set theory
topology
advanced fields
advanced fields unsolved

I have recently been reading some stuff about geometric measure theory, and I have become confused by some of the set terminology.

In Edgar and Miller's paper proving the Erdos-Volkmann ring conjecture, they say that analytic sets are the continuous images of Borel sets, and these are commonly called Suslin sets.

Mattila makes no mention of analytic sets but sometimes says "such and such requires the theory of Suslin sets" and doesn't ever explain more.

Mathworld also says that analytic sets are Suslin sets, but says that a Suslin set is a continuous image of a Polish space.

Wikipedia says a Suslin set is of the form
$F = \bigcup_{i_{1}, i_{2}, \ldots}\bigcup_{k=1}^\infty F_{i_{1}, i_{2}, \ldots, i_{k}}$
where the $F_{i_{1}, \ldots, i_{k}}$ are all closed and that if $X$ is a Polish space then $A \subset X$ is analytic if there is another Polish space $Y$ and a Borel set $B \in X \times Y$ such that $A$ is the projection of $B$...

So I'm totally lost at this point. Are these definitions all somehow equivalent? Or is there just no universally accepted convention? Is there anyone here familiar enough with this particular corner of descriptive set theory in order to enlighten me? Please take into account that I'm a noob. (I suspect that if I better understood the implications of "Polish space" I'd have less trouble with this. I don't want to clutter this with more definitions but I also can't see how Mathworld's definition of Polish space is equivalent to the Wikipedia one and I haven't ever heard that term before 15 minutes ago anyway... argh!)

I believe that the terms "Suslin" and "analytic" are interchangeable. Such sets can be defined in several ways: by the $(\mathcal A)$-operation (as in Wikipedia), by continuous images, or projections. Federer discusses Suslin sets on pp.65-68 of "Geometric measure theory". In a nutshell, Suslin sets form the smallest $\sigma$-ring* that contains all Borel sets and is closed under continuous maps.

(*) not $\sigma$-algebra: the complement of an analytic set $A$ is analytic if and only if $A$ is Borel.

I am closely studying a GMT text now, and I wanted to understand analytic sets a little better. I have convinced myself that all Borel sets are analytic, but there's another critical result that I can't seem to figure out:

How would one prove that the continuous image of an analytic set in a compact metric space is still analytic, assuming that we use the Wikipedia definition rather than the continuous image definition?

I would appreciate hints or a brief sketch, so that I can fill in the details myself.


geometry
Putnam
geometric transformation
reflection
perimeter

I know I've posted this problem before, but I just cannot for the life of me get it. It's out of Larson's book. If someone could post a solution that would be excellent. Thank you.

ABC is an acute-angled triangle. Let D be on the interior of AB. Locate points E on AC and F on CB such that the inscribed triangle DEF will have minimum perimeter.

The solution I'm looking for involves reflecting D over AC (D') then again over CB(D'') and analyzing the length of D'D''.

Thank you again for your help. I really need to start studying geometry for the Putnam, and I've heard (and read) that reflections are very important.

Reflection is the way to go for this problem.

[hide]Like you said, reflect D over AC to D', and reflect D over D'' to BC. For any choice of E and F, notice that ED = ED' because of the reflection, and likewise FD=FD''. So, perimeter = DE+EF+FD = D'E+EF+FD''. The path D'EFD'' is just a broken path from D' to D'', and we want to minimize this length. Well, thats easy... the shortest distance between D' and D'' is a straight line from D' to D''! So, the desired E and F that we want are the intersection of segment D'D'' with AC and BC, respectively. Do you see why minimizing the length of path from D' to D'' will also minimize the perimeter of triangle DEF?[/hide]

By the way, what book are you referring to?

Here are three similar problems:

1.
Given triangle A, B, C. Find points D, E, F, which lie on segments BC, CA, AB, respectively, such that the perimenter of the triangle DEF is minimized/maximized.

2.
Find a point P inside triangle ABC such that if X, Y, Z are the feet of the perpendiculars of P onto the three sides, then the perimeter of the triangle XYZ is minimized/maximized.

3.
Find a point P inside triangle ABC such that PA + PB + PC is minimized/maximized.

(one of min and max is more interesting than the other, but since I've done these problem quite a while ago, I can't recall which is the more interesting one. Oh well)

Zabelman, thank you very much for your help. I see it quite clearly now! I got it from Problem Solving Through Problems, by Loren C. Larson. I've heard it is excellent prep for the Putnam (what I am studying it for), and is relatively cheap as well. You can get it off ebay used for around 40$ US.

I will take a look at those other problems too. Thanks!

Or you can get it brand new from Amazon for $40. At least, you could three years ago (when I got it).

Or you can do what I did.  Take it out from your university's library at the beginning of the school year, and continually renew it for an entire year  8-)

You can renew books for a year? We're only allowed to renew books five times, which is a bit more than a quarter.

I can only renew thrice, actually :-/ Anyway, I wanted it for over the summer, and I live far away from my school *shrugs*. Plus my local library's crap :P


Find the greatest real k such that for any positive a, b, c with a^2 > bc we have
          (a^2-bc)^2 > k(b^2-ca)(c^2-ab)

Namdung

First let us take b=c. We have |a^2-b^2|>sqrt(k)*|b|*|a-b|, from where a+b>ksqrt(b). Take b close to a and find that k<=4. We prove that k=4 works. The inequality becomes f(a)>=0, where f(x)=x^4-6x^2bc+4x(b^3+c^3)-3b^2c^2. We have f''(x)=12(x^2-bc)>0 for all x>sqrt(bc). So, f' is increasing for x in (sqrt(bc), infinity), so f'(x)>f'(sqrt(bc))=4(b^3+c^3-bcsqrt(bc))>=0. So, f is stricly increasing and f(a)>f(sqrt(bc))=4sqrt(bc)(b^3+c^3)-8b^2c^2>=0.

As Harazi said, we can see that k<=4. To show k=4 work, my student NamTran (is not my son :) ) suggested very nice solution

We will show that if a^2>bc then (a^2-bc)^2 > 4(b^2-ac)(c^2-ab)

If b^2 > ca then a^2.b^2 > bc.ca => ab > c^2, the R.H.S < 0 and the result is trivial. Similarly with the case c^2 > ab.

Suppose that b^2 < ca, c^2 < ab then

(a^2-bc)^2/4(ac-b^2)(ab-c^2) > (a^2-bc)^2/(ab+ac-b^2-c^2)^2 >=1

(AM-GM and ab+ac-b^2-c^2<=a^2-bc)

Namdung

Wow, that really is a magnificient solution. Congratulations to your student. I bet he has lots to learn from you.


vector trigonometry limit integration function calculus

Vector of velocity of points $A$ $B$ are $\vec{V_{1}}$ and $\vec{V_{2}}$. We have that $|\vec{V_{1}}|=|\vec{V_{2}}|=V$. Line of vector $\vec{V_{2}}$ in every moment contains point $A$ . Prove that for time $t$ we have that $t \to \infty \implies \vec{V_{1}}\to \vec{V_{2}}$.

Hopefully I interpreted what you were saying correctly. Let's call the axis $A$ moves on the $x$ axis. Thus the $\Delta x = V t$, where $t$ starts at $0$ by convention. Effectively $A$ is always looking at $B$ and is hunting it down.

I won't assume that the two vectors are perpendicular to begin with, I'll say there is an angle of $90+\theta$ to begin with, so that if it actually turns out that the vectors are perpendicular, then it's a quick change.

Thus $\vec{V_{2}}_{x}= V \frac{Vt+a\sin{\theta}}{\sqrt{(Vt+a\sin{\theta})^{2}}+(a\cos{\theta})^{2}}$ and $\vec{V_{2}}_{y}= V \frac{a\cos{\theta}}{\sqrt{(Vt+a\sin{\theta})^{2}}+(a\cos{\theta})^{2}}$.  Thus when you take the limit there is a $\frac1{t}$ effect for $\vec{V_{2}}_{y}$, making it go to $0$. Then $\vec{V_{2}}_{x}$ approaches $V$, which means it is the same as $\vec{V_{1}}$ overall. That wasn't too rigorous, but the intermediate steps can be filled in and the wording can be improved.

Ok. Let $d(t)$ be the distance between $A$ and $B$ at the moment $t$. Find $\lim_{t \to \infty}{d(t)}$.

[quote="phucnv87"]$\lim_{t\to\infty}d(t)=\frac{a}{2}$[/quote]

How, about some prove? :) or, an experiment   :wink:

Your hypothesis is incorrect, my friend.

First make no assumptions about anything not stated in the problem.

We have $d\left(t\right) = |\vec{r}_{A}-\vec{r}_{B}| = a.$

Also $|\vec{V}_{1}| = |\vec{V}_{2}| = V.$

Now taking the limit as t approaches $\infty,$ of $\frac{\Delta d}{\Delta t}$ gives

$\lim_{t \to \infty}\frac{\Delta d}{\Delta t}= \frac{|\vec{V}_{1}(t+\Delta t)-\vec{V}_{2}(t+\Delta t)|-|\vec{V}_{1}t-\vec{V}_{2}t|}{\Delta t}= \lim_{t \to \infty}|\vec{V}_{1}-\vec{V}_{2}| \frac{|t+\Delta t|-|t|}{\Delta t}= |\vec{V}_{1}-\vec{V}_{2}|.$

The above result doesn't imply that $\vec{V}_{1}$ converges to $\vec{V}_{2}$ however.

$\lim_{t \to \infty}\frac{|t+\Delta t|-|t|}{\Delta t}= 1.$

While $\lim_{t \to \infty}|\vec{V}_{1}-\vec{V}_{2}| = 2 V |\sin \left( \frac{\theta}{2}\right)|,$ where $\theta$ is the angle between the two vectors.

$|\vec{V}_{1}-\vec{V}_{2}| \equiv \sqrt{\left(\vec{V}_{1}-\vec{V}_{2}\right)\cdot\left(\vec{V}_{1}-\vec{V}_{2}\right)}$

$\left(\vec{V}_{1}-\vec{V}_{2}\right)\cdot\left(\vec{V}_{1}-\vec{V}_{2}\right) = V_{1}^{2}-2\vec{V}_{1}\cdot \vec{V}_{2}+V_{2}^{2}$

$V_{1}^{2}-2\vec{V}_{1}\cdot \vec{V}_{2}+V_{2}^{2}= V_{1}^{2}-2 V_{1}V_{2}\cos\theta+V_{2}^{2}= V^{2}-2 V^{2}\cos\theta+V^{2}$

$= 2 V^{2}\left(1-\cos\theta\right).$

$1-\cos\theta = 2 \sin^{2}\left(\frac{\theta}{2}\right)$

$\therefore 2 V^{2}\left(1-\cos\theta\right) = 4 V^{2}\sin^{2}\left(\frac{\theta}{2}\right).$

$\lim_{t \to \infty}|\vec{V}_{1}-\vec{V}_{2}| =\sqrt{4 V^{2}\sin^{2}\left(\frac{\theta}{2}\right)}= 2 V |\sin\left(\frac{\theta}{2}\right)|.$

[quote="Anurag Agrawal"][quote]Dr. No wrote:
We have $d(t)=l\vec{r}_{A}-\vec{r}_{B}l=a$

[/quote]
How can you assume this in general,when it is given only as an initial condition?[/quote]
The way the problem was worded I had thought at first that dule_00 was saying that $d\left(t\right) = |r_{A}-r_{B}| = a$ was constant. But as you'll note, that is not the case. Rather $d\left(t\right) = |r_{A}-r_{B}| = a$ is only a statement of the separation distance between the two position vectors at time $t.$ The separation distance is changing with respect to time as the rest of my analysis shows. The only case where $d\left(t\right)$ is constant is one in which the velocity vectors are headed in the same direction. As $\frac{\Delta d}{\Delta t}= 0 \iff |\sin\left(\frac{\theta}{2}\right)| = 0.$ This condition implies $\frac{\theta}{2}= n\pi\;\therefore \theta = 2n\pi.$

To be candid, I edited my original post. The fact that the problem said to prove that $\vec{V}_{1}\to \vec{V}_{2}$ lead me to conclude erroneously, by inspection that $\lim_{t \to \infty}\frac{|t+\Delta t|-|t|}{\Delta t}= 0.$ When in fact, $t \to \infty \implies t > 0$ and $t+\Delta t > 0$ irrespective of the sign of $\Delta t.$ Thus $|t+\Delta t|-|t| = t+\Delta t-t = \Delta t.$ Therefore, $\frac{|t+\Delta t|-|t|}{\Delta t}= 1.$

Nevertheless, this correction does not change the result of the proof: $\vec{V}_{1}$ and $\vec{V}_{2}$ do not converge.

It seems to me that I've kind of discovered the required answer to the question. In my prior analysis I proved that the distance between the two position vectors cannot be constant unless $\theta = 2n\pi,$ i.e. said vectors are headed in the same direction. We already know that the velocity vectors $\vec{V}_{1}$ and $\vec{V}_{2}$ have the same magnitude; that's given in the problem statement.

By definition any two vectors having the same magnitude, headed in the same direction are equal to each other.

$\therefore \vec{V}_{1}= \vec{V}_{2}\, \text{ QED }.$

Let the velocity vector of B at any time t:
$\vec{V}_{2}=Vcos\theta\vec{i}+Vsin\theta\vec{j}$
since the initial seperation between A and B is a,
from small analysis of a diagram of an intermedita stage at time t,
$\frac{a-V\int_{0}^{t}sin\left(\theta\right)dt}{Vt}=tan\theta$
where $\theta$ is a function of t and hence cannot be taken out of the integral sign,
hence,continuing with Dr.No's proof we have
$\lim_{t\to\infty}l\vec{V}_{1}-\vec{V}_{2}l=2V\lim_{t\to\infty}lsin\frac{\theta}{2}l$
now,from the relation between $t$ and $\theta$ if you can prove that at $t\to\infty$, $\theta\to0$ then we are done.


Source: Italy 1999
Prove that for each prime $p$ the equation
$2^{p}+3^{p}=a^{n}$ has no intger solutions $(a,n)$ with $a,n>1$.

[hide]Equation obviously has no solutions with $p=2$, so we can say that $p$ is an odd number. \[2^{p}+3^{p}=(2+3)\cdot\sum_{i=0}^{p-1}2^{p-1-i}(-3)^{i}=a^{n}\] Thus $a$ must be divisible by $5$, but since $n>1$, LHS is divible by $25$, and it means that $\sum_{i=0}^{p-1}2^{p-1-i}(-3)^{i}$ is a multiple of $5$. \[\sum_{i=0}^{p-1}2^{p-1-i}(-3)^{i}\]  \[\equiv\sum_{i=0}^{p-1}2^{p-1-i}(2)^{i}\mod5\]  \[\equiv\sum_{i=0}^{p-1}2^{p-1}\mod5\]  \[\equiv p\cdot 2^{p-1}\mod5\]  \[\equiv 0 \mod5 \iff p=5\] But our initial equation doesn't have solutions with $p=5$.[/hide]

This is incorrect since i didn't use the fact the $p$ is prime, but don't know where is my mistake....

We already know that $25|a$.
Let $p=2k+1$ and look the equation mod 5.
We have: $2^{2k}(2)+3^{2k}(3)=4^{k}(2)+9^{k}(3)\equiv4^{k}(2+3)mod5$. (*Now this is the step i am unsure about*): This congruence shows that the LHS has only one factor of 5.

?

ArcticMonkey solution is correct IMHO, and your is not. The step that you are unsure about it's false indeed.

[quote="ArcticMonkey"][hide]Equation obviously has no solutions with $p=2$, so we can say that $p$ is an odd number. \[2^{p}+3^{p}=(2+3)\cdot\sum_{i=0}^{p-1}2^{p-1-i}(-3)^{i}=a^{n}\] So $a$ must be divisible by $5$, but since $n>1$, LHS is divible by $25$, and it means that $\sum_{i=0}^{p-1}2^{p-1-i}(-3)^{i}$ is a multiple of $5$. \[\sum_{i=0}^{p-1}2^{p-1-i}(-3)^{i}\]  \[\equiv\sum_{i=0}^{p-1}2^{p-1-i}(2)^{i}\mod5\]  \[\equiv\sum_{i=0}^{p-1}2^{p-1}\mod5\]  \[\equiv p\cdot 2^{p-1}\mod5\]  \[\equiv 0 \mod5 \Leftrightarrow p=5\] But our initial equation doesn't have solutions with $p=5$.

P.S I have a very strong feeling that I made a silly mistake in my solution and wasted like 10 minutes writing all that crap. :D[/hide][/quote]

Can you explain about your factorization a bit more?

Here is another, closely related way.

We note that for odd p, $2^{p}+3^{p}$ is always divisible by 5.  Since we want the expression to be an integer raised to some power bigger than 1, we need $2^{p}+3^{p}$ to be divisible by 25.  However, constructing a quick table of the powers of 2 and 3 mod 25, it is easy to see that the only time this can happen is if p is congruent to 5 or 15 mod 25.  But then p must be divisible by 5, so p must be 5, but 5 fails and so we're done.

[quote="PenguinIntegral"]Can you explain about your factorization a bit more?[/quote]

For all positive integers $n$ the following identity is correct, \[a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+..+ab^{n-2}+b^{n-1})\] It can also be applied for $a^{n}+b^{n}$, but only for odd $n$, because $a^{n}-b^{n}=a^{n}+(-b)^{n}$. And that's all what I used in my solution. I hope it is clear now. :wink:

Yeah, that is me being dumb. Sorry, I just didn't recognize it in sigma notation.


calculus
integration
combinatorics proposed
combinatorics

In the book given in a source I found the following nice problem and I found two different proofs of it. According to the same book, there are many more proofs I have not found. Can you find more?

In a plane we have given a rectangular divided in smaller rectangulars. Every small rectangular has a length of some side integer. Prove that the original rectangular has at least one side length integer.

It's been discussed several times before, and several proofs have been presented (I can remember at least two, one by using a multiple integral in the plane, and one by coloring the plane and measuring black/white areas).

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6285
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24139
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34446

http://www.cs.toronto.edu/~mackay/rectangles.pdf
http://www.cs.chalmers.se/~coquand/tile.ps

More? :)


modular arithmetic
algebra
polynomial
number theory
number theory proposed

[quote="iura"]Consider $p$ a prime number and $p$ consecutive positive integers $m_{1}, m_{2}, \ldots, m_{p}$. Choose a permutation $\sigma$ of $1, 2, \ldots, p$. Show that there exist two different numbers $k,l \in \{1,2, \ldots, p\}$ such that $m_{k}m_{\sigma(k)}-m_{l}m_{\sigma(l)}$ is divisible by $p$.[/quote]
Let's see that ${m_{1},\hdots,m_{p}}=Z_{p}.$
If we assume that $m_{k}m_{\sigma(k)}\not\equiv m_{l}m_{\sigma(l)}\pmod{p}$ for all $m\not=k$ therefore $m_{k}m_{\sigma(k)}$ gives all remainders mod p for $k\in\mathbb{Z}_{p}.$ Thus if $m_{i}=p$ it follows that $m_{\sigma(i)}=p.$ Therefore \[\prod_{k\in\mathbb{Z}_{p}/i}m_{k}m_{\sigma(k)}=(p-1)!(p-1)!\equiv (p-1)!\pmod{p}\] and thus $(p-1)!\equiv 1\pmod{p}.$ But this is a conradiction with Wilson's theorem. For proof of Wilson's Theorem consider the polynomial $x^{p-1}-1$ in $\mathbb{Z}_{p}.$ It has roots $1,2,...,p-1$ and degree $p-1$ and therefore all its roots are $1,2,...,p-1.$ Therefore $-1=1.2.\hdots(p-1)$ in $\mathbb{Z}_{p}.$

It is not true for p=2 if $\sigma (k)=k$. Let $p>2$.
If $p|m{}_{k}{}_{0}$ and $p\not |m_{\sigma (k_{0})}$, then we have
$p| m{}_{k}{}_{0}m_{\sigma (k_{0})}-m_{\sigma^{-1}(k_{0})}m{}_{k}{}_{0}.$
Else $p|m_{k}$ and $\sigma (k)=k$. For all terms $p\not |n_{k}=m_{k}m_{\sigma (k)}$. But $\prod_{k\not =k_{0}}n_{k}=(\prod_{k\not =k_{0}}m_{k})^{2}=(-1)(-1)=1.$
Therefore exist m, suth that $n_{k}\not =m(mod p)$ (else $\prod_{k\not =k_{0}}n_{k}=-1(mod p).$ From p-1 terms $n_{k}$ with no more p-2 means we can chose k and l, suth that $n_{k}=n_{l}(mod \ p)$.

I really like this problem!
Here you can find one related to this:
http://www.mathlinks.ro/Forum/viewtopic.php?p=494833#p494833

Note that the thesis is still true also when p is not a prime number. (Well, in this case, we should call it n!)
Prove it!


geometry
circumcircle
geometric transformation
geometry proposed

Prove that the two Brocard points, the Lemoine point and the circumcenter
of any trianlgle are concyclic.

More to that point, the two Brocard points of a triangle ABC lie symmetrically to each other with respect to the line OK and both lie on the circle with diameter OK, where O is the circumcenter and K is the symmedian point (= Lemoine point) of the triangle ABC. Of course I can just copy the proof from some book, but most likely you preferred to see somebody posting his own solution...

  Darij

I'm sure I'll skip some details (it's always like that: you can see the solution for a certain geometry problem, but when it comes to explaining what you see, it's a bit harder :)).

The perpendiculars through $A$ to $CA$, through $B$ to $AB$ and through $C$ to $BC$ intersect at three points $A',B',C'$ s.t. $A'B'C'$ and $ABC$ are similar in this order of the vertices. In the same way we define $A\"B\"C\"$ (the perpendicular through $A$ to $AB$ etc.).

Here are some facts (not hard to prove, so I won't do it :)):

$A'B'C'\equiv A\"B\"C\"$ and they're symmetric wrt $O$ (the circumcenter of $ABC$).

$O$ is Lemoine's point for each of $A'B'C',A\"B\"C\"$.

The circles $(AA'B),(BB'C),(CC'A)$ intersect at $W_1$, the first Brocard point. The second Brocard point is defined similarly, by using $A\"B\"C\"$.

By some well-known results in pivot theory the point $W_1$ must also be the first Brocard point of $A'B'C'$, and $A'B'C'$ is obtained from $ABC$ by a spiral similarity of center $W_1$. From here we can derive that $AB\perp A'B'\Rightarrow W_1O\perp W_1O'$, where $O',O\"$ are the circumcenters of $A'B'C',A\"B\"C\"$ respectively. Taking this back to $ABC$, we have shown that $W_1O\perp W_1K\ (*)$ ($K$ is the Lemoine point). Of course, something similar works for $W_2$.

We must now show that $W_1,W_2$ are symmetric to each other wrt $OK$. To do this, simply observe that $OK$ is a median in $OO'O\"$ and it's also an altitude (from $(*)$).


geometry
3D geometry
sphere
calculus
integration
perimeter
calculus computations

i am asking why does the integral from -r to r of 2:pi: :sqrt:( r 2 -t  2  )dt isn't equal to the surface of a sphere.
 r is the ray of the sphere.
  2 :pi:  (:sqrt:r 2 -t 2 ) is the perimeter of a cercle whose ray is (:sqrt:r 2 -t 2.
  I AM NOT ASKING TO CALCULATE THIS INTEGRAL.BECAUSE I AM SURE THAT IT ISN'T EQUAL TO 4 :pi: R 2 . SO WHERE IS THE ERROR???/

You're not accounting for the tilt of the surface; that's only the horizontal component of the area. Look at what happens near the poles; there's a lot more area there than  in your formula. You're probably better off parametrizing by latitude; that way the fact that the surface is perpendicular to the radius helps you.

thanks for your reply. but  i didn't understand what you mean.please tru to clarify your answer ;)

You have two copies of this thread.  If you go look at my response in the other copy (which is also on this forum, further down), you will see my response which has a more detailed explanation.


Alyssa rode her bicycle for seven hours. She started on level ground and rode at a rate of 8 mph. She came to a hill that slowed her down to a rate of 5 mph. Upon reaching the top of the hill, she turned around and descended at a rate of 20 mph. Finally, she returned home on the flat portion at 8 mph. What is the total number of miles that she rode?

[quote="math154"]Her average speed was $ \frac4{\frac18\plus{}\frac15\plus{}\frac1{20}\plus{}\frac18}\equal{}\frac4{\frac14\plus{}\frac14}\equal{}\frac4{\frac12}\equal{}8$ miles per hour, so the answer is $ 7\cdot8\equal{}\boxed{56}$.[/quote]
[quote="PhireKaLk6781"]There are 4 different speeds to calculate into the harmonic mean.[/quote]
Is there any way to solve this problem other than using the harmonic mean?

Wouldn't that only be her average speed though if she traveled each speed for the same amount of time?

Harmonic mean should only be used for same distances.  The trip going uphill and downhill has an average speed of $\frac{2}{\frac{1}{5}+\frac{1}{20}}=8$ miles per hour, which is the same as the speed going on flat ground, so the answer ends up being the same.  You could also do a degenerate version of this question and assume that Alyssa goes uphill and downhill for 0 miles and still get the correct answer, since the problem doesn't depend on how long she travels uphill/downhill.


Rules

Players may join at any time. If you want to join, participate in the next round.

Each round, players PM me with a number. The number you choose will be added to your score. However, if you choose the highest number that round, you lose that many points instead of getting them! (If multiple members tie, they all lose points.)

There are no limits to what number you choose.

You do not have to post to say you are joining. Just PM me with your chosen number, and please put "High Number Death Game", "HNDG" or something similar in the subject line, as well as the round number.

Players can go into negative scores.

The rounds end when the next round begins. This will be shown by me posting to say so in this thread, and posting the updated scoreboard. I will aim to do a new round about every 24 hours, but if it goes on longer because I can't get to the computer, the round will be extended and it won't really affect the gameplay.

You may post on this thread to discuss the game.

After round 30, the player with the most points is announced the winner for that game, and we'll start again from round 1!

Round 1 begins now!

Scoresheet:

Hall of Fame:

@Klebian: Can you let me edit this post a bunch of times? Cause I'm going to have to and I figured out that you could only edit 6 times on this site.

Well actually it's just that you cant edit after 24 hours since you posted. Only mods can. Go bump the topic in Questions and Announcements and stuff at the bottom of the site about GFF. Catalyst and I wanted to be able to be temp mods here for games.

I join the game.

However, I will be gone for ARML on May 30-31

Yeah, first round won't start till June 1, cause of ARML

Sure, but you won't get any points if you're the highest  :lol:

Oh, just imagine the opportunities for collusion  :roll:

This makes me wonder what would happen if everyone picked negative numbers.... :P

For ease of score, are we doing integer submissions only?

Yeah. Interger's only since I dislike math.

I join!

The message is too small. Please GET OUT OF HERE! :rotfl:

You guys can send in your numbers!

I've only gotten 4...

I'll close this round in 15 hours.

Scores for round 1.

[b][color=green]xpmath: 50[/color][/b]
lotsofmath: 41
Wickedestjr: 20
Shaggy75: 10
pythag011: 7
Chenhsi: 4
[b][color=red]Perfect628: -999[/color][/b]

You guys are so cautious. Look at this game:

[url]http://forums.mtgsalvation.com/showthread.php?t=115947[/url]

It is now Round 2!

[b]Deadline is Monday June 2 at 10:AM EST.[/b]

I realize some of you have ARML, but the consequences of missing a few days mean nothing. One day is all you need to come back! So, I'll post a round each day.

This is an hour early, but I'm going to be busy in an hour.

Perfect628: [b][color=green]123,455,790[/color][/b]
BOGTRO: 999,999
Brut3Forc3: 1,337
pythag011: 107
lotsofmath: 90
Chenhsi: 54
xpmath: 50
Wickedestjr: 20
Shaggy75: 10
azjips:[b][color=red]-15,900,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000[/color][/b]

And with a sudden push, Perfect takes the lead!

[b]New Rule - You must write your answer out. No scientific notation. That way I can copy and paste it![/b]

[b]Deadline is Tuesday June 3 at 9:AM EST. [/b]

[quote="iLord"] No scientific notation. [/quote]

Darn. Hmm there's no "i" in my username though, though it seems like many people like to insert one there.  :lol:

Looks like I'm losing by a bit.

This kinda reminds me of that Nash thingy. Like we want to go as high as possible for highest profit, but so does the other person. If you're highest, then that's bad, so you go 1 lower. But the other person is thinking the same thing. So you drop your number 2 lower. And so on, so you should choose 0.  :P .

Soooo.... do we pm a next round's number now?

Nah, I have to find time to add these numbers together, or find someone to help me add them.

I'll do it tonight, if I have time.

[quote="iLord"]"Can't let you do that, mz94"

/obscure reference.[/quote]

Don't tell me...
[hide]Wolf from Star Fox 64?[/hide]

Yeah, I forgot about this.

I'll put it up in a sec.

ooooooooooooooooh i wanna join!!
that sounds like so much funn =D
gahhhhhhh im spazzing againnn

 :rotfl:  :rotfl:  :rotfl:

[quote=dragoon]/bump $      $[/quote]

You bumped this 13-year old game that hasn't had a post in it for 13 years...

[quote=dragoon]/bump $      $[/quote]

gg 13 year old bump

wonder what these guys are doing right now in real life

This was before I was born lol

[quote=dragoon]/bump $      $[/quote]

bumping the other dragoon


OMG, did u guys hear about Wade Steffey? that was so sad. we need something to revive this forum, so just like post ur thoughts or something...

[quote="indianamath"] so much for my Purdue scholarship... [/quote]

don't tell me u seriously though about going to [i]Purdue[/i]!

what is number one on your list indianamath, harvard? :rotfl:

yeah probably. actaully, it is ITT tech or LIncoln Tech. it's my dream to have the "education of the future"

lol... indianamath lol -_-... go to MIT or Princeton -_-, or maybe harvard -_-... or Caltec lol... w/e, just don't go to purdue!! Wade Steffey, u mean the guy who wandered around into the eletric place and got ownt by the wires?


[quote="Pirkuliyev Rovsen"]Let a,b,c>0. prove that $ \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {a}}{b \plus{} c} \plus{} \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {b}}{c \plus{} a} \plus{} \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {c}}{a \plus{} b} > \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {a \plus{} b \plus{} c}}$           !! :oops:[/quote]

It's easy by Am-Gm.
*Note
$ \sqrt{a\frac{(b\plus{}c)^2}{4}} \le \frac{a\plus{}b\plus{}c}{3}\sqrt{\frac{a\plus{}b\plus{}c}{3}}$

use $ \frac{1\plus{}\sqrt{x}}{1\minus{}x}\geq \frac{9}{4}x\plus{}\frac{3\plus{}2\sqrt{3}}{4}$

WLOG, since the inequality is homogeneous; assume $ a\plus{}b\plus{}c\equal{}1$; then it equivalents
$ \sum \frac{1\plus{}\sqrt{a}}{1\minus{}a}\geq \frac {9\plus{}3\sqrt{3}}{2}$.
Now We have, from CS that
$ \sum \frac 1{1\minus{}a}\geq \frac 92$; hence our inequality is equivalent to
\[ \sum\frac{\sqrt{a}}{1\minus{}a} \geq \frac{3\sqrt{3}}{2}\]
Now substitute $ x^2\equal{}a; y^2\equal{}b; z^2\equal{}c$; $ x^2\plus{}y^2\plus{}z^2\equal{}1$.
Then we only have to prove $ \sum \frac x{1\minus{}x^2}\geq \frac{3\sqrt{3}}{2}$.
This is an old and very well-known problem. :)


search
advanced fields
advanced fields unsolved

Prove that the set $ \{k\cdot\alpha\minus{}[k\cdot\alpha], k\in{N}\}$, $ \alpha$ an irrational number, is dense in $ [0,1]$


If $p^{\alpha}||a-1$ and $p^{\alpha+\beta}| a^n-1$ prove that $p^{\beta} | n$.

$p^x||a$ means $p^x|a$, and $p^{x+1}$ doesn't divide $a$.

Can somebody help, I realy need this as fast as possible.

thanx for every future replay

This is from a book, and there is something that I don't understand but maybe you will:

Let $a=1+p^\alpha b$, where $p\not |b$, and let $p^\gamma||n$. Then \[ a^n-1=(1+p^\alpha b)^n-1=np^\alpha b + \frac{n(n-1)}2p^{2\alpha}b^2+\cdots + p^{n\alpha}b^n.  \]
And now it says that every number in this sum except the first ($np^\alpha b$) is divisible with $p^{\alpha+\gamma +1}$ (why?? it doesn't have to be $p^\gamma | \binom{n}{k}$ for every $k$) and the first one is divisible only with $p^{\alpha+\gamma}$ and not with $p^{\alpha+\gamma +1}$ so $p^{\alpha+\gamma}||a^n-1$.

I hope this will help you...

well it need to be (acording to proof from your book) that $p^{\gamma}|{n \choose k}p^{k\alpha}$, but i had no succes in proving this. :blush:

Is this true:
$p^{\alpha}||a-1$

$p^{\beta}||n \Leftrightarrow p^{\alpha+\beta}||a^n-1$?

According to my book, that statement is true, because this theorem in that book is in IFF form.

Can somebody provide a proof of this theorem? :)

Suppose that $p^\gamma|n$ we want to prove that $p^{\gamma +\alpha +1}| \binom{n}{k} p^{k\alpha}$ for $k>1$.

First one simple lemma: $p^x\geq x+2$ for every natural $x$, and
$p>2$. This is trivial to prove by induction.

Let $\delta$ be such number that $p^\delta \leq k < p^{\delta +1}$.
If we suppose that $\delta\geq \gamma$ then $p^{k\alpha}\geq p^{p^\delta \alpha}\geq p^{p^\gamma \alpha} > p^{(\gamma +2)\alpha}\geq p^{\gamma + \alpha +1}$ so problem is done.

Let $\delta<\gamma$. See that
$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$. Let $p^a||k$ then $a\leq \delta <\gamma$ so $p^{\gamma -a}|\frac{n}{k}|\binom{n}{k}$. From
that we see that $p^{k\alpha}\cdot p^{\gamma -a}=p^{k\alpha + \gamma -a }|\binom{n}{k}p^{k\alpha}$. $k\geq p^a$ so $k\alpha+\gamma -a\geq p^a\alpha + \gamma -a$ and by lemma last one is greater or equal
with $\alpha + \gamma + 1$ so $p^{\alpha+\gamma +1}|\binom{n}{k}p^k\alpha$.

I get it :roll: . you proved $p^{\beta}||n \Longrightarrow p^{\alpha+\beta}||a^n-1$, but $\Longleftarrow$ is still unsolved

I proved that if $p^\gamma||n$ then $p^{\alpha+\gamma}||a^n-1$, and
these are both directions, $\Rightarrow$ and $\Leftarrow$. Because,
let's suppose that $p^{\alpha+\beta}||a^n-1$ and $p^\gamma||n$. I
proved that then $p^{\alpha+\gamma}||a^n-1$, but we suppose that
$p^{\alpha+\beta}||a^n-1$ so from that $p^\beta=p^\gamma||n$, get it? :)

I wonder , does this formulas holds for +, i mean $p^{\alpha}||a+1$ is this true

$p^{\beta}||n \Longleftrightarrow p^{\alpha+\beta}||a^n+1$?

Well, if $p^\alpha||a+1$ then $a=p^\alpha b-1$ where $p\not | b$. Then $a^n+1=(p^\alpha b -1)^n +1$ so if $n$ is odd, nothing will change, only in that sum you will have numbers with signs $+$ and $-$, but if $n$ is even then $a^n+1=p\cdot A +2$, so $p\not | a^n+1$ :)

read the problem statement ..it's very clearly mentioned

oh...so the biggest power of p wich divides something


let $m$ and $l$ be the natural numbers .Find the smallest $n$ so that $2^{l}|m^{n}-1$

n=0.
I think you want find n>0.
If m is even, and l>0 have not solution n>0.
If m is odd and $k=ord_2(m-1),$ then  $n=1 \ if \ l\le k$, else $n=2^r,$ were  $r=min(l-k,l-2)$.

I dont get the question. In terms of $l$ or what?


induction
number theory theorems
number theory

hello every body!!!
i would ask you for a discovery i did one week ago (maybe it's not the apropriate place to speak about that  :blush: ) : recurrence in Q+* and we can extend to Q but it's not so important , content us with Q+*
let's P be a proposition
if P(1) is true
 P(1/n)==>P(1/(n+1)) for all n £ N
 P(m/n)==>P((m+1)/n) for all (m,n) £ N^2
than the proposition is true for all r£Q+*
is this recurrence exist? my teacher sayed no !!if yes , ok ,if no what's it's value? can we use it to solve some problems?
please respond me ,I realy need your help!!

I don't remember any problem being solvable like this, but at least it's not impossible. It's similar to an induction in two variables, and there are indeed examples.
One other usefull induction on $ \mathbb Q^\plus{}$:
a) $ P(1)$ is true.
b) $ P(x) \implies P(x \plus{} 1)$
c) $ P(x) \implies P\left( \frac 1x \right)$

hi ZETAX
if
a) $ P(1)$ is true.
b) $ P(x) \implies P(x \plus{} 1)$
c) $ P(x) \implies P\left( \frac 1x \right)$
than P is true .yes?

Yes, then $ P(x)$ is true for all $ x \in \mathbb Q$.

thnx ZETAx that can reduce a big part of work in some prblems  :thumbup:

It solve proplem only for positive rationals $ r\in Q, r>0$ by continiosly fractions.

[quote="alain-connes"]please RUSt what does it mean continiosly fraction?[/quote]

I guess Rust meant simple continued fractions (since those are sufficient)

Any positive rational can be expressed uniquely in the form
\[ [a_1, a_2, a_3, ..., a_n] \equal{} a_1 \plus{} \frac {1}{a_2 \plus{} \frac {1}{a_3 \plus{} \frac {1}{\ddots \plus{} \frac {1}{a_n}}}}
\]

for $ a_2, ...$ natural numbers, $ a_1$ a nonnegative integer.


let $(x_n)$ such that :
 $x_{1}=7$&$x_{n+1} =2x_{n^2}-1$$\forall x \geq 1$)
 Prove that :
 forall n , we have x_{n} isn't divisitor 2003

There's something wrong there... Infinitely many $x_n$ aren't defined by the given equation.

[quote="TheBB"]I'm guessing $x_{n+1}=2x_n^2-1$ instead.[/quote]
if 2003|x(n) Lejander simvol  (2/2003)=1. Because (2/2003)=-1 2003 is not divisor x(n) for all n.


[quote="randomdragoon"]One day, I came across a photograph of a woman. I only vaguely know the woman, but we are not friends or anything like that. She is in a location totally unfamiliar to me. There was nothing written on the photograph. However, I immediately knew who the photographer was. How?[/quote]

is this by any chance the mona lisa we're talking about

it's a bit of a stretch

[hide="Loophole"]
Nothing had better been [i]printed[/i] on it...
[/hide]

[hide="Something else"]
It's on a magazine or something where the picture-taker is written... below it?
[/hide]

Guessing was never a talent of mine.

ROFL it has to be you be because you took the picture!!

Nightmare, then that would mean you'd at least met the person before. Was it green-screen where you were interviewed for a news channel, of some sort?

[quote="i_like_pie"]This probably isn't what you're looking for, but it's worth a try...

[hide="My theory"]The woman took a picture of herself, and this was evident because you could see her harm extending toward the camera. This would make her the photographer.[/hide][/quote]

Yes that is correct.

Man, that was fast.


you have a balance beam.
you are shown 27 marbles.

one of them is either heavier or lighter .
you have to weigh the marbles by putting them on both sides of the balance.

what is the least number of weighings(is that the right word) needed to find the faulty marble assuming you have the worst luck available
please explain each weighing.

[hide="Solution"]I am confused do you want the least possible guesses (which if you are lucky could be one but there is only a 1/702 chance of you getting it right). But if you want the greatest possible weighings it would be 26 because you would have 1 marble on one side of the scale which you would always use in every weighing and one of the other 26 marbles. [/hide]
Note I may be wrong because I didnt fully understand what the problem was asking.

[hide="Solution?"] Divide the $ 27$ marbles into $ 3$ groupings of $ 9$. Pick any two of these groupings and weigh them against each other. Since we are told that we have the worst luck, let's assume they are equal. (We could probably just as easily assume they were unequal, but let's continue with what we are doing). Then we pick one of the groupings we just used and weigh it against the one we didn't uise, so as to determine whether the odd one is lighter or heavier. WLOG assume it is heavier. Next take the grouping of $ 9$ which was unequal to the other two and divide it into $ 3$ groupings of $ 3$. Pick any two random groupings and weigh them against each other. If they are equal, use the one you did not weigh. If they are not equal, use the heavier grouping of $ 3$ (since we assumed the odd one out was heavier). Finally, take this last grouping of $ 3$ and pick $ 2$ of them and weigh them against each other. if they are equal, use the odd one out. If they are not equal use the heavier one. This gives us a total of $ \boxed{4}$ weighings.[/hide]

hi gaussintraining
i think you are right
but a friend said he could do it in 3
i dont think so though

thanks

Well, if you know beforehand that the odd marble out is either heavier or lighter, then you can eliminate the weighing to see whether it is heavier or lighter than the others. That would give $ 3$ weighings. But your question specifically said that you were unsure as to whether it was heavier or lighter, only that it was unequal. That might be it.


number theory
prime numbers
relatively prime

The definition of a pair of relative prime numbers is stated that two numbers that don't have any common factor greater than 1. My question is, is the number pair (1,1) stated as relatively prime?


calculus
integration
geometry
LaTeX
calculus computations

Hi everyone,

Can somone tell me if this is correct, please?

I'm trying to find the area under the curves x=ysquared-4y and x=2y-ysquared (-3,3) (0,0)

I took 2y-y^2-y^2-4y dy from -3 to 0.

I got 62/2

Thank you very much

well seriously speaking it's quite not good to see maths here without $ \text{\LaTeX}$ . just enclose 'dollar' sign between what u write
for
$ \int\frac{1}{x^{2}-1}\text{dx}$
write
 [code]$\int \frac{1}{x^{2}-1} \text{dx}$[/code]


$\sum \frac{a^2b^2}{c^2} \geq \sum (\frac{ab}{c})(\frac{bc}{a}) = \sum a^2$

from whence it is clear.

you are right, it is my fault, and I am wrong.


geometry
linear algebra
matrix
congruent triangles
area of a triangle
Heron\u0027s formula
number theory proposed

[quote="Andreas"]Prove that there is an infinite number of non-congruent triangles $T$ such that
     1. the lengths of the sides of $T$ are consecutive integers.
     2. the area of $T$ is an integer.[/quote]
(edited from yesterday's version, which was slightly inaccurate)
Heron's formula in this case will be
\[ 16A^2=3x(x-2)x(x+2)  \]
where $A$ is the area and $x$ is the length of the "middle" side. $x$ must be even, say $x=2y$. We will need
\[ 3(y^2-1)=z^2.  \]
for some $z$. Borrowing an idea from the theory of Pell's equation, if $(y,z)$ is one solution, then we can get another by applying the square matrix
\[ \left(\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right)  \]
I.e. if $(y,z)$ is a solution then so is $(2y+z,3y+2z)$. We can start the sequence of solutions with a 3-4-5 triangle: $x=4$ so $(y,z)=(2,3)$.


geometry
3D geometry
pyramid
geometry proposed

Consider $ (OA,(OB,(OC$ in space such that: $ \angle AOB\equal{} \angle BOC \equal{}\angle COA\equal{}60^{\circ}$. If $ AO\equal{}6$, then find the length of $ (AO')$, where $ O'$ is the feet of the altitude from $ A$ to $ (OBC)$.

Take B' and C' on (OB, (OC respectively, so that OB' = OC' = OA = 6. OAB'C' is a regular pyramid and it's altitude will be 2 sqrt(6).

Best regards,
sunken rock


calculus
integration
parameterization
function
limit
calculus computations

You can do this one in your head if you use asymptotics.

For large $ n$ the integrand is asymptotic to:

$ \sqrt{(n+1)^{2}x^{2n}}=(n+1)x^{n}$ and $ \int_{0}^{1}(n+1)x^{n}\hspace{1 mm}dx=1$

That asymptotic estimate is false, and that's the wrong answer. The integrand converges pointwise to 1 at every point of the integral except 1.

I think it is incumbent on you to explain why the asymptotic estimate is false.  For what its worth, I just verified that Mathematica suggests the same answer for the problem as I provided above (possibly for different reasons).

By the way, pointwise convergence does not make any sense to me here since the integrand is with respect to $ dx$ and not $ dn$.  I think you mean that if I substitute $ x=1$ in the integral as if it were a parameter, then it would diverge, whereas substitution of any other value of $ x$ with $ 0 \leq x <1$, would converge.  But you just can't do that as if it were a parameter.  That rationale just says the integrand blows up on a set of measure zero.  So what.

This may come as a shock to you, but the answer is $ 2$. See this: http://www.mathlinks.ro/Forum/viewtopic.php?t=84846
It's actually a variation, but the idea is the same; we want to find a function that resembles $ \sqrt{1+x^{2}}$ (in your words, an "asymptotic"), such that the difference of the two is [i]well-behaved[/i] wrt taking that limit.
The link I gave provides a description of that vague "well-behaved" and the correct function is easy to spot.

About your comments on jmerry's post: What do you mean by [b]so what[/b]? That was exactly his point. We have pointwise convergence (wrt taking $ n \to \infty$) and only on $ [0,1)$. If we had (for example) uniform convergence, then by a well-known theorem we could have changed the limit with the integral. However, we can't even talk here about (bounded) convergence on $ [0,1]$, so it really matters how [b]bad[/b] it blows up near $ 1$.

$ \text{Touch\'e(e) Perfect-Radio}$   :blush:

I don't get how
$ | f(x) | = \left| \frac1{\sqrt{1+x^{2}}+x}-1 \right| \leq 1$. Can you explain?

@edoo: No need to blush, everybody on the forum has had his/her fair share of mistakes.

@mdk: $ f$ is a decreasing function and $ f(0) = 0, \, \lim_{x \to \infty}f(x) =-1$.

okay, but how do you get that expression?

Another estimate that works: $ 0\le(1+u)-\sqrt{1+u^{2}}<1$ for $ 0\le u$. Now $ \int_{0}^{1}1+(n+1)x^{n}\,dx=2$, so $ \int_{0}^{1}\sqrt{1+(n+1)^{2}x^{2n}}\le 2$. As $ n\to\infty$, the difference $ 1+(n+1)x^{n}-\sqrt{1+(n+1)^{2}x^{2n}}$ goes to zero pointwise almost everywhere (everywhere except $ x=1$), and it is bounded by 1. By the Dominated Convergence Theorem, its integral goes to zero.


geometry inradius circumcircle incenter trigonometry calculus

a,b,c - sides of traingle, r - inradius. We have:
$4r^2 = \frac{abc}{a+b+c}$
Prove that a=b=c.

It's obvious that 2r = R, but what can we do with that fact ? any ideas?

The relation between the inradius $r$, the circumradius $R$ and the distance $d = IO$ between the incenter $I$ and the circumcenter $O$ is

$\frac{1}{R - d} + \frac{1}{R + d} = \frac 1 r$

Only if 2 circles satisfy this relation, they can be the incircle and the circumcircle of some triangle, in fact, of an infinite number of triangles (browse for Poncelet's porism). Hence,

$2Rr + rd - rd = 2Rr = R^2 - d^2$

and if $R = 2r$, then

$4r^2 = 4r^2 - d^2$
$d = 0$

Then the incenter $I$ and the circumcenter $O$ are identical, $I \equiv O$, which is true only for an equilateral triangle.

Same, if you see the well know formulae

Area = $\frac {abc}{4R} =  r ((a+b+c)/2)$

Also, obviously,   if   two centers are at the same point ,   sin (A/2) =  1/2  whcih gives any angle = 60 degrees)

[quote="Gyan"]Same, if you see the well know formulae

Area = $\frac {abc}{4R} =  r ((a+b+c)/2)$

Also, obviously, if two centers are at the same point, sin (A/2) = 1/2 which gives any angle = 60 degrees).
[/quote]

I do not follow your reasoning. Most likely, Big Bolzano concluded from this formula and from the given formula that $R = 2r$ and you did not show that the two centers are at the same point.

Substituting $R = 2r$, $a = 2R \sin\alpha$, $b = 2R \sin\beta$, $c = 2R \sin\gamma = 2R \sin{(\alpha + \beta)}$,

$\frac {8R^3 \sin\alpha \sin\beta \sin{(\alpha + \beta)}}{4R} = \frac{2R[\sin\alpha + \sin\beta + \sin{(\alpha + \beta)}]}{4}$

$4 \sin\alpha \sin\beta \sin{(\alpha + \beta) = \sin\alpha + \sin\beta + \sin{(\alpha + \beta)}}$

${8 \sin\alpha \sin\beta \sin{\frac{\alpha + \beta}{2}} \cos\frac{\alpha + \beta}{2}} = 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{(\alpha - \beta)}{2}} + 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{(\alpha + \beta)}{2}}$

Since $0 < \frac{\alpha + \beta}{2} < 90^o$,

${4 \sin\alpha \sin\beta \cos\frac{\alpha + \beta}{2}} = \cos{\frac{(\alpha - \beta)}{2}} + \cos{\frac{(\alpha + \beta)}{2}}$

${16 \sin{\frac \alpha 2} \cos{\frac \alpha 2} \sin{\frac \beta 2} \cos{\frac \beta 2}\cos\frac{\alpha + \beta}{2}} = 2\cos{\frac \alpha 2} \cos{\frac \beta 2}$

Since $0 < \frac \alpha 2, \frac \beta 2 < 90^o$,

${{8 \sin{\frac \alpha 2} \sin{\frac \beta 2} \cos\frac{\alpha + \beta}{2}} = 8 \sin{\frac \alpha 2} \sin{\frac \beta 2} \sin\frac \gamma 2} = 1$

Now, you would have to show that

$8 \sin{\frac \alpha 2} \sin{\frac \beta 2} \sin{\frac \gamma 2} \le 1$

with equality only for an equilateral triangle, which I would do with Lagrange multiplier $\lambda$. On the other hand, the relation

$\frac{1}{R - d} + \frac{1}{R + d} = \frac 1 r$

can be proved by an inversion, with no calculus involved.

Yetti -

Sorry for being so terse.  I was just adding to yours (using what you have shown .. that's why I had  word  "same" )

  Use  the  Following standard formulae (Fairly popular) :
1.              Area =  abc/4R  =   r*s    to get  R=2r
2.               Distance between Incenter and Outcenter =  sqrt(R^2-2r*R)  to  get this distance as  zero.  -
                  (-That is,  they are at the  same point)  (No I did not explicitly shown that but assumed that , as it is a fairly obvious conclusion from a well-known formula -  and is posted in your post...)
3.               Use the right triangle  O (in-or outcenter) A (Corner) X (where in-center touches line AB)  and see the angle OAX = 30 degrees.


[quote="GameBot"]What is the arithmetic mean of the integers from -4 through 5, inclusive? Express your answer as a decimal to the nearest tenth.[/quote]

The sum is obviously $ 5$. We divide by $ 10$, which is the number of integers.

So, $ .5$

If we add the numbers up, -1 to -14 cancel out with 1 to 4, so the sum is 5. Then we divide by 10 (since there are 10 numbers), so the mean is 0.5.


calculus
integration
calculus computations

Calculate with at most 10% relative error

$\int_{-\infty}^{+\infty}\frac{1}{(x^4+4x+4)^{100}}dx$

Here's a first stab at it, noting that $x^4+4x+4$ has a minumum at $x=-1$ and using the substitution $u=x+1$:

$\int_{-\infty}^{\infty}\frac{dx}{(x^4+4x+4)^{100}} = \int_{-\infty}^{\infty}\frac{du}{(1+6u^2-4u^3+u^4)^{100}}$

$\approx\int_{-\infty}^{\infty}\frac{du}{(1+6u^2)^{100}} \approx \int_{-\infty}^{\infty}e^{-600u^2}\,du = \sqrt{\frac{\pi}{600}}\approx 0.072$

I'd have to work harder to prove that that is within 10%, but it shouldn't be too far off.


geometry
trapezoid
inequalities
ratio
parallelogram
triangle inequality
geometry proposed

In triangle $ABC$ let $F$ and $E$ be the intersections of the angles bisectors starting from $B$ and $C$ with $AC$ and $AB$. Consider the parallel lines to $EF$ though $B$ and $C$ which intersect lines $AC$ and $AB$ at $M$ and $N$. Show that in trapezoid $BNCM$ we have the following inequality

$BC \geq \frac{BM + CN}{2}$

Label the triangle sides $a = BC, b = CA, c = AB$. If the triangle $\triangle ABC$ is isosceles with $AB = AC$, then (by continuity) the angle bisectors $BE = CF$ are equal and consequently, the lines $EF \parallel BC$ are parallel. But then $BM = CN = 0$, $BC > 0$ and the inequality is trivial.

WLOG, assume that $c = AB > AC = b$. then the point $M$ lies inside of the segment $AB$ and the point $N$ lies on the extended ray $AC$ outside of the segment $AC$. The angle bisectors $BE, CF$ cut the opposite sides $c = AB, b = CA$ in the ratios of the adjacent sides $c, a$ and $a, b$, respectively.

$AE = \frac{bc}{c + a}$

$AF = \frac{bc}{a + b}$

Since the lines $EF \parallel CM \parallel BN$ are parallel,

$AM = AF \cdot \frac{AC}{AE} = \frac{bc}{a + b} \cdot \frac{b(c + a)}{bc} = b\ \frac{c + a}{a + b}$

$AN = AE \cdot \frac{AB}{AF} = \frac{bc}{c + a} \cdot \frac{c(a + b)}{bc} = c\ \frac{a + b}{c + a}$

Because we assumed $AB > AC$,

$BM = AB - AM = c - b\ \frac{c + a}{a + b} = \frac{a(c - b)}{a + b}$

$CN = AN - AC = c\ \frac{a + b}{c + a} - b = \frac{a(c - b)}{c + a}$

$\frac{BM + CN}{2} = \frac{a(c - b)}{2} \left(\frac{1}{a + b} + \frac{1}{c + a}\right)$

We have to show that

$(?)\ \ \ \frac{a(c - b)}{2} \left(\frac{1}{a + b} + \frac{1}{c + a}\right) < a$

$(?)\ \ \ (c - b)(2a + b + c) < 2(a + b)(c + a)$

$(?)\ \ \ 2ca + bc + c^2 - 2ab - b^2 - bc < 2ca + 2a^2 + 2bc + 2ab$

$(?)\ \ \ c^2 - 2bc - b^2 < 2a^2 + 4ab$

$(?)\ \ \ c^2 - 2bc + b^2 < 2(a^2 + 2ab + b^2)$

$(?)\ \ \ (c - b)^2 < 2(a + b)^2$

By assumption, $c - b > 0$ and we can take a square root of the inequality in question:

$(?)\ \ \ c - b < a \sqrt 2 + b \sqrt 2$

$(?)\ \ \ c < a \sqrt 2 + b(1 + \sqrt 2)$

The last inequality is certainly correct - due to the triangle inequality,

$c < a + b < a \sqrt 2 + b(1 + \sqrt 2)$

with a plenty of room to spare.

[quote="darij grinberg"]Manlio, have you noted that Yetti proved a completely different problem? And, I must say, a much easier one. (His points M and N are your points N and M, respectively.)

Sorry Yetti, just wanted to have things put right...

  Darij[/quote]
No problem, I appreciate that you pointed it out. I was fooled by the unexpected labeling of the angle bisectors as $BF, CE$, normally, I would expect $BE, CF$. In the figure of the 1st reply, the labels of the points $E \longleftrightarrow F$ and $M \longleftrightarrow N$ have to be exchanged and we have to show that in the trapezoid $BMCN$, the mean $\frac{BM + CN}{2} \le BC$ of the parallel bases (rather than the mean of the shoulder sides) is less than or equal to its diagonal $BC$.

If the triangle $\triangle ABC$ is isosceles, the lines $EF \parallel BC$ are parallel. Thus the points $M \equiv C$ and $B \equiv N$ are identical and the inequality becomes a trivial equality. WLOG, assume that $c = AB > b = CA$.

$AE = \frac{bc}{a + b}$

$AF = \frac{bc}{c + a}$

Since by assumption $c > b$, $AF < AE$. From similarity of the triangles $\triangle AEF \sim \triangle ANC \sim \triangle ABM$, we then deduce that $CM < BN$. It follows that $MN < BC$ is the shorter diagonal of the trapezoid $BMCN$. Hence, it is sufficient to show that

$\frac{BM + CN}{2} < \frac{BC + MN}{2} < BC$

Thus we have to prove the following lemma: In any trapezoid, the sum of the diagonals is greater than the sum of the parallel bases.

Let $ABCD$ be an arbitrary trapezoid with the bases $AB \parallel CD$. WLOG, assume $AB \ge CD$ and $BC \ge DA$. If $AB = CD$, the trapezoid is a parallelogram and $BC = DA$, a case not acceptable in the original problem, because the sides $BN \equiv AB$, $CM \equiv CA$ of the original triangle $\triangle ABC$ cannot be parallel, but the lemma holds anyway. Let $E \in BC, F \in DA$ be the midpoints of the shoulder sides $BC, DA$ and $EF \parallel AB \parallel CD$ the midline parallel to the bases. Since $AB \ge CD$, the diagonals $AC, BD$ intersect at a point $X$ on (for a parallelogram) or above the midline (i.e., on the same side of the midline as the shorter base $CD$). Let the diagonals $AC, BD$ intersect the midline $EF$ at points $Y, Z$. The points $E, Z,  Y, F$ then follow on the midline $EF$ in this order. Drop normals from the vertices $A, B$ to the midline $EF$ and let $G, H \in EF$ be the feet of these normals. From the right angle triangles $\triangle AYG, \triangle BZH$ with hypotenuses $AY = \frac{AC}{2}$, $BZ = \frac{BD}{2}$, we have

$\frac{AC + BD}{2} = AY + BZ > GY + HZ$

Clearly, $\measuredangle ABC \le \frac \pi 2$ (because by assumptions $AB \ge CD$ and $BC \ge DA$). If $\measuredangle DAB \le \frac \pi 2$, then

$\frac{AB + CD}{2} = EF = GY - GF + YZ + HZ - HE$

and we have to show that $GF + HE = YZ$, because then $GY + HZ = EF$ and we are done. If $\measuredangle DAB > \frac \pi 2$, then

$\frac{AB + CD}{2} = EF = GY + GF + YZ + HZ - HE$

and we have to show that $HE - GF = YZ$, because then again $GY + HZ = EF$ and we are done.

Consider the first case: Drop normals from the points $Y, Z$ to the longer base $AB$ and let $U, V \in AB$ be the feet of this normals. Let $M$ be the midpoint of the shorter base $CD$. drop a normal from this midpoint to the midline $EF$ and let $W \in EF$ be the foot of this normal. Draw lines $y \parallel DA, z \parallel BC$ through the points $Y, Z$ parallel to the shoulder sides $DA, BC$. Because $Y, Z$ are the midpoints of the diagonals $AC, BD$, the lines $y, z$ meet at the midpoint $M$ of the base $CD$. Let $P, Q$ be the intersections of the lines $y, z$ with the base $AB$. From the congruence of the triangles $\triangle AFG \cong \triangle YPU \cong MYW$, $\triangle BEH \cong \triangle ZQV \cong MZW$, we get $GF = PU = YW$, $HE = QV = ZW$. Consequently, $GF + HE = YW + ZW = YZ$, which is what was necessary to prove. The second case $\measuredangle DAB > \frac \pi 2$ is handled in an entirely similar way as the first one $\measuredangle DAB \le \frac \pi 2$.

[quote="yetti"]
Thus we have to prove the following lemma: In any trapezoid, the sum of the diagonals is greater than the sum of the parallel bases.
[/quote]
You can also prove this with the triangular inequality: let $P=AC\cap BD$. Then $AP+BP\geq AB$ and $CP+DP\geq CD$ therefor $AC+BD = AP+CP+BP+DP\geq AB+CD$ and we are done..

btw, today is my birthday!! happy birthday to me!!  :D


A liquid A is reacted with hot aqueous sodium carbonate solution. A mixture of 2 salts B and C are produced. The mixture on acidification with sulphuric acid and distillation produces the liquid A again. Identify A, B, C and write the equations involved.

:huuh:

Excuse me, but isn't this to vague?

[quote="Da Vinci"]this is the only data given[/quote]

A typical problem from an Indian book.

Also, why would A be an acid anhydride?


inequalities
calculus
number theory
Diophantine equation

How would you even get started on this one?  I don't have any experience solving exponential inequalities.  I think I should be spending more time with my Diophantine equation book, which covers equations like these....  :(

I got it but I don't see a non calculus approach. I used a limit and the EVT to show that the minimum was one and that this was greater than that. I'll post my solution tomorrow since it's late right now. Any non-calculus solutions?

This one posted before and it was a solution with extended Bernoulli ineq. But i cannot find it so I will post the solution posted here in a maybe different way.

We know that $0<x,y<1$ so we can we can write $x =\frac{1}{1+s}$  and

$y=\frac{1}{1+t}$

with $s>0$ and $t>0$.

We know from Bernoulli extended that  For $0<y<1$ and for any $s >0$ we have

$(1+s)^y < 1+sy$

Now

$x^y =(\frac{1}{1+s})^y > \frac{1}{1+sy} =\frac{1+t}{1+s+t}$

$y^x > \frac{1+s}{1+s+t}$

$x^y + y^x > \frac{1+s}{1+s+t} +\frac{1+t}{1+s+t} = 1 + \frac{1}{1+s+t} >1$

and we are done

[quote="silouan"]This one posted before and it was a solution with extended Bernoulli ineq. But i cannot find it so I will post the solution posted here in a maybe different way.

[/quote]

You can find it here :) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=15062
http://www.mathlinks.ro/Forum/viewtopic.php?t=484


Here's an easy, yet nice problem that I thought of.  In the following figure, determine the ratio of the area of the large circle C to the small circle c$\rightarrow C: c$.  $AB$ is the diameter of the large circle. $\angle A,B=45^{\circ}$.
[img]http://www.boxedexe.tamranda.com/images/circle.JPG[/img]
Masoud Zargar

Answer: 3 + 2 root 2
Solution:
Let R be the radius of a bigger circle and r be the radius of a smaller circle.
Then, AB = 2R, AX = root 2 R, and r = (root 2 - 1)R.
Hence, (ratio of areas of circle) = (pi)(R^2) over (pi)(r^2) = (pi)(R^2) over (pi)((root 2 -1)R)^2 = 1 over (3 - 2 root2) = 3 + 2 root 2.

Say $r=2$, therefore $C=4\pi$, now using $45,45,90$ triangles we see that the semiperimeter of the triangle is $2\sqrt{2}+2$ and the area is $4$, now use $rs=A$ and we see that the area of $c=12-8\sqrt{2}\pi$. So the ratio is $\boxed{1: 3-2\sqrt{2}}$.

well, if u call the big diameter 2r, and the solve for teh other sides of the triangle, then find teh triangles area, divide by its semiperimeter, you have the inradius. just square teh inradius and multiply by pi, and then put that under r^2/pi

Here's what I have. Let $AX=BX=1$; $r$ be the radius of $c$, and $R$ the radius of $C$.  We know that $A_{\triangle ABX}=rs=1/2$, and therefore, $r=\frac 12/s=\frac{1}{a+b+c}$.  It is obvious that $a=1,b=1$, and $c$ or $AB=\sqrt{2}$.  As a result, $r=\frac{1}{a+b+c}=\frac{1}{2+\sqrt{2}}$.  $R=\frac{\sqrt{2}}{2}$. Hence,
\[ c=\pi r^2\\ =\pi\left(\frac{1}{2+\sqrt{2}}\right)^2\\ =\pi\left(\frac{1}{4\sqrt{2}+6}\right)  \]
\[ C=\pi\left(\frac{\sqrt{2}}{2}\right)^2\\ =\pi\frac 12  \]
$C: c=\frac{\pi\frac 12}{\pi\left(\frac{1}{4\sqrt{2}+6}\right)}\\ =\frac{\frac 12}{\frac{1}{4\sqrt{2}+6}}\\ =\frac{4\sqrt{2}+6}{2}=2\sqrt{2}+3: 1$

:)

I have another solution that doesn't deal with $A=rs$.

Masoud Zargar


$ p \ge 5$
given that $ q \equal{} p^2 \minus{} p$
then, can $ q^2 \plus{} q$ be divisible by anyone of $ 30$,$ 12$ or $ 17$?

please give a solution if this is divisible by anyone of them...

thank you... :)

for $ p\equal{}12\times 30\times 17$ we have $ 17,12,30|q^2\plus{}q$

[quote="beparoa"]this is a special case.
i want to know whether it is always divisible by 12, 14, and 30?[/quote]
Is that 14 or 17?

Also, $ q^2\plus{}q\equal{}p(p\minus{}1)(p(p\minus{}1)\plus{}1)$.

It is not always divisible by 12.  If $ p$ is of the form $ 4k \plus{} 2$ or $ 4k \plus{} 3$, then the first two factors will only be divisible by one power of 2.  The third factor is always odd (as $ p(p \minus{} 1)$ is always even - one of the two factors must be even), hence it will never be divisible by a power of 2.

As for 14, since 2 must divide the expression we only have to worry about 7.  Now if $ p \equal{} 7k \plus{} 2$ then the product is not divisible by 7.  The same logic applies to 17 - if $ p \equal{} 17k \plus{} 2$ then the product is not divisible by 17.

We can pull the same trick for 5 - if $ p \equal{} 5k \plus{} 2$ then the expression is not divisible by 5.  As for 3, the expression is always divisible by 3.  If $ p \equal{} 3k$, then we're done.  If $ p \equal{} 3k \plus{} 1$, then $ p \minus{} 1 \equal{} 3k$ and we're done.  If $ p \equal{} 3k \plus{} 2$, then $ p(p \minus{} 1) \plus{} 1 \equal{} (3k \plus{} 2)(3k \plus{} 1) \plus{} 1 \equal{} 9k^2 \plus{} 9k \plus{} 3$ which is obviously always divisible by 3.

Hence we can guarantee that 6 always divides the expression.

More general problem:

Let $ P_i \in \mathbb{Z}[x] \; (1 \le i \le n)$ be distinct monic polynomials of degree at least one.  Is it always the case that $ n! \mid \prod_{i \equal{} 1}^n P(x)$ for at least one integer $ x$?  What if we restrict $ P_i$ to degree exactly one?

[hide="answer"]No.  For example, let $ P_i \equal{} x^2 \minus{} x \plus{} 2\cdot 2^i \plus{} 1$.  Then it is fairly easy to see that $ 2 \nmid P(x)$ for all $ i \ge 1$, $ x \in \mathbb{Z}$.  As for the second case, let $ P_i \equal{} x \minus{} (1 \plus{} 2i)$.  Then obviously $ 2 \nmid P(x)$ for even integers $ x$.[/hide]


number theory
greatest common divisor
number theory proposed

Assume that $ a,b,c>0$, $ \gcd(a,b)\equal{}\gcd(b,c)\equal{}\gcd(c,a)\equal{}1$, $ a\plus{}b|b^{2}\plus{}c^{2}$, $ b\plus{}c|c^{2}\plus{}a^{2}$, and $ c\plus{}a|a^{2}\plus{}b^{2}$.
Prove that $ a\equal{}b\equal{}c\equal{}1$.

Denote $ a\plus{}b\equal{}z$; $ b\plus{}c\equal{}x$;$ c\plus{}a\equal{}y$ where $ x,y,z$ are three integers and are also three sides of a triangle,.Because
$ \gcd(a,b)\equal{}\gcd(b,c)\equal{}\gcd(c,a)\equal{}1$ hence $ \gcd(x,y,z)\equal{}1$ or $ 2$
We have $ (a\plus{}b) | (b^{2}\plus{}c^{2})$ or $ 2(a\plus{}b) | ((b\plus{}c)^{2}\plus{}(c\minus{}b)^{2}$ or $ 2y | (x^{2}\plus{}(y\minus{}z)^{2})$
hence $ 2y | (x^{2}\plus{}y^{2}\plus{}z^{2})$
Similary $ 2x |(x^{2}\plus{}y^{2}\plus{}z^{2})$
             $ 2z | (x^{2}\plus{}y^{2}\plus{}z^{2})$
if $ gcd (x,y,z) \equal{}1$ then three above result is eqivalent to $ (x^{2}\plus{}y^{2}\plus{}z^{2})$ is divisible by $ xyz$
if $ gcd(x,y,z) \equal{} 2$ ,let $ x\equal{}2x_{1}$; $ y\equal{}2y_{1}$;$ z\equal{}2z_{1}$ ,then $ (x_{1}^{2}\plus{}y_{1}^{2}\plus{}z_{1}^{2})$ is divisible by$ x_{1}y_{1}z_{1}$
We need to find $ a,b,c$ are positive integers ,and are three sides of a triangle such that $ (a^{2}\plus{}b^{2}\plus{}c^{2})$ is divisible
by $ abc$ .
Suppose that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}k abc$ with $ k\in Z_{\plus{}}$
Because $ a,b,c$ are three sides of a triangle then $ a^{2}\plus{}b^{2}\plus{}c^{2}< 2(ab\plus{}bc\plus{}ca)$
or $ kabc < 2(ab\plus{}bc\plus{}ca)$ or $ \frac{k}{2}<\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\leq 3$
hence $ 1\leq k\leq 5$
Suppose that $ a\leq b\leq c$   then $ \frac{k}{2}<\frac{3}{a}$
Three cases $ k\equal{}5,4,3$ is simple
Cases $ k\equal{}2,1$ it's longer but straight forward.

[quote="Nbach"]Denote $ a\plus{}b \equal{} z$; $ b\plus{}c \equal{} x$;$ c\plus{}a \equal{} y$ where $ x,y,z$ are three integers and are also three sides of a triangle,.Because
$ \gcd(a,b) \equal{}\gcd(b,c) \equal{}\gcd(c,a) \equal{} 1$ hence $ \gcd(x,y,z) \equal{} 1$ or $ 2$
We have $ (a\plus{}b) | (b^{2}\plus{}c^{2})$ or $ 2(a\plus{}b) | ((b\plus{}c)^{2}\plus{}(c\minus{}b)^{2}$ or $ 2y | (x^{2}\plus{}(y\minus{}z)^{2})$
hence $ 2y | (x^{2}\plus{}y^{2}\plus{}z^{2})$
Similary $ 2x |(x^{2}\plus{}y^{2}\plus{}z^{2})$
             $ 2z | (x^{2}\plus{}y^{2}\plus{}z^{2})$
if $ gcd (x,y,z) \equal{} 1$ then three above result is eqivalent to $ (x^{2}\plus{}y^{2}\plus{}z^{2})$ is divisible by $ xyz$
if $ gcd(x,y,z) \equal{} 2$ ,let $ x \equal{} 2x_{1}$; $ y \equal{} 2y_{1}$;$ z \equal{} 2z_{1}$ ,then $ (x_{1}^{2}\plus{}y_{1}^{2}\plus{}z_{1}^{2})$ is divisible by$ x_{1}y_{1}z_{1}$
We need to find $ a,b,c$ are positive integers ,and are three sides of a triangle such that $ (a^{2}\plus{}b^{2}\plus{}c^{2})$ is divisible
by $ abc$ .
Suppose that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} k abc$ with $ k\in Z_{\plus{}}$
Because $ a,b,c$ are three sides of a triangle then $ a^{2}\plus{}b^{2}\plus{}c^{2}< 2(ab\plus{}bc\plus{}ca)$
or $ kabc < 2(ab\plus{}bc\plus{}ca)$ or $ \frac{k}{2}<\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\leq 3$
hence $ 1\leq k\leq 5$
Suppose that $ a\leq b\leq c$   then $ \frac{k}{2}<\frac{3}{a}$
Three cases $ k \equal{} 5,4,3$ is simple
Cases $ k \equal{} 2,1$ it's longer but straight forward.[/quote]
Very nice Nbach :)


geometry
conics
ellipse
ratio
projective geometry

How can you find the area of an ellipse? From 2003 3-D Geometry stretch.

I wikipedia-ed it, but apparently, wikipedia doesn't have it T_T

Thank you! :)

In an ellipse, you have a semi major axis $ a$, and a semi minor axis $ b$. The area of an ellipse is equivalent to $ \pi ab$ (Note that if $ a\equal{}b$, this turns into $ \pi r^2$  :P ).

is there any way to prove this using Euclidean geometry?
I know for area of circle you can get a rough proof by cutting the circle up into very small sectors, and connecting them to make a rough rectangle...
Could you do something like that for an ellipse?

hello, here you will find the answer
http://math.ucsd.edu/~wgarner/math20b/area_ellipse.htm
Sonnhard.

I don't think you can prove it using purely Euclidean geometry. I think you need to use projective geometry to prove it if you don't want to use calculus. The proof is mentioned in AoPS Volume 2.

[url=http://mathforum.org/library/drmath/view/54979.html]It's a bit long...[/url]

Use a stereographic projection that maps the ellipse to a circle, then use area ratios.

Suppose we have semimajor axis $ a$ and semiminor axis $ b$ and perform a projection that maps the ellipse to a circle of radius $ b$. We have that $ \frac{\text{[ellipse]}}{\pi b^2}\equal{}\frac{a}{b}\implies\text{[ellipse]}\equal{}\boxed{\pi ab}$.

This ventures outside of strictly Euclidean geometry, though.

[quote="grn_trtle"] $ \frac {\text{[ellipse]}}{\pi b^2} \equal{} \frac {a}{b}\implies\text{[ellipse]} \equal{} \boxed{\pi ab}$.[/quote]

Hello , could you explain geometrically why this area ratio ?? I relate this ratio with parallel projection but not a steriographic projection

[quote="Tiger100"][quote="grn_trtle"] $ \frac {\text{[ellipse]}}{\pi b^2} \equal{} \frac {a}{b}\implies\text{[ellipse]} \equal{} \boxed{\pi ab}$.[/quote]

Hello , could you explain geometrically why this area ratio ??[/quote]

It's not particular to that proportion, is it?

What about $ \frac{[ellipse]}{\pi a^2} \equal{} \frac ba$

leads to $ [ellipse] \equal{} \pi ab?$

No, not particular to one proportion. You can scale/stretch by any factor and the ratio should work out.

Look up affine transformations and projections and such. AoPS vol 2 provides a little bit of detail (and the ellipse area proof).

(And sorry, not stereographic projection, that's a different kind of transformation. I tend to get the names mixed up)


Find all natural pairs of $a$ and $b$ such that $2a^b-b^a=1$.

$a^b=\frac{1+b^a}{2}$ so that means that $b \ odd$
and $a \ even$
so $a=2m,b=2n+1$
so we have further $2^{2n+1}m^{2n+1}=\frac{1+(2n+1)^{2m}}{2}$
now $4^{n+1}\cdot m^{2n+1}=1+(2n+1)^{2m}$
so $4^{n+1}\cdot m^{2n+1}=2+\sum_{k=0}^{2m-1} C_{2m}^{k} (2n)^{2m-1-k}$
$2^{2n+1}\cdot m^{2n+1}=1+\sum_{k=0}^{2m-1} C_{2m}^{k} (2n)^{2m-2-k}$

so LHS is multiple of 2 but RHS is not.
so there are no solutions
i am probably wrong because if it was that simple someone would have
answered already
i'll post anyway...hoping

sorry,i will stop posting stupid stuff.i promise this is my last nonsense "proof".
u'r right a is not necessarily even


I dont know if this is the correct place to post this message but: Can anyone post the imo 2003 shortlist or send it to me please?
  My email is: gausseuler2001@yahoo.com

If anyone has the full Shortlist, could someone also post it to me? My e-mail address is vinoth_90_2004@hotmail.com.
Thanks.

if anyone have it, can you please also send imo shortlist 2003 to mathsworld2001@yahoo.com.hk
thx!

Orlando has kindly posted the IMO Shortlist 2003 problems on MathLinks:

  http://www.mathlinks.ro/Forum/viewtopic.php?p=111045

  Darij


algorithm
induction
algebra unsolved
algebra

Mr. Fat and Ms. Taf play a game. Mr. Fat chooses a sequence of positive integers $ k_1, k_2, \ldots , k_n$. Ms. Taf must guess this sequence of integers. She is allowed to give Mr. Fat a red card and a blue card, each with an integer written on it. Mr. Fat replaces the number on the red card with $ k_1$ times the number on the red card plus the number on the blue card, and replaces the number on the blue card with the number originally on the red card. He repeats this process with number $ k_2$. (That is, he replaces the number on the red card with $ k_2$ times the number now on the red card plus the number now on the blue card, and replaces the number on the blue card with the number that was just placed on the red card.) He then repeats this process with each of the numbers $ k_3, \ldots k_n$, in this order. After has has gone through the sequence of integers, Mr. Fat then gives the cards back to Ms. Taf. How many times must Ms. Taf submit the red and blue cards in order to be able to determine the sequence of integers $ k_1, k_2, \ldots k_n$?

Ms. Taf only needs to submit the cards once.
Let $ r_i$, $ b_i$ denote the numbers on the red and blue cards at stage i (after Mr. Fat has just carried out the rewriting process with number $ k_i$), where $ r_0$, $ b_0$ are the initial numbers.
Ms. Taf can choose $ r_0 > b_0$.
The given algorithm translates to ($ 0 \leq i < n$)
$ r_{i \plus{} 1} \equal{} k_{i \plus{} 1}r_i \plus{} b_i$
$ b_{i \plus{} 1} \equal{} r_i$
It can be proved by induction that $ r_i > b_i$.
Now show by backwards induction that Ms. Taf can determine $ r_i$, $ b_i$ for $ 0 \leq i \leq n$. We further show if she knows $ r_i,b_i$ then she can find $ k_i$. She knows $ r_n,b_n$. Suppose she knows $ r_i,b_i$. By the division algorithm there is a unique integer k, and $ r < b_i$ such that
$ r_i \equal{} kb_i \plus{} r \equal{} kr_{i \minus{} 1} \plus{} r$
But $ r_i \equal{} k_ir_{i \minus{} 1} \plus{} b_{i \minus{} 1}$ with integer $ k_i$ and $ b_{i\minus{}1} < r_{i\minus{}1}$. Thus $ k \equal{} k_i$, and Ms. Taf can find $ k_i$ if she can divide. She can find $ r_{i \minus{} 1} \equal{} b_i$ and $ b_{i \minus{} 1} \equal{} r_i \minus{} k_ir_{i \minus{} 1}$, finishing the induction step.
Thus she can determine all numbers from 1 submission.

Let $x_i,y_i$ be the numbers on the red, blue card at the $i$th step. So Mrs. Taf initially chooses $x_0,y_0$, and receives $x_n,y_n$ at the end, where for all $i$ $x_i=k_ix_{i-1}+y_{i-1}$ and $y_i=x_{i-1}$. We claim that only $1$ card is needed: choose $x_0=2,y_0=1$ (or any initial values satisfying $x_0>y_0>0$). Note that $x_i=k_ix_{i-1}+y_{i-1}>x_{i-1}=y_i$ for all $i$. Now if we are given $x_n,y_n$, then $x_n=k_{n}y_n+y_{n-1}$ for $y_{n-1}<y_n$, so $k_n$ can be uniquely determined by performing the division algorithm on $x_n$ and $y_n$. From this we also recover $y_{n-1}$ and of course $x_{n-1}$, so inductively we can determine all $x_i,y_i,$ and $k_i$ from this one initial choice.


AMC
AIME
geometry
search
function
trigonometry
number theory

What geometry topics do I need to know for the AIME (I'm taking the aime 2)? Do I need to know modular arithmetic? What topics outside of the highschool curriculum (basic algebra, pre-cal) would be useful for the AIME?

I have little-no experience with contest math and got a 4 on a practice aime. On the test, I would like to get a reasonably good score (maybe like a 7). Your advice is greatly appreciated, thanks in advance  :lol:

you should know everything up to precalc. and be strong in counting and number theory. after that, all you can do is to solve many, many aime problems. also, the search function really helps.

Many of the upper level problems are geometry, so be sure to have a good grasp on that as well.

i think u have to know a bit of trigonometry

I'm pretty sure most of the hard geometry problems have solutions that are more elegant when proven without trig, but you can sometimes trig bash successfully.

is three hours actually enough to do the aime?

Three hours is more than enough to do the AIME if you're good enough.

I've been told that the AIME is all about not spending an hour on a problem solvable in 15 minutes once you get past the initial 3.


geometry
3D geometry
IMO
packing
combinatorics
imo 1976

A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent of the volume of the box is occupied. Determine the possible dimensions of the box.

We name a,b,c the sides of the parallelepiped, which are positive integers. We also put
\[ x \equal{} \left\lfloor\frac{a}{\sqrt[3]{2}}\right\rfloor \ \ \ \
  y \equal{} \left\lfloor\frac{b}{\sqrt[3]{2}}\right\rfloor \ \ \ \
  z \equal{} \left\lfloor\frac{c}{\sqrt[3]{2}}\right\rfloor \ \ \ \]
It is clear that $ xyz$ is the maximal number of cubes with sides of length $ \sqrt[3]{2}$ that
can be put into the parallelepiped with sides parallels to the sides of the box.
Hence the corresponding volume is $ V_2\equal{}2\cdot xyz$. We need $ V_2\equal{}0.4\cdot V_1\equal{}0.4\cdot abc$,
hence \[ \frac ax\cdot \frac by\cdot \frac cz\equal{}5\ \ \ \ \ \ \ \ (1)\]
We give the  values of $ x$ and $ a/x$ for $ a\equal{}1,\dots ,8$. The same table is valid for $ b,y$ and $ c,z$.
\[ \begin{tabular}{|c|c|c|}
  \hline
  a & x & a/x \\ \hline
  1 & 0 & \minus{} \\ \hline
  2 & 1 & 2 \\ \hline
  3 & 2 &  3/2 \\ \hline
  4 & 3 & 4/3 \\ \hline
  5 & 3 & 5/3 \\ \hline
  6 & 4 & 3/2 \\ \hline
  7 & 5 & 7/5 \\ \hline
  8 & 6 &  4/3 \\
  \hline
\end{tabular}\]
By simple inspection we obtain two solutions of $ (1)$: $ \{a,b,c\}\equal{}\{2,5,3\}$ and $ \{a,b,c\}\equal{}\{2,5,6\}$.
We now show that they are the only solutions.

We can assume $ \frac ax\ge \frac by \ge \frac cz$. So necessarily $ \frac ax\ge \sqrt[3]{5}$. Note that
the definition of $ x$ implies \[ x< a/\sqrt[3]2 < x\plus{}1,\]
hence \[ \sqrt[3]2< a/x < \sqrt[3]2(1\plus{}\frac 1x)\]
If $ a\ge 4$ then $ x\ge 3$ and $ \frac ax<\sqrt[3]2(1\plus{}\frac 1x)\le \sqrt[3]2(\frac 43)<\sqrt[3]5$
since $ 2\cdot \frac {4^3}{3^3}<5$. So we have only left the cases $ a\equal{}2$ and $ a\equal{}3$. But for $ a\equal{}3$
we have $ a/x\equal{}3/2<\sqrt[3]5$ and so necessarily $ a\equal{}2$ and $ a/x\equal{}2$.
It follows
\[ \frac by \cdot \frac cz \equal{}\frac 52 \ \ \ \ \ \ (2)\]

Note that the definitions of $ y,z$ imply \[ y< b/\sqrt[3]2 < y\plus{}1,\ \
\textrm{and} \ \ z< c/\sqrt[3]2 < z\plus{}1.\ \ \ \ (3)\]
Moreover we have from (2) and from $ b/y\ge c/z$ that
\[ \frac by \ge \sqrt{5/2}\ \ \ \ \ (4)\]

If $ b\equal{}2$ then $ b/y\equal{}2$ and we would have $ c/z\equal{}5/4<\sqrt[3]2$, which contradicts $ (3)$.

On the other hand, if $ b>5$ then $ y>4$ and $ \frac by<\sqrt[3]2(1\plus{}\frac 1y)\le \sqrt[3]2(\frac 54)<\sqrt{5/2}$
since $ 2^2\cdot \frac {5^6}{4^6}<\frac{5^3}{2^3}$ as $ 5^3<2^7$. So we have only left the
cases $ b\equal{}3,4,5$. But for $ b\equal{}3$ we have $ b/y\equal{}3/2<\sqrt{5/2}$ and for $ b\equal{}4$ we have
$ b/y\equal{}4/3<\sqrt{5/2}$ and so necessarily $ b\equal{}5$ and $ b/y\equal{}5/3$ ($ >\sqrt{5/2}$)

 So we arrive finally at $ a\equal{}2,b\equal{}5$ and $ c/z\equal{}3/2$. If $ c\ge 8$
 then $ z\ge 6$ and $ \frac cz<\sqrt[3]2(1\plus{}\frac 1z)\le \sqrt[3]2(\frac 76)<\frac 32$
since $ 2\cdot \frac {7^3}{6^3}<\frac{3^3}{2^3}$. On the other hand, for $ c\le 7$ there are the only two possible values
$ c\equal{}3$ and $ c\equal{}6$ which yield the known solutions.

How can you fill a parallelpiped completely with cubes of dimension 1? Isn't this only possible when the faces are all either parallel or perpendicular? (i.e. a rectangular prism). Not sure why the wording includes "parallelpiped" rather than "rectangular prism." Am I misunderstanding something here?

The official wording is a rectangular box.


Se da functia $\ f: (1,\infty)\rightarrow\mathbb{R}$, definita prin \[f(x)=\ln(\ln x)\] Demonstrati ca $\ f$ este bijectiva si sa se calculeze $\ f^{-1}$

Am stiut si eu sa calculez inversa functiei, dar n-am stiut(ieri) sa demonstrez ca functia e surjectie..dar acuma stiu. :)

Tu ce clasa esti de sti logaritmi?Eu parca stiam ca treci pe a VIIIa...

Intr-adevar, am trecut in clasa a VIII-a.Nu stiu numai logaritmii, mai stiu si alte lucruri de nivel liceal.Vezi tu, cand iau o pauza(ca mai obosesc si eu), m-apuc sa fac de liceu.Ca anul asta am capacitate, asa ca nu mai am timp asa mult sa fac la mate..astea sunt ultimele clipe de fericire. :)


The way I did it is to write $\cos(tx)=\frac{\exp(itx)+\exp(-itx)}{2}$, and expand the whole thing, and use the fact that $\int^{2\pi}_0 \exp(inx)dx=2\pi\delta_n$.  There are better ways but this is straightforward.

Oh.  A single-subscript kronecker delta.

[quote="justdudxit"]Let $I_m = \int^{2\pi}_0 cos(x)cos(2x)...cos(mx)\,dx$.  For which integers m, $1 \leq m \leq 10$ is $I_m$ nonzero?[/quote]

The way I partially solved the problem was by considering the following identities:
$\int^a_0g(x)\,dx = \int^a_0g(a-x)\,dx$ ... (I), and $\int^{2a}_ag(x)\,dx = \int^a_0g(x+a)\,dx$ ... (II)
(I will leave these as exercises for the reader.)

Now,
$I_m = \int^{\pi}_0 cos(x)cos(2x)...cos(mx)\,dx + \int^{2\pi}_{\pi} cos(x)cos(2x)...cos(mx)\,dx$
$= I' + I''$ (say)
Using identity (II) above, we get,
$I'' = \int^{\pi}_0 cos(x+\pi)cos(2(x+\pi))...cos(m(x+\pi))\,dx$
$\therefore I'' = \int^{\pi}_0(-cos(x))cos(2x)...(-1)^mcos(mx)\,dx$
Now, we note that $I'' = -I'$ for $m=1,2,5,6,9,10,$ and $I'' = I'$ for $m=3,4,7,8.$

So, $I_1 = I_2 = I_5 = I_6 = I_9 = I_{10} = 0.$

This is my partial solution. Note that I still haven't used identity (I) in the solution above.

Could someone figure out if any of $I_3, I_4, I_7$ or $I_8$ is $0$?


Find all mappings $ f: \mathbb{N}\to\mathbb{N}$ where $ \gcd(f(a),f(b))\equal{}\gcd(a,b)$ for all $ a\not\equal{} b \in \mathbb{N}$

[hide]The proof below is based on the observation that $ \gcd(x,y)|x$ and that $ m|n\implies\gcd(m,n)\equal{}m$.

Let $ P(a,b)$ be the assertion that $ \gcd(f(a),f(b))\equal{}\gcd(a,b)$. $ P(a,2a)\implies \gcd(f(a),f(2a))\equal{}\gcd(a,2a)\equal{}a\implies a|f(a)$. $ P(a,f(a))\implies\gcd(f(a),f(f(a)))\equal{}\gcd(a,f(a))\implies f(a)\equal{}a$.[/hide]


[quote="espark52"]When not on the moving sidewalk, Kelsey can walk the legnth of the sidewalk in 3 minutes. If she stands on the sidewalk as it moves, she can travel the legnth in 2 minutes. If kelsey walks on the sidewalk as it moves, how many minutes will it take her to travel the same distance? Assume she always walks at the same speed, and express your answer as a decimal to the nearest tenth.[/quote]
[hide]
the speed of walking in sidewalks per minute is $\frac{1}{3}$
the speed of moving in sidewalks per minute is $\frac{1}{2}$
the speed of walking as it moves is $\frac{1}{3}+\frac{1}{2}$
the sum is $\frac{5}{6}$
so it would take $\frac{6}{5}$ minutes to walk 1 sidewalk as it moves.
the answer is $1.2$ minutes.
[/hide]

[quote="espark52"]When not on the moving sidewalk, Kelsey can walk the legnth of the sidewalk in 3 minutes. If she stands on the sidewalk as it moves, she can travel the legnth in 2 minutes. If kelsey walks on the sidewalk as it moves, how many minutes will it take her to travel the same distance? Assume she always walks at the same speed, and express your answer as a decimal to the nearest tenth.[/quote]
[hide]Walking rate: $\frac13$

Moving rate: $\frac12$

Walking + moving rate: $\frac56$

She can walk + move $5$ sidewalks in $6$ minutes, or $1$ sidewalk in $\boxed{1.2}$ minutes.[/hide]


superior algebra
superior algebra unsolved

Let $ (\frac {a,b}{K})$ denote the Hilbert symbol (which is equal to $ 1$ if $ ax^2 \plus{} b y^2 \equal{} 1$ has a solution in $ K$ and is $ \minus{} 1$ otherwise). Let $ L/K$ be an algebraic extension of $ K$. If $ (\frac {a,b}{K}) \equal{} \minus{} 1$ can we find $ (\frac {a,b}{L})$ with information of the extension $ L/K$ (i.e., the degree, what kind of algebraic element we are adding, etc...)

Stated otherwise: if $ ax^2 \plus{} by^2\equal{}1$ does not have solutions over $ K$. Can we "always" construct a field $ L \supset K$ such that it has a solution in $ L$

Ow, of course

and this are the only extension which work?

No there are more. Take any extension of type $ K(\sqrt s)$ for $ s\equal{}\frac{1\minus{}by^2}{a}$ (and arbitrary choosen $ y \in K$).Any bigger extension containing one of those works, too.


[quote="santosguzella"][quote="Rust"]$x_{n}=q^{n}+r^{n},q=\frac{a+\sqrt{a^{2}-4}}{2},r=\frac{a-\sqrt{a^{2}-4}}{2}, \ (qr=1)$.[/quote]
Why?  :blush:[/quote]
Induction on $n$  :wink:


function
algebra
domain
calculus
calculus computations

Suppose that $ f$ is a real-valued function whose domain contains the interval $ [a,b]$. Suppose that $ f$ is Riemann integrable on $ [a,b]$. Prove that $ f$ is bounded on $ [a,b]$. The problem wants us to go through the following steps by contraposition:

(a) Prove that there exists a sequence $ (y_n)$ in $ [a,b]$ that converges to $ y \in [a,b]$ such that for every $ n \in \mathbb{N}$, $ |f(y_n)| > n$.

(b) Let $ P$ be any partition of $ [a,b]$. Prove that the Riemann sums of $ f$ corresponding to $ P$ is an unbounded set of real numbers.

(c) Show that $ f$ cannot be Riemann integrable on $ [a,b]$.

So for (a), how would you show the existence? Would you use the compactness of $ [a,b]$? Also how would you construct the sequence?

For (b) would you consider $ \mathcal{R}(f,P) \equal{} \sum_{i\equal{}1}^{n} f(x_{i}^{*})(x_{i}\minus{}x_{i\minus{}1})$?

And then part (c) follows immediately?

[quote="t0rajir0u"]a)  Without convergence, this is the definition of unboundedness.  (Make sure you understand this.)  You can find a convergent subsequence by compactness.

b)  The upper Darboux sum is undefined at the interval in the partition containing $ y$.[/quote]

We don't have darboux definition.

So you can prove part (a) by just noting that it is the definition of unboundedness? You dont need to come up with an actual sequence?

[quote="t0rajir0u"]What do you mean by "come up with an actual sequence"?  A function is unbounded if there exist counterexamples to the claim "this function is bounded with bound $ M$" for every $ M$.  These counterexamples are the sequence.

The Darboux answer should clue you in to the Riemann answer.  If the upper Darboux sum isn't defined, you can take $ x_i^{*} \equal{} y_n$ for appropriate $ i$ and arbitrary $ n$.[/quote]

Do you need the actual counterexamples to complete the proof? What are some counterexamples? How do you come up with them?

[quote="t0rajir0u"]They exist [b]by definition[/b], so no.[/quote]

I know they exist. So thats it? You can just say they exist and that proves part (a)? Then for part (b) you don't need to actually work with one of those sequences?


No discussion on the 2005 AMC10B/12B should take place until Friday, Feb. 18, at 12:00 EST.

To be completely clear, discussion even includes writing about how you think you did, stating 'that was really easy', 'I made so many stupid mistakes!', etc. To be safe, in any and all posts, pretend that the 2005 AMC 10B/12B test never existed.

If you have any questions about this rule, PM me or one of the administrators.

That's right. Be fair to yourself, and be fair to all participants by not mentioning any aspect of the "B" date AMC tests until after the designated time here, by which time all test answer sheets should be in the mail to AMC headquarters to be scored. And good luck on the tests to all of you AoPSers.

Yes, good providence to everyone who will be taking the B tests!

We will be locking down the AMC forum tomorrow AM until Friday as we did with the AMC A test.

In that case, I'd like to say good luck to everyone now.

Awww man... another lockdown....

Good luck to all taking the AMC 10B then...

Tomorrow's AMC12-B is my last chance to qualify for the AIME, so I need all the luck I can get =D I'd wish you all good luck, but then there'd be less of it left for me  :lol:

Just kidding  :blush: Good luck to all writing it!

Good luck everyone! Wish me luck in getting a perfect AMC 12!!!! jk jk. thats not gonna happen for a couple years a least =)


[quote="ifai"]Let f(x+y)=f(x)f(y) and f(1)=2. Calculate f(2)/f(1)+f(4)/f(3)+f(6)/f(5)+...+f(2006)/f(2005)[/quote]

[hide]
by letting $y=1$ we can easily derive $\frac{f(x+1)}{f(x)}=2$ , and using these we may obtain the original equation is equal to $2\times 1003=2006$

 :)
[/hide]

ifai:

tip...you dont have to post the problem as the subject ;)

name this one "functions" or somethin

[hide]Notice $\frac{f(2)}{f(1)}=\frac{f(1)f(1)}{f(1)}$, $\frac{f(4)}{f(3)}=\frac{f(3)f(1)}{f(3)}$, etc.

So each of the terms is equal to $f(1)$ after everything cancels.  There are $\frac{2006-2}{2}+1=1003$ of the $f(1)'s$, so the answer is $2\cdot 1003=\boxed{2006}$.[/hide]


Problems gotten from here: http://www.aoshoo.com/bbs1/dispbbs.asp?boardID=15&ID=5828&page=1

Second Chinese North MO:

Day 1
Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?p=601275#601275
$AB$ is the diameter of circle $O$, $CD$ is a non-diameter chord that is perpendicular to $AB$. Let $E$ be the midpoint of $OC$, connect $AE$ and extend it to meet the circle at point $P$. Let $DP$ and $BC$ meet at $F$. Prove that $F$ is the midpoint of $BC$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=601277#601277
$p$ is a prime number that is greater than $2$. Let $\{ a_{n}\}$ be a sequence such that $ na_{n+1}= (n+1) a_{n}-\left( \frac{p}{2}\right)^{4}$.

Show that if $a_{1}=5$, the $16 \mid a_{81}$.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=601278#601278
$AD$ is the altitude of triangle $ABC$ at side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?p=601279#601279
Given a function $f(x)=x^{2}+ax+b$ with $a,b \in R$, if there exists a real number $m$ such that $\left| f(m) \right| \leq \frac{1}{4}$ and $\left| f(m+1) \right| \leq \frac{1}{4}$, then find the maximum and minimum of the value of $\Delta=a^{2}-4b$.

Day 2
Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?p=601280#601280
$a,b,c$ are positive numbers such that $a+b+c=3$, show that: \[\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2b^{2}+(a+c)^{2}}+\frac{c^{2}+9}{2c^{2}+(a+b)^{2}}\leq 5\] Problem 6: Canceled.

Problem 7: http://www.mathlinks.ro/Forum/viewtopic.php?p=601281#601281
Can we put positive integers $1,2,3, \cdots 64$ into $8 \times 8$ grids such that the sum of the numbers in any $4$ grids that have the form like $T$ ( $3$ on top and $1$ under the middle one on the top, this can be rotate to any direction) can be divided by $5$?

Problem 8: http://www.mathlinks.ro/Forum/viewtopic.php?p=601283#601283
Given a sequence $\{ a_{n}\}$ such that $a_{n+1}=a_{n}+\frac{1}{2006}a_{n}^{2}$ , $n \in N$, $a_{0}=\frac{1}{2}$.

Prove that $1-\frac{1}{2008}< a_{2006}< 1$.

[quote="José"]What happened with p6?[/quote]
I don't know......  :(   :maybe:


[quote="Treething"]Tubgirl loves orange juice. She drinks it at the rate of one pint per minute. If her bathtub holds 30 gallons of liquid and is filled to the brim with orange juice, how many minutes will it take her to drink all of it if the tub is drained at the rate of one cup per minute?[/quote]

[hide]$x(\frac{1}{16}+\frac{1}{8})=30$
$(\frac{3}{16})x=30$
$x=\boxed{160}$ minutes.[/hide]


calculus
integration
trigonometry
derivative
geometry
calculus computations

To Jimmy,, I'm sorry, I have made a miss type.

I will edit.

kunny

$\displaystyle \int \sqrt{2ax-x^2}\ dx =\frac{x \sqrt{2ax-x^2}}{2}+ax \arctan({\frac{x}{2ax-x^2}})$

Putting in the limits, we get.....
a whole lotta number crunching

$\int \sqrt{2ax-x^2} \ dx, a>0 = \frac{a^2\sin^{-1}\left(\frac{x-a}{a}\right)}{2}+\frac{(x-a)\sqrt{2ax-x^2}}{2}$

Plugging in the limits we get

$\frac{a^2\pi}{4}$

seamusoboyle, did you use a computer to get that anti-derivative, because if you treat $ax$ as a variable, not as $a\cdot x$, then

the answer you get is $ax\tan^{-1}\left(\frac{x}{\sqrt{2ax-x^2}}\right)+\frac{x\sqrt{2ax-x^2}}{2}$.

So when typing the integral in you have to type $a * x$.

Yes, that's right, Jimmy. Judging from the answer, you are to imagine something like.... ;)

Japanese highschool students don't study the integral of arc sine,cosine and tangent in highschool's class.

[quote="kunny"]Evaluate

\[\int_0^a \sqrt{2ax-x^2}\ dx \ (a>0)\][/quote]

I think the way tough in high school should be as follow:
Let $u = a-x\/$. Then we have
$\int_0^a \sqrt{2ax-x^2}\ dx = \int_0^a \sqrt{a^2 - (a-x)^2}\ dx =\int_0^a \sqrt{a^2 - u^2}\ du = \frac{a^2\pi}{4}$, since it is just a quarter area of a circle with radius $a\/$.

Yes, you are right, pipi.

$y=\sqrt{2ax-x^2} \Longleftrightarrow y^2=2ax-x^2\ and\ y\geq 0\Longleftrightarrow (x-a)^2+y^2=a^2\ and\ y\geq 0$ ,yielding pipi's reasult.


What do you guys think on this issue? Do humans have free willed or is it all predestined in the first place. Some others might argue that in the universe one event leads to another, and humans respond accordingly to outside stimuli using their genetically predetermined responses, leading to an endless chain of predetermined events in the universe, etc. So what's your point of view on this concept?

Actually, I've posted this video before on AoPS, but I'll post it again.
[youtube]_VxQuPBX1_U[/youtube]
I you would rather like something to read instead of a movie excerpt, just click the video and the guy who posted it wrote it all down in the description.

My opinion... Of course, the physics stuff makes sense, but... I don't know. I don't really have a side yet, although if you asked me to decide right now I would probably say predertermined.

[quote="jimhu"]Do humans have free willed or is it all predestined in the first place?[/quote]

Both  :)

It seems like the free will of humanity is mostly a philosophical question at this point and may start being scientific once neurobiology makes huge strides; predestination all depends on which interpretation of quantum mechanics you accept. The Copenhagen interpretation demands inherently nondeterministic and discontinuous wave function collapse as one of the two modes of time-evolution of the wave state; I personally don't like this, not so much due to the probabilistic notions as due to the discontinuity (I'm an analyst). Unfortunately, there doesn't seem to be any way around the discontinuity that I know of at least, but some of the nonprobabilistic interpretations still feel more comfortable to me, and with these interpretations one can have such things as a completely predetermined system nonetheless providing entirely random data to any observers (which I believe could not happen from the viewpoint of computational complexity for purely classical systems). For a discussion on this, see the latterpart of "quantum mechanics in your face" available on, say, googlevideo.

Although it seems like at least one half of the question is inherently a matter of the philosophy of physics, it's probably not necessary to delve so deeply into physical theories to explain my position. Even supposing humans were purely classical systems and thus entirely predetermined in the classical sense, why could not free will and predestination coexist? Even if the choices we made are predetermined, our thoughts about them would be predetermined as well, and surely some entity clever enough could set up the initial conditions so that the choices we make are always in accord with the thoughts that comprise our decision-making process. It certainly seems that this has held true for as much of my life as I can remember so far (well, discounting all the times when I do totally crazy things just because I'm me), so I don't have much trouble accepting that it'll continue to be true.

I've talked to some philosophy people on this, and "not having free will" does not necessarily mean your fate is "predetermined" (think: chance).

I do not believe in free will because I think every choice you make has its roots in a decision out of your hands.

This is a major major philosophical debate. My own opinion is that the topic should be narrowed slightly. There are probably more published pages on this topic than there are published pages that fall under the category "Real Analysis."

I don't like the term "predetermined" that keeps getting used. You don't have to believe something is predetermined to believe in determinism. Most forms of determinism just say that everything has a cause. Predeterminism seems to have the connotation that something has set up how everything is going to run ahead of time.

I believe we have free will, but that freedom is restricted by cultural influences, chemical balances in the brain, soicio-economic status, etc. I believe in free will almost by default. The more interesting question to me is: if determinism is true, how do we assign moral responsibility?

If really interested in this subject, I suggest looking up either Martin Luther's book [i]Bondage of the Will[/i].  This is still in publication, written by the founder of the reform movement in Germany in the 1300s.  It's in English and probably at your local library.

This is the guy that Martin Luther King and Martin Luther King, Jr. were named after.

An easier, faster source of discussion on this is to look up "Calvinism TULIP" on google.  The T stands for total depravity - the loss of human will.  This also talks about the issues the video talks about just in passing.

Hope this helps.


[quote="mr. math"]How would u do this problem?

Sam and Jed are racing around a track. Since Jed is faster than Sam, he decided to give Sam a 10 second head start. If Sam runs at a constant rate of 15 meters per second and Jed runs at a constant rate of 25 meters per second, how long will it take ,in seconds, for Jed to catch Sam?[/quote]

[hide] The 10 second head start gives Sam $15\cdot10=\mathbf{150}$ meters. Since Jed runs $25-15=\mathbf{10}$ meters per second faster than Sam, it takes $\frac{150}{10}=\boxed{15}$ seconds. [/hide]

Basically, utilize the formula Distance=Rate X Time.

[hide]In 10 sec, Sam gets 150 meters ahead of Jed. Every sec, Jed closes the gap by 10 meters, so it takes 15 sec for Jed to catch Sam.[/hide]

Generalization:
Sam runs at $a$ meters per second
Jed runs at $b$ meters per second
$a<b$
Jed gives Sam $s$ second head start.

$s$ second menas $as$ meters head start.

Jed gains at $b-a$ meters per second.

It takes $\frac{as}{b-a}$ seconds for Jed to catch up to Sam.

Think about what happens if $b-a<0$. :P


Write $ \sqrt{8}\equal{}2\sqrt{2}$ and $ \sqrt{18}\equal{}3\sqrt{2}$.  Hence the sum is $ 5\sqrt{2}$.


If $ f(x) \equal{} x^{2}\plus{}x$, prove that the equation $ 4f(a) \equal{} f(b)$ has no solutions in positive integers $ a$ and $ b$.

$ 4a^2\plus{}4a\equal{}b^2\plus{}b$

$ (2a\plus{}1)^2\equal{}b^2\plus{}b\plus{}1$

$ b^2<b^2\plus{}b\plus{}1<(b\plus{}1)^2$ though, and there can be no perfect square between 2 consecutive perfect squares, so we cannot have $ b^2\plus{}b\plus{}1\equal{}(2a\plus{}1)^2$.

Cheers,

Rofler

[hide="Solution"]

$ 4f(a) \equal{} 4a^2 \plus{} 4a$
$ f(b) \equal{} b^2 \plus{} b$

We need to find when equality of these two values occurs.

$ 4f(a) \equal{} 4a^2 \plus{} 4a \equal{} c$, for $ c \equal{} f(b) \equal{} b^2 \plus{} b$.

$ 4a^2 \plus{} 4a \minus{} c \equal{} 0$ From the quadratic formula, we must have the inside of the radicand to be a perfect square, such that the condition of a being a positive integer is met.

$ \sqrt {16 \plus{} 16c} \implies 4\sqrt {1 \plus{} c} \implies 4\sqrt {1 \plus{} b^2 \plus{} b} \implies 4\sqrt {(b \plus{} 1)^2 \minus{} b}$, and $ (b\plus{}1)^2 \minus{} b$ cannot be a perfect square for any positive integers.

Thus, there are no positive integer solutions $ a$ and $ b$ for the given equality.

[/hide]


geometry
algebra
polynomial
function
advanced fields
advanced fields unsolved

The question comes from the proof of Propositon 3.5, I don't understand why the equality below holds

$ f(\xi_1(P),...,\xi_n(P)) = h(f(\={x}_1,...,\={x}_n))(P) = 0$

The equality is at the bottom of that page.

I didn't expect this to stand unanswered for more than 1 day...

Since $ f$ is a polynomial, and $ h$ is a homomorphism of $ k$-algebras, we have

$ f\left(h\left(\overline{x_1}\right),...,h\left(\overline{x_n}\right)\right)\equal{}h\left(f\left(\overline{x_1},...,\overline{x}_n\right)\right)$.

In other words,

$ f\left(\xi_1,...,\xi_n\right)\equal{}h\left(f\left(\overline{x_1},...,\overline{x_n}\right)\right)$.

Both sides of this equation are regular functions on $ X$, so evaluating them at $ P$, we obtain

$ f\left(\xi_1\left(P\right),...,\xi_n\left(P\right)\right)\equal{}h\left(f\left(\overline{x_1},...,\overline{x_n}\right)\right)\left(P\right)$.

  darij


Hey guys, i'm new to this website so this may very well be a nooby question.  How can I prove that:

a1a2a3...an <= {1/n(a1 + a2 + a3 + ... + an)}^n

Thank you =)

hello, do you mean $ a_1 \cdot a_2\cdot a_3\cdots a_n\le \left(\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\right)^{\frac{1}{n}}$?
Sonnhard.

It is an AM-GM INEQUALITY.
Proof is given in AoPSWiki (see in left). There search for Proofs of AM-GM.
I remembered there are more than 4 proofs for this there.
check that.

I posted that, but I deleted it after rereading. I don't believe this is necessarily AM-GM, since if you raise both sides to the nth power, you have the arithmetic mean of a group of integers on one side, but then then a product of integers [i]raised[/i] to a power [i]greater than 1[/i]. Well, I guess that might not be so, since there is nothing mentioned about any of the $ a_i$ or n. However, it also doesn't say that the $ a_i$ are positive, which is a requirement for AM-GM.

Isnt that Caunchy's result, whereby, the arithmetic mean is less than the geometic mean?...

Its not, it looks VERY similar to AM-GM but I believe that is just to trick you. I don't see how you would use induction here.

I think the author meant for the right side to be raised to the nth power and not 1/n, making the equation basically AM-GM.

[quote="gurmies"]Hey guys, i'm new to this website so this may very well be a nooby question.  How can I prove that:

$ a_1 \cdot a_2\cdot a_3\cdots a_n\le \left(\frac{1}{n(a_1\plus{}a_2\plus{}a_3\plus{}\cdots\plus{} a_n)}\right)^n$

Thank you =)[/quote]

I can't prove it, but I can disprove it.

Let $ a_k\equal{}1\ \forall\ k\equal{}1, 2, 3 \cdots n$

Now we have:

$ a_1 \cdot a_2\cdot a_3\cdots a_n\le \left(\frac{1}{n(a_1\plus{}a_2\plus{}a_3\plus{}\cdots\plus{} a_n)}\right)^n\\
1\le \left(\frac{1}{n^2}\right)^n$

which can't be true since $ n$ is positive integer.

Thank you, I did mean to the power of "n".


Alice earned a total of $65 for working five days after school. Each day she earned $2 more than the previous day. How much did she earn on the first day?

 $9

problem taken from Math Olympiad Contest Problems For Elementary And Middle Schools. also available in the aops book store.[hide][/hide]

I would put this in classroom math, it's too easy for MATHCOUNTS.

[quote="Ihatepie"]It's too easy for MATHCOUNTS.[/quote]

What is the value of 6 times one seventh?

@dilberto: Hide your answer.

It was directed at elementary math students, given there is no such forum, and the olympiad problems are usually too hard for elementary students, this seemed like a perfect place but apparently not.

If someone can start a elementary math forum, I would moderate it, but none of us can start one. So scrap that. :whistling: [/hide]

Classroom math.
Or some other site other than AoPS.

is this really middle school math..... wow.... easy

Please don't revive old topics, unless you have something important to say, such as another way of solving the problem or some question.

That being said, another way that you could do it is notice that the third day is the average, or 13. Thus, the first day is 13-4=9.