff of Beckwith School are planning a trip. Each person can go by bus or train. Using the information below, what is the least possible average cost per person for them to make this trip?
Trailmore Bus Lines Arco Trains
Large Bus; $1025: holds up to 64 passengers         $20 per person for fIrst 100

Medium Bus: $792: holds up to 36 passengers        $16 per person for next 300

                                                 $16 per person for all passengers over 400

4.When a survey of 800 people was done, 55% of those surveyed said they ate candy, 65% said they ate chips and 90% said they ate hotdogs. What is the smallest possible number of people who said they ate candy, chips and hotdogs?
5.A right circular cone has a base circumference of 24pi cm and a volume of 1512pi cm3. The cone is cut parallel to its base, and the newly formed shorter cone has a volume of Sóir cm3. What is the height of the shorter cone? Express your answer as a decimal to the nearest tenth.
6.The front view of a metal sculpture is the region ABC shown, with
AG: BC: 22 feet and segment AG perpendicular to segment SC.
Points A and C are the endpoints of a quarter-circle. What is the area
of region ABC? Express your answer to the nearest whole number.
7.Leonard wishes to purchase an air purifier for a small building. He needs to determine the volume of air inside the building in order to decide which model to purchase. The building is a triangular- based right prism on top of a right rectangular prism, with measurements as shown. How many cubic feet of air are in the entire building? The deminesion are L=24'  H=9'  D=15'  The triangle angles are 45,45,90
9.Six pipes each hoving a radius of 0.5 feet are stacked in a triangular pile with three pipes on the ground tangent to each other, two in the next row and then one on top. What is the height of the pile? Express your answer in simplest radical form.
10.The side lengths of a right triangle are each an integral number of units. If one of the legs is 13 units, what is the perimeter of the triangle?

Umm just to tell you, most of these were in the threads that happyme posted awhile back, you should go look at them people have posted their solutions there.
And also, the wording on #3 isn't that great, same thing with the spelling throughout the entire entry.

2. The formula for the total interior angle is $ 180(n \minus{} 2)$ where n is the number of sides. So we set up an equation:

$ 9720 \equal{} 180(n \minus{} 2)$
$ 54 \equal{} n \minus{} 2$
$ 56 \equal{} n$

3. You find that the average price per person for bus is just over 16 while the train is about 20, so you try to put most people in the large bus. Once you cannot fill a large bus, you find that the average for putting the remainder in a large bus is 20 while you can put them all in the train for cheaper. Thus dividing total cost by total people you get 16.30 a person.

4. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180990

5. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180884

6 Basically, you can make a square with sides 22 feet, and the sculpture is the square minus a quarter circle that has a radius of 22 ft. so the area of the circle is $ \frac {1}{4}\pi22^2$ and the area of ABC is 104 square feet.

7. A 45, 45, 90 triangle has side lengths in ratio of 1, 1, $ \sqrt {2}$ So the height is the same as the base of any of the two triangles, or 7.5. Area is base times height, so the area of the top isoceles triangle is 15 times 7.5 times 1/2, and the area of the whole prism is 15*7.5*24. The area of the bottom prism is length * width * height and adding the two, the area is 4590.

9. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180884

10.  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180990

EDIT: Your first sentence is declarative not interrogatory, work on your grammar and spelling.


hi jus posting to revive the forum  :P .
prove that for positive reals $ a,b,c,d$ such that $ a \plus{} b \plus{} c \plus{} d \equal{} 1$,
$ \sum_{cyclic}\frac{a^2}{a \plus{} b} \ge \frac {1}{2}$

By Cauchy Schwartz, The $ LHS$ is greater than or equal to  $ \frac{(\sum a)^2 }{2\sum a} \equal{} \frac{1}{2}$.


summer program
Mathcamp
AwesomeMath
inequalities
AMC
AIME
USA(J)MO

Hey guys, I just found out that the MathCamp 2006 Qualifying Quiz is now online.

http://www.mathcamp.org/

Anyone applying???  I know a lot of American friends applying!  Trust me, they're nice people!  They won't bite! :rotfl:

Aherm, anyone got anything regarding CMO and/or APMO?  It's approaching the end of January, and I'm getting a little concerned...

No one from our school has heard anything yet.

Along similar lines, those who are looking for something to do in the summer may also consider the new summer camp [url=http://www.awesomemath.org/]AwesomeMath[/url].

This program is run by Titu Andreescu and Zuming Feng, whose names speak for themselves. They are both very famous and important figures in the training of the U.S. IMO team. So if you're looking for more of an Olympiad-oriented program, then this would be it.

Hey Yufei.

I heard you were going to be at AwesomeMath.  I'll look more into it.

The Ultra-conservative (Republican) Thomas Mildorf's going to be at Awesomemath.  Haha.  He also taught quite a bit in the first AoPs Olympiad Inequalities class.  It seems he's an Inequalities guy???  :?: :lol:

He says that IMO 2005 #3 was "ridiculously easy" and "should not have been on the IMO."  He Muirheaded the question I believe.  Lol Conservatives....

And they just won the election too.  Well...you guys have fun with Harpy or whoever's your Prime Minister now.  I'll uh...

Apparently I got my APMO email.  Monday March 13.  I'm extremely psyched!

One thing Yufei, isn't AwesomeMath more or less focused around AIME-introductory USAMO type problem solving?  Or are there opportunities there above and beyond?

Yeah. Thomas does seem to be quite an expert at inequalities.

lowest_reynolds, as far as I know, the AwesomeMath classes are divided into different levels to suit the different backgrounds of the students. Although it is up to the faculty to decide what to teach, by looking at the [url=http://www.awesomemath.org/sample_admission_test.pdf]sample admission test[/url] as well as the program's [url=http://reflections.awesomemath.org/]journal[/url], I would say that the material could potential reach very challenging levels. And, considering the reputation of Mr. Feng, I must say that there's virtually no upper limit to how hard his problem sets can get.

This MathJam [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=137]Transcript[/url] may provide you with more information about the program.

[quote="Singular"]Mr. Mildorf is not just an expert on inequalities.  From reading some of his posts, he also seems to be quite solid in geometry.[/quote]

He's good at Geometry.  But he's known for his skill with Inequalities.  I don't know who the GeoMasta is...

[quote="lowest_reynolds"]The Ultra-conservative (Republican) Thomas Mildorf's going to be at Awesomemath.  Haha.[/quote]

Ultra-conservative? :lol: In the entirety of the US population, I would be considered mildly conservative.

If the statement I recall from a friend on the Canadian IMO team is accurate (I believe it was "American Republicans make even the most conservative Canadian look like a commie!" or something like that), then perhaps on a Canadian spectrum I really am ultra-conservative. :)

For that matter, the MA spectrum probably isn't too different... ;)

[quote="lowest_reynolds"]He also taught quite a bit in the first AoPs Olympiad Inequalities class.  It seems he's an Inequalities guy???  :?: :lol: [/quote]

Yup. Geometry is my second specialty, but not quite to the same extent.

[quote="lowest_reynolds"]He says that IMO 2005 #3 was "ridiculously easy" and "should not have been on the IMO."  He Muirheaded the question I believe.[/quote]

I really said it was "ridiculously easy"? I suppose I was astonished that the most difficult (based on scores) IMO 2005 problem could be solved by straight expansion and AM-GM. I will definitely stand by it "should not have been on the IMO," seeing as how the vast majority of the solvers expanded and used similar approaches. The AM-GM I used was not quite equivalent to Muirhead, combining very strong terms with very weak terms to form two medium strong terms. (Something like (9,0,0)+(5,2,2) -> 2 * (7,1,1).) To be kosher I decided to list all of the needed applications. I am not sure whether they accepted statements to the effect of "because <foo> majorizes <bar>, this follows from Muirhead."

[quote="Mildorf"]I really said it was "ridiculously easy"? I suppose I was astonished that the most difficult (based on scores) IMO 2005 problem could be solved by straight expansion and AM-GM. I will definitely stand by it "should not have been on the IMO," seeing as how the vast majority of the solvers expanded and used similar approaches. The AM-GM I used was not quite equivalent to Muirhead, combining very strong terms with very weak terms to form two medium strong terms. (Something like (9,0,0)+(5,2,2) -> 2 * (7,1,1).) To be kosher I decided to list all of the needed applications. I am not sure whether they accepted statements to the effect of "because <foo> majorizes <bar>, this follows from Muirhead."[/quote]

That's also how I basically did it last year. But I quoted Muirhead. However, I included a two-page appendix at the end of my solution clarifying my notation and stating what Muirhead is (in retrospect, I probably should have proven the 3-variable case).

It was quite a frightening moment when I heard rumours that the jury was going to disallow the use of Muirhead ...

I wouldn't call #3 "easy" per se, especially considering it has the lowest average (0.912) of all six problems. Though, the nature of the problem meant that it was really a test of mechanical experience (say, someone who learned Muirhead just a month before probably would not have solved it). And, I noticed that there was quite a bit of correlation between whether one solved #3 and whether he/she got gold (though this observation itself may not be very meaningful at all). I do agree with you (and I said it many times during the IMO) that #3 should not have been on the IMO.

[quote="Mildorf"][quote="billzhao"]It was quite a frightening moment when I heard rumours that the jury was going to disallow the use of Muirhead ...[/quote] Just clarifying, they ultimately did allow it?[/quote]

I'm not sure of the details, but I think they allowed it under the condition that Muirhead was [i]clearly[/i] quoted in the solution. (so not like, [4,0,0] majorizes [2,1,1], so blah)


Find the shortest route from X to Y assuming that the red sagments are deleted.

HELP: The drawing is suppose to be shown.

I suppose even if the red segments were there any path from x to y would be the same length (assuming we're following the lines) as going around the edge of the whole rectangle.

How is the metric space defined over here? Is it standard Euclidean metric or Taxicab metric?

its just euclidean plane. say, the lines are roads and the red lines are closed roads.

[quote="matt276eagles"]Even with the red segments deleted there are many ways to get from the bottom left to the top right without ever going down or left.  All of these paths have the same distance - maybe you mean the number of these paths?[/quote]

note that I said the exact same thing above.

let's try not to repeat things that have already been said.

Yes, that is, the number of shoftest path from X to Y.

[hide]X=(0,0), Y=(9,4).
It's all of the paths not passing through (3,1). There are $ \binom{4}{1}\binom{9}{3}$ paths passing through (3,1). So the number of paths that work is $ \binom{13}{4}\minus{}\binom{3}{1}\binom{9}{4}$.[/hide]

How do you get that the total number of paths is $ \binom{13}{4}$?

Ok no need to answer that, i suddenly see a relation to binomial coefficients.  :P


If I roll three fair 6 sided dice, what is the probability that the sum of the dots facing up on the first two is greater than the number of dots face up on the third? (as usual, the sides have 1, 2, 3, 4, 5, and 6 dots on the various sides)

We work backwords. We can find P(sum of first and second dice > third dice) = 1 - P (sum of first and second dice <= the third dice)
I'm not really sure about this solution, but  I'll go ahead with it.
These are the combinations of dice rolls to fits in P(sum of first and second dice <= 3rd dice)

1st dice     2nd dice     3rd dice
_____________________________
1              1               2
1              1               3
1              2               3
2              1               3
1              1               4
1              2               4
2              1               4
1              3               4
3              1               4
2              2               4
1              1               5
2              1               5
1              1               5
1              2               5
2              1               5
1              3               5
3              1               5
2              2               5
1              4               5
4              1               5
2              3               5
3              2               5
1              1               6
2              1               6
1              1               6
1              2               6
2              1               6
1              3               6
3              1               6
2              2               6
1              4               6
4              1               6
2              3               6
3              2               6
1              5               6
5              1               6
2              4               6
4              2               6
3              3               6
That is 35 combinations, so P(1st dice+2nd dice > 3rd dice) = 216/216 - 35/216 = 181/216 So I came up with the answer of 181/216. Please tell me if I'm wrong.

Well, I only have 7 days left of 8th grade, but...

(6)(6)(6)=216  This is the denominator

all combinations 7 and higher for the first 2 numbers will work.  There are 21.  Multiplying this by 6 we get 126.

So far, 126+y/216 is the answer, y being:

 5 combinations of 6 times the 5 smaller numbers:(5)(5)
                     Following this format,      (4)(4), (3)(3), (2)(2), (1)(1)
                                                        25+16+9+4+1=55
y=55 126+55/216=181/216

Thanks. Your way was fast!  Oh, yeah, forgive me for spelling backwards "backwords" on my last post. :oops:

One more way to approach this problem might be to do a case-by-case off the possible rolls on the last die and multiply each 1/6 chance by the chance that the sum of the other two is larger.

When it comes to dice and similar probability problems it is good to be able to view such problems from different angles.  Often the harder problems require a particular angle to solve, so it's good to know all the angles.


ARML email inequalities induction AMC AIME

Greetings,

This forum is for Alabama problem solvers to discuss problems, problem solving strategies, discuss ARML, make preparations for math competitions, and to make new friends.  If you know somebody who would like to join this forum, please email me at crawford@artofproblemsolving.com and I will add them.

Mathew Crawford

Hello..

I'm Ryan Muller, a sophomore at VHHS. I went to ARML last year (but didn't do very well) and MAO last summer. I'll try to get some of my friends who will probably go to ARML this year to join the group.

Thanks Ryan!

So far this Message Board has been pretty quiet, but I figured that as the Vestavia, Grissom, Hoover students and others find this board it can become a good home for math/problem solving discussion.  We have a few other local ARML/state Message Boards that have gotten going with some good discussions, so hopefully this one will catch on as well.

Just to mention: this is a private forum.  I added you when I noticed your post and saw you were from Vestavia.  I have to manually add members of this group, so if some of your friends join, they need to PM or email me (crawford@artofproblemsolving.com) and let me know their AoPS usernames so that I can add them.

Also, if you have requests for problem types that aren't normally covered on Alabama math team tests (optimization, inequalities, induction, etc.), let me know and I'll find some problems for the Alabama students to practice.  This will probably help AMC, AIME, and ARML scores and get more Alabama students into the USAMO.

[quote=MegaLucario1001]boi its spelled pizitz get it right[/quote]

lol

[quote=pinkbunny1228]yeah. [b]Pizits[/b]?[/quote]

I thought you were better than that

[quote=hurdler][quote=asdf334]hi
didn't richard ruscyzk grow up here[/quote]

I think many of the founders/initial employees at AoPS grew up in Alabama, including Richard. By many, I mean more than 1.[/quote]

Wow! I never knew Alabama used to be like the Math hotspot. Now? With asdf334, we're gonna rise to our former glory  :yup:

[quote=hurdler]Just going to do a quick plug, in case anyone is interested in the local area.

If you're interested in trying out for the Alabama ARML team, you can contact me (Xinke Guo-Xue) at xinkeguoxue@gmail.com. I help to run the practices and sign the team up each year.

ARML is a cool high school level competition. Each year, Alabama selects 15 people to be on their team.

Also, I happen to run math classes year-round, so if you are interested in those, you can also contact me. ;)[/quote]

Cool! Unfortunately, I am a middle schooler and have yet to get into the AIME. I might in 9th grade.

Sure thing. Note it is not necessary to be in high school or have qualified for the AIME to make the team. In fact, last year, there were students in each of those two categories who made the team.

I will likely send emails to middle school math team teachers again this year, but I cannot guarantee the teachers will all pass on the info to their students.


Hey guys, I have been struggling with this question and any help will be greatly appreciated

How many 7 digit even numbers less than 3,000,000 can be formed using all the digits (1,2,2,3,5,5,6)?

[hide="solution"]
Break down the problem into parts: the number is even, so the ones digit is either 2 or 6.

If the ones digit is 6, then the remaining 6 digits must be used to form the first 6 digits of the number, with 2 2's and 2 5's.  So $ \frac{6!}{2!2!} \equal{} 180$.

If the ones digit is 2, then the remaining 6 digits must be used to form the first 6 digits of the number, with 2 5's.  So $ \frac{6!}{2!} \equal{} 360$.

Add the two together to get 540.

[/hide]

You forgot the restriction that the number is less than $ 3,000,000$. You might have to do some casework with the first digit too.

my bad, probably easier to consider the possibilities for the first digit as separate cases:

1 for first digit (use 2233556 to form a 6 digit even)
2 for first digit (use 1233556 to form a 6 digit even)

then add.

Should be easier than considering the ones digit first and then the millions digit.

[hide]If the first digit is a $ 1$ and the last digit is a $ 2$, then there are $ 1\cdot \frac {5!}{2!} \cdot 1 \equal{} 60$ possibilities.
If the first digit is a $ 1$ and the last digit is a $ 6$, then there are $ 1\cdot \frac {5!}{2!\cdot 2!}\cdot 1 \equal{} 30$ possibilities.
If the first digit is a $ 2$, then there are $ 2\cdot \frac {5!}{2!} \cdot 2 \equal{} 240$ possibilities.

Adding them up, you have $ 60 \plus{} 30 \plus{} 240 \equal{} \boxed{330\ \text{seven\minus{}digit even numbers less than 3,000,000}}$[/hide]


What advantages do marsupials gain from the marsupial  fetuses leaving the uterus and going into a pouch instead?

None. That's why they can't live in competitive environments.  :P

Ahem...but on a more serious note, the only real advantage that I see is that the newly-born marsupials would avoid the hazards of pregnancy (i.e. severe disorders, miscarriage, perhaps evasion from possible attacks from the mother's immune system).


The value of 480.434/40.035 is closest to which integer?

Er.. : 48 = 4*12, so I'd tend to say the closest integer you're talking about is : 12  :)

hehe, cool :)
btw, when you wrote this problem, it was "3:14 pm" my time (3,14 LMAO)


[quote="kunny"]Let $a,\ b,\ c$ be positive numbers such that $(a^{3}+2)(b^{3}+2)(c^{3}+2)=27$.
Prove that $a^{2}+b^{2}+c^{2}\geq 3$.[/quote]
It's wrong, kunny. Try $a=b=1.1$ and $c=0.75677161.$ :wink:

[quote="me@home"]Does it hold if $a^{2}+b^{2}+c^{2}=3$ then $\prod a^{2}+2\geq 27$?[/quote]
Actually, the reverse holds, it's just AM-GM on $a^{2}+2,b^{2}+2,c^{2}+2$.


[quote="GameBot"]Each of the faces of a tetrahedral die is an equilateral triangle. The faces are numbered 1, 5, 6, and 8. When the die is rolled, the number on the bottom face is recorded. If the die is rolled twice, what is the probability that the sum of the numbers is a prime number? Express your answer as a common fraction.[/quote]

16 possibilites, $ 1\plus{}6,5\plus{}6,5\plus{}8$ are prime. 3/16.

Sir, are you sure you are right?

[hide] Prime possibilities : $ 1+6, 5+6, 5+8, 6+1, 6+5, 8+5, 1+1 $
Number of possibilities: $ 4^2 = 16 $
Probability: $ \frac{7}{16} $ [/hide]


Pozdrav omladino!

Od sutra kreće novi ciklus takmičenja/natjecanja [b]Oliforum Contest[/b] na italijanskom podforumu u organizaciji [i]Paola Leonettija (bboypa)[/i]. Više informacija o takmičenju i zvanični thread možete naći [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1640043#1640043]ovdje[/url].

Sretno!

Neko ucestvuje? Za sada sam resio samo prvi...

Ispravka, resih i drugi...


Hi everybody :) This is my fist post here, and it's pretty embarassing since it's a stupid question. Namely I can't understand the solution of a problem :(

Here is the text of the problem (sorry for mistakes in the language, I write it all since maybe someone may enjoy solving it):

Given a cifcumference, we draw a chord AB, and a point P on the circumference. We draw than H, which the projection of P on AB. Then we draw the two tangents at the circumference in A and B. We project P on those two tangents findind points E (on the tangent in A), G (on the tangent in B). Show that PH is the geometric mean of PE and PG.

Now...in the solution I can't understand this passage "by inscribed angles PAQ = PBS"...by inscribed angles, why? What does that mean? (I know what an inscribed angle is but...I can't see why PAQ and PBS should be inscribed angles) Can you help me?

You didn't define Q and S...

See http://en.wikipedia.org/wiki/Inscribed_angle


geometry
number theory proposed
number theory

For any $4n-3$ integer point $(x_{i},y_{i}),1 \le i \le 4n-3$,$x_{i},y_{i}\in \mathbb Z$,then we can select $n$ point such that the gravity central of the $n$ point is also a integer point.

Hmmm... it is a not easy result proved recently by Reiher, and $4n-3$ is best possible.
I don't know if Reiher has published his result, but it is the two-dimensional version of the famous Erdös-Ziv-Ginzburg result about the $2n-1$ integers. And Reiher's proof uses classical results on its area, namely Chevalley-Warning theorem and so on...

An olympiad problem? :wink:

Pierre.

[quote="pbornsztein"]
An olympiad problem? :wink:

Pierre.[/quote]
No.
And do you think it will be true for d-dimensional version the best possible will be $2^{d}n+1-2^{d}$

The problem posted in this topic is Kemnitz's conjecture. Although it is proved, it is surely not an olympiad problem.

Thank you,two.And I find the paper of the nice proof of Reiher on the net.Anyone interesting can download it :)


Hiii, this is a problem that I have encountered and I need help ASAP.
This is the figure:
[img]http://img404.imageshack.us/img404/1120/mathhelpppp6yk.gif[/img]
Thanks a lot!

$2AD=2BC=AD+BC=AB+CD=26 \Longrightarrow AD=BC=13$. Therefore the area is $156$.

Correct, but wonder how you got the answer  :roll: .

I did 12*13=156.

Oh, and this isn't really Pre-Olympiad...

Thanks for the replies! Could you please give me a little more details how you get 2AD=2BC=AD+BC=AB+CD=26 ?
Like what proves 2AD = 2BC = AD + BC = AB + CD =26?
Thank you very much!

To pieterminate,
Sorry I didnt know I post it in wrong forum again -sigh- =/ I am newb to this forum, keep posting at wrong forum. Sorry   :(

Because it has an incribed circle, so the sum of the opposite sides are equal. This involves the fact that from a point outside circle draw two tangent lines, then the length of the tangents are equal.

Just curious, but how did you get 156 though?
$(13^2-5^2)*13$ is what I did.

Hmm is there any name for the rule? I tried to look through my books =/ I don't see any explains the inscribed circle =s
Anyway, thanks a lot :D

Hmm I am not sure how shobber did it, but from what i do was form a rectangle in the trapezoid, then use pytherogra (spelling..) theorem. 13^2 = 5^2 + x^2 that gets the height of the trapezoid then use the trapezoid formula to find the area

Here's how you derive the property:

Let $E$, $F$, $G$ and $H$ be the tangent points on $AB$, $BC$, $CD$ and $DA$.
As shobber mentioned, the lengths of the two tangent lines to a circle from a point outside the circle must be equal.
So we can let $AE=AH=a$, $BE=BF=b$, $CF=CG=c$ and $DG=DH=d$.
Now express the length of each side in terms of $a$, $b$, $c$ and $d$ and see what happens ;)

AH HUH! I see! Thanks alot frt :D and everyone thank you soo much :D


[hide][quote="parinaz.sa"]$a,b,c$are positive real numbers prove it:
$\frac{a}{(b+c)^{2}}+\frac{b}{(a+c)^{2}}+\frac{c}{(b+a)^{2}}\ge \frac{9}{4(a+b+c)}.$[/quote]Cauchy-Schwarz gives:
\begin{eqnarray*}\left(\left(\frac{\sqrt{a}}{b+c}\right)^{2}+\left(\frac{\sqrt{b}}{c+a}\right)^{2}+\left(\frac{\sqrt{c}}{a+b}\right)^{2}\right)\left((\sqrt{a})^{2}+(\sqrt{b})^{2}+(\sqrt{c})^{2}\right)&\ge& \left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^{2}\\ &\ge& \frac{9}{4}\end{eqnarray*}
and $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$ because $\left((a+b)+(b+c)+(c+a)\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge 9$
[/hide]


What time of the year does high school usually end?

What about for college?

I'm actually asking because a summer residential program for HS students starts in June 1, and I want to ask the coordinators of the program to change that... I mean you really can't convince your high school to let you out in May if school gets out June 20ish, can you?

The middle/high schools in Oregon where I used to live three years ago had their summer break mid-June through early September.

The jr/sr high school in Indiana I moved from had their summer break mid-May through early August.

It varies by region I guess?

So they don't completely exclude people... but how do I get them to change the program date?

Aside from the obvious answer "you don't" or "you quit school" or something like that...

heh sorry, but I would have to answer "you don't." :oops: There'll be other people who're taking that program, and the coordinators probably won't change the date for just one applicant.

Are you in middle school or high school? Either way, if I really wanted to attend such a program, I'd talk to my counselors to see if I can test out of my current classes - something which should be easy since you'd only have a month left of them. I'm not sure if you'd get credit for the courses, though. Just a guess..

May I ask what program this is?

You could try to take your exams earlier and then use up as many of your free absent days, and then just be like whatever for the rest


calculus
integration
limit
complex analysis

Compute $ \int_{\minus{} \infty}^{\plus{} \infty} \frac{x^2 \plus{} 1}{x^4 \plus{} 2x^2\plus{}2}\,dx$.

Expand to this contour "$ \gamma_a$";
$ \setlength{\unitlength}{2cm} \begin{picture}(4,3) \linethickness{0.4mm} \qbezier(1,1)(1,2)(2,2) \qbezier(2,2)(3,2)(3,1) \qbezier(1,1)(2,1)(3,1) \linethickness{0.2mm} \put(2,0.75){\vector(0,1){1.5}} \put(.75,1){\vector(1,0){2.5}} \put(2.1,2.20){Imag} \put(3.30,0.95){Real} \put(.835,.85){\textit{ \minus{} a}} \put(2.95,.85){\textit{a}} \end{picture}$

Split
$ \ell_{\gamma_a} \equal{} \oint_{\gamma_a} \frac {z^2 \plus{} 1}{z^4 \plus{} 2z^2 \plus{} 2} \mathrm{d}z$
Up to,
$ \int_{\Gamma_a} \frac {z^2 \plus{} 1}{z^4 \plus{} 2z^2 \plus{} 2} \mathrm{d}z \plus{} \int_{ \minus{} \infty}^\infty \frac {z^2 \plus{} 1}{z^4 \plus{} 2z^2 \plus{} 2} \mathrm{d}z$

Where $ \Gamma_a$ is the open semi circle of radius [i]a[/i].

Since $ \left|\frac {z^2 \plus{} 1}{z^4 \plus{} 2z^2 \plus{} 2}\right| \equal{} \mathcal{O}(|z|^{ \minus{} 2})$ we have,

$ \lim_{|z|\to\infty} f(z) \equal{} 0$ hence, $ \ell_{\Gamma_a} \equal{} 0$

Which in turn means that,

$ \oint_{\gamma_a} \frac {z^2 \plus{} 1}{z^4 \plus{} 2z^2 \plus{} 2} \mathrm{d}z \equal{} \int_{ \minus{} \infty}^\infty \frac {z^2 \plus{} 1}{z^4 \plus{} 2z^2 \plus{} 2} \mathrm{d}z \equal{} 2\pi i \sum_{a_k} \text{Res}(f,a_k)$

Where $ a_k$ are the points where $ f$ has poles, do you know how to compute residues?

i don't think that's the only way(but its most likely the shortest)  but i don't do know much about integrals so i can't really say much here    :oops:


If $n = 10^{100} -1 $ has 100 digit $9$'s , find the number of digit $9$'s in $n^3$ :D

[hide="Click to reveal hidden content"]
199
looking at patterns:
93 has 1 zero at the end of the number and 0 at the beginning
993 has 2 zeros at the end of the number and 1 at the beginning
9993 3 end 2 beginning
This will hold true because when we look at values for numbers consisting of 9's squares (81, 9801,  998001...) we can see that when we square (10n-1) we get 102n - 2*10n + 1.  This will leave sufficient room to multiply the 1 by 9999... for each number so the rule holds true.  so n+(n-1)=
100 + 99 = 199[/hide]

Nice observation.... :)
Anyway , thanks for the answer...

[quote="Shogia"][hide="Click to reveal hidden content"]
199
looking at patterns:
93 has 1 zero at the end of the number and 0 at the beginning
993 has 2 zeros at the end of the number and 1 at the beginning
[/hide][/quote]

do you mean 1 nine at the end of the number and 0 at the beginning for 93? i agree with you guy's answer, btw.

I guess she meant $9^3$ has one $9$ in the ending and 0 in the beginning....$99^3$ = 970299 has 2 nine's in the later digits and one in the first few digits..... :) Anyway , that's the only way it made sense....

i would solve it this way:
$(10^{100}-1)^3=10^{300}-3\cdot 10^{200}+3\cdot 10^{100}-1$
this is a $300$ digit number, with $99$, $9$'s in the beginig(from left) and one $7$, $99$ zeros, one $2$ and $100$ $9$'s , so the total number of nines would be :$100+99=199$


When using Latex to submit solutions, should we type the code uncompiled in the e-mail to be cut-and-pasted, or should we compile it and send as a PDF or DVI (or some other format) attachment?

as pdf i guess

they need to view your sol, not edit it...

Thanks; that's what I thought. Sorry if the question sounded silly. I just wanted to be sure.

Peter, please do not answer in my place next time ;)

You need to submit and editable type of file. PDF is not such an type. You need to submit your .tex source file.

oh
ok...

(why do you need to edit our papers if I may ask???)

When I put up solutions, I put many solutions from contestants. Almost each time they need to be edited to make them nicer. Also, when I read the solutions, I like to format them with my own LaTeX settings.

These are the main reasons. Other reasons include the file size for dial-up people (.tex is much smaller than any other output).

Well, I do not have Latex. It is not possible for me to download that. So what will I do to submit solutions?

Will I send in PDF or DOC(MS WORD) format?

Any editable format is ok, so .doc is ok in case you do not wish to write up in $\LaTeX$.


I want to qualify for the AIME next year, so should I train myself so that I know how to solve all the problems (so I make a few stupid mistkaes), or should I train myself to solve only the ones that aren't very very hard (so that way I can make sure that I make no stupid mistakes)? I only have limited time to study, so I don't think that I can learn to solve all problems the most efficiently and without stupid mistakes.

Threads like this one are constantly posted and it gets annoying to keep reading through them and answering the same questions.  Please use the "Search" feature and if you can't find your question, or something similar, then post.


AMC
AIME
inequalities
trigonometry
geometry
similar triangles

Letters for the solutions are:

[hide="Check_this"]
1 E
2 B
3 D
4 D
5 B
6 D
7 D
8 C
9 D
10 E
11 B
12 E
13 D
14 B
15 D
16 B
17 D
18 C
19 E
20 D
21 B
22 C
23 A
24 B
25 B[/hide]

No perfect score occurred

Participants: (13)
captcha000
tetrahedr0n
DePopov
Lucky707
DanielKirchner
towersfreak2006
shake9991
EFuzzy
CODE_NAME_BOB
Optimosis
dinoman
alan
RC-7th
[b]Qualifiers to the Mock AIME (just ones I think that passed this one with the score[/b]

The highest score:

captcha00 and towersfreak2006 - 144

Rest:

tetrahedr0n - 138
Optimosis - 130
RC-7th - 127.5
DePopov - 126
alan - 125
Lucky707 - 121.5
----------------------

The qualifing score I set was 110 but other people scored below 110. (This was in level of mock AMC-11 and that's why :))

Also, take the poll.

[quote]11. In trapezoid ABCD, $\overline {AB}$  and $\overline {CD}$  are perpendicular to $\overline {AD}$ , with AB+CD = BC, AB<CD, and AD = 7. What is AB * CD?

A) 12
B) 12.25
C) 12.5
D) 12.75
E) 13 [/quote]

This was in one of AMC-10's (can't remember which nor I need to know). I liked this problem a lot.

Draw a perpendicular line from B to CD, and the call the intersection E. Call AB = x and then AB = DE = x and call the CE = y. Then, BC = AB+CD or x+(x+y) = 2x+y.

Also, BE has length of 7 because it's same as AD.

So,

$BE^2+CE^2 = BC^2\\
7^2+y^2 = (2x+y)^2\\
49+y^2 = 4x^2+4xy+y^2\\
49 = 4x^2+4xy\\
12.25 = x^2+xy\\
12.25 = x(x+y)$ and $(x+y)$ denote AB*CD.

was it: 2 or 2.5 pts for unanswered questions?

I don't think I miscalculated but I think it was 2.5

Did I count it wrong?  ;)

number 21:

[hide]
mode is 8 which ends up more than once,

x+8+8+y+(x+18)= 60

x+16+y+(x+18)=60

44-x=y+(x+18)  b=(x+18)

-b+(62)=y+b      where 8<y<b

y=62-2b      8=62-2b b=( 28 for b is a max and it has the ability to go down to 21 as b=62-2b)

y=62-2(21<b<28)  b cannot be 28 exactly or it will make y=8 and it wouldnt make it the second largest.

this was a cool problem.  [/hide]

Can someone make a PDF file with the questions?

I want to but I'm currently too busy now. I have LOTS of things to do and Mock AMC G was just one thing that I could do because I was on vacation.

I am not responsible for any mistakes or anything else

The problems are the copyrighted (intellectual?) property of Silverfalcon

i missed the contest but then took it on my own and got 16 probelms  right so thats around 120
ive never done that well on amc-12 so is this much easier than an amc-12? i took very little time to solve some of the later problems such as 18 and 25. The problem with thet two chords stumped me tho.

It was definitely easier.

It was in level of AMC-11ish. I could made it really hard but nah, didn't feel like it. :D

I was hoping that people would discuss the solutions and talk about the way of solving it. Since it doesn't seem like it, I'll just start off to encourage more people.

#11's already up there so I can skip that.

[quote]3. If x,y, and z are real numbers such that:

$(x-3)^2+(y-4)^2+(z-5)^2 = 0$

Then what does x+y+z =?

A) -12
B) 0
C) 8
D) 12
E) 50
[/quote]

To make this possible, all terms in the () should be zero. Thus,

x=3, y=4, z=5 resulting 3+4+5 or 12 or D.

[quote]9. Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with one penny. Which of the following amounts could Boris have after using the machine repeatedly?

A) 5.13 dollars
B) 3.63 dollars
C) 6.30 dollars
D) 7.45 dollars
E) 9.07 dollars
[/quote]

This problm has constant increase of 1.24 dollars. And you're start off by 1 cent. See if you can figure this out. :)

[quote]12. Let f(x) be a function with the two properties:

(a) for any two real numbers x and y, $\displaystyle f(x+y) = x + f(y)$, and
(b) f(0) = 2.

What is the value of $f(1998)$ ?

A) 0
B) 2
C) 1996
D) 1998
E) 2000
[/quote]

This can be separated into 1998 and f(0). So,

f(1998) = 1998+f(0)
= 1998+2
=2000

Okay. No one yet helping me out now. I'm just doing this so whenever people get something wrong, they can see why they got wrong and get right next time.

[quote]14. Mr. and Mrs. Zeta want to name baby Zeta so that its mongram (the first, middle, and last initial) will be in alphabetical order with no letter repeated. How many such monograms are possible?

A) 276
B) 300
C) 552
D) 600
E) 15600[/quote]

This question was missed by many people. This is actually $\binom {25}{2}$, so 300. But why?

Well, the last name has to be Z so you have 25 leftovers. The thing is that subsets of $\binom {25}{2}$ are in alphabetical order so it's 300.

There are other ways but this is what I can remember now.

sure why not
7. the sum of the roots is -1, since the second highest coefiicient is 1. Thus we could have the roots be (1, -2), (2, -3)...(9, -10), so 9 different values, D.

[quote="Silverfalcon"]23. A child has a set of 96 distinct blocks. Each block is one of 2 materials (plastic, wood), 3 sizes (small, medium, large), 4 colors (blue, green, red, yellow), 4 shapes (circle, hexagon, square, triangle). How many blocks in the set different from the "plastic medium red circle" in exactly two ways? (The "wood medium red square" is such a block.) A) 29 B) 39 C) 48 D) 56 E) 62[/quote]

Using casework based on how they differ and letters to represent how they're changed:
(S represents small and s represents square)
WS
WL
WB
WG
WY
WH
Ws
WT
SB
SG
SY
SH
ST
Ss
LB
LG
LY
LH
LT
Ls
BH
BT
Bs
GH
GT
Gs
YH
YT
Ys
On the test we would notice that the number of combinations beginning with B,Gand Y is the same and that the number of combinations beginning with S and L are the same.

So we have 29, (A).

[quote="Silverfalcon"]22. Call a 7-digit telephone number $d_1d_2d_3-d_4d_5d_6d_7$memorable if the prefix sequence $d_1d_2d_3$is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each can be any of the ten decimal digts 0,1,2,.....9, the number of different memorable telephone numbers is:
A) 19810 B) 19910 C) 19990 D) 20000 E) 20100[/quote]

There are 1000 prefix sequences. If it is the same as the sequence of the last three numbers, there are 10 choices for the middle number, and vice versa for the other sequence. So we have 20000. However, we have overcounted those ten sequences who consist of numbers who are all the same. So the answeer is 19990, (C.

For number 22, we can use Inclusion-Exclusion Principle also.

Can anyone solve it using it?

 :D

Too late to ask for AMC A, but not too late for AMC B :) .

Questions I have:

1. How do you do 9?

2. How do you do 24? I putzed around with similar triangles for a while and got nowhere. Everything I tried just kept giving me the equivalent of a=a!

3. For 21, I took the problem casewise, and found that 8 occured at most 2 times. I then set up equations and inequalities, listed cases and found 4 possible values for the second largest element of the set. Obviously, this failed miserably sinc eI got it wrong. Can someone help?

Thanks for the help.

All right. I'll give some explanations.

And thanks for trying out these  :)

Ok. Number 9 is my favorite question. This is because if you see it, you'll go like "AHHH!!! Should've seen it!"

[quote]9. Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with one penny. Which of the following amounts could Boris have after using the machine repeatedly?

A) 5.13 dollars
B) 3.63 dollars
C) 6.30 dollars
D) 7.45 dollars
E) 9.07 dollars
[/quote]

[hide="here"]Here is the important part. Everytime she changes, her money is increased by 1.24 dollars, and she is started with one penny, or simply 0.01 dollar.

So, the amount should be in form of:

0.01 + 1.24(n)

Where n is the integer. I don't think it matters that much but usually, we speak this way.

When n=6, 0.01+1.24(6) = 7.45 so D is our answer.[/hide]

Number 24, there was another person who asked for the answer on the Problem page so I posted up there. It's still there.

http://www.artofproblemsolving.com/Forum/topic-22141-30.html

If you want the trigonometry way, beta has posted one in Mock AMC J.

I belive the number is 15.

http://artofproblemsolving.com/Forum/topic-24437.html

Just note that mock AMC J was perhaps the HARDEST AMC ever made in this forum (I'm not sure for HARDEST but certainly not EASIEST)

For number 21, it seems that you miss it out on the first page. Look on tongos's reply:

http://www.artofproblemsolving.com/Forum/topic-22344.html

Hope this helps!


A line parallel to $ y \equal{} 4x \plus{} 6$ passes through $ (5, 10)$. What is the $ y$-coordinate of the point where this line crosses the $ y$-axis?

An equation of y=ax+b has slope a, and parallel lines have the same slope.
Thus, the line that passes through (5,10) has an equation y=4x+b.
Plug it in and get 10=4(5)+b.
Thus, -10=b.
A y intercept is when x=0.
Since y=4x-10, plug x=0 and get [b]-10.[/b]

Just a little, technical note: slope-intercept form is most commonly expressed as $ y \equal{} mx \plus{} b$ rather than $ a$.

When expressing a line in slope-intercept form ($ y\equal{}mx\plus{}b$), $ b$ is the y-intercept.


United  Parcel Service (UPS) requires that the sum of a package's length, $ x$, and girth, $ 2y \plus{} 2z$, not exceed 130 inches. If a package has
a width  of 20 inches and a length of 36 inches, what is the maximum
number   of inches in the height of the package if UPS is to accept it?

[asy]real width=15;
real height=5;
real depth=25;
draw((0,0)--(width,0)--(width,height)--(0,height)--cycle);
pair delta=(-2*width/3,4*height/3);
draw((0,0)+delta--(0,height)+delta--(width,height)+delta--(width,height));
draw((0,0)--(0,0)+delta);
draw((0,height)--(0,height)+delta);
label("$y$",(width/2,0),S);
label("$y$",(width/2,height),S);
label("$z$",(0,height/2),E);
label("$z$",(width,height/2),E);
label("$x$",(2*(width,height)+delta)/2,NE);[/asy]


analytic geometry
trigonometry
graphing lines
slope
calculus
calculus computations

A curve is given by

$ x^{2} \plus{}y^{2}\plus{}2axy\equal{}1$,

where $ a$ is a constant satisfying $ 0<a<1$. Show that the gradient of the curve at the point $ P$ with coordinates $ (x, y)$ is
$ \minus{}\frac{x\plus{}ay}{ax\plus{}y}$,

provided $ ax\plus{}y \neq 0$. Show that $ \theta$, the acute angle between $ OP$ and the normal to the curve at $ P$, satisfies

$ \tan \theta \equal{} a \mid y^{2}\minus{}x^{2} \mid$.

Show further that, if $ \frac{dy}{dx} \equal{}0$ at $ P$, then:

(i) $ a(x^{2}\plus{}y^{2}) \plus{}2xy\equal{}0$;

(ii) $ (1\plus{}a)(x^{2}\plus{}y^{2} \plus{}2xy) \equal{}1$;

(iii) $ \tan \theta \equal{} \frac{a}{\sqrt{1\minus{}a^{2}}}$.

[hide="First Part"]Implicitly differentiating this we have $ 2x\plus{}2yy'\plus{}2ay\plus{}y'2ax\equal{}0$ or $ x\plus{}yy'\plus{}ay\plus{}y'ax\equal{}0$. Moving all the terms to the RHS except the terms that have the $ y'$ we have $ y'(ax\plus{}y)\equal{}\minus{}(x\plus{}ay)$ and solving, $ y'\equal{}\minus{}\frac{x\plus{}ay}{ax\plus{}y}$[/hide].

Where's point $ O$?

Oh right... :blush:

[hide="Arg"]The slope of the normal line to the curve at $ P$ is $ \frac{ax\plus{}y}{x\plus{}ay}$. Thus the angle this makes with the x-axis is $ \tan^{\minus{}1}(\frac{ax\plus{}y}{x\plus{}ay})$. The slope of the line $ OP$ is $ \frac{y}{x}$ so the angle that makes with the x-axis is $ \tan^{\minus{}1}(\frac{y}{x})$. Therefore the angle between these two lines is $ \tan\theta\equal{}\frac{\frac{ax\plus{}y}{x\plus{}ay}\plus{}\frac{y}{x}}{1\minus{}(\frac{ax\plus{}y}{x\plus{}ay})(\frac{y}{x})}$. So $ \tan\theta\equal{}\frac{ax^2\plus{}xy\plus{}xy\plus{}ay^2}{x^2\plus{}axy\minus{}axy\minus{}y^2}$. $ \tan\theta\equal{}\frac{a(x^2\plus{}y^2)\plus{}2xy}{x^2\minus{}y^2}$...[/hide]
Can you help me with this? :D
I don't know what I did wrong or what to do next...

Hi ckck.

Slope of $ OP$ is $ m_{1} \equal{} \frac {y}{x}$

Slope of normal is $ m_{2} \equal{} \frac {ax \plus{} y}{x \plus{} ay}$

Therefore $ \tan \theta \equal{} \mid \frac {m_{1} \minus{} m_{2}}{1 \plus{} m_{1}m_{2}} \mid$

$ \equal{} \mid \frac {\frac {y}{x} \minus{} \frac {ax \plus{} y}{x \plus{} ay}}{1 \plus{} \frac {y}{x} \cdot \frac {ax \plus{} y}{x \plus{} ay}} \mid$

$ \equal{} \mid \frac {\frac {xy \plus{} ay^{2} \minus{} ax^{2} \minus{} xy}{x(x \plus{} ay)}}{\frac {x^{2} \plus{} axy \plus{} axy \plus{} y^{2}}{x(x \plus{} ay)}} \mid$

$ \frac {a\mid y^{2} \minus{} x^{2}\mid}{x^{2} \plus{} y^{2} \plus{} 2xy}$

$ \equal{} \frac {a \mid y^{2} \minus{} x^{2}\mid}{1}$

$ \equal{} a \mid y^{2} \minus{} x^{2} \mid$.

Hope that helps ckck. I could do the last three parts if you like, but I posted the problem becasuse I thought it would interest others.


a,b,c are positive numbers(a,b,c>0).
max{$\frac {a^3 \times b^2 \times c }{a^6+b^6+c^6}$}=? :D

I know that but i mean can you find max by number!
Thank you.Find the number! :huh:

What do you mean??

Do you agree with my sol?
Is it ok??

 :lol:  :)  :D

I know it's true but is it an open question? :D

probably this could be in the
algebra unsolved or proposed exercises.......

Thanks all of you but may be there a number(only number).
ANYWAY THANK YOU SO MUCH. :D

please read socrate's post for the number.

yas he is right. i am so sorry. :blush:
His solution is very cool! ;)

[quote="jensen"]a,b,c are positive numbers(a,b,c>0).
List: [list]text[/liFont Color:  Default


function
superior algebra
superior algebra unsolved

I have

\[ ∏(x) = \sum_{k=1}^{\infty} 1/k \phi({x^{1/k}})\]
then from this formulas how can I by using mobius inversion formula conclude the following formula:

\[ \phi(x) = \sum_{k=1}{\infty} \frac{\mu(k)}{k} ∏(x^{1/k})\]

here $ \mu$ is mobius function $ \phi(n)$ is prime number less then n.

The last one should be

\phi(x) = \sum_{k=1}^{\infty} \frac{\mu(k)}{k} ∏(x^{1/k})

It should be

\[ \phi(x) = \sum_{k=1}^{\infty} \frac{\mu(k)}{k} ∏(x^{1/k}) \]

the last A hat should not be there.

Sorry for so many mistakes.


Given $ x \equal{} \minus{}3$ and $ y \equal{} \frac{1}{2}$, evaluate \[ \frac{\minus{}y^{\minus{}3}}{(2x \plus{} y^{\minus{}1})^{\minus{}2}}\]


trigonometry
ratio
geometric sequence
arithmetic sequence

So that people don't have to download:

[b](1)[/b] Let $ \phi$ and $ \alpha$ be two real values such that $ \left \{\frac {\cos ^4 \phi}{\cos ^2 \alpha} , \; \frac {1}{2} , \; \frac {\cos ^4 \alpha}{\cos ^2 \phi} \right \}$ is an arithmetic progression. Show that $ \left \{\frac {\cos ^{2n \plus{} 2} \phi}{\cos ^{2n} \alpha} , \; \frac {1}{2} , \; \frac {\cos ^{2n \plus{} 2} \alpha}{\cos ^{2n} \phi} \right \}$.

[b](2)[/b] Find a three digit number $ n$ such that the digits of $ n$ are in a geometric progression and the digits of $ n \minus{} 400$ are in an arithmetic progression.

Well, answer is 931.  Not sure of any way of doing it other than guess & check.

The hundreds digit is greater than 4 and must be divisible by a square (why?) -- therefore, it must be 8 or 9

They correspond to geometric sequences 842 and 931. But 442 is not an arithmetic sequence and 531 is.

Suppose one integer is $ a$. Then the next integer is $ b$. Therefore, the common ratio is $ b/a$, which must be a rational number.

So now, let the common ratio be $ p/q$, in lowest terms.

Then we know the third term is an integer. Thus, $ ap^2/q^2$ is an integer. Since $ q^2$ has no common factors with $ p^2$ (because $ p/q$ is reduced), then in order for this to be an integer, $ q^2$ must divide $ a$.


I would appreciate anyone's help with this problem:

'At time $ t$ a particle is moving vertically downwards with speed $ v$ in a medium which exerts a resistance to the motion proportional to the square of the speed of the particle. If its terminal velocity is $ V$ , find the equation of the motion of the particle in terms of $ \frac{dv}{dt}, g, v^2$ and $ V^2$, and write down the equation of motion when the particle is moving vertically upwards.
A particle is projected vertically upwards in the medium with speed equal to the terminal velocity $ V$. Find the speed of the particle, in terms of $ V$, when it returns to the point of projection.'

[url=http://www.karlscalculus.org/l12.1.html]this page[/url] I think has what you're looking for, it will do the first part, and i think has enough info for you to adapt the calculations for the second part, if you have any questions on it I can try to help.

Thanks. I can solve it but was just wondering if my way was long-winded.
Appreciate your help.


What's the solution of this equation:

$ x^{2}\plus{}x\plus{}1\equal{}\frac{156}{x^{2}\plus{}x}$

I am kind of too lazy to solve it right now. But here's how I'd being.

We notice the $ x^2 \plus{} x$ term is common to both sides. So, let $ a \equal{} x^2 \plus{} x.$ In that case, we have $ a \plus{} 1 \equal{} \frac {156}{a} \implies a^2 \plus{} a \minus{} 156 \equal{} 0.$ Solving that for $ a$ should end you up with the roots of the original equation.

[color=darkblue][b]soory limac
i didnt notic your solution
here another idea[/b][/color]
 :play_ball:

Ah ok ;)
Thanks men. I understand now, thanks mr. mohamed, interested logic.

I did it the same way as limac, getting $ x^{2} \plus{} x \minus{} 12 \equal{} 0$ so that $ x \equal{} \minus{} 4$ or $ 3$ and $ x^{2} \plus{} x \plus{} 13 \equal{} 0$, so that $ x \equal{} \frac { \minus{} 1 \pm i\sqrt {51}}{2}$.

[quote="AndrewTom"]I did it the same way as limac, getting $ x^{2} \plus{} x \minus{} 12 \equal{} 0$ so that $ x \equal{} \minus{} 4$ or $ 3$ and $ x^{2} \plus{} x \plus{} 13 \equal{} 0$, so that $ x \equal{} \frac { \minus{} 1 \plus{} i\sqrt {51}}{2}$.[/quote]

[quote]$ x\equal{}\frac{\minus{}1\plus{}i\sqrt{51}}{2}$[/quote]

i think the S.S in R my dear
x2+x+13 = 0  has no solution n R
thanks

Sorry luiseduardo, since you asked "what's the solution of this equation?", I presumed you meant all the solutions.

No problem, that's ok, the original question don't tells it. If is in R or not.


geometry
parallelogram
trigonometry
geometry solved

Given a parallelogram ABCD, let M and N be the midpoints of the sides BC and CD.
Can the lines AM, AN divide the angle BAD into three equal angles?

Yes, a nice problem. In fact, the lines AM and AN cannot divide the angle BAD into three equal angles. Here is my proof:

Assume that the lines AM and AN divide the angle BAD into three equal angles, i. e. assume that we have t = < BAM = < MAN = < NAD. Then,

  $\measuredangle BAD=\measuredangle BAM+\measuredangle MAN+\measuredangle NAD=3t$.

Since < BAM = < NAD, and also < ABM = < ADN (this is because AB || DN and BM || AD), the triangles ABM and ADN are similar, so that $\displaystyle \frac{AB}{BM}=\frac{AD}{DN}$. Since BM = AD / 2 and DN = AB / 2, we can rewrite this as $\displaystyle \frac{AB}{AD}=\frac{AD}{AB}$, hence AB = AD, and our parallelogram ABCD is a rhombus.

Therefore, $\measuredangle BAC=\measuredangle CAD=\frac{1}{2}\measuredangle BAD=\frac{1}{2}\cdot 3t=\frac{3}{2}t$. Hence, $\measuredangle MAC=\measuredangle BAC-\measuredangle BAM=\frac{3}{2}t-t=\frac{1}{2}t$. Also, since BM || AD, we have $\measuredangle ABM=180^{\circ }-\measuredangle BAD=180^{\circ }-3t$, and, since the triangle ABC is isosceles, we have $\measuredangle BCA=\measuredangle BAC=\frac{3}{2}t$, or, in other words, $\measuredangle MCA=\frac{3}{2}t$. Thus, by the Sine Law in triangle ABM, we find

  $\displaystyle \frac{BM}{AM}=\frac{\sin \measuredangle BAM}{\sin \measuredangle ABM}=\frac{\sin t}{\sin \left( 180^{\circ }-3t\right) }=\frac{\sin t}{\sin \left( 3t\right) }$,

and by the Sine Law in triangle ACM, we find

  $\displaystyle \frac{CM}{AM}=\frac{\sin \measuredangle MAC}{\sin \measuredangle MCA}=\frac{\sin \left( \frac{1}{2}t\right) }{\sin \left( \frac{3}{2}t\right) }$.

Now, since M is the midpoint of BC, we have BM = CM, so that the left hand sides of the last two equations are equal. Therefore we conclude

  $\displaystyle \frac{\sin t}{\sin \left( 3t\right) }=\frac{\sin \left( \frac{1}{2}t\right) }{\sin \left( \frac{3}{2}t\right) }$,

or, equivalently,

  $\displaystyle \frac{\sin t}{\sin \left( \frac{1}{2}t\right) }=\frac{\sin \left( 3t\right) }{\sin \left( \frac{3}{2}t\right) }$.

Using the formula $\sin x=2\sin \left( \frac{1}{2}x\right) \cos \left( \frac{1}{2}x\right)$, this becomes

  $\displaystyle \cos \left( \frac{1}{2}t\right) =\cos \left( \frac{3}{2}t\right) $.

But $\frac{1}{2}t$ and $\frac{3}{2}t$ are acute angles (since 3t = < BAD is an angle in our parallelogram!). Hence, we have $\frac{1}{2}t=\frac{3}{2}t$, what is impossible. Proof complete.

  Darij

Actually, I also found  this solutiuon (alike) but I'm still a beginner (13 years) so I posted it to see if my solution is correct, or to see another solution.
Thanks.


inequalities
real analysis
real analysis unsolved

Let A be a set of complex numbers. (A "multivalued" set may also be allowed.) Prove that there is a subset $B \subset A$ such that
$\pi | \sum_{x \in B} x | \geq \sum_{x \in A} |x|$.

[quote="Alex-S"](A "multivalued" set may also be allowed.) [/quote]

What does this mean!

It means that we really have a list rather than the set. Elements can be repeated, but we don't care about order.

If $A$ is infinite, we really need the sum of elements of $A$ to converge absolutely. This goes bad in annoying ways when that's not true.

Ok, $\sum_{x \in A} |x|$ schould converge. (Although otherwise the claim still makes sense if appropriate interpreted.)
thank you, jmerry, for your remark.

And you can forget about infinite and multivalued sets if you don't like them.
It doesn't really matter for this problem.


The bed of a pond can be modeled by $30y=2x^2-25x+12$, wher x and y rare measured in meters and the x-axis mathces the water level of the pond.  WHat is the width of the pond?

If I understand it right, it is just the distance between the zeros.
So $2x^2-25x+12=0$
$(2x-1)(x-12)=0$
$x=\frac12,\ 12$
distance: $11\frac12\ \text{m}$

[quote="mathmanman"]I think it'd be 11.5 then, wouldn't it?[/quote]
oh yeah, wow, thanks, stupid mistakes

No problem :)
You've done all the reasoning, that was just a silly mistake.
"errare humanum est"  ;)


I was browsing through my old files last night and I found this problem, an Irish training problem during Y2K. I found that my solution is long and inelegant. I wonder if someone has a nice solution to it, instead of by cases, the way I did it back then.

Find all integers $x,y$ satisfying the equation
\[
x^3+y^3=(x-y)\cdot1997^2
\]

[quote="3X.lich"]
Find all integers $x,y$ satisfying the equation
\[ x^3+y^3=(x-y)\cdot1997^2  \][/quote]

 :D  :D

[hide="Not that elegant, but..."]
Rewrite $a = x + y, b = x - y$.  We have $x = \frac{a+b}{2}, y = \frac{a-b}{2}$, and the problem can be rewritten

$a \left( \frac{a^2 + 2ab + b^2 - a^2 + b^2 + a^2 - 2ab + b^2}{4} \right) = b (1997)^2$
$a (a^2 + 3b^2) = b (2 \times 1997)^2$

Denote $k = (a, b)$.  Rewrite $a = ck, b = dk$ and the equation becomes

$ck^2 (c^2 + 3d^2) = d (2 \times 1997)^2$

Where $(c, d) = 1$.  This means $c^2 + 3d^2 | 2^2 1997^2$.  Note that $1997 \equiv 2 \bmod 3$ and that $c^2 + 3d^2 \equiv c^2 \bmod 3 \equiv 0, 1 \bmod 3$ so $c^2 + 3d^2$ cannot equal $1997$, so it can equal $1, 4, 1997^2$.  The casework is relatively simple.

Case 1:  $c^2 + 3d^2 = 1$.  Then $c = 1, d = 0, k = 0, a = 0, b = 0, \boxed{x = 0, y = 0}$

Case 2:  $c^2 + 3d^2 = 4$.  Then $c = 1, d = 1, k = 1997, a = 1997, b = 1997, \boxed{x = 1997, y = 0}$

Case 3:  $c^2 + 3d^2 = 1997^2$.  Then $ck^2 = 4d$, and since we know $(c, d) = 1$ we must have $k = 2, c = d$ or $c = 1$, which produces no solutions.
[/hide]

Looking at the problem it looks like you forgot the solution $(-1997, 0)$. Just my 2 cents. Don't what didn't catch it in your answer.

Ah, sorry.  That last step should be $c, d = \pm 1$ which would give $(\pm 1997, 0)$.  My mistake.   :)


just wondering..
anyone here applied to ShadValley?

I went... it was mediocre at best.

- The people there were very nice; a lot of them still talk to each other, hang out, etc... I really don't, although there -are a few shads from my year at waterloo who I still see every once in a while.
- Lots of fun camp-type stuff... random games, trips out to do stuff, excursions.
- The lectures were 50% good, 50% bad... and the bad ones really really sucked.
- The mathematical content consisted of maybe 2 lectures that were way way below the level of anyone on this forum.
- The engineering lectures were completely devoid of math... I couldn't believe that all the engineering had been robbed of mathematical content.  I felt like I was in grade 8 or something.
- The big projects were fun, but we didn't benefit from the university's facilities at all... we never got to use the shops or anything.  Really, we just ended up buying parts from Canadian Tire and using tape, power-tools, etc to put stuff together.
- There was a lot of technical knowledge lacking among the people there. Many of them were musically talented, very social, gung-ho type-a leadership people, but the bright people were few and far between.  (In one group, not a single person knew how to build a circuit to power a motor.  It was somewhat pathetic.)
- They make you sign a contract that anything you invent while at Shad becomes the property of Shad.  So save your good ideas for when you get home.
- Bombardment with spam.  A lot of people who went to various campuses reported getting visitors from microsoft give them free water bottles and books, etc, and all kinds of crap.  For doing the project, some sponsor donated web hosting for forums, which we got after we had created our own private forum, and we had to move all our crap over to not piss off the sponsor (yea... I was mad about that... they make us do a lot of things to keep sponsors happy... it was very very political in that respect).

So yeah.  Shad was mediocre.  A lot of people liked it more than I did.  They did, however, hook me up with a nice job for the second half of the summer (with a certain symbolic computation software company :D ).

Hmm, despite Elyot's amusing story about his bad experience, which he told me in the summer, I still applied for this summer.  Not going though...don't have the money to dish out for it.