Let $ a$ be a real constant. Removing $ \theta$ in the equations we have:

$ \begin{cases}x\sin\theta \plus{} y\cos\theta \equal{} 2a\sin\theta \\
x\cos \theta \minus{} y\sin\theta \equal{} a\cos\theta\end{cases}$

We have:

$ a)\; (x \plus{} y)^{\frac {2}{3}} \plus{} (x \minus{} y)^{\frac {2}{3}} \equal{} 2a^{\frac {2}{3}} \\
b)\; (x \minus{} y)^{\frac {2}{3}} \minus{} (x \plus{} y)^{\frac {2}{3}} \equal{} 2a^{\frac {2}{3}} \\
c)\; (x \plus{} y)^{\frac {2}{3}} \plus{} (y \minus{} x)^{\frac {2}{3}} \equal{} a^{\frac {2}{3}} \\
d)\; (x \plus{} y)^{\frac {2}{3}} \plus{} (x \minus{} y)^{\frac {2}{3}} \equal{} \frac {a^{\frac {2}{3}}}{2} \\
e)\;\text{None}$

Apparently, this is too easy for everyone else, so I figured I'd take a stab at it.

$ (x \minus{} 2a)\sin\theta \equal{} \minus{}y \cos\theta$
$ (x \minus{} a)\cos\theta \equal{} y\sin\theta$
$ (x \minus{} 2a)(x \minus{} a) \equal{} \minus{}y^2$
$ 2a^2 \minus{} 3ax \plus{} x^2 \plus{} y^2 \equal{} 0$

The answer has a lot of cube roots and we don't have any cube roots. We plug in $ x \equal{} 4$ and $ y \equal{} 1$ and that gets rid of all the answer choices. So the answer is E.

I don't think this is too easy for everyone else, because I have posted really easy problems which always came with replies.

Differ, first of all the idea is to [i]solve[/i] the problem, even though in the exam we would not care about solving but care about getting it right. Here in the forums I am thinking the purpose of me posting the problem is because I want to see how to reach the correct alternative. Second of all, it seems you are wrong, for that is not the correct answer.

You are correct, of course. I should have solved the problem completely. There must be a restriction hidden within the first two equations that precludes $ x \equal{} 4$ and $ y \equal{} 1$; however, I cannot find it.

There must be something wrong with the statement (i.e. with the first four choices OR with the initial system). For $ \theta\equal{}{\pi\over 4}$, the system becomes $ x\plus{}y\equal{}2a\land x\minus{}y\equal{}a$, but those values don't satisfy either of the four given formulas.

ITA (Aeronautic Technological Institute) Exam, here in Brasil. I don't have the exact year though.

Is this problem wrong?


abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove that a transitive abelian subgroup of $S_n$ has exactly $n$ elements.

P.S.

This is somewhat related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46589]this other problem[/url].

anyone a solution? :) seems simple and elegant.

since nontrivial element $G$ ,transitive abelian subgroup,do not have fixed points we conclude ,cardinal of $G$ doesnt exceed $n$ .(should be exactly $n$ ,as stated in problem)

Take $f$ in $G$ and suppose that there exists some $a \in [[1;n]]$ such that $f(a)=a$.
We know that for any $g \in G$, $f=gfg^{-1}$. So if $f=c_1c_2..c_r$ is the cycle decomposition of $f$, any $g \in G$ maps the element of $c_i$ to the elements of $c_j$. And because $G$ is transitive, this shows that all the $c_i$ have the same length. Hence $f$ is the identity: any non trivial element of $G$ has no fixed point.

From here the conclusion is easy. Take any $a \in [[1;n]]$ : its orbit under the action of $G$ is $[[1;n]]$. Hence $n=\frac{Card(G)}{Card(Stab(a)))}=Card(G)$.

It's so easy :-)

Let $G \leq S_{n}$ be transitive and abelian. Consider

$\alpha : G \to \{1,...,n\}, g \to  g(1)$

Since $G$ is transitive, $\alpha$ is surjective. To show that $\alpha$ is injective, assume $g \neq h$ in $G$. There exists $j \in \{1,...,n\}$ s.t. $g(j) \neq h(j)$. Since $G$ is transitive, there exists $f \in G$ s.t. $j=f(1)$. Now, $G$ is abelian, hence

$f(g(1)) = g(f(1)) \neq h(f(1))=f(h(1)) \Rightarrow g(1) \neq h(1)$


Tamara knows that the arithmetic mean of her five quiz scores is 95%. However, she has misplaced one of these quizzes. The ones she can find have scores of 100%, 100%, 99% and 98%. What is her score on the misplaced quiz?

If the mean is 95, then the sum of the five scores must be $ 95\times 5\equal{}475$.  Now subtract off the other 4: $ 475\minus{}100\minus{}100\minus{}99\minus{}98\equal{}78$.  She must have messed up bad. :D

for these problems, i dont like the "normal" method since i dont like multiplying :P . so this is how i do these "mean"(no pun intended) problems

since her first score is 100, and her mean is 95, you have to subtract 5 to get to 95. same thing for the second quiz. for the third quiz, you have to subtract 4.

continuing in this fashion, you get -5,-5,-4,-3. add it all up and you get -17.

so now u do 95(her average) and -17. that get you 78 :wink:

That method is a nice method, but only when the numbers are easily worked with.  For most problems of this nature, you should probably stick to the usual method.


limit
integration
real analysis
real analysis unsolved

Let us denote $ p={10}^{10}\;  ,\; \;  X_n=\displaystyle \frac{1}{p} \left(1+
2\sum\limits_{k=1}^{n+p}\exp{\left(-\frac{k^2}{p}\right)}\right)^2\;
. $  [b] Prove or disprove that [/b]    $\; \; \lim\limits_{n\to \infty}n\cdot\sin(X_n)=0\; . $

It is obvious that $ X_{n}(p)$ converges and i found that it extemely close to $ \pi$ for big $ n$ and $ p$. And i think that $ \lim_{p\to\infty}{\lim_{n\to\infty}{X_{n}(p)}} \equal{} \pi$.

But $ \lim_{n\to\infty}{X_{n}(p)} \equal{} X(p)\ne\pi$.

$ p$ is very big, but not $ \infty$, so:

$ \lim\limits_{n\to \infty}n\cdot\sin(X_n) \equal{} \infty$

We should find a good approximation for the sum $ \sum\limits_{k \equal{} 1}^{n \plus{} p}\exp{\left( \minus{} \frac {k^2}{p}\right)}$.

For big $ n$ it should be $ \sum\limits_{k \equal{} 1}^{n \plus{} p}\exp{\left( \minus{} \frac {k^2}{p}\right)}\approx\frac {\sqrt {\pi p} \minus{} 1}{2}$.

By Euler's summation formula: $ \sum\limits_{k \equal{} 1}^{n \plus{} p}e^{ \minus{} \frac {k^2}{p}} \equal{} \int_{1}^{n \plus{} p}e^{ \minus{} \frac {x^2}{p}}\;dx \minus{} \frac {1}{2}\left(e^{ \minus{} \frac {(n \plus{} p)^2}{p}} \minus{} e^{ \minus{} \frac {1}{p}}\right) \plus{} R$, where $ R$ is small.


People generally say that Phenyl group has greater migratory aptitude than H.. Is there any specific reason for this or is this just reaction dependent???

Here is a nice link. Actually relates to Pardesi's thread on stability (Pinacol-Pinacolone Rearrangement):

http://www.cem.msu.edu/~reusch/VirtualText/rearrang.htm

actually there is a wonderful article on this in Peter Sykes...which clearly explains the various cases where migration aptitude is involved and how to predict which si the best :)

Actually H has greater migratory aptitude: the Bayer-Villiger rearrangement is a good example, where one cannot obtain formates from aldehydes, but get acids instead.

Phenyl still has the brgeater migratory apptitude coz in the transition state for migration , the phenyl carbon retains its tetravalency whereas in case of alkyl group it does not .... so phenyl grp has greatest migratory apptitude

@valerium nothing in general can be told about the migratory aptitudes it depemds on the reaction conditions and on steric factors also..

But we can still tell abt the transtion state during the transfer pardesi....
The product which is formed via the most stable transition state will be the most stable product...

what if the phenyl  group approaches already present two phenyl groups??

Then that only causes some steric hindrance...
and moreover u should not consider those special cases...


In the wiki, if you type asymptote code such as this :
[code]<asy>
size(200pt);
defaultpen(1);
pair A=(0,0), B=(220,0);
label("A",A,SW);
label("$B$",B,SE);
</asy>[/code]
you will get a correct image in preview, like this :
[img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/e/7/5/e75cbb71f340fd97e702d56a7764a4c9b3eab535.png[/img]
But when you save the page, you will see this :
[img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/2/9/0/290eb104186d0bd324d7275de0d2f0b9c349d012.png[/img]
Apparently, the dollar signs are converted to <math></math> tags, and in text mode in LaTeX, < and > become $ \text{ < }$ and $ \text{ > }$, so what appears is something like $ \text{ < math > B < /math > }$, which is what you get with this code :
[code]$\text{<math>B</math>}$[/code]
If you replace the $ \$$. . .$ \$$ with \( . . . \) or \begin{math} . . . \end{math}, the picture works fine, though.

I've previously posted this problem, but no one responded...


1. http://www.mathlinks.ro/viewtopic.php?t=5830
2. http://www.mathlinks.ro/viewtopic.php?t=5831
3. http://www.mathlinks.ro/viewtopic.php?t=5832
4. http://www.mathlinks.ro/viewtopic.php?t=5833
5. http://www.mathlinks.ro/viewtopic.php?t=5834
6. http://www.mathlinks.ro/viewtopic.php?t=5835

Thanks.

no, thank you for sharing the problems with us. What did you do in the contest? :)

Well I'm very lucky in this contest, and I'll participate 2004 IMO in Greece.
Sorry for my poor English in the above postings.. :blush:

don't worry, your english is fine :) all of us make small mistakes from time to time, but we don't apologize all the time for that :) once is enough :P

anyway, which other members of ML from Korea will be joining in Greece? :)

I have spent some time to make this pdf a collection of problems from this TST. I have a word for it: I would like to know if problem #4 is translated the way it's supposed to be. If there is anything else that is also wrong, please tell me so I can correct the pdf file. Please enjoy.

Well I'm not sure that there is another member in ML, because this homepage isn't famous well in Korea. sorry.

And thanks 3X.lich for making PDF type of Korean TST! I think your word isn't wrong in #4. Thanks a lot :)


Prove that if $ a,b,c,d\geq0$ then
\[ \sum_{\rm cyclic}\frac{d^{3}\plus{}abc}{(d\plus{}a)(d\plus{}b)(d\plus{}c)}\geq1.\]

Once we make a full expansion (In this case with realitively little computation) it reduces to $ [3,2,1,0]$ majorises $ [2,2,1,1]$
but I guess you have a nicer solution... :lol:

$ \frac{x}{x\plus{}y}\plus{}\frac{y}{x\plus{}y}\equal{}1$ gives proof. :D

Hello fjwcls. From the linke, I saw some of your inequalities for the first time. I am interested in the following inequality
\[ \sum_{\rm cyclic}(a^{3}\minus{}bcd)(a\minus{}b)(a\minus{}c)(a\minus{}d)\geq0\]
as well as your method to solve such kinds. It seems to me that you use software. Otherwise, what?

Another less direct approach is to show
\[ \sum_{\rm cyclic}\frac{d^{3}}{(d\plus{}a)(d\plus{}b)(d\plus{}c)}\geq\frac{1}2,\]
which was discussed already in this forum.

Next, we show that\[ \sum_{\rm cyclic}\frac{abc}{(d\plus{}a)(d\plus{}b)(d\plus{}c)}\geq\frac{1}2,\]
which is turned into the above one by substituitions $ a\equal{}1/x,$ ...


Find all $f: (1, \infty ) \to R$ such that : $f(x)-f(y)=(y-x)f(xy)$

$f(1)-f(y)=(y-1)f(y) \Rightarrow f(y)=\frac{f(1)}{y}\vee [ f(1)=0 \wedge f(y)=0] \;\; \forall y \in \mathbb{R}^{+}$
$f(1)=0 \Rightarrow 0-f(y)=(y-1)f(y) \Rightarrow f(y)=0 \Rightarrow f(y)=\frac{f(1)}{y}\;\;\forall y \in \mathbb{R}^{+}$
$f(x)=\frac{f(1)}{x}\Rightarrow f(1)(\frac{1}{x}-\frac{1}{y})=f(1)\frac{y-x}{xy}\Rightarrow f(x)-f(y)=(y-x)f(xy) \;\;$
$\forall x, y \in \mathbb{R}^{+}$.

No problemo. ¿?


In how many ways can 7 identical marbles be placed in a red bag, a white bag, and a blue bag if there must be at least one marble in each bag?

This is the same as distributing 4 identical marbles to the three bags, then placing 1 marble into each bag. By that balls and urns argument thing, there are $ \binom{6}{2}\equal{}15$ ways to do this.

1+2+5 x6
1+1+6 x3
1+3+3 x3
2+2+3 x3

= 6 + 3 +3 +3 = 15 ways

Brute Force can u explain your balls and urns argument?

[hide]Basically, we have $ 7$ marbles in a row, and we want to place two partitions in between them to separate them into three different bags. However, since each bag must have at least one marble, we remove one marble from each bag, and we are left with $ 4$ marbles to work with, with no restrictions. By Balls and Urns, the amount of ways to do it is $ \displaystyle \binom{6}{2}\equal{}\boxed{15}$.

One example of a balls and urns arrangement is {1, 2, 4}:
*|**|****
In this case, the * are the balls, and the | are the dividers. The diagram would be interpreted as 1 ball in the red bag, 2 in the white bag, and 4 in the blue bag.[/hide]

I understand about the partitions, but when u say balls and urns make it 6C2 is that a specific theorem and if so can you show the general rule and derivation, because I dont understand how the number of ways to put 4 balls into 3 different bags is 6C2

The search function is useful here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1603770934&t=137177

What if you were able to put one bag into another?!

(This should be interesting...)


Heres a fun problem. Not sure how difficult it is..

You have a straight edge that can:
a) Connect two points
b) Construct the perpendicular to a line from a given point on that line.

Show how to construct the perpendicular to a line from a given point *not* on that line.

P is the point we are trying to find the perpendicular to. See attached.

Edit: How can I put a picture up in non-spoiler mode?

Edit: Meh, I'll just leave it for someone else to do anyways.

-- James

I think I'm missing something, but:

[hide]construct any perpendicular to a perpendicular, giving a parallel.  Let that parallel go beyond the original line.  Then we can construct a perpendicular to the parallel beyond the original line.  Or maybe if this is allowed you can extend the original line...[/hide]

Edit: never mind this post; I misunderstood the problem.

I don't get what you are saying.. how do you make that go through the given point?

edit - maybe you misunderstood the question. You have a given line (line, not line segment - lines are infinitely long) and a given point somewhere thats not on the line. You want to construct the line that goes through that point and is perpendicular to the line. I don't see how what you have written works.. you haven't mentioned the point anywhere.

My solution hinges on the fact that we can construct midpoints, as follows.

From segment AB, draw perpendiculars at A and B, choose a point on one of these, and draw another perpendicular to create rectangle ABCD. Draw diagonals AC and BD, intersecting at E. Draw the line through A perpendicular to AC, and draw the line through B perpendicular to BD, and let these two lines intersect at F. Let segment EF intersect AB at M. M is the midpoint of AB. The proof that this works is straightforward.

So, here's the construction. From the point P, draw a line that intersects the line m at A, and draw the line perpendicular to PA through P, and let it intersect line m at B. (If this line doesn't intersect line m, choose a different A). Let C, D, E be the midpoints of AP, PB, and AB respectively, and let F be the midpoint of CP. Let lines CE and DF intersect at G. Draw PG. Draw the perpendicular to PG through P. This is the desired line.

For the proof that it works, just use similar triangles to prove that PG is parallel to CD, and so it is parallel to line m.

A slight variation, using only one midpoint:

Constructs perpendiculars m and n of AB, with A on m and B on n. Construct line AP and BP, meeting m and n at C and D, respectively. Suppose that |m| < |n|. Construct rectangle ABED (with E on n), thereby constructing the midpoint M of segment AB. Join EM and extend to meet the extension of m at R. Join CR. CR meets AB at S. PS is the desired line.

Proof:

Suppose that PQ is perpendicular to AB with foot Q. Then, PQ/AD = BQ/BA, BC/PQ = AB/AQ. Multiply to yield BC/AD = BQ/AQ.

Join DS. We have DA = EB (rectangle) and EB = AR (EBM :cong: RAM); thus DA = AR. Then, DAS :cong: RAS. Since RAS ~ CBS, we must have DAS ~ CBS. Thus BC/AD = BS/AS.

Thus BQ/AQ = BS/AS; we must conclude that Q and S are the same point, and thus PS is perpendicular to AB.

[quote="Osiris"]A slight variation, using only one midpoint:

Constructs perpendiculars m and n of AB, with A on m and B on n. Construct line AP and BP, meeting m and n at C and D, respectively. Suppose that |m| < |n|. Construct rectangle ABED (with E on n), thereby constructing the midpoint M of segment AB. Join EM and extend to meet the extension of m at R. Join CR. CR meets AB at S. PS is the desired line.
[/quote]

You can find the midpoint without breaking the rules.  Start with just P and A.  Draw line AP.  Draw a perpendicular to AP at P; call its intersection with the original line B.  Proceed with Osiris's construction until the midpoint is needed.  Now, draw perpendiculars to AP at A and BP at B; they meet at point F.  AFBP is a rectangle; its diagonals bisect each other.  Draw line FP, and call its intersection with AB point M.  Continue with the construction.

Yeah. I just took the midpoint construction from zabelman's post.

[quote="zabelman"]JamesMatthews, thats a really cool solution![/quote]

Hehe, thanks. :)

My bad.  I only skimmed the posts before replying.  Kudos to JamesMatthews for that simple solution.


[quote="guyinPA"]Prob. 7  How many ordered integral triples, (a,b, c) have the propery that each number is the product of the other two? I have the answer (which I can pm you) but need to know how it is arrived at, please.

 Prob. 8  Randomly choosing two numbers from the set (1,3, 5, 7, 9) with replacement, what is the probability that the product is greater than 40? Express your answer as a percent. (Again, I have answer, but need "how to" solution.[/quote]
Those are from the State test, not the Nat test.

[hide="#7"]
(0,0,0)
(1,1,1)
(1,-1,-1) and its variations, 3 in total

So 5 ORDERED pairs[/hide]

[hide="#8"]
There are 5x5 = 25 possible pairs.  The ones with a product greater than 40 are : 9x9, 9x7, 9x5, 7x9, 7x7, 5x9

Thats 6 out of 25 or 24%[/hide]

hope that helps,

frost

I got number 8.

[hide]24%.  The denominator is 25 because there are 5 numbers with replacement.  For the numerators of problems like these, I always draw a little chart.  So I just did a 5 by 5 grid, numbered the sides 1, 3, 5, 7, and 9, and put the products in the spaces, and then I shaded in the boxes with products higher than 40.  There were 6 (5 x 9, 7 x 7, 9 x 5, 9 x 7, 7 x 9, and 9 x 9), so the answer is 6/25, or 24%.[/hide]

This might not be the type of explanation you're looking for, and it might not even be right.  I did the problem really fast. :roll:

oops, problems were from the state test - sorry for the error -- but thanks for the help in finding the solution!


function
ratio
vector
algebra
rational function
superior algebra
superior algebra unsolved

Prove or disprove that $ \mathbb{Z}[[x]] \cap \mathbb{C}(x) \subset \mathbb{Z}(x)$.  In other words, if $ A(x) \equal{} a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{} ...$ is the generating function for an integer sequence and $ A(x)$ is also a rational function, then it must be the ratio of two integer polynomials.

[quote="t0rajir0u"]Prove or disprove that $ \mathbb{Z}[[x]] \cap \mathbb{C}(x) \subset \mathbb{Z}(x)$.  In other words, if $ A(x) = a_0 + a_1 x + a_2 x^2 + ...$ is the generating function for an integer sequence and $ A(x)$ is also a rational function, then it must be the ratio of two integer polynomials.[/quote]
We effectively have $ \mathbb{Z}[[x]] \cap \mathbb{C}(x) \subset \mathbb{Z}(x)$. Indeed, let us write $ A(x) = \sum_{n \geq 0}{a_n x^n} = \frac {P(x)}{Q(x)}$, where $ P,Q \in \mathbb{C}[X]$. We write $ Q(x) = \sum_{n = 0}^d{q_n x^n}$. Let $ (\omega_1,\dots,\omega_k)$ be a base of the vector space $ \mathbb{Q}(q_0,q_1,\dots,q_d)$ over the field $ \mathbb{Q}$. $ \forall i \in \{0,\dots,d\}$, we write $ q_i = \sum_{j = 1}^k{\lambda_{i,j}\omega_j}$.
Now, if $ D$ is the degree of $ P$, we know that, $ \forall n \geq D+1, \sum_{i = 0}^d{a_{n + i}q_{d - i}} = \sum_{j = 1}^k\left(\sum_{i = 0}^d{a_{n + i}\lambda_{d - i,j}}\right)\omega_j = 0$, which assures that $ \sum_{i = 0}^d{a_{n + 1}\lambda_{d - i,0}} = 0$.

Then, it follows that, if $ Q_0(x) = \sum_{i \geq 0}{\lambda_{i,0}x^i} \neq 0$ and $ P_0 = A Q_0 \in \mathbb{C}[[x]]$, $ P_0$ is in fact of degree at most $ d + D + 1$, which shows that $ P_0 \in \mathbb{C}[x]$. Moreover, $ A$ and $ Q_0$ have rational coefficients, and therefore $ P_0 \in \mathbb{Q}[X]$.
Therefore, $ A = \frac {P_0}{Q_0} \in \mathbb{Q}(X) = \mathbb{Z}(X)$.


ratio
geometry
trigonometry
modular arithmetic
trig identities
Law of Cosines
number theory

Ok, the first one I got and think is kinda nice, but the second one.. I have no idea how to approach.

1.  Quadrilateral $ABCD$ is inscribed in a circle with $AB=8, BC=9, CD=12$, and $DA=5$.  Diagonal $AC$ divides $ABCD$ into two triangles.  Find the ratio of the area of the smaller of the two triangles to the area of the larger one.

2.  (source: 2004 senior NSML) A sequence {$a_n$} is defined as follows: $a_n$ is the $n$th digit after the decimal point in the infinite decimal representation of the rational number $\frac{1}{n}$.  For example, $a_1=0$ since $\frac{1}{1} = 1.00000...$, and $a_6 = 6$ since $\frac{1}{6} = .1666666...$, and $a_8 = 0$ since $\frac{1}{8} = .125000000...$.
The first ten terms of this sequence are: $0,0,3,0,0,6,1,0,1,0$
How many of the first 20 terms of this sequence are non-zero?

The answer is [hide]8, but I don't know how to arrive at this logically.[/hide]

For number 1, all you have to do is simultaneously apply the law of cosines to find AC, then use Heron's. ;)

The answer to your quesiton is a little hairy.

A fact is that the multiplicative order of $10\pmod{n}$, where $n$ is an integer, is equal to the repeating places in the decimal expansion of the reciprocal of $n$.

Ever notice that a repeating decimal, for example, $0.2525\ldots = \frac{25}{99}$, or such that the repeating $n$ digits can be made into a fraction by placing them over $10^n - 1$.

So we have the multiplicative order being the smallest $e$ s.t. $10^e \equiv 1\pmod{n}$, where $n$ is given. In this case, $n$ is the number which we must find the repeating sequence for $\frac{1}{n}$.

But you would have to know the modular orders of all those number that unless you know them, I forget how you would derive them.

If you are desperate enough, just divide the number $n$ into a bunch of repeating 9's and the quotient is the multiplicative order. LOL.

[quote="eryaman"][hide="1"]I get $\boxed{\frac{5}{6}}$, by using the law of cosines from both triangles on $AC$ to find $\cos x$, then using that to find $\sin x$, then using the formula $\frac{1}{2}ab\sin C$ to find the areas of the two triangles (plus the fact that the angles are supplementary and that $\sin(180^{\circ}-x)=\sin x$.  Then divide the two areas to get the answer.  May have screwed up computation somewhere though.[/hide][/quote]

That's right, but keeping in mind that $\sin(180^{\circ}-x)=\sin x$ can you see a much quicker way through it just using ratios instead of calculating the areas?

About the second one, I guess 1/17 and 1/19 are the only ones that are terrible, and the rest of the contest was pretty easy that year. . . but if that's the way they intended it to be solved, I'd say that's a pretty bad problem.

[quote="djshowdown2"][quote="eryaman"][hide="1"]I get $\boxed{\frac{5}{6}}$, by using the law of cosines from both triangles on $AC$ to find $\cos x$, then using that to find $\sin x$, then using the formula $\frac{1}{2}ab\sin C$ to find the areas of the two triangles (plus the fact that the angles are supplementary and that $\sin(180^{\circ}-x)=\sin x$.  Then divide the two areas to get the answer.  May have screwed up computation somewhere though.[/hide][/quote]

That's right, but keeping in mind that $\sin(180^{\circ}-x)=\sin x$ can you see a much quicker way through it just using ratios instead of calculating the areas?[/quote]Oh I see, you can just use the $\frac{1}{2}ab\sin x$ on both triangles (since $\sin(180^{\circ}-x)=\sin x$, and divide the two, in which the $\sin x$ part cancels out, leaving you with your ratio.

[quote="Elemennop"][quote="white_horse_king88"][quote="Elemennop"]Just find the next ten terms for #2. It seems much easier to chug out a little bit of division for a minute or two than to try to find some sort of all-encompasing pattern.[/quote]

LOL. The multiplicative order of 17 is 16, so unless you divide alot! It's going to take alot of scrap paper. And for 19, it's 18. So raw division won't work out too nicely.[/quote]

Sure, but which do you think will generally end up being faster -- sitting for a couple minutes trying to find some distinct pattern that might point in the direction of a quicker way to handle this, or just flatout dividing in these numbers (and dividing 19 and 17 isn't terribly bad either)? I rarely get the most elegant approach my first time through a problem, especially during something like the AMC. There just isn't enough time to waste working on elegance when the solution can be brute forced just as quick or quicker most of the time, and I'm sure most other high scorers would agree.[/quote]

I'm pretty sure there's a way to calculate it with primitive roots, but I don't know if I would sit there and go through 18 divisions for 1/19. I mean, I guess one could do both in about 5 minutes, but you can't make an error either.

[quote="ritchjp"]for any numbers in the sequence to be zero $n$ must be be divisible by only $2$ and $5$ so the numbers $2, \ 4, \ 5, \ 8, \ 10, \ 16, \ 20$ end with a $0$ in the sequence. The rest are non-zero numbers.

The only exception is $1$[/quote]

This isn't true. Please read above.

[quote="ritchjp"]I did a little more investigating and fortunately I had access to a calculator that went out to about 30 decimal places.

What I found out was rather interesting.

In my last post I commented on the numbers that will yield a $0$ and I came up with numbers that were divisible by $2$ and $5$ only. That was only partially true.

Additional numbers I found that yielded a $0$ in the sequence are $11, \ 13, \ 17, \ 19$

The numbers with non-zero results were $12, \ 14, \ 15, \ 18$

Do you see a pattern?

The numbers that had non-zero results have two or more prime factors in them while the numbers that had results of $0$ in the sequence were unfactorable numbers. That may be why $1$ yields a $0$ in the sequence.

So  there are 8 non-zero numbers in the sequence.[/quote]

This pattern can't be taken on faith - it requires an explanation, which is relatively simple.  For prime numbers $> 10$ the first digit in the decimal expansion is $0$, and the $n^{th}$ digit will always be congruent to the $1^{st}$ digit $\bmod \varphi(p)$ (by 'congruent' I mean they occupy the same place in the repeating decimal expansion.)

For composite numbers $> 10$ it is untrue that $1 \equiv n \bmod \varphi(n)$ and hence there is not a guarantee for the digit to be $0$ (however, I suspect it's still possible - check $\frac{1}{34}$).


Ok, muchos me han preguntado cuando iba a volver: La respuesta esta a la vuelta de la esquina, el reto de la semana empieza de nuevo inmediatamente que reciba 15 posts de personas diferentes en este topic diciendo que participaran!!!

Espero que quienes posteen se comprometan a enviar sus soluciones!!!

Las reglas son sencillas: Cuelgo 4 problemas un dia, tienen una semana para enviarme sus soluciones a pascual2k1@hotmail.com en cualquier formato y luego de esa semana cuelgo otro conjunto de problemas, los resultados que llevaremos en un ranking y discutimos las soluciones...

Las inscripciones estan abiertas, para participar solo envie un mail con el nombre : Reto de la semana y sus respectivas soluciones especificando que problema resolvera.

Soluciones parciales seran calificadas tambien!!!

Ok, espero esas 15 posts que encesito para asegurarme que esta vez no sera un fracaso como ediciones pasadas en las que tuvimos solo 1 participante... Si no las hay el reto de la semana no se hara!!! Si no se ve reflejada una alta participacion el reto de la semana se detendra, esas son las reglas!!!

Eso es todo, espero ver respuestas...

Aprovechare la semana de descanso para participar :D .

Vale yo me anoto, una buena oportunidad para entrenar.

Yo también me les uno. Está perfecto pera no perder el ritmo

No hay limite de edad! asi que vamos a aprender todos de todos!

Solo va a haber un nivel....

[quote="Pascual2005"]
Solo va a haber un nivel....[/quote]

Si es así, yo también.

$Tipe$

tengo poca experiencia pero de todas manera hare el intento  :blush:

Va, yo le entro... Ya no soy olímpico pero necesito ver problemas, porque soy entrenador  :agent:  :D

Yo tambien, pero hay que trabajar enserio $\bigstar$

Hola a todos,
saludo la buena iniciativa .. Pascual ;)

veo q ya van 13 personas ...
solo hacen faltan  2  más ....

y si me permiten ... yo tambien me agrego ..
(solo faltaria uno :) )

Carlos Bravo
Lima  -PERU

Ok, mañana pondre la primera lista de problemas tratare de seleccioanrlos cuidadosmanente para que sean lo mas interesantes posibles y saquemos todo el provecho de ellos!!!

Chao y nos vemos mañana!!!

ok, los problemas ya estan posteados en la pagina de Olimpiadas de matemáticas!!! Suerte!!!

AKI EL ENLACE:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103444

SALUDOS,


Prove that
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9}{2}\sqrt{\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}}\geq\frac{9}{2}+3$
where $a,b,c$ are positive real numbers

The following is sharper
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{14}{3}\sqrt{\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}}\ge \frac{23}{3}\]
:)


Putnam
real analysis
real analysis unsolved

Hi,
here's a problem I think was not yet proposed here:

Evaluate

$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{m^2n}{3^m(3^m n+3^n m)}$

It has a pretty solution.

Ref: Putnam, 199?

Let ai=3 i/i.
We must evaluate  S=\sum 1/ai(ai+aj).
We have S= \sum 1/aj(1/ai-1/(ai+aj))= \sum 1/(ai*aj)- \sum 1/(aj*(ai+aj))[b](*)[/b]= \sum 1/(ai*aj)-S, so 2*S= \sum 1/(ai*aj)= \sum ij/3 i+j[b](*)[/b].
By grouping the terms with the same i+j [b](*)[/b], we get
2*S= \sum {n>1} (n+1)n(n-1)/(6*3 n), so
108*S= \sum (n+1)n(n-1)/3 n-2=f'''(1/3), where
f(x)= 1+x+x 2 +...=1/(1-x) for x in (-1,1).
f'''(x)=6/(1-x) 4, so we have 108S=3 5/8, so S=9/32.

Lets prove that  T=\sum {i+j=n} ij=n(n+1)(n-1)/6.
We have T= \sum i(n-i)= n*\sum i- \sum i 2 , and the result easily follows.

I used some uncertain facts. For them to be true I need to know that the sum is convergent. I hope it is. I don't know if I can prove it because the writing  \sum {ij} is not explicit. I need something like  \sum {i} \sum {j} or  \sum {i+j=n}.

If the sum is not known to be convergent then I can not group terms in the sum. It is a well known result of Riemman that states that is a series is such that it is not absolutely convergent then we can permute the terms of the series and obtain any sum for the series.
Where I used (*), the result depends on the convergence of the series.

I think that my proof holds if we have the sum in the form  \sum {i+j \geq 2} ...


inequalities
rearrangement inequality
inequalities proposed

Hey guys, i came across this one recently and and am drawing a blank. Ideas?
Let a1, a2, . . . , an be strictly positive real numbers. Show that

a1 + a2 + · · · + an−1 + an <= ((a1)^2)/a2 + ((a2)^2)/a3) +...+ ((an)^2)/a1

have fun!

Since the sequences $ (a_1^2,a_2^2,\dots,a_n^2)$ and $ \left(\frac{1}{a_1},\frac{1}{a_2}\dots,\frac{1}{a_n}\right)$ are oppositely sorted, from Rearrangement inequality we get
\[ \frac{a_1^2}{a_2}\plus{}\frac{a_2^2}{a_3}\plus{}\dots\plus{}\frac{a_n^2}{a_1}\ge\frac{a_1^2}{a_1}\plus{}\frac{a_2^2}{a_2}\dots\plus{}\frac{a_n^2}{a_n}\equal{}a_1\plus{}a_2\plus{}\dots\plus{}a_n\]


algebra
polynomial
search
quadratics
algebra unsolved

The real polynomials p(x), q(x), r(x) have degree 2, 3, 3 respectively and satisfy p(x) 2  + q(x) 2  = r(x) 2 . Show that either q(x) or r(x) has all its roots real.

As far as I remember, I have posted a solution to a more general problem posted by Namdung. Try to search for it....

Since $\deg r=3$ we conclude $r(x)$ has a root $a$, and it follows $p$ and $q$ also have this root.
Suppose the contrary that $q(x)=(x-a)Q(x)$ and $r(x)=(x-a)R(x)$ where $Q$ and $R$ are positive quadratic polynomials. Then $p(x)=c(x-a)(x-b)$ and $c^{2}(x-b)^{2}=(R-Q)(Q+R)$. As $\deg (Q+R)=2$ we conclude $\deg (Q-R)=0$, so $c^{2}(x-b)^{2}$ positive OR negative on the $R$. Contradiction.


Find all $P(x)\in R[x]$ such that
$P(x)P(x+1) = P(x^2)$

If $x$ is a root of the polynomial (as a complex polynomial if needed) then $x^2$ and $(x-1)^2$ are also roots.
Since there are a finite number of roots, either P is constant (and $P=0$ or$P=1$) either the only possible roots are $0$ and $1$ (by looking for instance to the norm of the complex roots)

$P = X^n (X-1)^p$ then it's easy to see that $n=p$

$0$, $1$, $P = X^n (X-1)^n$ are the solutions.


rotation
geometry
geometric transformation
trigonometry
trig identities
Law of Cosines

$P$ is inside an equalateral triangle $ABC$ such that $PA=2$, $PB=2\sqrt{3}$, $PC=4$.
Find the side length of $\triangle{ABC}$.

[hide="steps"]1. rotate about a vertex,
2. look at smaller equilateral triangle formed by the interior point, center of rotation and its image,
3. look for a right triangle with the lengths of the sides given in the problem(one side from the previously mentioned triangle, one from the original triangle, 1 from the image)
4. use the law of cosines to finish it
[/hide]

[hide="answer"]after doing this, i got $2\sqrt{7}$[/hide]

I just realized I made a mistake, forget it.

[quote="FMako"]For all triangles there is a convenient formula:
\[ 1+\frac{r}{R}= \sum \cos{A} \]
[/quote]
In this problem, $A=B=C=60^o$.

I was thinking somehow I could apply that to the interior angles, that sum to 360.

I'll write a complete solution for this problem based on [b]Altheman's[/b] idea.

We make a rotations transformation $R: (C, 60^o)$ so that $P \to P'$. Connect $PP'$, $PC$ and $PB$.

Since $\angle{P'CA}=60^o-\angle{ACP}=\angle{BCP}$, $BC=AC$, $CP=CP'$, so $\triangle{CBP} \cong \triangle{CP'A}$, which gives $PB=P'A$. Now observe $\triangle{CPP'}$, we find that CP=CP', $\angle{PCP'}=60^o$, hence $\triangle{CPP'}$ is an equilateral triangle. Therefore $PP'=CP=4$.

In triangle APP', we have $PA=2$, $AP'=BP=2\sqrt{3}$. Thus $\triangle{APP'}$ is an right triangle with $\angle{AP'P}=30^o$. Since $\angle{CP'P}=60^o$, we get that $\angle{CP'A}=90^o$. Therefore in right triangle CP'A, $CP'=4$, $AP'=2\sqrt{3}$, thus:
\[ AC=\sqrt{P'C^2+P'A^2}=\sqrt{16+12}=2\sqrt{7} \]

I was thinking of manipulating the fact that if D, E, and F are feet of altitudes from P on BC, AC, AC respectively, $|PD| + |PE| + |PF| = \sqrt{3} |AB|$, and use Pythagoras and/or Apollonius to find $|AF| + |BF|$, etc. to find $|AB|$, $|AC|$ and $|BC|$.

I'll come back to it later.


advanced fields
advanced fields unsolved

1. Self-reference statements are meaningless.

2. There is one or more true statements.

If 1 is true then 2 can not be true. If 2 were true then it would be self-reference. If 2 can not be true then 1 can not be true. If 1 were true then there would be a true statement. That would mean 2 is true. If 1 is true then 2 can not be true.

1 CAN NOT BE TRUE.

First, the word in English is "self-referential."  Second, it looks to me from your posting that you don't really understand what this means.  What do you think the definition of self-referential is?

If 2 says "There is one or more true statements."  and 2 is itself a true statement then 2 is a self reference statement.

I will explain it better.  2 refers to the group of all true statements.  2 says this group contains 1 or more members.  If 2 is a true statement then 2 is a member of the group it is referring to.  I came up with this paradox to show that if self-reference statements are meaningless then a contradiction occurs if there are any true statements (including 1).  On one hand you have a true statement on the other hand 2 is not true.
The purpose of 1 was to save the truth.  1 prevents there from being any truths.  You can replace 1 with "no statement about all members of a group can be true"  The same problem will occur.

"no statement about all members of a group can be true"

The above statement is about all members of the group of statements about all members of a group. The statement itself is a statement about all members of a group. It is a member of the group it is referring to. It is a self-reference statement that declares itself to be untrue.

by your explaination, statement 2 is not self-referential.(my own oppinion^!^) it says something about this "group" but itself. And (even) if self-referential can be defined in this "group", statement 1 is not self-referential too.

I think You should make a difference between "our nature language" and "logic language".

I can write it   " statements 1,2,3... are true statements.  1,2,3... are all the true statements.  If this statement is true then it has a number. It is declaring itself true.

I can write it " 1 or more true statements exist"  If it is true then it says it exists.  This is the same statement I wrote.

sorry
I can write it " statements 1,2,3... are true statements." 1,2,3... are all the true statements. If this statement is true then it has a number. It is declaring itself true.

No 1 does not refer to itself.  I wish it were that easy.  It took along time for me to come up with a problem with 1.  I understand my problem will not end all disputes.  If 1 is  " no statement about all members of a group can be true" then 1 is self reference.

a.no statement about all members of a group can be true.
b.self-reference statements are meaningless.
If a is true then b is not.  if b is true then a is not.

What do you mean? It seems to me that you think others thought self-reference statements are meaningless, you disagree and point out how this can be "proved"?

Whether a sentence is self-reference depends on "us", i.e depends on how do you make the connection between the structure and the theory. You could not tell (via syntactic knowledge) whether a Turing machine use its own code, you could not tell (via syntactic knowledge) whether a turing machine is solving 3-SAT or other NP-complete problem, because the content of the problem is explained by "us", i.e another theory (connected with the structure in some way) .

I found this problem was already discovered it is the Russell-Myhill paradox.


linear algebra
matrix
group theory
abstract algebra
linear algebra solved

Let $G$ be a finite subgroup of $\mathrm{GL}_n(\mathbb C)$ (this is the group of all $n\times n$ matrices with complex coefficients with determinant different from $0$).

The sum of the matrices $g$ in $G$ is zero iff the sum of their traces $\mathrm{tr}(g)$ is zero.

Suppose that $A=\sum_{g\in G}g$ has trace zero. Since $Ag=A$ for any $g\in G$, summation yields $A^{2}=|G|A$. Therefore, each eigenvalue of $A$ is equal to either $0$ or $|G|$. But the trace of $A$ is zero, hence all eigenvalues are $0$, which means that $A$ is nilpotent. Since also $A^{n}=|G|^{n-1}A$, we conclude that $A$ is the zero matrix.

can some explain why for all g $ Ag\equal{}A$ ? and why if A has trace 0 then his all eingen value are 0?  :blush:

$ A$ is the sum of all elements of $ G$. Multiplying each element of a finite group by $ g$ on the right, you get the same elements written in a different order. In other words, multiplication by $ g$ amounts to permutation of group elements.

Knowing that (i) each eigenvalue of $ A$ is either $ 0$ or $ |G|$, and
(ii) the sum of all eigenvalues (i.e. the trace) is $ 0$,
it should not be hard to see that each eigenvalue is $ 0$.


agha inam az nazarie adaad...
mikhaym ye shoori be pa konim...hamaeye masale haye khafano inja ba ham hal konim,...
faghat man hamin tor ke goftam momkene ye 2,3 rooz dige ta ye zamaane namaloom beram va on nasham (be khaatere masaaele siyasi :rotfl:  :ninja: )...
bedrood

[quote="pirhoosh"]bashe masalato begoo bad boro!!
baba siasi[/quote]
agha zire pato negah koni ma ro mibini...
mokhlese hamegiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii..iiiiiii.iiii....iiiii....(sookhtam tamoom shod :rotfl: )
ta hafateye bad khodahaafez...

[quote="mehdi"]first problem:
1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.[/quote]
inam az javab:
farz konim $K^2$ avalin morabae kamel dar donbale bashad va $d$ ghadr nesbate donbale bashad pas:$K^2+d(2Km+dm^2)=(K+md)^2$ be ezaye tamame adde tabiyie $m$ ham morabae kamel ast va ham ozvi az donbale. ;)

[color=cyan][size=150] Problem 2 [/color][/size]
inam ye masaleye sade az short list:
$s(n)$ majmoe maghsoomo alah haye adade tabiyiye $n$ minamim.
sabet konid namotanahi $n$ vojood darad be goonei ke $\frac{s(n)}{n}\geq \frac{s(m)}{m}$ be ezaye tamame $m$ haye tabi'ye kochektar az $n$.

[quote="mehdi"][quote="doogh khore herfei"][quote="mehdi"]first problem:
1)farz konid yeki az jomalaate tasaaodi hesaabi va namotanahi az adaade tabii,morabbae kaamel bashad.saabet konid dar in tasaaod bi nahaayat jomle vojood daarad ke morabbae kamel and.[/quote]
in soal ro man hal kardam ke kheili sakht bood albate shoma nemitoonin halesh konin chon bayad mese man herfei bashin  :mad:    :arrow:  :idea:  :huh:  :P  :blush: hjfhgjhjg:[/quote]
bror baba az key ta hala kombooze oomade ghati miveha?
age ye zare nega koni mibini ke soalesh hamchinam sakht nabood va dar zemn hame gheyr az to mitoonan hallesh konan va dar zemn mohammad hosein hallesh karde tamoomm.
badesh ham enghadr alaki khali bandi nakon ke man badam miyad az khali bandi.  :mad:[/quote]
man ke khali nabastam.nemidoonam chera migi bastam.badesh ham ba to naboodam ke bem begi.badesh ham man moaddealam shod 17.30 pas ba man injoori harf nazan chon ke man kheili ba hoosham.vali chon midoonam ke to tazekari bet chizi nemigam.fghzgfhfghfhzfhghfhfhfhfh

:rotfl:  :rotfl:
ye nokta ro eshare nakardi azizam...
to tooye doogh khordan ham kheili maharat dari :rotfl:
kheili ham bahooshi chon ke hatman emtehan varzeshi chizi az 15 balatar shode :rotfl:
badesh ham age oghde type kardan dari boro vaghti computer khamooshe bezan rooye keyboard :lol:
dige ham inja chertopert nanvesi vagarna be moderator migam ke bokonadet biroon.
chon inja jaye bahse elmie na injoor araajif :mad:

doostan behtare bepardaazim be soale agha hosein


Hi,
I'm currently living in Canada, but I'm technically American (cause I was born in California...)

So, I was wondering, am I eligible for the USA's IMO team as long as I do good on the AMC, AIME and the USAMO??

Or, do I have to be LIVING in the USA to be eligible?

 :?:  :?:

There was  Canadian at MOP in my time; he was a citizen of both countries, and later competed on Canada's IMO team. I think you're eligible, although you're likely also eligible for the Canadian team, and that one's easier to reach.

The only place that I could find that answers this question is wikipedia, however they don't cite any sources for it

[quote]Only U.S. residents and [b]citizens[/b] may join the American IMO team[/quote]


geometry
parallelogram
analytic geometry
graphing lines
slope
AMC
USA(J)MO

If you are given any quadrilateral and you connect the midpoints of each of its sides, a parallelogram is formed. Prove why this phenomenon works.

I will give my solution after people have posted theirs.

Proving's kinda hard

but my explanation is " isn't it obivious" ?

if you have a quadrialteral, and connect each sides' midpoints you get a shape of diamond....
( wow it's really hard to explain... arg!! )

I drew a picture instead

undcerstand this

I don't know the proof, so I'll make one up...

You have 4 points, $(a_1,a_2),(a_3,a_4),(a_5,a_6),(a_7,a_8)$... By connecting the midpoints, you obtain these points:

$(\frac{a_1+a_3}{2},\frac{a_2+a_4}{2}),(\frac{a_3+a_5}{2},\frac{a_4+a_6}{2}),(\frac{a_5+a_7}{2},\frac{a_6+a_8}{2}),(\frac{a_7+a_1}{2},\frac{a_8+a_2}{2})$

To make it a paralellogram, both pairs of opposite sides must be congruent and parallel...

Since using the distance formula would be a little messy here, I will find the slopes...

Slope between first and second point above:

$\frac{\frac{a_6-a_2}{2}}{\frac{a_5-a_1}{2}}=\frac{a_6-a_2}{a_5-a_1}$

Slope between third and fourth point above:

$\frac{\frac{a_6-a_2}{2}}{\frac{a_5-a_1}{2}}=\frac{a_6-a_2}{a_5-a_1}$

And what do you know? It's the same slope!

Now let's try doing the other pair of opposite sides:

The first pair (midpoint of 2nd and 3rd point):

$\frac{\frac{a_8-a_4}{2}}{\frac{a_7-a_3}{2}}=\frac{a_8-a_4}{a_7-a_3}$

Now for the other pair:

$\frac{\frac{a_8-a_4}{2}}{\frac{a_7-a_3}{2}}=\frac{a_8-a_4}{a_7-a_3}$

Yay!!!

Both pairs of opposite sides are parallel!!! That means we have ourselves a parallelogram!

:coolspeak:

This is actually a theorem called Varignon's Theorem (1654-1722).

There are many ways and one easy way I can think is using mass point. But that's way above this level so I'll not post the solution using it.

Refer to shinwoo's picture, because I'm too lazy to draw one myself. Label the four vertices A, B, C, and D in order (aren't I creative?). Draw auxiliary segments AC and BD. Now in triangles ABC and BCD, the side of the new quadrilateral connects the midpoints of the two sides. Therefore we should probably use the midpoint connector theorem (I wonder why...) and we can prove that the opposite sides of the small quadrilateral are both parallel to auxiliary segment AC, and likewise for the other pair. Therefore the small quadrilateral is a parallelogram, QED.

I don't know if that helps. I could have cleared that up considerably if I had drawn a picture with labeled points.

[quote="mathnerd314"]I don't know if that helps. I could have cleared that up considerably if I had drawn a picture with labeled points.[/quote]

do it!

thanks...im pretty sure i get it now...

do u  guys just draw things in paint?

Ok, here's how I proved it. First, i connected the opposite midpoints, which formed two diagonals. Then, I used the sides of the "parallelogram" as midsegments of triangles. Eventually I proved that each of them was parallel to the diagonal and using the transitive property, they are congruent to each other.

I draw my figures with Macromedia Fireworks.  :lol:

[quote="mathnerd314"]Here's the pic. I still didn't label the points  :D  :lol:  :blush:  :mad:[/quote]

I did it the exact same way as you.  Its the easiest since it doesn't involve messy algebra

[quote="nat mc"]DOn't know.  Can someone tell me what "QED" means and what its for?[/quote]

It means Quantum Electrodynamics. Just kidding, most mathematicians would most likely think of 'Quod erat demonstrandum,' "which was to be proved" in Latin. You write it at the end of proofs.

ok

i was wondering y that is in ur siggy

Do people put QED at the end of their USAMO proofs?

He quoted it cause he can't quote the image.

[quote]Ok, here's how I proved it. First, i connected the opposite midpoints, which formed two diagonals. Then, I used the sides of the "parallelogram" as midsegments of triangles. Eventually I proved that each of them was parallel to the diagonal and using the transitive property, they are congruent to each other.

I draw my figures with Macromedia Fireworks. Very Happy[/quote]

that's not actually a proof

that's how it works

what Bob said is a full proof

Of course its a proof, just because it doesn't have algebra doesn't mean its not a proof


trigonometry
integration
real analysis
probability
function
expected value
absolute value

version of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=162601]counterexample to law of large numbers[/url] without details of $ X$.

Let $ X_1, X_2, \cdots$ be I.I.D. random variables and $ S_n \equal{} \sum_{j\equal{}1}^{n} X_j$ and $ \overline{X}_n \equal{} S_n / n$ for $ n \ge 1$.
If $ \mathbb{E}(|X_1|) \equal{} \infty$ then $ \limsup |\overline{X}_n| \equal{} \infty$ almost surely.

A little programming fun: try simulating this. One possibility: assume that $ \text{rand}()$ calls a random number generator that gives a uniform $ (0,1)$, and is independent each time called (well, maybe not, but we're going to act like that's what it does). Let $ X_i: \equal{}\tan(\pi(\text{rand}()\minus{}.5)).$

This gives $ X_i$ the distribution with density $ \frac{1}{\pi(x^2\plus{}1)}$ on $ \mathbb{R}$ - I think it's called the Cauchy distribution, although I prefer to think of it as the Poisson kernel. Just run the simulation with various values of $ n$ and print out a selection of values of $ \overline{X}_n.$

Let $ c>0$. Using $ \sum P(|X_n| > cn) \ge E(|X_1|) /c \equal{} \infty$ and Borel-Cantelli, we have $ |X_n| > cn$ infinitely often a.e. This means $ \limsup \frac{|X_n|}{n} \equal{} \infty$. Since $ \frac{S_{n\minus{}1}}{n} \plus{} \frac{X_n}{n} \equal{} \frac{S_n}{n}$ we also have $ \limsup \frac{|S_n|}{n} \equal{} \infty$

Since $ \frac{S_{n\minus{}1}}{n} \plus{} \frac{X_n}{n} \equal{} \frac{S_n}{n}$ we also have $ \limsup \frac{|S_n|}{n} \equal{} \infty$.

Can you justify this?

[quote="limsup"]Since $ \frac {S_{n \minus{} 1}}{n} \plus{} \frac {X_n}{n} \equal{} \frac {S_n}{n}$ we also have $ \limsup \frac {|S_n|}{n} \equal{} \infty$.

Can you justify this?[/quote]

Claim : The event [$ \limsup \frac {|S_n|}{n} < \infty$] implies the event [$ \limsup \frac{|X_n|}{n} < \infty$].
Proof : Let $ c$ be a positive number. The event [$ \limsup \frac {|S_n|}{n} \le c$] implies the event [$ \limsup \frac{|S_{n\minus{}1}|}{n} \equal{} \limsup \frac{|S_{n\minus{}1}|}{n\minus{}1} \cdot \frac{n\minus{}1}{n} \le c$ ] and hence also implies the event [$ \limsup \frac {|X_n|}{n} \le 2c$] because $ \frac {S_{n \minus{} 1}}{n} \plus{} \frac {X_n}{n} \equal{} \frac {S_n}{n}$.

Thanks Redkimchi. Mathlinks really helping me very much.

[quote="Kent Merryfield"] Let $ X_i: \equal{} \tan(\pi(\text{rand}() \minus{} .5)).$This gives $ X_i$ the distribution with density $ \frac {1}{\pi(x^2 \plus{} 1)}$ on $ \mathbb{R}$ [/quote]
Could you please explain this?

compute $ E[\phi(\tan(\pi(U\minus{}\frac{1}{2})))] \equal{} \int_{[0;1]} \phi(\tan(\pi(x\minus{}\frac{1}{2}))) dx$ where $ \phi$ is an arbitrary function. You should find that this is equal to $ \int_{\mathbb{R}} \phi(x) \frac{dx}{\pi(1\plus{}x^2)}$

alekk:
Yes I agree that $ \tan(\pi(U\minus{}\frac{1}{2}))$ follows a cauchy distribution. But I just want to know the motivation for that particular expression, I think $ \tan(\pi(U))$ would also follow Cauchy distribution. Isn't it?

alekk, one more question: What for Prof Kent Merryfield's programming intended?

[quote="limsup"]But I just want to know the motivation for that particular expression, I think $ \tan(\pi(U))$ would also follow Cauchy distribution.[/quote]

where is $ x\to\tan(\pi x)$ discontinuous?

blahblahblah and limsup: I admit that $ \tan(\pi(U\minus{}1/2))$ and $ \tan(\pi U)$ have exactly the same distribution, as would $ \tan(\pi(5U\minus{}.374)).$ Hitting the discontinuity is a event of probability zero; we may safely disregard it. I used the first form primarily because I thought it was the most intuitive.

[quote="alekk"]compute $ E[\phi(\tan(\pi(U \minus{} \frac {1}{2})))] \equal{} \int_{[0;1]} \phi(\tan(\pi(x \minus{} \frac {1}{2}))) dx$ where $ \phi$ is an arbitrary function. You should find that this is equal to $ \int_{\mathbb{R}} \phi(x) \frac {dx}{\pi(1 \plus{} x^2)}$[/quote]
There are distinct limitations on "arbitrary function" here - this only makes sense if that expected value makes sense. In particular, it does not make sense for $ \phi(x)\equal{}x,$ since $ \frac{x}{\pi(1\plus{}x^2)}$ is not absolutely integrable over $ \mathbb{R}.$

And that was the point of my programming exercise. It was an illustration of the principle that redkimchi started this thread with. The Cauchy distribution is a readily accessible distribution whose expectation is undefined (because the expectation of the absolute value is infinite). If you run various samples of that and print out various sample means, you should find that those sample means jump around in an unpredictable fashion.

The simulated example would probably not by itself convince you of redkimchi's result - that the sample means can become arbitrarily large - but you would at least suspect that something fishy is going on with those sample means.

blahblahblah: Yes you are right!

To Kent Merryfield: Is my following program correct?

[b]n=100;
s=0;
for i=1:n
    x=rand(1,1);
    x=tan(pi*(x-0.5));
    s=s+x;
    s1(i)=(1/i)*abs(s);
end
plot(s1)  [/b]

I don't know the syntax of the language you're using so I don't know whether "rand(1,1)" is the right way to say that. And I don't know what syntax you need to close a loop is this actual code in some language or system or is it "pseudocode?" Also: why say "abs(s)"? You can let $ s$ be negative as well as positive.

I might push the numbers a little harder - say, the following.

n=100;
m=1000;
s=0;
for i=1:n
for j=1:m
x=rand(1,1);
x=tan(pi*(x-0.5));
s=s+x;
(next  j? something to close the loop?)
s1(i)=(1/(m*i))*s;
(next i?)
end
plot(s1)


induction
combinatorics proposed
combinatorics

let $ n$ is positive integer number and $ n\geq 2$ and $ a_{1},a_{2},...,a_{n}$ are positive integers number and for any $ k \equal{} 1,2,..,n$ we have:$ 0 < a_{k}\leq k$ and $ a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}$ is a even number
prove that: We put $ \plus{}$ or $ \minus{}$ in order that $ \pm a_{1}\pm a_{2}\pm ...\pm a_{n} \equal{} 0$

We can use induction, no?

Take $ a_{n\minus{}1},a_n$, replace it with $ |a_n\minus{}a_{n\minus{}1}|$.


number theory
prime factorization
number theory unsolved

Show that the number of Pythagorean triples x, y, z (x^2+y^2=z^2) with a fixed integer x is
1) ( tau(x^2) - 1  ) / 2 if x is odd
2) ( tau(x^2/4) - 1  ) / 2 if x is even.
( tau(n) being the the number of divisors of n )

This must be not too difficult, but I can't find the solution and I have more than 200 problems to do in 2 days.
So could someone give me a hand? thank you.

factor: $ x^2 \equal{} (z\minus{}y)(z\plus{}y)$

Can you elaborate please...

if x is odd, gcd(x,y,z)=d
x = d(m^2-n^2) = d(m+n)(m-n)
m>n
m+n=a, m-n=b
d , a, b are all odd.
a = 1 (mod 4) and b = 3 (mod 4) or a = 3 (mod 4) and b = 1 (mod 4)
So a>b, the prime factorization of a has an odd number of prime factors of which remainders are 3 modulo 4 and
the prime factorization of b has an even number of primes of which remainders are 3 modulo 4, or vice versa.
Is this a right approach?[/code]

$ x^2 \equal{} (z \minus{} y)(z \plus{} y)$. so we're looking to write $ x^2 \equal{} mn$ for integers $ m$, $ n$, such that $ m \equal{} z \minus{} y$, $ n \equal{} z \plus{} y$. the restrictions on $ m,n$ are that $ m < n$ (b/c $ z \minus{} y < z \plus{} y$) and $ m \plus{} n$ be even (b/c $ m \plus{} n \equal{} 2z$). and notice that any such $ m,n$ will give you a solution $ (y,z)$ when you solve the system $ m \equal{} z \minus{} y$, $ n \equal{} z \plus{} y$. so you need to find the number of ways to factor $ x^2$ as $ mn$, where $ m < n$, $ m \plus{} n$ even.

1. if $ x$ is odd, any factorization will have $ m \plus{} n$ be even b/c $ m,n$ are odd in this case. so you only need to make sure $ m < n$. $ m < n$, $ mn \equal{} x^2$ mean that $ m < x$. the number of factors of $ x^2$ that are less than $ x$ is $ \frac {\tau(x^2) \minus{} 1}{2}$.

2. if $ x$ is even, at least one factor must be even, so in light of $ m \plus{} n$ having to be even, we have to make sure *both* $ m,n$ are even. that is, $ m \equal{} 2M,n \equal{} 2N$. your conditions become $ x^2 \equal{} 4MN$, $ M < N$. (the parity of $ m \plus{} n$ is taken care of.) $ \frac {x^2}{4} \equal{} MN$, $ M < N$, mean that $ M < \frac {x}{2}$. and the number of factors of $ \frac {x^2}{4}$ less than $ \frac {x}{2}$ is $ \frac {\tau(\frac {x^2}{4}) \minus{} 1}{2}$.


I don't understand these kind of problems. Can someone explain them to me?

The annoyance level of a particular room is directly proportional to the frequency of Wei's comments and inversely proportional to Ashley's whining. The annoyance level is 660 when Wei talks 550 times per minute and Ashley's whining intensity is $ \frac{5}{6}$. What is the annoyance level if Wei talks 550 times per minute and Ashley's whining intensity is .05?

first of all you don't have to swear to get our attention.

the basics for inverse and direct proportion problems are these: when two values are inversely proportional they have a constant product, so as one increases by a certain ratio, the other value has to decrease by the same ratio to maintiain a constant product-- $ xy\equal{}k$ such that k is a constant
two directly proportional values are as such $ \dfrac{x}{y}\equal{}k$ for a constant k. as the top or bottom increases by a certain ration the top or bottom has to increase by that same ratio as well.

that should help with the problem.

[quote="mihail911"]first of all you don't have to swear to get our attention.
[/quote]
How is crap a swear word? ....


summer program
MathPath
analytic geometry

Source:MathPath From Dr. Zeitz

3 frogs are placed on 3 vertiecs of a sqrare. Every minute, one frog leaps over another frog, in such a way that teh "leapee" is at the midpoint of the line segment whose endpoints are the starting and ending position of the "leaper". Will a frog ever occupy the vertex of the square that was originally unoccupied.

The Names of the the frogs are Albatross, Bobert, and Cattlebot.

 [hide]Use Parity, [/hide] :P

ok....i don't understand the problem, can someone explain it differently?

O.K

there are 3 frogs on the verticies of a square like this.
f       f

f       ()

every minute, a frog jumps over another frog and goes the same distance on both sides.

so a sample move would be:
()      f       f
f       ()

Now we need to prove that the one vertex of the original triangle cannot be landed on by a frog (occupied)

In this case it is the lower right vertex that we need to prove that cannot be landed on.

[quote="behemoth571"]I think that version is not the same as the original. Intrepid said that the "leapee" was at the midpoint of the side on which the leaper leaps (so many leaps!). I also think that the frogs are not meant to go out of the square.

In any case, for the first leap, the frog on the top left corner can choose 2 directions, right or down, while the other 2 frogs are restricted to one direction each, left and up. A frog must be able to get to either the bottom or right side of the square in order to get to the unused vertex, which poses a sort of impossibility. Maybe it has something to do with combinations? After all, 3C2 or ( 3 2 ) = 3. But how can you prove that the unused vertex is the missing one?[/quote]

The leaper goes over one of the frogs.
The frogs can leave the square.
If we put this on coordinates with the lower left frog being the origin, and the frogs making a unit square, if it jumps over the lower right frog, it will go to the poin (2,0).

One more hint (yes, coordinates are the insight), use the parity of the points.

I hope that this helps.


geometry
perimeter
inequalities
triangle inequality
AMC

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15.  What is the greatest possible perimeter of the triangle?

$ \textbf{(A) } 43 \qquad \textbf{(B) } 44 \qquad \textbf{(C) } 45 \qquad \textbf{(D) } 46 \qquad \textbf{(E) } 47$

[hide="Answer"]By triangle inequality, $15$ has to have the middle length. The largest value of $x$ which satisfies $x+15>3x$ is $x=7$, so we have $7+15+21=43\Rightarrow \boxed{A}$.[/hide]

I can't believe I forgot that two sides added must be greater than the third...ahhhh stupid mistake!! on such an easy question

[hide=Solution]
Let the side lengths be $x$ and $3x$. By Triangle Inequality, $15+x>3x$ so $x\le 7$. $7+21+15=\boxed{\textbf{(A)} 43}$
[/hide]


geometry
3D geometry
rotation
complex numbers

As a matter of interest, what is the QEDMO. Suitable name for a maths contest!

[quote="ZetaX"]Prove that if $n$ can be written as $n=a^2+ab+b^2$, then also $7n$ can be written that way.[/quote]
$\{a,b\}\subset\math N?$

FWIW  a  related Question.  Why Would  any number of  the form $a^2+3b^2$ , instead of $7=2^2+3\cdot 1^2$ would work here?

More generally:
Let $A$ be the set of integers of type $a^2+ab+b^2$ and $B$ be the set of integers of type $a^2+3b^2$.
Prove for $x,y \in A ; w,z \in B$:
$xy \in A ; wz \in B , xz \in A,B$
or more exactly: prove $A=B$

[quote="scorpius119"][hide]Let $x=2a+3b$, $y=a-2b$. Then $x^2=4a^2+12ab+9b^2$, $y^2=a^2-4ab+4b^2$, $xy=2a^2-ab-6b^2$, so that $x^2+xy+y^2=7a^2+7ab+7b^2=7n$.[/hide][/quote]

How could you take $x=2a+3b$ and others by yourself. Is it some kind of a rule or something like that?

For such problems, trying linear combinations $x=ua+vb$ and similar for $y$ work pretty well.
But there is a more general method using complex numbers.

[quote]How could you take $x=2a+3b$  and others by yourself. Is it some kind of a rule or something like that? [/quote]

One useful (related to  multiplication of complex numbers) generalization of:
$(a^2+b^2)(c^2+d^2) = (ac +bd)^2 +(ad-bc)^2$ (     Can also be written as  (   $(ac-bd)^2+(ad+bc)^2$    )

Is
$(a^2+nb^2)(c^2+nd^2) = (ac+nbd)^2 +n (ad-bc)^2$ (    Can also be written as ( $(ac-nbd)^2+n(ad+bc)^2$ )

What this means, is product of two numbers of the form $(a^2+nb^2)$ can also be written  in the same form.

Of course, $(a^2+ab+b^2)$ fits in this form . (this is nothing but $( (\frac{a-b} 2)^2 + 3 (\frac{a+b} 2)^2 ))$

(This is why I  wanted to point out  that $7=2^2+3\cdot 1^2$ etc ... )

(You write the product of two terms of  the form $x^2+3y^2$ as $x^2+3y^2$ etc ..)

Hope this  helps.

Alternately, $a^2+ab+b^2=|a+\omega b|^2$, where $\omega$ is a cube root of unity. The numbers of the form $a+\omega b$ with $a,b$ integers are closed under addition and multiplication.
To get to the $a^2+3b^2$ form, rotate $60^\circ$.


integration
calculus
trigonometry
calculus computations

Find  $\int ln(\sqrt{x}+\sqrt{x+1})dx$

       $\int \frac{1}{x-\sqrt{1-x^{2}}}dx$

For the second one first make the substitution x=siny and satisfy yourself thet you now need to evaluate the integral of (cosy)/(siny-cosy). To do this let I be the integral I've just mentioned and let J be the integral of (siny)/(siny-cosy). Then simply evaluate J+I and J-I and solve simultaneously for I and J. (Both I+J and I-J are easily evaluated)

For the first one, change variables according to $\sqrt{x}= t$ and then integrate by parts. The answer should then be

$(x+\frac{1}{2})\ln|\sqrt{x}+\sqrt{1+x}|-\frac{\sqrt{x}\sqrt{1+x}}{2}+C.$

For the second one, start by multiplying and dividing by $x+\sqrt{1-x^{2}}$. Then split in two terms, and in the second one make $x=\sin t$. The final answer should be easy to reach.

Carcul


AMC
AIME
geometry
ARML
3D geometry
tetrahedron
AMC 12

Let the difficulty rating of the 2004 A  AIME be 8.
From there, we may build upon the others. So for example, if a difficulty rating of an AIME is 10, a person can expect to solve 2 more problems on the 2004 AIME  than that AIME. So lets start it off:
1983 AIME: 9
1984 AIME: 5
1985 AIME: 4
.
.
.

Keep the ratings coming.

If you ask me, those last two ratings were way too low.  I find that AIMEs are fairly consistent in difficulty, except the AIME I last year which seemed exceptionally easy (I got 13 on a practice ????).

It's impossible to rate an AIME for the following two reasons:

Say some AIME have easier #1-5s but harder #11-15s, that AIME would've been easier for people who are aiming for a 4 or 5 but harder for a person who are aiming for a 10+.

Also, say 1996 AIME, which had a lot of counting questions and very few geometry problems, it would've been an advantage for people who are good at counting but bad at geometry, but a disadvantage for people who are good at geometry but bad at counting(me).

btw I disagree with your rating for 1983, 84, 85.

1994 was very hard. At least in my opinion, it was the hardest of all.

The system of classification seems a little bit counterintuitive to me. An easier way to codify it (from my POV) might be just a plain 1-10 scale, without any of the precision problem equivalence in the scale.

A question I had: How would you guys rank the 1987 AIME? I just got 5/6 in 50 minutes, and had all but one case of the 1 OK. This is my first year doing the AIME; from what others at my school said, my goal was going to be a nonzero score. So I was wondering: is 1987 just one of those beginning easy, end hard AIME's that beta mentioned? Overall, how is the difficulty compared to more recent AIMES?
Thanks

First of all, [b]your goal is not to get a nonzero score[/b].  My first year taking the AIME, I could pretty consistently solve the first 7 or so.  So it's not a fluke.  I usually finished all the problems that I could do (relatively) quickly with an hour or so left.  Then I worked on the rest, usually just helping to fry my brain but occasionally squeezing out one more correct answer :).

Really, I usually check my answers for the last hour.

I ought to, but I tend to get bored that way :D.  Instead, I try to check often while finding the answer and once again right after I bubble in.

I'm still working frantically by the end of the three hours.

I usually work for about 1-2 hours and manage to solve 8-10 of them, i can usually squeeze in one of the 12-15 in the last hour.

Usually, I do 6 or 7 problems in the first hour. In the second hour I can do like 3 problems. Third hour I pull off one more problem, and rest of the time checking.

If I get better at counting/probability and cut down # of careless mistakes, I can do pretty well on AIME. Are these two goals achievable in the next month? How should I do them?

I'll hopefully do 6-7 in the first hour, then 3-4 in the next hour, and 1 in the next 30 minutes. Then I'll check for the last 30 minutes. The maximum I'll answer is probably 12 because I won't know how to do at least three of them. Consequently, when you consider stupid mistakes, I'm hoping to get a 10+.

I noticed that almost always one of the last 5 problems is counting (permutations of something). I find those very annoying and can never get them, either. Grr.

No one else does the suicidal last hour squeeze-out-one-more-problem? I unfortunately do not have the patience to spend time checking answers, so I just try to do #14 or #15 for the last hour and fail miserably. :D

[quote="beta"]If I get better at counting/probability and cut down # of careless mistakes, I can do pretty well on AIME. Are these two goals achievable in the next month? How should I do them?[/quote]

Cut down careless mistakes is easy if you become a paranoid person like me who checks every step 3-4 times on tests like this.  For getting better at counting/probability, I feel that I tend to improve slowly but steadily, naturally gaining a slightly deeper understanding every year.  Your best bet I think is to read solutions to hard counting/probability problems and try to learn to get into the right mindset for these problems.  Also, don't be afraid to do a little casework; all elegance is defenestrated in the mindless score-oriented test taking mode for AIME :).

1999 was brutal. I solved like ten of them and had a stupid mistake on [i]each[/i] one. Score: 0. Lovely eh? so happy. go lucky. little angry. hey, happy rhymes with angry.

On the aime, and when i take it in march, im hoping that the problems are good, like geometry type (i like geometry), and three- dimensional geometry stuff like tetrahedrons and soccer ball shapes.  If there is not atleast 4 geometry problems on the aime, then i dont think i'll make the United States of America Mathematical Olympiad.  BTW, how do people say AIME? Do people say each letter A-I-M-E, or do they just say like you would say," my [i]aim[/i] is to make the USAMO."  [/i]

I most often hear it said "Amy" and occasionally "Ay Me" but nobody I know says each letter.

Ugh... geometry. I hope there are as few geometry problems in the last 5 as possible. Anything before that will probably be solvable for me (I hope).

LOL, 10 stupid mistakes on 10 problems...that requires SKILL.

I got skill. ANd they were the really stupid mistakes like I added m and n wrong in the end or wrote down a case but forgot to include it. I need to organize my work.

In eight grade, right before the AMC 8, I kept on dividing 11/2 and getting 6.5. Really weird how many times I get it wrong before it sinks in.

[quote="beta"]Hey recently I thought 2004/6 = 338

Just wondering, was 1988 AIME generally considered the easiest AIME ever? Because I hit my record 11 on it...[/quote]

I don't distinctly remember, but again, I don't think it has to do with the test itself.  You probably improved or were in the right mood (this makes a pretty big difference sometimes).  Except for the 2004 AIME I.  That was weirdly easy...  :?

[quote="probability1.01"]
I don't distinctly remember, but again, I don't think it has to do with the test itself.  You probably improved or were in the right mood (this makes a pretty big difference sometimes).  Except for the 2004 AIME I.  That was weirdly easy...  :?[/quote]

Right mood? What's the "right mood"? I guess 1988 has only two counting questions, one of them was a easy #1, so it didn't hit my weakness too bad.

But still I'm not sure about the "right mood", and I know I definitely didn't improve.

Eh, I found the 2004 AIME to be about average, maybe a little easier than average. I got a 7 on it, and I've gotten 6-9 on most of the AIMEs I've practiced with since. So...

Ah I got a 10 on 1992 AIME in practice (I still made 2 careless mistakes :( ), was that year really easy too?

Wait I think I got a 10, kalva doesn't have an answer for #14, can someone confirm the answer to #14, I believe the answer is 094.

14.  ABC is a triangle. The points A', B', C' are on sides BC, CA, AB and AA', BB', CC' meet at O. Also AO/A'O + BO/B'O + CO/C'O = 92. Find (AO/A'O)(BO/B'O)(CO/C'O).

[quote="beta"]Hey recently I thought 2004/6 = 338

[/quote]

whoaa... at arml i divided 2004 by 6 and ALSO kept getting 338! hence... i got the wrong answer.  :( freaky, eh?

[quote]1999 was brutal.[/quote]
From what I recall, the scores in 1999 ran lower than they usually do. In particular, I track Southern California scores, and the second highest from that half of the state was a 7.

what do i need on the aime to move on if im a sophmore?
and, if I took the amc12 but i am a sophmore, does it matter?

Where do you all find these AIMEs to take as practice tests?  Is there somewhere online or something?  It would be kinda useless to have the problems without an answer key.

[quote="everyday847"]Maybe I'm crazy, but I do better on #8-#15 than #1-#7.[/quote]

YES you are crazy.

http://www.kalva.demon.co.uk/aime.html
Lots of AIME problems, answers with solution outlines too.


Support
induction
number theory
prime factorization

Here are a couple of random problems that I need major assistance on:

(Well, the fact that this browser doesn't support Microsoft Equation Editor makes the next question hard to read but, try your best)

1. Prove that [(N+k) Choose (k)]*(1/2^k)=2^N, where k is an integer starting from zero and increasing until N. (Its like sigma notation, where N is the limit, [(N+k) Choose (k)]*(1/2^k) is the conditional, and k starts from zero to N.
 ***well, I know you're supposed to use induction for this, and the base case is easy enough, but the inductive step is really killing me. Any sugestions?***

2. Given m,n,p are postive integers, and m<=n<=p, and that
   2mnp=(m+2)(n+2)(p+2). Find Max(P).

***I have no idea.....***

3.Let T(N) equal the number of ways to represent a positive integer N as the sum of one or more consecutive integers. Let "p" and "q" equal distinct odd primes, and let "a" and "b" represent positive integers. Prove:
   a) T(p)=2
   b)T(2^a)=1
   c)T(p^a)=a+1
   d)T(p^a*q^b)=T(p^a)*T(q^b)

***OK. Parts a) and b) are easily proved. However, I haven't been able to get very far with parts c) and d). Perhaps given the properties of the first two parts, you can prove the last two parts. Any suggestions?

you do know that if you press the "view more emoticons" button on the left you bring up mathematical symbols and such. thats all i can help.

happy solving

I got #2:

Rewrite the equation as (m+2)/m * (n+2)/n * (p+2)/p = 2. We want (p+2)/p to be as small as possible. So we try a greedy approach: since (m+2)/m >= (n+2)/n >= (p+2)/p, we assume that (m+2)/m is as large as possible. So we try m=3. Then we need n>= 13. But (3, 13) doesn't give a solution, and neither does (3, 14). But (3, 15) does: it gives p=50. But we have no guarantee that it's the largest. So we try m=4. Then n>=7. With n=7, we get p=54, which is better. Continuing, we get (5, 5, 98). We must continue until (m+2)/m<= 2^(1/3), so we get (6, 6, 16), m=7 doesn't work with n<=7, so it must be worse, and if m=8, then (m+2)/m < 2^(1/3). Thus the optimal solution is (5, 5, 98). I'm not sure if that argument is particularly comprehensible, but it should be right if I didn't do anything horribly wrong.

For #3 I think you mean the sum of consecutive [b]positive[/b] integers. Otherwise some of them are false. (That's why I got confused at first.) I think it's easier to look at the number of ways to write N as the sum of one or more consecutive integers that need not be positive. Let's call that U(N). If we write N=2^a*3^b*5^c... (the prime factorization), then U(N)=2*(b+1)*(c+1)*(d+1)... (see if you can prove it; I think it's easier than tackling (c) and (d) directly). Remember that the average of the consecutive numbers must be an integer if there are an odd number of them and either an integer or a half-integer if there are an even number of them.

yea, ur rite. My mistake. Its the sum of consecutive postivie integers. Sorry about that.

Oops, COmplexZeta, Im afraid your answer is incorrect. I did the problem this morning, and I found out that the answer is 130.

2mnp=(m+2)(n+2)(p+2)
2mnp=mnp+2(mn+np+mp)+4(m+n+p)+8
mnp-2(mn+np+mp)-4(m+n+p)-8=0.

mnp-2(mn+np+mp)+4(m+n+p)-8=8(m+n+p)
(m-2)(n-2)(p-2)=8(m+n+p)

Let m-2=a
     n-2=b
     p-2=c

abc=8(a+s+b+2+c+2)
abc=8a+8b+8c+48
c=[8(a+b+6)]/(ab-8)

Thus, in order to maximize c, we must minimize ab. Since m,n,p are all natural numbers (sorry bout that, i wrote they were integers before), the minimum ab-8 can be is 1. Therefore, (a,b)=(1,9) or (3,3). Plugging that back in, we see that c is greater when (a,b)=(1,9), which is 128. Thus, p is equal to 130.

But does anyone have ideas on the other two problems?

Oops. When I did the case m=3, I got (n+2)/n<6/5, which would imply n>10, not n>12. So that's why mine was wrong. Good catch.

[color=cyan]For the T(n) question, it is worth considering that for odd n, n|T(n) and for even n n|2T(n) always.

I also have this distinct memeory that number 1 isn't that hard, but I'll have to go and actually try it and I don't have the time right now, sorry.  Try using simple things, like nCr + nC(r+1) = (n+1)C(r+1).[/color]


If you have a function $ f(x)$, and $ f(0)\equal{}1$, is it always true that $ f(x\plus{}y)\equal{}f(x) \cdot f(y)$. If so, why?

[quote="t0rajir0u"]Er... try $ f(x) \equal{} x \plus{} 1$.  Or... any function passing through $ (0, 1)$ that isn't of the form $ e^{f(x)}$ where $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ (in other words, $ f(x)$ is a solution to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url]).[/quote]

heh. sorry. sometimes I ask a question that I don't even look for the simplest in-my face solution. for some reason, someone in a class used f(0)=1 as sufficient proof that f(x) has to be exponential and therefore followed f(x+y)=f(x)f(y) like f(x)=2^x.

[quote="mihail911"]someone in a class used f(0)=1 as sufficient proof that f(x) has to be exponential[/quote]

If I was a car mechanic, this would be the point where I stop, pick my head out from under your hood, and yell:
Yep, that's your problem right there. That'll be five hundred, please.


In the thetrahedron ABCD the sum of the areas of faces which share the same side AB equals to the sum of areas of faces which share the same side CD.Prove that the midpoints of BC,AD,AC,BD lie on a plane that contains the inecenter of the inscribed sphere.

Let K, L,M and N be the midpoints of BC,AD,AC,BD respectively and let I be the center of the inscribed sphere.

First of all is it immediate to see that K,L,M and N are the vertex of a parallelogram and then they lie on the same plain. This plain is, of course, parallel and at the same  distance wrt AB and DC. It holds that ABLNKM and DCLNKM have the same volume.  Indeed as (CMK)=(CAB)/4 and (NKB)=(BCD)/4, (DMKC) = (ANKB)=(ABCD)/4 and, as A and D are at the same distance from LNKM, (ALNKM) = (DLNKM).

Since, from hypothesis, (IBCD)+(IACD) = (ICAB)+(IDAB), and  as 4(ICKM) = 4(IABKM)/3 = (IABC) and similar relations hold, then it results that (DCLNKMI) = (ABLNKMI) and this implies that I lies on LNKM.


Let $\sin x+\cos x =t,$ we can rewrite as $t+\frac{t^{2}-1}{2}=1\Longleftrightarrow t^{2}+2t-3=0\Longleftrightarrow t=1,-3.$Since$(\sin x+\cos x)^{2}+(\sin x-\cos x)^{2}=2\Longleftrightarrow t^{2}\leq 2.$Thus$t=1.$that is to say,$\sin x+\cos x=1$Solving this by unit circle, yielding$x=2n\pi , \frac{\pi}{2}+2n\pi\ (n: integer).$

[quote="maokid7"]$(\sin(x)+\cos(x))^{2}=(1-\sin(x)\cos(x)^{2}$

$1+2\sin(x)\cos(x)=1-2\sin(x)\cos(x)+\sin^{2}(x)\cos^{2}(x)$

$4\sin(x)\cos(x)=\sin^{2}(x)\cos^{2}(x)$  now for $\sin(x)\cos(x)\neq{0}$ we have

$4=\sin(x)\cos(x)\Longrightarrow 8=\sin(2x)$ so there are no solutons with $x\in\mathbb{R}$. If you want complex solutions then solve $8=\frac{e^{2ix}-e^{-2ix}}{2i}$.

now if $\sin(x)\cos(x)=0$ we have that $x=\frac{n\pi}{2}$ for $n\in\mathbb{Z}$[/quote]
$sinxcosx=0\rightarrow sinx+cosx=1$
So the answer must be $x=2k\pi$ or $x=\frac{\pi}{2}+2k\pi$ with $k\in\mathbb Z$


[quote="VANCHANH123"]If $x,y,z>0$  and $xyz=1$ .then prove that $\frac{x}{1+xy}+\frac{y}{1+zy}+\frac{z}{1+xz}\le \frac{3}{2}$[/quote]
It's wrong: $x=y=2$ and $z=\frac{1}{4}.$ :wink:


Suppse we have an ideal vase and we do the following:
at 1minute befor 1 o'clock we  put the numbers from 1 to 10 in the vase and take out the number 1;
at half a minute before 1 o'clock we put the numbers from 11 to 20 and take out 2
at 1/3 min befor 1 o'clock we put numbers 21-30 and take out 3...
AND SO ON...

QUESTION:
HOW MANY NUMBERS WILL THERE BE IN THE VASE AT EXACTLY 1 O'CLOCK?

i just want to point out a minor error
[quote]and take out the number 1; [/quote]
should be [b]and take out 1 number;[/b]
have fun!

No, the original version is-for the first step-take out the number one, for the second-take out the number two...     but 3X.lich's version seems more interesting!!!

:? I would say 0 of course, same for both versions :?

(as when it's like 1/20 before 1, you'll put in 10 and take out 20, and so on

Hehe, this is classic. It's been heavily discussed on the sci.math newsgroup, if anybody else hangs around there.

[BTW, I hope I'm understanding it right. My impression was that on the third step we take out the number 3, not 3 numbers, as Peter VDD (and 3x.lich) are assuming, because if we do it their way, there would soon be a negative number of numbers!]

[edit: didn't read carefully enough. I was right, the original intention, as clarified by xxxxtt himself, was that you take out 1 number each turn. Thus, all of the below is relevant. I'll do another reply on the other method, though.]

Now, if we do it xxxxtt's way, there'll be no numbers left. This may seem a bit odd, but consider it like this: try to tell me one number that is still there at the end. No matter what number N you say, that number was taken out on the Nth turn, 1/N minutes before 1 o'clock. So there are no numbers in the vase. There are an infinite number of steps, so there are an infinite number of numbers taken out.

Now, let's consider another way, which is more general--you get to choose which number to take out. Now it depends on what number you chose to take out. If each time you decide to take out the lowest number, than it is the same as the original. However, say that each time you chose to take out the highest number. Then, the numbers 1-9 will be left after the first step, 11-19 after the second step, 21-29 after the third step, etc. Since there are still an infinite number of steps and there is a net gain of 9 numbers after each step, there will be an infinite number of balls left.

You can also make it so that there will be, say, exactly 5 numbers left. On the first step, take out the number 6. On the second step, take out the number 7, and so on. The only numbers left will be 1-5.

The dangers of infinity...

.dzhonatan

The first impression that I saw was that 'take out 2' meant literally taking out two numbers but of course it meant taking out '2' because after I replied I realized that you can't take more than k numbers out after k steps, for some large k... As for the solution of swiftstream, it looks definitely like that story of a race between Achilles and the tortoise... Umm! paradoxes are fun. :D

The other method...

This is an analysis of the method 3x.lich and Peter VDD are suggesting, where on the first turn you take out 1 number, the second turn you take out 2 numbers, etc.

Now, after N turns, the total number of numbers is  \sum (from 1 to N) of (10-N). Or, (10-1)+(10-2)+(10-3)+...+(10-N), i.e., 10N- \sum (from 1 to N) of (N), or, applying the sum formula,

10N-(N^2+N)/2

=-N^2/2+19N/2

which is a parabola opening downward,  \lim N-->(infinity)=(-infinity), so that after infinity steps there will be negative infinity numbers left, which just doesn't make much sense :-)

.dzhonatan

Grrr... ascii is frustrating. I look forward to the day when you will be able to post on forums in true mathematical form, whether it be handwritten scrawl (now that would be cool :-) or just a good typesetting interface.

Not quite like achilles and the tortoise, because that has an understandable resolution with limits and infinitesimals.

This is a bit wacky, because it all depends on which number you take out. It doesn't seem like that should make a difference. A lot of people just can't accept it at all...

Anyway, though, I'm going to go make a thread about paradoxes, because I agree, they are fun :-)

.dzhonatan

ya I know, then it gets negative after a while. That's also why i found it that easy :D

but still, the question is confusing :? so I'm not surprised to see more misunderstandings around this.


calculus
derivative
real analysis
real analysis unsolved

By looking at a graphing calculator, is there a way to find inflection points for complex Calculus problems? or is there a more simplistic way of solving difficult second derivatives, besides setting the second derivative to 0?

I can't think of one.  You're better off just doing it by hand anyway since often times, you won't be able to use a calculator.


Show that if $f(x)$ is a polynomial whose degree is less than $n$, then the fraction
$\frac{f(x)}{(x-x_{1})(x-x_{2})...(x-x_{n})}$,
where $x_{1},x_{2}, ...,x_{n}$ are $n$ distinct numbers, can be represented as a sum of $n$ partial fractions \[\frac{A_{1}}{x-x_{1}}+\frac{A_{2}}{x-x_{2}}+\ldots+\frac{A_{n}}{x-x_{n}}\] Hint: use [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange[/url]

[hide="solution(please tell me if it is incorrect)"]
By Lagrange we know that $f(x)$ can be expressed as $\sum \frac{(x-x_{2})(x-x_{2})...(x-x_{n})}{(x_{1}-x_{2})(x_{1}-x_{3})...(x_{1}-x_{n})}y_{1}+...$
Where $x_{i}, y_{i}\in \mathbb{R}$. Thus we write

$\frac{f(x)}{(x-x_{1})(x-x_{2})...(x-x_{n})}= \frac{ \sum y_{1}\frac{(x-x_{2})(x-x_{2})...(x-x_{n})}{(x_{1}-x_{2})(x_{1}-x_{3})...(x_{1}-x_{n})+...}}{(x-x_{1})(x-x_{2})...(x-x_{n})}$

We can divide by $\sum_{i=1}^{n}x-x_{i}$ to get

$\frac{y_{1}}{(x-x_{1})(x_{1}-x_{2})(x_{1}-x_{3})...(x_{1}-x_{n})}+...$

Since $y_{i}, x_{i}$ are arbitrary numbers, we have found the desired partial fractions.
[/hide]

So Lagrange is just the closed form of the nth degree polynomial passing through n+1 specific points...

Sorry, I can't tell if your post is just a comment or if you're trying to point out something wrong with my solution.

Looks OK to me if you mean $\prod_{i=2}^{n}(x-x_{i})$ instead of $\sum_{i=1}^{n}x-x_{i}$,and if you put
$A_{1}=\frac{y_{1}}{(x_{1}-x_{2})(x_{1}-x_{3})\hdots(x_{1}-x_{n})}$

EDIT: you have to divide both numerator and denominator by
$\prod_{i\neq k}^{n}(x-x_{i})$
for the k$^{th}$ term.


What is the maximum number of points of intersection when $ 5$ lines intersect each other?

There are $ 4\plus{}3\plus{}2\plus{}1\equal{}\boxed{10}$ points of intersection.

Assuming no 2 lines are parallel, we need 2 lines to determine an intersection point, so $ \binom{5}{2}\equal{}10$

@usernameyourself, please, [b]please[/b], [i]please[/i] stop bumping up random threads to not say anything useful

There should be an award for that 10 year bump though


I'm in the UK and the problem solving book looks perfect for me... will it be shipped to the UK though? I know the site is American (or I guessed it is, no offence intended if it isn't), ad all the prices are in dollars which is probably the reason it looks so expensive at first glance. (I think $40 = about 25?)

Also, what is the solutions book? I guessed it was solutions to problems, but I didn't know what problems it was the solutions to? Perhaps there are problems in the book... this uncertainty is why I ask. :cool:

Thanks, I really want to get betetr at these maths problems but my only book "The Mathematical Olympiad Handbook" (for BMO-British maths olympiads) has a very breif teaching session and assumes I am already pretty good at it.

AoPS is American.  Well now that they have joined with mathlinks it's more international.  If you get the books you will really want the solution books.  The books consist of about 30 chapters each.  In the chapters they teach you the stuff you need to know to do the problems.  They have a few examples of problems in each chapter where they show how to do a question.  Then they give you exercises to do which help your understanding of each section within a chapter.  Finally at the end of each chapter they have a bunch of problems to do that pertain in some way to what was taught in the previous chapter.  The solution manual is very useful for these problems at the end of each chapter.

We are able to ship the Art of Problem Solving books to any country.  There is an additional charge for international shipping that is not listed on the online order form ([url=http://artofproblemsolving.com/Books/AoPS_B_Order.php]click here[/url]).  There are instructions at the bottom of the order form that tell you how to find out how much the shipping charges are.

I would also encourage you to buy the solutions along with the books.  Most of the time when people just buy the book they end up coming back for the solutions.  The solution manual are valuable teaching tools in that they show how to solve each problem, not just the answer.


1. There are 27 candidates in elections and $ n$ citizens that vote for them.
If a candidate gets $ m$ votes, then $ 100m/n\le m\minus{}1$. What is the smallest possible value of $ n$?
[hide="answer"]134[/hide]

2. The set A consists of $ m$ consecutive integers with sum $ 2m$. The set B consists of $ n$ consecutive integers with sum $ 2n$. The difference between the largest elements of $ m$ and $ n$ is 99. Find $ m$.
[hide="answer"]201[/hide]

[quote] 2. The set A consists of m consecutive integers with sum 2m. The set B consists of n consecutive integers with sum 2n. The difference between the largest elements of m and n is 99. Find m.  [/quote]

[color=darkblue] Hey! Are the above two problems inter-linked or are they independent? [/color]

"Find m", Does this mean we have to calculate the number of elements of set 'A' which conforms to the rules as stated in the problem, right?

If it's so, here's a solution:

Consider the set A as follows : A={-1, 0 , 1 , 2 , 3 , 4 , 5}

Here set A consists of "seven" consecutive integers with sum =14.

So, [b] m =7 [/b] and [b] sum =2m = 14 [/b]

Now, coming to the second part of the problem.

Consider the set B as follows: B= {-100,-99,-98,..0...+100,101,102,103,104}

In set B, "n" i.e. the number of consecutive integers is [b] 205 [/b] and the sum (2n) of  these integers is [b] 410 [/b].

(The number of consecutive integers in set B is: 205)

-100 to 0 --> No of digits (including 0) is: [b] 101 [/b]

1 to 104 --> No of digits (excluding 0, since it has already been counted) is :[b]104 [/b]

The largest number of set 'A' is: 5, and the largest number of set 'B' is :104 (Hence, the difference between the largest numbers of 'A' and 'B' is 99).

Hence, [b] 'm' is = '7' [/b] or if we consider it the other way round that is -- Set A= (-100,-99,-98.....104} and Set B = {-1, 0 , 1, 2, 3, 4, 5} then [b] 'm' = 205 [/b]

[quote="mathwizarddude"]
2. The set A consists of $ m$ consecutive integers with sum $ 2m$. The set B consists of $ n$ consecutive integers with sum $ 2n$. The difference between the largest elements of $ m$ and $ n$ is 99. Find $ m$.
[hide="answer"]201[/hide][/quote]

Wouldn't any set of consecutive integers centered around 2 work?
$ 2\equal{}2(1)$

$ 1\plus{}2\plus{}3\equal{}2(3)$

$ 0\plus{}1\plus{}2\plus{}3\plus{}4\equal{}2(5)$

So, couldn't M be any positive odd number?

This problem was taken from the imomath test [url]http://test.imomath.com/iop_logovanje.php?ImModerator=t&req=gradeTest&subjwName=gen[/url]
only the answer is given but without solution to this problem and that's why I posted it here (I'm baffled by the multiple answers, too) but does the first problem make sense?

For the 2nd problem if the two sequences of consecutive integers start with $ a,b$, then $ |a\minus{}b|\equal{}99$.
Does anyone have a solution for the first problem?
(@zserf: yours doesn't work because 4-3 does not equal to 99 :P )

But because the largest term can be any odd number, M can be any odd number that is atleast 199.

If N is 1
M is 199

If N is 3
M is 201

and so on.

$ M\equal{}N\plus{}198$


Compute the number of  3 digit positive integers $b$, such that $\log_8 b$ is a rational number.

This is how I solved it:

$8= 2^3$

Therefore, the number has to be multiples of $2^3$, so:

$(2^3)^3 =  512$
$(2^3*2)^2 = 256$
$(2^3*3)^2 = 576$

Which means the only numbers that $b$ can be are $512$, $256$ or $576$. However, I got this through guess and check, and I was wondering if there were any other better, more "mathematical" ways of getting this answer, rather than randomly guessing.

Are you sure log  (base 8) of 576 is rational?  [hide](Log 9 to the base 8 is not rational, and  you still will not have a rational number if you   you add 2 to it..)

Now 128,  256, 512 are another matter ..
hint: any power of 2 would do..
.[/hide]

Yep, I just realized that 128, 256 and 512 are the only values that would work, because the other two are not rational, thanks for pointing this out :blush:


Construct a triangle ABC given the  three radios of the excircles.....

$ \frac{a}{S} \equal{} \frac{2}{h_a} \equal{} \frac{1}{r_b} \plus{} \frac{1}{r_c}$ and cyclic exchange. Let $ A,Y,Z$ be points on a ray $ (Ox,$ such that $ OA \equal{} r_a, OY \equal{} r_b, OZ \equal{} r_c.$ Inversions with center $ O$ and powers $ \plus{}OA^2, \minus{}OA^2$ take $ Y, Z$ into $ Y_1, Z_1$ and $ Y_2, Z_2,$ respectively. Denote $ c' \equal{} AY_2.$ Circles $ (A), (Y_2)$ with radii $ b' \equal{} AZ_2, a' \equal{} Y_2Z_1$ intersect at $ Z_3$ and $ \triangle AY_2Z_3 \sim \triangle ABC.$ Construct any exradius of $ \triangle AY_2Z_3,$ say $ r_a'$ against $ A,$ and scale $ \triangle AY_2Z_3$ by coefficient $ \frac{r_a}{r_a'}.$


This is from 2004-2005 handbook warmup 18
There are four cookies and four chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie?

[hide]
I'm sorry I can't find a more algebraic way to do it, but you can take all ones, and then a two, two ones, and a 0 and rearrange them $4!/2!$ times, or 12. Finally, take two zeroes and two 2s and rearrange them $\frac{4!}{2!2!}$ times, or 6. Add together to get $19$ ways. I hope I did that right.
[/hide]

The answer is 51/64, but I don't know how to get it. Can someone help?

[hide]
Let's count the ways to put 3 or 4 chips in each cookie.

First of all, there are $4^{4}$ ways to put 4 chips in 4 cookies.

There are 4 ways to put 4 chips in one cookie.

For 3 chips in one cookie and 1 chip in another, there are 4 ways to choose the cookie with 3 chips in it and 3 ways to choose the cookie with one chip in it, for a total of 12 ways to choose the cookies. There are $\binom{4}{3}$ ways to choose which 3 chips go into the cookie with three chips in it. So there are a total of 48 ways for this case.

So there are $256-4-48$ ways, for a probability of $\frac{204}{256}= \frac{51}{64}$[/hide]


[img]http://img525.imageshack.us/img525/4653/augquizwv3.jpg[/img]

solve this and mensa will accept you.

Let $ AB \equal{} x,\ AD \equal{} y, EC \equal{} a,\ FC \equal{} b$, we have $ [CEF] \equal{} 3\Longleftrightarrow ab \equal{} 6\ \cdots [1]$, $ [ABE] \equal{} 4\Longleftrightarrow x(y \minus{} a) \equal{} 8\ \cdots [2]$,
$ \ [ADF] \equal{} 5\Longleftrightarrow y(x \minus{} b) \equal{} 10\ \cdots [3]$. From $ [2],\ [3]$, substituting $ a \equal{} \frac {xy \minus{} 8}{x},\ b \equal{} \frac {xy \minus{} 10}{y}$ gives to $ [1]$, yielding $ (xy)^2 \minus{} 24xy \plus{} 80 \equal{} 0\Longleftrightarrow xy \equal{} 4,\ 20$.
$ \therefore [AEF] \equal{} xy \minus{} (3 \plus{} 4 \plus{} 5) \equal{} xy \minus{} 12 \equal{} 20 \minus{} 12 \equal{} 8$.

Are you sure that Mensa will accept anyone for solving such an easy problem. I'm not sure...

[quote="Bugi"]Exactly like yours! Nice solution :lol: ! But still i think it's easy...[/quote]Oh really? There is a non-algebraic and shorter solution, much faster than kunny's, it's a slick solution and it took me a while to find it. I challenge you to find it, if you think it's easy...

I'm waiting.

[hide=" :rotfl: "]Well, this didn't take long but...
set EC=3, FC=2, AB=4, BE=2.  This gives the desired areas

Then, rectangle-triangles  :rotfl: [/hide]

[color=blue][u]click smiley to see hidden content[/u][/color]

The problem can be reduced to the figure where x is the area of the rectangle.
Now there are two possibilities.
a) (x-16) + (6) >= (24- x)+ (x-14) then
    (x-16) >= (24-x) gives x >= 20
    and 6>= (x-14) gives x<=20.

b) (x-16) + (6) <= (24- x)+ (x-14) then
    (x-16) <= (24-x) so x <= 20
    and 6<= (x-14) x>=20.

    The area of the rectangle is 20.


integration
function
calculus
calculus computations

Does the series $\sum_{n=1}^{\infty}(-1)^{n}\int_{2n}^{2n+1}\frac{\ln(1+3t)}{t}dt$ converges?

It converges. You show this is a Leibniz series.

Let $a_{n}=\int_{2n}^{2n+1}\frac{\ln(1+3t)}{t}dt=\frac{\ln(1+3c)}{c}$ for some $2n<c<2n+1$.

Now it is simply to see that $a_{n}\rightarrow 0$ as $n \rightarrow \infty$.

It remains to prove that $a_{n}$ is a decreasing sequence. It decreases in view of the fact that it converges to 0, however it can be proven to be decreasing by elementary calculations.

$a'(n)=2\left(\frac{\ln(1+3(2n+1))}{2n+1}-\frac{\ln(1+3(2n))}{2n}\right)$, hence it sufficient to show that

$x\rightarrow \frac{\ln(1+3x)}{x}$ is a decreasing function, which is.


Just wondering how fast you guys run...
List your times for mile/100yd dash or whatever else you do.

I run a 7:50 mile as of May/June , but I should be able to bring that down to atleast 6:40ish cuz I'm doing cross country this year(my first time-I'm in 7th grade).
My 100yd dash  was 15.9 secs I think as of May/June.
I'll update in about two weeks when we run in gym.

[quote="nonie"]Just wondering how fast you guys run...
List your times for mile/100yd dash or whatever else you do.

I run a 7:50 mile as of May/June , but I should be able to bring that down to atleast 6:40ish cuz I'm doing cross country this year(my first time).
My 100yd dash  was 15.9 secs I think as of May/June.
I'll update in about two weeks when we run in gym.[/quote]

Mile time: 6:20

100 METER dash time: 14 sec.

mile:   6:12

i dont know the other one

theres this guy in my grade who can run 5:53

wow. i just got my mile under 6:00 this spring, and i'm a high school senior...
then again, i only ever did a season of indoor track in 9th grade, then quit.

My mile time is about 5:50. I prolly could do better but I haven't timed myself for a while.
I forget my time for 100 yd dash though

Don't laugh at mine, please

10:45 :blush: :blush:

I'm serious. I'm not proud.

[quote="236factorial"]I'm at 6:08, but I should be faster since I'm running cross country, and I've never ran the 100m dash, but I did it once as a trial and got 17.2, or was it 16 something?[/quote]
hmm...
I've never done the 100 METER dash only the 100 YD one(and only like twice in gym)
any idea how big the difference is in seconds?i think 100 m is 9 yds more than 100yds rite?

[quote="nonie"]I've never done the 100 METER dash only the 100 YD one(and only like twice in gym)
any idea how big the difference is in seconds?i think 100 m is 9 yds more than 100yds rite?[/quote]

Yeah, it's a little more than 9 yards more.

I think you add about 1 second to your yards time to obtain meters time.

In 6th grade I ran 6:27...I havent done the mile run ever since. Im in 9th grade right now.

i dont know my endurance is horrible but my speed is good

Wow and I thought my 7 minute mile was fast...! Oh well, I'm only 13 & I don't train or anything. running is great though! I burn so many calories and I get to eat that extra slice of cheesecake at the end of the day :D

the fastest i've seen anyone run a mile is 5:20.

[quote="jhollenbeck"]theres a guy at my friends school that can run a mile in 4:30! :o hes fast. I can only run it in like 7 minutes[/quote]

4:30??? That's crazy! He'd be... 400 meters ahead. me and my crazy math skills :D

Around 6:30 for the first mile.

I don't run speed.


Prove that $ b^{x\plus{}y} \equal{} b^{x}b^{y}$ for all real $ x,y$.

So $ b^{x}b^{y} \equal{} \sup B(x) \sup B(y) \equal{} \sup(B(x)B(y))$ where $ B(x) \equal{} \{b^{t}| 0< t \leq x \}$.

Is this correct?

Again, what is your definition of exponentiation?

We can always find a monotone  secuence of rationals $ \{a_{n}\}$ so that this secuence have  $ x$ (real) as limit,  we define $ b^x$ as the $ \lim_{n\rightarrow \infty} b^{a_{n}}$. Finally if we do the same with $ y$ is easily to prove that $ b^{xy}\equal{}b^{x\plus{}y}$

Obviously you should take the sup over $ \{b^t | t\le x, t\in \mathbb{Q}\}$..