function quadratics calculus derivative logarithms conics parabola

 I was just wondering...... how do i find the function if they jz giv me the data of the graph .

Must i think hard or is there some sort of way to find the function?

well.....we need to know if it's a quadratic, cubic, linear, or else it could be really weird...

also, we need the points they gave u

As AIME15 said, if we have some kind of information as to what kind of function it is, we can use what we know about that kind of function to find an equation that fits the graph. For example, if we know for certain that the graph is associated with a quadratic function, then we can take any 3 points from the graph (2 points if one of them in the vertex) and find the equation.

Exercise:
1. Find the equation of a quadratic function which passes through the points (-1,2), (1,-2), and (3,4).

2. Find the equation of an exponential function which passes through (1,8) and (3, 2).

Some graphs, however, do not have any simple (elementary) function associated with them. So in general, it is not possible to write down an equation for any arbitrary graph using only elementary functions.

[quote="Poincare"]derivatives....
[hide]first, graph the data, then you can get a basic idea of the graph then you can bring the vertex  to the origin then try different derivatives but its pretty much brute force... and, if you have an infinite amount of data you can't really do it .....[/hide][/quote]

That doesn't work. Derivatives only give you the equation of one less power than the original. And you're going to be finding the equation of the derivative instead of the equation of the curve.

If you're into statistics, you could attempt to find the least squares regression line outfitted with squares, logarithms, etc, and the correlation will just be $ r\equal{}1$ instead of a decimal. But again, you won't know what extension you will use until you have a general sense of what you're looking for.

ok lets say this is the data x=-3 and y =-32        x=-2 and y =-11      x is -1 and y is -2       x is 0 and y is 1   x=1 and y=4      x =2 y =13                 x=3 y =34     Hw do i the function?

The only way i had thought of finding the function of the graph is by brute forcing my brain and ended up spending a long time

Putting the points in table form:

$ \begin{tabular}{|c| c |}
\hline
x&y\\
\hline
\hline
\minus{}3&\minus{}32\\
\hline
\minus{}2&\minus{}11\\
\hline
\minus{}1&\minus{}2\\
\hline
0&1\\
\hline
1&4\\
\hline
2&13\\
\hline
3&34\\
\hline
\end{tabular}$

Plot them into a stat graph in your TI and use linereg, quadreg, expreg, etc

[quote="cf249"]$ \begin{tabular}{|c| c |} \hline x & y \\
\hline \hline \minus{} 3 & \minus{} 32 \\
\hline \minus{} 2 & \minus{} 11 \\
\hline \minus{} 1 & \minus{} 2 \\
\hline 0 & 1 \\
\hline 1 & 4 \\
\hline 2 & 13 \\
\hline 3 & 34 \\
\hline \end{tabular}$[/quote]

Plotting the points on a TI, and hitting Zoom:9 Statistics, we see that there is a resemblence to a function where the degree is $ 3$. Therefore, we use CubicReg (cubic regression) and we find that our equation is $ f(x) \equal{} x^3 \plus{} 0x^2 \plus{} 2x \plus{} 1$, or $ \boxed{f(x) \equal{} x^3 \plus{} 2x \plus{} 1}$, with $ \boxed{R^2 \equal{} 1}$, which implies that this is truly a curve of best fit for the data.

common differences to find the degree:

-21 -12 -6
-9 -6 -6
-3 0 -6
-3 6 -6
-9 12
-21

It turns into a constant function in 3 differences, thus it has a degree of 3. Then we can let $ f(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$, plug in some values, and solve for the coefficients.

Some quick and easy methods to plot approximate graphs could be:

(1) All quadratics are parabolas and hence by knowing if coefficient of $ x^2$ is positive or negative and knowing the vertex, it can be easily sketcehd.
(2) All cubics are either strictly increasing, strictly decreasing or both increasing and decreasing. Since a cubic is surjective, by knowing the nature(real,complex,..) of the roots of its derivatives, we can sketch it roughly.
(3) For any general function, find its asymptotes, its values at these asymptotes($ \plus{} \infty$ or $ \minus{} \infty$), the intervals of increase and decrese (its extrema), points of discontinuity to plot the graph.
(4) Most importantly, you must know how to graph $ f(|x|)$, $ |f(x)|$, etc from f(x) and be able to apply the transformations to a known curve to get the equation of the given curve.


Let $z_{1},z_{2}\in\mathbb{C}$ with $|z_{1}|=1$ and $z_{1}\neq z_{2}$. Calculate $p=\left| \frac{z_{1}-z_{2}}{1-z_{1}\overline{z_{2}}}\right|$

I solved this way: Denote $c=\frac{z_{1}-z_{2}}{1-z_{1}\overline{z_{2}}}$ then $p=|c|$. Then I got $\overline{c}=\frac{1-z_{1}\overline{z_{2}}}{z_{1}-z_{2}}=\frac{1}{c}$. but we all konw that $c\overline{c}=|c|^{2}$ so $|c|^{2}=1\Longleftrightarrow |c|=1$ which means that p=1. Is this corect?


Hi, I'm new here and I wanted to take some classes over summer, any suggestions?

I scored a 24/25 on the AMC 8, 101 on the AMC 10 and I made State CDR in Washington.

My goal for '10 is to make AIME.

I've already worked through the Intro to C&P, Intro to NT and Intro to Geo books.

Any help is greatly appreciated

If you don't think you need a class to cover the books you mentioned, then I'd start with the AMC 10 Problem Series, and then move on to Algebra 3.


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
AMC 10 B

I PWNED myself...I should have gotten in but I'm 99.99999% sure I didn't :D

[quote="rclee21"]wohooo!!!!! i definitely got pwned!  :ninja:  there's something called 'stupid mistakes' you should be wary of...[/quote]

I agree. :(  :)

I most definitely got owned by the test. I got a 1.  :D Although I should've gotten 3...

I got owned, and yet i should have made it.  In fact, I still don't know, but I probably didn't make usamo.  I solved and answered 12 with confidence, guessed the others.  Yet, I didn't check enough, and my carelessness cost me 5 points.  I don't know if a 7 will make it...

[quote="probability1.01"]Did you qualify for the 2006 USAMO?[/quote]

YES BECAUSE I'M THE ONLY (CURRENTLY ALIVE) PERSON IN THE EASTERN UNITED STATES

Yes, if the floor score is less than a 13.

YES, BECAUSE ALL OF THE OTHER QUALIFIERS HAVE DIED OF A STRANGE DISEASE CALLED "UNITED STATES OF AMERICA MATHEMATICAL OLYMPIAD QUALIFICATION HEART ATTACK SYNDROME."

[quote="scorpius119"]I made it. Easily. The index is NOT going to be a perfect 5th power!

Now if only the expansion had been taken into account [i]last[/i] year...[/quote]
Come on now, we both know your 32 is not going to make it. Be realistic.
[hide=" :D"]Just kidding[/hide] [hide=" "]LOL[/hide]

[quote="randomdragoon"]You should probably remake this poll in light of the new annoucement.[/quote]

Ehh... too lazy.

190 index off the 10B
i dont call that OWNED
not for 8th grade anyway
t-2nd best 8th grade index in Texas
i think (based on online stats)
even though there were 4 seventh graders that got in
but i do know a guy who had a 284 index
he is amazing
no doubt he'll make...
whatever the next level is :huh:
i cant remember

[quote="sapphyre571"]
no doubt he'll make...
whatever the next level is :huh:
i cant remember[/quote]

That would be MOSP, I suppose.  284 index doesn't necessarily presume that he's good enough to make it; after all, as somebody else remarked, AMC/AIME require very different skills from USAMO ;)

It's just so easy to have a bad day, and mess it all up with carelessness and lack of focus.  USAMO isn't like that so much.  You could take a 30 minute power nap during the exam if you wanted to...

Yeah, it's much harder to make a stupid mistake on the USAMO... except that I managed to do that on #4 last year, didn't read "A table can have cut leg of length 0"

[quote="inyoung"]yup, I made it through the second cut off.. THANK YOU AMC!!! yeah, I'm from VA and really psyched, although I know I'll do horribly[/quote]

Lol, are you by any chance Yeonsik's sister? In that case you just need to get one question. (And partial credit on others).

[quote="randomdragoon"][quote="pieterminate"]Also, you have to "show work" on USAMO, plus a chance for partial credit.  And USAMO is harder... :)

BTW, how rigorous do the proofs have to be?  We don't have to prove any (well-known) formulas right?[/quote]
If you can name it, you don't have to prove it.
If you can't name it but you can prove it, then you better prove it.
If you can't name it and you can't prove it, try stating it like it's obvious and hope for the best.[/quote]

 :rotfl:  :rotfl:  :rotfl:  :rotfl:

[quote="randomdragoon"]
If you can name it, you don't have to prove it.
If you can't name it but you can prove it, then you better prove it.
If you can't name it and you can't prove it, try stating it like it's obvious and hope for the best.[/quote]

Well there are well-known theorems that do not have names. For example something like an odd prime $p$ can be written as sum of two squares iff p=1(mod 4) is well-known, but does not have a name(or does it?). Proving that on the USAMO will be a waste of time and paper.

Right, and we only get 1 piece of paper for each problem (looking at the teacher's manual)?

And...
[quote="randomdragon"]
If you can't name it and you can't prove it, try stating it like it's obvious and hope for the best.[/quote]

So say: "by this obvious well known theorem that I  don't recall the name of, $a^2+b^2=c^2$ " ?
 :rotfl:

[quote="beta"][quote="pieterminate"]Right, and we only get 1 piece of paper for each problem (looking at the teacher's manual)?
[/quote]

That is not true, you can use as many pages as you want. You will probably have to print it at home.[/quote]

Why would you have to print it at home?  Your teacher should provide it for you.  Just make sure that he or she knows that you can use as many sheets as you wish.


geometry
circumcircle
inradius
trigonometry
incenter
geometry proposed

[hide="Problem."][quote="Erken"]Let $ \triangle ABC$ be a triangle and let $ K$ be some point on the side $ AB$, so that the tangent line from $ K$ to the incircle of $ \triangle ABC$ intersects the ray $ AC$ at $ L$. Assume that $ \omega$ is tangent to sides $ AB$ and $ AC$,  and to the circumcircle of $ \triangle AKL$. Prove that $ \omega$ is tangent to the circumcircle of $ \triangle ABC$ as well.[/quote][/hide]
Generally, 4 circles tangent to the lines $ AB, AC$ and to the circumcircle of $ \triangle AKL$ exist, one identical with the mixtilinear A-incircle of $ \triangle ABC,$ which is tangent to its circumcircle $ (O).$ But the other 3 are generally not tangent to $ (O).$

Counterexample with both $ K, L$ and tangency points of $ \omega$ with the lines $ AB, AC$ on the segments $ (AB), (AC)$:

okay, let's count that the tangent line from $ K$ to the incircle intersects the ray $ AC$ at $ L$, so that $ C\in[AL]$.

Perpendicular to $ AI$ at $ I$ is tangent to the circumcircle of $ \triangle IBC$ $ \Longrightarrow$ it cuts the segments $ (AB), (AC)$ at $ F, E.$ Since $ \triangle AKL, \triangle ABC$ now have common incircle $ (I),$ $ F, E$ are tangency points of their common mixtilinear A-incircle with $ AB, AC,$ which is tangent to the circumcircles of both triangles. There is nothing else, the mixtilinear A-excircle of $ \triangle AKL$ does not touch the segment $ (AC)$ and none of the 2 circles centered on the external bisector of $ \angle A$ and tangent to the lines $ AB, AC$ and to the circumcircle of $ \triangle AKL$ touches both the segments $ (AC), (AB).$

[quote="Erken"]Let $ \triangle ABC$ be a triangle and let $ K$ be some point on the side $ AB$, so that the tangent line from $ K$ to the incircle of $ \triangle ABC$ intersects the ray $ AC$ at $ L$. Assume that $ \omega$ is tangent to sides $ AB$ and $ AC$,  and to the circumcircle of $ \triangle AKL$. Prove that $ \omega$ is tangent to the circumcircle of $ \triangle ABC$ as well.[/quote]
Use twice Lyness theorem we have desired !

[quote="Mr Green"]Use twice [b]Lyness theorem[/b] we have desired ![/quote]
What's that?

Btw (though my solution is actually equivalent to yetti's idea), the problem is just a simple consequence of the fact that the radius of the $ A$-mixtilinear incircle of a triangle $ ABC$ is something depending only on the triangle's inradius and the magnitude of the corresponding angle (more precisely, it is equal to $ r/ \cos^{2}{\frac {A}{2}}$, where $ r$ is the triangle's inradius).

[quote="pohoatza"][quote="Mr Green"]Use twice [b]Lyness theorem[/b] we have desired ![/quote]
What's that?[/quote]
Lyness theorem
If a circle is tangent internally (or esternally) to the circumcircle of a triangle and to two sides of the triangle,the line joining its points of contact with the sides passes through the incenter (os corresponding excenter) of the triangle

[quote="izat"]Can you prove it?(lynness theorm)[/quote]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=31739&]Incenter I is the midpoint of DE[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=597724&]Mixtilinear circle[/url].


Let $ W$ be a subspace of a finite-dimensional inner product space $ V$. Prove that $ (W^{\bot})^{\bot}\equal{}W$.

It is clear that $ W\subset(W^{\bot})^{\bot}$, but how about $ (W^{\bot})^{\bot}\subset W$?

Actually, you want to show that $ (W^{\perp})^{\perp}\subseteq W$ and $ W\subseteq (W^{\perp})^{\perp}$.  Then you get equality.

From my textbook, it hints to use the following: Let $ V$ be an inner product space, and let $ W$ be a finite-dimensional subspace of $ V$.  If $ x\notin W$, then $ \exists$ $ y\in V$ such that $ y\in W^{\perp}$, but $ \left\langle x,y\right\rangle\neq 0$.


function
search
combinatorics unsolved
combinatorics

Let S={1,2,3,4}. Find the number of functions $ f: S \to S$ such that $ f(f(x)) \equal{} 1$ for all $ x \in S$.

We'll do this generally for $ S \equal{} \{1, \ldots, n\}$.

Let $ z \equal{} f(1)$, and let's calculate $ f(f(z))$ in two ways. First, by the requirement on $ f$, we must have $ f(f(z))\equal{}1$. Second, we have $ f(z)\equal{}f(f(1))$ by the definition of $ z$, and because of the property of $ f$ we must therefore have $ f(z)\equal{}1$. Plugging this back into $ f(f(z))\equal{}1$ we get $ \boxed{f(1)\equal{}1}$.

Let $ A \equal{} f^{\minus{}1}(1)$ be the inverse image of $ \{1\}$ under $ f$, namely those elements that are mapped by $ f$ to $ 1$. The function $ f$ satisfies the requirements iff $ f(S) \subseteq A$. We just saw that $ 1 \in A$, so let $ A \equal{} B \cup \{1\}$ with $ 1 \not \in B$. Any element of $ S$ that lies outside $ A$ (if there is one) must be mapped by $ f$ into $ B$: it cannot map to $ 1$ lest it will actually be in $ A$, and it cannot map outside of $ A$ because that would violate the requirement.

So for any $ f$ we get a well-defined, nonempty set $ B \subseteq \{2, \ldots, n\}$ and the elements of $ S \smallsetminus (\{1\} \cup B)$ are mapped into $ B$. On the other hand, it's clear that any choice of $ B$ and any choice of mapping for the elements outside $ B \cup \{1\}$ is legitimate and will lead to a different $ f$. If $ B$ has size $ k > 0$, it can be chosen in $ \binom{n\minus{}1}{k}$, and there are $ n\minus{}1\minus{}k$ elements outside of $ A$ that can be mapped arbitrarily into $ B$. So overall the number of such functions is

$ \sum_{k\equal{}1}^{n\minus{}1} \binom{n\minus{}1}{k} k^{n\minus{}k\minus{}1} \equal{} \sum_{k\equal{}1}^m \binom{m}{k} k^{m\minus{}k} \equal{} \sum_{k\equal{}0}^m \binom{m}{k} k^{m\minus{}k}$.

Here $ m \equal{} n\minus{}1$. For $ n \equal{} 4$, this number is $ 10$. Here is the [url=http://www.research.att.com/~njas/sequences/?q=1%2C3%2C10%2C41%2C196&sort=0&fmt=0&language=english&go=Search]OEIS[/url]entry.

The problem requires the following (obvious) conditions:
1) f(1)=1.
2) No element other than 1 can be mapped onto itself.
3) Atleast one of f(2), f(3), f(4) is equal to 1.

So f(2) can take 3 possible values:

[u]Case 1[/u]: f(2)=1
If f(3)=1, f(4)=1 or 2 or 3
If f(3)=2, f(4)=1 or 2
If f(3)=4, f(4)=1

6 functions

[u]Case 2[/u]: f(2)=3
f(3)=1 (forced)
f(4)=1 or 3

2 functions

[u]Case 3[/u]: f(2)=4
f(4)=1 (forced)
f(3)=1 or 4

2 functions

Total=6+2+2=10 functions

Is that right?

Looks fine to me. The only drawback of this approach is that it doesn't scale very well (what would you do if the problem was about $ \{1, \ldots, 10\}$ instead of $ \{1, 2, 3, 4\}$?)


if n is an odd prime, p a prime such that p|2004^n + 1, and p=2 mod 501. Calculate p mod4

Because $ n$ is odd so we have :
$ p|2004.((2004)^{\frac{n\minus{}1}{2}})^2\plus{}1$
$ \Rightarrow (\frac{\minus{}2004}{p})\equal{}1$
$ \Rightarrow (\frac{\minus{}1}{p})(\frac{4}{p}).(\frac{501}{p})\equal{}1$
Because $ 501\equiv 1 (\mod 4)$ s0
$ (\frac{501}{p})\equal{}(\frac{p}{501})\equal{}(\frac{2}{501})\equal{}\minus{}1$
More than $ (\frac{4}{p})\equal{}1$
So $ (\frac{\minus{}1}{p})\equal{}\minus{}1$
$ \Rightarrow p\equiv 3 (\mod 4)$


A number $n$ is satisfying the conditions below

i) $n$ is a positive odd integer;

ii) there are some odd integers such that their squares' sum is equal to $n^{4}$.

Find all such numbers.


Prove that: If n is an integer  :ge: 170 then
n = a 2 +b 2 +c 2 +d 2 +e 2  (a,b,c,d,e is integers). [/quote]

Well, Lagrange's theorem states that any natural number is the sum of four squares, so..

[quote="zscool"]maybe he means a,b,c,d,e are positive integers?[/quote]

I think he does. But it can be sharper since 169 = 144 + 16 + 4 + 4 + 1.

Mr Grobber, I know that every natural number is the sum of four square.
But in this problem, we have to prove that every integer   :ge: 170 is the sum of five square. They are different. You will prove it again, won't you?
Xdragon

I don't even know how to prove the four square theorem, let alone this one :).

Hm,I think the key is number 169,that is the number satisfying the condition to be formed by sum of 1,2,3 or 4 squares.So let see how many positive squares n-169 can be formed by their sum.

It is a beautiful solution. But I want to ask you that why you thought about the number 169, not any different numbers and how you found 169 is the sum of 1,2,3,4 squares? It is not easy to think. Am I right?
Thanhk you for proving it.[/i]

I don't even get why it is a solution... :?

And how to find the number 169 that's quite obvious when you're stating all numbers $\geq 170$   :maybe:

Here's why it's a solution: if $n>169$ then $n=169+n-169$. If $n-169$ is written as the sum of $1$ (:)) square, then consider $n-169$ as the sum of $4$ squares. If $n-169$ is the sum of two squares, then consider $169$ as the sum of $3$ squares. If $n-169$ is the sum of three squares, then consider $169$ as the sum of $2$ squares. If $n-169$ is the sum of $4$ squares, then consider $169$ as the sum of $1$ square, and we are done.

By square I mean non-zero square, of course.


rebecca celebrated her graduation by going for a hot air balloon ride. the wind first blew the balloon $\frac{1}{2}$ mile due east. then the balloon was blown $\frac{3}{4}$ mi SW. finally, it was blown $\frac{1}{2}$ mi N and then landed. how many miles did they land from the pint where the balloon was launched? express your answer as a decimal to the nearest hundredth.

[hide="solution"]
When we draw this out, we get something that looks like an incomplete 45-45-90 triangle. (The angles between East/North and SW are both 45 degrees). So if we complete the triangle we get a 45-45-90 with side lengths $\frac{1}{2}+x$. ($x$ is the same on both sides). By our 45-45-90 identity special stuff (pythag, too) we know that $\left(\frac{1}{2}+x\right)\sqrt{2}=\frac{3}{4}$, so ${x=\frac{3\sqrt{2}-4}{8}}$. It is clear that the distance between the two endpoints is $x\sqrt{2}$, so multiplying it all out, we get that the distance is $\frac{3-2\sqrt{2}}{4}= \fbox{.04}$ [/hide]

I got $\frac{2\sqrt{2}-3}{4}$ as the answer... yours is negative.

This was on a team round. I got a negative number the first time. :rotfl:

Um, Ubemaya, $2\sqrt{2}$ is less than 3, so your answer is negative.

[quote="ritchjp"][hide]

I came up with an answer of 1.03 mi.

[/hide]

A solution will follow.[/quote]
 incorrect

I see what I did wrong. I went SE instead of SW.
if you trace the path of the balloon it will look like a 45-45-90 degree triangle with its corner missing. If you draw a line from the point of launch to the point of landing it will create an isosceles trapezoid where the two equal legs are given and the longer base is given. It turns out to be .042 miles between the two points.


c_ 1 ,c_2,c_3...is a sequence of circles whose radii r_1,r_2,r_3... form an increasing

sequence. Every c_i  touches c_i+1 externally and touches no c_i+j for j>1.The whole

sequence of    circles have the same pair of lines as direct common tangents.Then

r_1,r_2,r_3...    is a sequence in

a)AP    b)GP     c)HP     d)none of these

note:here underscore means a subscript and please tell me how to use it in latex

For subscripts in LaTeX, basically just do what you did, but put dollar signs ($\$$) around them. For example, $C_i$ is C_i with dollar signs. If there is more than one character in the subscript, like $C_{i+j}$, then you need braces around the subscript. For example, C_{i+j} with dollar signs around it.  :)

[quote="lightrhee"]Consider two diagram.
1) circle $O_1$, circle $O_2$ and the two common tangent lines.
2) circle $O_2$, circle $O_3$ and the two common tangent lines.

They are similar :) .
Therefore $r_1, r_2, \cdots$ is a geometric sequence.[/quote]

thanks for sending me the answer
and also how to use latex
i didnt get the relation between circle $O_1$ ,circle$O_2$ and circle$O_3$
and the tangents.but now after you told me igot that relation and understood the problem
this site is really very good
 :)


LaTeX
inequalities
inequalities proposed

please write it in latex  i don't know how!!!
let a,b,c  positive real number such that a^2+b^2+c^2<=3

find MIN of
1./  P(a,b,c)=  :Sigma: 1/(a^5.(b+1).(c+1))
2./  P(a,b,c)=  :Sigma: 1/(a^14.(ab+c^2).(ca+b^2);

1. Multiply together
$\sum \frac{1}{a^4}+\frac{1}{a^5} \geq 6$ and
$8 \geq (a+1)(b+1)(c+1)$ to get a bound of $\frac{3}{4}$

2. Note that $(a+c)^2 \geq 3ac$, and multiplying by $(a-c)^2$ we get $a^4+4a^2c^2+c^4 \geq 3a^3c+3ac^3$. Summing, we get $\sum a^3c \leq 6$, and hence $8 \geq (ab+c^2)(ac+b^2)(bc+a^2)$. Multiplying this by $\sum \frac{1}{a^{12}} + \frac {bc}{a^{14}} \geq 6$ yields a bound of $\frac{3}{4}$.

PS. You just use \$ signs to enter and exit latex.

i don't understand your solution for no.1.. can you explain why is the first inequality true?

[quote="Fiachra"]It is simply the AM-GM of the 6 terms mentioned (remembering that from the condition $abc \leq 1$).[/quote]

the question says a,b,c are real numbers  :?

[quote="Fiachra"]
2. Note that $(a+c)^2 \geq 3ac$, and multiplying by $(a-c)^2$ we get $a^4+4a^2c^2+c^4 \geq 3a^3c+3ac^3$. Summing, we get $\sum a^3c \leq 6$, and hence $8 \geq (ab+c^2)(ac+b^2)(bc+a^2)$. Multiplying this by $\sum \frac{1}{a^{12}} + \frac {bc}{a^{14}} \geq 6$ yields a bound of $\frac{3}{4}$.

[/quote]

Can you please explain to me why $(a+c)^2 \geq 3ac$

and why $\sum a^3c \leq 6$ implies$8 \geq (ab+c^2)(ac+b^2)(bc+a^2)$ ?

Thank you very much.

i think that the condition a^2+b^2+c^2 can be replace by abc<=1 but the problem still can be done


integration
function
calculus
Functional Analysis
advanced fields
advanced fields unsolved

Prove that $ C[0,1]$ is not a Hilbert space with respect to the following inner product

\[ (f,g)\equal{} \int_0^1 f(t)\overline{g(t)} dt\]

For $ n > 2,$ let $ f_n(x)$ be the piecewise linear continuous function with vertices at $ (0,0),\left(\frac12 \minus{} \frac1n,0\right),\left(\frac12 \plus{} \frac1n,1\right),$ and $ (1,1).$

This is a Cauchy sequence in the norm in question, but there is no continuous function that it can converge to.

The space in question is an inner product space. Only one property of a Hilbert space fails - it is not complete.

Can you please explain in more detail that what is vertices.

Get out a piece of paper. Draw axes. Plot the points I named as dots. Connect the dots, using line segments. That's the graph of my function.

I could have written this as

$ f_n(x)\equal{}\begin{cases}0,&0\le x\le\frac12\minus{}\frac1n\\ \text{(some linear formula)},&\frac12\minus{}\frac1n<x<\frac12\plus{}\frac1n\\ 1,&\frac12\plus{}\frac1n\le x\le 1\end{cases}.$

where "some linear formula" is something I'm too lazy to figure right now, but you could if you wanted.

You can build a lot of useful examples this way. The "connect the dots" approach is clearer to me than the branched-case formula would be.

I have found the function you are talking about but then what ? How to use this?

(1) Compute $ \|f_n\minus{}f_m\|$ to show that this is a Cauchy sequence.

The calculus on this is a little annoying, but you don't have to calculate exactly. Crude estimates will do: use that $ |f_n(x)\minus{}f_m(x)|\le\frac12$ for $ \frac12\minus{}\frac1n<x<\frac12\plus{}\frac1n.$ (This is assuming that $ m>n.$)

(2) Find a way of arguing that there is no continuous function that could be the limit of this sequence. (The fact that there's an "obvious" limit function which happens to be discontinuous at $ \frac12$ will have some bearing on this.)

But if I am not mistaken, they made quotient of this space with this relation $ f\sim g$ if and only if $ f\equal{}g, \mbox{a.e.}$. Then you will have a Hilbert space,

No, Omid - the statement was that this was the set of continuous functions. It is a normed space already (no need for quotients) but it's not complete.

The quotient you're talking about is part of the definition of $ L^2,$ which is complete, and which is the completion of $ C$ under this norm.


Point C is on the segment AB which has endpoints A(2,  1)
and B(11, 5).  Point C is twice as far from point A as it is from
point B.  What are the coordinates of point C?

We can draw a right triangle with hypotenuse AB. One leg is 11-2=9, and the other is 5-1=4. Therefore, if I build another right triangle using BC as hypoteuse this time, one leg would be 3 and the other would be 4/3. Two of the coordinates of the small triangle would be 11,5 and 11, 3 2/3, so C would be 8, 3 2/3 or $ (8, 3\frac{2}{3})$

The answer that dragon stated is perfectly acceptable, but I think it only accepts it as $ (8, \frac{11}{3})$.  Again, I am not sure.

Generally, if not specified, you should answer as a common fraction.

i dont think it matters, unless the problem states so (i think they said that in the rules)

I agree with ernie and izzy about putting it as a common fraction

Well, in any case, it's easier to express as a common fraction, because you type less.


[hide="#1"]If a number X consists of 66 digits of threes, and a number Y consists of 66 digits of sixes, name all the digits which will appear in the product of X and Y.[/hide]

[hide="#2"]Start polygons are created as follows: n points are equally distributed around the circumference of a unit circle.  An initial point P_0 is chosen and each of the remaining poitns are then consecutively labeled clockwise P1, P2, P3....P(n-1).  Segement P_0 P_m is constructed where 0<m<n. Then segment P_m P_(2m) is constructed; and so on skipping m-1 poitns each time you constuct a new segment until you arrive back at P0.  The figure is a star polygon only if all n points were used to construct the segments.  How many distinct star plygons are there for n=36?[/hide]

Thanks for your help!

[hide]For #1, the answer is 2, 1, 7, and 8.  The fact that there are 66 of each digit doesn't really matter, the pattern can be observed after only 3 or 4 of each digit.[/hide]

[hide="1"]
To be slightly more rigorous,
\begin{align*} \underbrace{\overline{333\cdots 3}}_{66} \times \underbrace{\overline{666\cdots 6}}_{66} & = \left(\overline{1\underbrace{000\cdots 0}_{66}} - 1\right) \times \underbrace{\overline{222\cdots 2}}_{66} \\
& = \overline{\underbrace{222 \cdots 2}_{66}\underbrace{000 \cdots 0}_{66}} - \underbrace{\overline{222\cdots 2}}_{66} \\
& = \overline{\underbrace{\boxed{2}22 \cdots 2}_{65}\boxed{1}\underbrace{\boxed{7}77 \cdots 7}_{65}\boxed{8}}
\end{align*}
[/hide]

[hide="2"]
Its not too hard to see that the only numbers $ m$ that work are relatively prime to $ n = 36$ (why?), so the answer is $ \phi(36) = 12$.

Edit, but wait $ m$ and $ 36-m$ give the same star polygon, so the answer is $ 12/2 = \boxed{6}$.  :oops:
[/hide]

oh right.  I got 12 too, but forgot about the symmetry.

thanks azjps, it probably took a lot of time to post that  :)


Thanks, and can you post all these problems when it is THE END? :D


I think this one easy to prove.: :)
And I knew this inequality very early,but I have never applied it even once.  :)

I used it here http://www.mathlinks.ro/Forum/viewtopic.php?p=457162#p457162
 ;)

And I had posted as theorem here http://www.mathlinks.ro/Forum/viewtopic.php?t=68963


algebra
polynomial
algorithm
calculus
integration
complex numbers
number theory

Vandermonde determinants

Let $a_{1},a_{2},\ldots a_{n}$ be complex numbers and let
\[
D_{n}=\left|
\begin{array}
[c]{cccc}
1 & 1 & \ldots & 1\\
a_{1} & a_{2} & \ldots &  a_{n}\\
a_{1}^{2} & a_{2}^{2} & \ldots &  a_{n}^{2}\\
\vdots &  &  & \\
a_{1}^{n-1} & a_{2}^{n-1} & \ldots &  a_{n}^{n-1}
\end{array}
\right|  .
\]
$D_{n}$ is called Vandermonde determinant and its value equals
\[
D_{n}=\underset{1\leq i<j\leq n}{\prod}\left(  a_{j}-a_{i}\right)  .
\]
The proof goes by induction. The base case is obvious, so assume the asertion true for $n-1.$ Consider the polynomial
\[
P\left(  x\right)  =\left|
\begin{array}
[c]{cccc}
1 & 1 & \ldots & 1\\
a_{1} & a_{2} & \ldots &  x\\
a_{1}^{2} & a_{2}^{2} & \ldots &  x^{2}\\
\vdots &  &  & \\
a_{1}^{n-1} & a_{2}^{n-1} & \ldots &  x^{n-1}
\end{array}
\right|  .
\]
Clearly, $\deg P=n-1$ and $P(a_{i})=0$ for $1\leq i\leq n-1.$ Thus, we have
\[
P(x)=c(x-a_{1})\ldots(x-a_{n-1}),
\]
for some constant $c.$ Expanding the determinant after the last column, we find that
\[
c=\left|
\begin{array}
[c]{cccc}
1 & 1 & \ldots & 1\\
a_{1} & a_{2} & \ldots &  a_{n-1}\\
a_{1}^{2} & a_{2}^{2} & \ldots &  a_{n-1}^{2}\\
\vdots &  &  & \\
a_{1}^{n-2} & a_{2}^{n-2} & \ldots &  a_{n-1}^{n-2}
\end{array}
\right|  =D_{n-1}.
\]
Finally, setting $x=a_{n}$ gives the result.

[quote="-oo-"]Nice proof. Of course, the $a_i$ don't have to be complex numbers. They may be elements of an arbitrary commutative unitary ring.[/quote]

Is this an Olympiad Problem?

[b]Moderators[/b] please move it away. thanks

[quote="conceive"][quote="-oo-"]Nice proof. Of course, the $a_{i}$ don't have to be complex numbers. They may be elements of an arbitrary commutative unitary ring.[/quote]

Is this an Olympiad Problem?

[b]Moderators[/b] please move it away. thanks[/quote]

Why do you you think that if a post doesn't contain an olympiad problem, it should be moved away? The reason of creating forums "... Theorems and Formulas" is just writing general statements (i. e. theorems and formulas) which can be used in olympiads after considering some particular cases. They are useful and sholudn't be moved away, for example field $\mathbb{Z}_{p}$, which is a bit stronger thing than a commutative unitary ring, is very useful in number theory, however it isn't required in any olympiad.

Total nitpicking, but I just want to keep stuff in Theorems & Formulas correct:

Enescu's proof requires $ a_1$, $ a_2$, ..., $ a_{n \minus{} 1}$ to be distinct (otherwise, we cannot conclude that $ P(x) \equal{} c(x \minus{} a_{1})\ldots(x \minus{} a_{n \minus{} 1})$ from $ P(a_{i}) \equal{} 0$ for $ 1\leq i\leq n \minus{} 1$). But the other case is even easier (both $ D_n$ and $ \prod_{1\leq i < j\leq n}\left(a_j \minus{} a_i\right)$ are zero), so the proof is not in danger.

Unfortunately, this trick doesn't work for arbitrary commutative rings. Instead one has to prove Vandermonde over $ \mathbb{Z}$ and conclude that it holds over an arbitrary commutative ring. Or proceed in a different way (I know of at least one good proof which works over any arbitrary commutative ring with $ 1$; can you find it?).

  darij

darij, the proof works over $ \mathbb{Z}$, since Euclidian algorithm works for monic divisors like $ x\minus{}a_i$, and $ \mathbb{Z}$ is an integral domain. The problem with other commutative rings $ R$, even with unit, I think is that the factorization in $ R[x]$ may no more be unique, e.g. in $ \mathbb{Z}_4[x]$ we have $ x^2 \equal{}(x\minus{}2)^2 \neq x(x\minus{}2)$.

Yes, that's what I meant. For $ \mathbb{Z}$ it follows from $ \mathbb{Q}$.

  darij


articles
geometry
geometric transformation
rotation
LaTeX

Hello,
I use yap2.5 viewer to read dvi files. I need to  include eps graphics file. The compilation to dvi goes OK, but yap displays only the Bounding Box for this picture.

[b]Do you know the other dvi viewer that can display eps pictures?[/b]

Is there the way to adjust yap2.5  to display eps graphics by adding some special package(s)?

Thank you

marajohns

Yap can show some eps picures; the latest versions of MikTeX includes a version of Ghostscript in order to do so. But it may not show all eps if they have special effects in them.

In Yap go to View, Render Method and tick Dvips to see if that helps. If not try this simple file:
[code]\documentclass[12pt]{article}
\usepackage{graphics}
\begin{document}
\includegraphics{test.eps}
\end{document}[/code]with the test.eps fle attached below. If you can see this in Yap then it could be your graphics has some effects like rotation that it can't show.

Hi Stevem,

Thank you very much for the very fast response.
Your test.eps tremendously helped me (my MikTex has the Ghostscript). I am preparing an article for scientific journal with a simple single line plot exported from Matlab with axes, ticks and  background with about 10kb (test.eps is 2kb). Because I could see test.eps but could not see my eps, I tried to simplify the plot by removing "fancy” axes, ticks, etc and decreasing the size to 7kb. So Yap now normally renders my eps plot in DVIps mode and I will continue adding one by one fancy details up to the Yap's limit.

But is this really important thing to do?
I mean that we send to scientific journals article.tex and picture.eps files. Maybe they have better tools to display the fancy pictures with axes, ticks, background than Yap we use, and there will be no problem to render my original plot?

Thank you very much again.
marajohns

It's only DVI viewers like Yap that are restricted so publishers will have no problems as they will use Postscript. You should be able to view the eps in Ghostscript itself - check that you can - if not then there's a problem with the file.

But the best way to view your complete article is to compile to pdf (possibly via ps) and then you should be able to see your original eps image and the text. In TexnicCenter there's a LaTeX=>PS=>PDF option if Ghostscript is installed. There's a profile you can import [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=702368#702368]here[/url].

DVI viewers generally aren't able to view eps images so have to call on an external program like Ghostscript whereas using pdf converts the eps to pdf first.


Triangle $ ABC$ with bisector $ AD,BE,CF$ ($ D,E,F$ on $ BC,CA,AB$ respectively). Call $ I,J,K$ are incenter point of triangle $ AEF,BFD,CDE$ respectively. Call $ I',J',K'$ are symetric point of $ I,J,K$ over midpoint of $ EF,FD,DE$ respectively. Prove that $ AI',BJ',CK'$ are concurrent!

could you post a solution for this problem...it seems I can't solve it  :(

Hi,dear  Erken,can you post your solution?thanks!

[quote="Erken"][quote="nsato"]In general, I don't think these lines are concurrent.[/quote]
But they are concurrent,indeed... :wink:
Do you know the source of this problem,tdl?[/quote]
As I can see by the drawing, I think also that the problem is not true generally.

I will check more carefully, before a definite answer.

Kostas Vittas.

For the figure I used geogebra....and the lines were concurent(If I remember well)....I will try again :(

It is my own problem! I found it by Geometer Sketchpad. Sorry because I haven't solved it, too! Anyone can prove or disprove it? :oops:

I think the problem is true,but i can't find a way to solve it

[quote="¬[ƒ(Gabriel)³²¹º]¼"]sorry but this problem is not true  :noo: (prove with 135, 35, 10 or similar)[/quote]
Sorry but I don't understand your idea! What is 135, 35, 10?

He says that you better can see by the drawing that the problem is not true, in the configuration of a triangle $ \bigtriangleup ABC,$ with $ \angle A \equal{}135^{o},\ \angle B\equal{}35^{o},\ \angle C\equal{}10^{o}.$

Kostas Vittas.

[quote="vittasko"]He says that you better can see by the drawing that the problem is not true, in the configuration of a triangle $ \bigtriangleup ABC,$ with $ \angle A \equal{} 135^{o},\ \angle B \equal{} 35^{o},\ \angle C \equal{} 10^{o}.$

Kostas Vittas.[/quote]

Please look at my figure! In this $ AI',BJ',CK'$ are concur at point $ T$!

[img]http://i227.photobucket.com/albums/dd72/TPFC/Concur.png[/img]

[quote="vittasko"]Good morning dear [b][size=100]tdl[/size][/b].

Check please your drawing as follows:

Denote the point $ T\equiv CK'\cap AI'$ and measure the segment $ CT.$

Delete, for a moment, the line segment $ AI'$ and denote the point $ T'\equiv CK'\cap BJ'$ and measure the segment $ CT'.$

If the points $ T,\ T'$ are coincided, then must be equal the measurements of $ CT,\ CT'.$

Are they equal? In my drawin they are not equal.

Also, I can see clear the result that $ T\neq T',$ by some enlargements of the drawing, near the area of these points.

Best regards, Kostas Vittas.[/quote]

Thanks vittasko! When enlargements of the drawing I can see that these aren't concurrent! What's a pity! Sorry! :oops:
By the way, could you please consider my below topic? It's also a problem which I found! Hope it's true!
[url]http://www.mathlinks.ro/viewtopic.php?t=202678[/url]


let r(k) be the sum of remainder k divided by 1, 2, ..., k.
prove that there are infinite positive integers n that r(n)=r(n+1).

Let's consider $ r(m)\minus{}r(m\minus{}1)$. The remainder on dividing $ m$ by a positive integer $ a\leq m\minus{}1$ is 1 more than the remainder on dividing $ m\minus{}1$ by $ a$ [i]unless[/i] $ a$ divides $ m$, in which case the remainder drops from $ a\minus{}1$ to 0. So we get
\[ r(m)\minus{}r(m\minus{}1)\equal{}m\minus{}1\minus{}\sum_{\stackrel{a|m}{0<a<m}} a\]
So this is asking for infinitely many $ m$ such that the sum of the proper divisors of $ m$ is $ m\minus{}1$. This is true for every power of 2, so we're done.


There are 2 piles of objects and 2 players. In turn, each player takes some objects from a pile or from 2 piles but if from 2 piles, the number of objects taken from each must be equal. The winer is the one who takes the last objects. Prove that there is always a way for the first player to win at all times.
My teacher give me this problem but his solution is long and difficult.Can anybody try this one?

I just have one little query(doubt). Are there any restrictions as to the amount that each player can take? If the answer is no, it is easy.

Suppose the two piles are the same amount. Then player 1 can take the entire both piles

But what happens when both piles have different amounts? Counterexample: let the two piles be 2 and 1. In every way you can think of (which is not many), second player wins.

Am I missing something?

If I remember correctly, the game is a heck of a lot more complicated than that. It's called [url=http://mathworld.wolfram.com/WythoffsGame.html]Wythoff's game[/url]. The problem has been posted before.


symmetry
trigonometry
modular arithmetic
geometry
cyclic quadrilateral

$ A_1 A_2...A_9$ is a regular 9-gon(which probably has a mroe common name). If $ A_1A_2\equal{}a$, and $ A_2A_4\equal{}b$, find $ A_1A_5$ in terms of a and b.

[hide]
$ A_1A_5\equal{}A_4A_8$ so consider the quadrilateral $ A_1A_2A_4A_8$. Now, by symmetry, $ A_2A_4\equal{}A_1A_8$ and $ \angle A_8A_1A_2\equal{}\angle A_1A_2A_4$ and $ \angle A_2A_4A_8\equal{}\angle A_4A_8A_1$. So $ A_1A_2A_4A_8$ is an isosceles trapezoid.

Note that $ \angle A_1A_2A_4\equal{}\angle A_1A_2A_3\minus{}A_3A_2A_4\equal{}140^\circ\minus{}20^\circ\equal{}120^\circ$. Therefore, $ \angle A_1A_8A_4\equal{}60^\circ$ and therefore $ A_4A_8\equal{}a\plus{}2(b/2)\equal{}a\plus{}b$.
[/hide]

Could someone please tell me what's wrong with this using Ptolemy theorem?

If $ A_1A_4\equal{}y$ then in cyclic quadrilateral $ A_1A_2A_3A_4, ay\plus{}a^2\equal{}b^2$
If $ A_1A_5\equal{}x$ then in cyclic quadrilateral $ A_1A_2A_4A_5, bx\plus{}a^2\equal{}y^2$

solving for $ x$ we get $ b(b^2\minus{}2a^2)/a^2$

There is nothing wrong... that is another way to express the same thing since $ a$ and $ b$ themselves are related.


I could do this problem by trying every number, but I'm going to need to explain how to do it without just plugging in, so if anybody could help, it would be greatly appreciated

[u]Problem[/u]
A third-grader showed his teacher a new way to reduce fractions:
16/64, cross out the 6's, you get 1/4
The teacher told him that it was nonsense, but the student said that it worked for other fractions.  Find all such fractions, assuming both numerator and denominator are two digits.

Again, thanks.

16/64, 19/95, 26/65, and there's another one that I forgot. There's a marvelous proof that those four are the only fractions that work (apart from the trivial 11/11, 22/22, 33/33, 44/44, ......), but I have forgotten it, it's in a number theory book I have upstairs, and this margin is too small to contain it anyway :D ...

Thanks.  It would be great if anyone had the 4th number, and if anyone was able to "paraphrase" the proof "to get it to fit" :lol:

Oh... from where is this problem? (For what do you need this problem?) Don't want to be giving answers to take-home exams or anything. Of course, it's a well-known mathematical joke that these numbers can be such reduced and the four fractions are well-known, but...

The proof probably belongs in a forum about number theory... if there was one.

No, a math teacher gave it to me for the summer, but even when I figure out his puzzles, he wants a mathematical way of solving them, and I hate to disappoint.  I've never had a take home test in my life.

The last one is 49/98. Actually, if you just write the condition of the problem as
ab/bc=a/c where a,b and c are digits and by ab we mean the two digits number formed with the digits a and b, then all you have to do is find all numbers b and c for which b*c/(10b-9c) is a positive integer, which, considering that you have 9 possibilities for b and 9 for c, and that b and c must be different so that you won't get the trivial solutions like 22/22, then you just have 72 choices, and that's not too much. I used a computer. :)

Okay, but how did you come up with that formula?

The non-trivial solutions are
16/64 = 1/4;    19/95 = 1/5;       26/65 = 2/5;  and     49/98 = 4/8 :lol:

I am explaining my approach.

a, b, and c are single digit decimal numbers that satisfy 'a''b'/'b''c' = a/c.

(i) There can be a set of trivial solutions when a = b = c.
11/11 = 1/1; 22/22 = 2/2; 33/33 = 3/3; ...... ; 99/99 = 9/9

(ii) Ignoring all types of trivial solutions, we have1 <= a, b, c <= 9.

The basic equation is  10a + b         a
                                   -----------   =  --
                                    10b + c         c

=> 10*a*(c - b)  = c*(a - b)   ... (1)

If a > b, then 'a''b'/'b''c' > 1 and hence, a/c > 1. Therefore, c > b as the L.H.S. of equation (1) is positive.
We have a > c > b.
(a - b) > 10*(c - b) > 0   as c < a   and this is not possible.

So, we are left with all the cases where a < b. This implies that c < b to keep the integrity of the sign of the equation (1).
      10*a*(c - b)  = c*(a - b)
=> a*(c - b) = c*(a - b)/(2*5)

To keep the integrity, we have either of the following requirements.
(1) 5 divides c and 2 divides (a - b)
(2) 2 divides c and 5 divides (a - b)

Suppose the case (1) holds. We have c = 5.
2*a*(5 - b) = (a - b) This means that a divides (a - b) or a divides b.
Therefore, a = {1, 2, 3, 4}
                 b = {9, 6, 27/5, 36/7}

We have a =1, b = 9, c = 5 and a = 2, b = 6, c = 5 as solutions for the case (1).
19/95 = 1/5
26/65 = 2/5

When case (2) holds, we have 2*a*(b - c)  = c .
Again, {a,b} = {(1,6); (2,7); (3,8); (4,9)}
                  c = {4, 28/5, 48/7, 8}
We have a= 1, b = 6, c = 4 and a = 4, b = 9, c = 8.
16/64 = 1/4
49/98 = 4/8

Thank you, Shamik Banerjee.

[quote="aidan"]Hmm how about the trivial $\frac {10}{20}, \frac {10}{30}, \frac {10}{40} \cdots$? But it's an interesting problem anyhow.  :)[/quote]

That wouldn't work. You'd have to cancel the digits diagonally, i.e. the second digit of the numerator and the first of the denominator.

No... you'd be cancelling the $0$'s in both numbers.

I appreciate your thanks :lol:  I just concentrate doing the basics right most of the time. Someone told me that this is the only way to survive in this world :P


$\triangle ABC$ has $AC=7$, $BC=24$, and $\angle C=90^\circ$. $M$ is the midpoint of $AB$, $D$ lies on the same side of $AB$ as $C$ and has $DA=DB=15$. Find the area of $\triangle CDM$.

[hide="Solution"]Let $\angle CAB=\alpha$
$AB^2=AC^2+BC^2\Rightarrow AB=\sqrt{65}$
$CM=AB/2=\sqrt{65}/2$
$DA=DB\Rightarrow DM\perp AB\Rightarrow DM^2=DA^2-AM^2\Rightarrow DM=\sqrt{835}/2$
$\angle CMD=90^\circ-\angle CMB=90^\circ-2\alpha\Rightarrow$
$\Rightarrow\sin\angle CMD=\cos{2\alpha}=\cos^2{\alpha}-\sin^2{\alpha}=(AC^2-BC^2)/AB^2=33/65$
$A(\triangle CMD)={{CM\cdot DM\cdot \sin\angle CMD} \over 2}={33\over 8}{\sqrt{167 \over 13}}$[/hide]

Oops. Sorry about that...I was wondering why your solution was so off.... :blush:

$BC=24$, not 4.

[hide="Answer"]By definition, $\overline{CM}=\frac{25}{2}$, and by Pythagorean, $\overline{DM}=\frac{5\sqrt{11}}{2}$. Say that $\angle ABC=\alpha$. Therefore, by similar triangles, $\angle AMC=2\alpha$, and $\angle CMD=90-2\alpha$. $\cos CMD=\cos (90-2\alpha)=\sin 2\alpha=2\sin \alpha \cos \alpha=2\left(\frac{7}{25}\right)\left(\frac{24}{25}\right)$, so $\cos CMD=\frac{336}{625}$. From here, we have $\sin CMD=\frac{527}{625}$, so we compare this to $DM$. Call the height of $\triangle CMD$ $h$. $\sin CMD=\frac{h}{\frac{5\sqrt{11}}{2}}=\frac{527}{625}$, so $h=\frac{527\sqrt{11}}{250}$, and the area of the triangle is $\frac{1}{2}\left(\frac{25}{2}\right)\left(\frac{527\sqrt{11}}{250}\right)=\boxed{\frac{527\sqrt{11}}{40}}$.[/hide]

[url=http://www.upal.ca/adeel/problems/2006/1/3.pdf]I have another trig solution... :D[/url]


Could someone define invariant? x is invariant to y?

have you tried lookingit up on Mathworld yet?

as for now, toodles

Invariant is not a Getting Started concept.

Moreover, a question concerning the term should be moved to Other Problem Solving Forum.

Moved.

Woops, mathworld, good idea. Sorry, not thinking, really tired kind of thing. Yeah, checked math world, not really getting it. My question came from AoPS book 1, proportions section, the word was heavly used in one question, and there is no real reference to what it means in the book ( checked the index and the whole chapter ).

Any help would be appriciated, untill then, the search continues.

Could you post the problem where you got that from?  I'm not exactly sure of what they meant from what you've said.

Actualy, it's in part of the solution to the problem that I encountered the word.

Problem:
It is stated that x varies directly as y and inversely as the square of z, and that x = 10, when y = 4 and z = 14. What is x when y = 16 and z = 7?

Now I can solve this, it's not a hard problem, x = 160.

But: in the first line of the solution in AoPS book: "We are given (X)(Z^2)/(Y) is invariant". Invariant? Value does not change as one varible changes? Is that right, thats what I need clarification on.

Invariant means that it doesn't change.  Therefore, that expression you had, stays the same for whatever values of X, Y, and Z you use.

Invariants are especially useful in problems involving algorithms.

It's pretty much anything that doesn't change after some operation.


Is it true that for every rational number $x$ which is greater than 1,
there exist a $y$ which is also a rational number, where $x^{2}-y^{2}= 1$?
why?

[hide="Well..."] We have $y = \sqrt{x^{2}-1}$.  This is clearly not always rational; take, for example, $x = \frac{3}{2}$, where we have $y = \sqrt{ \frac{5}{4}}$. [/hide]

A little easy for Pre-Olympiad, unless I mistake the question.

Yeah, this is too easy for pre-olympiad.....I can do it :rotfl:

[hide]If x=2, y=the square root of 3, which is irrational[/hide]

edit:forgot to hide my answer :blush:

hmm...i know we can prove that by counter example.
but i was looking for a more theoretical solution, like, can we prove it by other ways. develope a general solution?

thanks  :)

I don't really understand what you're asking for.  Disproving a statement just means that some set does not exhibit some structure.  The lack of structure is nothing mathematically significant  :huh:

[hide="@violetcraze"]See if you can sneak of changing the $=$ sign to an $\equiv$ and make it a mod problem... then say it was like that all along :lol: [/hide]

The problem kind of reminds me of the larger Mihalescu's Recently proven with $x^{p}-y^{q}=1$


geometry
geometric transformation
homothety
rotation
dilation
trigonometry
power of a point

I truly enjoyed assisting in the [i]Geometric Proofs[/i] course at AMSP 08.

Now that the class is over, a very important question arises: what should you do now to further your learning? The very first thing to do is get a strong base in the fundamentals. You should know how to prove all the basic theorems, how to do geometric constructions, and a host of other simple but nice topics. Many nice olympiad problems are created by analyzing the fundamental techniques well.

Now you need to to learn the standard techniques for olympiad geometry. I think that we covered certain topics better than others. I think that angle chasing, cyclic quadrilaterals, collinearity, concurrency, homothety, spiral similarity, and triangle centers went well enough. I gave a forum on power of a point and radical axes, not everybody attended.

Anyway, you should all read Geometry Revisited to solidify your knowledge of the topics. Also there is a ton of materials for you to read over in the folder that you got so you can really learn what we went over in class. Ivan and Mr. Muskarov gave you a huge amount of lecture notes that you should read. I also included by spiral similarity lecture.

This is all great, but accumulating knowledge is not what problem solving is about. The day that Mr. Feng came into class, he gave simple advice: do problems.

On the first day of class, I presented you all with my 'four pivots' framework for solving geometry problems. It is very important to develop your own ways to solve problems. But how do you do this? This is where Mr. Feng's advice comes in. There is no substitute for doing lots of problems. By spending lots of time struggling with difficult questions, you learn how to approach them. You start to recognize when an approach will be fruitful or not before you actually do it.

Now, where do you get problems to work on? Mr. Feng suggested that you all should go through the IMO shortlists and do G1 from each year, then proceed to G2, G3, etc until you have done all of them. I think that this is good advice, but it might be a bit advanced for some students in the class. It is very easy to spend lots and lots of time getting pleasure by 'pwning easy problems.' While this can be fun, it is not a good way to learn. I think that you should be saying: "well I solve them about half of the time." I suppose that it takes a certain level of maturity to deal with solving so few problems, but you learn best when you are struggling.

Related to this, on the last day, I gave you a take-home test. I'd enjoy looking at your solutions to the problems on the test.

Since I am going out of my way to grade these tests, I ask that if you are planning to send me solutions, that they be written up in a way that is easy for me to grade. That means: i) solutions should be latexed ii) solutions should have a diagram.

I hope that all of you who took the geometric proofs course learned a lot and that you will continue to enjoy geometry as I do!

Hi mr. accompanier!  :P

is there a copy of the problems available?

Alright, here is a copy...

Note that problems 3,4 were things that I found when I was trying to make the test. 4 is most likely to be known...

For getting a strong base in the fundamentals which books/articles can you recommend? Also can you recommend anything for people that plan to take the course this year?

I was introduced to geometry from just my high school textbook. Try to understand the proofs for basic things like angles in circles, congruency, power of a point, etc. Then I did the AoPS book geometry sections. The more that you know, the better.

Could you explain you four points of framework for solving geometry problems? also is there a copy of the notes available?  :)

Hello, so it's been like a year (almost?) so I guess it is OK to post solutions here. For the sole sake of making me feel good about myself. These are really nice problems (3 and 4)! Maybe I will post #5 later if I solve it.

[hide="1"]We have two directly similar figures, so a spiral similarity centered at P maps one to another. Each point O in ABCD is rotated and dilated about P, but assuming the spiral similarity is not degenerate (that is, the rotation is not by 0 and the dilation is not by 1), the image O' can only be the same point as O if O=P. Thus, only one such point can exist.

This proves that only one such O can exist, I don't think O [i]must[/i] exist, depending on your definition of 'superimpose'.[/hide]

[hide="2"]I had seen this problem before, so Solution 1 was the first one I came up with, Solution 2 follows the hint.

Solution 1: By the Law of Sines in $ \Delta ADE$, $ \dfrac{\sin{\angle{ADE}}}{\sin{\angle{AED}}} \equal{} \dfrac{AE}{AD}$, and by the Law of Sines in $ \Delta CDE$, $ \dfrac{\sin{\angle{EDC}}}{\sin{\angle{DEC}}} \equal{} \dfrac{EC}{ED}$. Note that $ \angle{AED}$ and $ \angle{DEC}$ are supplementary, so their sines are equal. It follows that $ \dfrac{\sin{\angle{ADE}}}{\sin{\angle{EDC}}} \equal{} \dfrac{AE\cdot ED}{AD\cdot EC}$. But $ \sin{\angle{EDC}} \equal{} \sin(\dfrac{\pi}{2} \minus{} \angle{ADE}) \equal{} \cos{\angle{ADE}}$, so we get $ \tan{\angle{ADE}} \equal{} \dfrac{AE\cdot CD}{AD\cdot EC}$. Similarly, $ \tan{\angle{ADF}} \equal{} \dfrac{AF\cdot BD}{AD\cdot FB}$. But it follows from Ceva that these two are equal, so we get $ \tan{\angle{ADE}} \equal{} \tan{\angle{ADF}}$, and since $ \tan$ is increasing on $ (0,\dfrac{\pi}{2})$, $ \angle{ADE} \equal{} \angle{ADF}$.

Solution 2: Construct the parallel to $ BC$ through $ A$, and extend $ DE$ and $ DF$ to meet the parallel at $ X$ and $ Y$, respectively. It suffices to prove that $ AX \equal{} AY$ since $ DA\perp XY$. But note that $ \Delta AEX\sim\Delta CED$, so $ \dfrac{AX}{CD} \equal{} \dfrac{AE}{CE}$, and $ AX \equal{} \dfrac{AE\cdot CD}{EC}$. Similarly, $ AY \equal{} \dfrac{AF\cdot BD}{FB}$. But these two quantities are equal by Ceva, so we're done.

(note that these two solutions are equivalent because $ AX \equal{} AD\tan{\angle{ADX}} \equal{} AD\tan{\angle{ADE}}$).[/hide]

[hide="3"]Let $ l$ be the radical axis of $ C_{1}$ and $ C_{2}$. $ M$ is on the $ l$ (as $ MQ_{1} \equal{} MQ_{2}$), so $ (MR_{1})(MP_{1}) \equal{} (MR_{2})(MP_{2})$. Thus, $ P_{1}P_{2}R_{2}R_{1}$ is cyclic. Let $ X$ be a point on $ P_{1}P_{2}$ such that $ P_{1}$ is between $ X$ and $ P_{2}$. Now, $ \angle{XP_{1}S_{1}} \equal{} \angle{P_{1}R_{1}S_{1}} \equal{} \angle{R_{2}R_{1}M} \equal{} \angle{R_{2}P_{2}P_{1}} \equal{} \angle{P_{2}S_{2}S_{1}}$. It follows that $ P_{1}P_{2}S_{1}S_{2}$ is cyclic. Thus, $ (AP_{1})(AS_{1}) \equal{} (AP_{2})(AS_{2})$, meaning $ A$ is on $ l$.

We have $ \angle{XP_{1}S_{1}} \equal{} \angle{P_{1}P_{2}R_{2}} \equal{} \angle{XP_{2}R_{2}}$, so it follows that $ P_{1}S_{1}\parallel{}P_{2}R_{2}$. Also, $ \angle{P_{1}R_{1}S_{1}} \equal{} \angle{P_{2}S_{2}S_{1}}$, so $ P_{1}R_{1}\parallel{}P_{2}S_{2}$.

Consider the external homothety taking $ C_{1}$ to $ C_{2}$. It takes $ O_{1}$ to $ O_{2}$, $ P_{1}$ to $ P_{2}$, and $ Q_{1}$ to $ Q_{2}$. Since $ P_{1}R_{2}\parallel{}P_{2}S_{2}$, it takes $ R_{2}$ to $ S_{2}$, and similarly it takes $ S_{1}$ to $ R_{2}$. Thus, it takes $ \Delta P_{1}S_{1}R_{1}$ to $ \Delta P_{2}R_{2}S_{2}$. It follows that $ \Delta O_{1}S_{1}R_{1}\sim\Delta O_{2}R_{2}S_{2}$.

(as an interesting note, using this homothety we easily find that $ P_{1}P_{2},O_{1}O_{2},S_{1}S_{2},Q_{1}Q_{2}$ are concurrent).

Now, we consider $ B$. We claim that $ BR_{1} \equal{} BR_{2}$, that is, $ \Delta BR_{1}R_{2}$ is isosceles. To do this, we prove that $ \angle{S_{1}R_{1}B} \equal{} \angle{S_{2}R_{2}B}$. We have $ \angle{S_{1}R_{1}B} \equal{} \angle{SR_{1}O} \plus{} \angle{BR_{1}O_{1}} \equal{} \dfrac{\pi}{2} \plus{} \angle{SR_{1}O} \equal{} \dfrac{\pi}{2} \plus{} \angle{O_{2}S_{2}R_{2}}$
$ \equal{} \angle{O_{2}R_{2}B} \plus{} \angle{O_{2}R_{2}S_{2}} \equal{} \angle{BR_{2}S_{2}}$. Thus, $ BR_{1} \equal{} BR_{2}$. But the powers of $ B$ to both circles are $ BR_{1}^{2}$ and $ BR_{2}^{2}$, so thus $ B$ is also on $ l$.

To finish, we have $ l\perp O_{1}O_{2}$ since $ l$ is the radical axis. Thus, since $ A,B\in l$, $ AB\equiv l\perp O_{1}O_{2}$, as desired.[/hide]

[hide="4"]Consider the homothety $ h$ taking $ \omega_{1}$ to $ \omega$, centered at $ T$. Extend $ CZ$ to meet $ \omega$ at $ Y$; note that $ Y$ is the midpoint of arc $ AB$. Construct the tangent $ l$ to $ \omega$ at $ Z$; it follows that $ l\parallel{}AB$. Now, $ AB$ is tangent to $ \omega_{1}$ at $ F$, so thus $ h$ takes $ AB$ to $ l$ and $ F$ to $ Z$, meaning $ T,F,Z$ are collinear. Now, $ \angle{FTA} \equal{} \angle{ETA} \equal{} \angle{ECA} \equal{} \dfrac{\gamma}{2}$ (using standard notation for the angle measures). Also, $ \angle{ATC} \equal{} \angle{ABC} \equal{} \beta$. This gives $ \angle{FTC} \equal{} \angle{FTA} \plus{} \angle{ATC} \equal{} \beta \plus{} \dfrac{\gamma}{2}$. Finally, $ \angle{FZC} \equal{} \angle{BZC} \equal{} \pi \minus{} \angle{ABC} \minus{} \angle{ZCB} \equal{} \pi \minus{} \beta \minus{} \dfrac{\gamma}{2} \equal{} \pi \minus{} \angle{FTC}$, so $ FZCT$ is cyclic, as desired.[/hide]

Solution to #5: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3552861&#p3552861

Problem 2 is also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2268953#p2268953]Canada 1994 #5[/url].


superior algebra
superior algebra unsolved

I have seen the following theorems:

Th1: Let $ F$ be a field, $ d,n\in\mathbb{N}$ and $ x^m\minus{}1,x^n\minus{}1\in F[x]$. Then $ x^m\minus{}1\mid x^n\minus{}1$ iff $ m\mid n$.

and

Th2: Let $ m,n\in\mathbb{N}$ such that $ m\mid n$. Then for every prime number $ p$ we have $ p^m\minus{}1|p^n\minus{}1$.

I was wondering if Th2 can be proven by using Th1. Is there some kind of evaluating homomorphism which does the job?

Th. 2 can be proved using th. 1 by setting $ x\equal{}p$, or if you wish, use the evaluation homomorphism $ e_p: \mathbb{Q}[x]\to \mathbb{Q}: x\mapsto p$.

Remember, $ x^{md}\minus{}1\equal{}(x^m\minus{}1)(1\plus{}x^m\plus{}\ldots\plus{}x^{m(d\minus{}1)})$


inequalities
trigonometry
inequalities proposed

1. (Me) Let $ a,b,c>0$. Prove that \[ \frac{(a\plus{}b\plus{}c)^3}{a^2b\plus{}b^2c\plus{}c^2a}\plus{}\frac{3(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\ge 10\]

2. (Posted also here: http://www.mathlinks.ro/viewtopic.php?t=169649 ) For any triangle $ ABC$, show that \[ \frac{\cos A}{\sin^2A}\plus{}\frac{\cos B}{\sin^2B}\plus{}\frac{\cos C}{\sin^2C}\ge \frac Rr\]

i think 1st question is wrong because substitute a=b=c=1 and we get $ 4\ge10$ which is not possible my friend....

Well, if we substitute $ a\equal{}b\equal{}c\equal{}1$, we have $ \frac{(1\plus{}1\plus{}1)^3}{1\plus{}1\plus{}1}\plus{}\frac{3(1\plus{}1\plus{}1)}{(1\plus{}1\plus{}1)^2}\equal{}9\plus{}1\ge 10$, which is correct I think. :wink:

[quote="nayel"]1. (Me) Let $ a,b,c > 0$. Prove that
\[ \frac {(a \plus{} b \plus{} c)^3}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}\ge 10
\]
[/quote]
I think, it's still wrong.
Try $ a \equal{} b \equal{} 1$ and $ c\rightarrow0^ \plus{} .$
By the way, $ \frac {(a \plus{} b \plus{} c)^3}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {6(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}\ge 11$ is wrong too. :wink:

Yes, I am very sorry. I found my mistake. But the following one is true, I think. :wink: \[ \frac {(a \plus{} b \plus{} c)^3}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {30(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}\ge 19\]

hello, for a proof of the inequality
$ \frac{\cos(A)}{\sin^2(A)}\plus{}\frac{\cos(B)}{\sin^2(B)}\plus{}\frac{\cos(C)}{\sin^2(C)}\ge\frac{R}{r}$ you must use the theorem of cosine and the term
$ \frac{R}{r}\equal{}\frac{abcs}{4A^2}$ with $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$.
Sonnhard.

[quote=nayel] But the following one is true, I think. :wink: \[ \frac {(a \plus{} b \plus{} c)^3}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {30(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}\ge 19\][/quote]
The following inequality is stronger.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+ba\neq0$. Prove that:
$$\frac{(a+b+c)^3}{a^2b+b^2c+c^2a}+\frac{12(a^2+b^2+c^2)}{(a+b+c)^2}\geq13$$

[quote=arqady][quote=nayel] But the following one is true, I think. :wink: \[ \frac {(a \plus{} b \plus{} c)^3}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {30(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}\ge 19\][/quote]
The following inequality is stronger.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+ba\neq0$. Prove that:
$$\frac{(a+b+c)^3}{a^2b+b^2c+c^2a}+\frac{12(a^2+b^2+c^2)}{(a+b+c)^3}\geq13$$[/quote]

easily proof


advanced fields
advanced fields unsolved

Hi! Trying to solve some problem, I used in my proof the fact that the eight figure (contained in $ R^2$) is a deformation retract of  $ R^2 \backslash \{a,b\}$, where $ a$ is inside of one the circles and $ b$ in the other.   Making a picture, this is clear for me, but I do not know how to give an explicit deformation retraction.  I have a picture in my mind of how this map acts over $ R^2 \backslash \{a,b\}$, but I do not know how to write this as a formula.    It is hard to explain the picture that I have in my mind, but what I can say is that if $ F$ is my desired map, then for some points $ x \in R^2 \backslash \{a,b\}$ the curve joining $ F(x,0)$ and $ F(x,1)$ cannot be a line (remark: for [i]some[/i] points).

Thank you!

It was very easy.   The curves that I described can be lines.   In fact, without loss of generality, we can assume that $ a \equal{} ( \minus{} 1,0)$, $ b \equal{} (1,0)$ and the eight figure is the union of the circles centered at $ ( \minus{} 1,0)$ and $ (1,0)$ whose radii are $ 1$ for each of them.   Hence, if $ (u,v) \equal{} x \in R^2 \backslash \{a,b\}$, then the curve that join $ F(x,0)$ and $ F(x,1)$ is the projection of the point $ x$ to one of the circle (actually, if $ u \equal{} > 0$ then $ x$ is projected to the circle $ C( (1,0);1)$ and similarly when $ u < 0$).

I do not see any problem with this, right?

Thank you!

EDIT: it is wrong  :mad: , so I need help again


calculus
inequalities
trigonometry
inequalities unsolved

Let $ x,y,z$ real numbers such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$. Find the minimum value of
\[ {x \over x^2 \plus{} 1} \plus{} {y\over y^2 \plus{} 1} \plus{} {z\over z^2 \plus{} 1}\]

[quote="cyshine"]Let $ x,y,z$ real numbers such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$. Find the minimum value of
\[ {x \over x^2 \plus{} 1} \plus{} {y\over y^2 \plus{} 1} \plus{} {z\over z^2 \plus{} 1}
\]
[/quote]
Note that we have
\[ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx \equal{} \left( { \minus{} x} \right)\left( { \minus{} y} \right) \plus{} \left( { \minus{} y} \right)\left( { \minus{} z} \right) \plus{} \left( { \minus{} z} \right)\left( { \minus{} x} \right) \equal{} \minus{} x \minus{} y \minus{} z
\]

\[ \Rightarrow x \plus{} y \plus{} z \equal{} 0
\]

\[ \Rightarrow \left\{\begin{array}{l} xy \plus{} yz \plus{} zx \equal{} 0 \\
xy \plus{} yz \plus{} zx \equal{} xy \minus{} \left( {x \plus{} y} \right)^2 \\
\end{array} \right.
\]

\[ \Rightarrow xy \equal{} \left( {x \plus{} y} \right)^2
\]
But we always have $ xy \le \frac {{\left( {x \plus{} y} \right)^2 }}{4}$. Therefor
\[ \Rightarrow xy \equal{} \left( {x \plus{} y} \right)^2 \le \frac {{\left( {x \plus{} y} \right)^2 }}{4} \Leftrightarrow x \plus{} y \equal{} 0
\]
Hence we got $ x \equal{}  \minus{} y$ and $ z\equal{}0$ then $ P \equal{} 0$.

Sorry, but $ 1 \plus{} (\minus{}1) \plus{} (\minus{}1) \equal{} 1\cdot (\minus{}1) \plus{} 1\cdot (\minus{}1) \plus{} (\minus{}1)\cdot (\minus{}1)$. This means that $ x \plus{} y \plus{} z$ is not necessarily $ 0$.

[quote="cyshine"]Sorry, but $ 1 \plus{} ( \minus{} 1) \plus{} ( \minus{} 1) \equal{} 1\cdot ( \minus{} 1) \plus{} 1\cdot ( \minus{} 1) \plus{} ( \minus{} 1)\cdot ( \minus{} 1)$. This means that $ x \plus{} y \plus{} z$ is not necessarily $ 0$.[/quote]
Thank you, cyshine. I am misunderstanding here
[quote="Honey_S"]
\[ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx \equal{} \left( { \minus{} x} \right)\left( { \minus{} y} \right) \plus{} \left( { \minus{} y} \right)\left( { \minus{} z} \right) \plus{} \left( { \minus{} z} \right)\left( { \minus{} x} \right) \equal{}  \minus{} x \minus{} y \minus{} z \Rightarrow x \plus{} y \plus{} z \equal{} 0
\]
[/quote]

We will prove that this minimum value is $ \minus{}\frac{1}{2}$.

      If we take $ x\equal{}y\equal{}\minus{}1$ , $ z\equal{}1$ , the value is $ \minus{}\frac{1}{2}$.

      Let's prove that $ \frac{x}{1\plus{}x^2}\plus{}\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}\plus{}\frac{1}{2}\geq0$

      $ \frac{x}{1\plus{}x^2}\plus{}\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}\plus{}\frac{1}{2}\equal{}\frac{(1\plus{}x)^2}{2\plus{}2x^2}\plus{}\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}\geq\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}$

      If $ \frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}<0$ then $ (y\plus{}z)(yz\plus{}1)<0$ and by similar way  $ (x\plus{}z)(xz\plus{}1)<0$ and $ (y\plus{}z)(yz\plus{}1)<0$ . Let all of $ x$ , $ y$ , $ z$ are different from $ 0$.

      $ i)$ All of  $ x\plus{}y$ , $ y\plus{}z$ , $ x\plus{}z$ are $ \geq0$ then $ x\plus{}y\plus{}z\geq0$ and $ xy\plus{}yz\plus{}xz\leq\minus{}3$ it's a contradiction.

      $ ii)$ Exactly one of the $ (x\plus{}y)$ , $ y\plus{}z$ , $ x\plus{}z$ is $ <0$. (Without the loss of generality) $ y\plus{}z<0$. Because of  $ x\plus{}z>0$ and $ x\plus{}y>0$ so $ x>0$.

      $ xz\plus{}1<0$ and $ xy\plus{}1<0$ hence $ y$ and $ z$ are $ <0$. Let $ y\equal{}\minus{}a$ and $ z\equal{}\minus{}b$. $ x\equal{}\frac{ab\plus{}a\plus{}b}{a\plus{}b\plus{}1}$ and $ x>a>\frac{1}{x}$ and $ x>b>\frac{1}{x}$. So $ x>1$ and $ ab>1$.

      Otherwise because of $ x>a$ and $ x>b$ hence $ \frac{ab\plus{}a\plus{}b}{a\plus{}b\plus{}1}>a$ and $ \frac{ab\plus{}a\plus{}b}{a\plus{}b\plus{}1}>b$. So $ b>a^2$ and $ a>b^2$. So $ ab<1$. It's a contradiction.

      $ iii)$ Exactly two of them $ (x\plus{}y)$,$ (y\plus{}z)$,$ (x\plus{}z)$ are $ <0$. (Without the loss of generality) $ y\plus{}z$ and $ x\plus{}z$ are $ <0$. Because of $ x\plus{}y>0$ so $ z<0$.Because of $ xy<0$  without the loss of generality again $ x<0$ and $ y>0$. Let $ x\equal{}\minus{}a$ and $ z\equal{}\minus{}c$ and because of $ xz\plus{}1>0$ and $ xy\plus{}1<0$ so $ c<\frac{1}{y}$ and $ a>\frac{1}{y}$. Because of $ y\plus{}z<0$ and $ x\plus{}y>0$ hence $ a<y<c$ and so $ \frac{1}{y}<a<y<c<\frac{1}{y}$. It's a contradiction.

      $ iv)$ All of them are $ <0$ .So $ x\plus{}y\plus{}z<0$ and $ xy\plus{}yz\plus{}xz>0$ . It's a contradiction.

      $ v)$ Some of $ x$ , $ y$ , $ z$ are $ \equal{}0$ (without the loss of generality) $ x\equal{}0$. So $ y\plus{}z\equal{}yz\equal{}K$  and   $ \frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}\equal{}\frac{K^2\plus{}K}{2K^2\minus{}2K\plus{}1}\geq\minus{}\frac{1}{2} \Longleftrightarrow 4K^2\plus{}1\geq0$ which is obviously true. The
 proof is ended.

Let $ t \equal{} \frac {x}{x^2 \plus{} 1} \plus{} \frac {y}{y^2 \plus{} 1} \plus{} \frac {z}{z^2 \plus{} 1}$.

Now, we have $ 2t \equal{} \frac {2x}{x^2 \plus{} 1} \plus{} \frac {2y}{y^2 \plus{} 1} \plus{} \frac {2z}{z^2 \plus{} 1}$.

Then $ 2t \plus{} 3 \equal{} \frac {x^2 \plus{} 2x \plus{} 1}{x^2 \plus{} 1} \plus{} \frac {y^2 \plus{} 2y \plus{} 1}{y^2 \plus{} 1} \plus{} \frac {z^2 \plus{} 2z \plus{} 1}{z^2 \plus{} 1}$.

Cauhcy schwarz, $ 2t \plus{} 3 \ge \frac {(x \plus{} y \plus{} z \plus{} 3)^2}{x^2 \plus{} y^2 \plus{} z^2 \plus{} 3} \ge 0$ with equality iff $ x \plus{} y \plus{} z \equal{} \minus{} 3$

So we have $ t \ge \minus{} \frac {3}{2}$.

Check $ x \equal{} \minus{} 1$, $ y \equal{} \minus{} 1 \plus{} \sqrt {6}$, and $ z \equal{} \minus{} 1 \minus{} \sqrt {6}$.

So the minimum value is $ \minus{} \frac {3}{2}$.

Edit: sorry, I've made big mistake... I should not prove $ \frac{(x\plus{}y\plus{}z\plus{}3)^2}{x^2\plus{}y^2\plus{}z^2\plus{}3} \ge 0$, but $ 0$ should be replaced by $ \frac{1}{2}$, but I haven't tried it... could someone help?

[quote="cyshine"]Let $ x,y,z$ real numbers such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$. Find the minimum value of
\[ {x \over x^2 \plus{} 1} \plus{} {y\over y^2 \plus{} 1} \plus{} {z\over z^2 \plus{} 1}
\]
[/quote]
The answer is $ \minus{}\frac{1}{2}$, and there is a very nice solution by Cauchy Schwarz. ;) :)

[quote="efejzullahu"][quote="erdugan"]dear can_Hang 2007

Could you send your solution?[/quote]

  I think he is just trying to get up his reputation , and he doesn't solv the problem !
     Best Wishes : E.F

P.S - Sorry if I'm Wrong ! :|[/quote]

Wrong thinking, new guy!

Careful with yours language.

[quote="efejzullahu"][quote="erdugan"]dear can_Hang 2007

Could you send your solution?[/quote]

  I think he is just trying to get up his reputation , and he doesn't solv the problem !
     Best Wishes : E.F

P.S - Sorry if I'm Wrong ! :|[/quote]
Sorry, but I am not a spammer! Sometimes, I just type some words about the solution of a problem, since they maybe relate to my book, and because of the security, I cannot post the full solution. I hope you understand! About this problem, here is the full solution:
We need to prove that
\[ \frac {x}{x^2 \plus{} 1} \plus{} \frac {y}{y^2 \plus{} 1} \plus{} \frac {z}{z^2 \plus{} 1} \ge \minus{} \frac {1}{2}
\]

\[ \Leftrightarrow \frac {(x \plus{} 1)^2}{x^2 \plus{} 1} \plus{} \frac {(y \plus{} 1)^2}{y^2 \plus{} 1} \ge \frac {(z \minus{} 1)^2}{z^2 \plus{} 1}
\]
From the given condition, we have that
\[ z(x \plus{} y \minus{} 1) \equal{} x \plus{} y \minus{} xy
\]
Notice that we can't have $ x \plus{} y \equal{} 1$. Otherwise, if $ x \plus{} y \equal{} 1$, then $ x \plus{} y \minus{} xy \equal{} 0$ or $ xy \equal{} 1$, which is contradition since $ xy \le \frac {(x \plus{} y)^2}{4} \equal{} \frac {1}{4}$, therefore $ x \plus{} y \neq 1$, hence
\[ z \equal{} \frac {x \plus{} y \minus{} xy}{x \plus{} y \minus{} 1}
\]
Then, we can easily rewrite our inequality as
\[ \frac {(x \plus{} 1)^2}{x^2 \plus{} 1} \plus{} \frac {(y \plus{} 1)^2}{y^2 \plus{} 1} \ge \frac {(xy \minus{} 1)^2}{(x \plus{} y \minus{} 1)^2 \plus{} (x \plus{} y \minus{} xy)^2}
\]
Now, using Cauchy Schwarz Inequality, we have
\[ \frac {(x \plus{} 1)^2}{x^2 \plus{} 1} \plus{} \frac {(y \plus{} 1)^2}{y^2 \plus{} 1} \ge \frac {[(1 \plus{} x)(1 \minus{} y) \plus{} (1 \plus{} y)(1 \minus{} x)]^2}{(1 \plus{} x^2)(1 \minus{} y)^2 \plus{} (1 \plus{} y^2)(1 \minus{} x)^2} \equal{} \frac {4(xy \minus{} 1)^2}{(1 \plus{} x^2)(1 \minus{} y)^2 \plus{} (1 \plus{} y^2)(1 \minus{} x)^2}
\]
Hence, it suffices to prove that
\[ 4(x \plus{} y \minus{} 1)^2 \plus{} 4(x \plus{} y \minus{} xy)^2 \ge (1 \plus{} x^2)(1 \minus{} y)^2 \plus{} (1 \plus{} y^2)(1 \minus{} x)^2
\]

\[ \Leftrightarrow f(x) \equal{} (y^2 \minus{} 3y \plus{} 3)x^2 \minus{} (3y^2 \minus{} 8y \plus{} 3)x \plus{} 3y^2 \minus{} 3y \plus{} 1 \ge 0
\]
Now, we have
\[ \Delta_{f} \equal{} (3y^2 \minus{} 8y \plus{} 3)^2 \minus{} 4(y^2 \minus{} 3y \plus{} 3)(3y^2 \minus{} 3y \plus{} 1) \equal{} \minus{} 3(y^2 \minus{} 1)^2 \le 0
\]
Thus $ f(x) \ge 0$ and we have done!

Is it okay now, my friend?

[quote="efejzullahu"][quote="erdugan"]dear can_Hang 2007

Could you send your solution?[/quote]

  I think he is just trying to get up his reputation , and he doesn't solv the problem !
     Best Wishes : E.F

P.S - Sorry if I'm Wrong ! :|[/quote]

Very nice solution, can_hang2007. :)

P.S. Please, efejzullahu, be carefull with your words.

A nice solution given by a Brazilian student during the test: in short, he proved that
\[ \frac x{x^2\plus{}1} \plus{} \frac y{y^2\plus{}1} \plus{} \frac z{z^2\plus{}1} \leq \minus{}\frac12
\iff (x\minus{}yz)^2\plus{}(y\minus{}zx)^2\plus{}(z\minus{}xy)^2\plus{}(x\plus{}y\plus{}z\plus{}1)^2\plus{}(x\plus{}y\plus{}z\plus{}xyz)^2\geq0\]

He expanded the inequality and used $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$ two times.

My idea is to use a funny transformation, $ x \equal{} (1 \minus{} X)/2$. So
$ x \plus{} y \plus{} z \equal{} xy \plus{} xz \plus{} yz$ will be equivalent to $ XY \plus{} XZ \plus{} YZ \equal{} 3$ and
the sum will be $ \frac {2(1 \minus{} X)}{(1 \minus{} X)^2 \plus{} 4}$

But it seems to be a bit more complicated...

How do we go about proving this without assuming the value is $ \minus{}\frac{1}{2}$ ??

Hmm this got me thinking... couldn't there be a very nice solution by trigonometry? I mean substituting $ x\equal{}\tan \alpha$, $ y\equal{}\tan \beta$, $ z\equal{} \tan\gamma$, the problem becomes proving that if $ \tan\alpha\plus{}\tan\beta\plus{}\tan\gamma\equal{}\tan\alpha\tan\beta\plus{}\tan\beta\tan\gamma\plus{}\tan\gamma\tan\alpha$, then $ \sin 2\alpha\plus{} \sin 2\beta\plus{} \sin 2\gamma\geq \minus{}1$. The problem here is finding a closed value for $ \tan(\alpha\plus{}\beta\plus{}\gamma)$, since $ \tan(\alpha\plus{}\beta\plus{}\gamma)\equal{}\frac{\tan\alpha\plus{}\tan\beta\plus{}\tan\gamma\minus{}\tan\alpha\tan\beta\tan\gamma}{1\minus{}\tan\alpha\tan\beta\minus{}\tan\beta\tan\gamma\minus{}\tan\gamma\tan\alpha}$


Has anyone else had this problem?

You can try [url]http://www.cutepdf.com/[/url].  The image quality doesn't seem to me quite as high, but it works.

The same thing happened to me, so I uninstalled go2pdf and downloaded cutepdf. The cutepdf said it was freeware, so there is no purchase necessary. I suggest you take the go2pdf thing out of the LaTeX guide, since it stops working after 15 days if you don't pay them money. Although, with cutepdf you also have to download ghostscript, I think. Also, the image quality for cutepdf seems to be exactly the same as go2pdf!

Go2PDF moved their freeware to a different page.  I have updated the link so you can find the free (forever) version now.


algebra
polynomial
quadratics
complex numbers

[quote="missunique"]factor.
64x^3+343

x^4+8x^2-9[/quote]

$ 64x^3\plus{}343\equal{}(4x\plus{}7)(16x^2\minus{}28x\plus{}49)$ Note: You can factor this further in complex numbers, and you can find it quite easy with quadratic furmula

$ x^4\plus{}8x^2\minus{}9\equal{}(x^2\plus{}9)(x^2\minus{}1)\equal{}(x\plus{}3i)(x\minus{}3i)(x\minus{}1)(x\plus{}1)$


If $a,b$ and $c$ are three (not necessarily different) numbers chosen randomly and with replacement from the set $\{1,2,3,4,5 \}$, the probability that $ab+c$ is even is

$\textbf{(A)}\ \dfrac{2}{5} \qquad
\textbf{(B)}\ \dfrac{59}{125} \qquad
\textbf{(C)}\ \dfrac{1}{2} \qquad
\textbf{(D)}\ \dfrac{64}{125} \qquad
\textbf{(E)}\ \dfrac{3}{5}$

[hide]If C is odd, which there is a $\frac{3}{5}$ possibility,
ab has to be odd,
so a and b must both be odd,

odd x odd= $\frac{3}{5}\cdot\frac{3}{5}=\frac{9}{25}$

(chance c is odd) x (chance ab is odd)=$\frac{3}{5}\cdot\frac{9}{25}=\frac{27}{125}$

If C is even, which there is a $\frac{2}{5}$ possibility,
ab must be even,
so a and b must both not be odd

$1-\frac{9}{25}=\frac{16}{25}$

$\frac{16}{25}\cdot\frac{2}{5}=\frac{32}{125}$

$\frac{27}{125}+\frac{32}{125}=\frac{59}{125}$, B[/hide]

[hide="Answer"]The chance of either of the first two numbers being even is $\frac{16}{25}$, and the probability of the last being even is $\frac{2}{5}$, so we have $\frac{16}{25}\cdot \frac{2}{5}=\frac{32}{125}$ there.

The chance of both of the first two numbers being odd is $\frac{9}{25}$, and the probability of the last being odd is $\frac{3}{5}$, so we have $\frac{9}{25}\cdot \frac{3}{5}=\frac{27}{125}$ there.

Adding the probabilities, we have $\frac{32}{125}+\frac{27}{125}=\frac{59}{125}\Rightarrow \boxed{B}$}.[/hide]


AMC
AIME
linear algebra
matrix
geometry
geometric transformation
rotation

Solve the recursion, along with proof:

$ x_n \equal{} \frac {\sqrt {2}}{2}(x_{n \minus{} 1} \minus{} y_{n \minus{} 1}) \plus{} 10$

$ y_n \equal{} \frac {\sqrt {2}}{2}(x_{n \minus{} 1} \plus{} y_{n \minus{} 1}), \ n \in \mathbb{N}$

Write the explicit formula for $ x_n, y_n$ in terms of $ x_0, y_0$.

hello, what is with the initial values of the recursions?
Sonnhard.

I don't think I understand what you meant.  Just let $ x_0 \equal{} a, y_0 \equal{} b; \ a,b \in \mathbb{R}$ and write the closed-forms in terms of $ a,b$.

hello, ok, that is what i meant.  Thanks.
Sonnhard.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1038337#1038337]AMC 2008A #25[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187900]Turkey TST 1997[/url].  The technique only requires one modification; the keyword is [url=http://en.wikipedia.org/wiki/Affine_transformation]affine transformation[/url].

This is essentially [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088469#1088469]AIME II 2008, #9[/url].

There is a nice variety of solutions in that post, but let me show you
[hide="this one!"] It is possible to write this recursion as follows:

$ \left[ \begin{array}{c} x_n \\
y_n \\
1 \end{array} \right] = \left[ \begin{array}{ccc} \frac {\sqrt {2} }{ 2} & - \frac {\sqrt {2} }{2} & 10 \\
\frac {\sqrt {2} }{2} & \frac {\sqrt {2} }{2} & 0 \\
0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} x_{n - 1} & y_{n - 1} & 1 \end{array} \right].$

From a high-concept point of view, what we are doing is something like writing $ x_n = \frac {X_n}{Z_n}, y_n = \frac {Y_n}{Z_n}$ and consider the system as a recursion in three instead of two dimensions, and then taking an affine slice by keeping $ Z_n$ fixed; see [url=http://en.wikipedia.org/wiki/Projective_space]projective space[/url].

It remains to diagonalize the above matrix, which involves finding a suitable set of eigenvectors (and that calculation corresponds to finding the center of rotation in one of the solutions TZF linked to).  The calculations are still somewhat tedious but at least now they can be completely mechanized; moreover, the eigenvalues are the same as if the $ 10$ were not there, which is a good way to see that the result is still a rotation and a translation. [/hide]


trigonometry
parameterization
calculus
derivative
calculus computations

If $x = 2 \cos \theta$, $y = \sin 2\theta$, how would you eliminate the parameter $\theta$? I am trying to find the points on the curve where the tangent is horizontal and vertical. I figure that the cartesian equation is an ellipse. But I tried this:

$x = 2\cos \theta$, $y = 2\sin \theta \cos \theta$, but I cant use the identity $\sin^{2}\theta+\cos^{2}\theta = 1$. I took the derivative and got:

$\frac{-\cos 2\theta}{\sin \theta}$ but I dont see how this would help if I am trying to find the specific points.

Any help is appreciated.

Thanks

[quote="mlok"]Since $y=2\sin \theta\cos\theta$, it follows that $(x/2)^{2}+(y/x)^{2}=1$.

I think the curve looks like this: $\infty$[/quote]

Very keen guess:

[url=http://allyoucanupload.webshots.com/v/2005945213650995389][img]http://aycu34.webshots.com/image/6753/2005945213650995389_rs.jpg[/img][/url]

Note that since $y = 2\sin \theta \cos \theta$ and $\frac{x}{2}= \cos \theta$, $y = (2\sin \theta )\frac{x}{2}= x \sin \theta$.

Therefore, $\sin \theta = \frac{y}{x}$.

As a result, $\sin^{2}\theta+\cos^{2}\theta = (\frac{y}{x})^{2}+(\frac{x}{2})^{2}= 1$.

That's where mlok's equation came from.


For $a,b,c,d$ positive reals prove that

$\frac{d+a^2}{d+bc}+\frac{d+b^2}{d+ca}+\frac{d+c^2}{d+ab}\geq3$

Very nice!!! You apply AM-GM on the left side to the 3 fractions. Then You have to prove: (d+a^2)(d+b^2)(d+c^2)=>(d+ab)(d+bc)(d+ac). After multipliing the terms you have: d^2(a^2+b^2+c^2)+d(a^2b^2+b^2c^2+c^2a^2)=>d^2(ab+bc+ca)+d(a^2bc+ab^2c+abc^2) which follows from the inequalities a^2+b^2+c^2=>ab+bc+ca and a^2b^2+b^2c^2+c^2a^2=>a^2bc+b^2ac+c^2ba.
The first is obvious the second follows from AM-GM: a^2b^2+b^2c^2=>2ab^2c.

Misha


[quote="triplebig"]Let $ x$ and $ y$ be two real numbers whose sum is $ 8$ . Show that:

$ \left(x \plus{} \frac {1}{y}\right)^2 \plus{} \left(y \plus{} \frac {1}{x}\right)^2\geq \frac {289}{8}$[/quote]
Try $ x\equal{}\minus{}0.1$ and $ y\equal{}8.1.$ :wink:

[quote="arqady"][quote="triplebig"]Let $ x$ and $ y$ be two real numbers whose sum is $ 8$ . Show that:

$ \left(x \plus{} \frac {1}{y}\right)^2 \plus{} \left(y \plus{} \frac {1}{x}\right)^2\geq \frac {289}{8}$[/quote]
Try $ x \equal{} \minus{} 0.1$ and $ y \equal{} 8.1.$ :wink:[/quote]
This inequality is true with $ x,y>0$ and $ x\plus{}y\equal{}8$

[quote="mathson"][quote="triplebig"]Let $ x$ and $ y$ be two real numbers whose sum is $ 8$ . Show that:

$ \left(x \plus{} \frac {1}{y}\right)^2 \plus{} \left(y \plus{} \frac {1}{x}\right)^2\geq \frac {289}{8}$[/quote]

Using calculus make the prove easier.[/quote]
Are you sure? this inequality is false :wink:

using the cauchy shwartz inequality :

$ (x \plus{} \frac {1}{y})^2 \plus{} (y \plus{} \frac {1}{x})^2\geq \frac {(x \plus{} y \plus{} \frac {1}{x} \plus{} \frac {1}{y})^2}{2}\geq\frac {(x \plus{} y \plus{} \frac {4}{x \plus{} y})^2}{2} \equal{} \frac {289}{8}$

[quote="enndb0x"]Its not clear how you used the Cauchy .Be more clear ![/quote]

he multiplied by $ \frac{(1\plus{}1)} {2}$


a compound A gives the following reaction
1)decolourises  acidic $ \ce{KMnO4}$ with evolution of $ \ce{O2}$
2)decolourises  alkaline  $ \ce{KMnO4}$ with evolution of $ \ce{O2}$
3)Removes black colour from old oil painitings
identify the compound with necessary reactions

h2o2.
h2o2 reacts with both acidic n alkaline kmno4 giving mnso4 and mno2 respectively and reacts with pbs present in oil paintings to pbso4

my prof said ( and notice he always suffixes a 'man' term )
See man if you ever find this description
' Removes black colour from old oil painitings '
then directly write H2O2 man  :D
Sorry i couldnt get this wasnt online for 2 days  :o


There are 500 students in a middle school. The principal has to choose 4 for the mathcounts team. How many different possible teams could he pick? [hide=" obvious hint"] Order does not count[/hide]

oops! :oops:  I must of punched in the wrong numbers on the calulator. And I'm usually bad with these kind of problems.

[quote="biffanddoc"][quote="pianoforte"][hide]
500x499x498x497=58444626500? :? [/hide][/quote]
You did permutations. And $500*499*498*497=61752747000$. The answer is...
[hide="click to see solution and answer"]
The formula for combinations is : $ \frac{n!}{(n-r)!r!}$
where $n$ is the number to choose from and $r$ is the number to choose. Substituting we get $ \frac{500!}{496!4!}$ or $ \frac{500*499*498*497}{4!}$ or $\frac{500*499*498*497}{24}$ Use a calculator and get $2573031125$. WOW thats a lot of options.[/hide][/quote]

Yes i think you are correct! :) [hide=" basically all you are doing is "]  500 choose 4[/hide]

This is just a normal combination question.

[hide]500 choose 4[/hide]

[quote="236factorial"]Why is this highschool?

Is there a quick way to do 500 choose 4 though?[/quote]

I remember putting it in basics. :o

[quote="IntrepidMath"][hide]
$\displaystyle {500\choose4}=(\frac{500!}{496!4!})=\frac{500*499*498*496}{4*3*2*1}=2573031125$

[/hide][/quote]

I'm sure this is what you meant. did you use a calculator?  :P

even TI30's series have Combination and Permutations keys


I'm sure you're all excited about your AMC scores, but please don't discuss them here until tomorrow.  To keep your mind occupied, I have created a second mock AMC.  Since I happen to like TeXing, I created them in LaTeX and generated a PDF, which you can view or download.  Have fun.

The PDF and other file formats are [url=http://mathcontests.0catch.com/mock_amc_2.htm]here[/url], or below.  Solutions are incomplete at this time.

Enjoy,
mathfanatic

uh do you have solutions or the answers at least?

ask questions here.  (assuming we're not sending in answers to mathfanatic).  anyways, it'd take really long time to tex up solutions...

Yeah, I have answers to most of these.

I hadn't given much thought to how this mock AMC would be run, so go ahead and post your solutions here.

Very decent questions.

Just a comment.

You guys really have lots of spare time doing all this work.(mathfanatic)

Another comment.

You guys are really nice people to be helping others this much. Thank You.

I took off one problem yesterday after someone noted its similarity to an actual one on the AMC.  I added it back now.

The answer to the last problem is [color=white]500[/color], is it not?

Are these questions AMC 10 or AMC 12 level?

#25 is an olympiad level problem. I have seen a very similar problem in the 1995 Canadian Math Olympiad.

Hard, yes -- probably harder than most of the things that ever appear on the AMC.  But not olympiad level, I don't think -- it could be mid-to-hard question on the AIME.  Actually, it looks like it might be an AIME question.

Well, it's sort of unusual, but there's only one thing to do, so I don't think it's that hard.

This is nothing compare to USAMO,but here is the first of a Canadian Math Olympiad:
1.  Find g(1/1996) + g(2/1996) + g(3/1996) + ... + g(1995/1996) where g(x) = 9^x/(3 + 9^x).

I agree that this is like an AIME 10th or 9th problem

I have a feeling that it is 1, but I have not found a complete solution for it yet.

Nope, you're a bit low there :)

Ignobilium, your answer is correct.

Oh, I don't know if you're taking correction for your questions, but you have a typo in question 1.  As does number 14.

Number 19 is nice.

What is this FAU/Stuy Alumni Math Competition?

[quote="ml2"]mathfanatic, i belive that your solution for no. 3, factors to x^4-x^2*y^2+y^4 and not the asked for x^4[b]+[/b]x^2*y^2+y^4,
Is there a solution to the question which is stated? (factor x^4+x^2*y^2+y^4 into two quadratics?)[/quote]

My solution (x^2 + xy + y^2)(x^2 - xy + y^2) is correct, or so says my TI-89.

Joel - I'll fix those typos.  Thanks.  As for what FAU-Stuyvesant Alumni is, I found it [url=http://www.math.fau.edu/MathematicsCompetition/Home.htm]here[/url].   Interestingly enough, the Stuyvesant part refers to the Stuy alumni association in Florida.  The competition is for high school students - it's just run by Stuy alumni in Florida as well as Florida Atlantic University, I think.

Ah, okay.  Yes, there's a fairly active alumni association branch in Florida -- goodness knows why.  I certainly don't.  Thanks.

[quote="JBL"]Yes, there's a fairly active alumni association branch in Florida.[/quote]

Maybe because the weather is generally better in Florida than in New York? Actually, several times that I have visited varying cities in Florida I have been amazed at how many New Yorkers I have encountered there. Some are what we call "snow birds" here in Minnesota, people who leave the north only in the winter. (Snow birds from Minnesota mostly go to Arizona in the winter.) Others are one-time New Yorkers who now live in Florida year-round. I could well believe that a New York high school might have a more active alumni association in Florida than in New York.

[quote="Tare"]13.) (3:^2:*:rt5:):^2:=405[/quote]

Right as rain.  Hrm.  I used to say that a lot.

Perhaps a simpler solution to (3) is [color=white]to note that

x^4 + x^2y^2 + y^4 = (x^2+y^2)^2 - x^2y^2

then use difference of two squares.[/color]

Quote:If , find .[hide]
Since log3(x) = 2, then x = 32 = 9.

Since logy(5) = 2, then y2 = 5 . . . y = sqrt{5}.

Therefore: (xy)2 = [9sqrt{5}]2 = 405[/hide]

Quote:The first three terms of a geometric sequence are:
, , and  for some value of .
Find the sum of all possible values for the fourth term.[hide]

We have:

 [1] a1 = 2x - 4

[2] a2 = 5x - 7 = r(2x - 4)

[3] a3 = 10x + 4 = 4(5x - 7)

From [2] and [3], we have: x = (4r - 7)/(2r - 5) = (7r + 4)/(5r - 10)

which simplifies to the quadratic: r2 - 8r + 15 = 0 ... with roots: r = 3 and  r = 5.

If r = 3, then x = 5 ... and and the progression is: 6, 18, 54, 162, . . .

If r = 5, then x = 13/5 ... and he progression is: 6/5, 6. 30. 150, ...

Therefore, the sum of the fourth terms is: 162+ 150 = 312.[/hide]

[quote="denizuzun"]what is the solution for no.25[/quote]
Sorry to bring back an old topic.

Here's how I did it (and I did not think it was even Medium AIME level):

[hide]
consider the pairs of terms for which the numerators of the arguments sum to 1001.  We have
$
f\left( \frac{n}{1001}\right)+f\left( \frac{1001-n}{1001}\right)
=\frac{4^{n/1001}}{4^{n/1001}+2}+\frac{4^{1001-n/1001}}{4^{1001-n/1001}+2}
=\frac{4+2(4^{n/1001})+4+2(4^{1001-n/1001})}{4+2(4^{n/1001})+2(4^{1001-n/1001})+4}
\\=1
$
There are 500 such pairs of terms, and thus the desired sum is [b]500[/b]
[/hide]

does 24 diverge? if not then i'm doing something wrong. can someone explain?

No, it doesn't diverge.  Look at beta's solution (you have to highlight it) to see his solution...ask again if you still don't understand.

Is the answer for number 21 [hide]$x=1334667 ?$[/hide]

I thought 21 was the dice problem.  The answer should have been a fraction.  I wrote a calculator program to figure out 24.  :blush:

Hi, sorry I know this mock contest came up a long time ago, but I could really use this as I hope to make it to the USAMO this year.

Are the solutions to this thing up somewhere now?

[quote="kool_dudy"]Try first post on this topic.[/quote]

They only have the solutions to the first 7 problems

I decided to go ahead and write up a solution packet.  I'm not finished yet though...I should have it up tonight or maybe tomorrow...it depends on how busy I am.  :)

Here ya go.  Mock AMC 2 Solutions :lol: (hopefully they are all correct...)

Looks  good. Nice solutions and formatting. :)

But you might want to fix #21's arrangement.

[quote=asdf334]24 is really darn similar to another q i've seen (sorry for bumping a 14 year old post)[/quote]

Yes, a problem [i]very[/i] similar to this showed up on the 2018 AMC 12A. :)

[quote=2018 AMC 12A #19]Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum\[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots\]of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?[/quote]


Prove that if $ \{x,y,z\}\subset\left(0,\infty\right)$ and $ \sum\frac {1}{2x\plus{}1}\equal{}1$ , then $ 1\plus{}(1\plus{}x)(1\plus{}y)(1\plus{}z)\ge 9xyz$ .

[quote="Virgil Nicula"]Prove that if $ \{x,y,z\}\subset\left(0,\infty\right)$ and $ \sum\frac {1}{2x \plus{} 1} \equal{} 1$ , then $ 1 \plus{} (1 \plus{} x)(1 \plus{} y)(1 \plus{} z)\ge 9xyz$ .[/quote]

It's Schur inequality.  :wink:

Let $ \frac{1}{2x\plus{}1}\equal{}a,\frac{1}{2y\plus{}1}\equal{}b,\frac{1}{2z\plus{}1}\equal{}c$. Then $ a\plus{}b\plus{}c\equal{}1$ and our inequality is equivalent to
\[ 1\plus{}\prod\left(\frac{1\plus{}a}{2a}\right)\ge 9\prod\left(\frac{1\minus{}a}{2a}\right)\\
\Leftrightarrow 1\plus{}9abc\ge 4(ab\plus{}bc\plus{}ca)\]
which is Schur's inequality.

[quote="Virgil Nicula"]Prove that if $ \{x,y,z\}\subset\left(0,\infty\right)$ and $ \sum\frac {1}{2x \plus{} 1} \equal{} 1$ , then $ 1 \plus{} (1 \plus{} x)(1 \plus{} y)(1 \plus{} z)\ge 9xyz$ .[/quote]
By Nayels method we easily find that it's equivalent to:
$ 1 \plus{} 9abc\ge 4(ab \plus{} bc \plus{} ca)$. Homogenizing gives and multiplying by 2 gives: $ 2(a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} a^2c \minus{} b^2a \minus{} b^2c \minus{} c^2a \minus{} c^2b \plus{} 3abc) \equal{} (a \plus{} b \minus{} c)(a \minus{} b)^2 \plus{} (a \minus{} b \plus{} c)(a \minus{} c)^2 \plus{} ( \minus{} a \plus{} b \plus{} c)(b \minus{} c)^2$
Because of symmetri we can assume $ a \ge b \ge c$. Then $ (a \minus{} c)^2 \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2$. So:
$ (a \plus{} b \minus{} c)(a \minus{} b)^2 \plus{} (a \minus{} b \plus{} c)(a \minus{} c)^2 \plus{} ( \minus{} a \plus{} b \plus{} c)(b \minus{} c)^2 \ge (a \minus{} b)^2(a \plus{} b \minus{} c \plus{} a \minus{} b \plus{} c) \plus{} (b \minus{} c)^2( \minus{} a \plus{} b \plus{} c \plus{} a \minus{} b \plus{} c) \equal{} 2a(a \minus{} b)^2 \plus{} 2c(a \minus{} b)^2 \ge 0$. With equality iff $ a \equal{} b \equal{} c$. (Or $ a \equal{} 0$ and $ c \equal{} 0$ but $ a \ge b \ge c$ so $ b \equal{} 0$ giving $ a \equal{} b \equal{} c \equal{} 0$ :) )