algebra
IMO Shortlist
modular arithmetic
number theory

Let $ k$ be a positive integer. Show that there are infinitely many perfect squares of the form $ n \cdot 2^k \minus{} 7$ where $ n$ is a positive integer.

Very easy to see (and well known) that if a\equiv s^2 (mod 8)
than for each k tjere exist r such that a\equiv r^2 (mod 2^k)
(If a is a quadratic res mod 8 than it is a quadratic res mod 2^k)

-7 -- quadratic mod 8.

Cool! Here's what I did:

I used induction. Let f(n) be the greatest power of 2 which divides n.
We prove the property for k=1 and then we assume it's true for k from 1 to t and we try to prove it for t+1:

There are infinitely many b s.t. b2+7=n*2t for some n. Let's assume the sentence is false for t+1, so there are infinitely many b s.t. f(b2+7)=2t. We can obviously find infinitely many a s.t. f(a(a+2b))=2t (we just take a s.t. f(a)=2t-1 ). We get that f(b2+a(a+2b))=2t+1, so f((a+b)2)=2t+1, so we have a contradiction because we assumed the sentence was false for k=t+1, so the sentence is true for k=t+1, and the induction is over.

A silly overkill with Hensel's Lemma.It suffices to show there is at least one integer $x$ for each $k$ such that $n\cdot 2^k-7=x^2(\star)$.See that if $x$ works then $(2^k+x)$ works too.Since $(2^k+x)^2=2^k(2^k+2x)+x^2=(n+2^k+2x)\cdot 2^k-7$
We consider the polynomial $f(x)=x^2+7$.
We will try to find a root of $f(x)$ modulo $2$.To rephrase:
$x^2+7\equiv 0\pmod{2}$ we get $x=1$ works.But here $f^{\prime}(x)\equiv 2x \equiv 0\pmod{2}$ so Hensel's Lemma doesnot directly work.Indeed $f(x)=x^2+7\equiv (x-1)^2$ in $\mathbb{F}_2[x]$.But see that $f(1)=8=4\cdot 2$ so for each $k>1$ there is a lifting of the root modulo $2^k$.Hence for each $k$ there exists an integer $x_k$ such that $x_k^2+7\equiv 0\pmod{2^k}\implies x_k^2+7=n_k\cdot 2^k$ and hence we are done.Because we wanted to find (at least) one integer $x$ such that $(\star)$ is satisfied.

[quote=Math1331Math]Set $a=7a', b=7b'$

which gives $7a'-b'2^k=-1$

Thus if we can show that perfect squares can be written in the form $b'2^k$ the result immediately follows by Pell
But this is trivial to show so we're done[/quote]

I too got sorta same.. but is it correct?

it may be but it is incomplete. this only tests for multiples of $7$.

All numbers of the form $((2^k-7)^2)m^2$are solutions .Or I am not understanding the question.

The problem is equivalent to showing that for a fixed $k,$ there are infinitely many positive integers $a,$ such that $2^k|a^2+7$. It is obvious for $k=1, 2, 3$ that we can just choose $a \equiv 1 (\mod 8)$.

Assume now the statement is true for all integers up to some $k$. We shall prove it for $k+1$. Let's look at some $a$ that is solution of the divisibility for $k$. Obviously $a$ is odd.

If $2^{k+1}|a^2+7$, we can just pick the same $a$ as a solution for $k+1$.

Let's see what happens if $2^k$ divides $a^2+7,$ but $2^{k+1}$ doesn't. We want to find an integer $r,$ such that $a+r$ is a solution. In other words, we want $a^2+7+2ar+r^2$ to be divisible by $2^{k+1}$. Note that $r$ must be even. We have $r(r+2a) = r^2+2ar \equiv-a^2-7 \equiv 2^k (\mod 2^{k+1})$. Since $a$ is odd, $r+2a$ will be only divisible by $2$ and so we should look for $r$ divisible by $2^{k-1},$ but not by $2^k$. So we decide to go with $r=2^{k-1}$. Now $a+r$ is indeed a solution, since $(a+r)^2+7 = a^2+7+r^2+2ar \equiv 2^k+0+2^k \equiv 0 (\mod 2^{k+1})$ and we are done $\blacksquare$.

Remark: For the $r^2$ term to vanish, we need $2k-2 \geq k+1,$ or $k \geq 3$, so for our solution it is crucial to check the cases $k=1, 2, 3$. One can easily prove wrong statements like: For all $k,$ there is an integer $a,$ such that $a^2+3$ is divisible by $2^k$ with the same lifting method if he or she doesn't check the base cases.

[b][color=#00f]Claim: [/color][/b] Suppose $x^2\equiv -7\pmod{2^k}$ for some $k\ge 3$. Then
\[ x^2\equiv -7 \pmod{2^{k+1}} \quad \text{or} \quad (x+2^{k-1})^2 \equiv -7 \pmod{2^{k+1}}.\]
[color=#00f][i]Proof:[/i][/color] We know $\ell:= \tfrac{x^2+7}{2^k} \in \mathbb{Z}$ from the condition.
[list]
[*] [color=#00f][i]Case 1:[/i][/color] $\ell$ is even. Then \[\frac{\ell}{2} = \frac{x^2+7}{2^{k+1}} \in \mathbb{Z},\] so $x^2\equiv -7 \pmod{2^{k+1}}$.
[*]  [color=#00f][i]Case 2:[/i][/color] $\ell$ is odd. Then
    \[ \frac{x^2+2^kx+7}{2^{k+1}} = \frac{x^2+7}{2^{k+1}} + \frac{x}{2} =\frac{\ell+x}{2} \in \mathbb{Z}\]
    since $x$ and $\ell$ are both odd. Therefore, $-7\equiv x^2+2^kx \equiv (x+2^{k-1})^2\pmod{2^{k+1}}$.
[/list]
This proves the claim. $\blacksquare$

We have $(1+8\ell)^2\equiv -7\pmod8$ for any $\ell\ge 0$. So we can build up an infinite family of pairwise distinct values using the claim that are $-7\pmod{2^k}$, increasing the exponent of the power of $2$ one each time, by initially having the $\ell$'s sufficiently far apart.

[hide=Remark]
The lifting process in the Claim is similar to Hensel's Lemma. [/hide]

We show by induction on $k$ that there exists a positive integer $z$ for which $z^2 \equiv -7 \pmod{2^{k}}$. The problem statement will follow then as $z+r\cdot 2^{k}$ will also work for any $r \in \mathbb{Z^{+}}$\\

[color=#f0f][b]Base Case:-[/b][/color] for $k=1,2,3$ we can take $z=1$ and base case trivially works.\\
Now,\\

[color=#f00][b]Claim:-[/b][/color] suppose $z^2 \equiv -7 \pmod{2^{k}}$ for $k \geqslant 3$ . Then:
$\[ z^2\equiv -7 \pmod{2^{k+1}} \quad \text{or} \quad (z+2^{k-1})^2 \equiv -7 \pmod{2^{k+1}}.\]$\\

[color=#00f][b]Proof:-[/b][/color] set $\mathcal{F}=\frac{z^2+7}{2^{k}} \in \mathbb{Z}$\\

[list]
[*] [color=#073763][b]Case 1:-[/b][/color] if $\mathcal{F}$ is even then,\\

$\frac{\mathcal{F}}{2} \in \mathbb{Z} \implies \frac{z^2+7}{2^{k+1}} \in \mathbb{Z}$\\

which gives $z^2 \equiv -7 \pmod{2^{k+1}}$\\

[*][color=#073763][b]Case 2:-[/b][/color] if $\mathcal{F}$ is odd. Then we notice:\\

$\frac{z^2+7}{2^{k+1}}+\frac{z}{2}=\frac{\mathcal{F}+z}{2} \in \mathbb{Z}$ as $\mathcal{F}$ and $z$ are odd.\\

hence we get $\frac{z^2+2^{k}z+7}{2^{k+1}} \in \mathbb{Z}$\\

i.e. $z^2+2^{k}z \equiv -7 \pmod{2^{k+1}} \implies z^2+2^{k}z+2^{k+1}\cdot 2^{k-3} \equiv -7 \pmod{2^{k+1}} \implies (z+2^{k-1})^2 \equiv -7 \pmod{2^{k+1}}$\\
[/list]

hence claim proved $\blacksquare$\\

now clearly from our setup of $z+r\cdot 2^{k}$ we get infinite such square numbers as it's true for  any $r \in \mathbb{Z^{+}}$\\

$\blacksquare$


If $ \displaystyle\sum_{i\equal{}1}^n{v_i^2}\equal{}1$, prove the following inequality:
\[ \sqrt{\sum_{i\equal{}1}^n{(v_id_i)^2}}\leq\max{\{|d_1|,|d_2|,\ldots,|d_n|\}}\]

$ \sqrt {\sum_{i \equal{} 1}^n v_i ^2 d_i^2} \leq \sqrt {\sum_{i \equal{} 1}^n v_i ^2 d^2_{\mathbf{max}}} \equal{} \sqrt {d^2 _{\mathbf{max}} \sum_{i \equal{} 1}^n v_i ^2} \equal{} \sqrt {d^2_{\mathbf{max}}} \equal{} | d |_{\mathbf{max}}$


Derive the Quadratic formula.

Starting with..... $f(x) = ax^{2}+bx+c$

 :)

That, or solve for x in a S.F. Parabola so that $x = h \pm\sqrt{-k}$   !!!!~!

:10:

or, do the textbook way, where you multiply everything by 4a, factor and solve for x.

[hide]
Let $ax^{2}+bx+c = 0$ and $a \neq 0$.  Dividing by $a$ gives
$x^{2}+\frac{bx}{a}+\frac{c}{a}= 0$
$\implies x^{2}+\frac{bx}{a}=-\frac{c}{a}$
Now complete the square.
$x^{2}+\frac{bx}{a}+\left(\frac{b}{2a}\right)^{2}= \left(\frac{b}{2a}\right)^{2}-\frac{c}{a}$.
$\implies \left(x+\frac{b}{2a}\right)^{2}= \frac{b^{2}-4ac}{4a^{2}}$
$\implies x+\frac{b}{2a}= \sqrt{\frac{b^{2}-4ac}{4a^{2}}}$
$\implies x+\frac{b}{2a}= \frac{\sqrt{b^{2}-4ac}}{2a}$
$\implies x = \boxed{\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}}$.

[/hide]


Find root of system: $ \begin {cases} x^4\minus{}y^4\equal{}1\\ |x\plus{}y| \plus{} |x\minus{}y| \equal{} 2 \end {cases}$

Brute force yields (1,0). Unless you need an algebraic proof, it's easy to figure it out. Think of some 4th powers whose differences are 1. (Most people would start with 1 and 0.)

[hide]Square the second equation, giving $ (x \plus{} y)^2 \plus{} (x \minus{} y)^2 \plus{} 2\cdot|x \plus{} y|\cdot|x \minus{} y| \equal{} 4 \Rightarrow (x^2 \plus{} y^2) \plus{} |x^2 \minus{} y^2| \equal{} 2$. From the first equation, we have $ (x^2 \plus{} y^2)(x^2 \minus{} y^2) \equal{} 1$, implying that $ x^2 \minus{} y^2 > 0$. Thus, we have $ x^2 \plus{} y^2 \plus{} x^2 \minus{} y^2 \equal{} 2 \Rightarrow x^2 \equal{} 1$. This gives the solutions $ (1,0), ( \minus{} 1,0)$.[/hide]
EDIT: must hide!

Nice. Another approach: the second equation is just a square with vertices $ (\pm 1, \pm 1)$ (consider cases $ x \ge y \ge 0,\ y\ge x \ge 0$, and the negatives follow by symmetry), but the first equation implies $ y \equal{} \sqrt[4]{x^4 \minus{} 1} \Longrightarrow |x| \ge 1$, so $ (x,y) \equal{} (\pm 1,0)$.

[quote="azjps"]Nice. Another approach: the second equation is just a square with vertices $ (\pm 1, \pm 1)$ (consider cases $ x \ge y \ge 0,\ y\ge x \ge 0$, and the negatives follow by symmetry), but the first equation implies $ y \equal{} \sqrt [4]{x^4 \minus{} 1} \Longrightarrow |x| \ge 1$, so $ (x,y) \equal{} (\pm 1,0)$.[/quote]

[color=darkblue]$ \bullet$ $ x^4\minus{}y^4\equal{}1\Longleftrightarrow (x\minus{}1)(x\plus{}1)(x^2\plus{}y^2)\equal{}1\Longrightarrow (x\minus{}1)(x\plus{}1)>0$

$ \Longrightarrow \begin {cases}x\minus{}1>0\\x\plus{}1>0 \end {cases}$ OR $ \begin {cases}x\minus{}1<0\\x\plus{}1<0 \end {cases}$

$ \bullet$ If $ \begin {cases}x\minus{}1>0\\x\plus{}1>0 \end {cases}$, system $ \Longleftrightarrow\begin {cases} x^4\minus{}y^4\equal{}1\\(x\plus{}y) \plus{} (x\minus{}y) \equal{} 2 \end {cases}$ $ \Longleftrightarrow \begin {cases} x^4\minus{}y^4\equal{}1\\x\equal{}1 \end {cases}$ $ \Longleftrightarrow \begin {cases} x\equal{}1\\y\equal{}1 \end {cases}$

$ \bullet$ If $ \begin {cases}x\minus{}1<0\\x\plus{}1<0 \end {cases}$, system $ \Longleftrightarrow \begin {cases} x^4\minus{}y^4\equal{}1\\(\minus{}x\minus{}y) \plus{} (\minus{}x\plus{}y) \equal{} 2 \end {cases}$ $ \Longleftrightarrow \begin {cases} x^4\minus{}y^4\equal{}1\\x\equal{}\minus{}1 \end {cases}$ $ \Longleftrightarrow \begin {cases} x\equal{}\minus{}1\\y\equal{}1 \end {cases}$[/color]


[quote="mihail911"][b]When there is no wind, a plane traveling at a constant rate can cover the 1000 kilometer round-trip distance from A to B and back in 10 hours. If the wind blows from A toward B at $ k$ km/hr, it adds 25 minutes to the round trip. Find $ k$.[/b]

enjoy![/quote]

[hide="solution"]Let the speed of he plane with no wind be $ r$.  So:

$ r\cdot 10 \equal{} 1000\Rightarrow r \equal{} 100$ km/hr.

Solve for $ t$ to get:

$ t \equal{}\frac{d}{r}$.  Thus,

$ \frac{500}{100\plus{}k}\plus{}\frac{500}{100\minus{}k}\equal{} 10\plus{}\frac{25}{60}$

$ 500\left(\frac{200}{10000\minus{}k^{2}}\right) \equal{}\frac{125}{12}$

$ 10000\minus{}k^{2}\equal{} 9600$

$ k \equal{}\boxed{20}$.[/hide]

oops...sry about that

how'd u get $ r\equal{}200 mph$. the distance isn't 2000 miles, it's 1000? :huh:

[quote="mihail911"]how'd u get $ r \equal{} 200 mph$. the distance isn't 2000 miles, it's 1000? :huh:[/quote]
Is 1000 the round trip distance or the one way distance?


induction
number theory
number theory unsolved

Can anyone help prove this:

$F_{2n}=F_{n}L_{n}$

where $F_{n}$ is the $nth$ Fibonacci term and Lucas terms are define as follow

$L_{n}=F_{n+1}+F_{n-1}$

I found this in the book of $Number$ $Theory$ by $George$ $E.$ $Andrews$ which is a good book, is in the section about $Induction$ so this proof should be using it.

This problem have been posted right now in Number Theory Proposed Probem.


HI,
I am solving "Problem solving strategies " by ARthur Engel.I am able to solve all the chapters , save one i.e. Extremal principle.The approaches seem to me very ingenious and hard to guess.Do you know any of the sites or material where this principle is explained in detail ?
thanks
kishore

Is this what you where looking for?[url]http://en.wikipedia.org/wiki/Hertz's_principle_of_least_curvature[/url]

No.  The Extremal Principle is just a problem-solving strategy which dictates that when considering a problem involving some set of elements, especially one which asks you to prove some existence property, it is beneficial to look at the "extremal" (the greatest, the least) members of the set (according to some ordering), ex. to prove Bezout's Lemma.

The application process is similar to that of Pigeonhole - what sets can we consider?  How can we order those sets to find "extremal" elements?

the above explanation of the extremal principle is correct


calculus
integration
trigonometry
calculus computations

Extremely difficult problem from a competition:

Solving the ODE:

dy/dx= sqrt(A+a*y*y)

 ("A" is a positive constant, "a" is also a constant, the range of y can ensure A+a*y^2 is always non-negative)

a*y*y=a*(y^2)

I worked out it to be:

a1*x=arcsin(a2*y)+C,    a1,a2,C are all constants

But I want to know how can we simplify the solution by reducing the Integration contants.

Please try to Latex. Do you mean $ \frac{dy}{dx} \equal{} \sqrt{A \plus{} a^{y^y}}$ or $ \frac{dy}{dx} \equal{} \sqrt{A \plus{} a \cdot y \cdot y}$?

In either case, it's a separable equation. We can write $ x\equal{}g(y)$ where $ g$ is some integral expression. In the second case, the integral is routine; solve it with a trig or hyperbolic substitution.

If you want to express everything in real terms, there are two cases depending on the sign of $ a$.


For $ a,b,c$ positive reals

$ \sum \frac{a}{\sqrt{b^2\plus{}bc\plus{}c^2}} \geq \frac{\sum a}{\sqrt{\sum ab}}$

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {a}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \geq \frac {\sum a}{\sqrt {\sum ab}}$[/quote]
But it's easy Holder.   :maybe:

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum \frac {a}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \geq \frac {\sum a}{\sqrt {\sum ab}}$[/quote]

Yes;It's easy holder
$ LHS^2[\sum a(b^2 \plus{} bc \plus{} c^2)] \ge (a \plus{} b \plus{} c)^3$
We have:
$ \sum a(b^2 \plus{} bc \plus{} c^2) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$  :)


geometry
similar triangles
angle bisector
projective geometry
geometry proposed

[quote="Prime factorization"][asy]unitsize(100); pair A,B,C,D,F,G,H,O; A = dir(0); B = dir(30); C = dir(60); D = (1.3660254,0); O = origin; F = (0.1830127,-0.6830127); G = (0.57735027,0); draw(anglemark(D,O,C,18,20),red); draw(unitcircle); draw(O--(0.4,0.69282032),MidArrow); draw(O--(0.45,0.77942286),MidArrow); draw(O--C,blue); draw(O--D,blue); draw(O--F); draw(O--B,MidArrow); draw(O--B,blue); draw(C--D,blue); draw(A--B); draw(F--D,MidArrow); draw(F--(0.75,0.29903811),MidArrow); draw(F--(0.70,0.21243557),MidArrow); draw(F--B); label("\small { $O$} ",O,W); label("\small { $A$} ",A,S); label("\small { $B$} ",B,NE); label("\small { $C$} ",C,NE); label("\small { $D$} ",D,E); label("\small { $E$} ",F,S); label("\small { $F$} ",(0.57,0),SW);[/asy]
$ \angle COD$ is bisected into $ \angle COB$ and $ \angle BOD$. In addition, $ \overline{OC}// \overline{EB}$ and $ \overline{OB}// \overline{ED}$. Prove that

$ \frac{\overline{OD}}{\overline{OC}}= \frac{\overline{BD}}{\overline{CB}}$[/quote]

Isn't this true without the presence of $ E$, as a simple theorem about angle bisectors?  Or am I misinterpreting the problem?

Try proving it only using similar triangles.

The proof of this theorem with only similar triangles is as follows.  As in the diagra, let the triangle be $ OCD$, and let $ B$ be the foot of the internal angle bisector of $ \angle{COD}$.  Draw a line through $ D$ parallel to $ OB$, and let it intersect line $ OC$ at $ P$.  Then $ m\angle{CPD}=m\angle{COB}=m\angle{BOD}=m\angle{ODP}\implies OD=OP$.  Cleary, $ \frac{OC}{OP}=\frac{BC}{BD}$, so then the result follows.

The proof for external angle bisectors is completely analogous.

Interesting... seems shorter than the proof I used.

$ \angle COB \cong \angle BOD$ (given)
$ \overline{OB}// \overline{ED}$  (given)  $ \therefore$
$ \angle BOD \cong \angle EDO$ (alternate interior angles)
$ \overline{OC}// \overline{EB}$  (given) $ \therefore$
$ \angle COB \cong \angle OBE$  (alternate interior angles)
$ \angle OBE \cong \angle BED$  (alternate interior angles)
$ \triangle OFB \sim \triangle EFD (A.A.)$
$ \overline{OF}= \overline{FB}$ thus  $ \triangle OFB$ is isoceles
$ \overline{EF}= \overline{FD}$ thus  $ \triangle EFD$ is isoceles
$ \overline{EF}+\overline{FB}= \overline{OF}+\overline{FD}\therefore$
$ \overline{EB}= \overline{OD}$
$ \angle OCD \cong \angle EBD$ (corresponding angles)
$ \triangle OCB \sim \triangle EBD (A.A.)$
$ \frac{\overline{OC}}{\overline{EB}}= \frac{\overline{OB}}{\overline{ED}}= \frac{\overline{CB}}{\overline{BD}}$

Substitute $ \overline{EB}$ by $ \overline{OD}$ into the previous equation to reach the harmonic division theorem.

Given any triangle $ OCD$, if angle $ COD$ is bisected, then:

$ \boxed{ \frac{\overline{OC}}{\overline{OD}}= \frac{\overline{CB}}{\overline{BD}}}$


Find all values of n such that the number $ 10101...0101$ where we have $ n$ ones and $ n\minus{}1$  zeroes  is prime.

This number is just a geometric sum: $ 1\plus{}100\plus{}\cdots\plus{}100^{n\minus{}1}\equal{}\frac{100^n\minus{}1}{99}\equal{}a_n$. If $ n|m$ then $ a_n | a_m, a_n < a_m$, so $ a_n$ is a prime $ \implies n$ is a prime. Also, $ n \neq 3$, otherwise the number is divisible by 3.
I think this is all we can know, it's an open problem like finding prime Mersenne numbers ($ 2^n\minus{}1$).

$ \frac {100^n \minus{} 1}{99} \equal{} \frac {(10^n \plus{} 1)(10^n \minus{} 1)}{99}$ so it is only true for n=1, is this right?
$ 10^n \plus{} 1$ will always be divisible by 11 but grow large enough to have enough factor, so will $ 10^n \minus{} 1$ but with 9 instead of 11, so it must have at least two other factors than  one 9 and  one 11.

[quote="v235711"][quote="bambaman"]This number is just a geometric sum: $ 1 \plus{} 100 \plus{} \cdots \plus{} 100^{n \minus{} 1} \equal{} \frac {100^n \minus{} 1}{99} \equal{} a_n$. If $ n|m$ then $ a_n | a_m, a_n < a_m$, so $ a_n$ is a prime $ \implies n$ is a prime. Also, $ n \neq 3$, otherwise the number is divisible by 3.
I think this is all we can know, it's an open problem like finding prime Mersenne numbers ($ 2^n \minus{} 1$).[/quote]

this problem is from a math olympiad called olimpiada de maio(it is for contries that speak languages with origin form spanish and portuguese).therefore, it has a solution.
they wouldn´t post an open problem .[/quote]

Luckily, $ 100^n\minus{}1\equal{}(10^n\minus{}1)(10^n\plus{}1)$, so when n>2, we have both factors larger than 99, therefore the number must be a composite. So the only prime is n=2 which is 101.

Sorry that I rushed into conclusions, that's completely right :o


Given $ \overline{A'C} \equal{} \dfrac{1}{3}\cdot\overline{BC}, \overline{B'A} \equal{} \dfrac{1}{3}\cdot\overline{CA},$ and $ \overline{C'B} \equal{} \dfrac{1}{3}\cdot\overline{AB},$ find $ \dfrac{\overline{MN}}{\overline{AA'}}$.

[hide="Answer"] Edit: $ \dfrac{3}{7}$[/hide]

I don't think so...
[hide]
By Menelaus' Theorem, in triangle $ A'AB$, there is $ \frac {BC}{CA'}\cdot \frac {A'N}{NA}\cdot \frac {AC'}{C'B} \equal{} 1$, so $ \frac {A'N}{NA} \equal{} \frac {1}{6}$. Therefore $ \frac {A'N}{AA'} \equal{} \frac {1}{7}$.
Similarly, in triangle $ A'CA$, there also is $ \frac {CB}{BA'}\cdot \frac {A'M}{MA}\cdot \frac {AB'}{B'C} \equal{} 1$, so $ \frac {A'M}{MA} \equal{} \frac {4}{3}$. Therefore $ \frac {MN \plus{} A'N}{AA'} \equal{} \frac {4}{7}$. Thus $ \frac {MN}{AA'} \equal{} \frac {4}{7} \minus{} \frac {1}{7} \equal{} \frac {3}{7}$.
[/hide]

You're right timwu, mistake was mine. Edited.


Let $x,y,z,w$ be positive integers and $y,z,w$ odd. Show that $17$ divides $x^{y^{z^{w}}}-x^{y^{z}}$.

It is true if $x=1(mod 17)$ or $y=1$. If $x\not =1(mod 17)$ and $y>1$ sufficiently $y^{z^{w}}=y^{z}(mod 16)$. Because z,w are odd $z^{w}=z(mod 8)$, it give $y^{z^{w}}=y^{z}(mod 32)$.


Find the minimum value of $\int_0^{\pi} (x-y)^2 (\sin x)|\cos x|dx$.

I'm sure this is wrong but I got:

$\frac{3\pi^2}{8}-\pi+\frac{1}{2}$

If this is right (which I doubt) I'll explain my logic.

Thank you for your reply,Jimmy.

You are wrong.

kunny


Hola,
¿Qué es diferencia entre para y por? No sé... Ayudamé. Muchas gracias.

[color=blue]La principal diferencia radica en lo siguiente:
[b]Para[/b] se utiliza para indicar un fin, un objetivo. [i]Estudió [b]para[/b] aprobar el examen.[/i]
[b]Por[/b], en cambio, indica un medio, lo que se utiliza. [i][b]Por[/b] medio del estudio, se aprueban los exámenes.[/i]
Hay veces que [b]por[/b] tiene otras connotaciones. [i]Se apoyó [b]por[/b] la silla.[/i]
Inclusive, muchas veces se dice [b]por[/b] en lugar de [b]hacia[/b]. [i][b]Por[/b] aquellos páramos… / [b]Hacia[/b] aquellos páramos…[/i]

Avísame, por favor, si te resulta útil esta información.

¡Sigue adelante![/color]

Gracias pero es no suficiente.

Por ejemplo, [b]por[/b] también puede incidar un fin:

Estudio química [b]por[/b] parecerme interesante. (es de mi libro; libro dice que necesita usar [b]por[/b])

 :maybe:

[quote="Silverfalcon"]
Estudio química [b]por[/b] parecerme interesante. (es de mi libro; libro dice que necesita usar [b]por[/b])[/quote]
[color=blue]Interesante ejemplo.
No es común ver una oración expresada de esa forma.
Nota que el verbo [b]parecer[/b] está conjugado en forma reflexiva.
En este caso, expresas un motivo, un porqué, no un fin.
[i]Estudio química [b]porque[/b] me parece interesante[/i] es una oración con el mismo significado.
No es tu objetivo que te parezca interesante.

¿Te ayuda ésto?[/color]

Por estar con ella, daría lo que sea.
El uso de "por" indicaría un fin, un objetivo.
Con el objetivo de estar con ella, daría lo que sea.

Hmm, confunde bastante tan estudiaré más y regresaré con lo que sé. Gracias. Yo lo debo anunciar otra vez alguna vez hoy.

Soy de Corea del Sur (South Korea) pero estoy en Estados Unidos ahora.

[color=blue]¿Cuáles son todos los idiomas que te interesaría dominar?[/color]

Pienso que cuando estás hablando de un longitud del tiempo, usas [b]por[/b] también.

Lo siento [u]por[/u] mi gramatica mala, no hablo muy bien.  :wink:


i just took the AMC 12 A (2005)

i wanted to get some context as to where i am at, what would be a good score on that contest? and what is qualifying for the aime? how did you guys do?, etc

Anything above 100 is an AIME qualifying score - I think I took the B version of that exam, so I can't comment on its particular difficulty.  How you score really depends on how much experience you have with the problems, and how familiar you are with a pretty broad variety of math concepts.

To be honest, this was rather easy.  A good score would be 135+.

I didnt take it originally, took the 10.

I got a 136 on it recently.

[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05scoredist-grade12A.html]Here[/url] is a link to the score distribution of the test, if you're  interested :)

interesting, that is good to look at, but at the same time, i would definately like to improve a bit before the actual test, there was one problem that i should have gotten

Wouldn't they care more what your aime score was.

For some reason I thought AMC 12A 2005 was hard and I did pretty badly on it and I got a 114.5 when I took it (I suck at counting).  Then again, I made USAMO.

[quote="mna851"]If you're a freshman/sophmore who really cares whether you get a 150 or a 101, your eligibility for the USAMO will be determined by your relative comparison with the cutoff score rather than your USAMO index.[/quote]

I care what my score is because I want to know approximately how well I can expect to do as a junior, by comparing my freshman and sophomore scores.


a,b,c nonnegative real numbers: prove that (if it's true)
$ \sqrt{a^{3}b^{3}}\plus{}\sqrt{b^{3}c^{3}}\plus{}\sqrt{a^{3}c^{3}} \leqslant c^{3}\plus{}a^{2}b\plus{}b^{2}a$

hello, your inequality is wrong, consider the example $ a\equal{}1,b\equal{}\frac{1}{20},c\equal{}\frac{1}{4}$.
Sonnhard.

suppose that $ a>>c$ and $ b\rightarrow0$

[b]This is a cheating attempt --- correct version of this inequality is a problem from the first round of Polish MO which is currently running. Please block this topic and do not reply to it.[/b]


[quote="parisoft"]and please tell me how you answered it not just the answer!!!

Joe is rich. He decides to take 900 dollars and invest part of it at 12% and the rest at 9%. If his total return was 96 dollars, how much did he invest at each rate?

Please help soon!![/quote]
translation

[quote="parisoft"]and please tell me how you answered it not just the answer!!!

Joe is rich. He decides to take $900 and invest part of it at 12% and the rest at 9%. If his total return was
$96 how much did he invest at each rate?

Please help soon!![/quote]
One thing you're doing to get this error is putting your latex signs all around the whole sentence. And not that but its also around percentage signs. That's one thing yu should never put latex signs around.

I dont even know what a latex sign is does any one know the answer to the actual question I fixed it by turning off the code.

[quote="parisoft"]and please tell me how you answered it not just the answer!!!

Joe is rich. He decides to take $900 and invest part of it at 12% and the rest at 9%. If his total return was
$96 how much did he invest at each rate?

Please help soon!![/quote]
Hope I did this right.
[hide]So lets say he invested x amount of money at 12%. That means he invested 900-x at 9%. So now we have an equation. So .12x+(900-x).09=96. Solving for x we get 500. So Joe invested 500 dollars at 12% and 400 dollars at 9%.[/hide]

[quote="parisoft"]Joe is rich. He decides to take $900 and invest part of it at 12% and the rest at 9%. If his total return was
$96 how much did he invest at each rate?[/quote]
[hide]$1.12x+1.09(900-x)=996$

$1.12x+981-1.09x=996$

$0.03x=15$

$x=500$

He invested $\$500$ at $12\%
$ and $\$400$ at $9\%
$.[/hide]

[quote="parisoft"]and please tell me how you answered it not just the answer!!!

Joe is rich. He decides to take $900 and invest part of it at 12% and the rest at 9%. If his total return was
$96 how much did he invest at each rate?

Please help soon!![/quote]

[hide]
$x+y=900$

$0.12x+0.09y=96$

substitution:

$y=900-x$

$0.12x+0.09(900-x)=96$

$0.12x+81-0.09x=96$

$0.03x=15$

$x=500$

$900-500=400$

$y=400$

answer: 500 dollars at 12%, 400 dollars at 9%
[/hide]


linear algebra
matrix
inequalities
vector
algebra
polynomial
linear algebra unsolved

Let $ A \in M_{n}(\mathbb{C})$ such that : $ Rank(A) + Rank(I_n - A) + Rank(I_n + A) = 2n$
Prove or disprove that :
                                            $ A^3 = A$
$ I_n$ is $ nxn$ identity matrix
[b]Thanks ![/b]  :help:

Let $ V\equal{}\ker (A)\oplus \ker(A\minus{}I_n)\oplus \ker (A\plus{}I_n)$. Direct calculation yields $ \mathbb{C}^n\equal{}V$ and $ A$ is similar to a matrix of the form $ 0_r\oplus I_s\oplus \minus{}I_{n\minus{}r\minus{}s}$, since the inequality $ n\equal{}3n\minus{}2n\equal{}\dim\ker (A) \plus{}\dim\ker (A\minus{}I_n) \plus{}\dim \ker (A\plus{}I_n)\equal{}\dim (V)\leq n$ holds. But then $ A^3\equal{}A$.

Thanks hjbrasch !  :roll:
  [quote] Direct calculation yields $ \mathbb{C}^n \equal{} V$ [/quote]
I don't see . $ V \subset \mathbb{C}^n$  and $ \dim(V)\equal{}\dim(\mathbb{C}^n )$ then $ V\equal{} \mathbb{C}^n$
Can prove $ A^3 \equal{} A$ by $ \forall x \in \mathbb{C}^n$ We have $ A^3x\equal{}Ax$

As for your first line/question: That should be basic, e.g., if $ \mathbb{C}^n\not\equal{}V$ is non-empty, choose a vector $ \bar{0}\not\equal{}\bar{v}\in\mathbb{C}^n\setminus V$, then $ V\oplus <\bar{v}>\subseteq\mathbb{C}^n$, but $ \dim (V \oplus <\bar{v}>)\equal{}n\plus{}1>n\equal{}\dim(\mathbb{C}^n)$, a contradiction.

I do not get your second line at all. $ A\sim 0_r\oplus I_s\oplus \minus{}I_{n\minus{}r\minus{}s}$ implies $ A^3\equal{}A$, so you do not need any test vectors for the evaluation of this identity.

From the rank-nullity theorem, $ n\minus{}Rank(A)$ is the geometric multiplicity of 0 as an eigenvalue of $ A$, while $ n\minus{}Rank(I_n\minus{}A)$ is that of 1 and $ n\minus{}Rank(I_n\plus{}A)$ is that of -1. The given equation states that the sum of these is $ n$, so the three bases for the eigenspaces span the entire space $ \mathbb C^n$. But then $ A$ has minimal polynomial $ x(x\minus{}1)(x\plus{}1)$, so $ A^3\equal{}A$, or alternatively, $ A$ is diagonalizable with only eigenvalues $ 0,\pm1$ which makes $ A^3\equal{}A$. (The diagonalizability follows from the fact that the geometric multiplicities are always at most the algebraic, but in this case their sum is $ n$, which is the sum of the algebraic multiplicities, making them equal. It is well-known that matrices are diagonalizable iff the geometric and algebraic multiplicities of their eigenvalues are equal. Then cubing a diagonal matrix is easy.)

Hope that helped.


Ok. Egg Toss Game #2! I know that Another Egg Toss is going on already, but there's only some number of players playing it. Hopefully the number of players playing this will be able to fill up.
This is just another chance for those who have "died" in the last game to have revenge in this game :P
Anyway, it's the same rules as jjfun, I will make a story.and well lets start. But, First we need a list of players. I allow 14 players :)

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14
GOOD LUCK AND HAVE FUN!

Woah! Everytime I check it, people keep joining!!! :D
yay!

haha. We still need 10 people though...-.-
:(
Well, I'll try to get people to join

Hmmm... almost all the people joining were in the last egg toss. :|
[hide="Player List"]1. meewhee009
2. eqjj168
3. spaceguy524
4. piegraph
5. bluecarneal
6. Yoshi
7. andrewjjiang97
8.
9.
10.
11.
12.
13.
14. [/hide]

wow...!
only 7 more people need to join :)

wait....we still need 6 more people
lol.
But this is good progress
8 people joined in a day!

[color=white]become king of the bog! lol[/color]

[b]/throws innocent-looking egg decorated with pink bunnies at meewhee[/b]

Good.
[hide]cf249 throws an egg at meewhee.Meewhee returns the egg back at cf249. 1=2 pitches an egg at cf249, since he hit meewhee, who was after all 1=2's girlfriend. cf249 sinks into bog, and eqjj accepts him as a worker :P Westiepaw casts an egg into the bog...but it gets swallowed[/hide]
THE REST OF THE PPL R DEAD.
player list
1.1=2
2.meewhee
3.westie

you know the dead players, lol[/list]

[b]Throw at 1=2[/b]

For obvious reasons.  :D

Ahh.
Don't want either of you to die.
[b]Throws egg @ meewhee009[/b]

humbug.
GOOD LUCK TO CHAMPIONS.
NO story this time xD.
sorry!
ok
Let's all congratulate the 3rd place winner, meewhee!
Sorry meewhee...
STUPID WEBSITE.
:(
Anyway, CONGRATULATION!!!
(we give you a standing ovation!)
 :clap:  :winner_third:
OH well
now its ur turn
Final champions
We'll have runner up, and grand finalist :D
PIck a number form 1-30

PM please
Im not partial i wrote down the number rofl, and..well
GO FOR IT!

WE HAVE OUR WINNER
[hide="and our winner is "][b]1=2[b]...sorry westie!
My number was 13
and he picked 17..
:( SO CLOSE PPL!!!!
[/b][/hide]
yea
and our runner up CLAP CLAP CLAP [hide="you know who"] westiepaw![/hide]

Yay!  What a beast game!
Good job 1=2, westiepaw and meewhee!

Some host egg toss game #3! :P

WOO!!!

I was predicting that westie would pick 15 and the number would be lower than 15.

Good game, westie, meewhee, cf249, y'all.


geometry
geometric transformation
reflection
conics

As Altheman pointed out in the other thread, we had at least one wrong answer in our solutions for the Geometry test.

The answer to #4, the 6 tangent circle one, is actually $\frac{5-2\sqrt{3}}{13}$. You can use the Descartes Circle Theorem.

The answer to #9, the locus problem, should actually be $25x^{2}-36xy+13y^{2}= 4$ rather than $25x^{2}-36xy-13y^{2}= 4$. (There is not yet complete consensus about this, so I welcome anyone to post a detailed solution here.)

Unfortunately, it is now too late for us to do anything about the incorrect grading. We have already received some suggestions about how to give contestants a chance to challenge our answers in the future, and if you have any ideas about this, please post them here or in the "Afterthoughts" thread.

I think that's the answer I got for #9, solution is by taking some point (x, y) that's on the curve; reflect the y = x line over it and find where it intersects the y = 2x line; this point must be a distance 2 (or whatever it was) away from our point so set up that equation. Probably the locus isn't the whole of that conic, but whatever, all of the locus is on the conic.

the steps reverse...

in particular, for any point $(x,y)$ on the locus, there exists unique real numbers $a$ and $b$ such that $x=\frac{a+b}{2}$ and $y=\frac{2a+b}{2}$,

the condition implies that $(a-b)^{2}=(2a-b)^{2}=4^{2}$, that is, the distance from $(a,2a)$ and $(b,b)$ is equal to 4


I just put nine years' worth of MML questions online at
http://www.matharchive.org/mml/

There had been seven years' worth of WOCOMAL varsity and WOCOMAL freshman questions online; now there are 15 and 10, respectively:
http://www.matharchive.org/wocomal/varsity/
http://www.matharchive.org/wocomal/freshman/

Thirteen MAML playoffs:
http://www.matharchive.org/maml/

Eighteen NEAML playoffs:
http://www.matharchive.org/neaml/

Most of the files are scans (bitmaps) so the quality is what it is, but in most cases it's pretty easy to read the questions.

Have fun.

Rick Yanco
Worcester Academy

Many thanks for these problems.

Do you have past years of level1 or level2 of the Massachusetts math Olympiad by any chance?

I do, and fairly soon I'll put some online.  Right now I'm working on another nine years of MML (back to 1988) because I have a notebook to scan from and that makes it fairly easy.

OK, now there are 25 years of level one Olympiads, and (just) two years of level two Olympiads at
http://www.matharchive.org/olympiad/

I'll upload a bunch of level twos during winter break.

Study hard!

Rick Yanco
Worcester Academy

P.S. I also added nine more years of MML for a total of eighteen:
http://www.matharchive.org/mml/

Do you have any of the solutions to the Massachusetts Math Olympiad level 1 tests? I haven't been able to locate them anywhere. Thanks in advance!

No, I don't have many solution packets, and in the very few cases where I had a solution packet I didn't put it online.  However, the [i]answer keys[/i] are located at the back of the PDF files whenever I had them.  If you get stuck on the solution you can always post the question to AOPS.

One recent improvement is that there is now a MAML website, although anyone who has been to the WOCOMAL website (or the unofficial MML website) will notice an uncanny similarity.

Rick Yanco
Worcester Academy

I didn't see the answer keys in the back before. Sorry to bother you. Many Thanks!

[quote="rjyanco"]OK, now there are 25 years of level one Olympiads, and (just) two years of level two Olympiads at
http://www.matharchive.org/olympiad/

I'll upload a bunch of level twos during winter break.
[/quote]

Level twos? Fire!

Seems like the MAML II is significantly easier than other local olympiads like the Colorado Math Olympiad and BAMO.

Sorry, I wasn't able to get ahold of the L2 exams at the MAML board meeting.  (It's easier to get the L1 exams, since most schools have extra each year, than the L2 exams, which are only given to ~100 students.)

Awww, I guess I'll just have to study BAMO or something.

Would you by any chance happen to be able to upload the rest of the 06-07 year for MML?

Thank you very much


Which positive integers have the following property? If the final (units) digit is deleted, the original integer is divisible by the resulting integer.

Interesting question.

[hide]If it's in the form 10n.  :lol: [/hide]

[hide]i came up with the multiples of 10 and the numbers 10-19. i'm sure there are more[/hide]

I cannot find a [i]numerical[/i] answer, but there is a condition that all those numbers follow:
EDIT: Never mind... I overlooked something.
[hide]
Represent the number in the form 10a+b.

We want (10a+b)/a to be an integer.

.: (10a+b)/a=(10a)/a + (b)/a = 10 + b/a.

Therefore b/a must be an integer, and therefore b must be divisible by a.
But if b is divisible by a, then a<=b. Since b is a single digit, thus a must be a single digit, and therefore only two digit numbers are capable of this property.

We can do case-by-case pretty easily. There are 10 integers where a=1, 5 integers where a =2, 4 integers where a=3,  3 integers where a = 4, 2 integers where a = 5, 2 integers where a = 6, 2 integers where a = 7, 2 integers where a =8, and 2 integers where a =9, for a grand total of 32 integers.
[/hide]

No... not only two digit numbers are capable of this property.  Any multiple of ten works.
You could make this a separate case from finding a numerical answer, although since there are infinite multiples of ten, the answer will be infinity anyway.

[hide]Yes, like everyone has said, any multiple of 10 will work. The way I got that result was:

We can write any positive number $x$ with $n$ digits in the form $x=a_n10^{n-1}+a_{n-1}10^{n-2}+...+a_110^{0}$

If we take away the units digit, the resulting number would be:

$y=a_n10^{n-2}+a_{n-1}10^{n-3}+...+a_210^{0}$

We're looking for the cases where $y|x$.

If we make the units digit of $x$ 0, then it is clear that $y$ divides $x$ and the quotient is always 10. For example, if we have the number 153, it is clear to see that 15 divides 150 easily and that 150/15=10.

Then, that means $y|a_1$ in order for the condition to be fulfilled. Of course, when $a_1=0$, $y|a_1$ since everything divides 0. This is when the number is a multiple of 10, like everyone is saying.

When, $x$ is more than 2 digits, it is clear to see that $y\geq10$ if $a_1>0$ so for numbers more than 2 digits, $y|x$ only if $a_1=0$.

For single digit numbers, it never holds because division by 0 is not allowed.

But for two digit numbers, we will look for cases where $a_1\neq0$ and $y=a_2|a_1$.

We'll go through cases:

If $a_2=1$, then $a_1$ can be any digit 0-9, therefore, the numbers 10-19 work.
If $a_2=2$, then $a_1$ can be 0,2,4,6,8, therefore the numbers 20, 22, 24, 26, 28 work.
If $a_2=3$, then $a_1$ can be 0,3,6,9, therefore the numbers 30, 33, 36, 39 work, etc.

After doing all the cases, we end up with the final answer of:

$\text{Any multiple of }10\text{ and the numbers }11-19,22,24,26,28,33,36,39,44,48,55,66,77,88,99$
[/hide]


Let triangle $ABC (AB=AC)$, circle$(O)$ passthrough $,A,B$intersect $AH$at $K,(AH \perp BC), L\in AB$ such $KL//AC, CL \cap BK \equiv E ,EA\cap (O) \equiv F.$Prove that $\widehat{LFA}=\widehat{BAC}$.

What is the point K? Isn't AH the altitude from A to BC, and AB is the side, so they interesct at A... but this is what you call K?

EDIT: sorry, I read your problem wrong... I understand it now.

[img]6884[/img]
I will try to solve when I have time... expect a couple days.

ok thanks.
i hope you have nice solution  :lol:


An engine pumps water continuously through a hose.
If the speed at which water passes through the nozzle is V, prove that the rate at which kinetic energy is being imparted to the water is proportional to V^3

The work $\text dA$ done by a power P in a time $\text dt$ is $\text dA = P\ \text dt$, which is equal to the kinetic energy $\text dE$ of the water leaving the the nozzle during this time. This water column has the length $\text dx = v\ \text dt$, volume $\text dV = S\ \text dx = S v\ \text dt$, mass $\text dm = \rho\ \text dV = \rho S\ \text dx = \rho S v\ \text dt$, and the kinetic energy $\text dE = \frac 1 2\ \text dm\ v^2 = \frac 1 2\ \rho S v^3\ \text dt$, where v is the water velocity, S the area of the nozzle opening, $\rho$ the water density. Thus

$P\ \text dt = \text dA = \text dE = \frac 1 2\ \rho S v^3\ \text dt$

$P = \frac 1 2\ \rho S v^3$


Rezolvati in N ecuatia: (p-1)^q+(q-1)^p=3m+2,m natural. :D

Se pare ca nu intereseaza pe nimeni.E chiar asa de simpla? :( E pentru clasa a VII-a, dar merita postata(I think so) :D


If "a/b" has a quotient with a period of "b-1" digits long, even in cases in which the period has hundreds of digits, is it possible with a fast calculation to find the value of any unknown digit knowing all the others?
I can do it with a conjecture, but I would like to know if there is a rule to resolve the problem.

I will try to "quantify" what I intend for fast calculation: for instance 2/541 =
0.0036968576709796672828096118299445 47134935304990757
8558225508 3179297597042513863216266173 75231053604436
2292051756007 3937153419593345656192236 59889094269870
609981515711 6451 01663585951940850277264 3253234750462
10720887245841 0351201478743068391866913123 84473197781
88539741219963031 4232902033271719038817005545 28650646
95009242144177449168 2070240295748613678373382624 76894
63955637707948243992606 2846580406654343807763401109 05
73012939001848428835489833 6414048059149722735674676524
9537892791127541589648798521 256931608 1330868761552680
22 181146025878

The period is of 540 digits long, with a period of this length, using my conjecture it only takes 2 minutes to find the value of any unknown digit, without any calculator and using only pen and paper. I could use another rule to find the value of any unknown digit, easily demonstrable and also without calculator, which takes about 5-10 minutes.

What method, the simplest and fastest possible, would you use for finding any unknown digit of the quotient of 2/541?

The period doesn't need to be $ b\minus{}1$ digits long; it's enough that we know the period $ p$ and that $ \gcd(b, 10) \equal{} 1$.

Let $ k \equal{} \frac{10^p \minus{} 1}{b}$.  The digits that make up the repeating block can be shown to be the number $ ak$, whose remainder $ \bmod 11$ is then easily calculated.

[quote="t0rajir0u"]The period doesn't need to be $ b \minus{} 1$ digits long; it's enough that we know the period $ p$ and that $ \gcd(b, 10) \equal{} 1$.

Let $ k \equal{} \frac {10^p \minus{} 1}{b}$.  The digits that make up the repeating block can be shown to be the number $ ak$, whose remainder $ \bmod 11$ is then easily calculated.[/quote]

So, for finding the unknown digit of 2/541 you have to add all the 539 known digits?

[quote="t0rajir0u"]Yes (well, adding alternately; that's how you find remainders $ \bmod 11$).  Were you looking for an algorithm that found you a digit in a particular location without knowledge of the others?[/quote]

As matter of fact, I have found two algorithms that allow me to find the
missing digit with some faster operations and, since I am not
an expert in mathematics, I wanted to see if the already known algorithms
are simpler or coincide with those that I have found.

[quote="t0rajir0u"]Care to share?  The $ \bmod 11$ strategy seems pretty fast to me.[/quote]

I would like to share it, however, for the moment I prefer not to tell it, since I have posted the same thread in other forums and I want to verify the difficulty of the algorithms that I have found.
Thanks again for your contribution and compliments for your ability.

[quote="t0rajir0u"]Care to share?  The $ \bmod 11$ strategy seems pretty fast to me.[/quote]

This is the fast method that is always valid for the maximal decimal period of “b-1”; in the case of 2/541 I take the three digits that precede the unknown digit “242x” , but you can select which number of digits you wish, for instance “9242x”; I calculate the number that added to “9242” has the result “9999”, in this case the number is “0757”. I look for the sequence “0757” among the known digits and I discover the value of the following digit that is “8” and therefore I can conclude that the unknown digit is “1” because this added to “8” results “9”. Therefore, in the case in which the period is “b-1” long, it’s enough for me to know only half of the digits in order to determine quickly all the others, “0” corresponds to “9”, “1” corresponds to “8”, “2” corresponds to “7” and so on.

In order to easily determine the digit, one can apply the two halves of the number one above another, for instance,
0,11764705
   88x35294

and one can find easier the value of the unknown digit “x” which is “2”.

I would surely appreciate your opinion regarding his method.

Nice; I'd forgotten that there were extra patterns in the digits.


limit
logarithms
MATHCOUNTS
calculus
geometry
3D geometry
tetrahedron

I thought $\text {lim}$ was from Calculus or something.

Yeah, lim stands for limit, and it's generally considered calculus.  In this case it's saying that as x approaches infinity, x/lnx becomes an arbitrarily good approximation for the number of primes less than or equal to x.  I don't know how to show this though; I imagine that it is very advanced.  If you want to understand this, the college playground forum might give you more luck :D.

Moderator, bump this thread into Intermediate Forum (Skip Getting Started since Calculus is above Getting Started).

Or Calculs Forum.

Yeah, this is wayyy beyond Mathcounts! You might wanna bump this up to some other forum.

$\pi(n)$ is the number of primes less than or equal to n.

Usually, it's stated as the following: $\pi(x) \approx \displaystyle\frac{x}{ln(x)}$

So it makes sense that as $x$ gets bigger and bigger, this theorem will be more and more true, so once $x$ gets up to inifinity, you can assume equality.  Therefore, $\pi(x) = \displaystyle\frac{x}{ln(x)}$.  Then by multiplying by $ln(x)$ and dividing by $x$, you will get the RHS to 1, so you'll get the theorem chess64 originally stated.

It's basically saying that there are infinitely many primes, but 0% of the natural numbers are prime. I don't think this is covered at even the high school level, let alone MathCounts.

I don't see why 0% of the natural numbers are prime. Clearly both sets (prime numbers and natural numbers are infinite) and I'm tempted to say both are countable so both have equal cardinality?

i think maybe the percentage of numbers under a certain integer that are prime as you get higher gets closer and closer to zero?

Well, I don't know if this helps any, but the number of consecutive composite numbers increases. (Find x consecutive composite numbers...)

[quote="chess64"]I was reading some paper on number theory when I came across something that looked like this:

\[\lim \pi(x)\frac{\ln x}{x} = 1\]

but under the $\lim$, there was: $x \rightarrow \text {(infinity symbol)}$. It's the Prime Number Theorem. Because I don't understand the $\lim$ thing, I don't know if it's mathcounts level or not, sorry. Does anyone know what it means? Thanks.[/quote]This can be written\[\lim_{x\to\infty}\pi(x)\frac{\ln{x}}{x}=1.\]This says that as $x$ grows larger, the expression $\pi(x)\frac{\ln{x}}{x}$ approaches $1$. Or, $\pi(x)$ and $\frac{x}{\ln{x}}$ come closer to eachother.

A little more precisely, we can make $\pi(x)\frac{\ln{x}}{x}$ as close to $1$ as we like by choosing $x$ sufficiently large. Even more precisely, for any $\epsilon$ there exists $x_1$ such that $|\pi(x)\frac{\ln{x}}{x}-1|<\epsilon$ when $x>x_1$.

Let me know if I didn't say that right.. I haven't taken a course in analysis and AP Calculus doesn't really teach calculus.

Tetrahedron, cardinality is not the same as density.  Consider, for example, the 10,000th powers of the integers -- they are clearly very, very spread out, and your chances of picking one "at random" are very very small over any large interval of integers.  Their density approaches 0, so it does make sense to say that they make up 0% of the natural numbers, even though they are the same size.  An interesting thing about primes, of course, is that they aren't nearly so regular as the 10,000th powers.  So, they are still sufficiently dense that $\sum_{p \text{ prime}}\frac{1}{p} \to \infty$, which I think is a result of Euler although I'm not positive of that.

Yes, the last one is due to Euler.

The first mentioned problem is surely about ALL competitions you will take part in the next years (even above IMO!), I have read the proof and it uses a lot of advanced calculus (complex analysis), nothing you ever have to know for such competitions.

But already Tschebychew showed a more weaker result:
There are constants $0<a<1<b$ (close to $1$), so that $a < \pi (x) \cdot \frac{\ln (x)}{x} < b$ for all sufficient large $x$, you can try if you want to, it's not that easy but possible to get it!

Yes, thanks for your explanation Joel.

The last theorem ZetaX mentioned is proved in Andrews' Number Theory.


Prove the orthocenter of an obtuse-angled triangle is an excenter of its orthic triangle. =)

[quote="neelnal"]Can you please define orthic and excenter and orthocenter plz?

thanks in advanced.[/quote]
um you could do a search....
[url]http://mathworld.wolfram.com/Excircles.html[/url]
[url]http://mathworld.wolfram.com/OrthicTriangle.html[/url]

[quote="Inspired By Nature"][quote="computer_nuke"]Well, if you could prove that the altitudes bisect the exterior angles of the orthic triangle....[/quote]
Try it. Anyone have a proof?[/quote]
See attached picture.
Let the triangle be $ABC$, the orthocenter be $H$ and points of intersection $HA$ with $BC$, $HB$ with $AC$ and $HC$ with $AB$ respectively $A'$, $B'$ and $C'$. Then obviously $\left|\measuredangle BCC'\right|= \left|\measuredangle BAA'\right|$. Of course the quadrilaterals $AB'HC'$ and $CB'HA'$ are cyclical, so $\left|\measuredangle A'B'H\right|= \left|\measuredangle A'CH\right|= \left|\measuredangle HAC'\right|= \left|\measuredangle HB'C'\right|$. Thus $B'H$ is the bisector of angle $A'B'C'$. It can be analogously proved for the other two angles.

[quote="computer_nuke"]Rzeszut, I think the original question was to prove it for an obtuse-angled triangle. And, did you draw your excenters correctly?[/quote]
I misread the problem  :wallbash_red:  :wallbash_red:  :wallbash_red: . In fact, I proved that the altitudes of an acute-angled triangle are the angle bisectors in the orthic triangle. I hope that this solution is correct:

Let $ABC$ be an obtuse-angled triangle such that $\left|\measuredangle ACB\right|> 90^{\circ}$ and $H$ be its orthocenter. Furthermore let $AC$ and $BH$ intersect at $A'$, $BC$ and $AH$ intersect at $B'$, $AB$ and $CH$ intersect at $C'$. We have to prove that the point $H$ is one of the excenters of the orthic triangle of the triangle $ABC$, i.e. $A'B'C'$. It's easy to see that $ABH$ is an acute-angled triangle, $C$ is its orthocenter and $A'B'C'$ is its orthic triangle. Thus from my previous post we have $\left|\measuredangle A'C'H\right|= \left|\measuredangle HC'B'\right|$ and the point $h$ lies on the bisector of the angle $A'C'B'$. Sincethe situation is symmetric in $A$ and $B$, it sufficies to prove that $h$ lies on the bisector of the angle $B'A'C''$, i.e. $\left|\measuredangle C''A'H\right|= \left|\measuredangle HA'B'\right|$. But applying the result of my previous post one more time, we get $\left|\measuredangle C''A'H\right|= \left|\measuredangle BA'C'\right|= 90^{\circ}-\left|\measuredangle C'A'C\right|= 90^{\circ}-\left|\measuredangle CA'B'\right|=$ $90^{\circ}-\left(90^{\circ}-\left|\measuredangle HA'B'\right|\right)= \left|\measuredangle HA'B'\right|$ which ends the proof.


vector
algebra
polynomial
linear algebra
matrix

I'm having some difficulties interpreting this question:

Let $ S \equal{} \{u_1,u_2,...,u_n\}$ be a linearly independent subset of a vector space $ V$ over the field $ Z_2$ (The field consisting of 0 and 1). How many vectors are there in span(S)?

I'm not sure what $ S$ actually is. If $ n \equal{} 1$, does this mean that $ S \equal{} \{0\}\cup\{1\}$? Does that mean that this spans itself?

If $ n \equal{} 2$, does that mean that $ S \equal{} \{0,0\},\{0,1\},\{1,0\},\{1,1\}$? So S is always its own span?

Is addition in this $ Z_2$ like this: 0+0=1, 0+1=1, 1+1=0?

Yes, you have the operations in this field correct.

$ S$ is some [b]linearly independent[/b] set. It can't contain the zero vector, so it is never equal to its span. The space $ V$ doesn't have any intrinsic relation to $ n$; if $ n\equal{}2$, we could have $ S\equal{}\{(0,1),(1,0)\}$ in $ F_2^2$ or $ S\equal{}\{(1,1,1),(1,0,1)\}$ in $ F_2^3$ or $ S\equal{}\{x\plus{}x^2,x^3\plus{}x^8\}$ in the space of polynomials over $ F_2$ or any of infinitely many other possibilities.

The span of $ S$ is a vector space. What is its dimension?

Ah, thanks a lot. That distinction clears things up a lot.

Is it $ 2^n$?

Since if there are n terms, you can multiply each term by 1 of 2 numbers (0 or 1) and sum them, resulting in $ 2^n$ possible vectors, which are the span?

Thanks a lot again btw, that's incredibly useful.

Yes, we have $ 2^n$ vectors in the span. A $ d$-dimensional vector space over the field with $ q$ elements has $ q^d$ elements.

I'm struggling with another problem...some help would be very much appreciated.

Let V be a vector space having dimension n, and let S be a subset of V that generates V.

a. Prove that there is a subset of S that is a basis for V. Be careful not to assume that S is finite.
b. Prove that S contains at least n vectors.

For part A, does the fact that $ S\subseteq V$ which has dim(v)=n mean that dim(S)≤dim(V)? Or does this theorem only work if S is a subspace? If so, do I need to show that S is a subspace? Or is there some other approach?

Assuming the above is true, we go on to say that since S generates V, some linearly independent subset of S will be a basis, correct? If S is not already linearly independent, then we can continue expressing one term in terms of the others until it is independent, and we then have a linearly independent subset of S that spans V, or a basis, correct?

For part B, if S did not have at least n vectors, then it would be unable to span V, as the replacement theorem states that any basis of V (which has dim(V)=n) must have at least n elements. Therefore S must have n vectors.

Does this make sense?

Also,

Let $ V = M_{2\cdot2}(F)$

$ W_1 = \left\{\left(\begin{array}{ccc} a & b \\
c & a\end{array}\right)\in V: a,b,c\in F\right\}$,
and
$ W_2=\left\{\left(\begin{array}{ccc} 0 & a \\ -a & b\end{array}\right) \in V: a,b \in F\right\}$.

Prove that $ W_1$ and $ W_2$ are subspaces of V, and find the dimensions of $ W_1$, $ W_2$, $ W_1+W_2$ and $ W_1 \cap W_2$.

W1 is a subspace of V because it is closed under addition and scalar multiplication, and is a subset of V, correct? The same is true of W2. The dimension of W1 is 3, dimension of W2 is 2, correct?

Would $ W_1+W_2$ just be $ \left(\begin{array}{ccc} a & b+a \\ c-a & a+b\end{array}\right)$ ? The basis of this is just:

$ \left\{\left(\begin{array}{ccc} 1 & 0 \\ 0 & 0\end{array}\right), \left(\begin{array}{ccc} 0 & 1 \\ 0 & 1\end{array}\right),\left(\begin{array}{ccc} 0 & 0 \\ 1 & 0\end{array}\right)\right\}$, correct? Then the dimension is 3?

For the last one, the dimension is just 2, right? All 0's, with one matrix having a -1 in the bottom left, the other having a 1 in the top right?

You're better off giving new questions new threads.

On that last question- those aren't the same $ a$ and $ b$. The union is larger than either space and the intersection is smaller than either space, so their dimensions should be as well.


[quote="GameBot"]A rectangular field is half as wide as it is long, and it is completely enclosed by 54 meters of fencing. What is the number of square meters in the area of the field?[/quote]

Let $ x$ be the width. The length is then $ 2x$. Since the perimeter is 54, the sum of the length and width is $ 54 \div 2 \equal{} 27$. Thus $ 2x \plus{} x \equal{} 27 \implies x \equal{} 9$ and the length is then 18, and so the area is 162.


1.Form the integer 24 using the numbers 2, 4, 5, 7. Only addition, subtraction, division, and multiplication can be used. Every number can only be used once (no exponents or anything above a 4th grade level, just simple +,-,x, /.

2. do the same as above using the numbers 2,3,4,7

I assume you can use parentheses however you want.

In that case...

1. [hide](5+7)x4/2[/hide]
2. [hide]2(7+3)+4[/hide]

Hi. A nice one I came across about a year ago is to form $ 24$ using $ 1, 3, 4$ and $ 6$.

[quote="fishythefish"]How about:
\[ \frac {6}{{(}1 - \frac {3}{4}{)}} = 24\]
To simplify:
\[ \frac {6}{{(}1 - \frac {3}{4}{)}} = \frac {6}{{(}\frac {1}{4}{)}} = {(}6{)}{(}4{)} = 24\]
[/quote]

That is the idea, but your answer does not count because you used
vertical fraction bars, which are not part of the allowed symbols.
Instead, use the horizontal division symbol as aforementioned.

You can use this as a solution:

6/(1 - 3/4)

The idea of using the horizontal bar was to make it look neater. Since it's the same thing as using a slash, the answer is acceptable. Furthermore, if you read the original post, it doesn't say anything about what symbols can be used, only which operations. :)

[quote="fishythefish"]Furthermore, if you read the original post, it doesn't say anything about what symbols can be used,
only which operations. :)[/quote]

Yes, it did mention these symbols:

"... just simple +,-,x, /."

Oh, *but I did read the original post*, and that's precisely *why* I posted
as I did.  But just keep on posting in the form as you like.

------------------------------------------------------------------------------------------------

[quote="Ihatepie"]They mean the exact same thing, so his answer is right too.
[/quote]

No, that is not correct reasoning.  His answer *may* be allowable this time,
but just because an answer is the same thing (equivalent), does not mean
it is correct, too, especially as there are allowable forms for answers for
certain problems and not for others as per the instructions.

------------------------------------------------------------------------------------------------

[quote="basketball"]
1. (7+5) x 4 / 2

[/quote]

I already included this in my "And change the order of "5" and "7"
to get more." statement from my post prior in this thread.

When it said that, it was referring to the operations. Read the line above it. And you showed yourself that my solution is correct, just by simply changing one of my symbols to an equivalent one, which you seem to think is allowed, when mine is not.

Also tan, if you want to be that critical, all of your answers are wrong. You put a number next to parenthesis to imply multiplication. He explicitly said to use "x".

And parentheses were not one of the allowed symbols either, remember?


For wich $ n\in \mathbb{N},n\ge 2$ , $ (n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}$ is divisible with $ n^{n}$?


algorithm
function
induction
algebra
polynomial
advanced fields
advanced fields unsolved

hi,

the below algorithm needs to be expressed as a number of assignments in terms of n (required for the algorithm to solve an instance of the problem), and the time complexity function has to be deduced... Not sure how to do it. An example of a similar type of a problem will be most appreciated.

thanx in advance.

Input: positive integer n, real number y
a:=1, b:=1
    for i =1 to n do
     a:= (a*y)/i,
       b:=b+a
   end do
output: b

I can't say I fully understand what the problem is asking you to find (what does "expressing an algorithm as a number of assignments" mean?). Perhaps the question is to give a simple formula for the output $ b$ in terms of the inputs $ y$ and $ n$.

The part of the problem that asks about "time complexity" cannot be answered without a clear definition of the basic steps that are assumed to take $ O(1)$ time; this algorithm deals with real numbers, so presumably we are talking about a (highly fictitious) computational model that allows real addition, multiplication and division in constant time? Is that the context?

Anyway, if I'm guessing the intention correctly, what I would suggest is to start with analyzing the simple cases $ n\equal{}1,2,3,4$ first. For each of those, what is the output $ b$ (as a function of $ y$)? How many assignments take place? How many multiplications, divisions and additions? Can you guess the general formulas for arbitrary $ n$, and prove it by induction on $ n$?

If you replace division by $ i$ with multiplication by some arbitrary sequence (which we would input as an array), this is an efficient method of evaluating a polynomial in which the coefficients are given as products with a certain form. It's particularly computationally nice when we expect the low-degree terms to be largest, for numerical reasons. In this case, the polynomials are Taylor polynomials for $ e^y$, a best-case scenario for the method.

The actual question you asked is trivially easy- you can easily count the operations in one round through the looped block, and also easily count the number of rounds through it.


geometry
parallelogram
geometry unsolved

Given by triangle ABC. Find locus the points M inside triangle such that dintance from M to one side is equal with sum distances from M to other sides

It is not difficult. The locus of all points P whose distance to the side BC of triangle ABC equals the sum of their distances to the sides CA and AB is the line EF, where E and F are the points of intersection of the angle bisectors of the angles ABC and BCA with the lines CA and AB, respectively.

Proof... may be next month.

  darij

Let $MD,ME,MF$ be the distances from $M$ to the side $BC,CA,AB$ respectively.

Suppose that $MD=MF+ME$

Since $M$ is inside the triangle $ABC$ we have

$(MBC)+(MCA)+(MAB) = (ABC) \Leftrightarrow$

$a\cdot MD + b\cdot ME + c\cdot MF = 2(ABC) \Leftrightarrow$

$a\cdot (MF+ME) + b\cdot ME + c\cdot MF = 2(ABC) \Leftrightarrow$

$(a+b)\cdot ME + (a+c)\cdot MF = 2(ABC) \ \ (1)$

Let $BP,CN,AQ$ be the internal angle bisectors of the triangle $ABC$

$\frac{AN}{NB}=\frac{b}{a}\Rightarrow \frac{AN}{c}=\frac{b}{a+b}\Rightarrow \boxed{a+b=\frac{bc}{AN}}$

$\frac{AP}{PC}=\frac{c}{a}\Rightarrow \frac{AP}{b}=\frac{c}{a+c}\Rightarrow \boxed{a+c=\frac{bc}{AP}}$

$(1) \Leftrightarrow \frac{bc}{AN} \cdot ME + \frac{bc}{AP}\cdot MF = \frac{abc}{2R}\Leftrightarrow$

$\frac{ME}{AN} + \frac{MF}{AP} = \frac{a}{2R} =sinA \Leftrightarrow$

$ME \cdot AP + MF\cdot AN = AN\cdot AP \cdot sinA \Leftrightarrow$

$\frac{1}{2} \cdot ME \cdot AP + \frac{1}{2} \cdot MF\cdot AN = \frac{1}{2} \cdot AN\cdot AP \cdot sinA \Leftrightarrow$

$(AMP)+(AMN) = (ANP) \ \ (2)$

If $M$ is inside $\triangle ANP$, then $(ANP)=(AMP)+(AMN)+(MNP)>(AMP)+(AMN)$
If $M$ is inside $BCPN$ , then $(ANP)=(AMP)+(AMN)-(MNP)<(AMP)+(AMN)$

So, the only solution to $(2)$ is when $M$ is on the segment $NP$

Similarly we have

$ME = MD+MF \Rightarrow M \in NQ$
$MF = MD+ME \Rightarrow M \in PQ$

So the locus is the triangle $PNQ$

[hide="An extension"]
If we take a point on the ray $PN$, such that $M$ is not on the segment $NP$, then we have

$(AMP)-(AMN)=(ANP)$ , so, if we go inversely we'll get

$(a+b)ME - a \cdot MF - c\cdot MF = 2(ABC) \Rightarrow$

$a\cdot (ME-MF)+b\cdot ME - c\cdot MF = 2(ABC) \Rightarrow$

$\frac{1}{2} \cdot a\cdot (ME-MF)+(MCA) - (MAB) = (ABC)$

But we have $(MBC)+(MCA)-(MAB) = (ABC)$, because $M$ is in the interior of $\angle BCA$ and out of $\triangle ABC$

and $(MBC)=\frac{1}{2} \cdot a\cdot MD$

Hence $ME-MF=MD$

Similarly, we can see that the set of the [u]lines[/u] $PN,NQ,QP$ is the locus of all the points of the plane (not necessary inside ABC) with the same properties[/hide]

[hide="Here is a remarkable geometrical locus."][color=darkred]
[b]Lemma.[/b] Let $d_1,d_2$ be two concurrent lines (in the point $V$) and let $k,k_1,k_2$ be three fixed (constant) real positive numbers. Denote for a mobile point $L\in (d_1,d_2)$ (the plane defined by the concurrent or parallel lines $d_1$, $d_2$) the its distancies $x_1,x_2$ to the lines $d_1,d_2$ respectively. Then the geometrical locus of the point $L$ for which $k_1x_1+k_2x_2=k$ is a parallelogram for which the point $V$ is the intersection of its diagonals.

[b]Proof.[/b] Denote: the projections $L_1,L_2$ of the point $L$ on the lines $d_1,d_2$ respectively; the points $K_1\in VL_1$, $K_2\in VL_2$ so that $VK_1=k_1$, $VK_2=k_2$. Thus, $2[VK_1K_2]+2[LK_1K_2]=2[VL_1LL_2]=k_1x_1+k_2x_2=k$ (the area $[VK_1K_2]$ is constant) $\Longrightarrow$ the area $[LK_1K_2]$ is constant and the points $K_1$, $K_2$ are fixed $\Longrightarrow$ the mobile point $L$ belongs to a parallel segment with the line $K_1K_2$ and is comprised in the interior of the angle $\widehat {K_1VK_2}$ a.s.o.[/color]

[b]Proof of the proposed problem.[/b] Denote: the feet $E,F$ of the angle-bisectors from the vertices $B,C$ respectively; the distancies $x,y$ of the mobile inside point $L$ to the sides $[AB],[AC]$ respectively. Prove easily that $(a+c)\cdot x+(a+b)\cdot y=2\cdot S\ \ (*)$ and prove easily that the points $E,F$ verify the relation $(*)$ a.s.o.

[color=darkblue][b]Exercises.[/b]
$1.\blacktriangleright$ Let $ABCD$ be a convex quadrilateral which is circumscribed to the circle $w=C(I,r)$. Denote the middlepoints $M$, $N$ of the diagonals $AC$, $BD$. Prove that $I\in MN$ ([u]the Newton's line[/u]).
$2.\blacktriangleright$ [u]Generalization of the lemma.[/u] Let $A,B,C$ be three noncollinear points and let $k,k_1,k_2,k_3$ be four fixed (constant) real numbers. For a mobile point $L$ from the plane of the triangle $ABC$ denote the distancies $x_1,x_2,x_3$ to the lines $BC$, $CA$, $AB$ respectively. Ascertain the geometrical locus of the point $L$ for which $k_1x_1+k_2x_2+k_3x_3=k$.
[/color][/hide]

Now I don't know how to say. I only say thanhks to Pontios and Virgil Nicula. You show me the mathematics is very beatiful. Thank you again

I came back. See the last my reply, where I added the proof of the lemma and two proposed exercises !


rotation
number theory unsolved
number theory

Let $ n\geq2$. Determine how many sequences $ (a_k)_{0\leq k\leq 2n}$ are there such that $ 0 \equal{} a_0 \equal{} a_{2n}$, $ a_k\neq0$ for $ k\in\{1,\ldots,2k \minus{} 1\}$ and $ |a_{k \plus{} 1} \minus{} a_k| \equal{} 1$ for $ k\in\{0,\ldots,2k \minus{} 1\}$.

P.S. If $ N_n$ is the number we are looking for I got $ N_2 \equal{} 1$, $ N_3 \equal{} 2$, $ N_4 \equal{} 5$, $ N_5 \equal{} 14$.

If I'm not mistaken (and I very well may be >_<)...

[hide]Set up a recursion, and you'll get the recurrence that generates the [url=http://en.wikipedia.org/wiki/Catalan_number]Catalan numbers[/url].

EDIT: You're looking at a sequence that either goes up 1 or down 1. Look at this: http://mathworld.wolfram.com/DyckPath.html. Rotate it 45 degrees and scale it so that the the line $ x \equal{} y$ becomes the x-axis. There, in your path, you either rise up 1 or go down 1; the number of such paths is given by the Catalan numbers.

That site should give you more information about the Catalan numbers. [/hide]


The inside jokes game!
(By the way after several posts have passed feel free to explain your jokes to us, but let us be bewildered for a couple minutes)

Inside joke #1
Psyduck and Mudkip. LOL.

Does that have something to do with Pokemon?

I think i saw that episode

Psyduck is a yellow pokemon duck, and the other one is an advanced version of psyduck

they dont really have special powers, unless they are trained well

and all they say is "psyduck" and "mudkip"

No, it's an inside joke between my friends. Post some of yours as well!

[quote="now a ranger"]milkshake, yard, girls[/quote]

my milkshake bring all the boys to the yard, damn right its betta than yours

[quote="now a ranger"]milkshake, yard, girls[/quote]

that's the least inside joke ever

it's like going "hey guys my inside joke is about a dumb blonde, a tv, and a microwave"

1. Where's my backpack?
2. Does anyone have a bandaid?
3. Oh no! A deer!
4. Close the door!!!
5. You have two noses.
6. 99 cases of blush on the wall........ -62 cases of blush on the wall!
7. You can see the roof access from here!!
8. I wonder what human flesh tastes like...
9. *THUD* Aw, man! I didn't know the wall was there...
10. hahahahahahahahaha... *10 minutes later*... hahahahahahahaha
11. turtle igloo pigeon wake up glasses fruitalicious!
12. I was being silent... you just didn't hear.
13. The oh so modest Conway.
14. Chainsaw the giraffe!
15. The laughing disease
16. Hey Albert, where's your silicon spray?
17. Earthquake!
18. Fire Drill!
19. OMFGD
20. I have herpees!

Wow, I have a lot of these... Probably more though, I just can't remember all of them. I'll probably post some more later.


trigonometry
algebra
polynomial
geometry
3D geometry
tetrahedron
probability

Here is a series of problems. Some I wasn't able to do, and some are really interesting (2 main reasons why I post :lol:).

1. Show that in a $\bigtriangleup$ABC, $\sin(\frac{A}{2})=\sqrt{\frac{(s-b)(s-c)}{bc}}$.
[b]a[/b], [b]b[/b], and [b]c[/b] would be the sides and s stands for semiperimeter.

2. In $\bigtriangleup$ABC, CD is the bisector of $\angle C$, with D on AB. If $\frac{\cos(C)}{2}=\frac{1}{3}$ and $CD=6$, then find $\frac{1}{a}+\frac{1}{b}$.

3. Show that if AD is a median of $\bigtriangleup$ABC, then $4AD^{2}=AB^{2}+AC^{2}+2(AB)(AC)\cos\angle BAC$.

4. Prove that the equation $4a^{2}+4a=b^{2}+b$ has no positive integer solutions $(a,b)$.

5. Find the positive integer solutions $(x,y)$ of the equation $x^{2}+3=y(x+2)$.

6. Let $g(x)=x^{5}+x^{4}+x^{3}+x^{2}+x+1$. What is the remainder when the polynomial $g(x{}^{1}{}^{2})$ is divided by the polynomial $g(x)$.

7. Let $ABCD$ be a tetrahedron having each sum of opposite edges equal to 1. Prove that $r_{A}+r_{B}+r_{C}+r_{D}\leqslant\frac{\sqrt{3}}{3}$.

8. A sequence of numbers $a_{1}, a_{2}, a_{3},...$ satisfies $a_{1}=\frac{1}{2}$ and $a_{1}+a_{2}+...+a_{n}=n^{2}a_{n}$ for $n\geqslant1$. Find $a_{n}$ in terms of [b]n[/b].

9. 3 points A, B, and C are selected randomly on the circumference of a circle. Find the probability that the points lie on the semicircle.

10. 3 numbers are chosen at random between 0 and 1. What is the probability that the difference between the greatest and the least is less than $\frac{1}{3}$.

[quote="PenguinIntegral"]It would be best if you put these is separate threads like in the AoPS classes. One thread with many problems tends to get cluttered and is hard to find solutions for specific problems.

I'll be happy to repost these if you decide it is too much work.[/quote]
Thanks PenguinIntegral :)

[hide="#5"]
we solve for $y$:
$y=\frac{x^{2}+3}{x+2}$.
Here we use the long division and get:
$y=x-2+\frac{7}{x+2}$.
Since $x,y$ are integers $\frac{7}{x+2}$ should be an integer.
So the only solution (unless i didn't make any mistake) is
$(5,4)$.
[/hide]

[quote="drunner2007"]your first problem should have $\sin(\frac{A}{2})$. What you have written implies that $\cos(A)=1$.[/quote]
Thanks for catching that Drunner2007. :)

I'm a little busy right now, but I will post some solutions later.

[hide="#6"]
Writing out what we've got:
$x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1=(x^{5}+x^{4}+x^{3}+x^{2}+x+1)Q(x)+R(x)$, for $Q(x)$ and $R(x)$ stand for quotient and remainder, respectively. Also note that $R(x)$ has a degree of 4, by some theorem (is it remainder theorem?).
Now we find the sixth roots of unity, namely $w,w^{2},w^{3},w^{4},w^{5}$ (exclude $w^{6}$ since it's equal to 1).
By subbing in those five values, we find that
$R(w)=R(w^{2})=R(w^{3})=R(w^{4})=R(x^{5})=6$. However, $R(x)$ has a deg 4! So it must be a constant value i.e. $6$!
Thus, the answer is $6$.[/hide]

#9
I hope there are no mistakes. Its a nice problem  :)
[hide]There exists a unique diameter for every point chosen on the circle. It doesn't matter where the first point is placed. After the second point is placed, imagine drawing their diameters and finding the acute angle of intersection. This angle, $\theta$, can range from 0 to 90 degrees. Also, from simple geometry, it is easy to verify that the third point can lie anywhere on an arc subtended by angle $360-\theta$ so that the three points lie on a semi-circle.
Thus the probability that they lie on a semicircle is $\frac{360-\theta}{360}$. The average of this function as $\theta$ ranges from 0 to 90 degrees should gives us our answer. Using basic calculus we find that the probability is $\frac{7}{8}$.[/hide]

#10

[hide]Let $x$,$y$, and $z$ be the numbers chosen at random. Without loss of generality, assume $x\leq y \leq z$. Another way to think of the problem is to imagine selected $x$ and $z$ first such that $d(x,z)<\frac{1}{3}$ and then multiplying this by the probability that $y$ will fall in interval between $x$ and $z$.
To find the probability that $d(x,y)<\frac{1}{3}$, we need to find the probability that $|z-x|<\frac{1}{3}$. Then we need to multiply this by the probability that $y$ will land in the interval between $x$ and $z$, which is $\frac{|z-x|}{1}$. So we need to find the probability that $|z-x|^{2}<\frac{1}{3}$. This is not too hard to figure out (draw two coordinate axes, one z and one x and then come up with geometric interpretation to find the probability). If I havn't made any algebra errors, I believe it comes out to be $\frac{3\sqrt{3}-1}{6}$.[/hide]

for number 7, do those $r$'s denote inradii?

[quote="drunner2007"]for number 7, do those $r$'s denote inradii?[/quote]
Yes, all four are the inradii of the faces. Also, equality holds when ABCD is regular.
[quote="drunner2007"]#9
I hope there are no mistakes. Its a nice problem  :)
[hide]There exists a unique diameter for every point chosen on the circle. It doesn't matter where the first point is placed. After the second point is placed, imagine drawing their diameters and finding the acute angle of intersection. This angle, $\theta$, can range from 0 to 90 degrees. Also, from simple geometry, it is easy to verify that the third point can lie anywhere on an arc subtended by angle $360-\theta$ so that the three points lie on a semi-circle.
Thus the probability that they lie on a semicircle is $\frac{360-\theta}{360}$. The average of this function as $\theta$ ranges from 0 to 90 degrees should gives us our answer. Using basic calculus we find that the probability is $\frac{7}{8}$.[/hide]

#10

[hide]Let $x$,$y$, and $z$ be the numbers chosen at random. Without loss of generality, assume $x\leq y \leq z$. Another way to think of the problem is to imagine selected $x$ and $z$ first such that $d(x,z)<\frac{1}{3}$ and then multiplying this by the probability that $y$ will fall in interval between $x$ and $z$.
To find the probability that $d(x,y)<\frac{1}{3}$, we need to find the probability that $|z-x|<\frac{1}{3}$. Then we need to multiply this by the probability that $y$ will land in the interval between $x$ and $z$, which is $\frac{|z-x|}{1}$. So we need to find the probability that $|z-x|^{2}<\frac{1}{3}$. This is not too hard to figure out (draw two coordinate axes, one z and one x and then come up with geometric interpretation to find the probability). If I havn't made any algebra errors, I believe it comes out to be $\frac{3\sqrt{3}-1}{6}$.[/hide][/quote]
Your solutions for both 9 and 10 are very brilliant and unique, but your answers aren't right. For 9, you should use arcs such that $AB\geq BC\geq AC$. That should make the question quite easy. As for 10, you are on the right path using a graph and such. But I won't give any hints except:
-let the 3 numbers be $x,y, \& z$.
-$x,y \geq z$
-suppose z=a for some a such that $0<a<1$...
-Now use a graph, and I think you may be able to do the rest. :P
PM me if you have any questions.

[quote="lotrgreengrapes7926"][quote="drunner2007"]#9
I hope there are no mistakes. Its a nice problem  :)
[hide]There exists a unique diameter for every point chosen on the circle. It doesn't matter where the first point is placed. After the second point is placed, imagine drawing their diameters and finding the acute angle of intersection. This angle, $\theta$, can range from 0 to 90 degrees. Also, from simple geometry, it is easy to verify that the third point can lie anywhere on an arc subtended by angle $360-\theta$ so that the three points lie on a semi-circle.
Thus the probability that they lie on a semicircle is $\frac{360-\theta}{360}$. The average of this function as $\theta$ ranges from 0 to 90 degrees should gives us our answer. Using basic calculus we find that the probability is $\frac{7}{8}$.[/hide]
[/quote][hide]Shouldn't $\theta$ range from $0^\circ$ to $180^\circ$? This would give the answer $\boxed{\frac{3}{4}}$. [/hide][/quote]
Yes, that's the answer. :)

yeah 0 to 90 is what messed it up. i was using the acute angle in the problem but I guess the acute angle of intersection can be the same while the orientation is different.

#10 was a stupid mistake. you need to multiply those two probabilities together, but not the equations. I think that the answer might be the expected value of $|z-x|$ times the probability that $|z-x|< \frac{1}{3}$. I found that the expected value of $|z-x|$ is [hide]$\frac{1}{2}$[/hide] (which was harder to verify than I thought) so maybe the answer is [hide]$\frac{4}{9}*\frac{1}{2}=\frac{2}{9}$[/hide]?

Anybody have an easy way to find the expected value of $|z-x|$?.

I thought about letting $z=b$ and $x$ a variable so that you can draw a function which represents the values of $|z-x|$ as $x$ ranges from $0$ to $1$. The function is $|b-x|$. You find the average value of this function, which is $\frac{1}{2}$ (b cancles out)
edit: nvm, that was kind of easy, but it seems like there is something wrong.

Your answer is very close the real answer Drunner2007. And your expected value might be hard to find, so I suggest a different approach.

[quote="kimby_102"]I'll take #7  :)
[hide="#7"]
first we note that the area of triangle $ABC$ is
$rs=\sqrt{s(s-a)(s-b)(s-c)}$ for $s$ denotes the semiperimeter of the triangle.
And by AM-GM, we get:
$\sqrt[3]{(s-a)(s-b)(s-c)}\leq \frac{s}{3}$.
By cubing and taking square root of the above inequality we get:
$\sqrt{(s-a)(s-b)(s-c)}\leq \sqrt{\frac{s^{3}}{27}}$.
Thus, the area of the triangle is:
$rs\leq \frac{s^{2}}{\sqrt{27}}$, or simplifying
$r\leq \frac{s}{3\sqrt{3}}$.
now we just apply this to a tetrahedron.
We denote the six edges of the tetrahedron as $a,b,c,d,e,f$.
So $r_{A}+r_{B}+r_{C}+r_{D}\leq \frac{s_{A}+s_{B}+s_{C}+s_{D}}{3\sqrt{3}}$, for $s_{A}, s_{B}, s_{C}, s_{D}$ means semiperimeter of the faces of the tetra.
But note that
$s_{A}+s_{B}+s_{C}+s_{D}=a+b+c+d+e+f$.
However, we are given that the sum of opposite edges equal to 1, so $a+b+c+d+e+f=3$.
Hence, $r_{A}+r_{B}+r_{C}+r_{D}\leq \frac{1}{\sqrt{3}}$. whew,:D

sorry for this long and drowning solution.....

[/hide][/quote]
That's a nice solution. Ingenious, with a bit of brute force. :lol:


linear algebra
matrix
algebra
system of equations

[quote="felipesa"]Be

$ax-y+2z=1$
$-4x+ay+4z=2$
$x+2y+z=-2a$

Find $a$ for the system be impossible[/quote]

from first two $(a^2-4)x+2(a+2)z=a+2$ which is also $(a-2)x+2z=1$-----(i)
from first and third , $(2a+1)x +5z=2-2a)$-----(ii)

For the system to be indeterminable , $\frac{2a+1}{5}=\frac{a-2}{2}$ .Hence $a=12$

[quote="felipesa"]Be

$ax-y+2z=1$
$-4x+ay+4z=2$
$x+2y+z=-2a$

Find $a$ for the system be impossible[/quote]
The determinant of the coefficient matrix must be zero.
 \[
\left(
\begin{array}{ccc}
a & -1 & 2 \\
-4 & a & 4 \\
1 & 2 & 1 \end{array}
\right)\]$-4-2a-2(4a+8)+a^2-4=a^2-10a-24=(a-12)(a+2)=0$
$a=12$ or $a=-2$.
If $a=-2$, then the second equation is just a multiple of the first equation, so that value of $a$ actually gives an infinite number of solutions.
If $a=12$, then we have the following system of equations:\begin{eqnarray*}
12x-y+2z&=&1\\
-4x+12y+4z&=&2\\
x+2y+z&=&24\end{eqnarray*}Adding three times the second equation to the first equation and four times the third equation to the first equation, we get: \begin{eqnarray*}
35y+14z&=&7\\
20y+8z&=&98\\
x+2y+z&=&24\end{eqnarray*}
Note that the first equation states $5y+2z=1$ while the second says $5y+2z=24.5$. Thus, $a=12$ makes the system impossible.


geometry
parallelogram
congruent triangles
Pythagorean Theorem

1. Prove that if AB, BC, CD, AD are the sides, and AC and BD are diagonals of a parallelogram (ABCD), then Prove that

2(AB^2 + BC^2) = AC^2 + BD^2

2. Draw a Parallelogram ABCD, draw the diagonals. How many sets of congruent triangles do you have?

3. Prove that a parallelogram is formed when you connect consecutive midpoints of the sides of a quadrilateral.

Please excuse my bad MS Paint skills.

1.[hide]  Let AC = e_2 and BD = e_1.  We want to prove that

[tex]2(AB^2 + BC^2) = e_2^2 + e_1^2.[/tex]

By the Pythagorean Theorem, the RHS becomes

[tex]

(DC+CF)^2 + BF^2 + AH^2 + HC^2  = [/tex]

[tex] DC^2 + 2DC\cdot CF+CF^2 + BF^2 + BF^2 + HC^2[/tex]

[tex] =   DC^2 + 2DC CF + CF^2 + 2BF^2 + (DC-CF)^2[/tex]

[tex] =  DC^2 + CF^2 + 2BF^2 + DC^2 + CF^2[/tex]

[tex] =  2DC^2 + 2CF^2 + 2BF^2[/tex]

[tex] =  2(DC^2 + CF^2 + BF^2)[/tex]

[tex] = 2(DC^2 + BC^2)[/tex]

[tex] = 2(AB^2 + BC^2)[/tex]

Since DH = CF = GB and DC = AB and AH = GC = BF.

[/hide]

2.  (using the same diagram)

    [hide] Triangle AOD is congruent to COB

     AOB is congruent to COD

     ABC is congruent to CDA

     DCB is congruent to BAD

That makes four pairs.[/hide]


What's the rarest name you've ever heard of, just first name.

For me it's Shravas.  According to a search on some people finding thing, there is no one named Shravas that is an adult in the U.S.

[quote="bleumoose"]
Shetheed
[/quote]

Urban legend.  everyone claims to know someone who knows someone with that name, but none of you actually know someone with this name.  should set off your skepticism meter..

same thing for the names Orangejello (supposedly pronounced like "or-an'-je-lo" or Nosmo King..

Agi Fatou Kah.  (Fatou is just basically pronounced photo...and she hates her middle name.)  I don't know how "rare" that is...

[quote="TripleM"]Thinh. I don't know if this is common in another country, but I've never heard of it (other than once.) It seems the place to find the weirdest names is in school textbooks! I'm sure theres some rule about what names you are allowed to include etc.. In this textbook I was reading, I saw a Vinh, thought an h on the end looked funny, then came across this thinh..[/quote]
That name looks Vietnamese, so it might not really be very rare.

Actually, I have no idea how rare this is, but: Pafnuty. (Chebyshev's first name.) It's kind of fun to say, though, even though I'm not sure how to pronounce it. Pafnuty Lvovich Chebyshev.  :) Cool name. I mean, my first name isn't exceedingly common (not rare, either), but my last name is...Brown. Yeah.

Edit: I Googled "Pafnuty" and excluded as many spellings of "Chebyshev" as I could think of, and got just over 600 pages, with most of those still about Tchebyscheff, Cebychev, Tschebyshew, Tchebyshiov, Cebysev, etc.

[quote="bleumoose"]did you hear about that guy that, instead of naming his kid "junior" he named him "2.0"?[/quote]

Coulda named him Junior 2.0. Wait, no, then he'd sound like a chess program.

Virgal, Josiea, Tru Love and Spring Dew (their parents were hippies), Neah Kye (indian parents), and Keorprich (german). Those are some of the weirdest name choices to me. Theyre unique and yet they all have a certain commoness to them.

thinh, vinh, anything like that is vietnamese (thanh, khanh, etc.)

i know someone whose mother's name is Pebble...

Sky?  Sortof leftover from the 70s...

Non american names don't count as odd sounding names...of course they sound odd to you...

konstantine spelt with a k

the only place i've seen it is in a something corprate song.

but hey, out of the 200 million or so people in the US, someone's bound to have replaced the "c" with a "k" right?

i knew a Star
anyone whose ACTUAL name is "Bubba" (i'm pretty sure this has happened)

[quote="Fermatprime"]Edit: I Googled "Pafnuty" and excluded as many spellings of "Chebyshev" as I could think of, and got just over 600 pages, with most of those still about Tchebyscheff, Cebychev, Tschebyshew, Tchebyshiov, Cebysev, etc.[/quote]

What about Cebasev? I've read this somewhere...

And yes, I think Pafnuty is not a really Russian name... sounds like something Greek spelt in a cyrillic way, but I actually don't know...

German names can be sometimes strange. Try to find out the genders of the following ones: [b]Hauke[/b], [b]Irmgard[/b], [b]Ingmar[/b], [b]Selmar[/b], [b]Sren[/b], [b]Hedwig[/b], [b]Henning[/b], [b]Horst[/b], [b]Dankrad[/b]. (I know, some of these are not really German, but they are used in Germany. I have really seen people with all of these names!)

And then look at

http://www.vornamenlexikon.de/

(the red letters stand for female names, the blue letters for male ones).

  Darij

Chinese language is different from Western Language. So there is no common and unique name in China.

fhqwagadshgnshjsdbkhjsbkhjsbkhjsbdbkveybvi

lol homestarrunner.com

homestar runner.com is so funny!! :rotfl:

Do you people think "Stellar" is a wierd name for a korean girl?

[quote="pkothari13"]the rarest name i've heard or seen?  [b]!xobile.[/b]

lol, tell me if anyone recognizes that name.  :rotfl:

or how about Apple for gwyneth paltrow's daughter?[/quote]

Please do not yell in the casino.  :P

He is one of the funniest guys I have ever seen.
I'd say !xobile is the rares name I've heard as well. He's got a click in his name!!!  :D

i know a rare name, my brother is named [b]hairy monkey [/b]:D

supposedly, there is a person who is named Gaylord Focker, as in the movie, Meet the Parents.

Penn, from Penn & Teller, recently named his son's middle name as 'Crimefighter'

hairy monkey is actually a self-made nickname :lol by the way, he is nothing close to being a hairy monkey, hes a cool person ;)

[quote=mathleticguyyy]Vinjai...[/quote]

NOT AGAIN -.-

[quote=mikimoto12][quote=bleumoose]my english teacher a few years back told me about this little kid that used to be in the preschool class she had whose name was (im going to spell it phonetically):
Shetheed

thats how it was pronounced. but replace the first e with an i and the last e with an a for the spelling.

did you hear about that guy that, instead of naming his kid "junior" he named him "2.0"?[/quote]

HAHAHAHAHAHAHAHAHAHA Sh*thead seriously HAHAHAHAHAHA[/quote]

yeah like really!!!!! hahaha  :rotfl:
do u know anyone named sotowa

[quote=Sotowa]
do u know anyone named sotowa[/quote]

no ;)


geometry
3D geometry
sphere
combinatorics proposed
combinatorics

A space is divided into 2 parts by a plain in the space, call the other (not encluding plane) [i]semi plane[/i].
Let $ S$ be the set with 10 sets in space such that any 4 points are not coplanar.Find the number of common part
which is sub set of $ S$ and become the common part of $ S$ and [i]semi plane[/i].

[url=http://mathworld.wolfram.com/SpaceDivisionbySpheres.html]Space Division by Spheres[/url] - MathWorld
[url=http://www.research.att.com/~njas/sequences/A046127]A046127[/url] - OEIS

[hide="Answe"]260[/hide]


number theory
Diophantine equation
relatively prime

This is a little brief but we can rule out the case where $p=1$ or $q=1$ immediatly, and then we get the result straight away from Mihalescu's theorem.

[quote=Vulch]Why q is odd??
Would anyone solve it with Mihalescu's theorem?
Thank you![/quote]

If $2\mid q$ , then $2\nmid q^2-1=2^p$ , where $p$ is a positive integer . Contradiction.

Sorry I'm bad , but what is Mihalescu's theorem?

You aren't bad. It's just that the theorem is too fancy!.

It states that the only positive integer solution of the equation $a^m - b^n = 1$ is $(a, b, m, n) \equiv (3, 2, 2, 3)$.

Just use Mihaelescu's!

Using Mihaelescu's theorem, we know that the only two consecutive powers are $8$ and $9$. So, $2^p = 8$ and $q^2 = 9$. It follows that $p = 3$ and $q = 3$ (which is the only pair).

[quote=Tiago Guerreiro]hi :)

Find all pairs $(p,q)$ of positve integers such that: $2^{p}+1=q^{2}$[/quote]

\[2^p = (q+1)(q-1)\]
So,
\[q+1 = 2^x \quad \text{and}\quad q-1 = 2^y\]
where $x>y$ and $x+y=p$. Eliminate, we have \[2 = 2^x-2^y = 2^y\left (2^{x-y} - 1\right ) \]
Because $2^{x-y}-1$ odd numbers, we have $y=1$. Then, $2^{x-1} = 2\Rightarrow x=2$. Then, $(p,q) = (3,3)$.


AMC
AIME
geometry
trigonometry
trig identities
Law of Sines
angle bisector

The triangle ABC has ∠A = 60 :^o:, ∠B = 45:^o:. The bisector of ∠A meets BC at T where AT = 24. Find area ABC.

hrmph all the ways I have depend on calculating sin (5:pi:/12). This could be done with the addition formula, but it's still not very nice. The simplest way that I've found is observing that TAC is isosceles, meaning AC=24, using the law of sines to find AB in terms of AT, and then calculating the area by 1/2(AC)(AB)(sin BAC).

Bob, no matter what, I think you're going to have to use trig for this problem.  It's not hard at all.

you could also make some 30-60-90 and 45-45 triangles to get AB and stuff.

Yeah, that could also possibly work, but trig is still a valuable shortcut for this problem.

Yeah, I know, it's easy to solve the problem, and I sketched the solution above. I was just wondering whether there was possibly a nicer way. I suppose not.

I went through the problem and found you can use 30-60-90s, 45-45-90s, and isoceles triangles the whole way through.  For instance, if you draw the altitude to T in triangle ATB, you get a 45-45-90 and a 30-60-90, so the triangle has dimensions 12:rt2:, 24, and 12(1+:rt3:).  Since TC/AC = BT/AB (since AT is an angle bisector, we know that TC/AC = :rt2:/(1+:rt3:).  Triangle ATC is isoceles with two sides of length 24, so you can draw the altitue to A to produce two right triangles.  We then express TC in terms of AC (TC = 24:rt2:/(1+:rt3:)). Finally, we compute the area of ATC, which I'm too lazy to do, and add it to the area of ABC, which is easily computed using those special right triangles, to get the total area.

You could use trig, but considering that the AIME doesn't allow calculators, it might be a pain.

not really more of a pain that doing it your way, methinks.

Instead of just saying that, why don't you show your method so we can see whether or not it is more of a pain?


Hi every one . I'm trying to study Russian  ,but i doesn't have any experience in this problem . So I create this topic to take some ideas about how to study Russian . How should I start learning ?
Thanks for your help. :)

hmm...according to what english-speaking people,who studied russian language,said,it is one of the hardest languages due to huge number of different phrasal verbs(like get on,shut up,take off or turn on in english language)...
also the grammar in russian language is really very hard,I can even bet that chinese language or whatelse is much easier.
Even though I have studied russian for 11 years,I am afraid that I still haven't studied it completely.
well...if grammar is not this important for you,then you can easily bite just on speaking skills,though I doubt it is possible in non-russian-speaking environment.
I guess the first word you need to study in russian is "hello" and it can be translated as "здраствуйте" and can be pronounced as
"zdravstvuitye"-really hard if comparing it with hello,isn't it?or you may also use привет instead of "zdravstvuitye",but it is rather informal greeting word,the same as "hi".it can be pronounced as "preevyet".btw can you see my inscriptions in english?
I would suggest you to study english at first and only after that to start learning russian...
If you have any questions or misconceptions then feel free to ask.

Thanks for your help  :wink:. Does there any shorter ways to study Russian ? Now ,I study  Russian to read some books in Russian ,for example Kvant  and in the future I can study deeper to speaking ,reading ...

you can do some on line websites that teach you russian and youtube helps also.
[url]http://www.masterrussian.com/[/url]
this site may help.

Hello everyone

I'm trying to learn Russian. I know that some words in Russian have some realtions with English

[hide=" For example "]аэропорт --- airport

бар --- bar

Геометрия---geometry

алгебра--Algebra

Анализ--analysis

Москва--Moskva

Путин--Putin

директор --- director

Интернет --- Internet

 кафе --- cafe

класс --- class

компьютер --- computer

 лифт --- lift (elevator)

океан --- ocean

опера --- opera

паспорт --- passport

президент --- president

робот --- robot

студент --- student
[/hide]

Can anyone tell me some tips like that to learn Russian ? :)

Thank you in advance !


A gas station has 12 vehicles per hour go into for gas pumps at average. find the probability of the event: in one hour there are at least 8 vehicles go into for gas pumps

without any additional hypothesis, the answer can be almost any number between $ 0$ and $ 1$. I suspect that you are modelling the arrival of new customers as a Poisson process, or something like that ?

Yes, I would assume that the intent behind supplying just one piece of information, with that piece being a rate (12 cars per hour), is that this should be represented by a Poisson distribution.

$ P(X\ge 8)\equal{}\sum_{k\equal{}8}^{\infty}\frac{e^{\minus{}12}12^k}{k!}\equal{}1\minus{}\sum_{k\equal{}0}^7\frac{e^{\minus{}12}12^k}{k!}\approxeq .8450$

For efficient computation on a spreadsheet, compute probabilities recursively.

$ P(X\equal{}0)\equal{}e^{\minus{}12}$ and after that, $ P(X\equal{}k)\equal{}P(X\equal{}k\minus{}1)\cdot\frac{12}{k}.$ The spreadsheet can also easily keep track of the running sum.

Other items from the same sheet:
$ P(X\ge 12)\approxeq .4240$
$ P(X\equal{}0)\approxeq 6.14\times 10^{\minus{}6}$
$ P(X\ge 16)\approxeq .1013$
$ P(X\ge 30)\approxeq 3.37\times 10^{\minus{}6}$

oh, thank you so much! your result is the same to the hint in my book  :)


algebra
polynomial
number theory unsolved
number theory

Given p be a prime and $ p \ge 5$.Prove that exist $ q_1$,$ q_2$ such that $ 1 < q_1 < q_2 < p$ simultaneously $ q_1 ^{p - 1} - 1$ and $ \q_2 ^{p - 1} - 1$ non divisible $ p^2$[/quote]

[hide]
Suppose that at most one value of $ x$ from $ 1$ to $ p\minus{}1$ does not satisfy $ x^{p\minus{}1}\minus{}1\equiv 0\mod p^2$. (Then there are at least $ p\minus{}2$ solutions to the equation between $ 1$ and $ p\minus{}1$.) Now let $ a$ and $ b$ be among $ p\minus{}1$, $ p\minus{}2$, and $ p\minus{}3$ such that $ a>b$ and $ a$ and $ b$ both satisfy $ x^{p\minus{}1}\minus{}1\equiv 0\mod p^2$. Then $ a^2$ and $ ab$ both satisfy the equation, but they are greater than $ p$ and less than $ p^2$. Thus, there at least $ p$ solutions to the equation mod $ p^2$. However, the equation has degree $ p\minus{}1$, so it can have at most $ p\minus{}1$ solutions, a contradiction.
[/hide]

[quote="tjhance"][hide]
Suppose that at most one value of $ x$ from $ 1$ to $ p \minus{} 1$ does not satisfy $ x^{p \minus{} 1} \minus{} 1\equiv 0\mod p^2$. (Then there are at least $ p \minus{} 2$ solutions to the equation between $ 1$ and $ p \minus{} 1$.) Now let $ a$ and $ b$ be among $ p \minus{} 1$, $ p \minus{} 2$, and $ p \minus{} 3$ such that $ a > b$ and $ a$ and $ b$ both satisfy $ x^{p \minus{} 1} \minus{} 1\equiv 0\mod p^2$. Then $ a^2$ and $ ab$ both satisfy the equation, but they are greater than $ p$ and less than $ p^2$. Thus, there at least $ p$ solutions to the equation mod $ p^2$. However, the equation has degree $ p \minus{} 1$, so it can have at most $ p \minus{} 1$ solutions, a contradiction.
[/hide][/quote]
A very good solution ! . I have an other solution .
For each prime $ p$ ,call a natural number $ a$  is good (to p ) iff $ a^{p \minus{} 1}\equiv 1 (\mod p^2)$ .Others are called ''bad number''
[u]Result 1 :  [/u]
$ p \plus{} 1,p \minus{} 1$ are not good numbers .
It is obvious from lifting exponent $ p\parallel{}(p \plus{} 1)^{p \minus{} 1} \minus{} 1$ and $ p\parallel{}(p \minus{} 1)^{p \minus{} 1} \minus{} 1$
[u]Result 2 [/u] Product of two good numbers is an good number and product of a good and a bad number is a bad number .
Result 3 : $ p^2|(p^2 \minus{} k)^{p \minus{} 1} \minus{} k^{p \minus{} 1}$ ,therefore if $ k$ is an good number then $ p^2 \minus{} k$ is also .
Now consider our problem . Because $ p \plus{} 1,p \minus{} 1$ are bad numbers so each of them has at least a prime divisor that is ''bad '' . If this number is different from $ 2$  then each of number $ \frac {p \plus{} 1}{2} , \frac {p \minus{} 1}{2}$ is a bad number and they are relative prime .Thus we can chose two different prime divisors of two number . They will satisfy condition .
For the hard case , $ 2$ is an bad number  we will prove that it has at least an odd number (different from 1 ) less than $ p$ which is bad .  Suppose all odd numbers are good . From result 3 $ p^2 \minus{} (2j \plus{} 1),j \equal{} 1,...,\frac {p \minus{} 3}{2}$ is good number . Use the fact that the equation $ x^{p \minus{} 1}\equiv 1 (\mod p^2)$ has at most $ p \minus{} 1$ solution it will give contradiction . Thus at least an odd number $ 1 < c < p$ is a bad number .$ c$ has at least bad prime divisor $ q$ . The set $ (q_1,q_2)\equal{}(2,q)$ satisfy condition .
This problem has been proved .

great solutions guys. i wonder what else can be deduced from that method. can u find $ q_1,q_2,q_3$ for sufficiently large $ p$
and i was also wondering... i was trying to proove this before i saw your solutions:
there exists no prime $ p$ such that $ 2^p\minus{}1$ is devisable by $ p^2$

Well, I just realized that because if $ x^{p\minus{}1}\equiv 1\mod p^2$, then $ (\minus{}x)^{p\minus{}1}\equiv 1\mod p^2$. So since there are at most $ p\minus{}1$ roots, there are at most $ \frac{p\minus{}1}{2}$ roots below $ p$, so you can find $ q_1,q_2,...,q_{\frac{p\minus{}1}{2}}$.

On second thought, I think my solutions are invalid. My "the maximum number of solutions is the degree of the polynomial" technique doesn't really work because its not mod a prime. For instance, $ x^2 \minus{} 1\equiv 0\mod 8$ has more than $ 2$ solutions...

true  :blush: only holds for prime moduls

well here are my ideas. but it's all based on the idea that $ 2$ is always a "bad" number. can someone proove that? i think it's true but i can't prove it.

since $ 2$ is bad and a product of good and bad number is a bad number we have:
one of the numbers $ 3$ and $ 6$ is a bad number. because if they were both good than we would have a product of good and bad number be a good number which is a contradiction. so unleast one of them must be "bad".

we may conclude the same thing for the pairs of numbers $ (5,10)\ ; \ (7,14)\ ; \ (9,18)\ ......\ (2k \plus{} 1,4k \plus{} 2)$ (these pairs are all disjnctive).
and we may add here also the pair $ (4,8)$.

therefore the number of bad numbers between smaller than $ p$ is unleast $ \left[ \frac {p \plus{} 1}{4}\right] \plus{} 1$ for every prime larger than $ 7$ (every prime that includes the pair 4,8)

[quote="tjhance"]On second thought, I think my solutions are invalid. My "the maximum number of solutions is the degree of the polynomial" technique doesn't really work because its not mod a prime. For instance, $ x^2 - 1\equiv 0\mod 8$ has more than $ 2$ solutions...[/quote]
From Helsen lemma we can prove following problem .
Let $ P(x)$ is a polynomial with integer coefficient has exactly $ m$ solution $ x_1,..,x_m$ mod p (I mean $ x_j\in \mathbb{F}_p$ and p is odd prime ) and $ \gcd(P'(x_j),p) = 1$ then for each positive integer $ k$ the congruence equation : $ P(x)\equiv 0 (\mod p^k)$ has exactly $ m$ solution in ${ \mathbb{F}_{p^k}}$
For the equation $ x^{p - 1} - 1\equiv 0 (\mod p)$ , it is clear that $ \gcd(f'(x),p) = 1$ for all $ x$ doesn't divide by $ p$ .

Sorry I have searched the Internet by Google and I just don't know what is a "Helsen 's lemma" you mentioned and how to prove it. :blush:
Could you please introduce more?Thanks!

http://en.wikipedia.org/wiki/Hensel's_lemma

It's Hensel's lemma

Daniel