geometry
3D geometry
analytic geometry
trigonometry
LaTeX
logarithms
tetrahedron

These are from Vestavia 01

1. How many values of x satisfy the equation: log(x^2)=cos(x)

2. Find the value of x in the following system of equations:
   3x+2y+z=3
   3y+2q+w=2
   3z+2w+x=1
   3w+2x+y=0

3. Two circles have an external tangent with length 36 cm. The shortest distance between teh circles is 14 cm. Find the radius, in cm, of the larger circle if its radius is four times that of the smaller circle.

4. If A is the fourth perfect number, B is the least integer number greater than one which is both a cube and a square, and C ist he twelfth triangular number, find the value of sqr[(A/B)-C]

5. The graph of 2x^2-21xy+29x-62y+10y^2=0 is a(n)....

6. Find the volume of a hexahedron with vertices (1,0,5) (4,0,2) (0,0,0) (-1,-3,-1) (2,3,3)

7. Find the sum of the abscissas of all (x,y) coordinates which satisfy
2x^2+xy-y^2+3y-2=0 and 3x^2+3xy+y^2-6x-5y+4=0

8. If f(n) = n^8-2n^7+4n^6-8n^5+16n^4-32n^3+64n^2-128n+1 then what is f(1/128)?

9. Find the distance between teh point (5,1,3) and the line defined by the parametric equations
x=3, y=7, and z=1+t

10. Two ships leave port at the same time. One travels 20 mph at a bearing of N32degreesW. The second ship travels 30mph at a bearing of S28W. Find the distance in miles between the ships after 2 hours.

Sorry I didn't use LATEX.

These are all individual questions I am having trouble with.

Please post solution as soon as you know how to do one.

Hint for #10
[hide]Notice that the angles, when added, and considering the axis, equal $ 120$ $ degrees$. Use the [b]Law of Cosines[/b].[/hide]
Answer for #10 (I don't have Latex yet so this will look ugly) (This is corrected)
[hide]The [i]Law of Cosines[/i], for triangles with sides [b]$ A$[/b], [b]$ B$[/b], [b]$ C$[/b], and the given angle between sides $ A$ and $ B$, can be used to find the length of [b]$ C^2$[/b]. The formula is:
[b]$ A^2 \plus{} B^2 \minus{} 2ABcos(angle)\equal{}C^2$[/b]. Replace the values [b]$ A$[/b] and [b]$ B$[/b] with $ 40$ and $ 60 miles$ ($ 20 mph*2(1/h) \equal{} 40 miles$ and $ 30 mph*2(1/h) \equal{} 60 miles$) and let [b]$ C$[/b] equal the distance of the boats after $ 2$ hours had passed.

Also, $ cos(120)\equal{}\minus{} (1/2)$.

After replacing the values, and squaring both sides and simplifying to get [b]$ C$[/b], the distance between the boats at this moment is:
$ 20sqrt(19)$.[/hide]

Something is wrong

[hide]I worked it and got 120 as the angle. Using the law of cosines, it comes to 20*sqr(7)[/hide]

Whoops, misread the given directions.  :blush:

It's actually
[hide]$ 20sqrt(19)$[/hide]

[hide="#3"]
[asy]size(200); pair A,B,C,D,F,G,H,I,J,K; A=origin; B=(78,0); C=(78,10); D=(0,40); draw(A--B--C--D--cycle); F=(0,10); draw(C--F); G=(0,80); H=(40,40); draw(circumcircle(A,G,H)); I=(78,20); J=(68,10); draw(circumcircle(B,I,J)); label("x",(78,5),E); label("x",(0,5),W); label("3x",(0,25),W); label("14",(45,22),NE); label("36",(36,0),S); label("x",(70.7,11.8),NE); label("4x",(15,35),NE);[/asy]

From the diagram, we have $ (3x)^2\plus{}(36)^2\equal{}(5x\plus{}14)^2$. Which, after solving (I leave it for you to solve) and taking the positive root, we get $ x\equal{}5$. Thus the radius of the larger circle is $ 4x\equal{}20$.

[/hide]

That is correct.

Can anyone get the other problems?
I know the answers to check them, I just wanted to post some challenging problems to see if yall could solve them.

Hints:

[hide="1"] First of all, $ 2 \log x \equal{} \cos x$.  Second of all, on what interval is $ \cos x$ concave? [/hide]
[hide="2"] The equations are symmetrical.  Look to symmetric sums. [/hide]
[hide="4"] All you really need to know for this is the definition of a [url=http://en.wikipedia.org/wiki/Perfect_number]perfect number[/url] and a [url=http://mathworld.wolfram.com/TriangularNumber.html]triangular number[/url].  Oh, and if a number is a cube and a square, it's a sixth power. [/hide]
[hide="5"] Complete the square.  A lot.  Try to get rid of the $ \minus{}21xy$ term. [/hide]
[hide="6"] Sketch it out.  Split it up into tetrahedrons, maybe.  The volume of a tetrahedron defined by the points $ (0, 0, 0), (a, b, c), (d, e, f), (g, h, i)$ is one-sixth the determinant $ \left| \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i \end{array} \right|$. [/hide]
[hide="7"] "Abscissa" means $ x$-coordinate (if you didn't know).  There should be four points.  Consider a quartic polynomial and Vieta's. [/hide]
[hide="8"] What is $ f(2n) \plus{} 255$? [/hide]
[hide="9"] Don't be intimidated.  Use the distance formula.  The distance between a point and a line is the length of the altitude from the point to the line, which is the shortest distance from a point on the line to the point. [/hide]

Can you clarify 2, 7, and 8?

For #6, is there a general method for finding the area of 3-d objects similar to the shoelace method used for 2d polygons in the Cartesian plane?

Thanks

[hide="2 (solution)"] Add all of the equations.  We get $ 6(x \plus{} y \plus{} z \plus{} w) \equal{} 6 \Leftrightarrow x \plus{} y \plus{} z \plus{} w \equal{} 1$.  Next, add the first and third equations and the second and fourth equations.  We get $ 2(2x \plus{} y \plus{} 2z \plus{} w) \equal{} 4 \Leftrightarrow 2x \plus{} y \plus{} 2z \plus{} w \equal{} 2$ and $ 2(x \plus{} 2y \plus{} z \plus{} 2w) \equal{} 2 \Leftrightarrow x \plus{} 2y \plus{} z \plus{} 2w \equal{} 0$.

Along with the first equation, we can now deduce that $ x \plus{} z \equal{} 1, y \plus{} w \equal{} 0$.  The first equation simplifies to $ x \plus{} y \equal{} 1$, the second to $ y \plus{} z \equal{} 1$, the third to $ z \plus{} w \equal{} 0$, and the fourth to $ x \plus{} w \equal{} 0$.

Almost there!  $ 2x \plus{} y \plus{} 2z \plus{} w \equal{} 2x \plus{} (y \plus{} z) \plus{} (z \plus{} w) \equal{} 2x \plus{} 1 \equal{} 2 \implies x \equal{} \boxed{ \frac{1}{2} }$.

To check for consistency, we now have $ y \equal{} \frac{1}{2}, z \equal{} \frac{1}{2}, w \equal{} \minus{} \frac{1}{2}$.  You can check that this satisfies the original system.

Alternate method:  Cramer's rule, although this is overkill. [/hide]
[hide="4 (solution)"] If $ 2^n \minus{} 1$ is prime, then $ (2^n \minus{} 1)2^{n\minus{}1}$ is a perfect number.  Euclid proved that all even perfect numbers were of this form, but it is not known whether odd perfect numbers exist.

It is not hard to show that $ 2^n \minus{} 1$ is prime only if $ n$ is prime.  Primes of this form are known as [b]Mersenne primes.[/b]  The first few perfect numbers occur when $ n \equal{} 2, 3, 5, 7$, giving $ 6, 28, 496, 8128$.

The $ n^{th}$ triangular number is given by

$ T_n \equal{} 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} n \equal{} \frac{n(n\plus{}1)}{2}$

The twelfth triangular number is therefore $ \frac{12 \cdot 13}{2} \equal{} 78$.

The first sixth power greater than $ 1$ is $ 2^6 \equal{} 64$.

Our answer is $ \sqrt{ \frac{8128}{64} \minus{} 78 } \equal{} \sqrt{ 127 \minus{} 78} \equal{} \sqrt{49} \equal{} \boxed{7}$. [/hide]
[hide="6 (comment)"] There is no "Shoelace Formula" in three dimensions, to my knowledge.  The tetrahedron volume formula is the closest thing you're going to get. [/hide]
[hide="7 (comment)"] The approach I am suggesting (by no means the best one, but the only nice approach I can think of is beyond the scope of this forum) is to use the quadratic formula on one of the conics to solve for $ y$ in terms of $ x$ and plug that into the other one.  When you simplify everything (which could be really nasty), you'll end up with a quartic polynomial whose roots are exactly the abscissas of the intersections.[/hide]
[hide="8 (comment)"] $ g(n) \equal{} f(2n) \plus{} 255 \equal{} 256(n^8 \minus{} n^7 \plus{} n^6 \minus{} n^5 \plus{} n^4 \minus{} n^3 \plus{} n^2 \minus{} n \plus{} 1) \equal{} 256 \frac{n^9 \plus{} 1}{n \plus{} 1}$.  :) [/hide]

Another comment on #8
[hide]I find resimplifying $ g(n)$ to $ f(n)$ mind straining for some reason. A way I did it was this:
for $ f(n) \equal{} n^8 \minus{} 2n^7 \plus{} 4n^6 \minus{} 8n^5 \plus{} 16n^4 \minus{} 32n^3 \plus{} 64n^2 \minus{} 128n \plus{} 1$, Notice that $ n^8 \plus{} ... \minus{} 128n$ follows a geometric sum series. I instead rewrote it as $ f(n) \equal{} S \plus{} 1$ where $ S$ is the geometric summed portion.
$ S \equal{} n^8 \minus{} 2n^7 \plus{} 4n^6 \minus{} 8n^5 \plus{} 16n^4 \minus{} 32n^3 \plus{} 64n^2 \minus{} 128n$.
Then, $ S(n/2) \equal{} (n^9/2) \minus{} n^8 \plus{} 2n^7 \minus{} 4n^6 \plus{} 8n^5 \minus{} 16n^4 \plus{} 32n^3 \minus{} 64n^2$.
From this I got $ S \plus{} S(n/2) \equal{} (n^9/2) \minus{} 128n \minus{}> S(1 \plus{} (n/2)) \equal{} (n^9/2) \minus{} 128n.$
Dividing both sides and simplifying I got $ S \equal{} (n^9 \minus{} 256n)/(n \plus{} 2)$. Now, adding $ S$ by $ 1$, I got this:
$ f(n) \equal{} (n^9 \minus{} 255n \plus{} 2)/(n \plus{} 2)$ Next comes plugging in the value $ 1/128$, or $ 1/2^7$ for $ n$.[/hide]
I think something's wrong with the solution #2
[hide]There's one variable missing: $ q$.
Notice that adding all these instead gives $ 6w \plus{} 6x \plus{} 6y \plus{} 4z \plus{} 2q \equal{} 6$, unless $ q$ is actually $ z$. The question in improved font is:
Find the value of $ x$ in the following system of equations:
$ 3x \plus{} 2y \plus{} z \equal{} 3$
$ 3y \plus{} 2q \plus{} w \equal{} 2$
$ 3z \plus{} 2w \plus{} x \equal{} 1$
$ 3w \plus{} 2x \plus{} y \equal{} 0$[/hide]

$ q$ is almost certainly $ z$.  Your system is a system of five variables in four equations, and there won't be a unique value of $ x$.

You don't need to download LaTeX to post it on these forums; just type the code you want inside dollar signs.

[quote="brianb"]1. How many values of x satisfy the equation: log(x^2)=cos(x)[/quote]

[hide]The first solution is approximately $ x\equal{}1.324$ (using my calculator's intersect tool). Since $ 2 \log x$ is strictly increasing, and $ \cos x$ is limited to the range $ [\minus{}1,1]$, a simple inspection shows us that the two curves will not meet again. (The next time when $ \cos x$ reaches $ 1$ is at $ x\equal{}2\pi$, and at this point $ 2 \log x \equal{} 1.5963 > 1$[/hide]

On number 6, are you missing a few vertices or am I imagining things? (Unless you mean something like "regular hexahedron")

The answer choices to the function question are in the form:  $ \frac {128}{257}$$ \minus{} \frac {1}{2}$ $ (\frac {1}{256})^7$ Simliar to that form with slightly varying numbers.

Also, can you work out the one with parametric equations?

[hide="number six"]
The first three points on your list are all in the x-z plane (the y value is 0)
I will use the x-z plane to split the hexahedron into two pyramids/tetrahedrons

The area of the base of each pyramid is the area of the triangle formed by $ (0,0,0) , (1,0,5) , (4,0,2)$
using a vector cross product

$ A_{triangle} \equal{} \frac{1}{2}ab\sin\theta \equal{} \frac{1}{2}|\vec{a} \times \vec{b}|$
$ A \equal{} \frac{1}{2} | <1,0,5> \times <4,0,2> |$
$ A \equal{} \frac{1}{2} | <0,18,0> |$
$ A \equal{} \frac{3\sqrt2}{2}$

The height of each pyramid is the distance from the 4th point to the x-z plane, which is the y value. The y values of the other two points are $ \minus{}3$ and $ 3$, so the height of each pyramid is 3.

$ V_{pyramid} \equal{} \frac{1}{3}Ah \equal{} \frac{1}{3} \frac{3\sqrt2}{2} (3) \equal{} \frac{3\sqrt2}{2}$

since there are two pyramids

$ V_{hexahedron} \equal{} 3\sqrt2$
[/hide]

Sorry about the double post

9. Find the distance between teh point (5,1,3) and the line defined by the parametric equations
x=3, y=7, and z=1+t

[hide="number 9"]

By the distance formula, the distance between the point and a generic point on that line is

$ d \equal{} \sqrt{ (x\minus{}5)^2 \plus{} (y\minus{}1)^2 \plus{} (z\minus{}3)^2}$

$ d \equal{} \sqrt{ (\minus{}2)^2 \plus{} (6)^2 \plus{} (t\minus{}2)^2 }$

$ d \equal{} \sqrt{ 4 \plus{} 36 \plus{} t^2 \minus{} 4t \plus{} 4}$

$ d \equal{} \sqrt{t^2 \minus{} 4t \plus{} 44}$

The shortest of these distances is the perpendicular distance from the point to the line

The shortest distance occurs at $ t \equal{} \minus{}\frac{\minus{}4}{2} \equal{} 2$ (using parabola vertex formula)

the distance at $ t\equal{}2$ is
$ \sqrt{4 \minus{} 8 \plus{} 44} \equal{} \sqrt{40} \equal{} 2\sqrt{10}$
[/hide]


limit
logarithms
calculus
integration
trigonometry
calculus computations

$(1) \lim_{x \to \infty}\sqrt[x]{x!}$
$(2) \lim_{x \to \infty}\left(1-\left(\frac{ln(x)}{ln(x+1)}\right)^{x}\right)ln(x)$
$(3) \lim_{n \to \infty}\sqrt[n]{\prod_{k=1}^{n}sin\left(\frac{k\pi}{2n}\right)}$
$(4) \lim_{n \to \infty}\prod_{k=1}^{n}cot^{2}\left(\frac{k\pi}{2n+1}\right)$
$(5) \lim_{n \to \infty}\sum_{k=1}^{n}tan\left(\frac{k}{n^\alpha}\right)$

I found these on SOSMath and can't solve them all for certain. I have a hunch that $(1)$ is divergant, because plugging in stirling's formula yields an infinite answer; $(2)$ I'm pretty clueless since l'hopital is too messy for it; $(3)$ I think you need to convert to series with ln's and make it a reimann sum, i'm not certain but my answer is $2^{-\frac{\pi}{2}}$; $(4)$ and $(5)$ I have no clue other than it looks like a reimann sum

#1 is infinite. Stirling's formula is more than you need.

#2: $\frac{\ln x}{\ln(x+1)}=\frac1{1+\frac{\ln(1+1/x)}{\ln x}}$. Raised to the $x$ power, we have
${\left(1+\frac{\ln(1+\frac1x)}{\ln x}\right)^{-x}=\left(1+\frac1{x\ln x}+O(\frac1{x^{2}\ln x})\right)^{-x}=1-\frac1{\ln x}+O(\frac1{x\ln x}})$.
Subtracting  from 1 and multiplying by $\ln x$, the limit is 1.

#3: You've got the right idea. Work out the details yourself: what integral is the logarithm of the limit?

#4 and 5 are not Riemann sums.

are you sure 2 is right, what about the original 1 added to it and the ln(x) multipied by it?
my solution for 3

$\ln(I) = \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\ln\left(\sin\left(\frac{k\pi}{2n}\right)\right)$
$\ln(I) = \int_{0}^{1}\ln\left(\sin\left(\frac{\pi}{2}x\right)\right)$
$\ln(I) = \frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\ln(\sin(u))du$
$\ln(I) = \frac{2}{\pi}\left(-\frac{\pi}{2}\right)\ln(2) = \ln\left(\frac{1}{2}\right)$
$I=\frac{1}{2}$  :)

may I have a hint for 4 and 5...

3) $\prod_{k-1}^{n}\sin \frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$.

Therefore the limit is $\frac{1}{2}$.

4) $\prod_{k=1}^{n}\sin \frac{k\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^{n}}$

$\prod_{k=1}^{n}\cos \frac{k\pi}{2n+1}=\frac{1}{2^{n}}$

 The product of cotangents equals $\frac{1}{2n+1}$, hence the limit is 0.

5) I assume that $\alpha>1$, otherwive it makes no sense to study such limit. In this case we have

a) $\alpha \in (1,2)$ the limit is $\infty$.

b) $\alpha=2$ the limit is $\frac{1}{2}$.

c) $\alpha >2$ the limit is 0.

how did you convert those trigonometric products to radicals/exponents, can you please explain?

Use the following identity:

$\prod_{k=1}^{n}\left(1-2x\cos \frac{2k\pi}{2n+1}+x^{2}\right)=\frac{x^{2n+1}-1}{x-1}$, then set $x=1$ and $x=i$, and you should get those two identities.

Solve the equation $x^{2n+1}-1=0$, and then observe that

$x^{2n+1}-1=(1-x)\prod_{k=1}^{n}\left(x-e^{\frac{ik\pi}{(2n+1)}\right)\left(x-e^{-i\frac{k\pi}{(2n+1)}\right)}}$.

in #2 whats the problem in applying l´hospital in the
$\frac{ln(x)}{ln(x+1)}$ then you use this result in the formula? you will arrive in complete different solutions ( it will diverge)

There is a much simpler aprouch to #1 :
$n!^{1/n}\geq [(\frac{n}{2}-1)!(\frac{n}{2})^\frac{n}{2}]^{1/n}=\sqrt{\frac{n}{2}}(\frac{n}{2}-1)!^{1/n}\geq \sqrt{\frac{n}{2}}$
This last term, oviously tends to infty


I'm new to latex, and i've been trying it out by myself and its pretty weird to me.  :o  I got like 60 errors, but it still printed out the way i wanted it to. Is that ok?

yea it's ok, when I first started LaTeX i didn't get anything and i keep getting errors

to solve the chronic error problems just keep practicing, you'll get better with each attempt you make.

[quote="stevem"]I assume you mean Warnings as distinct from Errors (errors would stop the compilation). Many are not worth worrying about though it's good to learn what is going wrong.

However, you are right - the final printout is the most important thing, unless you are sending the original tex document to be compiled by someone else.

To learn about the warnings, cut down your document to see if you can see where the problem lies.[/quote]
I have errors but it still compiles (though it looks really bad)...

If you are using TeXnicCenter it will show errors if you haven't got any citations, biography or index files. But these can be ignored if you don't have these things.

The last line is the important one which says:
LaTeX-Result: k Error(s), l Warning(s), m Bad Box(es), n Page(s)
where k, l, m, n are integers.

You should try and fix the k Error(s) first and then the l Warnings. The m Bad Boxes are less important. Again in TeXnicCenter, if you click on the button at the top right marked ->! (or just press F9) it will show you where the error is. A common one is ! Missing $\$$ inserted which means you've missed an $\$$ but LaTeX has sorted it out for you, but it shouldn't be difficult to correct. Other errors must be relatively minor like these as most errors stop the compilation (not all of them as I suggested above :)).

What sort of errors do you get? What editor are you using? Do you want to get rid of the errors?

In texnix, say I type in sin instead of \sin it says 1 error, but it still compiles. Yes I want to fix errors.

That's odd, because if you type sin x you get $sin x$ where it thinks you mean $s\times i\times n\times x$ but it doesn't give an error. So something else must be happening.
[quote="bpms"]Yes I want to fix errors.[/quote]OK then, make a small document with just one of the errors, post it here and we'll see if we can help. Fixing one error may help eliminate others, so it may not mean posting one document for every error.

Well... I guess i'm very lucky im using Texnic Center because i have 260 errors now... and yes, errors. I never learned latex before and don't get how the "grammar" of LaTeX goes. :( All of my errors seem very simple, but i don't know how to fix them at all because i never learned. For example:
Missing $ Inserted.
Missing } Inserted.
Missing \endgroup inserted.
Missing \cr inserted.
Missing { Inserted.
Missing $$ Inserted.
LaTeX Error: Lonely \item--perhaps a missing list environment
Misplaced align tab character &.
Misplaced \cr.
Misplaced \noalign

I think its basically those... repeated about 100x each -____-

At the bottom:
LaTeX-Result: 262 Error(s), 0 Warning(s), 0 Bad Box(es), 2 Page(s)

Fortunately, those errors are minor and easily fixable. But how? Well the errors are an ideal opportunity to learn. When you have learnt how to fix them you'll be so much more confident with LaTeX.

Many problems are to do with 'balancing'. So { needs to be matched with a }. The problem comes when you have a lot of these but here TeXNicCenter is very helpful. Put the cursor just after a { or a } and it will show the one it is matched with and then you can see if it's right.

262 errors is a bit mind-blowing, so start small. Take one of the offending line(s) and put it into a new document, compile, get the errors and see if you can fix it. If you can't post it here and we will help.

You will find, like mathematics, it gets easier to LaTeX the more you practice.

the problem is... i don't know where to start at all since i don't know where to put brackets regularly.  :blush:

I assume your document is quite long given that there are so many errors. So cut out all but the first page and attach the document here or copy and paste the code (including all the stuff before \begin{document}) here.

If you can't do that, then the only option is to attach the complete document.

\documentclass{article}
\begin{document}
\begin{eqnarray*}
\begin{enumerate}
 \item 1. Since BC $\cong$ AC, $\angle$ A $\cong$ $\angle$ B because of the Isosceles Triangle Theorem. Since m$\angle$ B = 43, m$\angle$ A = 43, and all of the angles of a triangle adds up to 180:\\
m\angle C &=& 180 - m\angle B - m\angle A \\
&=& 180 - 43 - 43 \\
&=& 180 - 86 \\
&=& 94 \\
.... truncated.... then in the end...
\end{enumerate}
\end{eqnarray*}
\end{document}

Wow! That gives 52 errors. But you'll be delighted that all the errors can be eliminated in one go.
1. The problem is that the eqnarray environment is [i]only[/i] for equations not for text or lists. So move the eqnarray to be after \item and all the errors vanish!.
2. You do need to add dollar signs for the bits of equations you have following \item as it is no longer covered by eqnarray
3. No need for 1. following \item as LaTeX will add the 1 itself
4. No need for \\ at the end of the \item line (it will give a Bad Box warning otherwise)
5. I have also taken the liberty of adding the degree symbol $^{\circ}$, which you can remove if you don't like it. (Can I ask what the m in m\angle is for?)
This is what you get after the changes[code]\documentclass{article}
\begin{document}
\begin{enumerate}
	\item Since $BC=AC$,  $A=B$ because of the Isosceles Triangle Theorem. Since $m\angle B = 43^{\circ}, m\angle B = 43^{\circ}$, and all of the angles of a triangle adds up to $180^{\circ}$:
	\begin{eqnarray*}
		m\angle C &=& 180^{\circ} - m\angle B - m\angle A \\
		&=& 180^{\circ} - 43^{\circ} - 43^{\circ} \\
		&=& 180^{\circ} - 86^{\circ} \\
		&=& 94^{\circ} \\
	\end{eqnarray*}
	.... truncated.... then in the end...
\end{enumerate}[/code]
Indentation of the code is purely optional (and ignored by LaTeX) but is a good programming habit to acquire as it clearly shows where environments begin and end.

Oh wow.... i thought eqnarray included everything AND equations, not just equations :lol:. THANKS A LOT!!! wow this get rid of so many errors :) No wonder whatever i write comes out as an error...

The m before \angle is a common notation for denoting the measure of the angle.


Find all the values of $a$ and $b$ such that the following system has just one solution:

$xyz+z=a$
$xyz^2+z=b$
$x^2+y^2+z^2=4$

$\{ a;b;x;y;z \} \in \mathbb R$

$xyz+z=a$---(1)
$xyz^2+z=b$---(2)
$x^2+y^2+z^2=4$---(3)
From (1): $xyz=a-z$
Sub. into (2): $(a-z)z+z=b$
That is $z^2-z+(b-a)=0$.
Suppose the system has a unique solution, then
$\Delta=(-1)^2-4(1)(b-a)=0 \implies 4a -4b = -1$---(4)
Besides, $z=\frac 1 2$ because $\Delta = 0$.
Sub. back into (1) and (2):
$\frac 1 2 {xy} +\frac 1 2 = a$ and $\frac 14 {xy} +\frac 1 2=b$
Eliminate xy: $4b-2a = 1$---(5)
Solving (4) and (5), $a = 0$ and $b=\frac 1 4$.

What remains to do is substituting back the values of a , b and z to check whether the system has a unique solution.

hhmmm....what happened to condition (3)?
It's not used   :?

Substitute the values of z, a and b into (1) and (3).
Then solve the equations and see what will happen.
It isn't very interesting, so I didn't complete the work.

Without thinking about luimichael's solution at all, here's mine:

Notice that if $(x, y, z)$ is a solution then so is $(-x, -y, z)$.  These two solutions will be different unless $x = y = 0$.  So our solution must be with $x = y = 0$.  So the system $z = a; z = b; z^2 = 4$ must have a unique solution, so the only choices are $a = b = 2$ or $a = b = -2$.  Checking, however, we see that there are other real solutions when $a = b = 2$, so the only possibility is $a = b = -2$.  (There are also some complex solutions in this case, but no real solutions.)

Solving with luimichael's values of $a, b$ gives a number of solutions.  Someone else will have to figure out where he went wrong.

[b]Yes, JBL ![/b] It is and my solution. Here is some similar examples from a old (romanian) regional mathematical competition :

$1.\blacktriangleright$ [color=darkred]Find all the values of $a\in R$ so that the system
$2^{|x|}+|x|=y+x^2+a\ ;\ x^2+y^2=1$ has just only solution.[/color]
[hide="Here is a indication."] Denote the set $S$ of all solutions.
Then $(x,y)\in S\Longrightarrow (-x,y)\in S$ a.s.o.[/hide] $2.\blacktriangleright$ [color=darkblue]Let $f: [0,1]\rightarrow [0,1]\ ,\ f(x)=\frac{(x+1)(x+a)}{x^2+1}$ be a function.
Ascertain $a\in R$ and prove that the function $f$ is injectively.[/color]
[hide="Here is a indication."] $f(0)=a\in [0,1]$ and $f(1)=a+1\in [0,1]\Longrightarrow a=0$ a.s.o.[/hide] $3.\blacktriangleright$ Let $f: R\rightarrow R\ ,\ f(x)=|x^2-3x+2|+|x-a|\ .$
Ascertain $a\in R$ so that $f(R)\equiv Im\ f=[0,\infty )\ .$

[quote="luimichael"]
Sub. into (2): $(a-z)z+z=b$
That is $z^2-z+(b-a)=0$.
[/quote]

This is his mistake :P


let $ a$,$ b$,$ c$ and $ d$ >=0  and  $ a \plus{} b \plus{} c \plus{} d \leq 5$
$ 5\sqrt [25]{\frac {a(a \plus{} b)(a \plus{} b \plus{} c)(a \plus{} b \plus{} c \plus{} d)}{24}} \plus{} \sqrt [5]{\frac {25bcd(5 \minus{} a \minus{} b \minus{} c)}{(a \plus{} b)(a \plus{} b \plus{} c)(a \plus{} b \plus{} c \plus{} d)} }\leq6$

any one can solve it or what .there is a small malicious  thecnic to do.


geometry
Extremal combinatorics
combinatorial geometry
polygon
right angle
angles
IMO Shortlist

Let $n \geq 5$ be a given integer.  Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles.

[i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Call $ f(n)\equal{}k$. I call $ g(n)\equal{}\lceil\frac{2n\plus{}1}{3}\rceil$, and claim $ f(n)\equal{}g(n)$ for all $ n>5$. Also, $ f(5)\equal{}3$.

To prove this, I will need two lemmas:
1. $ f(n\plus{}3)\ge f(n)\plus{}2$
2. $ f(n)\le g(n)$

Proof of lemma 1. Take any non-right angle ABC is an n-gon with f(n) right angles. There exists such a non-right angle since n>4. If we replace B with two points arbitrarily close to B on AB, BC respectively (call them D and G), and then draw a square DEFG away from the inside of the polygon, then by replacing ...ABC... by ...ADEFGC..., we have an (n+3)-gon with at least f(n)+2 right angles. Note that ...ADEFGC... is not self-intersecting in general since we can make square DEFG arbitrarily small. Hence $ f(n\plus{}3)\ge f(n)\plus{}2$.

Proof of lemma 2. An n-gon with f(n) right angles has sum of non-right angles $ 180(n\minus{}2)\minus{}90f(n)<360(n\minus{}f(n))$. Hence $ f(n)<\frac{2n\plus{}4}{3}$, and $ f(n)\le\lceil\frac{2n\plus{}1}{3}\rceil\equal{}g(n)$, as desired.

According to lemma 2, $ f(n)\le g(n)$ for $ n\equal{}6$, $ 7$, $ 8$. That $ f(n)\equal{}g(n)$ is actually achievable in these cases, and $ f(5)\equal{}3$, can be easily checked. Now assume that $ f(n)\equal{}g(n)$. Then lemma 2 says that $ f(n\plus{}3)\le g(n\plus{}3)$. Also, lemma 1 says that $ f(n\plus{}3)\ge f(n)\plus{}2\equal{}g(n)\plus{}2\equal{}g(n\plus{}3)$. Consequently, $ f(n\plus{}3)\equal{}g(n\plus{}3)$, and we are done by induction on n.

I shall show that when $ n = 5 $ we have $ k = 3 $ and when $ n > 5 $ we have $ k = \left\lceil\frac{2n + 1}{3}\right\rceil $.

Lemma 1: When $ n = 5 $ we have $ k = 3 $

[i]Proof:[/i]  First I shall construct an example where the number of interior right angles is $ 3 $.  Consider a square $ ABCD $.  Let $ E $ be a point not in the interior of $ ABCD $ such that $ AED $ is an isosceles right triangle with hypotenuse $ AD. $  Then the pentagon $ ABCDE $ clearly has $ 3 $ interior right angles as desired.  Now, I shall show that there may not be $ 4 $ interior right angles.  Assume the contrary.  Then since the sum of interior angles must be equal to $ 540 $, we have that the non-right interior angle in the pentagon has angle measure of 180 degrees, contradiction.  Similarly $ 5 $ interior right angles are impossible so in this case $ k = 3 $ as desired.

Lemma 2: For all integers $ n \geq 5 $, we have that $ k \leq \left\lceil\frac{2n + 1}{3}\right\rceil $.

[i]Proof:[/i]  Consider a $ n $-gon with $ k $ interior right angles.  Then the sum of the non-right interior angles is $ 180(n - 2) - 90k $.  But every such angle has angle measure less than 360 degrees so we have the inequality $ 180(n - 2) - 90k < 360(n - k) $ which yields the desired result.

Lemma 3: When $ n = 6 $ we can construct a figure with $ 5 $ interior right angles.

[i]Proof:[/i]  Consider a square $ ABCD $.  Construct points $ E, F, G, H $ all distinct from $ A, B, C, D $ such that quadrilaterals $ ABEF $ and $ BCGH $ are squares.  Then the hexagon $ BEFDGH $ has $ 5 $ interior right angles as desired.

Lemma 4:  When $ n = 7 $ we can construct a figure with $ 5 $ interior right angles.

[i]Proof:[/i]  Consider hexagon $ BEFDGH $ from the $ n = 6 $ case.  Let $ M, N $ be the midpoints of segments $ BE, BH $ respectively.  Then heptagon $ DGHNMEF $ has $ 5 $ interior right angles as desired.

Lemma 5: When $ n = 8 $ we can construct a figure with $ 6 $ interior right angles.

[i]Proof:[/i]  Consider hexagon $ BEFDGH $ from the $ n = 6 $ case.  Let $ M $ be the midpoint of segment $ EF $.  Construct points $ R, S $ not in the interior of the hexagon such that quadrilateral $ FMRS $ is a square.  Then octagon $ DGHBEMRS $ has $ 6 $ interior right angles as desired.

Lemma 6:  For any $ n $-gon with $ k $ interior right angles, we can construct an $ (n + 3) $-gon with $ k + 2 $ interior right angles.

[i]Proof:[/i]  Let the $ n $-gon be denoted $ A_{1}A_{2} \dots A_{n} $.  Assume WLOG that $ \angle{A_{1}A_{2}A_{3}} \ne 90 $.  Let $ X $ be a point distinct from $ A_2 $ on segment $ A_{1}A_{2} $ and let $ Y $ be a point distinct from $ A_2 $ on segment $ A_{3}A_{2} $.  Construct points $ V, W $ not in the interior of the $ n $-gon such that quadrilateral $ XYVW $ is a square.  Then the $ (n + 3) $-gon $ A_{1}XWVYA_{3}A_{4} \dots A_n $ has $ k + 2 $ interior right angles as desired.  By making the distances $ A_{2}X $ and $ A_{2}Y $ sufficiently small we can force the resulting $ (n + 3) $-gon to not self-intersect and so we have the desired result.

The combination of these lemmas clearly proves my initial claim.

$ k_n = \left\lceil\frac{2n + 4}{3}\right\rceil -1$ is the answer. Using sum-of-angles formula, one gets this as upepr bound. I prove by induction that is also lower bound. For $5,6,7$ is easy.

now I prove we can get $n+3$-gon with 2 more 90 angles than $n$-gon. For $n$-gon with $k_n$ 90 angles, there are two consecutive ones since $k_n > n/2$. If you change those two consecutive ones like in the figure, we are done! (Notice we can make the change very close to original figure, so that the polygon is still not self intersects!)

(original polygon is right, result polygon is left)

I claim that $ k_n = \left\lceil\frac{2n+1}{3}\right\rceil - 1$

[b]UPPER BOUND[/b]:
Consider a $n$-gin with $k$ right angles.
Then the sum of the [i]non-interior[/i] angles will be $180(n-2) - 90k$. Since every [i]non-interior[/i] angle is less than 360, we have $180(n-2) - 90k < 360(n-k)$. Our desired conclusion follows from this.

[i]It is trivial to check our claim for $n \leq 8$.[/i]

[b]CONSTRUCTION[/b]:
We will construct by induction.
Suppose we have $k_n$ right angled $n$-gon.
Let $A,B$ be points on the $n$-gon such that $\angle{A} = 90,  \angle{B} < 90$. It is easy to see that such points exist.
Now consider points $C,D$ such that $C$ is outside the $n$-gon and $D$ is on the $n$-gon such that their inclusion into the polygon will add $2$ more right angles. Obviously $2$ such points exist.
Hence we are done by induction.

Let $f(n)$ be the maximum number of right angles for a polygon with $n$ vertices. We claim
\[ f(n) =\begin{cases} 3 &\text{ if } n=5 \\ 2m+1 & \text{ if } n=3m \\ 2m+1 &\text{ if } n=3m+1 \\ 2m+2 &\text{ if } n=3m+2.\end{cases} \]
[color=#00f][b]Proof of Bound[/b][/color] It is well-known that the sum of the internal angles in a polygon (be it non-convex) with $n$ vertices is $180(n-2)$. Every internal angle is less than $360^\circ$, so
\[ 90f(n)+360(n-f(n)) > 180(n-2) \implies f(n) < \frac{2n+4}{3}. \]
Since $f(n)$ is an integer, this implies $f(n) \le \left\lfloor \frac{2n+3}{3} \right\rfloor$.

[color=#00f][b]Construction[/b][/color] The idea is that by the above bound, we want the non-$90^\circ$ angles to be very close to $360^\circ$. We use different constructions based on $n\pmod 3$. Firstly, for $n=5$, it is easy to see that we actually cannot have $\lfloor \tfrac{2\cdot 5+3}{3} \rfloor = 4$ right angles, we can only have 3. For $n\ge 6$, the following constructions work (drawn in red):
[img]
https://lh3.googleusercontent.com/-IvN209oBUzU/YFesdaRhMmI/AAAAAAAAHUk/EQlBQ_VqXsU2i8mmdUybsgRt87milv6agCLcBGAsYHQ/s0/ISL_2003_C3.PNG?authuser=0[/img]
We start with a right angle at the ``wedge'', and draw two segments radially outward. Then mark points along the arc created, and erect squares on each of these segments. Connecting all the points gives the above shape for $n\equiv 0 \pmod3$. For $n\equiv 1 \pmod3$, we don't need any more right angles, so the filler segment is added. For $n\equiv 2 \pmod3$, we add two right angles in the wedge as shown (removing the central right angle), which adds a net of 1 more right angle, as needed.

[hide=Remarks]
Finding the answer is not hard with the proof of the bound, but the constructions are not guaranteed to exist. ``Nice'' looking constructions with staircases and so on are not optimal since the non-$90^\circ$ angles are not as close to $360^\circ$ as possible, which is needed as per our bounds for an equality case.
[/hide]

Consider our exterior angles, which sum to $360^\circ$. Note that each right angle has an exterior angle of $90^\circ$, and each other angle has an exterior angle of less than $-180^\circ$ so an $n$-gon has $k$ right angles then $360-90k< -180(n-k)$ so
$3k<2n+4$ meaning that \[k\le \left\lfloor \frac{2k+3}3 \right\rfloor\]
It is easy to check that for $n=5,6,7,8$ the corresponding $k_{\text{max}}$ are $3,5,5,6$. For higher $n$, we can do something funny like this
[img width=70]https://media.discordapp.net/attachments/1091890356796280852/1094282259256848534/Screenshot_2023-04-08_at_11.26.53.png?width=1898&height=550[/img]
 to get $+3$ sides but $+2$ right angles.

Let $k(n)$ denote the maximum number of right angles in a polygon with $n$ vertices which is not self-intersecting. We claim that, $k(5)=3$ and
\[k(n)=\lceil{\frac{2n+4}{3}}\rceil-1\]
for all $n\geq 6$.

For $n=5$ it is clear that we cannot have $5$ right angles as the sum of internal angles in $540^\circ$. Further, we cannot have $4$ right angles as this implies that the other angle is $180^\circ$ which is a clear contradiction. It is clear that we can have 3 right angles - simply consider a rectangle with a slice cut off from one corner.

Now, consider $n\geq 6$. We first deal with the bound. Say we have $k$ right angles in the polygon. Let $S$ be the sum of internal angles of this $n$ vertex polygon. We count $S$ in two ways.
\[180(n-2)=S < 90k + 360(n-k) \implies k < \frac{2n+4}{3}\]
From this, it follows that,
\[k \leq \lceil{\frac{2n+4}{3}\rceil}-1\]
as desired.

Now, we deal with the difficult part of this problem - the construction. When $n>5$ and $n\equiv 0 \pmod{3}$ ($n=3a+6$ for $a\geq 0$), we simply consider the following construction. We construct a [i]petal[/i] - which is a L shaped structure, with 3 edges and two right angles between them. Further, we draw $n$ of these, and connect the top most edge of the first petal and the bottom edge of the last petal to form a rectangular structure. This requires $3a+6$ edges and forms $2a+5$ right angles. Thus, for all $n>5$ such that $n\equiv 0 \pmod{3}$ we can achieve the desired bound with this construction. For $n\equiv 1 \pmod{3}$ we simply break the edge from the end of the last petal into two edges which have no right angles between them. This achieves the desired bound. When $n\equiv 2 \pmod{3}$ we draw an additional right angle before connecting to the bottom petal adding one more right angle and one more side. Thus, this bound is achievable in all cases and we are done.

[i]This proof is very hard to explain without plenty of diagrams - I add the cases [/i] $n=15,16,17$ [i]to make this clear.[/i]


Let's use military time; it's a lot easier.

Without any lunch break, you simply add $ 8: 00$ to $ 7: 35$, receiving $ 15: 35$.  Adding $ 15$ minutes would result in $ 15: 50$, or $ \boxed{3: 50}$.

I got that too, but the answer given was 4:50.

What if we slightly alter the problem? Changes in bold.

MaryJo must work 8 hours a day at her job [b]as a truck driver[/b]. She begins at 7:35am [b]in St. Louis[/b], and she takes a 15-minute lunch break at 11:30a.m. [b]She ends in Washington D.C.[/b] At what time in the afternoon does her working day end?

That would be a [b]time zone[/b] problem.

You change time zones, adding one hour.

But we don't know if she changed time zones.

[quote]Look at a map of the US. :wink: [/quote]

Mewto substituted those locations in as an example. The actual problem never specified whether her location changes or anything like that.

@Bugi: It's very possible it was a typo.

MaryJo must work 8 hours a day at her job. She begins at 7:35am, and she takes a 15-minute lunch break at 11:30a.m. At what time in the afternoon does her working day end?

Hello! Where is time zone in the original problem?

Bugi we found the answer is 1 hour behind the "correct" answer and were trying to account for the discrepency.


number theory
greatest common divisor
number theory unsolved

Let $a$ be an odd integer and $n,m$ be unequal natural numbers.
Prove that $gcd(a^{2^n}+2^{2^n}, a^{2^m}+2^{2^m})=1$

[quote="ZetaX"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=67642 ...
(substitution $p=a^{2^n} , q=2^{2^n}$ for $n < m$ will be necessary)

Edit:
it's even simpler:
$\gcd(p+q , p^k+q^k)$ is ($k=2^{m-n}$) what we looking for.
Thus: $p^k+q^k \equiv 2p^k \mod (p+q)$, which gives the result since $\gcd(p,p+q)=1=\gcd(2,p+q)$.[/quote]
Thanks very much.I also was thinking the solution was smth. like these methods.,but i couldn't solve.Maybe I could ,if I had stayed thinking on it longer.


Let $ a$,$ b$,$ c$,$ d$, and $ e$ be five consecutive terms in an
arithmetic sequence, and suppose that $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}30$.  Which of the
following can be found?\\
$ \text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e$ \\ \\

Hello
$ c \minus{} 2d' \plus{} c \minus{} d' \plus{} c \plus{} c \plus{} d' \plus{} c \plus{} 2d' \equal{} 30$
Thus $ c \equal{} 6$
Thank u.

You can't use d for common difference as you have it in the terms of the sequence.
Otherwise, correct.


If the election were held today and you could vote, who would you vote for?

This is going to become heated, I'm sure. Try to back up your posts with real evidence and such please - it will make the debate better and more interesting.

cite it??? Please tell me what you mean max.

I was actually thinking of leaving the debate out of this. I just want to see how we stand currently on Bush vs. Kerry. But, if you would like another debate; then go ahead.


Find all function $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that
\[ xf(x) \minus{} yf(y) \equal{} (x \minus{} y)f(x \plus{} y)
\]

An old problem .
Consider $ g(x)\equal{}f(x)\minus{}f(0)$
Then $ xg(x)\minus{}yg(y)\equal{}(x\minus{}y)g(x\plus{}y)$
Take $ y\to \minus{}x$ then
$ g(y)\equal{}\minus{}g(y)$ so $ g(y)$ is an odd function .
Take $ y\equal{}\to \minus{}y$ then
$ (x\plus{}y)g(x\minus{}y)\equal{}(x\minus{}y)g(x\plus{}y)$
So $ g(x)\equal{}ax$
It mean that $ f(x)\equal{}ax\plus{}b$ :wink:


[quote="yedi"]Does there exist an infinite set M of natural numbers which has property that arithmetic mean of any nonempty subset of M is integer?[/quote]
Obvious ,there is  no set satisfy condition .
Suppose $ x_1<...<x_n<....$ is the set satisfy condition .
Consider two set $ (x_{i_1},...,x_{i_k}) ,(x_{i_2},...,x_{i_{k\plus{}1}})$
From the condition we have :
$ k|x_{i_1}\plus{}...\plus{}x_{i_k}$ and $ k|x_{i_2}\plus{}...\plus{}x_{i_{k\plus{}1}}$
Therefore $ k|x_{i_1}\minus{}x_{i_{k\plus{}1}}$
It follows that for all n then $ n|x_i\minus{}x_j$ ,it gives contradiction .
So there is not a set satisfy condition of problem .


[quote="Ihawk007"]Prove that n^4 - 1 is composite for every n > 1.[/quote]

[hide]$ n^4\minus{}1 \equal{} (n\minus{}1)(n\plus{}1)(n^2\plus{}1)$[/hide]


Hi,

Which software did AoPS used for making this forum?

What's the best and most famous forum software out there? I believe the most famous is vBulletin but the best? I am not so sure about that. For one thing, I know College Confidential uses vB.

SMF ;-)

Mr. Vornicu modified phpBB so much that this is nowhere near phpBB so its like not something you can find on teh internet

SMF? huh? I don't get it.

And, how do you know it is phpBB?

I might be giving you the wrong link, but click [url=http://en.wikipedia.org/wiki/Service_Management_Framework]here[/url].

Yes, this is a very heavily modified version of phpBB.


function
analytic geometry
calculus
derivative
real analysis
real analysis solved

let $f:R^n -> R$ be a continuous function. Prove that the set of local maxima of $f$ is countable.

Every local maximum is the maximum of the function in a ball with rational radius centered at a point with rational cordinates :).

You may choose the same ball for many local maxima.

According to [url=http://mathworld.wolfram.com/LocalMaximum.html]this[/url], the local maximum is the value of $f$, not the point where it achieves the maximum. If we assume otherwise, a constant function would be a counterexample, so I think this is what alekk meant.

Ops...

Anyway, I want to refresh some things which was already discussed on the forum.

Let $f:%Error. "mathh" is a bad command.
R\to\mathbb{R}$ be a differentiable function. Prove that the set $\{f(x)\mid f'(x)=0\}$ is a zero measure set.

Really nice?

I think it's Sard's Lemma, but wasn't there also the assumption that the derivative was continuous? :?

It's terrible I cannot find this topic with my proof :(

Here it is:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=neglijable&t=20251[/url]

It looks like $f$ was indeed only differentiable, not $C^1$, but Harazi called it Sard's Lemma, and Sard's Lemma includes $C^1$ in its hypotheses, as far as I know. :?

It seems non one read my proof, so it is possible there is some mistake :?
But in my opinion the $C^1$ condition is superfluous.

No, I don't think so. We may define $f$ to be $0$ on Cantor's set, for instance, and then we can define it as we please, so that it's always $>0$, on each of the countably many maximal inetrvals which form the complement of Cantor's set.

This question was not addressed to you  ;) (I wrote "A very easy additional question...")

[quote="Myth"]A very easy additional question, related to my first understanding of alekk's problem.

Suppose that $f$ is a non-constant on every interval. Is it true that the sets of points of local minima is countable?[/quote]

And another one:

What if we drop the condition that $f$ is non-constant, and replace "local minima" with "strict local minima"?

And again the idea with rational intervals. So answer is yes.


are any of you guys in [url=http://www.firstlegoleague.org]First Lego League?[/url]

Im a designer

I did Lego League a few years ago. The theme when I did it was The Arctic. I mainly was a designer, but since we didn't have much people in our team I helped with some of the programming. Lego League was fun.

It doesn't look to fun or even about legos


[quote="madatmath"]For a given circle $x^2 + y^2 + 2fx + 2gy + c = 0$
 determine and interpret the Power of a point $(x_1, y_1)$
 lying within the circle. The determine part is easy, it's the interpretation
 i'm interested in.[/quote]

If C(-f, -g) is the center of the circle, $(x_1, y_1)$ is P and r is the radius of the circle,
the number $CP^2 - r^2$ is called the power of the point P with respect to the circle.

When CP>r, i.e., the point is outside the circle, its power is positive.
When CP=r, the point is on the circle, its power is zero.
When CP<r. the point is inside the circle, its power is negative.

When the point is outside the circle, draw a tangent to the circle from P and let it touch the circle at A.  Then, $CP^2 - r^2=PA^2$, which shows, the power of the point is square of the length of the tangent when the point is outside.

For the above given equation of circle and the point, the power is given by
$x_1^2 + y_1^2 + 2fx_1 + 2gy_1 + c$

[quote="madatmath"]Interpret it//
 why is the equation that of the circle???[/quote]

That is not the "equation".  It is just a number

C(-f, -g), $r^2 = f^2 + g^2 - c$ and P is $(x_1 ,\,y_1 )$

$CP^2 - r^2 = (x_1 + f)^2 + (y_1 + g)^2 - f^2 - g^2 + c$
simplification will give you the required expression for power of point.


function
calculus
derivative
integration
algebra
polynomial
advanced fields

I was just wondering a few things about unitary transformations and their relationship to the bases of the relevant spaces.

First, let's generally suppose $U: H\rightarrow H$ where $H$ is an infinite-dimensional Hilbert space. Pick an orthonormal basis for the space, say $\{e_n\}_{n=0}^\infty$. I know it is necessarily true that $\{Ue_n\}$ is an orthonormal basis for $H$ as well, but is it necessary that the $e_n$ be eigenvectors of $U$? (It doesn't seem like it should be to me, but I'm not certain how you'd show it, besides by example, and I'm not certain if my example is legit either. I'd like a better understanding, or an explanation of the reason if I'm wrong.)

Example: consider the transformation $Ue_n = \frac{\sqrt{2}}{2}e_n + \frac{\sqrt{2}}{2}e_{n+1}$. Isn't this unitary? Or if not, why not?

This leads to the question: if not all orthonormal bases for $H$ consist of eigenvectors of $U$, then must there exist some orthonormal basis for $H$ which does consist of eigenvectors of $U$?

I'll conclude with an application. It isn't obvious what an orthonormal basis for $L^2(R)$ might be, even though we know such entities exist. But my prof gave us a problem from Stein's text that took us through several steps to prove that

$\left\{\frac{1}{\pi^{1/2}}\left(\frac{i-x}{i+x}\right)^n\frac{1}{i+x}\right\}_{n=-\infty}^\infty$

is an orthonormal basis for $L^2(R)$. This is pretty cool, I've gotta admit. But now here's what I'm interested in -- I could probably calculate this but I haven't tried yet -- what happens when we use this basis and look at the Fourier transform from $L^2(R)$ to itself? Can we get a different basis for $L^2(R)$ from that? Or are these eigenvectors of that transformation? Is there any good way to tell without calculating directly?

$\left\{\frac{1}{\pi^{1/2}}\left(\frac{i-x}{i+x}\right)^n \frac{1}{i+x}\right\}_{n=-\infty}^\infty$

What happens to this sequence of functions under the Fourier transform? Note that each of these functions is $C^\infty(\mathbb{R})$ - in fact, real analytic (but not entire). Furthermore, I'm pretty sure that each derivative is in $L^2(\mathbb{R}.$ That means that each of these functions will have a Fourier transform that vanishes at infinity faster than any power of the variable. (In particular, I would expect them to decay at $\infty$ like some exponential.)

But each of these functions $f_n$ decays slowly at infinity - only as fast as $\frac1x.$ That means that $xf_n(x)\not\in L^2(\mathbb{R}).$ That means that the Fourier transforms fail to even have one derivative in $L^2.$

Summary: this is a sequence of smooth, slowly decaying functions, so its Fourier transforms would be a sequence of non-smooth (probably discontinuous) rapidly decaying functions.

Not eigenvalues.

In general, it is seldom particularly productive to deal with the eigenvalues or eigenvectors of a unitary operator.

This is a little too complex for me (heh heh) but I tried evaluating the Fourier transforms of these functions in Maple and I seem to get

$\widehat{e_n}(\xi) = -2i\sqrt{\pi}P_n(\pi\xi)e^{-2\pi|\xi|}$

When $n < 0$, this is good for $\xi < 0$, and when $n \ge 0$, this works for $\xi > 0$. The integral doesn't make sense when $\xi = 0$, but I suppose you could just define $\widehat{e_n}(0)$ as the limit as $\xi \to 0$ from the appropriate side. I guess here is the discontinuity Professor Merryfield predicted.

A few $P_n$ near 0 are

$P_{-3}(x) = 8x^2 + 8x + 1 \\ P_{-2}(x) = -4x - 1 \\ P_{-1}(x) = 1\\ P_0(x) = 1\\ P_1(x) = 4x - 1 \\ P_2(x) = 8x^2 - 8x + 1 \\ P_3(x) = 32x^3 - 72x^2 + 36x - 3 \\ P_4(x) = 32x^4 -128x^3 + 144x^2 - 48x + 3$

This seems potentially interesting to me. Do these polynomials look familiar to anyone? I don't have a lot of spare time but I'll try to see if there are any interesting recursive relations or anything between them. Avast!

edit: Now that I think about it, this integral is probably best evaluated by the residue method, so I'd just be expanding the $e_n$ into the series around $\pm i$ (depending on $n$) and finding the appropriate coefficient. I'll do it within a few days and see what happens (maybe the nature of the polynomials will be more apparent from that) unless someone else cares to try it first. Have at it, men! And women, too! Yarrrrr!

(sorry, there was a pirate-themed event on campus recently...)


If $ a, b, c \geq 0$ and $ a\plus{}b\plus{}c\equal{}3$ then

$ \frac{2}{3}(ab\plus{}bc\plus{}ca) \leq 1\plus{} \frac{ab}{a\plus{}b\plus{}1}\plus{} \frac{bc}{b\plus{}c\plus{}1}\plus{} \frac{ca}{c\plus{}a\plus{}1}$

[quote="Nak"]If $ a, b, c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$ then

$ \frac {2}{3}(ab \plus{} bc \plus{} ca) \leq 1 \plus{} \frac {ab}{a \plus{} b \plus{} 1} \plus{} \frac {bc}{b \plus{} c \plus{} 1} \plus{} \frac {ca}{c \plus{} a \plus{} 1}$[/quote]
By Hoder's ineq:
$ \left( \frac {ab}{a \plus{} b \plus{} 1} \plus{} \frac {bc}{b \plus{} c \plus{} 1} \plus{} \frac {ca}{c \plus{} a \plus{} 1}\right)( ab(a \plus{} b \plus{} 1) \plus{} bc(b \plus{} c \plus{} 1) \plus{} ca(c \plus{} a \plus{} 1)) \geq (ab \plus{} bc \plus{} ca)^2$

Or $ \frac {ab}{a \plus{} b \plus{} 1} \plus{} \frac {bc}{b \plus{} c \plus{} 1} \plus{} \frac {ca}{c \plus{} a \plus{} 1} \geq \frac {(ab \plus{} bc \plus{} ca)^2}{9}$
Let be $ ab \plus{} bc \plus{} ca \equal{} q$
The inequality is equivalent to:
$ \frac {q^2}{9} \plus{} 1 \geq \frac {2q}{3}$
$ \frac {(q \minus{} 3)^2}{9} \geq 0$ ( obvious right)

Equality occurs iff $ a \equal{} b \equal{} c \equal{} 1$

Ok, how about this?
If $ a, b, c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 2$ then

$ \frac {2}{3}(ab \plus{} bc \plus{} ca) \leq \frac{1}{3} \plus{} \frac {ab}{a \plus{} b \plus{} 1} \plus{} \frac {bc}{b \plus{} c \plus{} 1} \plus{} \frac {ca}{c \plus{} a \plus{} 1}$


geometry
circumcircle
geometric transformation
reflection
rectangle
linear algebra
matrix

IMO 2004 Leader Information and IMO ShortList 2004:

[i]Update: Geometry file with embedded images now![/i]

[b]Links where the problems should be discussed can be found below.[/b]

Orlando Döhring

[quote="ryanchua"]how about the 2003 shortlist? does anyone know where i can get one? :)[/quote]

Try this here: http://www.mathlinks.ro/Forum/viewtopic.php?t=15621

  dg

[quote="Templar"]Try contestants from IMO 2004. They should have done some of them in preparation for IMO.[/quote]It's like the second time I tell you: the shortlists are on the site, right in this very own sub-forum. ;)

The thread was updated as all teams must have been announced to the host until 10th of June.

[quote="nntrkien"]Sorry for the wrong place of my question. I have a problem, I can't see my download dvi file  :?
So can you post a PDF file or please show me the software to read DVI files, my old software is miktex.
Thank you very much.[/quote]
I think pdf files are attached already.

Using orl's Tex files, I extrated the problems manually. Those people who want to try the problems by themselves first might find this file helpful.

Thank you, orl, for uploading the shortlist for us.

[quote="Valentin Vornicu"]Ergh, I was already posting the same file, so I just modified yours.[/quote]

My first joint paper with Valentin.  :D

You are welcome, PengShi. :)

Thanks to Darij for helping me to find the links to the problems that were already on ML.  :)

[b][size=150]Problems on the board: [/size][/b]

[u][size=150]By category:[/size][/u]

 [b]GEOMETRY[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14020]IMO ShortList 2004, geometry problem 1[/url]

[color=red]Let $ABC$ be an acute-angled triangle with ${AB\ne AC}$. The circle with diameter $BC$ intersects the sides$AB$ and  $AC$ at $M$ and $N$, respectively. Denote by $O$ the midpoint of $BC$. The bisectors of the angles $BAC$ and $MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the line segment $BC$.[/color]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30771]IMO ShortList 2004, geometry problem 2, first thread[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24078]IMO ShortList 2004, geometry problem 2, second thread[/url]

The circle $\Gamma$ and the line $\ell$ do not intersect. Let $AB$ be the diameter of $\Gamma$ perpendicular to $\ell$, with $B$ closer to $\ell$ than $A$. An arbitrary point ${C\ne A,B}$ is chosen on $\Gamma$. The line $AC$ intersects $\ell$ at $D$. The line $DE$ is tangent to $\Gamma$ at $E$, with $B$ and $E$ on the same side of $AC$. Let $BE$ intersect $\ell$ at $F$, and let $AF$ intersect $\Gamma$ at ${G\ne A}$. Prove that the reflection of $G$ in $AB$ lies on the line $CF$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33052]IMO ShortList 2004, geometry problem 3[/url]

Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14092]IMO ShortList 2004, geometry problem 4[/url]

[color=red]In a convex quadrilateral $ABCD$ the diagonal $BD$ does not bisect the angles $ABC$ and $CDA$. The point $P$ lies inside $ABCD$ and satisfies

$\begin{matrix*} \angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA. \end{matrix*}$

Prove that $ABCD$ is a cyclic quadrilateral if and only if ${AP=CP}$.[/color]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40116]IMO ShortList 2004, geometry problem 5[/url]

Let $A_{1}A_{2}\dots A_{n}$ be a regular $n$-gon. The points $B_{1},\dots ,B_{n-1}$ are defined as follows:

- If ${i=1}$ or ${i=n-1}$, then $B_i$ is the midpoint of the side $A_iA_{i+1}$;

- If ${i\neq 1}$, ${i\neq n-1}$ and $S$ is the intersection point of $A_{1}A_{i+1}$ and $A_{n}A_{i}$, then $B_{i}$ is the intersection point of the bisector of the angle $A_{i}SA_{i+1}$ with $A_{i}A_{i+1}$.

Prove the equality

$\begin{matrix*} \angle A_1B_1A_n+\angle A_1B_2A_n+\cdots+\angle A_1B_{n-1}A_n=180^\circ. \end{matrix*}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22296]IMO ShortList 2004, geometry problem 6[/url]

Let $\mathcal{P}$ be a convex polygon. Prove that there is a convex hexagon which is contained in $\mathcal{P}$ and which occupies at least 75 percent of the area of $\mathcal{P}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41033]IMO ShortList 2004, geometry problem 7[/url]

For a given triangle $ABC$, let $X$ be a variable point on the line $BC$ such that $C$ lies between $B$ and $X$ and the incircles of the triangles $ABX$ and $ACX$ intersect at two distinct points $P$ and $Q$. Prove that the line $PQ$  passes through a point independent of $X$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39093]IMO ShortList 2004, geometry problem 8[/url]

A cyclic quadrilateral $ABCD$ is given. The lines $AD$ and $BC$ intersect at $E$, with $C$ between $B$ and $E$; the diagonals $AC$ and $BD$ intersect at $F$. Let $M$ be the midpoint of the side $CD$, and let ${N\neq M}$ be a point on the circumcircle of the triangle $ABM$ such that ${AN/BN=AM/BM}$. Prove that the points $E,F$ and $N$ are collinear.

 [b]NUMBER THEORY[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34224]IMO ShortList 2004, number theory problem 1[/url]

Let $\tau(n)$ denote the number of positive divisors of the positive integer $n$. Prove that there exist infinitely many positive integers $a$ such that the equation

$\begin{matrix*} \tau(an)=n \end{matrix*}$

does not have a positive integer solution $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30906]IMO ShortList 2004, number theory problem 2[/url]

The function $\psi$ from the set $\mathbf{N}$ of positive integers into itself is defined by the equality

$\begin{matrix*} \psi(n)=\sum_{k=1}^{n}(k,n),\qquad n\in \mathbf{N}, \end{matrix*}$

where $(k,n)$ denotes the greatest common divisor of $k$ and $n$.

[i]a)[/i] Prove that ${\psi(mn)=\psi(m)\psi(n)}$ for every two relatively prime ${m,n\in\mathbf{N}}$.

[i]b)[/i] Prove that for each ${a\in\mathbf{N}}$ the equation ${\psi(x)=ax}$ has a solution.

[i]c)[/i] Find all ${a\in\mathbf{N}}$ such that the equation ${\psi(x)=ax}$ has a unique solution.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27188]IMO ShortList 2004, number theory problem 3[/url]

A function $f$ from the set of positive integers $\mathbf{N}$ into itself is such that for all ${m,n\in\mathbf{N}}$ the number ${(m^2+n)^2}$ is divisible by ${f^2(m)+f(n)}$. Prove that ${f(n)=n}$ for each ${n\in\mathbf{N}}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33810]IMO ShortList 2004, number theory problem 4[/url]

Let $k$ be a fixed integer greater than 1, and let ${m=4k^2-5}$. Show that there exist positive integers $a$ and $b$ such that the sequence $(x_n)$ defined by

$\begin{matrix*} x_0=a,\quad x_1=b,\quad x_{n+2}=x_{n+1}+x_n\quad\text{for}\quad n=0,1,2,\dots, \end{matrix*}$

has all of its terms relatively prime to $m$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14093]IMO ShortList 2004, number theory problem 5[/url]

[color=red]We call a positive integer [i]alternative[/i] if its decimal digits are alternatively odd and even. Find all positive integers $n$ such that $n$ has an alternative multiple.[/color]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41072]IMO ShortList 2004, number theory problem 6[/url]

Given an integer ${n>1}$, denote by $P_{n}$ the product of all positive integers $x$ less than $n$ and such that $n$ divides ${x^2-1}$. For each ${n>1}$, find the remainder of $P_{n}$ on division by $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=258097#258097]IMO ShortList 2004, number theory problem 7[/url]

Let $p$ be an odd prime and $n$ a positive integer. In the coordinate plane, eight distinct points with integer coordinates lie on a circle with diameter of length $p^{n}$. Prove that there exists a triangle with vertices at three of the given points such that the squares of its side lengths are integers divisible by $p^{n+1}$.

[b]ALGEBRA[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14091]IMO ShortList 2004, algebra problem 1[/url]

[color=red] Let ${n\ge 3}$ be an integer and ${t_1,t_2,\dots,t_n}$ positive real numbers such that

$\begin{matrix*} n^2+1>(t_1+t_2+\cdots +t_n)\left(\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n}\right). \end{matrix*}$

Show that $t_i,t_j,t_k$ are the side lengths of a triangle for all $i,j,k$ with ${1\le i<j<k\le n}$.[/color]

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24075]IMO ShortList 2004, algebra problem 2[/url]

 An infinite sequence ${a_{0},a_{1},a_{2},\dots }$ of real numbers satisfies the condition

$\begin{matrix*} a_{n}=|a_{n+1}-a_{n+2}| \end{matrix*}$

for every ${n\geq 0}$, with $a_{0}$ and  $a_{1}$ positive and distinct. Can this sequence be bounded?

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30462]IMO ShortList 2004, algebra problem 3[/url]

 Does there exist a function $s\colon \mathbf{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer.

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14021]IMO ShortList 2004, algebra problem 4[/url]

[color=red] Find all polynomials $P(x)$ with real coefficients which satisfy the equality

$\begin{matrix*} P(a-b)+P(b-c)+P(c-a)=2P(a+b+c) \end{matrix*}$

for all triples $a,b,c$ of real numbers such that ${ab+bc+ca=0}$.[/color]

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30865]IMO ShortList 2004, algebra problem 5[/url]

 Let ${a,b,c>0}$ and ${ab+bc+ca=1}$. Prove the inequality

$\begin{matrix*} \root3\of{\frac 1a+6b}+\root3\of{\frac 1b+6c}+\root3\of{\frac 1c+6a}\leq \frac{1}{abc}. \end{matrix*}$

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33734]IMO ShortList 2004, algebra problem 6[/url]

 Find all functions ${f\colon\mathbf{R} \rightarrow\mathbf{R}}$ satisfying the equation

$\begin{matrix*} f\bigl(x^2+y^2+2f(xy)\bigr)=\bigl(f(x+y)\bigr)^2 \end{matrix*}$

for all ${x,y\in \mathbf{R}}$.

 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41115]IMO ShortList 2004, algebra problem 7[/url]

 Let ${a_1,a_2,\dots,a_n}$ be positive real numbers, ${n>1}$. Denote by $g_n$ their geometric mean, and by $% A_1,A_2,\dots,A_n
$ the sequence of arithmetic means defined by

$\begin{matrix*} A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \end{matrix*}$

Let $G_n$ be the geometric mean of $A_1,A_2,\dots,A_n$. Prove the inequality

$\begin{matrix*} n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \end{matrix*}$

and establish the cases of equality.

[b]COMBINATORICS[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41116]IMO ShortList 2004, combinatorics problem 1[/url]

There are $10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold:

[i]i.)[/i] Each pair of students are in exactly one club.

[i]ii.)[/i] For each student and each society, the student is in exactly one club of the society.

[i]iii.)[/i] Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is
in exactly $m$ societies.

Find all possible values of $k$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=152742#152742]IMO ShortList 2004, combinatorics problem 2[/url]

Let ${n}$ and $k$ be positive integers. There are given ${n}$circles in the plane. Every two of them intersect at two distinct points, and all points of intersection they determine are pairwise distinct. no three have a point in common. ach intersection point must be colored with one of $n$ distinct colors so that each color is used at least once and exactly $k$ distinct colors occur on each circle. Find all values of ${n\geq 2}$ and $k$ for which there exists such a coloring is possible.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40115]IMO ShortList 2004, combinatorics problem 3[/url]

The following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer ${n\ge 4}$, find the least number of edges of a graph that can be obtained by repeated applications of this operation from the complete graph on $n$ vertices (where each pair of vertices are joined by an edge).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41119]IMO ShortList 2004, combinatorics problem 4[/url]

Consider a matrix of size $n\times n$ whose entries are real numbers of absolute value not exceeding 1, and the sum of all entries is 0. Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40197]IMO ShortList 2004, combinatorics problem 5[/url]

Let $N$ be a positive integer. Two players $A$ and $B$, taking turns, write numbers from the set $\{1,\dots ,N\}$ on a blackboard. $A$ begins the game by writing 1 on his first move. Then, if a player has written $n$ on a certain move, his adversary is allowed to write ${n+1}$ or $2n$ (provided the number he writes does not exceed $N$). The player who writes $N$ wins. We say that $N$ is of type $A$ or of type $B$ according as $A$ or $B$ has a winning strategy.

[i] (a)[/i] Determine whether ${N=2004}$ is of type $A$ or of type $B$.

[i] (b)[/i] Find the least ${N>2004}$ whose type is different from the one of $2004$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30903]IMO ShortList 2004, combinatorics problem 6[/url]

For an ${n\times n}$ matrix $A$, let $X_{i}$ be the set of entries in row $i$, and $Y_{j}$ the set of entries in column $j$, ${1\leq i,j\leq n}$. We say that $A$ is [i]golden[/i] if ${X_{1},\dots ,X_{n},Y_{1},\dots ,Y_{n}}$ are distinct sets. Find the least integer $n$ such that there exists a ${2004\times 2004}$ golden matrix with entries in the set ${\{1,2,\dots ,n\}}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14023]IMO ShortList 2004, combinatorics problem 7[/url]

[color=red]Determine all $m\times n$ rectangles that can be covered with "hooks" made up of $6$ unit squares, as in the figure. See the hook below.

Rotations and reflections of hooks are allowed. The rectangle must be covered without gaps and overlaps. No part of a hook may cover area outside the rectangle. [/color]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41118]IMO ShortList 2004, combinatorics problem 8[/url]

For a finite graph $G$, let $f(G)$ be the number of triangles and $g(G)$ the number of tetrahedra formed by edges of $G$. Find the least constant $c$ such that

$\begin{matrix*} g(G)^3\le c\cdot f(G)^4 \end{matrix*}$

for every graph $G$.

[b]Remark: ISL problems G1, G4, N5, A1, A4, C7 were chosen for IMO.[/b]

Orlando Döhring

[quote="N.T.TUAN"]ORL hasn't solution of this problems? :lol:[/quote]:huh: The solutions have been posted! The problems were also (most of them) solved before on the Forum.

Thank you very much for posting and editing this shortlist problems, orl and Peng. :omighty:


Let A_n=(2^(2n))(2^(2n+1) -1)
Find the last digit of A_n and the digit which lies before the last digit of A_n.

If n is odd A_n ends by 28
If n is even A_n end by 96, 16, 36, 56, 76, 96 ... periodically.
It can be proven by induction.

Namdung


P.1
find all k satisfying:for a,b,c>0 abc=1 then f(a)+f(b)+f(c)>=3f(1)
P.2
Find all k satisfying:fora,b,c>0 abc=1then f(a)+f(b)+f(c)<=3f(1)
f(x)=(1+x)^(-k)

Let $a,b,c$ be positive reals with $abc=1$.

Let $f(x) = \frac{1}{(1+x)^k}$.
---

If $k$ is such that $f$ is increasing and convex, or decreasing and concave, then by Jensen's /AMGM we are done.

This happens when $k$ is negative, and it is easy to find appropriate range for $k$ (I am lazy)

--

Now, if $k$ is positive, we have http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32031  .


Balance the redox reaction:

NO2- + Al(s) goes to NH3 (aq) + Al(OH)4-

Start by splitting it up into its half-reactions.
NO2- + H2O -> NH3
Al + H2O -> Al(OH)4-


[tex]\displaystyle \mbox{Find all ordered triples of primes (p, q, r) such that}[/tex]

[tex]\displaystyle p \ | \ q^r+1, \ q \ | \ r^p+1,  \ r \ | \ p^q+1[/tex]

assume p,q,r > 2
what we are given can be rewriten as:
q^r = -1 (mod p)
r^p = -1 (mod q)
p^q = -1 (mod r)

which yields
q^(2r) = 1 (mod p)
r^(2p) = 1 (mod q)
p^(2q) = 1 (mod r)

fermat's little theorem says that (because p,q,r are prime)

q^(p-1) = 1 (mod p)
r^ (q-1) = 1 (mod q)
p^(r-1) = 1 (mod r)

because q^(2r) = 1 (mod p) and q^r = -1 (mod p), we know that the order of q modulo p is either 2r or 2. similarly, the order of r modulo q is 2p or 2 and the order of p modulo r is 2q or 2.

note that if the order of q modulo p is 2r, then:
2r | p-1  (implying that 2r <= p-1)
while if the order of q modulo p is 2,
q^2 = 1 (mod p) =>  q = +/-1 (mod p).  (implying p<=q+1).

if the order of q modulo p is 2r,  the order of r modulo q is 2p and the order of p modulo r is 2q, then we get a contradiction by substituting into the inequalities.
similarly, if the order of q modulo p is 2, the order of r modulo q is 2, and the order of p modulo r is 2, then we get another contradiction by substitution.

so, we must have one that has order 2 and another that has order 2*prime.
WLOG, let q have order 2 modulo p and let r have order 2p modulo q. then we have
2p<=q-1
p<=q+1
yielding our desired contradiction.

so, if p,q,r>2, there are no solutions.  i haven't been able to find any if we allow some of them to be two, but i don't have a proof of that quite yet.

so, we want to consider the case when at least one of the primes is 2.

WLOG let p=2. we can write the system as:

q^r = -1 (mod 2)
r^2 = -1 (mod q)
2^q = -1 (mod r)

the first equation implies q is odd. because q is odd, we can write the third equation as:

(-2)^q = 1 (mod r), so the order of r-2 modulo r is q or 1.

if the order of r-2 modulo r is 1, then -2 = 1 (mod r), so r=3. now, the second equation says that q divides r^2+1 = 10. recalling that q is odd, we see that q=5. this yields the triplet (2,5,3), as JBL earlier posted.

now, if the order of r-2 modulo r is q, we have that q|r-1 by fermat's little theorem.this implies that q|(r^2-1). however, the second equation implies that q|(r^2+1). combining these two results, we see that q|2. contradiction, as q is odd. no solutions for this case.

thus, the only solution to the given problem is (2,5,3).


Let $ ABC$ be a equilateral triangle.Find all point $ M$ in this triangle such that $ \angle{MAB}\plus{}\angle{MBC}\plus{}\angle{MCA}\equal{}90^o$

[quote="chien than"]Let $ ABC$ be a equilateral triangle.Find all point $ M$ in this triangle such that $ \angle{MAB} \plus{} \angle{MBC} \plus{} \angle{MCA} \equal{} 90^o$[/quote]
It has already been posted before in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=123822]Geometric locus[/url].

Kostas Vittas.


geometry
rectangle
parallelogram
geometry unsolved

A circle can be inscribed in the convex quadrilateral $ ABCD$.  The incircle touches the sides $ AB, BC, CD, DA$ at $ A_1, B_1, C_1, D_1$ respectively.  The points $ E, F, G, H$ are the midpoints of $ A_1B_1, B_1C_1, C_1D_1, D_1A_1$ respectively. Prove that the quadrilateral $ EFGH$ is a rectangle if and only if $ A, B, C, D$ are concyclic.

Why is this problem still unsolved?

Let $ X \equal{} AC\cap BD$ and $ Y \equal{} A_1C_1\cap B_1D_1$

Then it is well known that $ X \equal{} Y$

We also known $ EFGH$ is always parallelogram.

Now $ EFGH$ is rectangel $ \Leftrightarrow$ $ A_1C_1\bot B_1D_1$.

Let $ T \equal{} AB\cap CD$

Now angle chase ...

Since $A_1C_1||EF$ and $B_1D_1||EH$, we have:
$\angle BAD=\pi-2\angle AA_1D_1=\pi-2\angle A_1B_1D_1=\pi-2\angle A_1EH$
$\angle BCD=\pi-2\angle CB_1C_1=\pi-2\angle B_1A_1C_1=\pi-2\angle B_1EF$
Suppose $EFGH$ is a rectangle. So we have $\angle FEH=\frac{\pi}{2}$, therefore:
$\angle A_1EH+\angle B_1EF=\frac{\pi}{2}\Leftrightarrow-2\angle A_1EH-2\angle B_1EF=-\pi\Leftrightarrow$
$\Leftrightarrow(\pi-2\angle A_1EH)+(\pi-2\angle B_1EF)=\pi\Leftrightarrow\angle BAD+\angle BCD=\pi$
Therefore $ABCD$ is cyclic, QED.
Suppose $ABCD$ is cyclic. So we have $\angle BAD+\angle BCD=\pi$, therefore:
$\angle BAD+\angle BCD=\pi\Leftrightarrow(\pi-2\angle A_1EH)+(\pi-2\angle B_1EF)=\pi\Leftrightarrow$
$\Leftrightarrow-2\angle A_1EH-2\angle B_1EF=-\pi\Leftrightarrow\angle A_1EH+\angle B_1EF=\frac{\pi}{2}$
Therefore $\angle FEH=\frac{\pi}{2}$ and analogously $\angle EFG=\angle FGH=\angle EHG=\frac{\pi}{2}$, therefore $EFGH$ is a rectangle, QED.

Invert about the inscribed circle. Now $A$ inverts to $E$, $B$ to $F$, $C$ to $G$ and $D$ to $H$.
$A,B,C,D$ are cyclic if and only if $E,F,G,H$ are cyclic. Also note that $E,F,G,H$ is a parallelogram by Varigon's Theorem. The rest is trivial. :)

[quote="genxium"]$ABCD$ concyclic suffices that, $\angle A +\angle C=\pi , $ suffices $A_1C_1 \perp B_1D_1$,  suffices $EFGH$ is a rectangle.[/quote]
why A1C1 perpendicular to B1D1

I think it's because $\angle C_1D_1B_1 +\angle D_1C_1A_1 = \frac{\pi}{2},$ derived by $\angle A +\angle C=\pi$. BTW, the derivation can be easily reversed so that I didn't state the inverse version.


Let $ m_1,m_2,m_3,\dots,m_n$ be a rearrangement of the numbers $ 1,2,\dots,n$. Suppose that $ n$ is odd. Prove that the product
\[ \left(m_1\minus{}1\right)\left(m_2\minus{}2\right)\cdots \left(m_n\minus{}n\right)\]
is an even integer.

We need to prove that at least one of the factors is even. Suppose the contrary, then there are more even than odd numbers in 1,2,...,n, but that contradicts the n being odd.

Those factors sum up to zero. Since there's an odd number of them they can't all be odd.

[quote="apratimgtr"]how you are getting these historic inmo problems in italy?Even we indians do not have them.[/quote]

I have find them on the WEB  :)

[quote="RAGHAV GROVER"]Where tell us the link :?:[/quote]

The link is

http://madanpur.in/inmo/inmo.htm

[quote="Maths Mechanic"]Where tell us the link :?:[/quote]
very good


Let we call a natural number "good number" which can be divided by every positive number that is equal or less than its square root.Find the greatest good number.

Maybe this will help a little.
Let $n$ be a good number and $m = [\sqrt{n}]$
Let $r_1$ be the biggest positive integer such that: $p_1^{r_1} \leqslant m$ for some prime $p$
Show $p_1^{r_1}p_2^{r_2}...p_k^{r_k} \geqslant (m+1) > n$ for big enough m ($p_k$ is the biggest prime such that $p_k \leqslant m$), so $p_1^{r_1}p_2^{r_2}...p_k^{r_k}$ can't divide $n$.

simple.
n is divisible by a=[  n] and b=[ n]-1 and thus by their product.
hence either n=ab or n>=2ab>2(  n -1)(  n -2)>n so long as n - 6  n +4 >0 which is satified once n>36.
Now if n=ab. then [  n -2] needs to be either one or two. (why?) hence implying a set of answers which are easy to obtain.

simple.
n is divisible by a=[   :sqrt: n] and b=[  :sqrt: n]-1 and thus by their product.
hence either n=ab or n>=2ab>2(  :sqrt:  n -1)(  :sqrt:  n -2)>n so long as n - 6   :sqrt: n +4 >0 which is satified once n>36.
Now if n=ab. then [   :sqrt: n -2] needs to be either one or two. (why?) hence implying a set of answers which are easy to obtain.

Here's another hint: it's the title of a popular show on Fox...


Given f(x) = (x-1) (x-2) (x+1), find c so that the only zero of f(x) + c is -2. Assume that c is a constant

i see your a new member so "WELCOME TO AOPS" it this  problem for a homework assignment?


MIT college Princeton Stanford search Duke blogs

I've always [b]dreamt of MIT[/b] and want to go there more than anything in the world (even if  I get Caltech or Princeton or Stanford etc.)I live in India and somebody told me just 2 students are going to be taken from india this year.Is this true?????????

I've come among the [b]top 40 Young Scientists in India[/b] under the KVPY Scholarship conducted by the [b]IIT's[/b](the best and respected worldwide)and the IISc , BAngalore in India(it';s conducted by the govt.).I'm also a National Talent search Scholar and have also been selected for the Singapore International Airlines Scholarship.I've always been top in class and scored 94% in ALL India Board CBSE exams.(regarding very high)
Ive also got 2280 in SAT(new)(800-math CR-700 :(  and weiting - 780 (10/12))and am quite confident about SAT II's i'll give next year.I'm also expecting near 100% in TOEFl
I was also adjudged the best All round student in my school in Grade 10.(i had to leave that scholl coz it did'nt have classes beyond grade 10 I go to another school now)

Yet that 2 students taking thing is really a blow to me .Can u guys help me out.This is  a long post but plz rate my chances and what else you think need to be done to reach my goaL.I'd like to know the achievements etc. of other international students who have got in.

[b]THANK YOU[/b]

[quote="xxreddevilzxx"]First of all, please do not double post. It is pretty anoying, and everyone is pretty annoyed by the possible spamming (not that you have done it).

Second, what do you mean by "yet that 2 students taking thing is really a blow to me"[/quote]

Regarding double posts I did not do it on purpose. It happened without my knowledge and I have not been able to correct it (I've been trying so for the past 2 hours)> (I didn't know where to get help)
Any ways I'm really[b] sorry [/b]for any inconvenience caused and want some help to correct it. [b]Thanks!!![/b]
The "2 students taking thing" refers to the fact that someone in India told me that just 2 indian national will probably be admitted in fall 2006.I wanted to know your opinion on it and want to know what u make of my CV

[b]Thanks again[/b]

[quote="Rushil"]I've always [b]dreamt of MIT[/b][/quote]I'm glad to hear that.

[quote="Rushil"]I've come among the [b]top 40 Young Scientists in India[/b] under the KVPY Scholarship conducted by the [b]IIT's[/b](the best and respected worldwide)and the IISc , BAngalore in India(it';s conducted by the govt.).I'm also a National Talent search Scholar and have also been selected for the Singapore International Airlines Scholarship.I've always been top in class and scored 94% in ALL India Board CBSE exams.(regarding very high)
Ive also got 2280 in SAT(new)(800-math CR-700 :(  and weiting - 780 (10/12))and am quite confident about SAT II's i'll give next year.I'm also expecting near 100% in TOEFl
I was also adjudged the best All round student in my school in Grade 10.[/quote]that is very impressive. However, you [b]cannot[/b] get into MIT just from the stats. Especially for international students, I've seen IMO medalists and/or SAT 1600s getting rejected. There are way too many well-qualified students and not enough spots to accept them all. So, what you need to do will be - just basic - "do what you love, love what you do" (quote by someone whom I admire alot  :D )

Anyway, I wish you the very best of luck. If you think it is possible, it will be(that is from experience!!) And if you have any further questions, you can PM me - I'll be more than happy to answer them.

To correct a double post you simply delete the post without any replies. I'm pretty sure each user can always delete a post without replies on any thread on any forum on this site. Now that both threads have replies   :o   I'll see what I can do with moderator powers to keep only one thread active.

ok, so from the MIT website [url=http://admissions.mit.edu/AdmissionsWeb/appmanager/AdmissionsWeb/Main;jsessionid=CGWc4TBQwSBS8vhbNgQsNBv2bZhMC3B4Bh2JPVtknJvFvr6YbWzx!774035993!453319266?_nfpb=true&_pageLabel=pageWhosAtMIT]2004 MIT admissions stats[/url]
10,464 students applied
1,664 admitted
...
2165 international students applied
104 admitted

these numbers show that international students are at a disadvantage.
in total, about 7% of the MIT undergrad population is international students. it's pretty hard to get in as international student (i know only one person got in from egypt this year), but as somone said, if you don't apply you make sure you dont' get in. just do your best and don't let go of your goal :starwars:

are puertoricans considered international students??
i don't know but for puertoricans its not that hard to get into mit.. i would say tha half of my friends that have applied to mit have benn accepted.. and with no exceptionals sats (1300 and higher) or big things.. can somebody explain me why this happens in my country?

Puerto Ricans are an American minority, and thus it's supposedly easier for them to get in
[quote]"To that end, the mission of the Office of Minority Education is to provide effective academic enrichment programs to enhance matriculation, promote higher retention and greater excellence in underrepresented minority (African American, Mexican American, Native American and Puerto Rican/Hispanic) students' academic and general educational achievements, and to encourage their pursuits of higher degrees and professional careers. OME's mission embrace a strategy to address academic and graduation gaps between underrepresented minority and non-minority students on MIT campus."  [/quote]

oh i see...
so im glad to be a minority, i think .. heh

Hello Rushil

To be honest your CV was quite impressive. But...
I know 3 people who went/are going to MIT. All of them participated at some kind of International Olympiad, and had different succes(Bronze, Silver). Their SATs/TOEFL were not perfect, and I'm sure that there are other things they are not the best at. Yet, they got to MIT!  It is impossible to say for sure why did they get in. I can only say that  these people are really nice,  smart, interesting, hard working, AND, they were lucky. (now it's important to realise that luck is an important factor, but not the most important.) So, the best thing to do is be yourself, do your best, tackle the application process carefully and... hope.

Ah, one more thing: As far as I know MIT tries to select people from different regions of our planet,  consequently it is not very probable that they will admit many students from the same region in the same year.

And the last thing: Remember, MIT is not the only option, life goes on, and maybe not going to MIT might be better on the long run, only God knows.
Good Luck!

no MIT doesn't have quotas like that. If say 4 people are well qualified from ur region, they will very likely take all 4 of them.

Since you're international, you do need to get national level recognitions to stand you out. It's a pretty tough competition, but don't stress over it too much. Just be yourself in your application and show who you're as a person behind the numbers. If they don't want you for who you are, then that's probably not a school you would want to go to anyway.

Best of luck  :)

Rushil  -  FWIW (Personal views follows – some statements, obviously, could be inaccurate as they are opinions of mine – so take for what ever its worth – and check out the facts)

MIT website  is quite good  so please visit it, you will  find  lot  a of useful  information there.

Also in a PM, I  sent you my son's mit e-mail address,. You may    ask him (or others you know ) if you have any specific questions. – I have also included  my perspective there, hope that is helpful.

As you already know,  unlike  graduate students,  there are relatively small number of Indian Undergraduate students at MIT. (If you do not count American-Indians or  those who live in US – there are indeed a small number of foreign undergraduate students from India) . Guess , like most parents, Indian parents also do not want to send their kids far away!

Another factor is financial consideration.  There  are no “merit scholarships” for undergraduates in MIT . All scholarships are considered  “financial aid” . One way it is good, that is, if you are admitted, you are  provided with  all the financial “need”. (What they think you need may be different from  what you – or your parents- think you need )  This also means , in most cases, if you have any outside scholarship, unless it covers all, is not likely to make any difference in how much you/your parents have to spend.

You may like to check out some other schools, similar to MIT .  Duke has excellent ( eg AB Duke scholarship) scholarships and incentives (they seriously recruit math stars) to attract good students. (Last year’s IMO participant from India  went to Duke) from  all over   the world.  From what I know, math  (physics too) is quite good at Duke.

Cal-Tech also has good scholarships to attract really  good students and I know some Indian students who are  happy there ( without scholarships, they could not have gone there)

There are many other places in US    which you or others may like  to check out.  Rice,  Washington University in St Louis come to mind. Many other schools (Eg University of Indiana ( Wells scholarship) , University of Southern California (with full tuition scholarship) and many others have  prestigious “scholarship” programs which apart from full tuition also provide many goodies (Eg    programs in Oxford/Cambridge etc..) .. (For example, if you a “scholar”  they may even  pay for you to go to MIT   (or school of your choice) for a semester or two..)

 Of course, I certainly think MIT is an excellent school so I am sorry to
  talk about other schools here but  I do  hope  that this information is useful  Best of luck.  When  (which year) do you think you  are applying to college?

You  may already know this, but apply early action to MIT (also other shcools which take early action) and you may  hear back from them in  December  05.  Even the schools which  formally do not have early action, applying (specially for foreign students) as early as possible is good because some such  schools will let you know  (some time unofficially) if you got admitted or not if you ask.

Added later: I think somebody already answered it but anyway if you have not gotten the answer to this.
[quote]somebody told me just 2 students are going to be taken from india this year.Is this true?????????
[/quote]
I don't think there is any truth in that.    There is no quota  etc.

I checked but MIT doesn't allow international students to apply Early action...

(This is my first post on AoPS!)

To clear up a few things:
* We consider all international applicants (those who do not possess US citizenship or a Green card) together during one process, apart from both the early action and regular action pools.
* Applying early action does not give an advantage over applying regualar action; we treat each pool using the same criteria.
* Being an international applicant (again: those who do not possess US citizenship or a Green card) are (unfortunately) at somewhat of a disadvantage, as we are quota-limited in the number of international students we may admit each year.
* We have no quotas by country, but we try to admit students from many different countries, meaning it is unlikely that any country would have a large number (>10) of admitted students.
* Puerto Rican applicants are US citizens, and hence are not considered international applicants.
* We will notify early action applicants in December, and regular action applicants (including international applicants) in March.

For more information:
* I'll be hosting an MIT Admissions [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url] here on Wednesday, October 12th.
* I have a blog which discusses admissions, MIT, and MIT admissions at [url]http://matt.mitblogs.com[/url], and there are more admissions & financial aid officer blogs as well as student blogs at our website, [url]http://my.mit.edu[/url].
* I'll try to visit the College forum here from time to time as well...

Hi Geehoon and Maria!  Look forward to seeing you on campus.

-- Matt McGann
Assistant Director of Admissions, MIT

Matt McGann: Wecome to the Forum! Thanks ! Hope you keep us honest.

They've got about a 10% acceptance rate as someone pointed out.

[ [i]Moderator edit: Here on the College Forum we try to be polite and constructive, as befits college-bound or college-attending young people.[/i]  ]

Hey Cyberman, you might want to tone down the language.  You comments are quite offensive...

[quote="MattMcGannMIT"](This is my first post on AoPS!)

To clear up a few things:
* We consider all international applicants (those who do not possess US citizenship or a Green card) together during one process, apart from both the early action and regular action pools.

. . . .

For more information:
* I'll be hosting an MIT Admissions [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url] here on Wednesday, October 12th.
* I have a blog which discusses admissions, MIT, and MIT admissions at [url]http://matt.mitblogs.com[/url], and there are more admissions & financial aid officer blogs as well as student blogs at our website, [url]http://my.mit.edu[/url].
* I'll try to visit the College forum here from time to time as well. [/quote]

Welcome aboard. It's good to see you here.

I'm guessing this is a G forum.

Use Google.

[quote="Treething"]Dude, cyberman, only I use that language.[/quote]

 :rotfl:  :rotfl:  :rotfl:

thnx to Rushil who started this forum and others who are posting very valuable comments regarding MIT.this forum is worth it :lol:

Hello Matt. It's indeed nice to have you on this forum. I hope I can gain a lot from you and interact with you here besides on your wonderful blog at MIT's website.

There's just one thing. I am an international student and as you would know more than ANYONE else, the competiton for international admission is THE toughest.However, I couldn't find much info either on the MIT site or on your blog regarding international undergrad admissions.

SO I set up this forum mainly with the purpose of getting info , tips and guidelines regarding international undergrad admissions. I know that you WILL NOT divulge much. I do support that for I am strong in my ethics and believe in such matters. But anyways, I'd really appreciate if you can offer some tips.

Thanks!!!!


1) If $ \lim\sup_{n\rightarrow\infty}a_{n}\equal{}\minus{}\infty$, show that $ (a_{n})$ diverges to $ \minus{}\infty$.  If $ \lim\sup_{n\rightarrow\infty}a_{n}\equal{}\infty$, show that $ (a_{n})$ diverges to $ \infty$.

2) If $ (a_{n_{k}})$ is a convergent subsequence of $ (a_{n})$, show that $ \lim\inf_{n\rightarrow\infty}a_{n}\leq\lim_{k\rightarrow\infty}a_{n_{k}}\leq\lim\sup_{n\rightarrow\infty}a_{n}$.

^^I know how to prove that there is a subsequence that converges to the limit inferior and the limit superior but I'm not sure how to show that these are bounds on the sequence.

definitions?

$ \lim_{n \to \infty} a_n \equal{} A \in [\minus{}\infty,\infty]$
  iff
 $ \lim \inf a_n \equal{} A \equal{} \lim \sup a_n$

$ \minus{}\infty \le  \inf_{n \ge N} a_n \le a_N \le \sup_{n \ge N} a_n \le \infty$

$ \inf_{n \ge N} a_n \le \inf_{n_k \ge N} a_{n_k}$

$ \sup_{n \ge N} a_n \ge \sup_{n_k \ge N} a_{n_k}$

Thanks for that one.  I really need the first one.  I can prove the first part, but when I turn around the proof for the second part, the logic fails.

first question is incorrect as stated, or at least the second part is.

for the second question, if $ \lim_{k\to\infty} a_{n_k} < \liminf_{n\to\infty} a_n$, then there is $ \varepsilon > 0$ such that for all $ k\geq K$, $ a_{n_k} \plus{} \varepsilon < \liminf_{n\to\infty} a_n$.  does this lead to a contradiction?

Yes, it's a subsequence that diverges.  Forgot to type it apparently.

it's still wrong.  if $ \liminf_{n\to\infty} a_n \equal{} \infty$, then you can conclude that $ a_n\to\infty$, but if $ \limsup_{n\to\infty} a_n \equal{} \infty$ then you can't (what if, for example, $ a_n\equal{}0$ for even $ n$ and $ a_n\equal{}n$ for odd $ n$?)

How is it wrong now? There is a subsequence that converges to $ \infty$

i guess i misunderstood his last post.  if he is saying that it should be 'if $ \limsup_{n\to\infty} a_n\equal{}\infty$, [i]then there is a subsequence diverging to $ \infty$[/i]', then of course it's fine.

Yes, I was referring to the limit superior part.  And if $ \lim_{n\rightarrow\infty}\inf a_{n}\equal{}\minus{}\infty$, then there is a subsequence that diverges to $ \minus{}\infty$.


50. 1/3e^-3t=0.9 I divided everything by 1/3 and I got- e^-3t=2.7 and for my answer I got t=-.3311

52. 4e^t-1=4 I divided everything by 4 and I got e^t-1=1 and then I got t-1=ln1 and the answer came out to be 1.

54. 12-e^0.4=3 I subtracted the 12 and got -e^0.4=-9, but this is whats giving me a problem because you can't take the ln of a negative number so i'm stuck.

56. 200/1+3e^-0.3t=100 and for my answer I got ln 1/3=-0.3t and then I got t=3.66

58. A/1+Be^t/2=c for my answer I got up to lnA/B=2t but I don't know where to go from here.

I know it's a lot but if you could please check it you would be doing me a huge favor.

In the first problem it appears you multiplied (you said you divided), but it's correct ((edit): not that your others are necessarily wrong though). The use of latex is preferable for clarity, and can be achieved by adding dollar signs to the left and right sides of your code.

[hide="#58"]$ a + be^{\frac {t}{2}} = c$
$ be^{\frac {t}{2}} = c - a$
$ e^{\frac {t}{2}} = \frac {c - a}{b}$
$ \frac {t}{2} = \ln{|c - a|} - \ln{|b|}$
$ t = \ln{(c - a)^2 - \ln{b^2}}$[/hide]
Where $ \ln{x}$ denotes the natural log of x. Note that I used logarithmic properties in lines 4 and 5 to simplify the answer. (edit): It might be worth mentioning that you'll want to isolate your exponential term before you take the log of both sides and cancel it out.

for #54 just notice that both sides of the equation are negative

not sure if this is considered spammy, but for fractions, you have to put dollar signs in front and behind:
\frac{x}{y}

to get $ \frac {x}{y}$
and for exponents, you have to put dollar signs in front of and behind:
a^{b}
to get $ a^{b}$
the reason the braces must be there is for exponents that are more than one character long
maybe ill try some of the problems now that im here


inequalities
floor function
number theory
prime numbers
number theory proposed

If $\{p_n\}$ is the sequence of prime numbers ($p_{n-1}<p_n$ and all the primes are in the list) prove that for $i=1,2,\cdots,\left\lfloor\frac{n}{2}\right\rfloor$ \[p_{n-i}+p_i\leq p_n\] with equality iff $i=1$ and $p_{n-1}, p_n$ are twin primes.

EDIT: hehe, this was my post 911  :P

[quote="pbornsztein"][quote="djimenez"]EDIT: hehe, this was my post 911  :P[/quote]

Cool! But you posted it in a wrong section  ;)
I thought that after 910 posts you would have understand the right way of posting  :P

I move it...

Pierre.[/quote]

No Pierre, it was just that I posted the problem and after that I realize that it was my 911 post, and turned to be funny for me (The Emergency post :D , I wouldn't like to relate it to 9/11/2001). But, does anyone have given a thought to the problem?


[quote="José"]Prove that there are infinitly interger solutions to the equation $xy(x-y)+yz(y-z)+zx(z-x)=6$.[/quote]
Do you mean infinitely many?
Such as: [quote]Prove that there are infinitly many interger solutions to the equation $xy(x-y)+yz(y-z)+zx(z-x)=6$.[/quote]
I'm working on a solution...

oh silly mistake, of course infinite solutions satisfy this.

Write $x-y = 1, y-z = 2, z-x = 3$ (for instance).
Actually, we see the above does not fit as a solution because adding all three together yields $0=6$. There for we must choose three numbers whose sum is 0. We write:
$x-y = -1, y-z = -2, z-x = 3$
Which yields 0=0, or infinite solutions.

[quote="me@home"]we can rewrite the rhs as:
$x^2y - xy^2 + y^2z - yz^2 + z^2x - zx^2$
$= (x-y)(y-z)(z-x)$
So we see that the product of three integers is 6.
How can infinite integer solutions satisfy this?[/quote]

Wow I am making a lot of mistakes, actually it is the negative, $= -(x-y)(y-z)(z-x) = 6$
This doesn't affect the solution, just negate everything so the final equation is: $x-y=1, y-z=2, z-x=-3$
(which also has infinite solutions)


[quote="a_vakilian"]Prove that :
$(a,m)=1 \Rightarrow a^{m-\phi(m)} \equiv a^m \mod m$
i.e if m is a prime number it is obviosly correct![/quote]

Isnt that obvious ? :huh:

since we know that $a^{\phi(m)} \equiv 1\mod m$ and hence

$\frac{a^m}{a^{\phi(m)}}\equiv \frac{a^m}{1} \equiv a^m \mod m$

 :ninja:

Yes, I wouldn't call it a "generalisation".

oooo!! since it is Mr.jalali question i wanted to say something about it !!  actually shyong has written a compelet solution !!  :lol:

[quote="Soarer"]it's a generalization without $(a,m)=1$. It's from niven's book.[/quote]
Oh,yes you are right I was wrong. I checked and see that the generalization is:
Prove that $a^{m-\phi{m}} \equiv a^m \mod m$.
So shyong your answer is not coorect for this!


geometry
rectangle
videos
modular arithmetic
AMC 10

Counting with Casework Part 2

We tackle two more problems with casework, one involving counting rectangles in a diagram, and the other a problem from the AMC 10.

Sorry for bumping this and all but in the AMC 10 if you had a really large number like 10 digits or something like 10 digits how would you approach it?
Also a suggestion that I think we should have if you have one of these topics that are open for a while they get deleted/locked and when you discuss the video you can make the new one.

Way to attempt the amc 10 problem:
[hide="My really cool solution"]
Using casework:
We find the hundred digit, and the unit digit:
Our possibilities are:
$\mathbb{P}={1x0}->{9x9}$
We have to find the numbers that have the middle digit being the average of the first and last digits.
Let's start listing some numbers:
1-0 will not have a middle digit, because the average of $1$ and $0$ is $\frac{1}{2}$
1-1 will have a middle digit, because the average of $1$ and $1$ is $\frac{2}{2}=1$
1-2 will not have a middle digit, because the average is $\frac{3}{2}$
and so on for the numbers that have a hundred's digit.

That is going to be the set of numbers, because the numbers that have an odd hundred's digit, will have the unit digit of being $>10$ and $\equiv 1\pmod{2}$
$\mathbb{P}={111,123, 135, 147, 159}$ or 5 numbers.  That will also work for all the odd hundred's digit, or $mathbb{P}={1,3,5,7,9}$ or 5 numbers.  $5*5=25$ numbers for having an odd hundred's digit.

For an even hundred's digit, we would do the same casework as above:
2-0 will have a middle digit, because $\frac{2+0}{2}\implies \frac{2}{2}=1$
2-1 will not have a middle digit, because $\frac{2+1}{2}\implies \frac{3}{2}$
2-2 will have a middle digit, because $\frac{2+2}{2}\implies \frac{4}{2}=2$
and so forth.
The numbers that have an even hundred's digit, have the unit's digit of numbers that are $>10$ and $\equiv 0\pmod{2}$
$\mathbb{P}={0,2,4,6,8}$ or once again 5 numbers.

The set of even hundred digit, is the same case above, of numbers that are $>10$ and $\equiv 0\pmod{2}$ so the set is once again:
$\mathbb{P}={2,4,6,8}$ or 4 numbers.
5*5=20

Our cases are odd hundred digit, and even hundred digit.
Odd hundred digit=25, and even hundred digit=20.
$20+25=\boxed{45}$
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Edit:  Sorry for the double post, the last post was kind of stupid, and I can't edit/delete it while I still want to show my solution.
Edit 2:  Just watched rrusycyzk video, and learned my solution of 50 was wrong.  I went through my solution again, and figured out that I said that the hundred digit was the set:
$\mathbb{P}={0,2,4,6,8}$ to get 5*5=25, instead of the set $\mathbb{P}={2,4,6,8}$ to get 5*4=20.
Fail