Hey, could someone please explain (with a very amount of detail) how to solve problem #25 on the AMC 12-A.  I'd really appreciate it, its been bugging me.  thanks

if you forget the question is
" For each integer n greater than or equal to 4, let a-sub-n denote the base-n number .133(repeating)n.  The product a4a5a6....a99 can be expressed as m/n!, where m and n are positive integers and n is as small as possible.   What is the value of m?
a) 98 b) 101 c) 132 d) 798 e) 962

The problem was discussed in a math jam. The transcript is here:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=25

Also, note that we will be hosting a Math Jam this coming Saturday where we will be discussing in detail how to attack the problems from the B exams.

http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php

Welcome to the board.  I think you will find a lot of good discussion relating to AMC and AIME problems.

Thanks for the link, the reason I had so much trouble solving it was because I didn't know about infinite geometric sums....which makes it much easier. anyways, thanks again


calculus
integration
geometry
limit
calculus computations

For $ 0<a<1$, let $ S(a)$ is the area of the figure bounded by three curves $ y\equal{}e^x,\ y\equal{}e^{\frac{1\plus{}a}{1\minus{}a}x}$ and $ y\equal{}e^{2\minus{}x}$.
Find $ \lim_{a\rightarrow 0} \frac{S(a)}{a}$.

$ e^x\equal{}e^{\frac{1\plus{}a}{1\minus{}a}x} \to x\equal{}0, y\equal{}1$
$ e^x\equal{}e^{2\minus{}x} \to x\equal{}1, y\equal{}e$
$ e^{2\minus{}x}\equal{}e^{\frac{1\plus{}a}{1\minus{}a}x} \to x\equal{}1\minus{}a, y\equal{}e^{1\plus{}a}$

$ S(a)\equal{}\int_0^{1\minus{}a}e^{\frac{1\plus{}a}{1\minus{}a}x} \minus{}e^x dx \plus{} \int_{1\minus{}a}^{1} e^{2\minus{}x}\minus{}e^xdx\equal{}$

$ \left( \frac{1\minus{}a}{1\plus{}a}e^{\frac{1\plus{}a}{1\minus{}a}x} \minus{}e^x \right)_0^{1\minus{}a}\plus{}\left( \minus{}e^{2\minus{}x}\minus{}e^x \right)_{1\minus{}a}^1\equal{}$

$ \frac{1\minus{}a}{1\plus{}a}e^{1\plus{}a}\minus{}e^{1\minus{}a}\minus{} \frac{1\minus{}a}{1\plus{}a}\plus{}1 \minus{}2e\plus{}e^{1\plus{}a}\plus{}e^{1\minus{}a}\equal{} \frac{2}{1\plus{}a}e^{1\plus{}a}\minus{}\frac{2a}{1\plus{}a}\minus{}2e$

$ \lim_{a\to 0} \frac{S(a)}{a} \equal{}\lim_{a\to 0} \frac{\frac{2}{1\plus{}a}e^{1\plus{}a}\minus{}\frac{2a}{1\plus{}a}\minus{}2e}{a} \equal{}$ (using L'H)

$ \lim_{a\to 0} \frac{\partial}{\partial a}(\frac{2}{1\plus{}a}e^{1\plus{}a}\minus{}\frac{2a}{1\plus{}a}\minus{}2e)\equal{}$

$ \lim_{a\to 0} \left( \frac{2}{1\plus{}a}\minus{}\frac{2}{(1\plus{}a)^2} \right)e^{1\plus{}a}\minus{}\frac{2}{1\plus{}a}\plus{}\frac{2a}{(1\plus{}a)^2}\equal{}$

$ \lim_{a\to 0}2\frac{ae^{1\plus{}a}\minus{}1}{(1\plus{}a)^2}\equal{}\minus{}2$

$ S(a)$ is right, but the answer is incorrect.


probability
function
algebra
floor function

A four-digit number is chosen at random from all four-digit numbers.  Express as a common fraction the probability that the number is divisible by $ 2, 3, 4,$ and $ 5$.

[hide="ANS."]$ \frac{1}{60}$[/hide]

... Just find how many multiples of 60 there are between $ 1000\minus{}9999$

[hide]$ [\frac{9999}{60}]\minus{}[\frac{1000}{60}]\equal{}166\minus{}16\equal{}150$
($ [x]$ denotes the greatest integer function
$ \frac{150}{9000}\equal{}\frac{1}{60}$[/hide]


If $ a$ and $ b$ are natural numbers such that:

$ \sqrt[a\minus{}b]{\frac{\sqrt[b]{b.a^{a}}}{\sqrt[a]{b.a^{b}}}}\equal{}\frac{3}{4}$

Then, find the value of $ a^{2}\plus{}b^{2}$

hello, are you sure that there is no typo in your problem?
Sonnhard.

I think that this a trigonometric problem, but I don't know how to do. No, the question is it.


calculus
integration
trigonometry
function
calculus computations

Why do the integral has two result?

$ *\ \ I \equal{} \int\limits_{a}^{b}{\sqrt {\left( x \minus{} a \right)\left( b \minus{} x \right)}.dx}$

$ put\ x \equal{} a \plus{} \left( b \minus{} a \right){{\sin }^{2}}t\Rightarrow dx \equal{} 2\left( b \minus{} a \right)\sin t.\cos t.dt$

$ x \equal{} a\Rightarrow \left( b \minus{} a \right){{\sin }^{2}}t \equal{} 0\Leftrightarrow \sin t \equal{} 0\Leftrightarrow t \equal{} 0\ \ \ \ x \equal{} b\Rightarrow {{\sin }^{2}}t \equal{} 1\Leftrightarrow t \equal{} \pm \frac {\pi }{2}$

$ \Rightarrow I \equal{} \left[ \frac {{{\left( b \minus{} a \right)}^{2}}t}{4} \minus{} \frac {{{\left( b \minus{} a \right)}^{2}}\sin 4t}{16} \right]_{0}^{\pm \frac {\pi }{2}} \equal{} \frac {\pm {{\left( b \minus{} a \right)}^{2}}\pi }{8}$

Because you don't take both $ \pm\frac{\pi}{2}$.  Just take $ \frac{\pi}{2}$.  I find it easier to not change the limits of integration when doing the substitution-make the substitution, do the integral, convert back to the original variable, and then plug in the limits.  I think it's easier this way-just my opinion.

The question is meant to be this:

Why, if I choose $ \minus{} \frac {\pi}{2}$ as the upper bound, do I get a different answer (even though I'm integrating the same function over the same interval)?

The answer is:

You don't get a different answer.  When simplifying the expression $ \sqrt {(x \minus{} a)(b \minus{} x)}$ after the substitution, you change it to $ \sqrt {(a \minus{} b)^2 \sin^2 t (1 \minus{} \sin^2 t)}$.  This is equal to $ (b \minus{} a) \sin t \cos t$ [i]provided this expression is positive[/i]; if you choose to have $ t$ run over the interval $ \left[ \minus{} \frac {\pi}{2}, 0\right]$, then this expression will be negative and so instead you have $ \sqrt {(a \minus{} b)^2 \sin^2 t (1 \minus{} \sin^2 t)} \equal{} \minus{} (b \minus{} a)\sin t \cos t$.  This exactly cancels out the other minus sign, giving you the same answer.

The lesson is:

Don't be sloppy!  Be careful when considering functions like the square root that potentially are multiply-valued to make sure you have the branch you wanted.


function
logarithms
number theory unsolved
number theory

Let $ d(n)$ denote the number of (positive) divisors of n. Thus $ d(6) \equal{} 4$.

(a) Show that d(n) is a multiplicative function, and compute $ d(p^j)$ for $ p$ a prime, $ j\geq 1$.

(b) Prove that there is a constant $ C$ such that
$ d(n) \leq exp(C\frac{log n}{log log n})$
holds for $ n \geq 3$.

1. $ d(n)$ is multiplicative by the fact that if $ n \equal{} rs$, where $ (r,s) \equal{} 1$, every factor of $ n$ can be uniquely expressed as $ ab$, where $ a|r$ and $ b|s$. $ d(p^j) \equal{} j \plus{} 1$ because the factors of $ p^j$ are $ 1, p, p^2, ..., p^j$.

2. let $ n \equal{} p_1^{r_1}\cdots p_k^{r_k}$ and let $ r$ be the biggest of the $ r_i$ and $ p$ the corresponding prime. we have $ p^r\leq n$, so that $ r\leq \frac {\log n}{\log p} \leq \frac {\log n}{\log 2}$. also, $ 2^{k} \leq n$, so that $ k\leq \frac {\log n}{\log 2}$. so

$ d(n)\leq (r \plus{} 1)^k \leq (\frac {\log n}{\log 2} \plus{} 1)^{\frac {\log n}{\log 2}} \leq 1 \plus{} \left(\frac {\log n}{\log 2}\right)^2 < \frac {2}{(\log 2)^2}(\log n)^2 \equal{} \exp(a \plus{} 2\log\log n) \equal{} \exp((\log \log n)(\frac {a \plus{} 2\log\log n}{\log\log n})) < \exp(\frac {3(\log\log n)^2}{\log\log n}) < \exp(\frac {3\log n}{\log\log n}),$

where $ a \equal{} \log\frac {2}{(\log 2)^2}$


Let $a,b,c > 0$ real numbers,
such that: $a+b+c = 1$.

Prove that:
\[ \frac{1}{2} \leq \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a} < 1. \]

Carlos Bravo :)
LIMA-PERU

I guess this belongs to pre-olympiad stuff, anyway,
$\sum \frac{a^2}{a+b} \ge \frac{(\sum a)^2}{2(\sum a)} = \frac{\sum a}{2} = \frac{1}{2}$

$\sum \frac{a^2}{a+b} \le \sum \frac{a^2+ab}{a+b} = \sum a = 1$


Let z=(1+i√3)/(α-(α+1)i). Find all α"∊" R, where z"∊"R

Let $ z\equal{}\frac{1\plus{}\cdot i\sqrt{3}}{a\minus{}(a\plus{}1)\cdot i}$ Find $ a\in\mathbb R$ such that $ z\in\mathbb R$
$ z\equal{}\frac{(1\plus{}i\sqrt{3})(a\plus{}(a\plus{}1)i)}{a^{2}\plus{}(a\plus{}1)^{2}}$ $ \Leftrightarrow$ $ z\equal{}\frac{(a\minus{}(a\plus{}1)\sqrt{3})\plus{}i(a\sqrt{3}\plus{}a\plus{}1)}{a^{2}\plus{}(a\plus{}1)^{2}}\in\mathbb R$
$ \Leftrightarrow$
$ a(\sqrt{3}\plus{}1)\plus{}1\equal{}0$ $ \Leftrightarrow$ $ a\equal{}\minus{}\frac{1}{1\plus{}\sqrt{3}}$

I think you should have posted in highschool basics.


Euler
trigonometry
LaTeX
complex numbers

Proof that sin x = cos(x - :pi:/2) by using the Euler Formula.

[quote="d343seven"]Euler's Formula says
$e^{ix}=\cos x+i\sin x$
Plugging in $(x-\pi/2)$ for $x$ we get
$e^{i(x-\frac{\pi}2)}=\cos (x-\frac{\pi}2)+i\sin (x-\frac{\pi}2)$
$\frac{e^{ix}}{e^{i \frac{\pi}2}}=\frac{\cos x+i\sin x}{0+i} \Rightarrow \sin x-i \cos x$

Comparing this to the original equation and setting the real part equal to the real part, we get $\sin x = \cos{x-\frac{\pi}2}$[/quote]

My proof was slightly more simple:
\begin{eqnarray*}
\sin(x+\pi) &=& \frac{e^{ix+\frac{\pi}{2}} - e^{-ix-\pi}}{2i}\\
&=& \frac{e^{ix+\frac{\pi}{2}} + e^{i\pi}\cdot e^{ -ix-\frac{\pi}{2}} }{2i}\\
&=& \frac{ e^{ix}+e^{-ix} }{2}\\
&=& \cos x
\end{eqnarray*}

But I must agree yours is welfound  ;)

PS: if anyone can put this in LaTeX, be my guest...

[i][edit: i put it into $\text{\LaTeX}$ for you...joml88][/i]

i think it should be:
\[\sin{(x+\pi)}=-\cos{x}\]


Given 1, 4, 13, 40...n, find an equation to describe this pattern.

I know that it increases by powers of 3, but the only equation I can find is:

when A0=0,

An= An-1 + 3^n-1

Is there another equation that wouldn't depend on knowing the previous value?

[hide="Alternatively"]you can use geometric sequence:

$A_{1}=3^{0}$

$A_{2}=A_{1}+3^{1}=3^{0}+3^{1}$

$A_{3}=A_{2}+3^{2}=3^{0}+3^{1}+3^{2}$

etc. hence

$A_{n}=3^{0}+3^{1}+\dots+3^{n-1}$

Then use the summation formula for the last expression[/hide]

the summation formula is derived as follows:
\[\text{Let }S=\sum_{i=0}^{n}3^{n}\]
Write an equation in $S$ and solve for it.
[hide="click only if stuck"]
\[3S=\sum_{i=1}^{n+1}3^{n}=S+3^{n+1}-1\implies 2S=3^{n+1}-1\implies \boxed{\frac{S=3^{n+1}-1}{2}}\]
[/hide]
[hide]nice solution t0r, nice method (more realistic for harder problems of this type as opposed to the standard method i gave)[/hide]


geometry
inequalities
inequalities unsolved

For a triangle of sides $a,b,c$ and area $A$  let

$ K=ab+bc+ca $    and  $ H=1/2(a^2+b^2+c^2) $.

Prove the inequality

$ \frac{(K-H)^2}{12} \geq A^2 \geq \frac {(K-H)(3K-5H)}{12} $

The second inequality is of no value in case $ 5H \geq 3K $, since the lower estimate which it gives for $ A^2 $ is not even positive.

Take [tex]x=p-a;y=p-b;z=p-c [/tex] where [tex]p=\frac{a+b+c}{2}[/tex].Then
[tex]K-H=2(xy+xz+yz)[/tex];[tex]H=\frac12[(x+y)^2+(y+z)^2+(z+x)^2][/tex].
The left one is equal to:
[tex](xy+xz+yz)^2 \geq3xyz(x+y+z)[/tex]
[tex]\Leftrightarrow x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2\geq 0[/tex]
The right one is equal to:
[tex]\sum_{sym}x^3y\geq\sum_{sym}x^2y^2[/tex] which is right since
[tex] x^3y+y^3x \geq 2x^2y^2 [/tex] :)


Let $ O$ be the point of intersection between diagonals $ AC,BD$ of quadrilateral $ ABCD$ ($ AB\equal{}BC,CD\equal{}DA$). Let $ N,K$ be the feet of the perpendiculars from $ D,B$ to $ AB,CD$ respectively. Prove that $ N,O,K$ are collinear.

See the attached diagram.

$ \angle BAD\equal{}\alpha\implies\angle ADN\equal{}\alpha\minus{}90^\circ$

Since quadrilateral $ AODN$ is cyclic (opposite angles are right), we get $ \angle AON\equal{}\angle ADN\equal{}\alpha\minus{}90^\circ$

$ \angle BCD\equal{}\alpha\implies\angle CBK\equal{}\alpha\minus{}90^\circ$

Since quadrilateral $ COBK$ is cyclic (opposite angles are right), we get $ \angle COK\equal{}\angle CBK\equal{}\alpha\minus{}90^\circ$

Now $ \angle AON\equal{}\angle COK$ implies that $ N, O, K$ lie on the same line. QED


geometry
rectangle
floor function
trigonometry

dynamo729's aunt has a table with a diameter of 7 feet. She wants to know how many chairs she can fit around her table if the chairs seats are 23 inches wide and 29 inches long. She wants the chairs to be able to be pushed in.

Any help guys? An answer soon would be greatly appreciated.
.

[b]Essential Information:[/b] The table is circular.

Let's see... I'm guessing this would involve some nasty trigonometry. :)

How high are they? How high is the seat?  :P

That doesn't matter. The basic idea is that you have a circle of radius 42 (diameter 84 inches = 7 feet) and you need to inscribe a number of 23 by 29 rectangles such that

(1) Two of each rectangle's vertices are on the circumference of the circle.
(2) The other two vertices are shared with adjacent rectangles.

To get an idea, look at #25 on the AMC 10A 2008. :)

So what is the answer? I need it ASAP

EDIT: please. :)

Basically, you find the angle between two chairs, say $ \alpha$, then take $ \left \lfloor \frac {2\pi}{\alpha} \right \rfloor$ (or if $ \alpha$ is in degrees, $ \left \lfloor \frac {360}{\alpha} \right \rfloor$)

My current gift of introspection tells me that $ \alpha$ is somewhere around $ 72^\circ$ (or $ \frac {2\pi}5$) so the answer should be around $ 4$-$ 5$, although you would need to use trigonometry to figure out the exact angle.

But wait... I see a 20-21-29 triangle coming to play... :)

I'm sorry I have no idea what at all you said. So  you think 5? It is a circle table

I suggest to understand the solution u probably need trig so pm me or just post here if you want but here's the answer :
$ 4$ maybe $ 5$ the things will be "ovelapping" (touching) by only something like $ \frac {\pi}{8}$ inches


I forgot to mention that the entire fraction lies within the radicand.

What are the steps to find the answer?

I want more than the answer.  I want you to show me how you got the answer.

[quote="sharkman"]Simplify the cube root of{5x}/4y[/quote]
[hide]$\sqrt[3]{\frac{5x}{4y}}$
$\frac{\sqrt[3]{5x}}{\sqrt[3]{4y}}$
Rationalize the denominator by multiplying both the numerator and the denominator by $\sqrt[3]{16y}$. After that it becomes $\frac{\sqrt[3]{80xy}}{4y}$ An 8 can come out of 80 because it is a cube root. So the final answer is
$\frac{2\sqrt[3]{10xy}}{4y}$.

Edit: Also divide everything by 2 to get $\frac{\sqrt[3]{10xy}}{2y}$. oops, I didn't see that before.[/hide]

[quote="#H34N1"]Hide tags, please :!:[/quote]
Sorry, I was in a rush. Edited. :blush:


geometry
3D geometry
inequalities
logarithms
function
inequalities unsolved

It was created from: $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 24abc \leq\ (a \plus{} b \plus{} c)^3$
 :)

The following inequality also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ 3(a\plus{}b\plus{}c) \geq 7\sqrt[3]{abc}\plus{}2\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\]

[quote="arqady"]The following inequality also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ 3(a \plus{} b \plus{} c) \geq 7\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]

Arqady, how do you prove it? can you please give at least a hint?  :|

I proved it by the $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2,$ $ abc\equal{}w^3$'s technics and
by the following estimation:
$ 3uv^2\geq2u^3\plus{}w^3\minus{}2\sqrt{(u^2\minus{}w^2)^3}.$ :wink:

[quote="arqady"]I proved it by the $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$'s technics and
by the following estimation:
$ 3uv^2\geq2u^3 \plus{} w^3 \minus{} 2\sqrt {(u^2 \minus{} w^2)^3}.$ :wink:[/quote]

Can you please post your proof Arqady?

Thank you very much. :)

OK, here is my proof.
$ 3(a\plus{}b\plus{}c) \geq 7\sqrt[3]{abc}\plus{}2\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\Leftrightarrow9u\minus{}7w\geq2\sqrt[3]{9u^3\minus{}9uv^2\plus{}w^3}\Leftrightarrow$
$ \Leftrightarrow219u^3\minus{}567u^2w\plus{}441aw^2\minus{}117w^3\plus{}24uv^2\geq0.$
$ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\geq0$ gives $ w^3\leq3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}.$
Hence, $ 3uv^2\geq2u^3\plus{}w^3\minus{}2\sqrt{(u^2\minus{}v^2)^3}\geq2u^3\plus{}w^3\minus{}2\sqrt{(u^2\minus{}w^2)^3}.$
Thus, it remains to prove that
$ 235u^3\minus{}567u^2w\plus{}441uw^2\minus{}109w^3\geq16\sqrt{(u^2\minus{}w^2)^3}.$
Let $ u\equal{}tw.$ Hence, $ t\geq1.$
Hence, $ 235u^3\minus{}567u^2w\plus{}441uw^2\minus{}109w^3\geq16\sqrt{(u^2\minus{}w^2)^3}\Leftrightarrow$
$ 235t^3\minus{}567t^2\plus{}441t\minus{}109\geq16\sqrt{(t^2\minus{}1)^3}\Leftrightarrow$
$ \Leftrightarrow235t^2\minus{}332t\plus{}109\geq16\sqrt{(t\minus{}1)(t\plus{}1)^3},$ which is true for all $ t\geq1.$

Nice and hard your proof, Arqady.  :)  Congratulations!!

Do you think for the weaker ineq

[quote="apollo"]Let $ a,b,c$ be positive real numbers. Prove that
\[ 9(a \plus{} b \plus{} c) \geq 24\sqrt [3]{abc} \plus{} 3\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]

it is possible to find a simpler proof. Nguoivn said it is derived from

$ \sum a^3 \plus{}24abc \leq (\sum a)^3$ but I don't see it. :(

[quote="manlio"]Nice and hard your proof, Arqady.  :)  Congratulations!!

Do you think for the weaker ineq

[quote="apollo"]Let $ a,b,c$ be positive real numbers. Prove that
\[ 9(a \plus{} b \plus{} c) \geq 24\sqrt [3]{abc} \plus{} 3\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]

it is possible to find a simpler proof. Nguoivn said it is derived from

$ \sum a^3 \plus{} 24abc \leq (\sum a)^3$ but I don't see it. :([/quote]

from Nguoivn's hint ;
\[ 9(a \plus{} b \plus{} c) \geq 9\sqrt [3]{a^3 \plus{} b^3 \plus{} c^3 \plus{} 24abc}
\]
by holder ;
\[ (24\sqrt [3]{abc} \plus{} 3\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}})^3 \leq (24 \plus{} 3)(8 \plus{} 1)(72abc \plus{} 3(a^3 \plus{} b^3 \plus{} c^3))
\]
thus,
\[ 24\sqrt [3]{abc} \plus{} 3\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \leq 9\sqrt [3]{a^3 \plus{} b^3 \plus{} c^3 \plus{} 24abc}
\]
therefore,
\[ 24\sqrt [3]{abc} \plus{} 3\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \leq 9\sqrt [3]{a^3 \plus{} b^3 \plus{} c^3 \plus{} 24abc} \leq 9(a \plus{} b \plus{} c)
\]

$ \Leftrightarrow235t^2 \minus{} 332t \plus{} 109\geq16\sqrt {(t \minus{} 1)(t \plus{} 1)^3},$

How do you prove it, Arqady?

I tried applying AM-GM at rhs,

$ 2t^2 \plus{}2t \equal{} (t^2\minus{}1) \plus{} (t^2\plus{}2t\plus{}1) \geq2\sqrt {(t \minus{} 1)(t \plus{} 1)^3}$

but it doesn't work. :( as this inequality is very strong.

By your beautiful idea we obtain:
 $ 64(t^2\minus{}1)\plus{}(t\plus{}1)^2\geq16\sqrt{(t\minus{}1)(t\plus{}1)^3}$ and $ 170t^2\minus{}334t\plus{}172\geq0.$  :lol:

[quote="arqady"]By your beautiful idea we obtain:
 $ 64(t^2 \minus{} 1) \plus{} (t \plus{} 1)^2\geq16\sqrt {(t \minus{} 1)(t \plus{} 1)^3}$ and $ 170t^2 \minus{} 334t \plus{} 172\geq0.$  :lol:[/quote]

Arqady , your solution is so elegant. But, that is hard for many people to apply for other inequalities including me  :|

This is my solution :  (Correct me if you see any error)

$ 3(a \plus{} b \plus{} c) \geq 7 \sqrt [3] {abc} \plus{} 2 \sqrt [3] {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$    ----->     (1)

Observe that ,

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \geq abc$   from  AM - GM

Taking the cube of both sides and multiplying both sides by 2 will give :

$ 2 \sqrt [3] {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \geq 2 \sqrt [3] {abc}$   ----->  (2)

Adding inequalities (1) and (2) will produce:

$ 3(a \plus{} b \plus{} c) \geq 9 \sqrt [3] {abc}$

and that is  same as:

$ (a \plus{} b \plus{} c) \geq 3 \sqrt [3] {abc}$

which is obviously true from AM-GM  :D

[quote="shaaam"]
This is my solution :  (Correct me if you see any error)

$ 3(a \plus{} b \plus{} c) \geq 7 \sqrt [3] {abc} \plus{} 2 \sqrt [3] {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$    ----->     (1)

Observe that ,

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \geq abc$   from  AM - GM

Taking the cube of both sides and multiplying both sides by 2 will give :

$ 2 \sqrt [3] {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \geq 2 \sqrt [3] {abc}$   ----->  (2)

Adding inequalities (1) and (2) will produce:

$ 3(a \plus{} b \plus{} c) \geq 9 \sqrt [3] {abc}$

[/quote]
Here is mistake. $ (1)$ still not proved. :wink:

Arqady ,you are right. So , you mean  if start from the asked (given) inequality and finish with true result, then we cannot say the given is right. it is kind of reasoning problem i am having here, right?

The stronger seems not trues althought very nice :wink:
$ a\plus{}b\plus{}c \geq\ 2\sqrt[3]{abc} \plus{} \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}$

\[ 3(a\plus{}b\plus{}c) \ge 7\sqrt[3]{abc}\plus{}2\sqrt[3]{ \frac{a^3\plus{}b^3\plus{}c^3}{3}}\]
Assume :$ \sum a\equal{}3$
$ ( 7\sqrt[3]{abc}\plus{}2\sqrt[3]{ \frac{a^3\plus{}b^3\plus{}c^3}{3}}  )^3 \le   3[ 21\frac{abc}{ab\plus{}bc\plus{}ca}\plus{} 2\frac{ \sum a^3}{3} )[7(ab\plus{}bc\plus{}ca)\plus{}6 )$
By am-gm we have :$ abc \le \frac{(ab\plus{}bc\plus{}ca)^2}{9}$
And the last ineq is equivalent to :
$ (135\minus{}14(ab\plus{}bc\plus{}ca) )(ab\plus{}bc\plus{}ca\minus{}3)^2 \ge 0$

[quote="arqady"]The following inequality also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ 3(a \plus{} b \plus{} c) \geq 7\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]
The stronger one:
If $ a,b,c \ge 0,a\plus{}b\plus{}c\equal{}3$, then
\[ 5\sqrt[3]{abc}\plus{}2\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}} \le 7\]
:)

[quote="can_hang2007"][quote="arqady"]The following inequality also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ 3(a \plus{} b \plus{} c) \geq 7\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]
The stronger one:
If $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 3$, then
\[ 5\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \le 7
\]
:)[/quote]
We need to prove that
$ 7t\geq2\sqrt[3]{7t^3\minus{}6t\plus{}2\sqrt{(t^2\minus{}1)^3}}\plus{}5\sqrt[3]{3t\minus{}2t^3\plus{}2\sqrt{(t^2\minus{}1)^3}}$ is true for $ t\geq1,$ which is true.
I checked it by my calculator. :D

[quote="arqady"][quote="can_hang2007"][quote="arqady"]The following inequality also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ 3(a \plus{} b \plus{} c) \geq 7\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]
The stronger one:
If $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 3$, then
\[ 5\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \le 7
\]
:)[/quote]
We need to prove that
$ 7t\geq2\sqrt [3]{7t^3 \minus{} 6t \plus{} 2\sqrt {(t^2 \minus{} 1)^3}} \plus{} 5\sqrt [3]{3t \minus{} 2t^3 \plus{} 2\sqrt {(t^2 \minus{} 1)^3}}$ is true for $ t\geq1,$ which is true.
I checked it by my calculator. :D[/quote]
But how about not calculator?

Given $ a, b, c \geq\ 0.$ Prove that:
$ \sqrt [3]{9(a^3 \plus{} b^3 \plus{} c^3)} \plus{} 8\sqrt [3]{abc} \leq\ \frac {11}{3}(a \plus{} b \plus{} c)$
It can be solved easily by Am-Gm  :)
(it's stronger than arqady's ineq but weaker than can_hang's ineq)

[quote="can_hang2007"]
The stronger one:
If $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 3$, then
\[ 5\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \le 7\]
:)[/quote]
Here is my proof for your inequality, can_hang2007.
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ 5\sqrt [3]{abc} \plus{} 2\sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \le 7\Leftrightarrow f(w^3)\leq0,$
where $ f(w^3) \equal{} 5w \plus{} 2\sqrt [3]{9u^3 \minus{} 9uv^2 \plus{} w^3} \minus{} 7u.$
But $ f$ is an increasing function. Thus, $ f$ gets a maximal value when $ w^3$ gets a maximal value,
which holds when two numbers from $ \{a,b,c\}$ are equal.
Id est, for the proof we need to check only one case: $ b \equal{} c \equal{} 1.$
Let $ a \equal{} x^3.$
Hence, $ f(w^3)\leq0\Leftrightarrow5x \plus{} 2\sqrt [3]{\frac {x^9 \plus{} 2}{3}}\leq\frac {7(x^3 \plus{} 2)}{3}\Leftrightarrow$
$ \Leftrightarrow g(x)\geq0,$ where
$ g(x) \equal{} \ln(7x^3 \minus{} 15x \plus{} 14) \minus{} \frac {1}{3}\ln(x^9 \plus{} 2) \minus{} \ln\left(2\sqrt [3]{9}\right).$
But $ g'(x) \equal{} \frac {6(x \minus{} 1)(x^2 \plus{} x \plus{} 1)(5x^6 \minus{} 7x^5 \plus{} 5x^3 \minus{} 7x^2 \plus{} 5)}{(7x^3 \minus{} 15x \plus{} 14)(x^9 \plus{} 2)}.$
Since, $ 5x^6 \minus{} 7x^5 \plus{} 5x^3 \minus{} 7x^2 \plus{} 5 > 0$ we obtain: $ g(x)\geq g(1) \equal{} 0.$ Done! :)

See my proof here (#183) : http://www.maths.vn/forums/showthread.php?t=9244&page=19
And my proof also effect with the stronger of can_hang  :)

We should consider the best constant for which the following inequality holds for non-negative reals:

\[ \frac{a \plus{}b\plus{}c\minus{}3\sqrt[3]{abc} }{ \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\minus{}\sqrt[3]{abc}}\ge K\]

By computer, I calculate that $ K\equal{}0.938615...$

\[ \frac{3u\minus{}3w}{\sqrt[3]{9u^3\minus{}9uv^2\plus{}w^3}\minus{}w}\ge K\]

$ 3L\equal{}K$

$ u\ge w(1\minus{}L)\plus{}L \sqrt[3]{9u^3\minus{}9uv^2\plus{}w^3}$

The LHS is an increasing function of $ w$, (we can easily show $ L<1$) and we know

$ w^3\le 3uv^2\minus{}2u^3\plus{}2(u^2\minus{}w^2)^{\frac{3}{2}}$

And then we can plug this expression in for $ w$ and we have a function that we can maximize $ t\equal{}u/v$, etc

[quote="Altheman"]
$ w^3\le 3uv^2 \minus{} 2u^3 \plus{} 2(u^2 \minus{} w^2)^{\frac {3}{2}}$

And then we can plug this expression in for $ w$ and we have a function that we can maximize $ t \equal{} u/v$, etc[/quote]
I don't think you can plug this one in for $ w$ since your $ RHS$ still has the variable $ w$.Unless I misunderstand what you meant, this way won't work if you can't separate $ w$ from the rest of the inequality.
On the other hand, using $ p,q, r$ method with fixing $ a \plus{} b \plus{} c \equal{} 1$ may help a lot since we have the strong following lemma of can_hang 2007:
If $ a \plus{} b \plus{} c \equal{} 1$ then we have:

$ \frac {(1 \plus{} q)^2(1 \minus{} 2q)}{27}\leq r \leq \frac {(1 \minus{} q)^2(1 \plus{} 2q)}{27}$

Sorry, there was a typo, I meant to write the following:

$ w^3\le 3uv^2\minus{}2u^3\plus{}2(u^2\minus{}v^2)^{\frac{3}{2}}$

[quote="arqady"]Dear, Nguyên văn bởi!
I can't see your proof because I don't know the key.
Please, post your proof here. Thank you![/quote]
My proof is here, arqady:  :)
http://www.mathlinks.ro/viewtopic.php?t=251509
And indeed, this proof also solved the stronger of can_hang's result:http://www.mathlinks.ro/viewtopic.php?t=297102

[quote="apollo"]Let $ a,b,c$ be positive real numbers. Prove that \[ 9(a\plus{}b\plus{}c) \geq 24\sqrt[3]{abc}\plus{}3\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\][/quote]
$\iff 3(a\plus{}b\plus{}c) \geq 8\sqrt[3]{abc}\plus{} \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}} .$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=598549


conics
hyperbola
analytic geometry
graphing lines
slope
Asymptote

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then what are the possible numbers of places where the lines can intersect the hyperbola?

[hide]i did this problem and i get 1,2,3 and 4. however the answer book gives 2,3,4. i count 1 since i thought there could be two lines which both intersect the hyperbola in the same place, hence one place. why didn't it include this in the answers?[/hide]

They can't intersect at one point. try it by drawing or using graphmatica. if you try to make the two lines intersect at one place, only one will intersect the hyperbola, the other one will be only tangent.

[quote="Aradesh"][quote="SplashD"]The line with the negative slope will eventually hit the right branch again.[/quote]

no, both lines are parallel to the asymptotes. i should have mentioned that. (atleast in theory, whether the graph shows that accurately or not is a different matter).[/quote]

It's unlikely that the book is wrong.

Have you tried graphing a larger section than the one you posted above?

Why don't you test it algebraically?

Take hyperbola $x^{2}-y^{2}=1$ with asymptotes $y=-x$ and $y=x$

I think the book means the 2 lines intersect the hyperbola at distinct points.

The book is about 99.9999999999% of the time right, not everyone's perfect. I can't see anything wrong with the picture that Aradesh and claiming that there's only one intersection.

[quote="Inspired By Nature"]off-topic: nice avatar breez :D
Aradesh, use graphmatica and try to get one intersection. if you do, please post the image. remember to zoom in and out to make sure there is only one intersection. :)[/quote]
I followed my own advice, and i think Aradesh has a point. :)

[quote="Aradesh"]take $x^{2}-y^{2}= 1$
take lines $y = x-1$ and $y = 1-x$
lines intercept at $(1,0)$ and both intercept hyperbola at $(1,0)$[/quote]

He proved it right here.

Aradesh would be correct if the angle between the two intersecting lines are equal to what the angle between the two asymptotes of the hyperbola would be. But if the angle of intersection were less than or greater than what the angle between the two asymptotes are, then it would intersect in more than one place. It also doesn't state that the lines need to intersect at two distinct points.  I can't say that Aradesh is incorrect.


This is a challenge problem from the McDougal Littel Geometry Textbook.
[url=http://img107.imageshack.us/my.php?image=triangleuf3.jpg][img]http://img107.imageshack.us/img107/539/triangleuf3.th.jpg[/img][/url]

In triangle ABC, AB=8, BC=6, AC=12. Each of the 3 segments drawn through point K has a length x and is parallel to a side of the triangle. Find the value of x.

I have been working on this problem for days and still can't figure it out! Any help is appreciated. [/img]

i tried the problem out,ive made a new drawing with some
points there.

ok so,we can use thales mainly to solve the whole problem,with a
little trick.
we start out by seeing that $x=A'K+KC'=AA''+C''C$(because $AA'KA''$ and $CC'KC''$ are paralelograms)
now we apply thales to find out the following relations to find out
$AA''$ and $C''C$
$\frac{AA''}{AC}=\frac{AC-A''C}{AC}=1-\frac{A''B''}{AB}=1-\frac{x}{AB}$
$\frac{C''C}{AC}=\frac{AC-AC''}{AC}=1-\frac{B'C''}{BC}=1-\frac{x}{BC}$
so..
$AA''=AB-x$
$C''C=BC-x$
now $x=AB+BC-2x$  =>  $x=14/3$
i hope i have not made mistakes

sorry , i was sure i was going to make some stupid mistake.
i seem to always have a solution on paper,and when i write it here
i find mistakes along the way and beeing eager to post i am not
very cautious to replace the correct part all over where i have
mistaken.
i hope this doesnt happen again(as it has several times until now)

I would to thank both of you for helping me out. I was looking through the forums at other problems you guys solved and was shocked.  :D Anyways, thank you once again and maybe I'll be back here for another problem.  :lol:

[quote="yetti"]You made mistakes.

$x = A'K+KC' = AA''+C''C = AC \left(1-\frac{x}{AB}+1-\frac{x}{BC}\right)$

$x = \frac{2}{\frac{1}{AB}+\frac{1}{BC}+\frac{1}{AC}}= \frac{2\ AB \cdot BC \cdot AC}{BC \cdot AC+AB \cdot AC+AB \cdot BC}= \frac{16}{3}$[/quote]

Hi again. I need some clarification on this portion.
why does x= AC(1- x/AB + 1-x/BC) ??


There are 2 main points of this discussion
1. If anyone (or everyone) has done the FIITJEE Full Tests 1,2,7,8,9 which are posted to you, then please post the marks.
2. Has anyone bought the RANKERS TEST PAPERS FILE of FIITJEE, which has 3 Part Tests and 9 Full Tests, and if yes, we can discuss a few questions and marks.

P.S. this is my first post, so if i have done anything wrong mod, i seek thy forgivance!!

achyuth sanjay

Wait for Chappli to login. He's almost finished all of them. Some people have already posted their full test marks, so search for them.

FT1 - 215
FT2 - 228
FT7 - 237

FT8 - 522 (didnt see one qsn :()
FT9 - 524.5 (I cut half mark for bad handwriting  :rotfl: )
jus kiddin.. i havent written the las 2 yet :D

and i havent subscribed for that RTPF.

it seems that they are reducing the level of the paper as we go on.
my scores are as follows-

FT 1- 250
FT 2- 271
FT 7- 327
FT 8- 330

btw did you correct the solutions of some papers, where there are some absolutely obvious mistakes?? i did, and hence these marks!

I finished all the full tests(but for ft8- fiitjee screwed my question paper packing), and all the RTPF and GMP tests. But I am too lazy to post my RTPF marks, so am doing my FT 7 and 9 marks only..:P
FT 1- 254
FT 2- 261
FT 7- 334
FT 8 - 339

[quote="Chapplikka"]I finished all the full tests(but for ft8- fiitjee screwed my question paper packing), and all the RTPF and GMP tests. But I am too lazy to post my RTPF marks, so am doing my FT 7 and 9 marks only..:P
FT 1- 254
FT 2- 261
FT 7- 334
FT [b][u]8[/u][/b] - 339[/quote]

shouldnt that be 9? :D

[quote="achu_182"]
btw did you correct the solutions of some papers, where there are some absolutely obvious mistakes?? i did, and hence these marks![/quote]
no i didn't. none of my friends did, and i compare only with them. so i didn't.
but ironically, the number of errors in the home tests are much less in comparison to the ones we right at the centre.  :mad:

[quote="sachin a S"]this is absolute cheatin :mad: i didn't get 7 8 & 9 ful test papers
don't say i didn't send my jee acknowledgement photocopy i did it
may i know wen u ppl got these papers
its very sad dat i had to write al d tough ones and dont get a chance to attempt d easy ones :rotfl:  :rotfl:[/quote]
you win a few, you lose a few. :rotfl:  dint you get some extra books from fiitjee.

we got it a long time ago :D Well they decided that supernerds dont need it  :P

YAY 2000TH POST :D

CONGRATS MADNESS. :thumbup:
BTW, SACHIN HOW WAS YOUR SANSKRIT EXAM.

thanks varun and sachin :)

Well sachin what do we have here? Another anit? :D :D

[quote="sachin a S"]@VARUN NICE AVATAAR BUDDY
SANSKRIT EXAM WAS CHILLD'S PLAY
COULD NOT'VE BEEN EASIER TO WAT IT WAS

@MADNESS CHEERS  :clap:
ABOUT FIITJEE PAPERS U KNOW SUPERNERDS'VE GOT IT SAY U OR ODERS IN D
LIST ABOVE
THEY HAVEN'T SENT TO PPL LIKE ME COZ DEY KNOW I'M NOT GOOD ENOUGH TO SOLVE JEE PAPERS MAYBE STATE BOARDS WOULD BE COMPETITIVE[/quote]
swalpa over aghoiito. :wink:

[quote="sachin a S"]@varun hagantiya parvaagilla bidu adjust madkotare
nin hindi exam hegaitu??

@madness no way i'm not even near anit's foot
his scores were unmatchable  :D  neways u'l know d treuth wit jee results ?
where anit stands where oders like me do :wink:[/quote]
pass agtini bido. nan pm sikilva?

the only 2 words i know in kannada - kannada gothilla  :rotfl:   :rotfl:

no fear when shre is here :D :D Good ol' shre will translate it for us :D

kannada gothilla = I dont know kannada......

thnx for the info da :D

Muahahahaha varun n sachin, u cant have private convos again in public forums when we have cyber here :D :D

[quote="VARUNARASU"][quote="Cyber-Shreyas"][hide]Translations::

1) swalpa over aghoiito.
Doesn't this seem a bit too much for you?

2) varun hagantiya parvaagilla bidu adjust madkotare
nin hindi exam hegaitu??

Is that so? Leave it. When is your hindi exam?

3) pass agtini bido. nan pm sikilva?
 I'll pass. Leave it. Didn't you get my PM.

4) ega sikto
nanu tumba sites parallel agi open madidino
hagagi sakat slow agide broadband ashtaku aops modale slow
hagagi....

Now I got it. I've opened too many sites parallel. Already AOPS is slow for me, and now & then it runs slow.[/hide]

I'm a total embarrassment. My kannada is pathetic. I'm sure Varun or Eucl or Sachin would be ashamed of me, if I ever came that way.

If you guys want me to remove this, I will... I respect privacy but I can have the best of both worlds.[/quote]
hey, you're actually gr8 at kannada. where did u learn?its a 100 times better than my tamil. :P  that post deserved rates.[/quote]

He's a kannadian by birth :)

[quote="madness"]
He's a [b]kannadian[/b] by birth[/quote]
he's indian only, and not "canadian"  :rotfl:  :rotfl:
its supposed to be kannadiGA  :D

anyone here knows malayalam??

i am a mallu living in delhi!

aah well my mom and grandma are highly proficient in malayalam, so ennaku korachu korachu paranyu  :D (is that rute achu??)
and of course the 2 standard words : prandhu patti!  :lol:

[quote="VARUNARASU"][quote="Cyber-Shreyas"][hide]Translations::

1) swalpa over aghoiito.
Doesn't this seem a bit too much for you?

2) varun hagantiya parvaagilla bidu adjust madkotare
nin hindi exam hegaitu??

Is that so? Leave it. When is your hindi exam?

3) pass agtini bido. nan pm sikilva?
 I'll pass. Leave it. Didn't you get my PM.

4) ega sikto
nanu tumba sites parallel agi open madidino
hagagi sakat slow agide broadband ashtaku aops modale slow
hagagi....

Now I got it. I've opened too many sites parallel. Already AOPS is slow for me, and now & then it runs slow.[/hide]

I'm a total embarrassment. My kannada is pathetic. I'm sure Varun or Eucl or Sachin would be ashamed of me, if I ever came that way.

If you guys want me to remove this, I will... I respect privacy but I can have the best of both worlds.[/quote][/quote]

that was very good especially 4 someone who hasn't lived in karnataka for yrs
neways next time i'l try different version of kannada & let me know your ability to translate

p.s though i want to rate your post my post count doesn't allow me for dat rite now . maybe some days later

@ hash_include: i can understand what you are trying to say, but i think it should be
ennikku korachu malayalam parayan arayam
(i know how to speak a bit in malayalam)

are u guys giving FT 10 which is tomorrow??

they must b sitting 4 it rite now!!!!!!!!! :D

OMG :o I just finished FT7 and i got 265 :( well i thought it was atleast decent when i saw the marks here. Well can you post your marks guys? Chappli also?

I dunno how u guys got above 300 :o that would mean that u had atleast 2 hundred above totals. I had only 1

EDIT: Changed things :D I thought assertions in paper 2 were 3 marks :rotfl:

[quote="karthik0112358"]
By the way who art thou #include????[/quote]

You don't know him. He's the real IIT Bigshot in Chennai... he's name is shridhar.

[quote="karthik0112358"]:o When did u guys (Varun,sachin) join AOPS!!!????
I thought me, bonny, and pradeep were the only guys alive on AOPS from BAngalore!!!
By the way who art thou #include????[/quote]

That means you have been inactive for over 3 months odd :D :D

very funny shreyas, but i think you should read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1086027#1086027]this[/url] thread, posted by someone i daresay you know quite well, and LEARN from it!  :wink:  :P

but karthik, to answer your question, i'm shridhar from chennai, which was the only piece of accurate info dear chronoz managed to give :D

in my spare time, i work for R&AW, and sometimes, when i can manage to find time, i teach quantum chromodynamics (hail the Z) at CERN :D

for further knowledge on my interesting life, pls pm me :)

[quote="karthik0112358"]:o When did u guys (Varun,sachin) join AOPS!!!????
I thought me, bonny, and pradeep were the only guys alive on AOPS from BAngalore!!!
By the way who art thou #include????[/quote]
I JOINED AOPS A LONG TIME BACK. (11 DEC 2007 TO BE PRECISE).  N YOU GUYS DONT SEEM TO BE EXACTLY ALIVE ON AOPS, YOU KNOW.  EVEN RAGHAVENDRA'S ON AOPS NOW.

Pathetic again :( Calculated FT8 and again ended up with only 265 (132 + 133) :(
This is too discouraging man!

naah don't be discouraged!!

it might be flukes on our part!!

try to locate ur mistakes and make sure that u don't repeat them. JEE is competitive enough, even without our careless mistakes!!

but ofc i feel careless mistakes increase during actual exam & ability to guess is also hindered somehow :(  :(

[quote="madness"]
He's a kannadian by birth :)[/quote]

correction!......kannadiga.....(so am i)!

Gave only FT2 - 316!


geometry
vector
parallelogram
geometric transformation

Given $ P \equal{} (2,7), B \equal{} (6,11),$ and $ M \equal{} (5,2)$:

(a) Find a point $ D$ such that $ PB \equal{} DM.$
(b) What can be said of quadrilateral $ PBMD?$

Unless you mean the vectors PB and DM, Dimitris X is correct, all of the segments of length $ \sqrt{32}$ would suffice. However, if you meant the vectors, then PBMD would be a paralellogram, formed by the translation of either PB or MD.

[quote="Dimitris X"]$ D$ is a not a fixed point.
$ D$ is lying on a circle with center $ O(5,2)$ and radii $ \sqrt {32}$[/quote]

Thank you.  I had reached the same conclusion.  The problem appears to be flawed.


post here and people can help you out.  no one should give you the solution though.  probably just hints and such

Sounds to me like what you need is not answers or solutions but methods: if you don't know how to begin, review the types of geometry you're having trouble with. Also we on the forum are always glad to help.

I used to have trouble with the kind of geometry you're talking about.

I went back and worked through sections of a geometry book that I did not have a good grasp on. That allowed me to focus on more advanced geometry concepts/problems.

[quote="now a ranger"]I don't know where to find answers to questions I don't know how to do. I looked up some, but many are hard to find. IS there a good website.

ESpecially solving things with chords and secants and arcs, as well as lots of inscribed quadrilaterals.[/quote]

:welcome:

This is a pretty good website :)

The AoPS books are a good starting point for learning different problem solving methods.

ok

SO AoPS books and school textbooks are good ways to learn? AS well as asking here for hints, which i did earlier

Where do you check your answers i the AoPS books. I have one of them, and there aren't answers in teh back.

Try Bob Miller's Geometry for the Clueless
umm well it is basically al review and that's why i'm saying this! it has so many definitions though so it is worth buying if you don't have it at your library...

Thanks Funcia

I am taking Geometry this year, so I still have a lot to learn.

Some problems are still really hard, so I have trouble visualizing

[quote="now a ranger"]]
Where do you check your answers i the AoPS books. I have one of them, and there aren't answers in teh back.[/quote]

You have to buy the solution book seperately.

Haha! I am aspiring to take it...my algebra has 1 &2 combned so yeah I am reading the book right now and must admit I am pretty far and like it! (Note: you are spposed to be taking geometry while reading it i'm just not doing it the way people should do it) Byreview I meant things like that you learn in addition to algebra like angle supplements compliments bisectors circles, (the list goes n) but the book fills in the gaps of that sort of knowledge..  :wink:

NOTE: Funcia is not a spelling type of person and is reading a book *guess what* right now so things might be messed up...

Wow I need to get that book

Did you get your book at the library funcia?

i saw bob miller's book on amazon it has no reviews yet  :roll:

Yeah i got the book at the library....must review it!

EDIT: Must purchase whole series too.

oh umm.. IT's not at our library. I found a book called GEometry the Easy way by Lawrence LEff and it seems helpful. especially after reading reviews

Hmm...yeah it's not at my library...but i have two...do you? that might help...yeah I'll look for that book since i also am trying to get this stuff! (maybe it has other ways to prove and different axioms and post.(i know prob. not the way i should make it smaller) listed(thought this one says how you might do things differently))

haha my friend says that Geometry the Easy WAy helped him when he was taking Geometry  :)

Yeah I reviewed the geometry book as a kid! I really don't want people contacting me. :wink:
I will now try to get that book!

It's at like every library there is in the city I live in.

I need ramdoooooozles of improvement in geometry, as I still am horrible at inscribed things and chords. I still can't visualize what others can.


trigonometry
invariant
geometry
parallelogram
ratio
geometric transformation
reflection

Let ABCD be a convex quadrilateral such that:
i)   AD is not parallel to BC
ii)  There exists a circle W, which is tangent to the lines BC and AD at C and D respectively.
iii) The circle W intersects the segment AB in two different points K and L
      (K closer to A than L).
The lines AC and BD meet in the point P. Let M be the midpoint of CD.
Prove that if CL=DL, then the points K, P, M are collinear.

well.. it seems that nobody wants to answer this one, so let's post a not very elegant, but quite straightforward, solution:
 let $T=BC\cap AD$ and $K'=MP\cap AB$. I'll show that $K=K'$. First, by power of a point, and since both $AD$ and $BC$ are tangent to the circle, we have \[\frac{AK}{BK}=\frac{AD^2}{BC^2}\frac{BL}{AL}        (1)\]
 By combining some Law of Sines from the triangles $AKP,BKP,CPM,DPM$ and since $CM=DM$ we obtain \[\frac{AK'}{BK'}=\frac{AP}{CP}\frac{PD}{CP}    (2)\]
 Now the only thing that remains to do is to replace the point $P$ with some relations of the other points, and for that we use Menelaos in triangles $ACT$, points $B,P,D$, and $BDT$, points $A,P,C$, and we get $\frac{AP}{CP}\frac{PD}{CP} = \frac{AD^2}{BC^2}\frac{BT}{AT}$, so combining with (2) we have that \[\frac{AK'}{BK'}=\frac{AD^2}{BC^2}\frac{BT}{AT}         (3)\] But $TM$ is the angle bisector of $\angle BTA$, so by the angle bisector theorem we have that $\frac{BT}{AT}=\frac{BL}{AL}$, and now with (3) and (1) we get exactly what we wanted: $\frac{AK}{BK}=\frac{AK'}{BK'}$, so $K=K'$ and we are done...

btw, can you post your solution?? because i think there has to be a nice solution..

Sorry if this is a stupid question but why are the points $T,M $and$ L$ collinear?
Maybe I missunderstood something?! :?

it's ok, i didn't explained it..remember that $T=BC\cap AD$.  $TC$ and $TD$ are both tangents to the circle, and are therefor equal... So $\triangle CDT$ is isosceles. And since $CL=DL$, $\triangle CDL$ is isosceles too. Now $M$ is the midpoint of the unequal side of the isosceles tirangle $\triangle CDT$ hence $TM$ is perpendicular to $CD$. The same with $LM$, and the results follows..

I very sorry,that thing it's obvious, I though that CD=DL and I couldn't solve the problem :blush:

this was really nice problem and here is my solution:
define the points $S=AB\cap OP$ and $R=CD\cap OP$.
it is easy to check $Q,D,R,C$ are harmonic and $O,L,M,N$ too.
$KL$ is angle bisecter of $DKC$ so $\frac {QD}{QC}=\frac{DR}{RC}=\frac{KD}{KC}$ therefore $KR$ is angle bisector of $DKC$
$\Longrightarrow$ $K,R,N$ are collinear.
$(PORS)=-1$ and $(OMLN)=-1$ therfore $KR,KP,KS,KO$ are harmonic and $KR,KM,KL,KO$ are harmonic. $\Longrightarrow$ $K,P,M$ are collinear.
and problem solved.

I have a different solution.

We start like RaMlaF: Denote by T the point of intersection of the lines BC and AD; then, since the two tangents from a point to a circle are equal in length, we have TC = TD, so the point T lies on the perpendicular bisector of the segment CD. Since CL = DL, the point L also lies on the perpendicular bisector of the segment CD. Finally, the point M, being the midpoint of the segment CD, must obviously lie on the perpendicular bisector of the segment CD. Thus, the points T, L and M are collinear (since they all lie on the perpendicular bisector of the segment CD); hence, the point M is the point of intersection of the lines TL and CD.

Now, apply the Pascal theorem to the degenerate hexagon CCKLDD, which is inscribed into the circle W. We obtain that the points $CC\cap LD$, $CK\cap DD$ and $KL\cap DC$ are collinear. Hereby, the lines CC and DD are actually the tangents to the circle W at the points C and D, i. e. the lines TB and TA; also, clearly, the line KL is simply the line AB, and the line DC is the line CD. So we see that the points $TB\cap LD$, $CK\cap TA$ and $AB\cap CD$ are collinear.

Now the problem would trivially follow if we prove the following theorem:

[color=blue][b]Theorem 1.[/b] Let T, A, B be three points in the plane. Let C and D be two points on the lines TB and TA, respectively. Let K and L be two points on the line AB such that the points $TB\cap LD$, $CK\cap TA$ and $AB\cap CD$ are collinear. Finally, let M be the point of intersection of the lines TL and CD, and let P be the point of intersection of the lines AC and BD.

Then, the points K, P, M are collinear.[/color]

[i]Proof of Theorem 1.[/i] Theorem 1 is a projectively invariant theorem; thus, we can map the line CD to infinity by a projective transformation. The resulting configuration has the following nice properties:

- The lines TA and BP are parallel (since they meet at the infinite point D), and the lines TB and AP are parallel (since they meet at the infinite point C). Hence, the quadrilateral TAPB is a parallelogram. Since the diagonals of a parallelogram bisect each other, it follows that the segments AB and TP bisect each other; in other words, the midpoint of the segment AB is the midpoint of the segment TP.

- Denote $TB\cap LD=U$. Then, the line LU is parallel to the line TA (since both lines LU and TA pass through the infinite point D), and thus Thales yields $\frac{BL}{LA}=\frac{BU}{UT}$. Similarly, denote $CK\cap TA=V$; then, the line KV is parallel to the line TB (since both lines KV and TB pass through the infinite point C), and thus, by Thales, $\frac{AK}{KB}=\frac{AV}{VT}$.

- But by a condition of Theorem 1, the points $U=TB\cap LD$, $V=CK\cap TA$ and $AB\cap CD$ are collinear. In other words, the line UV passes through the point $AB\cap CD$; equivalently, the lines UV, AB and CD concur at one point. But since the line CD is the line at infinity, this point is an infinite point, and thus the two lines UV and AB, both passing through this point, must be parallel. Hence, by Thales, $\frac{BU}{UT}=\frac{AV}{VT}$.

Hence, $\frac{BL}{LA}=\frac{BU}{UT}=\frac{AV}{VT}=\frac{AK}{KB}$. Thus, the points L and K divide the segment AB in the same ratio but "from different sides"; in other words, the points L and K are symmetric to each other with respect to the midpoint of the segment AB. Hence, the reflection with respect to the midpoint of the segment AB maps the point L to the point K. On the other hand, this reflection maps the point T to the point P (since the midpoint of the segment AB is the midpoint of the segment TP). Since reflections map lines to parallel lines, this yields PK || TL. In other words, the lines PK and TL have a common infinite point. But the infinite point on the line TL is the point M (in fact, the point M is the point of intersection of the line TL with the line CD, i. e. with the line at infinity); thus, the line PK must also pass through the point M. In other words, the points K, P, M are collinear. This proves Theorem 1.

  Darij


geometry
circumcircle
incenter
angle bisector
geometry unsolved

Let $ ABC$ be a triangle and let $ D$ be a point on $ AB$ such that $ 4AD\equal{}AB$. The half-line $ l$ is drawn on the same side of $ AB$ as $ C$, starting from $ D$ and making an angle $ \theta$ with $ DA$ where $ \theta \equal{} \angle ACB$. If the circumcircle of $ ABC$ meets the half-line at $ P$, show that $ PB\equal{}2PD$.

Interesting!!
See that $AP=AM$, where $M$ is the midpoint of $AB$, hence, if $PD$ meets 2nd time the circle $\odot (ABC)$ at $Q$, then $M$ is the incenter of $\triangle BPQ$, or $PM$ the angle bisector of $\angle BPD$, consequently $\frac{PD}{PB}=\frac{DM}{MB}=\frac{1}{2}$, done.

Best regards,
sunken rock


Could someone please help me.  I am trying to re-work a formula and I can't remember what steps to take.

The original formula is:

e = (1 + n/p)^p - 1

I am trying to solve for n

Can someone please help me?

Jim.

[hide]
$e=\left(1+\frac{n}{p}\right)^{p}-1\implies \left(1+\frac{n}{p}\right)^{p}=e+1$

$\implies \ln\left(e+1\right)=p\ln\left(1+\frac{n}{p}\right)$

$\implies \left(e+1\right)^{\frac{1}{p}}=1+\frac{n}{p}$

$\implies n=p\left(e+1\right)^{\frac{1}{p}}-p$
[/hide]

Thank you so much, it is amazing how much you forget when you get out of school.

Thanks again.   :D


There are four pipes leading into a well. Individually, the four pipes can fill the well in $\frac{1}{2}\: , \frac{1}{3}\: ,\frac{1}{4}\: ,$ and $\frac{1}{5}$ of a day. How long would it take for the pipes to fill the well if they were all working simutaneously?

[hide="Solution"]
I think I have seen a problem like this before in my algebra class - so i think this is the procedure: First we take the inverses of each and we get $2,3,4,$ and $5$. I added them each and got $14$. I took the inverse of that and got $\frac{1}{14}$.
[/hide]

To explain why Anxerzia's solution works,
[hide]Pipe A can fill 2 wells in one day.
Pipe B can fill 3 in one day, and so on.
So, all 4 pipes can fill 14 wells in one day, or one well in 1/14 day.[/hide]


Hi guys,

Sorry for disturbing. but i need some help. How can i find out the speed on a random point on an elipse ?

(it's about an asteroid for example, orbiting on some elipse orbit)

Anybody?

Thanks

PS: Sorry for posting it in advaced section, i haven't seen that when i was posting.

Angular momentum and mechanical energy are conserved, so you can write

$U_g = \frac{1}{2}mv^2 - \frac{GMm}{r} = constant$
$\vec{L} = mvr = constant$

You can just evaluate these at $r = a$ (i.e. aphelion) and solve  :P

ok
here is the solution:
v$^2$=Gm{(2/r)-(1/a)}
if you want the proof tell me ! :lol:

I think i got it. Thanks, saggita i believe you will prove it using the fact the total energy of an orbit, E=-GMm/2a  equals with the same constant as Ug from blendecho. Anyhow , some specific particularisations

Thank you!


trigonometry
geometry
incenter
inequalities
inequalities unsolved

Show that $ \frac{1}{4}$ $ <$ $ \frac{IA}{AD}$ $ .$ $ \frac{IB}{BE}$ $ .$ $ \frac{IC}{CF}$ $ .$ $ \le$ $ \frac{8}{27}$.

please avoid trigonometry while giving the solution.

Maybe telling us what D,E,F,I are? I assume I is the incenter, but are D,E,F touching points of the circle or intersections of bisectors and sides?

AD,BE,CF are the  bisectors and I is the incentre.sorry for inadequate information.

Let $ BC \equal{} a$ , $ CA\equal{}b$ , $ AB\equal{}c$
By the triangle bisector theorem at triangle ABC,
$ BD: DC \equal{} BA: AC \equal{} c: b$
Since $ a \equal{} BC \equal{} BD\plus{}DC$, we have $ BD\equal{} \frac{ac}{c\plus{}b}$
By the triangle bisector theorem at triangle ABD (BI is the bisector)
$ AI: ID \equal{} AB: BD \equal{} c: \frac{ac}{c\plus{}b} \equal{} (c\plus{}b): a$

Therefore $ \frac{IA}{AD} \equal{} \frac{IA}{IA\plus{}ID} \equal{} \frac{c\plus{}b}{a\plus{}b\plus{}c}$

Analogously, we have

$ \frac{IB}{BE} \equal{} \frac{a\plus{}c}{a\plus{}b\plus{}c}$
$ \frac{IC}{CF} \equal{} \frac{a\plus{}b}{a\plus{}b\plus{}c}$

Therefore, the given inequality is equivalent to
$ \frac{1}{4} < \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a\plus{}b\plus{}c)^3} \leq \frac{8}{27}$

The left side of the inequality, after expanding and grouping terms, is equivalent to $ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)>0$, which is trivial for a,b,c 3 sides of a triangle.
The right side of the inequality is a trivial consequence of the AM-GM inequality on a+b, b+c, c+a.


Shouldn't this be posted in Games and Fun Factory?

[quote="gr8sk8r"]I'm sure this problem has been posted before, but i'll post again just for kicks.

three men go into a hotel and ask for a room. the manager tells them that the room costs 30 dollars. the men decide to split the cost evenly between themselves and pay 10 dollars each. however, later that day, the manager realized he had mispriced the room and that it only costed 25 dollars, so he asks the bellhop to take the extra 5 dollars up to the men. but the bellhop decided to keep 2 dollars out of the five and only returned 3 dollars to the men. therefore, each man only payed 9 dollars each which totals 27 dollars and the bellhop has 2 dollars which totals 29 dollars. So...where did the last dollar of the 30 dollars go?[/quote]

the last dollar would have been split between the 3 men wouldn't it?

[quote="gr8sk8r"]I'm sure this problem has been posted before, but i'll post again just for kicks.

three men go into a hotel and ask for a room. the manager tells them that the room costs 30 dollars. the men decide to split the cost evenly between themselves and pay 10 dollars each. however, later that day, the manager realized he had mispriced the room and that it only costed 25 dollars, so he asks the bellhop to take the extra 5 dollars up to the men. but the bellhop decided to keep 2 dollars out of the five and only returned 3 dollars to the men. therefore, each man only payed 9 dollars each which totals 27 dollars and the bellhop has 2 dollars which totals 29 dollars. So...where did the last dollar of the 30 dollars go?[/quote]

[hide="hint"]
How much did the men end up actually paying?

[hide="hint2"]
Where did the money that the men paid go?

[hide="hint3"]
When the men paid their 9 dollars each, what's included in those 27 dollars?

Well then figure it out for yourself.

The men paid 27 dollars: 25 went to the room and 2 went to the bellboy. The other 3 dollars of the 30 were given back to the man.

[/hide]
[/hide]
[/hide]

[hide="Hint"]Where did the $\$$27 figure come from?
[hide="Hint 2"]$25+3\neq27$[/hint]
[/hide][/hide]

[hide="i just want to know if this is correct"]
the bellhop gave them 3 dollars, so they each did pay 9 dollars each.
they paid 27 dollars total.
the 2 dollars that the bellhop needs to be subtracted from the 27 dollars to get the actual price of the room (25).
so there is no problem in the math [/hide]

They shouldn't be doing $3\cdot9+2=30$ they need to do $3\cdot9-2=25$, which was the real price of the room.


geometry
circumcircle
incenter
geometry proposed

[color=darkblue]Let $ ABC$ be a triangle with the the circumcircle $ C(O,R)$ and the $ A$-incenter $ I_a$ .

Consider the points $ \left\|\begin{array}{cccc} M\in AB & , & B\in (AM) \\
\ N\in AC & , & C\in (AN)\end{array}\right\|$ so that $ BM\equal{}CN\equal{}BC$ .

Prove that $ MO^2 \minus{} ac \equal{} NO^2 \minus{} ab \equal{} R^2 \plus{} a^2$ and $ MN\perp OI_a$ .

[u][b]Remark.[/b][/u] See and[/color] [url=http://www.mathlinks.ro/viewtopic.php?t=232923][color=red][b]here[/b][/color][/url] [color=darkblue], where is a similar problem for the incenter of $ \triangle ABC$ .[/color]

[quote="Virgil Nicula"][color=darkblue]Let $ ABC$ be a triangle with the the circumcircle $ C(O,R)$ and the $ A$-incenter $ I_a$ .

Consider the points $ \left\|\begin{array}{cccc} M\in AB & , & B\in (AM) \\
\ N\in AC & , & C\in (AN)\end{array}\right\|$ .

Prove that $ MO^2 \minus{} ac \equal{} NO^2 \minus{} ab \equal{} R^2 \plus{} a^2$ and $ MN\perp OI_a$ .

[u][b]Remark.[/b][/u] See and[/color] [url=http://www.mathlinks.ro/viewtopic.php?t=232923][color=red][b]here[/b][/color][/url] [color=darkblue], where is a similar problem for the incenter of $ \triangle ABC$ .[/color][/quote]
I think you forgot posting BM=CN=BC and $ I_a$ is the A-excenter, not incenter.


conics
ellipse
analytic geometry
inequalities

If $f(x)=1-x-x^3$, what are all real values of $x$ which satisfy
$1-f(x)-(f(x))^3>f(1-5x)$?

Let $a$, $b$, and $c$ be integers satisfying $1\leq a \leq 9$, $0\leq b \leq 5$, $0\leq c \leq 9$, and $a+b+c \geq 20$.  How many possible ordered triplets $(a,b,c)$ are there?

This is from AOPS V. 2:

An ellipse is drawn with major and minor axes of lengths $10$ and $8$ respectively.  Using one focus as the center, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse.  Compute the radius of the circle.

The solution basically said, the two foci and a point of intersection must be collinear since a ball thrown from the center of the circle will go directly to the other focus and back to the center of the circle (since the foci of a circle is its center), therefore the answer is $2$.

I did the question algebraically, but it was a bit "weird" I thought, so I was wondering if there were any holes in the solution.
[hide]Let the ellipse be centered at $(0,0)$, and the vertices at $(5,0)$, $(-5,0)$, $(0,4)$, $(0,-4)$.  The the foci of the ellipse are at $(3,0)$ and $(-3,0)$.  Assume the circle is at $(3,0)$.

Then we have the simultaneous equations $\frac{x^2}{25}+\frac{y^2}{16}=1$ and $(x-3)^2+y^2=r^2$.  We want to find $r^2$.

From the second eq. we have $y^2=-(x-3)^2+r^2$, plugging this into the first equation gives $\frac{x^2}{25}+\frac{r^2-(x-3)^2}{16}=1$.  Multiplying by $400$, expanding and combining like terms we have $16x^2+25r^2-25x^2+150x-225=400$.  Rearranging the terms yields $9x^2-150x=25r^2-625$.

We know $3\leq x \leq 5$ (since $x$ represents the $x$ coordinate of the intersection point, and it must be between the center of the circle and the right vertex) and $0\leq r \leq 2$ (if the radius were greater than two then the circle would be outside the ellipse).  Plugging in $3$ and $5$ into the LHS of the equation we find $-525\leq LHS \leq -379$, and plugging in $0$ and $2$ into the RHS we find $-600 \leq RHS \leq -525$.  We see that $-525$ is the only possible common value, and this value is given when $x=5$ and $r=2$, therefore our answer is $2$.
[/hide]

[quote="eryaman"]If $f(x)=1-x-x^3$, what are all real values of $x$ which satisfy
$1-f(x)-(f(x))^3>f(1-5x)$?[/quote]

$f(x)=1-x-x^3$

$f'(x) =  - 1 - 3x^2  < 0$ for all real x.

thererore, f(x) is decreasing function.

for

$1-f(x)-(f(x))^3>f(1-5x)$ i.e., $f(f(x))>f(1-5x)$ is possible when $f(x)<1-5x$

$\Rightarrow 1 - x - x^3  < 1 - 5x$

solving this inequality, we get

$x \in ( - 2,\,0) \cup (2,\,\infty )$

[quote="eryaman"] This is from AOPS V. 2:

An ellipse is drawn with major and minor axes of lengths $10$ and $8$ respectively.  Using one focus as the center, a circle is drawn that is tangent to the ellipse, [b]with no part of the circle being outside the ellipse[/b].  Compute the radius of the circle.

[hide]Let the ellipse be centered at $(0,0)$, and the vertices at $(5,0)$, $(-5,0)$, $(0,4)$, $(0,-4)$.  The the foci of the ellipse are at $(3,0)$ and $(-3,0)$.  Assume the circle is at $(3,0)$.

Then we have the simultaneous equations $\frac{x^2}{25}+\frac{y^2}{16}=1$ and $(x-3)^2+y^2=r^2$.  We want to find $r^2$.

From the second eq. we have $y^2=-(x-3)^2+r^2$, plugging this into the first equation gives $\frac{x^2}{25}+\frac{r^2-(x-3)^2}{16}=1$.  Multiplying by $400$, expanding and combining like terms we have $16x^2+25r^2-25x^2+150x-225=400$.  Rearranging the terms yields $9x^2-150x=25r^2-625$.

We know $3\leq x \leq 5$ (since $x$ represents the $x$ coordinate of the intersection point, and it must be between the center of the circle and the right vertex) and $0\leq r \leq 2$ (if the radius were greater than two then the circle would be outside the ellipse).  Plugging in $3$ and $5$ into the LHS of the equation we find $-525\leq LHS \leq -379$, and plugging in $0$ and $2$ into the RHS we find $-600 \leq RHS \leq -525$.  We see that $-525$ is the only possible common value, and this value is given when $x=5$ and $r=2$, therefore our answer is $2$.
[/hide][/quote]

I am sure this value of r violates the condition that no part of the circle being outside the ellipse.
as the centre is at (3, 0) and major axis is of length 5, this value of r=2 makes the circle to touch the ellipse externally at (5, 0) ....the vertex of the ellipse.  So, part of the circle has to be outside the ellipse.

I tried for the solution again, but it seems it is not going to satisfy this condition.

Hence I feel, this problem has no solution.


MATHCOUNTS geometry trapezoid parameterization

I think that the question is for any trapezoid with the given description, so a diagram is not necessary--any diagram that the student constructs with the given parameters should be correct.

[hide]

Similar Triangles. 11 = 3x, x = 11/3. [/hide]

Or you could continue sides AD and BC until they intersect at a point, lets say F.   We can see that in triangle ABF, C and D are midpoints of sides BF and AF respectively because DC is half of AB.  Then can't we see that AC and BD are medians so then E is the centroid of ABF.  AE/EC = 2/1 so 11-x/x = 2.  Solving from there, x = 11/3.

Syntax Error wrote:ABCD is a trapezoid with the measure of base AB twice the measure of base DC.  point E is the point of intersection of the diagonals.  the measure of diagonal AC is 11.  find the length of segment EC.  express your answer as a common fraction

a diagram is given, im pretty sure you can figure it out, but ill explain it anywho

so D is on the top left.  C is on the top right. A is on the bottom left, and B is on the bottom right. E is the intersection of diagonals, that was already given, but meh

[hide]

triangle DCE is similar to triangle ABE

CE:AE = DC: AB = 1:2

x: (11-x) = 1: 2

2x = 11 - x

x = 11/3[/hide]

yaya boi

follow up to the uns who didnt post solutions: what similarity thingie did they use, SS, SAS, or SSS

haha, if the fish swims, let it swim

wait, no, AAAAA. lol

on such conditions the diagonals trisect each other. it can be proved.


logarithms
combinatorics proposed
combinatorics

Find all set $A \neq \emptyset$ is a subset of the set of natural number such that :
1, if $x \in A$ then $4x \in A$
2,if $x \in A$ then $[\sqrt{x}] \in A$

Suppose $a \in A$.  Choose $k$ such that $2^{2^{k}}> a$.  Then by $k$ applications of rule 2, $1 \in A$ and so by repeated application of rule 1, $4^{n}\in A$ and so using rule 2 again $2^{n}\in A$ for all nonnegative integers $n$.

Now, choose any $n \in \bf N$.  Pick $k$ large enough that $2^{k}\log_{2}(n+1) \geq 1+2^{k}\log_{2}n$.  Then there exists some $l \in \bf N$ satisfying $2^{k}\log_{2}(n+1) > l \geq 2^{k}\log_{2}n$ and so $(n+1)^{2^{k}}> 2^{l}\geq n^{2^{k}}$.  $2^{l}\in A$, and applying rule 2 to it $k$ times gives $n$.  But $n$ was arbitrary, so every positive integer is in $A$.  The only such subset is $\bf N$ itself.


function
real analysis
real analysis unsolved

Suppose $f(\boldsymbol{x})$ is a real-value convex function, where $\boldsymbol{x}$ can be a vector. Suppose $f(\boldsymbol{x})$ has a unique subgradient at $\boldsymbol{x}_0$, denoted by $\boldsymbol{\mu}_0$.
Then $\boldsymbol{\mu}_0$ determines a supporting hyperplane $A=\{(\boldsymbol{x},y):y=f(\boldsymbol{x}_0)+ \boldsymbol{\mu}_0^T(\boldsymbol{x}-\boldsymbol{x}_0)\}$ of $epi f$ at point $(\boldsymbol{x}_0,f(\boldsymbol{x}_0)$(see the figure).
Now let $B=\{(\boldsymbol{x},y):y=f(\boldsymbol{x}_0)+ \boldsymbol{\mu}_0^T(\boldsymbol{x}-\boldsymbol{x}_0)+\delta \}$, which just move $A$ along y-axis for $\delta>0$. Let $\Phi(\boldsymbol{x}_0,\delta)$ denote the collection of all subgradients of the function part squeezed between $A$ and $B$ (i.e. the bold part of $f(\boldsymbol{x})$).

My problem is: Under what conditions can we conclude that
$\sup_{\boldsymbol{\mu} \in \Phi(\boldsymbol{x}_0,\delta)}\|\boldsymbol{\mu}-\boldsymbol{\mu}_0\| \to 0$ as $\delta \to 0$.

For example: strictly convexity, differentiability of the function etc. Sufficient conditions are fine. But the weaker
a condition is, the better. I think that this holds for most regular convex functions. I just hope to find a simple condition that is much easier to verify than this one.

Strict convexity in a neighborhood of $x_0$ is sufficient (if we are talking about finite dimensional spaces). The proof is like this: without loss of generality, $\mu_0=0$. Then $f>0$ for all $x\neq x_0$ and $f(x)=o(\|x\|)$ as $x\to x_0$. Now fix $\varepsilon>0$ and choose a ball of radius $r$ centered at $x_0$ so small that $f(x)\leq \varepsilon\|x\|$ in that ball. Let $\delta=\min_{\|x\|=r/2}f(x)$. Then, due to convexity, the norm of any subgradient $\mu$ at any point $x$ in the ball of radius $r/2$ is not greater than $2\varepsilon$. If you do not demand strict convexity, you'll need to work with some global rather than local condition. Hope this helps. :)

[quote="fedja"]$f(x)=o(\|x\|)$ as $x\to x_0$[/quote]
Why?

[quote] Then, due to convexity, the norm of any subgradient $\mu$ at any point $x$ in the ball of radius $r/2$ is not greater than $2\varepsilon$ [/quote]
How convexity leads to this conclusion?

Thanks!

Let me explain it in a simpler way then. If there exist $x$ arbitrarily close to $x_0$ such that there is a subgradient $\mu$ at $x$ of norm $\geq \gamma>0$, then we can choose a convergent subsequence of these $\mu$'s and get a subgradient at $x_0$ different from $0$, which contradicts the assumption that the subgradient at $x_0$ is unique. So it is enough to ensure that the condition $f(x)<\delta$ implies that $x-x_0$ is small. But, with our choice of $\delta$, we may conclude that $\|x-x_0\|\leq r/2$ (and this is the whole point of strict convexity: it allows to conclude that the part of the graph between $A$ and $B$ has small diameter if $\delta$ is small). Is this explanation easier to understand?

Thanks fedja. So basically there are two subproblems here.
1) The subgradient at $x_0$ is unique implies that for any given $\gamma>0$, there exists $r>0$ s.t. $\|\mu-\mu_0\|<\gamma$ if $\|x-x_0\|<r/2$
2) There exists $\delta>0$ s.t. $\|x-x_0\|<r/2$ if $f(x)<\delta$.

I still don't understand how the $2\varepsilon$ comes from in your first post.

And is this condition sufficent? The subgradients of any two distinct points at the neibourhood of $x_0$ are different.


I need help on the following. Find x:
1) $ \sqrt{1\plus{}x}\minus{}\sqrt{1\minus{}x} \ge 2\minus{}\frac{x^2}{4}$

2) $ \sqrt{x\minus{}1}\plus{}x\minus{}3 \ge \sqrt{2x^2\minus{}10x\plus{}16}$

3) $ \sqrt{1\minus{}x\plus{}x^2}\plus{}\sqrt{3\minus{}x^2}\le \sqrt{8\minus{}2x}$

4) $ \sqrt{5\plus{}4x\minus{}x^2}\plus{}\sqrt{3\minus{}x^2} \ge 2$

5) $ \sqrt{8\minus{}x^2}\minus{}\sqrt{25\minus{}x^2}>x$

6) $ \sqrt{x\minus{}\frac{1}{2}}\plus{}\frac{x\plus{}1}{4}<\sqrt{2x\minus{}1\plus{}\frac{(x\minus{}1)^2}{8}}$

[quote="kcstevonne"]
1) $ \sqrt {1 \plus{} x} \minus{} \sqrt {1 \minus{} x} \ge 2 \minus{} \frac {x^2}{4}$
[/quote]

By the definition of square root,we have $ \minus{} 1\le x\le 1$.
Thus we have
$ 2\le \frac {x^2}{4} \plus{} \sqrt {1 \plus{} x} \minus{} \sqrt {1 \minus{} x}\le \frac {1^2}{4} \plus{} \sqrt {1 \plus{} 1} \minus{} \sqrt {1 \minus{} 1} \equal{} \frac {1}{4} \plus{} \sqrt {2} < 2$,
so there are no solution.

[quote="kcstevonne"]
2) $ \sqrt {x \minus{} 1} \plus{} x \minus{} 3 \ge \sqrt {2x^2 \minus{} 10x \plus{} 16}$
[/quote]

$ x \minus{} 3 \ge \sqrt {2(x \minus{} 3)^2 \plus{} 2(x \minus{} 1)} \minus{} \sqrt {x \minus{} 1}\ge 0$,we have $ x\ge 3$.

By Cauchy-schwartz's inequality we have $ (1\plus{}1)(x \minus{} 1 \plus{} (x \minus{} 3)^2)\ge (\sqrt {x \minus{} 1} \plus{} x \minus{} 3)^2\Longleftrightarrow \sqrt {2x^2 \minus{} 10x \plus{} 16}\ge \sqrt {x \minus{} 1} \plus{} x \minus{} 3$.

So the equality must holds ,then  $ \sqrt {x \minus{} 1} \equal{} x \minus{} 3$.Since $ x\ge 3$ we have the only solution $ x \equal{} 5$.

[quote="kcstevonne"]I need help on the following. Find x:

3) $ \sqrt {1 \minus{} x \plus{} x^2} \plus{} \sqrt {3 \minus{} x^2}\le \sqrt {8 \minus{} 2x}$

[/quote]
Codition : $ \minus{} \sqrt {3} \le\ x \le\ \sqrt {3}$

We have:

$ (\sqrt {1 \minus{} x \plus{} x^2} \plus{} \sqrt {3 \minus{} x^2})^2 \le\ 2(1 \minus{} x \plus{} x^2 \plus{} 3 \minus{} x^2) \equal{} 8 \minus{} 2x$

$ \rightarrow \sqrt {1 \minus{} x \plus{} x^2} \plus{} \sqrt {3 \minus{} x^2}\le \sqrt {8 \minus{} 2x}$

So It is true for $ \minus{} \sqrt {3} \le\ x \le\ \sqrt {3}$

hello, let's solve number 4), we rewrite it in the form
$ \sqrt{5\plus{}4x\minus{}x^2}\geq2\minus{}\sqrt{3\minus{}x^2}$. It must be
$ 5\plus{}4x\minus{}x^2\geq 0$ and $ 2\minus{}\sqrt{3\minus{}x^2}\geq0$ this is fulfilled for all $ x$ with $ \minus{}1\le x\le \sqrt{3}$. After squaring both sides we have
$ 2\sqrt{3\minus{}x^2}\geq1\minus{}2x$
Squaring again we get $ \minus{}8x^2\plus{}4x\plus{}11\geq0$ which fulfilled for all $ x$ with $ \frac{1}{4}\minus{}\frac{1}{4}\sqrt{23} \le x\le \sqrt{3}$
Sonnhard.

hello, no real $ x$ fulfills the inequality 5), probably we have a typo.
Sonnhard.

Dr. Graubner, for #5, how can we check that no $ x$ satisfies the inequality?

hello, we will show that no real $ x$ satifies the iequality $ \sqrt{8\minus{}x^2}\minus{}\sqrt{25\minus{}x^2}>x$ this is equivalent with $ \sqrt{8\minus{}x^2}>x\plus{}\sqrt{25\minus{}x^2}$, with $ \minus{}2\cdot\sqrt{2} \le x \le 2\cdot\sqrt{2}$. After squaring both sides we have $ 17\plus{}x^2<\minus{}2x\sqrt{25\minus{}x^2}$. Here it must be $ \minus{}2\sqrt{2} \le x \le0$. After squaring again we get $ 5x^4\minus{}66x^2\plus{}289<0$ which is not fulfilled for any real $ x$.
Sonnhard.


geometry
circumcircle
perpendicular bisector
geometry proposed

[quote="rogue"]Construct triangle $ ABC$ given points $ O_A$ and $ O_B,$ which are symmetric to its circumcenter $ O$ with respect to $ BC$ and $ AC,$ and the straight line $ h_A,$ which contains its altitude to $ BC.$[/quote]
The perpendicular bisector of $ O_AO_B$ meets the line $ h_A$ at the orthocenter H. The circle through H centered $ O_A$ meets the circle through H centered $ O_B$ at C. The circle $ (O_A)$ meets the perp line from C to $ h_A$ at B.


calculus
integration
real analysis
function
probability

Define f(x)=x for rational x
                =0 for irrational x

Integrate f(x) from 0 to 1

Which integral are you considering ?

I think f is continuous only at 0, so is not Riemann-integrable.

Q being of null measure, f is Lebesgue-integrable with a integral = 0.

A high school student can be forgiven for not yet being up on theories of integration. Expanding on what belenos said: If you use the Darboux upper sum - lower sum construction and definition of the Riemann integral, and use any partition of [0,1], then since there are irrational numbers in every subinterval, the lower sum is zero. The least upper bound of f(x) on any subinterval is the same as the least upper bound of x on that subinterval, so the upper sum for the integral of f is the same as an upper sum for the integral of x, hence greater than the integral of x, which is 1/2. With all lower sums equal to zero and all upper sums greater than 1/2, the function is not Riemann integrable.

According to the Lebesgue theory of integration (taught at the senior undergraduate or the graduate level) it's not just that this function is integrable. Our whole attitude towards it changes. Since the set of rational numbers, being countable, is a set of measure zero, the function f is equal to zero "almost everywhere" and thus behaves towards the Lebesgue integral exactly as the function which is identically zero behaves. We no longer consider it as having an identity different from the zero function.

The comment that the rational numbers form a set of measure zero is essentially equivalent to the following: Choose a real number in [0,1] "at random", taking that to mean the uniform distribution. What is the probability that the number chosen is rational? Zero. There's also a zero probability that it is algebraic.

The function f is a standard textbook example. Here's another standard textbook example:

Let g(x) = 0 whenever x is irrational, and if x is rational, write x = p/q in lowest terms (q > 0), and let g(x) = g(p/q) = 1/q.

It is possible to show that the Riemann integral of g from 0 to 1 (or any other convenient interval) exists and is zero. You can do this quite directly, without quoting the theorem that belenos indirectly referred to.


We are given a 1962 digit number. It is divisible by 9. Let x be the sum of its digits. Let the sum of the digits of x be y. Let the sum of the digits of y be z.
Find z.

This is a really good one. :D

[hide]The number is divisible by 9, so x is divisible by 9 also
$9\leq x\leq9\cdot1962=17658$
$9\leq y\leq 36$
$9\leq z\leq 9$
so $z=9$ or something like that [/hide]

[hide="solution"]I is at most 9999.....999, and the maximum sum of it's digits is 9*1962=17658, so the maximum sum of the digits of the sum of the digits of the original number is 9+9+9+9=36, and the sum of the digits of that is 9, so z=9.[/hide]

[quote="mewto55555"]Obviously it sums to a multiple of 9, and if you add the digits of those it is another multiple of 9, and so on. At some point, we get 9. After this many sums, seeing as you implied there was only one answer, it has to be 9.[/quote]

 :rotfl:  :rotfl:  :rotfl:


[url=http://bildr.no/view/488489][img]http://bildr.no/thumb/488489.jpeg[/img][/url]

How to get formula of $ b_n$?
why there exists $ i\leq m\minus{}1$ that ... ?
thanks

hello, to solve the equation
$ b_n\equal{}6b_{n\minus{}1}\minus{}8b_{n\minus{}2}$
let  $ b_n\equal{}q^n$ and we get the equation
$ q^2\minus{}6q\plus{}8\equal{}0$
with $ q_1\equal{}4$ and $ q_2\equal{}2$ so our $ b_n$ is from the form
$ b_n\equal{}C_1\cdot2^n\plus{}C_2\cdot4^n$
Sonnhard.


logarithms
function
integration
calculus
calculus computations

Evaluate, with proof,

$\sum_{k=-\infty}^\infty \frac{1}{(4k+2)(3k+1)}$.

One (bad?) beginning:

[hide]
\[\sum_{k=-n-1}^{n}\frac{1}{(4k+2)(3k+1)}= \frac{1}{2}\sum_{k =-n-1}^{n}\frac{3}{3k+1}-\frac{2}{2k+1}= \frac{1}{2}\left(\sum_{k = 0}^{n}\frac{3}{3k+1}-\sum_{k = 0}^{n}\frac{3}{3k+2}\right). \]
Now, I know there are people on the forum who know how to calculate the difference between these sums and $\log n$ as $n \to \infty$, but I'm not one of them.  Obviously, that would be sufficient.  (I would try generating functions, but I don't know how to justify questions of convergence.)  Alternatively, one can recombine these sums into the expression $\frac{3}{2}\sum_{k = 0}^{n}\frac{1}{(3k+1)(3k+2)}$, which might be more amenable to computation.[/hide]

Explain how you obtained the last equality?


Prove or disprove: $ f(g(x)\plus{}h(x))\equal{}f(g(x))\plus{}f(h(x))$.

What are the restrictions on $ f$, $ g$, $ h$?

If there are none, then it is trivially false for $ f(x) \equal{} c\not \equal{} 0$

If so, then $ f(a\plus{}b) \equal{} f(a)\plus{}f(b)\forall (a,b)\in D^{2}_{f}$, which, in most cases, either has no solution or forces a linear of the form $ f(x) \equal{} ax$

When $ g(x), h(x)$ are arbitrary constants this is the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url], and without extra constraints there are actually infinitely many pathological solutions.


modular arithmetic
quadratics
LaTeX
number theory
divisibility tests

(1) How many natural numbers require $ 3$ digits when written in base $ 12$, but require $ 4$ digits when written in base $ 9$?

(2) For each of $ n \equal{} 84$ and $ n \equal{} 88$, find the smallest integer multiple of $ n$ whose base $ 10$ representation consists entirely of $ 6$'s and $ 7$'s.

(3) When the number $ n$ is written in base $ b$ its representation is the two-digit number $ AB$ where $ A \equal{} b \minus{} 2$ and $ B \equal{} 2$.  What is the representation of $ n$ in base $ (b \minus{} 1)$?

(4) if $ a\neq1$ and $ \sqrt [a]{10000_{a}}$, find $ a$.

(5) Denote by $ p_{k}$ the $ k$th prime number.  Show that $ p_{1}p_{2}\cdots p_{n} \plus{} 1$ cannot be the perfect square of an integer.

(6) Prove that it is impossible for three consecutive squares to sum to another perfect square.

Thanks in advance!

[quote="elcric"]
(2) For each of $ n \equal{} 84$ and $ n \equal{} 88$, find the smallest integer multiple of $ n$ whose base $ 10$ representation consists entirely of $ 6$'s and $ 6$'s.[/quote]

This should be "base $ 10$ representation consists entirely of $ 6$'s and $ 7$'s".

And I think the majority of people on these forums have already read through AoPS Volume 1, where these problems are being taken from. If you just want to see the correct solutions, you should think about buying the Solutions Manual.  :wink:

(1) How many natural numbers require 3 digits when written in base 12, but require 4 digits when written in base 9?

The smallest number which requires 3 digits in base 12 is $ 100_{12} \equal{} 12^2 \equal{} 144_{10}$. The largest number which requires 3 digits in base 12 is $ bbb_{12} \equal{} 1000_{12} \minus{} 1_{12} \equal{} 12^3 \minus{} 1 \equal{} 1727_{10}$. The smallest number which requires 4 digits in base 9 is $ 1000_9 \equal{} 9^3 \equal{} 729_{10}$ and the largest number is $ 8888_9 \equal{} 10000_9 \minus{} 1_9 \equal{} 9^4 \minus{} 1 \equal{} 6560_{10}$. Thus, all numbers from 729 to 1727 require 3 digits when written in base 12 but 4 digits when written in base 9. This is 1727-729+1 or exactly $ \boxed{999}$ natural numbers.

Edit: Thanks, jvenezuela716.

[quote="jvenezuela76"]If you just want to see the correct solutions, you should think about buying the Solutions Manual.  :wink: [/quote]

I do have the Solutions Manual, but sometimes, even after reading the solution, I'm still not clear on how to solve a problem.  :(

[quote="towersfreak2006"](6) Prove that it is impossible for three consecutive squares to sum to another perfect square.

Of the three consecutive numbers, exactly one must be divisible by three. Hence, the three consecutive perfect squares are congruent to 0, 1, 1 (mod 3) in some order. Their sum is congruent to 2 (mod 3), but perfect squares are only 0, 1 (mod 3). Hence, it is impossible for three consecutive squares to sum to another perfect square.[/quote]

I don't get the use of mods.  Would the three consecutive perfect squares have to congruent to 0, 1, 2, in some order too?

For example,
[list]$ 4\equiv1 \mod3$
$ 5\equiv2 \mod3$
$ 6\equiv0 \mod3$.[/list]

Given three consecutive numbers, one of them must be congruent to zero, one of them must be to one, and one of them must be to two. Also, we have that $ (3k)^2 \equiv 0 \pmod{3}$, $ (3k\plus{}1)^2 \equiv 1 \pmod{3}$, $ (3k\plus{}2)^2 \equiv (3k\minus{}1)^2 \equiv 1 \pmod{3}$. For example, with $ 4,5,6$, we have $ 4^2 \equal{} 16 \equiv 1 \pmod{3}, 5^2 \equal{} 25 \equiv 1 \pmod{3}, 6^2 \equiv 0 \pmod{3}$.

So, after summing the squares our our three consecutive numbers, the sum comes out to be $ \equiv 0 \plus{} 1 \plus{} 1 \equiv 2 \pmod{3}$. However, as we showed above, there are no numbers which when squared can come out to be $ 2 \pmod{3}$, which gives us the contradiction.

[quote="elcric"]
I don't get the use of mods.  Would the three consecutive perfect squares have to congruent to 0, 1, 2, in some order too?[/quote]

It's easier to see the logic here in our familiar $ \pmod{10}$. Suppose I gave you the number $ 2304928487$ and asked you to prove to me that this could not be a perfect square. What you could do is simply look at the squares of the numbers from $ 0$ to $ 9$; squaring them, we see that the squares are congruent to $ 0,1,4,9,6,5,6,9,4,1$, which are simply the last digits. Therefore no square is congruent to $ 7\pmod{10}$.

Similarly, no square can be congruent to $ 2\pmod{3}$, after examining the squares of $ 0,1,2$.

I still haven't got four of the questions.  Can anyone help me?

Thanks!

[u][b]Problem 5:[/b][/u] [i]Fast but nice - I am sure there are many other different approaches.[/i]

Let $ p_1 p_2 \ldots p_k \plus{} 1 \equal{} n^2$. We can factor $ p_1 p_2 \ldots p_k \equal{} n^2 \minus{} 1 (n \plus{} 1)(n \minus{} 1)$. Notice that both $ n \plus{} 1$ and $ n \minus{} 1$ have the same [i]parity[/i] -- that is, either both are even or both are odd.

However, $ p_1 p_2 \ldots p_k$ has exactly one factor of $ 2$. This means that that if we express $ p_1 p_2 \ldots p_k$ as the product of two numbers, then one must have the factor of two, meaning it is even, and the other must have no factors of two, meaning it is odd.

So, it is impossible to express $ p_1 p_2 \ldots p_k$ in the form $ (n \plus{} 1)(n \minus{} 1) \equal{} n^2 \minus{} 1$, and thus $ p_1 p_2 \ldots p_k \plus{} 1 \neq n^2$

That's essentially the argument that $ n^2 \equiv 0, 1 \bmod 4$ but $ p_1 p_2 ... p_k \plus{} 1 \equiv 3 \bmod 4$.

Problem 4

I assume the problem is meant to be, for which $ a \in \mathbb{N}$, $ a \neq 1$ is $ \sqrt[a]{10000_a} \equal{} a$

[hide="Hint"]
Write $ 10000_a$ in the form $ 1a^4\plus{}0a^3\plus{}0a^2\plus{}0a^1\plus{}0a^0$
[/hide]

[hide="Solution"]
$ \sqrt[a]{a^4} \equal{}a$

Therefore $ a^{\frac{4}{a}} \equal{} a$

Therefore $ a \equal{} 4$

[/hide]

[quote="SimonM"]Problem 4

I assume the problem is meant to be, for which $ a \in \mathbb{N}$, $ a \neq 1$ is $ \sqrt [a]{10000_a} \equal{} a$

[hide="Hint"]
Write $ 10000_a$ in the form $ 1a^4 \plus{} 0a^3 \plus{} 0a^2 \plus{} 0a^1 \plus{} 0a^0$
[/hide]

[hide="Solution"]
$ \sqrt [a]{a^4} \equal{} a$

Therefore $ a^{\frac {4}{a}} \equal{} a$

Therefore $ a \equal{} 4$

[/hide][/quote]

Actually, the correct question is:

If $ a\neq1$ and $ \sqrt[a]{10000_{a}} \equal{} 10_{a}$, find $ a$.

Sorry about that.

Thats the same question

$ 10_a \equal{} a_{10}$

I still haven't got numbers (2) and (3).  Does anyone have a solution to them?  Thanks.

[b]Problem 2[/b]

[hide="Hint"]Use divisibility checks for the prime factors of the two numbers[/hide]

[hide="Solution"]I'll solve one, the other is done using essentially the same method

$ 88 = 2^{3} \times 11$

Divisibility by $ 8$ means the last $ 3$ digits are divisible by $ 8$

Clearly the last digit is a $ 6$ Therefore we check the four cases to find the last $ 3$ digits are

$ 776$

Divisibility by 11 suggests that the difference in the alternate sums is $ 0$. Therefore if we set the first digit to $ 6$ we will have our number

$ 88|6776 \Rightarrow \boxed{6776}$[/hide]

[b]Problem 3[/b]

$ AB_n = CD_{n - 1}$ where $ A = n - 2$, $ B = 2$

[hide="Hint"]Use the notation $ abcd_n = an^3 + bn^2+cn^1+dn^0$[/hide]

[hide="Solution"]
\begin{align*} n & = (n - 2)n + 2 \\
& = n^2 - 2n + 2 \\
& = (n - 1)^2 + 1 \\
& = 1 \times (n - 1)^2 + 0\times(n - 1)^1 + 1\times(n - 1)^0 \end{align*}

Therefore $ AB_n = 101_{n - 1}$[/hide]

[quote="SimonM"][b]Problem 2[/b]

[hide="Hint"]Use divisibility checks for the prime factors of the two numbers[/hide]

[hide="Solution"]I'll solve one, the other is done using essentially the same method

$ 88 = 2^{3} \times 11$

Divisibility by $ 8$ means the last $ 3$ digits are divisible by $ 8$

Clearly the last digit is a $ 6$ Therefore we check the four cases to find the last $ 3$ digits are

$ 776$

Divisibility by 11 suggests that the difference in the alternate sums is $ 0$. Therefore if we set the first digit to $ 6$ we will have our number

$ 88|6776 \Rightarrow \boxed{6776}$[/hide]

[b]Problem 3[/b]

$ AB_n = CD_{n - 1}$ where $ A = n - 2$, $ B = 2$

[hide="Hint"]Use the notation $ abcd_n = an^3 + bn^2 + cn^1 + dn^0$[/hide]

[hide="Solution"]
\begin{align*} n & = (n - 2)n + 2 \\
& = n^2 - 2n + 2 \\
& = (n - 1)^2 + 1 \\
& = 1 \times (n - 1)^2 + 0\times(n - 1)^1 + 1\times(n - 1)^0 \end{align*}

Therefore $ AB_n = 101_{n - 1}$[/hide][/quote]

For number 3, why did you introduce the $ C$ and $ D$, and what exactly do they represent?

Thanks.

The digits in the new number, I hadn't realised I'd actually written that. It was meant to be in my head. Oh well

I don't get how you got the following:

$ abcd_n \equal{} an^3 \plus{} bn^2 \plus{} cn^1 \plus{} dn^0$

How did you know that it had to be a four-digit number with digits of $ a$, $ b$, $ c$, and $ d$?

Thanks.

[quote="SimonM"]I didn't, I was just showing what the definition is.[/quote]

Definition?  Could you explain the solution to this problem a bit more, please?  I don't think that I get it.

Thanks.

[quote="elcric"]
(3) When the number $ n$ is written in base $ b$ its representation is the two-digit number $ AB$ where $ A = b - 2$ and $ B = 2$.  What is the representation of $ n$ in base $ (b - 1)$?[/quote]

Remember what the representation in a base means. In base 10, $ 9835_{10} = 9000 + 800 + 30 + 5 = 9\cdot10^3 + 8\cdot10^2 + 3\cdot10^1 + 5\cdot10^0$.

We can generalize this: $ 2538_b = 2\cdot b^3 + 5\cdot b^2 + 3\cdot b^1 + 8\cdot b^0$.

Taking this concept, we can find the solution:

[hide="Solution."]
Using our above logic:
$ n = AB_b = A\cdot b^1 + B\cdot b^0 = Ab + B = (b - 2)b + 2$.

If we want to convert this to base $ (b - 1)$, we seek something of the form:
$ n = ...WXYZ_{(b - 1)} = ... + W\cdot(b - 1)^3 + X\cdot(b - 1)^2 + Y\cdot(b - 1)^1 + Z\cdot(b - 1)^0$.

Don't let this notation throw you off: we simply want to express $ AB_b$ as some number in base $ (b - 1)$, though [b]we don't know how many digits it will have[/b].

So now our goal is to write $ n = AB_b = (b - 2)b + 2$ in terms of powers of $ b - 1$:

\begin{align*} n & = (b - 2)b + 2 \\
& = b^2 - 2b + 2 \end{align*}

Now we complete the square (see the Quadratics section in AoPS Volume 1 for more on this):

\begin{align*} n & = (b - 2)b + 2 \\
& = b^2 - 2b + 1 + 1 \\
& = (b - 1)^2 + 1 \\
& = 1\cdot(b - 1)^2 + 0\cdot(b - 1)^1 + 1\cdot(b - 1)^0 \end{align*}

But the above is equal to $ 101_{(b - 1)}$! Thus:

$ n = \boxed{101_{(b - 1)}}$
[/hide]

Thanks so much for everyone's help.  I got all the problems now!  :lol:

[quote="jvenezuela716"]We can generalize this: $ 2538_b = 2\cdot b^3 + 5\cdot b^2 + 3\cdot b^1 + 8\cdot b^0$.[/quote]

Just a friendly reminder with $ \text{\LaTeX}$; make sure you preview your posts before submitting and remember to put spaces between variables and the code for symbols (i.e. \cdotb will not work)  :wink:

I fixed it for you, this time.

[quote="b-flat"]
Just a friendly reminder with $ \text{\LaTeX}$; make sure you preview your posts before submitting and remember to put spaces between variables and the code for symbols (i.e. \cdotb will not work)  :wink:

I fixed it for you, this time.[/quote]

Ack, sorry about that!  :blush:

Thanks for fixing it, though; I'll be more thorough in my previews in the future.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

[color=darkred][b]Let $ABC$ be a triangle with the incircle $C(I)$ and the circumcircle $C(O)$. Denote : $D\in BC$ for which $ID\perp BC$ ;
the reflection $F$ of the point $A$ w.r.t. the point $O$ ; the second intersection $L$ of the line $FI$ with the circumcircle ;
the second intersection $E$ of the line $AI$ with the circumcircle ; the intersections $\{\begin{array}{c}X\in AD\cap LO\\\ Y\in DF\cap EO\end{array}$. Prove that $I\in XY$.[/b][/color]

I think the basic thing here is that $D \in LE$ and then just Pascal.

For proving that $D \in LE$ see that the triangle $\triangle{IDE}$ and $\triangle{AIF}$ are similar, because: $\angle{IED}=\angle{IFA}$ ($AFEL$) is cyclic) and $\frac{r}{AT}=\frac{IE}{2R}\iff 2Rr = AI \cdot IE = p_{w}I=R^{2}-OI^{2}$, which is true because of $OI^{2}=R^{2}-2Rr$.

[edit: typing errors]

According to my diagram, $I$ doesn't lie on $LE$...

But $D$ does.
[hide="because..."]As pohoatza showed above, $\Delta AIF\sim \Delta IDE.$ So $\angle LIE=\angle AIF=\angle IDE=a$ so $\Delta IDE\sim\Delta LIE.$ So $\angle DLI=\angle DIE=b$
Then $\angle LID=\angle LIE-\angle DIE=a-b.$
So $\angle IDL=180^\circ-(a-b+b)=180^\circ-a\implies \angle LDC=a-90^\circ$ since $\angle CDI=90^\circ.$
But $\angle BDE=a-\angle IDB=a-90^\circ.$
Thus $L,D,E$ are collinear.
[/hide]
Then by Pappu's, $X,Y,I$ are collinear.

Yes, of course it was $D$,another typing error  :( [I edited it]


trigonometry
geometry
angle bisector
geometry unsolved

Let triangle $ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of its angle $\angle B$ meets the side $AC$ at a point $D$ and that $BC = BD+AD$.
Determine $\angle A$.

[hide="Here is a short trigonometrical solution."][u]Denote[/u] $B=C=2x\ \left[\ A=180^{\circ}-4x\ ,\ m(\widehat{ADB})=3x\ \right]\ .$
$BC=BD+DA\Longleftrightarrow \frac{BC}{BD}=1+\frac{DA}{DB}\Longleftrightarrow\frac{\sin 3x}{\sin 2x}=1+\frac{\sin x}{\sin 4x}\Longleftrightarrow$

$\sin 3x\sin 4x=\sin 2x\sin 4x+\sin x\sin 2x\Longleftrightarrow$

$\cos x-\cos 7x=\cos 2x-\cos 6x+\cos x-\cos 3x\Longleftrightarrow$

$\cos 6x+\cos 3x=\cos 2x+\cos 7x\Longleftrightarrow \cos \frac{9x}{2}\cos \frac{3x}{2}=\cos \frac{9x}{2}\cos \frac{5x}{2}\Longleftrightarrow$
$\cos \frac{9x}{2}=0\Longleftrightarrow x=20^{\circ}\Longleftrightarrow A=100^{\circ}\ (\ B=C=40^{\circ}\ )\ .$[/hide]

[Moderator edit: See http://www.kalva.demon.co.uk/canada/casoln/csol964.html for a non-trigonometric solution.]

I met a trig problem this week and I found it the same as this problem:

In $\triangle{ABC}$, $a=2b \cos{C}$, $\sin{A} \cdot \sin{(\frac{B}{2}+C)}=\sin{C} \cdot \left( \sin{\frac{B}{2}}+\sin{A}\right)$.
Determine the angles of the triangle.

Another solution:
[hide]Extend $BD$ to $E$ s.t. $AE=ED$. Since $BC=BE+EA=BC+ED=BD$, $BCD$ is isosceles. Let angle $ABC = 4x$. Angle $ACB=4x$, angle $BAC=180-8x$, angle $DAE=$ angle $DEA=3x$, angle $ABE=$ angle $EBC=2x$, angle $BCD=$ angle $BDC=90-x$, angle $ACD = 90-5x$. By ceva's theorem for quadrilaterals, $ \frac{\sin ABE\sin BCA\sin CEB\sin EAC}{\sin EBC\sin ACE\sin AEB\sin BAC}=1
\Rightarrow \frac{\sin4x\sin(90-x)}{\sin(90-5x)\sin(180-8x)}=1$

 $\Rightarrow \cos x=2\cos4x\cos5x \Rightarrow cos9x=0 \Rightarrow 9x=90 \Rightarrow x=10 \Rightarrow \angle BAC = 180-8x = 100^{\circ}$.[/hide]

[quote=shobber]A geometrical solution:

Let $E \in BC$ such that $BE=BD$. Then by $BC=BD+AD$, we get $AD=CE$. Now use the angle bisector theorem:
\[ \frac{CE}{CD}=\frac{AD}{CD}=\frac{AB}{CB}=\frac{AC}{CB} \]
Since $\angle{DCE}=\angle{BCA}$, so $\triangle{DCE} \sim \triangle{BCA}$, thus we get $\angle{DCE}=\angle{CDE}=\angle{ABC}$. Let $\angle{ABC}=2x$, then $\angle{BDE}=\angle{BED}=4x$, $\angle{CBD}=x$, so $x+2x+(2x+4x)=180^o \Longrightarrow x=20^o$. Therefore: $\angle{A}=100^o$[/quote]
You can construct F on the segment AB and you easily prove that the triangles DAF and  CED are similar .But CE=AD,therefore triangles are also congruent, but AD=AF from Thales's Theorem, therefore CE=DE.
Then, m(<CED)=180-m(<BED)=m(<DAF)
Note m<ABC=2x;, so m<CED=180-(180-x)/2=180-4x; then x =20 and m<A= 180-4*20=100;

see https://www.artofproblemsolving.com/community/u228044h1195598p5859910 for better

Let angle CBD=angle ABD=x
AB=AC
angle ABC=angle ACB=2x
angle BAC=180°-4x
In ∆BCD,
angle BDC=180°-3x
sin angle BDC / BC= sin angle  DCB / BD
BC/ BD= sin 3x/ sin 2x
In ∆ABD,
AD/ BD=sin x/ sin 4x
BC=BD+AD
BC/ BD= sin 3x/ sin 2x
1+ (BD/AD)=sin 3x/ sin 2x
1+ sin x/ sin 4x= sin 3x/ sin 2x
2sin 2x sin 4x=2sin 3x sin 4x- 2 sin x sin 2x
cos 2x-cos 6x=cos 3x- cos 7x
 cos 2x- cos 3x- cos 6x+cos 7x=0
sin (x/2){ sin (13x)/2 - sin(5x)/2}=0
sin (x/2)≠0
x≠0
sin (13x)/2 - sin(5x)/2}=0
cos (9x/2) sin 2x=0
sin 2x≠0, x≠0
cos (9x/2)=0
cos (9x/2)= cos 90°
(9x/2)=90°
x=20°
angle A=180°-4*20°=100°
@Krishijivi


[quote="Leonhard Euler"]Hi, i am a student.
Can anyone suggest some really good books about elementary number theory (in English) which degree of difficult is usamo~putnam and explain theorem include kummer's , Lucas's, Wolstenholme's , Thue's , Hensel's theorem.

which is better
 [u]Introduction to the Theory of Numbers (Hardcover)
by Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery  [/u]
or
[u]A Course in Number Theory (Oxford Science Publications) (Paperback)
by H. E. Rose?[/u][/quote]

please reply


geometry
3D geometry
linear algebra
matrix

Show that if two positive integers $ a, b$ can be written in the form $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$ with $ x, y, z$ integers, then so can $ ab$.

Edit:  [b]Follow-up:[/b]  What primes are of this form?

Alright, so maybe I shouldn't have posted this in Intermediate  :oops:

[hide="Huge hint 1"] What is the determinant of $ \left[ \begin{array}{ccc} x & y & z \\ z & x & y \\ y & z & x \end{array} \right]$? [/hide]
[hide="Huge hint 2"] What is the product of two matrices of the above form? [/hide]

[hide="Hax"]By cheating using the Huge hints  :blush:
The determinant of the three by three matrix is $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz$.

If we have $ \left[ \begin{array}{ccc} x & y & z \\
z & x & y \\
y & z & x \end{array} \right]$ and  $ \left[ \begin{array}{ccc} x_1 & y_1 & z_1 \\
z_1 & x_1 & y_1 \\
y_1 & z_1 & x_1 \end{array} \right]$

and let $ a$ and $ b$ be the determinants of these two matrices, we have that the determinant of the product of the two matrices is the product of the determinants, namely $ ab$.

Then simply multiplying the two matrices gives us $ \left[ \begin{array}{ccc} xx_1\plus{}yz_1\plus{}zy_1 & xy_1\plus{}yx_1\plus{}zz_1 & xz_1\plus{}yy_1\plus{}zx_1 \\
xz_1\plus{}yy_1\plus{}zx_1 & xx_1\plus{}yz_1\plus{}zy_1 & xy_1\plus{}yx_1\plus{}zz_1 \\
xy_1\plus{}yx_1\plus{}zz_1 & xz_1\plus{}yy_1\plus{}zx_1 & xx_1\plus{}yz_1\plus{}zy_1 \end{array} \right]$, which is of the same form.

Thus, the determinant ($ ab$) can be expressed in the desired form.[/hide]

We could factor it as $ (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{}zx)$ for the prime thing.

[quote="Differ"]We could factor it as $ (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)$ for the prime thing.[/quote]
Alright, but the $ x \plus{} y \plus{} z \equal{} \pm 1$ case isn't trivial, is it?  In any case, I'll admit that I haven't thought about the follow-up, so another more important question:  does there exist $ c \equal{} ab$ with $ a, b > 1$ such that $ c$ is of this form but $ a, b$ aren't?


Since my "butt vs boob" topic was deemed innapropiate, I figure this can't do much harm

So yeah, I took a few pictures before going to school in the morning, and as the title says just rate how it is.

http://img100.imageshack.us/img100/2640/m3663zv8.jpg

ban this guy

where is our good ol' american censorship?

heck yeah, can't post anything potentially offensive that won't even get a second glance in retrospect.

Last warning. Any other inappropriate thread created will be locked and I will personally inform the admins about this.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I posed myself the following question.

If you consider the group $\mathbb{Z}$ with its normal subgroups $k\mathbb{\mathbb{Z}}$, then you can form its quotientgroup out of these: $\mathbb{Z}/k\mathbb{\mathbb{Z}}= \mathbb{Z}_{k}$. But, since $k\mathbb{Z}\cong \mathbb{Z}, \forall k \in \mathbb{Z}$.

why is then not $\mathbb{Z}/k\mathbb{Z}\cong \mathbb{Z}/\mathbb{Z}={0}$.

Every step seems justified because *abstractly* $\mathbb{Z}$ and $k\mathbb{Z}$ are the same. However it is terribly *not* isomorphic.

*if* you were going to prove $X \cong Y \Rightarrow G/X \cong G/Y$, where would the proof go wrong?

[quote="ZetaX"]For proving $X \cong Y$, you have to give an isomorphism.
The prove will go wrong if you try to define that map.[/quote]

mmm if you take $X=\mathbb{Z}$ and $Y=k \mathbb{Z}$
and define the map $\eta$ as follows:

$\eta: \mathbb{Z}\to k\mathbb{Z}: x \mapsto kx$
then this is bijective and $\eta$ is a morphism, since $\eta(x+y)=k(x+y)=kx+ky=\eta(x)+\eta(y)$. Not? Or which map were you talking about?

The version which actually works: If $\phi$ maps $G$ to $G'$ and $H$ is a normal subgroup of $G$, there is an induced map $\phi^{*}$ from $G/H$ to $G'/\phi(H)$. When $\phi$ is an isomorphism, $\phi^{*}$ is also an isomorphism.
The key is that we must apply $\phi$ to $G$ as well as $H$, and its codomain is $G'$. In your example, $\eta$ maps $\mathbb{Z}$ to itself, but it's not an isomorphism.

[quote="jmerry"]
The key is that we must apply $\phi$ to $G$ as well as $H$, and its codomain is $G'$. In your example, $\eta$ maps $\mathbb{Z}$ to itself, but it's not an isomorphism.[/quote]

Ow I see, I missed that pitfall

thanks guys