calculus
integration
trigonometry
function
calculus computations

Please, can someone give any ideas how to find the sum:
$\sum_{n=1}^{\infty}\frac{1}{\sqrt n \left( {n + 1} \right)}$
I've done some calculations in Mathcad. It's very near to $1+\frac{\sqrt 3}{2}$, but it's just a guessed value and I doubt very much if it's true. Anyway I firstly need a method of solving, and then a result.

Careful use of the integral test gives the following partial sum estimate:

$\sum_{n=1}^{\infty}\frac1{\sqrt{n}(n+1)}= \sum_{n=1}^N\frac1{\sqrt{n}(n+1)}+\left(\pi -2\tan^{-1}\left(\sqrt{N+\frac12}\right)\right)+O(N^{-5/2})$

I ran that estimate out to $N=256,$ then used Richardson extrapolation to narrow down the estimate. My best estimate for the sum is $1.8600250792.$ This differs from $1+\frac{\sqrt{3}}2$ by $0.006$; they're clearly not the same number.

There's no [i]a priori[/i] reason to believe that this sum could be expressed in closed form using the usual list of functions.

[quote="sedated"]Is it possible to prove that it couldn't be expressed in closed form using the usual list of functions??[/quote] I don't think anybody in the world has the slightest idea of how to prove that it is irrational. It is, probably, possible to prove that $\frac 1{\sqrt n(n+1)}$ is not of the form $F(n)-F(n+1)$ where $F$ is an elementary function but even that is, most likely, beyond my abilities. I'll think of it though.


induction
binomial coefficients
number theory proposed
number theory

[b]Problem[/b]: Let $p$ is an prime number. Find all solution in positive integers $(x,y,z)$ of the following equation
\[ x^p+p^x=z^p \]

Assume that $x,z,p$ satisfy the given relation.
Then, $p^x = (z-x)(z^{p-1} + \cdots + x^{p-1})$ which leads to $z=x+p^k$ for some nonnegative integer $k$.
Thus $x^p + p^x = z^p = x^p + C_p^1x^{p-1}p^k + \cdots + p^{kp}.$ (1)
Since $p$ is prime, it is well-known that each of the binomial coefficients $C_p^r$ is divisible by $p$ for $r=1, \cdots, p-1.$ It follows that $p^{kp} = 0 \mod [p]$ so that $k>0.$

Moreover, from (1) again, we deduce that each prime dividing $x$ divides $p^x$ too, which forces $x=p^a$ for some nonnegative integer $a$. Direct checking shows that $a>0.$
Then $z=p^a+p^k$ and the equation may be rewritten as $p^{ap} + p^{p^a} = (p^a+p^k)^p.$ (2)

If $p=2$ and $a \leq 2$, direct checking shows that the given relation is not satisfied.
Thus, we may assume that $p>3$ or $p=2$ and $a>2$. In that case, an easy induction gives $p^a > ap.$
From (2), we deduce that $ap$ is the exponent of $p$ in the prime decomposition of the LHS of (2), so that $a$ is the exponent of $p$ in the prime decomposition of $p^a+p^k$, which forces $k > a$.

Thus, we may rewrite (2) as $1+p^{p^a-ap} = (1+p^{k-a})^p.$
But, using again that $p$ divides each of the $C_p^r$, it follows that  the RHS as form $1+p^{k-a+1}m$, where $m$ is a positive integer which is greater than $1$ and coprime with $p$. Therefore, it cannot be equal to the LHS. A contradiction.

Hence, the given equation has no solution.

I hope I didn't miss some cases...

Pierre.

I think it almost holds but take a look at the example $(p,x,z)=(2,6,10)$

Oops... I forgot the $p^{kp}$ in (1)...

Pierre.


Hi!

By combustion in oxygen of $ 178g$ of organic substance, $ A$, is obtaining $ 126g{\rm{ H}}_2 {\rm{O}}$and $ 156.8L$
(normal conditions) gases. After the condensation of the water and the pass of gases through a solution of $ NaOH$ it remains
$ 22.4L{\rm{ N}}_2$. Determine the molecular formula of the substance $ A$, knowing that in its molecule there are twice
more oxygen atoms than the azote atoms.

Thanks!

For a start, here's the balanced chemical equation:

$ C_nH_mN_xO_{2x} \plus{} \left(n \plus{} \frac{m}{4} \minus{} x\right)\,O_2 \longrightarrow n\, CO_2 \plus{}\, \frac{m}{2}\,H_2O \plus{}\, \frac{x}{2}\,N_2$

[quote="Sebi Crisan"]
... and $ 156.8L$(normal conditions) gases.
... it remains $ 22.4L{\rm{ N}}_2$.
[/quote]

What can I find from these?

[quote="Carcul"]At normal conditions the molar volume is 22.4 L/mol, so from that we know that 1 mol of nitrogen has formed.[/quote]

So, $ x \equal{} 2$.

156.8 L means 7 moles. Since we have 1 mol of $ N_2$ => 6 moles of $ CO_2$ => $ n \equal{} 6$.

Then, from the equation we obtain $ m \equal{} 14$.

=> $ C_6H_{14}N_2O_4$.

In my book the answer is stated as being $ C_3H_{7}NO_2$. Can you explian me which of the two is correct?

[quote="Carcul"][quote="Sebi Crisan"]So, x = 2[/quote]

That would be true if we had burned 1 mol of the unknown compound: that's why your answer is not correct.[/quote]

How can I find the number of burnt moles of unknown compound?

Try to use the "178 g of organic substance".


Let ABC be a triangle, suppose: $\large BC=a; CA=b; AB=c$. Prove that:
$\large \sqrt{a^4+b^4+c^4} + \sqrt{abc(a+b+c)} \geq  2 \sqrt{a^2b^2+b^2c^2+c^2a^2}$

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70159]http://www.mathlinks.ro/Forum/viewtopic.php?t=70159[/url].
There is a stronger inequality I proved :)

To SuperMath:From what I see,you haven't reply any of your post....You just post the inequality then you go away for that topic...I don't who you are,but please be responsible for your post. :)


Let $ p$ be prime number and $ p \equal{} 3k \plus{} 2$ ,               $ k \in N^*$
Prove that:
$ \prod_{k \equal{} 1}^{p}(k^2 \plus{} k \plus{} 1) \equiv 3 (mod p)$

Please help me :)

The product is equal to $ 3(p^2\plus{}p\plus{}1)\prod_{k\equal{}2}^{p\minus{}1} \frac{k^3\minus{}1}{k\minus{}1}$, so we need to show that last product is congruent to 1 modulo $ p$.

To do this, we can use Fermat's Little Theorem to show that cubing is a permutation of the nonzero elements modulo $ p$. (We can find an inverse for cubing, namely raising to the $ \frac{2p\minus{}1}{3}$ power.) As $ 1^3\equiv 1\pmod{p}$ of course, cubing must be a permutation of $ \{ 2,3,\cdots ,p\minus{}1\}$ modulo $ p$. This gives
\[ \prod_{k\equal{}2}^{p\minus{}1}(k^3\minus{}1)\equiv \prod_{k\equal{}2}^{p\minus{}1} (k\minus{}1)\not\equiv 0\pmod{p}\]
and the desired result follows.

[quote="scorpius119"]As $ 1^3\equiv 1\pmod{p}$ of course, cubing must be a permutation of $ \{ 2,3,\cdots ,p \minus{} 1\}$ modulo $ p$.[/quote]
I don't understand here  :P


If $ |x| \plus{} x \plus{} y \equal{} 10$ and $ x \plus{} |y| \minus{} y \equal{} 12$, find $ x \plus{} y$.

$ \textbf{(A)}\ \minus{}2 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \frac{18}{5} \qquad \textbf{(D)}\ \frac{22}{3} \qquad \textbf{(E)}\ 22$

Assume x is negative. If you put that in the first equation, you can see that y=10. Put y into the 2nd equation and you get x=12. Which is not negative so contradiction. X is positive. Now assume y is positive. If you put that in the 2nd equation, you can see that x=12. Now put that into the 1st equation and y=-34. So contradiction. Y is negative.

So the two equations can be rewritten as 2x+y=10 and x-2y=12
5y=-14 y=-14/5. x=32/5. X+y=18/5 or C


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

Take a sphere and draw on it a great circle. There are two regions created. Here, I am referring to regions on the surface of the sphere. Now draw another great circle: there are four regions. Now draw a third, not passing through the points of intersection of the first two. How many regions?

 Thanks

I THINK that you can only cut through the middle of the sphere, since they are great circles, they must run along the circumference.


Putnam
number theory proposed
number theory

Let $ a>0$ be such that $ n^a$ is an integer for all $ n\in\mathbb{N}^{\ast}$.
Prove that $ a$ is an integer.

This is problem A-6 of the Putnam 1971 competition.

Pierre.


Συμβολίζουμε με $a \bmod b$ το υπόλοιπο της ευκλείδιας διαίρεσης του  $a$ με το  $b$. Βρείτε όλα τα ζεύγη θετικών ακεραίων  $(a,p)$ έτσι ώστε ο  $p$ να είναι ένας πρώτος αριθμός και  \[ a \bmod p + a\bmod 2p + a\bmod 3p + a\bmod 4p = a + p.  \]

Θέτω a mod p = υ με 0<=υ<p

    Είναι εύκολο να δειχθεί ότι

   a mod 2p = υ ή υ+p
   a mod 3p = υ ή υ+p ή υ+2p
   a mod 4p = υ ή υ+p ή υ+2p ή υ+3p

   Άρα αν a=k*p + υ έχουμε ότι a+p = (k+1)*p + υ = 4υ + l*p   με 0<=l<=6 (1)
    p|3υ και άρα p=3 ή υ=0. Για υ=0 έχουμε ότι a=k*p και αφού ισχύει η (1) τότε 1<=k<=5 και διακρίνω
   περιπτώσεις...(εύκολο). Αν p=3 τότε προκύπτει ότι 3(υ+6)>=3(κ+1) και άρα υ+5>=κ με υ=1,2 αφού για
   υ=0 θα έχω ελένξει πιο πάνω...
   Ελέγχω τις περιπτώσεις και βρίσκω αν υπάρχουν λύσεις

[quote="jj-the-king"]
   έχουμε ότι a+p = (k+1)*p + υ = 4υ + l*p   με 0<=l<=6 (1)
[/quote] πώς βγήκε αυτό?

Σιλουανε ή οποιος αλλος θελει μπορει να δωσει την τελικη απαντηση? μονο αν ειναι ευκολο.... :)

Οι λύσεις αν δεν έχω ξεχάσει κάποια περίπτωση είναι $(p,3p)$ και  $(3,1)$, $(3,17)$


trigonometry
real analysis
real analysis unsolved

suppose that $x \in \mathbb{R}$ is a numbers such that there exists a real $\lambda > 0$ satisfying
$\sum_{0}^{\infty}\sin^{2}( \pi \lambda x^{n}) < \infty$

Prove that $x$ is an algebraic integer whose conjugates have modulus $< 1$.

Now it's (7) from http://mathworld.wolfram.com/PisotNumber.html
I can't prove it though.

(by this "now", I think about : http://www.mathlinks.ro/Forum/viewtopic.php?t=100067 )

a hint for this hard pb:

Take $x>1$ such a real and define $\lambda x^{n}= a_{n}+\epsilon_{n}$
where $a_{n}\in \mathbb{N}$ and $|\epsilon_{n}| \leq \frac{1}{2}$. Remark that $a_{n}$ is well-defined for large $n$.
Prove that:
$\sum_{0}^{\infty}a_{n}z^{n}= \frac{P(z)}{Q(z)}$
in a neighboorhood of $0$ where $P,Q \in \mathbb{Z}[X]$ and $Q(0)=1$.


So you decide to go trick-or-treating tonight and you come to a guy's house. He takes you in and leads you to a bookcase with 2 layers of shelves. Only, instead of books on the shelves, there are jars. This is what you can see of the jars:

1) There are 6 jars, 3 on each layer of the bookcase
2) Each jar is directly under (or above) another
3) The jars have labels NOT MACTHING the contents (but ARE representative of the contents of A jar) - the labels are from left to right, up to down-
[first row] "misery", "(your favorite candy)", "death",
[second row],"sadness","despair","doom".

The guy tells you that the candy is only in one jar, and if you open another jar its contents shall haunt you for life. In addition, the guy says that the jar containing doom is under the jar containing sadness and the jar containing death is immediately to the right of the jar containing misery. Assuming that you make the wisest choice, what is the probability that you will get your favorite candy?

As a good moderator, I have one comment on this problem:  :rotfl: . A Halloween problem!  :rotfl:

Very creative, haven't seen one yet today.  :)

That is a cool problem  :D !

[hide="ANSWER"]There are 2 that are to the right of the sadness and doom, so you know those are on the far left.  This leaves only 4 jars to choose from and 1 has candy as its label, so you know that can't be it, making the probability you choose correctly $\frac13$[/hide]

[quote="chuckzeduck"]Your chance of getting your favorite candy is

[hide]$1: 3$ The jars of the far left will contain sadness and doom. This is possible to figure out because of the second given. Then, you know that your favorite candy is not in the jar marked "Your Favorite Candy", so, this leaves 3 jars to choose from.[/hide][/quote]

$1: 3$ does not mean the same thing is $\frac{1}{3}$.  Using a colon means you are talking about odds.  Example: If I say that my odds against rolling a six on a standard die are $5: 1$, I mean that in the long run, for every time I roll a six, I will roll something other than a six 5 times.


Is speed of sound frame dependent?
Consider two persons $A$ and $B$ where $A$ is stationary and $B$ moves in a train at some speed with respect to the ground.Will both feel the speed of sound to be the same or different?

I feel that the speed of sound shouldn't be different since the speed of sound is given by the expression
${v_{sound}= \sqrt{\frac{{\gamma}RT}{M}}}$ where ${\gamma}$ is the adiabatic gas constant which doesn't change as the frame moves.All other terms in the right hand side are also constant and hence the speed of sound should be frame independent.
Please clarify.

The speed of the sound in air is about $340 \frac{m}{s}$.

The speed of the sound only will be different only if the temperature in the train is different of the temperatura outside the train.

What happen is Doppler Effect. Witch is on repect of the frequency and the wavelength. the sound velocity is equal.

since the air is equal on both. there is no logical reason that the speed changes.

Think about a example. If i make a experience on earth and after make the same experience in the same conditions on a Lab in the speed, the results will be the same if the initial conditions are the same, Newtonian example.

the speed of the sound in a element depends only of the density of the element and the coefficient of stiffness

see

http://en.wikipedia.org/wiki/Speed_of_sound

there is no Velocity of the train in this formula, hehe

i don't understand the topic.i found about 16 definitions for frame from which about 6 can have something to do with sound.can somebody explain better what's this about?


logarithms
graph theory
combinatorics unsolved
combinatorics

For any two guests on a party of $kn$ participants there is some guest who knows both of them. Prove that if $k<\frac{n}{2\ln n}$ then one can find $n$ guests who together know everyone else.

I just think that the $ln n$ smells like probabilistic method, or something similar

Is there a solution in the Hungarian journal?

Nope, and I haven't solved it myself either. That's why it is in unsolved problems... this looks hard to me. (maybe I'll solve this when I get graph theory on univ)

Does anyone know the solution to this?  Or does anyone know which journal this was published in (it was in a Hungarian journal in March 2005) and might be able to find the solution in a subsequent edition?

[quote=katzrockso]Don't go bumping 10 year old threads...[/quote]
Why not? It sounds interesting!
But there is something wrong with the statement. As written, apparently $n>k$?!
Perhaps it should be read as $n<\frac{k}{2\ln k}$ or something...
As for the Hungarian journal, maybe it's Komal.
If one has a patience to seaerch here it is: http://www.komal.hu/verseny/korabbi.e.shtml

[quote]If one has a patience to search here it is: http://www.komal.hu/verseny/korabbi.e.shtml[/quote]

It is indeed Komal.  It's problem A-370 from March 2005.  This problem is intended for high school students (albeit ones with special training).  They did publish solutions, so the solution is probably in an issue sometime later in 2005 and in Hungarian.  Any one have access to it?  lol.

It was solved by 3 high school students, and it actually lists their email addresses from 10 years ago.  I'm going to reach out to them (but I am expecting "returned to sender" or silence).  If they respond with a solution I will post it here.

All that being said, how are terms like $\ln n$ derived?  I have not studied work done by Erdos and Reyni but they do derive terms like $\ln n$ with probability arguments.  Their's were in numerators, so my guess is that the $\ln n$ term is produced on the left hand side before it is divided.

But in general where do these $\ln n$ terms come from?  Sometimes when you halve things repeatedly you get terms like $\log_2 n$ but not $\ln n$.  Do they typically appear from sums or integrals?  Is there a good book that introduces these types of probability arguments?

BTW - Does anyone find it mind blowing that there are kids in the world studying the work of Erdos and Reyni, meanwhile 99.99% of the population has no idea this stuff even exists until maybe they are in their 20's?  Maybe.  The situation is actually orders of magnitude more outrageous than that.  I'm just curious.  No one seems to talk about it.

I located current work emails of two of the solvers.  So hopefully one or both of them will respond.

If not, maybe whoever runs the Komal journal will be able to lookup the solution in an old edition.

[quote=MadCowChicken3]
All that being said, how are terms like $\ln n$ derived?  I have not studied work done by Erdos and Reyni but they do derive terms like $\ln n$ with probability arguments.  Their's were in numerators, so my guess is that the $\ln n$ term is produced on the left hand side before it is divided.

But in general where do these $\ln n$ terms come from?  Sometimes when you halve things repeatedly you get terms like $\log_2 n$ but not $\ln n$.  Do they typically appear from sums or integrals?  Is there a good book that introduces these types of probability arguments?
[/quote]

Yes indeed. One such result is the following:
[quote = Theorem]
Let $p = \dfrac{c\ln n}{n}$. Then, if $0<c<1$, $$\lim_{n\rightarrow\infty}P(G(n,p)\text{ connected }) = 0.$$
If $c>1$, $$\lim_{n\rightarrow\infty}P(G(n,p)\text{ connected }) = 1.$$
Finally, if $c=1$, then we know nothing.
[/quote]

Numbers of this form are known as thresholds. Bollobás proved that for most properties $X$, there exist such a threshold for them. Interestingly enough, the $\ln n$ are in the numerator rather than the denominator...

As for your second question, I believe it is the nice nature of the logarithms that is the cause of their widespread appearance. For example, in the proof of the first part of the above theorem, it boils down to showing that $p\rightarrow 0$ and that $\dfrac{1}{n(1-p)^{n-1}}\rightarrow 0$, both of which are possible because of the logarithm. However, I do not know too much about this...

Sorry, I misread the number of participants as $k$ instead of $kn$. That was the reason of my comment above.

Now, suppose for a moment the number of participant were $n^2$. Any pair of two guests we map to a guest that knows each of them. Since all such pairs are $n^2(n^2-1)/2$, there is a guest $v_1$ corresponding to at least $(n^2-1)/2$ pairs. It means $v_1$ knows at least $n$ other guests.
Thus removing $v_1$ and the guests he knows, we get another graph with at most $n^2-n$ vertices.
If at every next step we remove at least $n$ vertices, we would have got at the end at most $n$ guests knowing the rest of the participants. But it's slightly different: at every next step the number of the removed vertices slightly decreases. That's why, in order to get $n$ vertices at the end, we must "slightly" decrease the number of participants: instead of $n^2$, something like $n^2/(2\log n)$.
I think this is the main idea, all that remains are some calculations. :)

BTW a solver on the actual test game me the solution.  It's been a while, but I'll go ahead and post it (I'll just outline it actually).  BTW the $\ln n$ tern does not come from the combinatorics.  It's a big misdirection that threw everyone off.

Assume “knowing” is symmetric and translate the problem into graph theory.  Call an $n$-element subset bad if the neighborhood of that subset does not contain all the other vertices.

We will create an upper bound for the number of bad subsets.  If this is less than the total number of $n$-element subsets, there must be a good subset.

The following is an upper bound for the number of bad subsets:

$\binom{kn-1-d(v_1)}{n}+\binom{kn-1-d(v_2)}{n}+\cdots+\binom{kn-1-d(v_{kn})}{n}$

Now, if any vertex has degree $n$ or less, it's neighborhood can be used to make a good subset.  So assume for every vertex $v$ that $d(v)>n$.

Plugging this into our upper bound we can get the following new upper bound:

$kn\binom{(k-1)n}{n}$

If this upper bound is less than the total number of $n$-element subsets, there must be a good subset.  So we need to find values of $k$ that make the following inequality true:

$\binom{kn}{n}>kn\binom{(k-1)n}{n}$

If you treat these as polynomials of $k$ you will see that the one on the right has degree that is one more than the one on the left, so it will always overtake it eventually.

And $\binom{kn}{n}>kn\binom{(k-1)n}{n}$ is true if $kn<\left(\frac{k}{k-1}\right)^n$ is true.

And $kn<\left(\frac{k}{k-1}\right)^n$ is true if $n^2<\left(\frac{k}{k-1}\right)^n$ is true, since we are looking only for $k<n$ from the problem statement.

You can solve $n^2<\left(\frac{k}{k-1}\right)^n$ for $k$ and get $k<\frac{1}{n^{\frac{2}{n}}-1}+1$ which is stronger than the problem statement.  But, the problem statement requires a weaker inequality that uses logs.

And $n^2<\left(\frac{k}{k-1}\right)^n$ is true if $2 \ln n<\frac{n}{k}$ is true.  Use the fact that in general $\frac{1}{t}>\ln\left(1+\frac{1}{t}\right)>\frac{1}{t+1}$.

And $2 \ln n<\frac{n}{k}$ is equivalent to $k<\frac{n}{2\ln n}$ so the problem is complete.

As mentioned, the $\ln n$ has nothing to do with the combinatorics because $k<\frac{1}{n^{\frac{2}{n}}-1}+1$ is a better bound than $k<\frac{n}{2\ln n}$ and it has no $\ln n$ term.


geometry
rectangle
Pythagorean Theorem
power of a point

Draw two non-concentric, tangent circles A and B with diameters on the same axis. Circle A has radius 8, the other radius 4. Extend the axis on which both of the diameters to point C, which is also the intersection between that axis and a line segment tangent to circle A at E, and circle B at D. Find EA.

[hide="My First solution/Book solution/Obvious one"]Draw a rectangle within triangle ACE with vertices at B, E, D, and the midpoint of AE, let's call it F. Using the pythagorean theorem, we find the FB is 8$ \sqrt{2}$. Using similarity, we prove that CE = 2 ED, and since ED = FB, CE = 16$ \sqrt{2}$[/hide]

But then, I tried power of a point:

[hide="Is there a mistake here?"]
Since DC is a tangent to Circle B at D, radius BD and DC form a right angle at D. Thus, by the pythagorean theorem, we have:       $ BD^2 \plus{} DC^2 \equal{} BC^2$
Since BD = 4,   $ 4^2 \plus{} DC ^2 \equal{} BC^2$
By power of a point, $ DC^2 \equal{} 2BD(BC \minus{} BD) \equal{} 8(BC \minus{} 4)$
Substituting in yields $ 4^2 \plus{} 8BC \minus{} 32 \equal{} BC^2$
Solving for BC yields BC = 4. But this is obviously impossible! What did I do wrong?[/hide]

I don't really understand how you're applying Power of a Point.  Applying it to point $ C$ with respect to circle $ B$ gives $ DC^2 \equal{} (BC \plus{} BD)(BC \minus{} BD)$, which is exactly what the Pythagorean Theorem tells you.  Read the statement of Power of a Point again.

Oh wait.... :blush: It's the entire segment times the part of the segment inside the circle. Alright ,thanks!

[quote="georgiamathdude"]It's the entire segment times the part of the segment inside the circle. [/quote]

It's not, which is what I'm trying to tell you.

The [i]power[/i] of a point $ P$ relative to a circle $ \mathcal{C}$ is defined as follows:  draw a line through $ P$ intersecting $ \mathcal{C}$ in points $ A, B$ (which may be the same).  Then the power of $ P$ is $ PA \cdot PB$.

What Power of a Point says is that the above product is the same regardless of what line you draw; in other words, that the power is a well-defined quantity.


geometry
function
linear algebra
matrix
vector
trigonometry
circumcircle

$ 1)$How many ways are there to draw $ n \minus{} 3$ diagonals of regular $ n$-gon such that none of them intersect each other, and each of triangles have a common side with $ n$-gon.

$ 2)$let $ I_{a}$ be excenter of triangle $ ABC$ for $ A$.excircle of triangle $ ABC$ for $ A$ is tangent to $ AB$ and $ AC$ at point $ B'$ and $ C'$.let $ P$ and $ Q$ be intersects of lines $ BI_{a}$ and $ CI_{a}$ with $ B'C'$ respectively.and let $ M$ be intersect of $ PC$ and $ BQ$.prove that if $ L$ be foot of the prependicular drawn from $ M$ to $ BC$ then $ ML \equal{} r$(radius of incircle of triangle $ ABC$)

$ 3$)let $ a,b,c,d$ be real numbers such that one of $ c$ or $ d$ aren't zero.let $ f: R\rightarrow R$ suchthat $ f(x) \equal{} \frac {ax \plus{} b}{cx \plus{} d}$ and there isn't real $ x$,$ f(x) \equal{} x$.show that if there is a real $ a$ such that $ f(f(\dots f(a)\dots) \equal{} a$($ 1387$times) then for all $ x$ we have $ f(f(\dots f(x)\dots) \equal{} x$($ 1387$times)

Wow nice problems again this time... i guess there would be another challenging Combinatorial problem tommorow... :)

tanx Sepehr for posting the problems  :)

Composition of functions of the form $ \frac{ax\plus{}b}{cx\plus{}d}$ yields a function of the same form, since
\[ \frac{a'\frac{ax\plus{}b}{cx\plus{}d}\plus{}b'}{c'\frac{ax\plus{}b}{cx\plus{}d}\plus{}d'}\equal{}\frac{(aa'\plus{}cb')x\plus{}(ba'\plus{}db')}{(ac'\plus{}cd')x\plus{}(bc'\plus{}dd')}\equal{}\frac{a''x\plus{}b''}{c''x\plus{}d''}.\]
In fact, we see trivially that
\[ AA'\equal{}\begin{pmatrix}
a&b\\
c&d\end{pmatrix}\begin{pmatrix}a'&b'\\
c'&d'\end{pmatrix}\equal{}\begin{pmatrix}a''&b''\\
c''&d''\end{pmatrix}\equal{}A''.\]
Now, if $ f(x)\equal{}x$ for some real $ x_0$, then $ \frac{ax_0\plus{}b}{x_0}\equal{}cx_0\plus{}d\equal{}\lambda$ is real, or $ A\begin{pmatrix}x\\1\end{pmatrix}\equal{}\lambda\begin{pmatrix}x\\1\end{pmatrix}$, ie, vector $ \begin{pmatrix}x\\1\end{pmatrix}$ is an eigenvector of matrix $ A$ with real eigenvalues.  Therefore, $ A$ does not have real eigenvalues, but its eigenvalues are the roots of a second degree equation, hence two complex conjugate numbers, which must be the 1387-th root of a real number so that $ A^{1387}$ does have real eigenvalues in order to have a solution for $ f(f(...f(x_0)...))\equal{}x_0$.  Then, $ A$ has two identical eigenvalues (both are real, and both are complex conjugates of each other), or we can write $ A^{1387}\equal{}\rho I_2$, where $ I_2$ is the $ 2\times2$ identity matrix, and $ f(f(...f(x)...))\equal{}x$ identically, qed.

Please post them also in the appropriate forums and write down here the corresponding links.

It is well known that $ AB' \equal{} AC' \equal{} \frac {a \plus{} b \plus{} c}{2} \equal{} 4R\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2}$, and that $ B'I_a \equal{} C'I_a \equal{} AB'\tan\frac {A}{2} \equal{} 4R\sin\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2}$.  Since $ B'I_a \equal{} BI_a\sin\left(\frac {\pi \minus{} B}{2}\right) \equal{} BI_a\cos\frac {B}{2}$, then $ BI_a \equal{} 4R\sin\frac {A}{2}\cos\frac {C}{2}$, and similarly $ CI_a \equal{} 4R\sin\frac {A}{2}\cos\frac {B}{2}$.  Now, $ \angle BB'P \equal{} \angle AB'C' \equal{} \frac {\pi \minus{} A}{2}$, or $ BP \equal{} \frac {BB'\cos\frac {A}{2}}{\cos\frac {C}{2}} \equal{} \frac {BI_a\sin\frac {B}{2}\cos\frac {A}{2}}{\cos\frac {C}{2}} \equal{} a\sin\frac {B}{2}$, and $ CP$ is the altitude from $ C$ in $ BCI_a$.  Similarly, $ BQ$ is the altitude from $ B$, and $ M$ is the orthocenter of $ BCI_a$.  Since the distance from orthocenter to side $ BC$ is $ 2\rho\cos(\angle CBI_a)\cos(\angle BCI_a) \equal{} 2\rho\sin\frac {B}{2}\sin\frac {C}{2}$, where $ \rho \equal{} \frac {BI_a}{2\cos\frac {C}{2}} \equal{} 2R\sin\frac {A}{2}$ is the circumradius of $ BCI_a$, and $ LM \equal{} 4R\sin\frac {A}{2}\sin\frac {B}{2}\sin\frac {C}{2} \equal{} r$, qed.


geometry
ratio
AMC 10
algebra
system of equations

Here are three AMC 10 questions I didn't get perfectly:

18. Through a point on the hypotenuse of a right triangel, lines are drawn parallel to the legs of the triangle to that the triangle is divided into a square and two smaller right triangles. The area of one of the two smaller right triangles is m times the area of the square. The ratio of the area of the other small right triangel to the area of the square is

A) $\frac{1}{2m+1}$

B) $m$

C) $1-m$

D) $\frac{1}{4m}$

E) $\frac{1}{8m^2}$

21. If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?

I. All aligators are creepy crawlers.

II. Some ferocious creatures are creepy crawlers.

III. Some alligators are not creepy crawlers

A) I only

B) II only

C) III only

D) II and III only

E) none must be true

22. One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

A) 3

B) 4

C) 5

D) 6

E) 7

Thanks everyone!

[quote="236factorial"]Here are three AMC 10 questions I didn't get perfectly:

18. Through a point on the hypotenuse of a right triangel, lines are drawn parallel to the legs of the triangle to that the triangle is divided into a square and two smaller right triangles. The area of one of the two smaller right triangles is m times the area of the square. The ratio of the area of the other small right triangel to the area of the square is

A) $\frac{1}{2m+1}$

B) $m$

C) $1-m$

D) $\frac{1}{4m}$

E) $\frac{1}{8m^2}$

[/quote]

Call the area of the square $x$, that means that lenght of the sides of the square are $\sqrt{x}$. Since the area of the triangle is $m$ times the square, its area is $mx$, and since one side of the square is also a side of the triangle, the legs of the triangle are $2m\sqrt{x}$ and $\sqrt{x}$. Since the smaller triangle is similar to the larger triangle, the length of the legs of that triangle are $\sqrt{x}$ and $\frac{\sqrt{x}}{2m}$, so the area of the smaller triangle is $\frac{x}{4m}$, so the answer is $\frac{1}{4m}$ or (D)

18.
(For the labels of the points, see the attached figure.)

Let's say $BF=p$ and $DF=q$. Then $DE=DF=q$.
Since $\triangle BDF\sim\triangle DAE$, $\frac{AE}{DE}=\frac{DF}{BF}$.
So, $AE=q\cdot\frac{q}{p}=\frac{q^2}{p}$

For our convenience, let's multiply $p$ to all of these.
That gives $BF=p^2$, $DF=DE=pq$, $AE=q^2$.

Then,
$m=\frac{\frac{1}{2}(pq)(q^2)}{(pq)^2}=\frac{\frac{1}{2}pq^3}{p^2q^2}=\frac{q}{2p}$

The ratio we want to find is:
$\frac{\frac{1}{2}(p^2)(pq)}{(pq)^2}=\frac{\frac{1}{2}p^3q}{p^2q^2}=\frac{p}{2q}$

So, $\frac{p}{2q}=\frac{1}{\frac{2q}{p}}=\frac{1}{4\cdot\frac{q}{2p}}=\frac{1}{4m}$

$\therefore \boxed{D}$

EDIT- I guess I was late :P Besides, the solution GoBraves posted was easier...

21 should be D. as for the coffee one, i've narrowed it down to C,D, or E.

[quote="236factorial"]
21. If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?

I. All aligators are creepy crawlers.

II. Some ferocious creatures are creepy crawlers.

III. Some alligators are not creepy crawlers

A) I only

B) II only

C) III only

D) II and III only

E) none must be true

[/quote]

Its obvious that II is correct, so we can eliminate E), and the question asks which MUST be true, so with the information we're given, I dont think that we can say that I and III always have to be true, so I think the answer is B)

How do you know 21 is D?  We know that some creepy crawlers are alligators.  So, there could be a case where all alligators are creepy crawlers, and there are some other stuff that are creepy crawlers too.  I think the answer is B.

[quote="236factorial"]22. One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
[/quote]

[hide]She drank $8=c+m$. $c=\frac{C}{4}$ and $m=\frac{M}{6}$.

$8=\frac{3C+2M}{12}$
$3C + 2M =96$.

$C+M=8x$, where $x$ is the # of people, so substituting we see $16x+C=96$.

Time for some trial and error...

A) 3 - $(C,M)=(48,-24)$
B) 4 - $(64, -\text {something })$
C) 5 - $(16, 24)$. It works!
D) 6 - $(0, \text {something})$
E) 7 - $(-\text {something}, -\text {something else})$

$\therefore C$.[/hide]

Thanks chess, that was the clear answer I was looking for  :P :D
The test solution was short, but it was unclear.

[quote="236factorial"]Thanks for the help on number 18.

For number 21, the answer is B. I and III contradict each other, so they can't both be true. II is obviously true, so why is III false?[/quote]

It says which must be true.  There's not enough information given to determine three is true, only enough to verify statement two.

"all alligators are ferocious creatures and some creepy crawlers are alligators."  Just because some creepy crawlers are alligators doesn't mean some alligators aren't creepy crawlers.

Okay.  For 18, I plugged in values for the sides of the square and then solved to get D.  21, I got B.  22 I'm working on.

I did something like this for 22 (Sorry if it's a bit messy):
Okay.  I decided to do the system of equations, but you have to 2 equations, so here.
$6c + 4m = x$
$5c + 3m = x - 8$
Then
$-18c - 12m = -3x$
$20c + 12m = 4x -32$
----------------------------
$2c = x - 32$...
I did it somehow, but it's getting late and I don't remember how I did it.  I might get on on Monday to edit it, but if someone could fill in the holes before me, that might be nice.


limit
real analysis
real analysis unsolved

[quote="Xingyuan"]$ \lim_{(x,y)\rightarrow(0,0)}\frac {\sin(x^3 \plus{} y^3)}{x^2 \plus{} y^2}$ :D

$ \lim_{(x,y)\rightarrow(0,0)}\frac {x^2y^2}{x^3 \plus{} y^3}$ :lol:[/quote]

For the first one note $ |\sin(x^3 \plus{} y^3)| \leq |x|^3 \plus{} |y|^3$. For the second one, consider the behavior near the line $ y \equal{} \minus{} x$.


abstract algebra
superior algebra
superior algebra unsolved

Let K and M be two abelian groups |K|=k, |M|=m, (k,m)=1. Then every extension G of K by M is a semi-direct product.
I dont understand what "extension of K by M" exactly means. May be it is clean for anyone?

I'm pretty sure it's this: $ G/K\equal{}M$.

Usually, it is the other way round: $ G/M\equal{}K$. owk


AMC
AIME
USAMTS
USA(J)MO
USAMO
AMC 12 A
AMC 10

Just wondering..

I'm just taking AMC 12 A and look over some AMC's frequently tested concepts. This will be my only prep for AMC this year so I'm pretty worried but let's hope things'll work out.  :)

My father is a bit obsessed, so I did 3 AMC tests today, and I will probably do 3 or 4 tomorrow.  Between them, relaxing things...

you gotta prepare your lucky pencil...
you know... write a-e on the each side of a wooden pencil, then on the test...roll. What ever it lands on, it usually is right for me.

[quote="Altheman"]i always try to not get enough sleep because that way i can finish my english homework[/quote]

amen brotha

sleep, eat, go to school, eat some more, work on my history project, watch a movie, sleep again, take the test

Take a practice contest that's way harder than the AMC 10, so when I open up the contest I think "this is easy..."  :D

Stress out all the way about two projects I have to take care of on top of new planning which has popped up for HMMT...

Well, I didn't have school today, so I slept from 9PM to noon today. Tomorrow, I'll wake up, shower, drink a freakin' huge glass of orange juice, and head to school (we're on vacation, so only the hardcore math nerds like me will be there). There I'll take the 12, miss the cutoff by one problem, curse loudly, and go back home.
Then I'll work out the solutions to all the problems, post my solutions on AOPS, and hope that mysmartmouth likes them enough to database-ify them.

Work on my most recent mock AIME installment :).

Sharpen pencils, procrastinate on my hw by staring out the window.

I have a feeling I am going to do decent, qualify for the AIME in 3 different ways (USAMTS, AMC 10, AMC 12), and then screw up the AIME and miss the cutoff for USAMO by a slim margin.

I have a feeling I will have a good day tomorrow and make the AIME cutoff on my 2nd try.   :D


Hello,

$ i)$ Solve in $ \mathbb{R}$ the equation :  $ \arcsin (2x)\equal{}\arcsin (x)\plus{}\arcsin (x\sqrt{2}).$

$ ii)$Solve in $ \mathbb{R}$ the equation :  $ \arctan(x\minus{}3)\plus{} \arctan(x)\plus{}\arctan(x\plus{}3)\equal{}\frac{5\pi}{4}$

[quote="linda2005"]Hello,

$ i)$ Solve in $ \mathbb{R}$ the equation :  $ \arcsin (2x) = \arcsin (x) + \arcsin (x\sqrt {2}).$

$ ii)$Solve in $ \mathbb{R}$ the equation :  $ \arctan(x - 3) + \arctan(x) + \arctan(x + 3) = \frac {5\pi}{4}$[/quote]

[b]i)[/b] Let $ p = \arcsin (2x), q = \arcsin (x) , r = \arcsin (x\sqrt {2})$
Then $ \begin{cases} sin p sin q = sin^2 r \\
sin r = sin \sqrt {2} sin q \end{cases}$

$ \Leftrightarrow \begin{cases} sin (q + r) sin q = sin^2 r \\
sin r = sin \sqrt {2} sin q \end{cases}$

$ \Leftrightarrow sin^2 q cos r + sin q cos q sin r = sin^2 r \\
sin r = sin \sqrt {2} sin q \end{case}$

$ \Rightarrow sin^2 q (cos r + \sqrt {2} cos q) = 2sin^2 q$

So $ sin q = 0 \Leftrightarrow x=0$ as a solution or

$ cos r + \sqrt {2} cos q = 2$

$ cos^2 r =(2- \sqrt{2} cos q)^2=1- (sin r)^2=1-2sin^2 q=2cos^2 q-1$

So $ 4-4 \sqrt{2} cos q + 2 cos^2 q=2cos^2 q-1$

$ \Leftrightarrow cos q=\frac{5}{4 \sqrt{2}}$
$ \Rightarrow x=sin q= \pm \frac{1}{4}\sqrt{\frac{7}{2}}$

[quote="linda2005"]hello,

for ii)  x=5 ?  :wink:[/quote]

Sorry, forgot that $ tan(p\plus{}q)$ can be negative.

hello, for i) the solutions are $ x_1\equal{}\minus{}\frac{1}{4}\sqrt{\frac{7}{2}}$, $ x_2\equal{}0$, $ x_3\equal{}\frac{1}{4}\sqrt{\frac{7}{2}}$
Sonnhard.

[hide="#2"]Using $ \arctan a \plus{} \arctan b \equal{} \arctan {a \plus{} b\over 1 \minus{} ab}$, we get

$ {5\pi\over 4} \equal{} \arctan{2x\over 1 \minus{} (x^2 \minus{} 9)} \plus{} \arctan x \equal{} \arctan\frac {x \plus{} {2x\over 10 \minus{} x^2}}{1 \minus{} {2x^2\over 10 \minus{} x^2}} \equal{} \arctan{12x \minus{} x^3\over 10 \minus{} 3x^2}$

Hence the necessary condition for $ x$ is $ 12x \minus{} x^3 \equal{} 10 \minus{} 3x^2\iff x^3 \minus{} 3x^2 \minus{} 12x \plus{} 10 \equal{} 0$

We can write this as $ x^3 \minus{} 5x^2 \plus{} 2x^2 \minus{} 10x \minus{} 2x \plus{} 10 \equal{} 0\iff (x \minus{} 5)(x^2 \plus{} 2x \minus{} 2) \equal{} 0$

Therefore $ x\in\{ 5, \minus{} 1\pm\sqrt {3}\}$

Now we must check all the solutions, bearing in mind that only the principal values of $ \arctan$ must be taken, and those lie in $ \left( \minus{} {\pi\over 2},{\pi\over 2}\right)$

For $ x \equal{} 5$, all three terms on the RHS are positive, hence their sum exceeds $ \pi$ (which is shown by bearing in mind that $ \arctan 2\plus{}\arctan 5\equal{}\pi\minus{}\arctan{2\cdot 5\over 1\minus{}2\cdot 5}$)

For $ x \equal{} \minus{} 1 \minus{} \sqrt {3}$, we have two negative and one positive term, hence their sum can't exceed $ {\pi\over 2}$

For $ x \equal{} \minus{} 1 \plus{} \sqrt {3}$, we have two positive and one negative term, hence their sum can't exceed $ \pi$

Therefore the only solution is $ x \equal{} 5$[/hide]


[quote="cdymdcool"]We learned that a golf ball with one unit of momentum when striking a bowling ball and rebounding will impart a momentum to the bowling ball greater than its initial 1 unit.  How is this possible if momentum is conserved?[/quote]

well, some of the momentum of the bowling ball transfer to the smaller golf ball.  Momentum is still conserved, it's just that the golf ball has a higher velocity (p is the same, if m is smaller, v has to be bigger)...and the bowling ball isn't as affected.

Here another fun one that I had problem wrapping my brain around. If I throw a ball at the wall..and it bounce back..it recieved 2x the original speed. (which arises from sign difference)

[quote="cdymdcool"]We learned that a golf ball with one unit of momentum when striking a bowling ball and rebounding will impart a momentum to the bowling ball greater than its initial 1 unit.  How is this possible if momentum is conserved?[/quote]When the golf ball bounces, it will come out of the bounce with some momentum, different from the momentum it had before.
For us (3-dimensional beings), momentum is a vector in a 3-D space. Assuming that all the movement was along a line (1-D situation), we can just use signs for the direction. A golf ball moving to the right with a momentum of $ \plus{}1$, hitting a stationary bowling ball and rebounding with the same speed (momentum = $ \minus{}1$, to the left), will transfer $ \plus{}2$ units of (positive, towards the right) momentum to the bowling ball. Before collision, the total momentum was $ \plus{}1$ (or 1 towards the right). After collision, the bowling ball has a momentum of $ \plus{}2$ (2 to the right) and the golf ball has a momentum of $ \minus{}1$ (1 to the left), for a total momentum of
$ \plus{}2 \plus{} (\minus{}1) \equal{} \plus{}1$,
the same as before the collision.

[quote="cdymdcool"]We learned that a golf ball with one unit of momentum when striking a bowling ball and rebounding will impart a momentum to the bowling ball greater than its initial 1 unit.  How is this possible if momentum is conserved?[/quote]

It is possible precisely because momentum is conserved. (The momentum of the entire system of particles is conserved whereas the momenta of the individual particles is free to change.)

The conservation of momentum equation is this:
\[ \underbrace{\vec{p}_{g_1} \plus{} \vec{p}_{b_1}}_{\text{before collision}} \equal{} \underbrace{\vec{p}_{g_2} \plus{} \vec{p}_{b_2}}_{\text{after collision}}
\]
Given the above equation,
\[ \vec{p}_{b_2} \minus{} \vec{p}_{b_1} \equal{} \vec{p}_{g_1} \minus{} \vec{p}_{g_2}
\]
The projection of the golf ball's rebound path along the line of its approach to the bowling ball is , $ \vec{p}_{g_2} \equal{}  \vec{p}_{g_1}\cos\theta$ and therefore,
\[ \vec{p}_{b_2} \minus{} \vec{p}_{b_1} \equal{} \vec{p}_{g_1}(1 \minus{} \cos\theta)
\]
$ \theta$ is free to vary over the interval $ [\frac {\pi}{2},\,\pi]$.

Thus
\[ |\vec{p}_{g_1}|\le |\Delta \vec{p}_{b}|\le 2|\vec{p}_{g_1}|
\]

[quote="cdymdcool"]
We learned that a golf ball with one unit of momentum when striking a bowling ball and rebounding will impart a momentum to the bowling ball greater than its initial 1 unit.  How is this possible if momentum is conserved?

[/quote]

$ \text{The momentum is conserved.}
\\
\\
m \equal{} \text{golf ball}, \; M\equal{}\text{bowling ball}
\\
\\
M(0) \plus{} m(v_i) \equal{} M(v) \plus{} m(v_f)
\\
\\
v_f \equal{} \minus{}kv_i
\\
\\
\text{where "k" is a positive constant that depends on the elasticity.}
\\
\text{Hence,}
\\
p\equal{}mv_i \equal{} Mv \minus{} kmv_i
\\
\\
Mv \equal{} m(v_i \plus{} kmv_i)$

So, you see the momentum imparted to the bowling ball is greater than the initial momentum of the golf ball, although the momentum is conserved.

[quote="Rajiv"]

$ \text{The momentum is conserved.} \\
 \\
m \equal{} \text{golf ball}, \; M \equal{} \text{bowling ball} \\
 \\
M(0) \plus{} m(v_i) \equal{} M(v) \plus{} m(v_f) \\
 \\
v_f \equal{} \minus{} kv_i \\
 \\
\text{where "k" is a positive constant that depends on the elasticity.} \\
\text{Hence,} \\
p \equal{} mv_i \equal{} Mv \minus{} kmv_i \\
 \\
Mv \equal{} m(v_i \plus{} kmv_i)$

So, you see the momentum imparted to the bowling ball is greater than the initial momentum of the golf ball, although the momentum is conserved.[/quote]

I think you mean
\[ v_f \minus{} v \equal{} \minus{} kv_i
\]

[quote="Immanuel Bonfils"]
I think you mean
\[ v_f \minus{} v \equal{} \minus{} kv_i
\]

[/quote]

No. The final velocity, $ v_f \equal{} \minus{}kv_i$.

eg, if a ball collides inelastically with a wall, then, $ k\equal{}0$.

You said it right: [b] with a wall [/b]

In the present case it's with a moving ball; you have to take relative velocities.

[quote="Immanuel Bonfils"]
In the present case it's with a moving ball; you have to take relative velocities.

[/quote]

Relative velocities?

The golf ball collides with the bowling ball and moves backwards with some velocity $ kv_i, \; k<1$.

[quote="Immanuel Bonfils"]Let's see if we are spiking (rather writing...) the same English; is your k the COR?[/quote]

What is COR?

Here, the $ e \text{ (COR as you call it)}$ is $ 1$.

But that doesn't mean $ k \equal{} 1$.
The final velocities, $ v'_1 \; \& \; v'_2$, of two bodies after perfectly elastic collision is:

$ v'_1 \equal{} \left(\frac {m_1 \minus{} m_2}{m_1 \plus{} m_2}\right)v_1 \plus{} \left(\frac {2m_2}{m_1 \plus{} m_2}\right)v_2 \\
 \\
 \\
v'_2 \equal{} \left(\frac {2m_1}{m_1 \plus{} m_2}\right)v_1 \minus{} \left(\frac {m_1 \minus{} m_2}{m_1 \plus{} m_2}\right)v_2$

The $ k$ here is the $ \frac {v'_1}{v_1}.$

[quote="Immanuel Bonfils"]If so, what kind of a constant is k ? It can be positive or negative, it depends of the masses involve, it depends of the velocity of the other body...[/quote]

As Rajiv originally defined it:

[quote="Rajiv"]\[ v_f = - kv_i \\
\\
\text{where "k" is a positive constant that depends on the elasticity.}\][/quote]

[quote="Then you"]Let's see if we are spiking (rather writing...) the same English; is your k the COR?[/quote]

I submit that no, $ k\neq COR$.

The COR is defined as:
\[ \mathrm{e}\stackrel{\mathrm{def}}=\frac{v_{2_f} -v_{1_f}}{v_{1_i} - v_{2_i}}\].

Substituting into this definition:

\[ \mathrm{e} =\frac{v_{2_f} + k v_{1_i}}{v_{1_i} - v_{2_i}}\]

Further, in this case, $ v_{2_i} = 0$, and therefore,
\[ begin{align*}\mathrm{e} \&=\frac{v_{2_f} + kv_{1_i}}{v_{1_i}}\\ \&= \frac{v_{2_f}}{v_{1_i}} + k\end{align*}\]

\[ \therefore k =\mathrm{e} - \frac{v_{2_f}}{v_{1_i}}\]

\[ k =\mathrm{e}\, \iff \textrm{ the bowling ball is an immovable object.}\]

In as much as Rajiv has idealized the bowling ball as an immovable object, then $ k =\mathrm{e}$, the COR.

And, as the bowling is movable;
[quote="cdymdcool"]
We learned that a golf ball with one unit of momentum when striking a bowling ball and rebounding will impart a momentum to the bowling ball greater than its initial 1 unit. How is this possible if momentum is conserved?[/quote]

[quote="Immanuel Bonfils"]
it depends on the velocity of the other body...[/quote]


These were the 5th & 6th problems on a local contest in Nassau county (NCIML).  Students could not use a calculator and the time limit was 10 minutes:

$3^{11}-3^{9}+3^{7}-3^{6}= 81x \text{, compute }x$

The number shown, with the hundreds' digit missing,
62894_52, is divisible by 48. Determine the missing digit.

[hide="#6"]To be divisible by 48, a number must be divisible by 16 and 3. To be divisible by 3, a numbers digits must all add up to a multiple of 3. To be a multiple of 16, the last 4 digits of a number must be divisible by 16. Adding the digits that we have, we get 36. This means that the missing digit can be 0,3,6, or 9. Then, realizing that 4000 is divisible by 16, we realize that we only need the last 3 digits to be divisible by 16. Plugging in the numbers, we find that the answer is [b]3[/b][/hide]

i_like_pie, I like your method.
You could also factor in #5 into $3^{6}(3^{5}-3^{3}+3^{1}-3^{0})$.

[hide]5) $48=2^{4}\cdot 3$

therefore the digits have to add up to a multiple of 3 and the last 4 digits have to be divisible by 16. The digits that we have now add up to 36, so that means the missing digit is either 0,3,6,or 9. Plugging in and dividing we find the missing digit is $3$.

6)

$3^{11}-3^{9}+3^{7}-3^{6}=81x$
$3^{6}(3^{5}-3^{3}+3^{1}-1) =3^{4}\cdot x$
$3^{6}(243-27+3-1)=3^{4}\cdot x$
$3^{2}(218) =x$
$x=1962$

[/hide]

The second NCIML contest was held yesterday.
Students were given about 5 minutes per problem.  Here are two that can be solved VERY quickly if you approach them the right way.

10. Compute the arithmetic mean of all the odd multiples of 3 between 70 and 250.

12. $\text{ If }x+y = 5 \text{ and }xy=3 \text{, compute the value of }\frac{1}{x^{2}}+\frac{1}{y^{2}}$

You really should make a new topic.

[hide="10"]The terms clearly form an arithmetic sequence.  To take the average of all the terms in an arithmetic sequence, take the average of the first term and the last terms.
The first term is 75.  The last is 249.
$\frac{75+249}{2}=\boxed{162}$[/hide]

[hide="12"]
$\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{x^{2}+y^{2}}{(xy)^{2}}=\frac{x^{2}+y^{2}}{9}$
$x+y=5$
$xy=3$
$(x+y)^{2}=25=x^{2}+y^{2}+2xy$
$x^{2}+y^{2}+2(3)=25$
$x^{2}+y^{2}=19$
$\frac{x^{2}+y^{2}}{(xy)^{2}}=\boxed{\frac{19}{9}}$[/hide]

[quote="Thorn"]Nicely done!

They haven't compiled all of the results yet, but I think very few students had both of these problems correct.

Would anyone like to post similar problems on these two themes?[/quote]

Given $p$ and $q$ are the roots of $x^{2}+35x-72$, compute $\frac{1}{p}+\frac{1}{q}$

[hide]The sum of the roots are $-35$ and the product is $-72$.  This can easily be found by sum and product of roots.  Now observe: $\frac{1}{p}+\frac{1}{q}=\frac{q}{pq}+\frac{p}{pq}=\frac{p+q}{pq}=\frac{-35}{-72}=\boxed{\frac{35}{72}}$.[/hide]

New problems should generally go in new threads.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

ComA is the matrix wich verify $A.(ComA)^t = detA.I_n$
I think in Romania you write it as $A^* = (ComA)^t $
$A.A^* = detA.I_n $

Well, the characteristic polynomial of $A$ is $A^3-TrA\cdot A^2+TrA^*\cdot A-detA\cdot I_3=0_3$, so the conclusion follows.


inequalities
combinatorics unsolved
combinatorics

Seventeen football fans were planning to go to Korea to watch the World Cup football match. They selected 17 matches. The conditions of the admission tickets they booked were such that

- One person should book at most one admission ticket for one match;

- At most one match was same in the tickets booked by every two persons;

- There was one person who booked six tickets.

How many tickets did those football fans book at most?

Let $ F_{1},...,F_{17}$ be fans. Each of them visit some match $ M_{1},...,M_{17}$ and say $ F_{1}$ visit 6 match's.
Since every two visit at most 1 match in comon, we have
\[ \sum_{i\equal{}1}^{17}{f_{i}\choose 2}\leq{17\choose 2}\]
where $ f_{i}$ means number of games $ F_{i}$ visit.
We are interested in maximum value of $ n \equal{}\sum_{i \equal{} 1}^{17}f_{i}$.
By Jensen inequality we have:
\[ {{1\over 16}(n\minus{}6)^{2}\minus{}(n\minus{}6)\over 2}\leq{17\choose 2}\minus{}{6\choose 2}\]
So $ n\leq 76$. Now I can't find nor configuration for $ n \equal{} 76$ nor better uper bound.

[quote=jmerry][quote="Number1"]Can somebody find something better?[/quote]
$ 71$ is best possible.

Here's a configuration for 71:
$ 1,2,3,4,5,6$
$ 1,7,11,15$
$ 1,8,12,16$
$ 1,9,13,17$
$ 2,10,14,16$
$ 2,8,13,15$
$ 3,9,14,15$
$ 3,10,11,17$
$ 4,8,14,17$
$ 4,9,11,16$
$ 4,10,12,15$
$ 5,7,14$
$ 5,15,16,17$
$ 6,7,8,9,10$
$ 6,11,12,13,14$

The missing pairs are $ (8,11),(9,12)$, and $ (10,13)$.

A proof that we can't get $ 72$:
Using Number1's notation, the only ways $ 72$ is possible are if $ f_{i}'\equal{}4$ for two $ i$, and $ f_{i}'\equal{}3$ for all other $ i$, or if $ f_{i}'\equal{}4$ for three $ i$, $ f_{i}'\equal{}2$ for one $ i$, and $ f_{i}'\equal{}3$ for all other $ i$. Suppose we have such a configuration.
WLOG, order the fans so that the $ f_{i}'$ are in nonincreasing order.

Case 1 ($ 4\cdot 2\plus{}3\cdot 14$): Now, for each $ j\in\{7,8,\cdots,17\}$, let $ a_{j}$ be the number of fans attending match $ j$. We have $ a_{j}\le 5$ for each $ j$, since each fan attending match $ j$ attends at least two other matches from the $ 11$, and these do not overlap. In addition, if $ a_{j}\equal{}5$, match $ j$ was not attended by fans $ 2$ or $ 3$; with ten matches to spread among five fans, we can't afford to put three matches with one fan. Since $ \sum_{j\equal{}7}^{17}a_{j}\equal{}50$, at least six of the $ a_{j}$ are equal to $ 5$. The two fans with $ f_{i}'\equal{}4$ cover at least seven matches between them, leaving only four other $ j$ that can possibly have $ a_{j}\equal{}5$. This is a contradiction, and this configuration is impossible.

Case 2 ($ 4\cdot 3\plus{}3\cdot 12\plus{}2\cdot 1$): Define $ a_{j}$ as before. With only one fan that doesn't attend at least three of the $ 11$ matches, we still have $ a_{j}\le 5$ everywhere, and thus at least six of the $ a_{j}$ are equal to $ 5$. The only ways $ a_{j}$ can be $ 5$: either match $ j$ was attended by none of the fans $ 2,3,4$ or match $ j$ was attended by one of those fans and also by fan $ 17$. There are at most two matches in the first category (at least $ 4\cdot 3\minus{}3\equal{}9$ matches covered by $ 2,3,4$) and at most two in the second (everything fan $ 17$ attended), for a total of four. This is a contradiction, and this configuration is also impossible.

[Edit- proof of maximality added][/quote]

Why there cant be one 5 and rest 3-s?

Seventeen matches. If one match is attended by five and the rest by three each, that's a total of $5+3\cdot 16=53$ tickets. Possible, but nowhere close to the maximum we're looking for.

I had the same construction as jmerry but using different cases:
Fill in the six games that person $17$ went to last. In total at most $22$ tickets can be bought for these games.

If the most popular game was visited by $>6$ people then you can use simple bounding

If the most popular game was visited by $6$ people then we get the construction above.

If the most popular game was visited by $5$ people you get that at most $6$ games can be visited by five people by breaking up into the following cases for $3$ games visited by five people:
[hide]{1,2,3,4,5}{6,7,8,9,10}{11,12,13,14,15}
{1,2,3,4,5}{1,7,8,9,10}{11,12,13,14,15}
{1,2,3,4,5}{1,7,8,9,10}{1,12,13,14,15}
{1,2,4,5,6}{1,3,7,8,9}{2,3,10,11,12}
{1,2,4,5,6}{1,3,7,8,9}{2,10,11,12,13}[/hide=cases]
After working systematically with these cases you get at most $70<71$ tickets sold in total. So the answer is $71$.
*This type of question was very popular from 2000-2003 similar to USAMO 2001 P1 and IMO 2001 P3. The only difference is that this question required a very long BASH.


number theory
greatest common divisor
real analysis
real analysis unsolved

The sequence$(a_{n})$ of positive integers is determined by its first two members and
       the rule$a_{n+2}= (a_{n+1}+a_{n})/gcd(a_{n}, a_{n+1})$. For which values of a1 and a2 is it bounded?

now i've done this much. if at any stage $gcd(a_{n},a_{n+1})=1$, then after that the sequence will keep increasing, its pretty much something like the fibonacci. so gcd of a1 and a2>1. also i found that a1=a2=2 gives that the sequence is bounded. are there any other values? i have a feeling there is no other, but i could be wrong. could someone give me hints? i hope this qn is posted in the right forum, or should it be in the number theory section? :maybe:

Suppose $d>1$ divides both $a_{n+1}$ and $a_{n+2}$. Then it also divides $a_{n}+a_{n+1}$ and consequently $a_{n}$. Therefore, $\gcd(a_{n},a_{n+1})$ does not increase with $n$. Being a sequence of positive integers, it must stabilize eventually. So we can assume $\gcd(a_{n},a_{n+1})=D$ for all sufficiently large $n$. This is only possible for $D=1$ or $D=2$; otherwise the sequence $a_{n+2}=(a_{n+1}+a_{n})/D$ goes to $0$. With $D=2$, each subsequent term is the average of two preceding ones. For a sequence of integers, this can only happen when all the terms are the same.

Thus, $a_{n}=2$ for all large $n$ if the sequence is bounded. It should not be hard to show that in this case $a_{n}=2$ for [b]all[/b] $n$.


Hi everyone,

What's a good way to self-study AP Computer Science? My school doesn't offer it.

Get a prep book.  Either the Barron's one, or the Maria Litvin one should be good.  And if you really need it, maybe an introductory book to Java should be good.

I slept in class (almost) everyday, and its not hard at all to study by yourself.

You probably want a good textbook, too. I self-studied the AB portion of it (which is basically everything hard) and got a 5, so it can't be that hard :D

I really recommend getting this:
http://www.ict.org/apcs.html

It is good AP curriculum.  I did all of its lessons and then I got the Maria Litvin review book.  I don't know what I got on the test, but I finished all of the questions to the best of my ability (there were none that i flat out didnt know) so it must have worked fairly well.

i seriously think all you need is a Barron's guide

Barron covers everything, and similar problems showed up in real APs

it covers Map and sets more thoroughly than ICT

map/sets showed up on... say 1.5 out of 4 FR in 2005 ap

[quote]map/sets showed up on... say 1.5 out of 4 FR in 2005 ap[/quote]

Really, I seem to remember using a map/set at least once on every one of the FR questions except the the Tree question. But then again that may have been because I used a map on the salmon question when it was probably easier not to use a map.

i dont rememebr playing with data structures  for salmon question except for traversing some allobject arraylists


Are there there infinitely many $n$ for which $2^{n}-1$ and $3^{n}-1$ are relatively prime?

When $2^n-1$ is prime number, we call it Mersenne prime. It is conjectured that there are infinitely many Mersenne prime(it is not proven yet, I think).

So, if this conjeture is true, your problem will be extremely easy to solve.

Sorry to be thick, but why cant $2^{n}-1$ divide $3^{n}-1$?

(As with n = 1 or 2)

I mean, if $2^{n}-1$ is prime is see no reason to believe that it can't divide $3^{n}-1$, but I'm probably missing something.

Sorry. For some strange reason, I thought $2^n-1 > 3^n-1$ :blush: .

In fact, $2^n-1$ can't divide $3^n-1$.
For example, have a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=2%5En-1+%7C+3%5En-1&t=20518][u]here[/u][/url].


I have pretty good trouble on geometry.

I think it's the one of the weakest part on my math and well, I want to be good at it.

Anyone knows a good site who teaches geometry well other than AoPS class?

I'm trying to learn it..even I have that class this year.. but i can't wait that long..

I'll post this reply as a placeholder for a FAQ on studying geometry I hope to post soon. I'll edit the text here when I have my FAQ ready.


inequalities
search
inequalities proposed

Let $ a,b,x,y,z$ be pozitive real numbers satisfying the condition: $ xy\plus{}yz\plus{}zx\equal{}x\plus{}y\plus{}z.$
Prove that:
$ \frac{x}{(ay\plus{}bz)^2}\plus{}\frac{y}{(az\plus{}bx)^2}\plus{}\frac{z}{(ax\plus{}by)^2}\geq \frac{3}{(a\plus{}b)^2}$

[quote="marin.bancos"]Let $ a,b,x,y,z$ be pozitive real numbers satisfying the condition: $ xy \plus{} yz \plus{} zx \equal{} x \plus{} y \plus{} z.$
Prove that:
$ \frac {x}{(ay \plus{} bz)^2} \plus{} \frac {y}{(az \plus{} bx)^2} \plus{} \frac {z}{(ax \plus{} by)^2}\geq \frac {3}{(a \plus{} b)^2}$[/quote]

$ \Longleftrightarrow \frac {x}{(ay \plus{} bz)^2} \plus{} \frac {y}{(az \plus{} bx)^2} \plus{} \frac {z}{(ax \plus{} by)^2}\geq \frac {3(x\plus{}y\plus{}z)}{(a \plus{} b)^2(xy\plus{}yz\plus{}zx)}$

by holder inequality, we have

$ \frac {x}{(ay \plus{} bz)^2} \plus{} \frac {y}{(az \plus{} bx)^2} \plus{} \frac {z}{(ax \plus{} by)^2}$

$ \equal{}\frac {x^3}{(x(ay \plus{} bz))^2} \plus{} \frac {y^3}{(y(az \plus{} bx))^2} \plus{} \frac {z^3}{(z(ax \plus{} by))^2}$

$ \ge \frac{(x\plus{}y\plus{}z)^3}{(x(ay \plus{} bz)\plus{}y(az \plus{} bx)\plus{}z(ax \plus{} by))^2}$

$ \equal{}\frac{(x\plus{}y\plus{}z)^3}{(a\plus{}b)^2(xy\plus{}yz\plus{}zx)^2}$

so we just need to prove

$ \frac{(x\plus{}y\plus{}z)^2}{xy\plus{}yz\plus{}zx} \ge 3$

that's obvious ture...

so inequality holds...

Thank you, [b]kuing[/b]!
In fact, the inequality without any additional conditions (as you proved) is the following:

Let $ a,b,x,y,z$ be positive real numbers. Prove that:
$ \frac{x}{(ay\plus{}bz)^{2}}\plus{}\frac{y}{(az\plus{}bx)^{2}}\plus{}\frac{z}{(ax\plus{}by)^{2}}\geq \frac{3(x\plus{}y\plus{}z)}{(a\plus{}b)^{2}(xy\plus{}yz\plus{}zx)}$

[b]Some remarks:[/b]
1. It's obvious that:
$ RHS\equal{}\frac{3(x\plus{}y\plus{}z)}{(a\plus{}b)^{2}(xy\plus{}yz\plus{}zx)}\geq \frac{9}{(a\plus{}b)^{2}(x\plus{}y\plus{}z)}\equal{}RHS^*$
2. For $ a\equal{}1,b\equal{}1$ and $ RHS^*$ we get to a weaker inequality, which was disscused at http://www.mathlinks.ro/viewtopic.php?search_id=774358916&t=5084
3. My solution to inequality MB-4  (as I've already said) is based on my lemma, which is at http://www.mathlinks.ro/viewtopic.php?t=310721
$ LHS\equal{}\frac{x^3}{[x(ay\plus{}bz)]^{2}}\plus{}\frac{y^3}{[y(az\plus{}bx)]^{2}}\plus{}\frac{z^3}{[z(ax\plus{}by)]^{2}}\stackrel{My\ lemma}{\geq}$
$ \geq \frac{x\plus{}y\plus{}z}{x(ay\plus{}bz)\plus{}y(az\plus{}bx)\plus{}z(ax\plus{}by)}\cdot\left[\frac{x^2}{x(ay\plus{}bz)}\plus{}\frac{y^2}{y(az\plus{}bx)}\plus{}\frac{z^2}{z(ax\plus{}by)}\right]\equal{}$
$ \equal{}\frac{x\plus{}y\plus{}z}{(a\plus{}b)\cdot(xy\plus{}yz\plus{}zx)}\cdot\underbrace{\left(\frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by}\right)}_{\geq \frac{3}{a\plus{}b}}\geq RHS$


Exista $2007$ puncte pe un cerc de raza $1$ astfel incat lungimea coardei care uneste oricare doua din acestea sa fie numar rational?
Gata, am alimentat suficinet cu probleme. Have fun si astept solutiile voastre:D


logarithms
calculus
integration
number theory open
number theory

Hi this forum was recommended me in a german math forum.

I have a assumption concerning this series:

edit:

now correctly texed:

$ \sum_{n \equal{} 1}^{\infty}\frac{1}{p(p(n))}\equal{} 1$

In words: The sum from n=1 to infinity of 1/p(p(n)), where p(n) gives the n-th prime equals 1!

Numerically i summed it till 11250482, which gives the value: 0.994092770

Does anybody know something about the correct value? That the series converges is clear.
And p(p(11250482)) is nearly 2^32. So i cant calculate further.

Thank you!

:) Please edit your problem . I don't understand wht do you mean :(

p(n) denotes the n-th prime number. He likes to prove the equation

$ \sum_{n\equal{}1}^{\infty}\frac{1}{p(p(n))}\equal{} 1.$

Is it a well-known theory? I've never heard of it. I wonder if it is a true argument...

I guess, if it was well-known, michifold wouldn't have asked.  :wink:

the n-th prime asymptotically is $ n\log{n}$. Hence $ p(p(n)$ is asymptotically $ (n\log{n})*\log{n\log{n}}$.
The sum then is bounded from above by $ n(\log{n})^{2}$, hence we can estimate the integral and get $ \frac{1}{n(\log{n})^{2}}$.

So if you have that the sum up to 11250482 is 0.994092770 , then we can estimate the rest by $ \frac{1}{\log{112550482}}$ which is .061592.
So we know the value is between 0.994092770  and 1.05.

The result that is is actually 1 is beautiful. Is it proven or just conjectured?

michifold is conjecturing it.  Indeed it would be quite beautiful if it were true, but somehow I think it would be too spectacular  :oops:

Hi,

unfortunately, I have to disappoint you: my conjecture is wrong. One of the matheboard.de forum (Arthur Dent) wrote a program, which calculated the sum till n=261150360 , p(p(261150360))=137438950763

$ \sum_{n\equal{}1}^{261\,150\,360}\frac{1}{p(p(n))}\approx 1\,.\,001\,106 > 1$

[quote="michifold"]Hi,

unfortunately, I have to disappoint you: my conjecture is wrong. One of the matheboard.de forum (Arthur Dent) wrote a program, which calculated the sum till n=261150360 , p(p(261150360))=137438950763

$ \sum_{n \equal{} 1}^{261\,150\,360}\frac{1}{p(p(n))}\approx 1\,.\,001\,106 > 1$[/quote]

Too bad. Well, it would have been amazing if true. It would surprise me if it converges to a rational number.


geometry
perimeter
circumcircle
trigonometry
inequalities
trig identities
Law of Sines

In any triangle,prove that:

$ 3r\sqrt{3} \le p \le \frac{3R \sqrt{3}}{2}$
, where p is half of perimeter,R is the circumradius, and    r is the inradius.

by the law of sines, the second inequality becomes $ \sin A \plus{} \sin B \plus{} \sin C \leq \frac {3\sqrt {3}}{2}$, which is an immediate consequence of Jensen. for the first inequality, draw the incircle, and notice that $ p$ equals the sum of alternating segments around the triangle determined by the points of tangency. you then get that the inequality is equivalent to $ 3\sqrt {3} \leq \cot\frac {A}{2} \plus{} \cot\frac {B}{2} \plus{} \cot \frac {C}{2}$, which is also an immediate consequence of Jensen.


A cubic equation f(x) = 0 has three roots x = a, x = b and x = c. Prove that the tangent to the curve at the point where x = (a+b)/2 (i.e. the mid-point of two of the roots) crosses the x-axis at the third root x = c.

hello, let $ f(x)\equal{}(x\minus{}a)(x\minus{}b)(x\minus{}c)$ then we have $ f'(x)\equal{}(x\minus{}b)(x\minus{}c)\plus{}(x\minus{}a)(x\minus{}c)\plus{}(x\minus{}a)(x\minus{}b)$ and $ f'\left(\frac{a\plus{}b}{2}\right)\equal{}\minus{}\frac{1}{4}(a\minus{}b)^2$,  so our tangent line has the form $ y\equal{}\minus{}\frac{1}{4}(a\minus{}b)^2x\plus{}n$, further we get $ f\left(\frac{a\plus{}b}{2}\right)\equal{}\minus{}\frac{1}{8}(a\minus{}b)^2(a\plus{}b\minus{}2c)$, we get $ n$ from the equation
$ \minus{}\frac{1}{8}(a\minus{}b)^2(a\plus{}b\minus{}2c)\equal{}\minus{}\frac{1}{4}(a\minus{}b)^2\cdot\left(\frac{a\plus{}b}{2}\right)\plus{}n$, $ n\equal{}\frac{1}{4}(a\minus{}b)^2c$, and our tangent line has the form
$ y\equal{}\minus{}\frac{1}{4}(a\minus{}b)^2x\plus{}\frac{1}{4}(a\minus{}b)^2c$ and we get from $ y\equal{}0$ $ x\equal{}c$ as desired.
Sonnhard.

Thanks. Any thoughts on "sum of tans" or "matrices and APs and GPs" (in Pre-Olympiad)?

[quote="AndrewTom"] [color=darkred]A equation $ f(x)\equiv ax^3 + bx^2 + cx + d = 0\ ,\ a\ne 0$ has three real roots $ x_k$ , $  k\overline{1,3}$ . Prove that

the tangent $ TT$ to $ G_f$  at the point $ T\in G_f$ with $ x_T = \frac {x_1 + x_2}{2}$  cut the $ X$-axis in the point $ R(x_3,0)$ .[/color] [/quote]

[quote] [color=darkred][b][u]Lemma[/u].[/b] Let $ f(x) = ax^3 + bx^2 + cx + d\ ,\ x\in \mathbb R$ be a real polynomial function, where $ a\ne 0$ .

and the points $ P_k\left(x_k, f\left(x_k\right)\right)$ , $ k\in\overline{1,3}$ . Then $ P_3\in P_1P_2\ \Longleftrightarrow\ x_1 + x_2 + x_3 = - \frac ba$ .[/color]

[hide="Proof."] [color=darkblue]$ P_3\in P_1P_2\Longleftrightarrow \left|\begin{array}{ccc} x_1 & ax_1^3 + bx_1^2 + cx_1 + d & 1 \\
\ x_2 & ax_2^3 + bx_2^2 + cx_2 + d & 1 \\
\ x_3 & ax_3^3 + bx_3^2 + cx_3 + d & 1\end{array}\right| = 0$ $ \Longleftrightarrow$

$ (x_1 - x_2)(x_2 - x_3)(x_3 - x_1)\left[a(x_1 + x_2 + x_3) + b\right] = 0$ $ \Longleftrightarrow$ $ x_1 + x_2 + x_3 = - \frac ba$ .[/color] [/hide][/quote]

[color=darkblue][b][u]Particular case[/u].[/b] Using the above lemma in the proposed problem, $ R\in TT\ \Longleftrightarrow\ 2\cdot \frac {x_1 + x_2}{2} + x_3 = - \frac ba$ , what is truly.[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b] Let $ f(x) = ax^3 + bx^2 + cx + d\ ,\ x\in \mathbb R$ be a real polynomial function, where $ a\ne 0$ .

Let $ d$ be a line which cut the graph $ G_f$ of the function $ f$ in the points $ P_k\ ,\ k\in\overline{1,3}$ . For any $ k\in \overline {1,3}$ the tangent

 in the point $ P_k\in G_f$ cut again $ G_f$ in the point $ Q_k$ . Prove that the points $ Q_k$ , $ k\in\overline{1,3}$ are collinearly.[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ P_k\left(x_k,f(x_k)\right)\in G_f$  and $ Q_k\left(y_k,f(y_k)\right)\in G_f$ , $ k\in\overline{1,3}$ . Thus, from the upper lemma,

$ P_3\in P_1P_2$ $ \Longleftrightarrow$ $ x_1 + x_2 + x_3 = - \frac ba$ and for any $ k\in\overline{1,3}$ we have  $ Q_k\in P_kP_k$ $ \Longleftrightarrow$ $ 2x_k + y_k = - \frac ba$ .

Observe that $ \sum_{k = 1}^3 (2x_k + y_k) = - \frac {3b}{a}$ , i.e. $ y_1 + y_2 + y_3 = - \frac ba$ what means from the same lemma that $ Q_3\in Q_1Q_2$ .

[b][u]Particular case[/u].[/b] The extremum points, if they exist $ (\ f'(x) = 0\ )$  and the inflexion point of $ G_f\ (\ f''(x) = 0\ )$ are collinearly.[/color]


geometry
circumcircle
incenter
projective geometry
ratio
trigonometry
geometric transformation

Assume that in traingle $ABC$, $\angle A=90^\circ$. Incircle touches $AB$ and $AC$ at points $E$ and $F$. $M$ and $N$ are midpoints of $AB$ and $AC$ respectively. $MN$ intersects circumcircle in $P$ and $Q$. Prove that $E,F,P,Q$ lie one a circle.

Here's my dirty proof.

Let $I$ be incenter of $\triangle ABC$, $K$ be the intersection of $AI$ and $PQ$ and WLOG $PM<PN$

$T$ is a point on arc $BC$ not containing A satisfing $BT=CT$. $L$ is the intersection of $AI$ and $BC$.

We'll show $TP=TE=TF=TQ$ to prove this problem.

It is self-evident that $TP=TQ$ cuz $PQ \parallel BC$.

$TE=TF$ is also self-evident cuz $T$ is on line $AI$.

so It is enough to show that $TQ=TE$

[b]STEP 1.[/b] $TQ^{2}= TK \times TA$

$\angle TAQ = \angle TPQ$ (the angle at the circumcircle) $=\angle TPQ$ (cuz $TP=TQ$)

$\Rightarrow$ $\angle TAQ = \angle TQK$,  so $TQ^{2}= TK \times TA$

 [b]STEP 1[/b] is proved.

[b]STEP 2.[/b] $TE^{2}= TK \times TA$

$\angle TAE = 45\DEG$, so if we prove $\angle KET = 45\DEG$, $TE^{2}= TK \times TA$ will be clear.

Let's prove $\angle KET = 45\DEG$.

(Remark 1. $AEIF$ is a square. so $AE=EI=IF=FA=r$)

(Remark 2. $\triangle AMN$ is $\frac{1}{2}$ of $\triangle ABC$)

Define a point $R$ on segment $AI$ such that $AR=IK$.

Let's show $AE^{2}= AR \times AT$.

To know their lenth, let's define lenthes $r=AE, l=AL, BC=a, CA=b, AB=c$.

It is clear that $r=\frac{b+c-a}{2}$. (see Remark 1.)

Using Pappus Theorem, we get $l= \frac{\sqrt{bc}\sqrt{(b+c)^{2}-a^{2}}}{b+c}$

[Pappus Theorem : Given $\triangle ABC$, When $L$ divide $BC$ with ratio $m: n$, $mb^{2}+nc^{2}= (m+n) AL^{2}+mn(m+n)$]

(Here, just put $m \rightarrow c$, $n \rightarrow b$)

cuz $b^{2}+c^{2}=a^{2}$, $l=\frac{\sqrt{2}bc}{b+c}$.

Because $K$ is on the line $PQ$, $AK= \frac{1}{2}AL = \frac{l}{2}$.

$AR=AK=AI-AK= \sqrt{2}r-\frac{l}{2}$

It's clear that $\triangle ABT$ resembles $\triangle ALC$.

so $AB: AT=AL: AC$ or $c: AT= l: b$

Therefore, $AT= \frac{bc}{l}$.

We reach our goal $AE^{2}= AR \times AT$.

$\Leftrightarrow$ $r^{2}= (\sqrt{2}r-\frac{l}{2}) \frac{bc}{l}$.

$\Leftrightarrow$ $r^{2}= \sqrt{2}r \frac{bc}{l}-\frac{bc}{2}$.

$\Leftrightarrow$ $r^{2}= r(b+c)-\frac{bc}{2}$

$\Leftrightarrow$ $(\frac{b+c-a}{2})^{2}= \frac{b+c-a}{2}(b+c)-\frac{bc}{2}$

This is identical equation. so true.

Therefore, $AE^{2}= AR \times AT$ is true.

which means, $\angle AER= \angle ATE$.

cuz $\angle AER= \angle IEK$, $\angle IEK=\angle ATE$

Define $G$ be the intersection of $EI$ and $AC$.

cuz $AG \parallel EI$, $\angle IEK = \angle AGE$.

so we get $\angle AGE= \angle ATE$ which means $A, E, T, G$ is on a circle.

therefore $\angle KET= \angle KAG = 45 \DEG$.

 [b]STEP 2[/b] is proved.

From [b]STEP 1 & 2[/b], we get $TP=TE=TF=TQ$ which means $P,E,F,Q$ is on a circle with center $T$.

$Q.E.D.$

[color=darkblue][b]Very nice, easy and interesting problem ![/b]

[b]Lemma (well-known).[/b] Let $ABC$ be a triangle and let $M$ be the second intersection between the circumcircle $w=C(O,R)$ of the given triangle with the bisector of the angle $\widehat{BAC}$. Denote the intersection $D\in BC\cap AM$. Then $AD=\frac{2bc\cos\frac{A}{2}}{b+c}$ and $AD\cdot AM=bc$, i.e. $AM=\frac{b+c}{2\cos\frac{A}{2}}$.
Particularly,  $\boxed{\ A=90\Longleftrightarrow AD=\frac{bc\sqrt 2}{b+c}\Longleftrightarrow AM=\frac{b+c}{\sqrt 2}\ }$.

[b]Proof of the proposed problem.[/b] Prove easily that $ME=MQ=\sqrt{\frac{a^{2}+bc}{2}}$, i.e. the points $P,E,F,Q$ belong to the circle with the center in the point $M\ .$
$1.\blacktriangleright$ Denote the reflection $N$ of the point $M$ w.r.t. the center $O$ of the circumcircle $w$ and the intersection $S\in MN\cap PQ$.
Then $MQ^{2}=MN\cdot MS=$ $2R\left(R+\frac{h_{a}}{2}\right)=$ $a\left(\frac{a}{2}+\frac{bc}{a}\right)$ $\Longrightarrow$ $\boxed{\ MQ^{2}=\frac{a^{2}+bc}{2}\ }\ .$
$2.\blacktriangleright$ It is well-known that $AE=AF=p-a$. Apply the Pithagoras' theorem in the triangle $AEM$ to the side $EM$ : $ME^{2}=AE^{2}+AM^{2}-2\cdot AE\cdot AM\cdot\cos \widehat{EAM}=$ $(p-a)^{2}+\frac{(b+c)^{2}}{2}-(p-a)(b+c)$ $\Longrightarrow$ $4\cdot ME^{2}=(b+c-a)^{2}+2(b+c)^{2}-2(b+c-a)(b+c)=$ $-(b+c-a)(a+b+c)+2(b+c)^{2}=$
$a^{2}-(b+c)^{2}+2(b+c)^{2}=$ $a^{2}+(b+c)^{2}=$ $2(a^{2}+bc)$ $\Longrightarrow$ $\boxed{\ ME^{2}=\frac{a^{2}+bc}{2}\ }\ .$[/color]


function
search
algebra unsolved
algebra

Let $ f$ be a function from the real numbers to the real numbers such that $ f(1) \equal{} 1, f(a\plus{}b) \equal{} f(a)\plus{}f(b)$ for all $ a, b,$ and $ f(x)f \left( \frac{1}{x} \right) \equal{} 1$ for all $ x \neq 0.$ Prove that $ f(x) \equal{} x$ for all real numbers $ x.$

It seems to me that this problem was at some Ireland olympiad as well.
The main point is how to go from rational numbers to real,if I remember well,some very nice trick was used in the official solution or somewhere else,even probably it was posted here.

...I was right,here is it:
http://www.mathlinks.ro/viewtopic.php?search_id=909383798&t=185279


analytic geometry
linear algebra
matrix
vector
LaTeX
geometry
geometric transformation

I'm writing an exporter for 3d studio max. 3D studio max uses a coordinate system that has the X axis going from left to right, the Z Axis going from bottom to top and the Y axis going from front to back. Opengl uses a left handed coordinate system, whilst 3d max uses a right handed coordinate system. OpenGL's X axis goes from left to right, Y goes from bottom to top and Z goes from back to front.

I want to convert a 3d studio max matrix to use it with OpenGL. Matrices hold transformation information. This is a problem of basis changing. So if i can find the coordinate vectors of the basis vectors of 3d studio max's matrix expressed as a linear combinations of opengl's basis vectors i can construct a matrix that will transform a 3d studio max coordinate vector to an opengl coordinate vector. I found that the matrix to use for this is:

1 0 0
0 0 1
0 -1 0

So multiplying each of the coordinate vectors of the 3d studio max matrix with this matrix should give me the coordinate vectors of the opengl matrix.
But this does not work when implemented in code. I may got something wrong it seems. What am i doing/understanding wrong?

What does work is doing it this way:
For every vertex that comes from 3d studio max, switch the z with the y. For every matrix that comes from 3d studio max, multiply it by the following matrix and then swap the Y axis with the Z axis
1 0 0
0 0 1
0 1 0
The 3d model is animated correctly but it faces backwards. This is because Z wasn't negated. This does not cause any stretch/skew when the model is animated. Why is this so and how can i transform an 3d studio max matrix to an opengl matrix?

If I understood this all correctly, it essentially boils down to transforming the standard ordered basis $ \beta \equal{} \{(1,0,0), (0,1,0), (0,0,1)\}$ into $ \gamma \equal{} \{(1,0,0), (0,0,1), (0,1,0)\}$ the transform required should be $ [I]_{\beta}^{\gamma}$ which is (I wish I knew how to LaTeX this):

1 0 0
0 0 1
0 1 0

I hope this can help!

A transformation matrix is a 4x4 matrix that encodes, rotation,scaling and translation. How about trasnforming the translation part of the matrix? I thought that beta is {(1,0,0),(0,0,-1),(0,1,0)} Can you detail a little how you obtained beta?

Ah, I missed the fact that in 3dmax the Y axis goes from front to back and in OpenGL the Z axis goes from back to front. So essentially you want the matrix representation of the transformation $ T((x,y,z)) \equal{} (x,z, \minus{} y)$? So this time I'd take $ [T]_{\beta}$ using the standard basis which is actually the first matrix that you have. That is, if you have some vector $ v$ in 3dmax, take $ [v]_{\beta}$ (which in this case is exactly $ v$) and then compute $ [T]_{\beta}[v]_{\beta}$ it should work fine? Could you describe the issue you have when you multiply the matrices together? Are you doing it in the right order?

I think the matrix multiplication is fine, this part is done by the computer and i don't think there are any mistakes in the implementation. The output i'm getting is an animated model that is stretched/skewed in arbitrary directions(i'm implementing bone animation).  Now i thought this out and realized that if you want to transform a point using a matrix by moving it forward you would have to construct a matrix that moves the coordinate system backwards. It would be easier to think of the coordinate system as a man with a camera on his head. If he moves backwards everything else advances with respect to him, thus moving forward. This is detailed very good in Fletcher Dunn's and Ian Parberry's book: 3D Math Primer for Graphics and Game Development.

Now i know how to change a point from basis A to basis B by first expressing the basis vectors from A as linear combinations of the basis vectors from B and then constructing the transition matrix. But changing the basis of coordinate vectors is the opposite of changing the coordinates of basis vectors. So, instead of using this matrix

\[ \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & \minus{}1 \\
0 & 1 & 0 \end{array} \right)\]

I should use it's inverse. Because the matrix is orthogonal it's inverse simply equals it's transpose:

\[ \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & \minus{}1 & 0 \end{array} \right)\]

So i tried multiplying the axes with the second matrix and multypling the translation part with the first matrix. Still the results aren't yet correct. There is stretch and skewing  going on there. I wonder what am i missing?   :blush:


I have two problems left for my Trigonometry homework.
They are identities and I need to prove them algebraically.

1. sin(6x) + cos(4x) + sin(2x) = 4 sin(3x) x cos(2x) x cos(x)
So far, I have thought about using a Product-to-Sum Identity, but I only know how to do that with 2 factors.

2. .25cos(4x) + .75 = (cos(x))^4 + (sin(x))^4
I have no idea how to begin solving this one. I think I'm supposed to use a Double-Angle Identity, but I'm not sure where to begin.

Please help. Thanks!

[hide]$ 2\cos^2{x}\plus{}2\sin^2{x}\equal{}2$

$ 4\cos^{4}x\plus{}8\cos^2{x}\sin^2{x}\plus{}4\sin^{4}x\equal{}4$

now look at $ 8\cos^2{x}\sin^2{x}$ take the square root of that gives us

$ 2\sqrt{2}\cos{x}\sin{x}$

$ 2\sqrt{2}\frac{\sin{2x}}{2}\equal{}\sqrt{2}\cdot\sin{2x}$
and the square of that is just $ 2\sin^{2}x$
now going back to our equation:

$ 4\cos^{4}\plus{}2\sin^2{2x}\plus{}4\sin^{4}x\equal{}4$

$ \cos^{4}x\plus{}\sin^{4}x\equal{}\frac{4\minus{}2\sin^2{2x}}{4}$

now we need to prove that

$ \frac{\cos{4x}\plus{}3}{4}\equal{}\frac{2\minus{}\sin^2{2x}}{2}$

$ 2(\cos{4x}\plus{}3)\equal{}(2\minus{}\sin^2{2x})4$

$ 2\cos^2{2x}\minus{}2\sin^2{2x}\plus{}6\equal{}8\minus{}4\sin^2{2x}$
$ 2\cos^2{2x}\plus{}2\sin^2{2x}\equal{}2$ and we are done[/hide]

[quote="sonny"][hide]$ 2\cos^2{x} \plus{} 2\sin^2{x} \equal{} 2$

$ 4\cos^{4}x \plus{} 8\cos^2{x}\sin^2{x} \plus{} 4\sin^{4}x \equal{} 4$

now look at $ 8\cos^2{x}\sin^2{x}$ take the square root of that gives us

$ 2\sqrt {2}\cos{x}\sin{x}$

$ 2\sqrt {2}\frac {\sin{2x}}{2} \equal{} \sqrt {2}\cdot\sin{2x}$
and the square of that is just $ 2\sin^{2}x$
now going back to our equation:

$ 4\cos^{4} \plus{} 2\sin^2{2x} \plus{} 4\sin^{4}x \equal{} 4$

$ \cos^{4}x \plus{} \sin^{4}x \equal{} \frac {4 \minus{} 2\sin^2{2x}}{4}$

now we need to prove that

$ \frac {\cos{4x} \plus{} 3}{4} \equal{} \frac {2 \minus{} \sin^2{2x}}{2}$

$ 2(\cos{4x} \plus{} 3) \equal{} (2 \minus{} \sin^2{2x})4$

$ 2\cos^2{2x} \minus{} 2\sin^2{2x} \plus{} 6 \equal{} 8 \minus{} 4\sin^2{2x}$
$ 2\cos^2{2x} \plus{} 2\sin^2{2x} \equal{} 2$ and we are done[/hide][/quote]

Is that for number 1 or number 2?  I already solved number 2 actually, so I just need number 1.


Determine the max of K.
To any a,b,c>0
$ \sum{a^3} \plus{} 3\sum{a^2b} \plus{} 6abc$$ \ge{k\sum{ab^2}}$(cyc)

[quote="Cai Ni"]Determine the max of $ k.$ To any $ a,b,c > 0,$

$ \sum{a^3} \plus{} 3\sum_{cyc}{a^2b} \plus{} 6abc\ge k\sum_{cyc}{ab^2}.$[/quote]$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3\left(a^2b \plus{} b^2c \plus{} c^2a\right) \plus{} 6abc \minus{} \frac {15}{4}\left(ab^2 \plus{} bc^2 \plus{} ca^2\right)$

$ \equal{} \frac {a(b \plus{} 4c \minus{} 2a)^2 \plus{} b(c \plus{} 4a \minus{} 2b)^2 \plus{} c(a \plus{} 4b \minus{} 2c)^2}{4}\geq 0,$

with equality if $ a \equal{} 0,b \equal{} 1,c \equal{} 2.$

Can I ask you a question Ji Chan? How do you get such an expression. I mean how did it come to your mind to regroup the terms like that and make the inequality as SOS? Is there any general method for easy rewriting an inequality in SOS?

i have also the same question...how did you find this expression?and how can we ensure that this $ k$ is maximum?


Find all pairs $ (a,\ b)$ of positive integers with $ a\geq 2,\ b\geq 2$ such that $ a^{a^a}\equal{}b^b$.

It seems that it is H74 PEN problem
http://www.mathlinks.ro/Forum/viewtopic.php?=150684

Thanks for the link.

Ironically my post was after the IMO Problem.

What't the source of the problem I posted?


Particles $ P$, of mass 2, and $ Q$, of mass 1, move along a line. Their distances from a fixed point are $ x_1$ and $ x_2$, respectively where $ x_2 > x_1$. Each particle is subject to a repulsive force from the other of magnitude $ \frac {2}{z^3}$ (edited), where $ z \equal{} x_2 \minus{} x_1$.

Initially, $ x_1 \equal{} 0$, $ x_2 \equal{} 1$, $ Q$ is at rest and $ P$ moves towards $ Q$ with speed 1. Show that $ z$ obeys the equation $ \frac {\textup{d}^2z}{\textup{d}t^2} \equal{} \frac {3}{z^3}$.

By first writing $ \frac {\textup{d}^2z}{\textup{d}t^2} \equal{} v \frac {\textup{d}v}{\textup{d}z}$, where $ v \equal{} \frac {\textup{d}z}{\textup{d}t}$, show that $ z \equal{} \sqrt {4t^2 \minus{} 2t \plus{} 1}$.

By considering the equation satisfied by $ 2x_1 \plus{} x_2$, find $ x_1$ and $ x_2$ in terms of $ t$.

Shouldn't it be $ \frac{\textup{d}^2z}{\textup{d}t^2} \equal{}  \minus{}\frac {3}{z^2} \;\;$ ?

Sorry the repulsive force should be $ \frac {2}{z^3}$, it's been edited. Could u please show the steps :) I've not much of an idea how to start on this one.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a simple group of order $ 168$.
Prove that $ G$ has a subgroup of order $ 21$, and that $ G$ is isomorphic to a subgroup of $ A_8$.

Let's look at the number of $ 7$-Sylow subgroups. Upon examination, the only divisors of $ 168$ congruent to $ 1 \mod 7$ are $ 1$ and $ 8$, so there must be eight $ 7$-Sylow subgroups. This means that for any $ 7$-Sylow subgroup $ K$, we have $ |N(K)| \equal{} 21$.

Given that $ G$ has a subgroup $ H$ of order $ 21$, then we can let $ G$ act on the left cosets $ gH$ by $ x(gH) \equal{} xgH$. This gives a group homomorphism from $ G$ to $ A_8$ with kernel contained in $ H$. By simplicity of $ G$, the kernel is $ 0$, and this is an embedding.

Can we say more?  Namely, that the only such group is $ \text{PSL}_2(\mathbb{F}_7)$.  It's a nice way of explaining the result, since this group has the obvious action on the projective line over $ \mathbb{F}_7$.

Here is a lemma: If $ G\leq S_n$ is a simple group of order greater than 2, then $ G\leq A_n$.

Proof: Compose the injection with the sign map $ S_n\to Z_2$ and look at the kernel.

[quote="Jan"][quote="calc rulz"]Given that $ G$ has a subgroup $ H$ of order $ 21$, then we can let $ G$ act on the left cosets $ gH$ by $ x(gH) \equal{} xgH$. This gives a group homomorphism from $ G$ to $ A_8$ with kernel contained in $ H$.[/quote]

My solution is of course essentially the same (standard techniques) but the $ A_8$ is not standard, at least not to me, so I used a lemma to prove that it's not only an embedding in $ S_8$, but also in $ A_8$. You seem to think this is immediate, but I fail to see why. Can you shed some ligt on why you conclude that every permutation is even? Thanks.

torajirou, that would be very nice. Any progress on how to prove this?

Best regards,
Jan[/quote]

Nice catch - I should have explained that. Nonetheless, by LydianRain's argument, the image has to be contained within $ A_8$, since otherwise there would be a homomorphism from $ G$ onto $ C_2$ from the sign map. So it's not too difficult to prove that  :wink: .

Ok nice, now I see, thanks for explaining. I used a slightly more general lemma which says:

Let $ G$ be a group and $ N$ a subgroup of index $ 2$. Then for every subgroup $ H \leq G$, we have that $ H \cap N$ is a subgroup of index $ 1$ or $ 2$ in $ H$.


A six digit number (base 10) is squarish if it satisfies the following conditions:

(i) none of its digits are zero;
(ii) it is a perfect square; and
(iii) the first of two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers.

How many squarish numbers are there?

$\text{(A)} \ 0 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 8 \qquad \text{(E)} \ 9$

[quote="t0rajir0u"][hide="Solution"]
Call the number $n^2$.  Then $316 < n < 1000$ in order that $n^2$ is a six-digit number.  Next, consider the two-digit squares that don't have zeroes in them.  They are $16, 25, 36, 49, 64, 81$.

Call the three squares that make up the number $a^2, b^2, c^2$ such that $4 \le a, b, c \le 9$.  Then $n^2 = (100a)^2 + (10b)^2 + c^2$ and $n = 10k + c$ for some integer $31 \le k \le 99$.  Then we can also write $n^2 = (10k)^2 + 20kc + c^2$.  Therefore, one of $k$ or $c$ must be divisible by 5 or else the last two digits will not be equal.  Casework.

Case 1:  $c = 5$.  We have $n^2 = (10k)^2 + 100k + 25 = (100a)^2 + (10b)^2 + 25 \implies k^2 + k = (10a)^2 + b^2 \implies (k+10a)(k-10a) = b^2 - k$.  $k^2 + k$ is even, so $b$ must also be even, and we have $b^2 - k \le 64 - 31 = 33$, whereas for $k \neq 10a$ we have $k+10a \ge 31 + 10 = 41$, so the only solutions occur when $k-10a = 0, b^2 - k = 0 \implies 10a = b^2$ which still has no solution.

Case 2:  $k = 5m$.  We have $n^2 = (50m)^2 + 100mc + c^2 = (100a)^2 + (10b)^2 + c^2 \implies (5m)^2 + mc = (10a)^2 + b^2$.

Subcase 1:  $m = 2p$.  Then $(10p)^2 + 2pc = (10a)^2 + b^2$ so we set $p = a, 2pc = b^2 \implies 2ac = b^2$.  Remembering $a, c \ge 4$ we have $b^2 \ge 32, b = 6, 8$.  For $b = 6$ we have $ac = 18$ which has no solutions, and for $b = 8$ we have $ac = 32$ which has the solutions $a = 8, c= 4$ and $c = 8, a = 4$.  So there are $\boxed{2}$ solutions in this case.

Subcase 2:  $m = 2p+1$.  Then $(10p + 5)^2 +(2p+1)c = (10a)^2 + b^2 \implies (10p)^2 + 100p + 25 + 2pc + c = (10a)^2 + b^2$, so we have $p(p+1) (+1) = a^2$ which has no solutions (even when we allow for the possibility that $25 + 2pc + c > 100$ )

Hence, our answer is 2, or $\boxed{B}$.

Edit:  I left out a case, $c = 4, 6$, in which case it is not necessary that $n \equiv c \bmod 10$, but the casework is very annoying, could someone post a more elegant solution?
[/hide][/quote]
Squares found in Subcase 1 are not squares: 163664 and 643616 are not squares. There should be a simpler way. (AMC questions cannot be expected to take that long but tricky)

I said there weren't any solutions for $b = 6$ - those two solutions are for $b = 8$ and the squares are $166464, 646416$, which are $408^2$ and $804^2$ respectively.  I am, however, still missing a $n \not \equiv c \bmod 10$ case.

[hide]166464
646416
$B$
all i did was look at the table on my calc for x^2 at 100,200,300,400,etc, and stopped after the first 2 digits went past a perfect square, took maybe 3 mins[/hide]

Calculators weren't allowed on the AHSME.

[quote="PhireKaLk6781"]Sorry for reviving, but it is interesting to note that this question was written with a logical flaw, as the problem defined the squarish six-digit numbers, while it asked for a count of all six-digit numbers. For the problem to be precise it should have asked for the total count as "How many [i]six-digit[/i] squarish numbers are there?" as the problem never indicated that a number that doesn't have six digits isn't squarish.[/quote]

Oh, may i ask which year of AMC does that come from?

[quote="math154"][hide="Shorter Solution?"]
Let a squarish number be $ x^2\equal{}\overline{abcdef}$, such that $ \overline{ab},\overline{cd},\overline{ef}\in\{16,25,36,49,64\}$.

Then we can have:

$ 160000\le x^2\le169999\implies400\le x\le412$,
$ 25000\le x^2\le259999\implies500\le x\le509$,
$ 36000\le x^2\le369999\implies600\le x\le608$,
$ 49000\le x^2\le499999\implies700\le x\le707$,
$ 64000\le x^2\le649999\implies800\le x\le806$,
$ 81000\le x^2\le819999\implies900\le x\le905$.

$ x\equal{}410,411,412$ are easy to see false. Otherwise, we can express $ x\equal{}100n\plus{}k$ for $ 0\le k\le9$ so that $ x^2\equal{}10000n^2\plus{}200nk\plus{}k^2$. The last two digits are not influenced by the $ 200nk$, and since $ 200nk\le16200$, the first two digits will either be $ n^2$ or $ n^2\plus{}1$. We thus conclude that $ 4\le n,k\le9$, and $ 2nk$ must be a perfect square (this is necessary but not sufficient, as if it makes $ 200nk$ tip over and increase the $ n^2$ to $ n^2\plus{}1$, the first two digits can't be a perfect square). Analysis quickly yields that $ (n,k)\equal{}\{(4,8),(8,4),(8,9),(9,8)\}$ are the only possibilities, but the latter two make $ 2nk$ greater than two digits hence making the first two digits not a perfect square, giving an answer of $ 4\minus{}2\equal{}\boxed{\textbf{(B)}\ 2}$.
[/hide][/quote]

I don't understand why 16200 bigger or equal to 200nk.

[quote="applepi2000"]$k \le 9$
$n le 9$
Thus, $nk \le 81$ and $200nk \le 16200$[/quote]

So, actually there is a typo in that post:
missing a zero on those inequality?

[quote="math154"][hide="Shorter Solution?"]
Let a squarish number be $ x^2\equal{}\overline{abcdef}$, such that $ \overline{ab},\overline{cd},\overline{ef}\in\{16,25,36,49,64\}$.

Then we can have:

$ 160000\le x^2\le169999\implies400\le x\le412$,
$ 25000\le x^2\le259999\implies500\le x\le509$,
$ 36000\le x^2\le369999\implies600\le x\le608$,
$ 49000\le x^2\le499999\implies700\le x\le707$,
$ 64000\le x^2\le649999\implies800\le x\le806$,
$ 81000\le x^2\le819999\implies900\le x\le905$.

$ x\equal{}410,411,412$ are easy to see false. Otherwise, we can express $ x\equal{}100n\plus{}k$ for $ 0\le k\le9$ so that $ x^2\equal{}10000n^2\plus{}200nk\plus{}k^2$. The last two digits are not influenced by the $ 200nk$, and since $ 200nk\le16200$, the first two digits will either be $ n^2$ or $ n^2\plus{}1$. We thus conclude that $ 4\le n,k\le9$, and $ 2nk$ must be a perfect square (this is necessary but not sufficient, as if it makes $ 200nk$ tip over and increase the $ n^2$ to $ n^2\plus{}1$, the first two digits can't be a perfect square). Analysis quickly yields that $ (n,k)\equal{}\{(4,8),(8,4),(8,9),(9,8)\}$ are the only possibilities, but the latter two make $ 2nk$ greater than two digits hence making the first two digits not a perfect square, giving an answer of $ 4\minus{}2\equal{}\boxed{\textbf{(B)}\ 2}$.
[/hide][/quote]

Also, shouldn't 9 bigger or equal than n,why 4 bigger or equal to n?

@above, [b]math154[/b] wrote $4 \leq n$, so $n$ is bigger than or equal to $4$, not $4$ is bigger than or equal to $n$...

I think that there are a few minor flaws in [b]math154[/b]'s solution.  He's missing a $0$ on the left most side of almost every inequality (edit: almost every inequality on the left side, of $x^2$) that he has listed; instead of $25000 \leq x^2 \leq 259999$, for example, it should be $250000 \leq x^2 \leq 259999$.  That threw me off the first time I looked at it.

\bump (i know this is a REALLLY old thread).. can anyone provide a good, short and clear solution?

[hide=Solution]$(100a+10b+c)^2=10000(a^2)+1000(2ab)+100(2ac+b^2)+10(2bc)+c^2$. We know $a\ge 4$ and the first two digits of the number are $a^2$. Since we don't want the $1000(2ab)$ to affect the first two digits, $2ab<10$. This means either $b=0$ or $a=4,b=1$. Check the cases where $a=4,b=1$. None of the work. If $b=0$, $(100a+10b+c)^2=10000a^2+100(2ac)+c^2$. $2ac<100$, since we don't want to affect the first two digits. $2ac$ also must be an even perfect square. We know $c\ge4$ since there are no non-zero digits in $c^2$ when $c\ge4$. Because $a,c\ge4$, $2ac\ge32$. The only even perfect squares between $32$ and $99$ are $36$ and $64$. This means $ac=18,32$. $ac=18$ has no solutions with $a,c\ge4$. $ac=32$ has $\boxed{2}$ solutions namely $a=4,c=8$ and $a=8,c=4$.[/hide]
This is very similar to math154's solution.


calculus
symmetry
geometry
geometric transformation
reflection
similar triangles
AMC

Two poles of heights of 11 and 33 feet are placed with 240 feet between them.  Bucky needs to connect a wire between the tops of the two poles, but it also needs to touch the ground at a point between them.  What is the length in feet of the shortest wire that Bucky can use and also, how far away from the pole of 11 feet will the wire touch the ground?

Hint: [hide]Think about how you can draw an oval[/hide]

Umm...Tare?  I'm looking to see if somebody can solve this with no hints sorry.  And hmm...maybe you would like to give this a shot.  By the way, your hint was way different from what I was thinking of.

Are you suggesting [hide]using calculus to minimize the equation for a caternary? (y=h/w cosh ( w/h)x + C) If so....[/hide] ouch.

This is a beautiful problem, and it has a solution that uses no trace of calculus whatsoever.

I agree with Simon... this is a form of one of my favorite problems and one that I frequently elaborate upon in contests that I write.  Looked at properly it is very simple and elegant.

Several assumptions must be made:

[hide]- the rope is weightless

- it's straight when the shortest path is found.[/hide]

Then,

[hide]Using the ground as an axis of symmetry, reflect either pole. I'll reflect the shorter pole. Join the tip of the longer pole with the tip of the reflection. The point at which the rope touches the ground is the intersection between this line and the ground. Then join this point to the tip of the shorter pole. Using similar triangles, one finds that this point is 60 feet away from the shorter pole.[/hide]

Mmm hmm. First of all I'll apologize to WarpedKlown1335. I was kinda in a hurry...

Anyways[hide] the thought process is the same, if the polls have the same height then you can draw an oval using the tip of the poles as the foci. Same thing happens here. Just keep drawing ovals until you begin to notice that you get the least length of rope needed when the distance between the tangency and the two poles are proportional to the length of the poles. In this case it's 1:3 so the ground length is 60 and 180 feet. Thus the length of the rope is :sqrt:(60:^2:+11:^2:)+:sqrt:(180:^2:+33:^2:)=244 feet.[/hide]

Well, I made this problem up. And yes, Tare and Osiris, you are both correct on the parts of the problem that you did (Tare finding the length of the wire and Osiris finding the distance from the point the wire touches the ground to the 11-foot-pole).


Hello,

Solve in  $ \mathbb{R}$ the inequation :

\[ {\lfloor x \rfloor}^2 \minus{} x\lfloor x \rfloor \plus{} 3 \leq 0  \]

[hide]If $ x\equal{}n\plus{}\alpha$, where $ n\equal{}[x]$ and $ \alpha\equal{}\{x\}$, we get

$ n^2\minus{}n(n\plus{}\alpha)\plus{}3\leqslant 0\iff n\alpha\geqslant 3$

Hence $ n>0$ (because $ 0\leqslant\alpha<1)$, so we have $ \alpha\geqslant{3\over n}$, but that implies $ n\geqslant 4$, since otherwise $ \alpha<1$ couldn't be satisfied. Therefore the solution is

$ x\in\bigcup_{n\equal{}4}^{\infty}\left[n\plus{}{3\over n},n\plus{}1\right)$[/hide]


The decimal representation of m/n, where m and n are relatively prime positive integers and , contains the digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of  for which this is possible.


Imagine that humans have developed machines that can go underwater, under the crust and widstand hot temperatures i.e. it can emerge at the other end of the earth. Prove that this machine can enter and emerge from water.

I think what varahamitra means is that that machine dives into a ocean and emerges on the other side but this other side is also ocean, perhaps a different one.

Proving this is not too difficult. Let me give you a hint.
[hide="hint"]
The earth is 71% water and 71>50.
[/hide]

There's 71% percent water on Earth (more than 50%, that's important).

Suppose that it's not possible. Then where ever you enter in water, you will leave on ground. But that means there's less than 50% of water. Contradiction.


Click to reveal hidden content has no effect on my computer; however, using  $ (x\plus{}y\plus{}z)^{2} \equal{} x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 2(xy \plus{} yz \plus{} zx)$ and $ xy \equal{} z^2$, I probably did the same as you. Thanks for a nice little problem.

[quote="ThetaPi"]Though simple, this problem offers some mild excitement for me.

If $ x,y,z$ satisfy the equations $ x \plus{} y \plus{} z \equal{} 5, x^2 \plus{} y^2 \plus{} z^2 \equal{} 15, xy \equal{} z^2$, determine the value of $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}$. [hide] We have $ (x \plus{} y \plus{} z)^2 \equal{} 25 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} yz \plus{} xz)$, so $ xy \plus{} yz \plus{} xz \equal{} z^2 \plus{} yz \plus{} xz \equal{} z(x \plus{} y \plus{} z) \equal{} 5$. Since $ x \plus{} y \plus{} z \equal{} 5$, $ z \equal{} 1 \Rightarrow xy \equal{} 1$. The desired value is $ \frac {xy \plus{} yz \plus{} xz}{xyz} \equal{} \frac {z(x \plus{} y) \plus{} 1}{z(z^2)} \equal{} 5$. [/hide][/quote]
i have another solution
only one minite
almonkez

[quote="ThetaPi"]Though simple, this problem offers some mild excitement for me.

If $ x,y,z$ satisfy the equations $ x \plus{} y \plus{} z \equal{} 5, x^2 \plus{} y^2 \plus{} z^2 \equal{} 15, xy \equal{} z^2$, determine the value of $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}$. [hide] We have $ (x \plus{} y \plus{} z)^2 \equal{} 25 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} yz \plus{} xz)$, so $ xy \plus{} yz \plus{} xz \equal{} z^2 \plus{} yz \plus{} xz \equal{} z(x \plus{} y \plus{} z) \equal{} 5$. Since $ x \plus{} y \plus{} z \equal{} 5$, $ z \equal{} 1 \Rightarrow xy \equal{} 1$. The desired value is $ \frac {xy \plus{} yz \plus{} xz}{xyz} \equal{} \frac {z(x \plus{} y) \plus{} 1}{z(z^2)} \equal{} 5$. [/hide][/quote]
 :play_ball:

hello, i have found the following solutions
$ x\equal{}2\plus{}\sqrt{3},y\equal{}2\minus{}\sqrt{3},z\equal{}1$
$ x\equal{}2\minus{}\sqrt{3},y\equal{}2\plus{}\sqrt{3},z\equal{}1$
and we get
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}5$.
Sonnhard.

Interesting solution, almonkez. Next time, it would be easier for you to just type directly in LaTeX: $ x^2\plus{}2x$.

[quote="ThetaPi"]Interesting solution, almonkez. Next time, it would be easier for you to just type directly in LaTeX: $ x^2 \plus{} 2x$.[/quote]
thanks : thetapi
im regestered in 20 forum at least
in all of them i send my solutions writen with word and paint or by scaner
may i have troubles with latex
thanks for your reply

If $ x,y,z$ satisfy the equations $ x\plus{}y\plus{}z \equal{} 5, x^{2}\plus{}y^{2}\plus{}z^{2}\equal{} 15, xy \equal{} z^{2}$, determine the value of $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}$

well since you know that z^2 equals xy you sbstitute it into x^2 +y^2+xy

and you also subtitute it into x+y+ the square root of xy

if you square the equation
 you (x+ y+sqrt xy)^2=25

and since we know x^2 +y^2 + xy=15

that is all I know can wome one please continue my solution


algorithm
number theory
Euclidean algorithm
number theory unsolved

a) Adding together the n+1 identities in Euclids algorithm, the result is

$a + b + \sum_{i=1}^{n-1} r_i \;=\; \sum_{i=0}^n r_iq_{i+1} \: +\: \sum_{i=1}^n r_i$,

i.e.

$\sum_{i=0}^n r_iq_{i+1} \;=\; a + b - r_n = a + b - gcd(a,b)$.

b) Multiplying identity number i in Euclids algorithm with $r_{i-1}$ for each $i \in \{1,2,\ldots,n\}$, and then adding the $n+1$ resulting identities, we get

$ar_0 + \sum_{i=0}^{n-1} r_ir_{i+1} \;=\; \sum_{i=0}^n r_i^2q_{i+1} \: +\: \sum_{i=0}^{n-1} r_ir_{i+1}$,

i.e.

$\sum_{i=0}^n r_i^2q_{i+1} \,=\; ar_0 = ab$.