Prove: Let $ a_0,a_1,\ldots,a_n$ be numbers such that $ a_0\equal{}1$ and $ a_k\ne 0,\; 1\leq k\leq n.$ Then for each positive integer $ n,$
\[ \sum_{k\equal{}0}^{n\minus{}1}{\frac{a_0 a_1 \cdot\cdot\cdot a_k}{(x\plus{}a_1)(x\plus{}a_2) \cdot\cdot\cdot (x\plus{}a_{k\plus{}1})}}\equal{}\frac{1}{x}\minus{}\frac{1}{x(1\plus{}x/a_1)(1\plus{}x/a_2)\cdot\cdot\cdot (1\plus{}x/a_n)}.\]


Convex quadrilateral ABCD has AB=12, BC=6, CD=20. ABCD is inscribed inside a circle which is tangent to the midpoint of BC. What's the area of ABCD?

$f(n)$ is the product of all positive odd integers $\le$ n. For how many odd n's from 3 to 99 does $f(n-1)$ differ from a multiple of n by 1?

[i]Mod edit: What was with the title? [/i]

EDIT: If you move every letter 1 up in the alphabet you will get rnldsghmf

[quote="Treething"][quote="Totally Zealous"]Convex quadrilateral ABCD has AB=12, BC=6, CD=20. ABCD is inscribed inside a circle which is tangent to the midpoint of BC. What's the area of ABCD?[/quote]

This makes no sense.[/quote]

oh come on...he obviously means ABCD has an inscribed circle...

oops.... :blush:

for some reason I can't edit my post.


geometry
3D geometry
function
inequalities
calculus
derivative
integration

Find all positive integer sequences $ a_{k}$ such that

$ \left( \sum_{k \equal{} 1}^{n}a_{k}\right)^{2} \equal{} \sum_{k \equal{} 1}^{n}a_{k}^{3}$

[hide]The trick is to prove that the left side cannot exceed the right one. I found this excercise to the book an Introduction to Diophantine Equations by Titu Anndrescu, Dorin Andrica...[/hide]

Hmm this reminds me of $ \sum_{d|n}\tau(d)^3 \equal{} \left(\sum_{d|n}\tau(d)\right)^2$
where $ \tau$ is the number of divisors function (or $ \sigma_0$, so this is one such sequences (not sure about the uniqueness, though)...

(By the way does someone have a good proof on this identity?)

No, it's not.  The summation given needs to hold for all $ n$.  As for a proof of your identity, the standard way to show that two arithmetic functions are equal is to show that they are equal for $ n$ a prime power.

As for the original question, there is a fairly straightforward proof by induction.  The proof suggested by panos_lo, if done properly, extends to [hide="the following inequality"] Let $ f(x)$ be a function with a continuous derivative satisfying $ 0 \le f'(x) \le 1$.  Show that

$ \left( \int_{0}^{1} f(x) \, dx \right)^2 \ge \int_{0}^{1} f(x)^3 \, dx$. [/hide]


trigonometry
function
inequalities unsolved
inequalities

[quote="befuddlers"]Prove or disprove?

If A,C are angles of a triangle in which B is the smallest angle

prove that cos(A-C)>=cosA + cosC[/quote]
It's wrong.
Try $ A \equal{} 170^{\circ},$ $ B \equal{} 1^{\circ}$ and $ C \equal{} 9^{\circ}.$ :wink:


Let ABCD be a cyclic quadrilateral. Prove that
|AB − CD|/(AB + CD)+ |AD − BC|/(AD + BC) = |AC − BD|/(AC + BD).

[quote="kyyuanmathcount"]Let ABCD be a cyclic quadrilateral. Prove that

$ \frac {|AB \minus{} CD|}{AB \plus{} CD} \plus{} \frac {|AD \minus{} BC|}{AD \plus{} BC} \equal{} \frac {|AC \minus{} BD|}{AC \plus{} BD}$

[/quote]
[color=darkblue]
Rewrite it as follow
\[ \frac {|1 \minus{} \frac {CD}{AB}|}{1 \plus{} \frac {CD}{AB}} \plus{} \frac {|1 \minus{} \frac {BC}{AD}|}{1 \plus{} \frac {BC}{AD}} \equal{} \frac {|1 \minus{} \frac {BD}{AC}|}{1 \plus{} \frac {BD}{AC}}
\]

\[ \frac {|1 \minus{} k|}{1 \plus{} k} \plus{} \frac {|1 \minus{} m|}{1 \plus{} m} \equal{} \frac {|1 \minus{} n|}{1 \plus{} n} \ \ (*)
\]
From second Ptolemeu theorem:
\[ n \equal{} \dfrac{BD}{AC} \equal{} \dfrac{BA\cdot BC \plus{} DA\cdot DC}{ AB\cdot AD \plus{} CB\cdot CD} \equal{} \dfrac{1 \plus{} \dfrac{DA}{BC}\cdot \dfrac{DC}{AB}}{\dfrac{AD}{BC} \plus{} \dfrac{CD}{BA}} \equal{} \dfrac{1 \plus{} \dfrac{1}{km}}{\dfrac{1}{k} \plus{} \dfrac{1}{m}} \equal{} \dfrac{1 \plus{} km}{k \plus{} m}
\]
Then $ (*)$ transforms into
\[ \frac {|1 \minus{} k|}{1 \plus{} k} \plus{} \frac {|1 \minus{} m|}{1 \plus{} m} \equal{} \frac {|(k \minus{} 1)(m \minus{} 1)|}{(m \plus{} 1)(k \plus{} 1)}
\]
Now we only have to check the cases when $ m$ and $ k$ are $ \le$ or $ \ge 1$.[/color]

I am not so sure whether this equality is true now...

The problem should actually say
\[ \frac{|AB \minus{} CD|}{AB \plus{} CD} \cdot \frac{|AD \minus{} BC|}{AD \plus{} BC} \equal{} \frac{|AC \minus{} BD|}{AC \plus{} BD}.\]

Oh! Thanks that makes much more sense with the work I did.


inequalities
LaTeX
inequalities proposed

give $a_i,\alpha,\beta >0$ ; max{$a_i$}$\leq \alpha$
and $\displaystyle \sum_{i=1}^n a_i = \sum_{i=1}^n \frac{\beta^2}{a_i}$
prove that
$\displaystyle \sum_{i=1}^n \frac 1{\alpha +a_i} \leq \frac {n}{\alpha + \beta}$

 :D

actually this is easy,after making sum substitution ,jensen and the am-gm  does the trick,i cant post the whole proof coz i dont really know how to use latex :(

I have put up a latex tutorial in the How to use Latex topic ... but I guess no-one reads that ... :(


number theory
greatest common divisor
modular arithmetic
blogs
relatively prime

I read through the proof and I still don't understand these:

1) If gcd([i]b,m[/i]) > 1, then b does not have a modulo [i]m[/i] inverse.
2) A modulo m residue r has a single inverse if and only if gcd(m,r) = 1. Otherwise it has no inverse.

can you give me examples or explain how it came about? thanks!

A modular inverse,$a^{-1}$ is a number such that
$a\cdot a^{-1}\equiv 1 \pmod n$. A necessary condition for the existence of the inverse is that $\text{gcd}(a,n) = 1$.

This is the basis of internet public key encryption.  A message is assigned a number and then encoded.  To be decrypted, the inverse must be found - this is a task that requires factoring a number that may be several hundred digits long.
I'm going to bed now but I can elaborate on that if you'd like.

And usually, N is a really massive prime.

@illcrowflu

so basically 1) and 2) means the same thing...

that's a really interesting piece of info you gave me... anyways, I googled it but I don't get what "internet public key encryption" is

[quote="illcrowflu"]A modular inverse,$a^{-1}$ is a number such that
$a\cdot a^{-1}\equiv 1 \pmod n$. A necessary condition for the existence of the inverse is that $\text{gcd}(a,n) = 1$.[/quote]

This condition is both necessary and sufficient, which is the content of the second problem.  The second problem encapsulates the first.

[hide="Proof"] Necessity:  If $gcd(a, n) > 1$, then $gcd(a \cdot a^{-1}, n) > 1$, but $a \cdot a^{-1}\equiv 1 \bmod n$, contradiction.  Hence $a^{-1}$ does not exist when $gcd(a, n) > 1$.

Sufficiency:  One way to prove this is by Bezout's Lemma.  I discuss this in an [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=127224]early blog post[/url].  Once we have established that result, we can find integers $x, y$ such that

$ax+ny = 1$

From which it follows that

$ax \equiv 1 \bmod n$

So $x = a^{-1}$.

QED.

[b]Edit:[/b]  Perhaps this proof is unnecessarily complicated.  Instead, we will prove that for any given modulus $n$ and integer $gcd(a, n) = 1$ the numbers

$\{ 1, ... n-1 \}\equiv \{ a, ... (n-1) \}\bmod n$

Where the LHS is the set of numbers $b$ such that $gcd(b, n) = 1$, and the RHS is that set multiplied by $a \bmod n$.

Clearly the numbers in the right set are all relatively prime to $n$, and there are the same number of them as there are numbers on the left side; so if they are all distinct $\bmod n$, the two sets must be the same set.  Now, if $i, j$ such that $gcd(i, n) = gcd(j, n) = 1$ then

$ia \equiv ja \bmod n \implies$
$(i-j) a \equiv 0 \bmod n$

Since $a$ has no common factors with $n$, we conclude that $n | i-j$, but this is a contradiction since $i, j \in \{ 1, ... n-1 \}$.

Our desired result then follows by noting that one of the numbers on the right side must be congruent to $1 \bmod n$.  QED.  [/hide]

[quote="jeez123"] anyways, I googled it but I don't get what "internet public key encryption" is[/quote]

Understanding this requires at least understanding Euler's Totient Theorem, but the general idea is that we are converting information to numbers and then raising that number to large powers modulo large primes to encrypt and decrypt it.


Sorry, but I didn't find that funny at all.

Then I will explain what you likely missed.

(Remember if you're being sincere and at 4 stars' worth,
you used the phrase "didn't ... at all" which essentially means 0%.)

The alien was just mortified in a very public way, so to partially save
face, he tries to switch gears to another world's students.

You can't rate posts in this forum, can you?
Also, Arrange your tan is a troll, so ignore... ignore...

Brut3Forc3,

you rated poorly.  You are spam.

Move along.

[quote="Arrange your tan"]The alien was just mortified in a very public way, so to partially save face, he tries to switch gears to another world's students.[/quote]
And I still don't find it funny.

I don't even get it. At least, I don't think I get it, because what I think is supposed to be the joke doesn't seem really funny.

Sorry.

[quote="QuantumTiger"]Sorry, but it just isn't funny........I get what you're saying, Arrange Your Tan, but it just isn't funny.[/quote]

Don't ever speak for all others.  You don't 'happen' to find it funny.
Stop there.  I find it funny as it parallels arrogant people who think
they have the only correct perception about something, such as the
alien, but it takes a simple truth (such as from the boy) to deflate
them, because they don't have real wisdom.  They're book smart, but
they fall harder from off of their high horses.  That's pretty funny to me.

[quote="QuantumTiger"]Uh....ok.....but it's not funny to me....however, we know nothing about aliens, so your "rationale" about them being "book smart" and not having "real wisdom" is inherently flawed........[/quote]

That is the hypothetical situation I created for this comic strip, so they can
have most whatever kind of intelligence/attitude/etc. I describe for these
*fictional* characters.

This thread should really be locked... it's very old, and has degenerated into flaming.


Can someone please go over this in depth?

and in particular, what's so special about the characterstic of a field to be a prime number, such as 2?

thanks

Start with the axioms of a field: a field has addition and multiplication. Both are commutative and associative, and multiplication distributes over addition. There's an additive identity ($0$), a multiplicative identity ($1$), and $0\ne 1.$ There are additive inverses, and for everthing other than zero, multiplicative inverses.

You can prove that, in a field, $xy=0$ iff $x=0$ or $y=0.$ The proof is simple: if $x\ne 0,$ then multiply by $x^{-1}.$

Now, consider this sequence of numbers in the field: $1,1+1,1+1+1,1+1+1+1,\dots.$ Either this is never equal to zero (in which case, by convention, we say we have a "field of characteristic zero"), or it does equal zero for some value.

Assume the latter case (a "field of finite characteristic.") Define the characteristic of the field to be the smalled positive integer $n$ such that $\sum_{k=1}^{n}1=0.$ Note that $n>1$ by the axiom $0\ne 1.$

Is it possible for $n$ to be a composite number? What happens if it is? Suppose $n=ab$ for some integers $a>1,b>1.$ Now consider

$\left(\sum_{j=1}^{a}1\right)\left(\sum_{k=1}^{b}1\right)=\sum_{k=1}^{n}1$

The identity here can be proven by using the distributive law, along with the associativity and commutativity of addition. But if the sum on the right is zero, then at least one of the two sums on the left must be zero - and that would violate the assumption that $n$ was the minimal integer for with the sum equaled zero.

Hence, for a field of finite characteristic, the characteristic must be prime. Once we've established that, we have as a followup theorem in a field of characteristic $p:$

$\sum_{k=1}^{p}x=0$ for all $x$ in the field.


limit
trigonometry
function
algebra
polynomial
calculus
calculus computations

a) Compute the limit $ \displaystyle \lim_{n \to \infty} \sin^{2} ( \pi \sqrt{n^{2}+n} ).$

b) Compute the limit $ \displaystyle \lim_{n \to \infty} \cos^{2} ( \pi \sqrt{n^{2}+n} ).$

hi
i suppose you meant it as a sequence and not a function (n is integer right?)

the trick is to find the fractional part of sqrt(n*n+n)

for all n>=1 : we have

sqrt(n*n+n) = n+   sqrt(n*n+n ) -n =n+      n /   ( sqrt(n*n+n)+n )

the last term is positive yet smaller than one, so it is the fractional part

now we can write sin (Pi sqrt(n*n+n ))=  sin (  Pi*n +   Pi*n/(sqrt(n*n+n)+n)   )

=(-1)^n * sin(Pi*n/(sqrt(n*n+n)+n))

now the argument of the sinus obviously tends to Pi/2 , by continuity the sinus has 1 as limit, but that means we have no limit, because the subsequence of even n tends to 1, the other subsequence of odd n tends to -1

the situation is different with the cosinus , there we have :

(-1)^n * cos(Pi*n/(sqrt(n*n+n)+n))

the argument in the cosinus tends to Pi/2, the cosinus tends to zero, the changing of signs is no problem : it goes to zero !

come to think of it: saying that it really is the fractional part isn't actually necessary

Another place where big-O, little-o, Taylor polynomial thinking is absolutely the right argument.

P.S. - I fixed a typo in Moubinool's post, in his first (binomial expansion) line.

[quote="fredbel6"]...the sinus has 1 as limit, but that means we have no limit, because the subsequence of even n tends to 1, the other subsequence of odd n tends to -1 [/quote]
Almost; the only thing you overlooked was the square. After squaring, both subsequences tend to 1.

:( once again i was too fast
indead, when squaring the sign problem disappears

[quote="Kent Merryfield"]

P.S. - I fixed a typo in Moubinool's post, in his first (binomial expansion) line.

[/quote]

I corrected Kent, thank's


Your inequality is false with $a = 1, b=c=\frac{1}{2}$.

This is my solution in .pdf and I used a generic conclusion.

since $f(x)=ln(\frac{1}{x}-1)$ is convex over $(0,1/2)$ ( because $f''(x)=\frac{1-2x}{x^2(x-1)^2}>0$

Hence it follows from jensen that

$f(a)+f(b)+f(c)\geq f(a+b+c/3)$ .

the equality holds when $a=b=c$ ;)


What is Harmonic mean. How do you apply to find average rates.

See [url=http://www.artofproblemsolving.com/Wiki/index.php/Harmonic_mean]here[/url].

The harmonic mean of $ n$ numbers is $ n$ divided by the sum of the reciprocals of those numbers.

The harmonic mean can be used to find the average speed or rate of a "round trip". For example, if someone is going somewhere at the speed of $ x$ and coming back the same distance with a speed of $ y$, the overall average speed is the harmonic mean of $ x$ and $ y$, no matter what the distance.

A great example of a harmonic mean problem can be found in this year's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314298]2009 AMC 8 Problem 14[/url].

Is it just me or did they have a problem just like that for this year's amc 8?

[quote="PowerOfPi"]A great example of a harmonic mean problem can be found in this year's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314298]2009 AMC 8 Problem 14[/url].[/quote]

Last line in my post.

It is useful because the distance, rate, time stuff with a weighted average takes too long.   :)


Anyone know a good solution to this one? sorry, this is the 2003 one....

Have you looked at the Math Jam transcpript for that? http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

Also, there were 2 AIME's in 2004.  You should specify which one and/or give the problem.

yes, I would give the problem, but i'm not to good at texing, it'll take me hours.

This one?

When written as a decimal, the fraction m/n (with m < n) contains the consecutive digits 2, 5, 1 (in that order). Find the smallest possible n.

Yes, ok so it doesn't need texing, I thought it did... I coudn't find the less than sign. hahaha, found it now.

where do oyu find the previous test questions?


probability
geometry
trigonometry
integration
calculus
conditional probability
probability and stats

Pick two points A and B independently at random in a circle with centre O. Show that the probability that the triangle AOB is obtuse is $ \frac{3}{4}$.

Let the circle be the standard unit circle.

Choose the point $ A$ first, and denote length $ OA$ by $ r.$

Draw the circle whose diameter is $ \overline{OA}.$ If $ B$ lies inside that circle, then $ \angle OBA$ is obtuse. The conditional probability of this happening, given $ A$, is the area of the small circle divided by the are of the big circle, namely $ \frac {r^2}4.$

Draw the diameter of the big circle perpendicular to $ \overline{OA}$ If $ B$ lies on the side of that opposite to $ A,$ then $ \angle AOB$ is obtuse. The conditional probability of this happening, given $ A$, is $ \frac12.$

Draw the chord through $ A$ perpendicular to $ \overline{OA}.$ If $ B$ lies on the "outer" side of this chord, then $ \angle OAB$ is obtuse. We have work a little harder to find the probability of this. Take a wedge with angular opening $ 2\cos^{ \minus{} 1}r$ and subtract a triangle of area $ r\sqrt {1 \minus{} r^2}.$ We get that the conditional probability of $ \angle OAB$ being obtuse is $ \frac1{\pi}\left(\cos^{ \minus{} 1}r \minus{} r\sqrt {1 \minus{} r^2}\right).$

These three regions have no overlap, so we can add these probabilities. The conditional probability of an obtuse triangle, given $ A,$ is $ \frac12 \plus{} \frac {r^2}4 \plus{} \frac1{\pi}\left(\cos^{ \minus{} 1}r \minus{} r\sqrt {1 \minus{} r^2}\right).$

The distribution of $ r$ has density $ 2r$ on $ [0,1]$ then the overall probability of an obtuse angle should be

$ \int_0^12r\left[\frac12 \plus{} \frac {r^2}4 \plus{} \frac1{\pi} \left(\cos^{ \minus{} 1}r \minus{} r\sqrt {1 \minus{} r^2}\right)\right]\,dr.$

This integral can be done, but let me back up a moment and write the conditional probability that $ \angle OAB$ is obtuse as $ \frac1{\pi}\int_r^12\sqrt {1 \minus{} x^2}\,dx.$

Then the the integral I need is the first two terms above, which integrate to $ \frac12 \plus{} \frac18 \equal{} \frac58,$ plus the following:

$ \frac1{\pi}\int_0^12r\int_r^12\sqrt {1 \minus{} x^2}\,dx\,dr$

$ \equal{} \frac4{\pi}\int_0^1\sqrt {1 \minus{} x^2}\int_0^xr\,dr\,dx \equal{} \frac2{\pi}\int_0^1x\sqrt {1 \minus{} x^2}\,dx \equal{} \frac {2}{3\pi}.$

So I'm getting the probability of an obtuse angle as $ \frac58 \plus{} \frac2{3\pi}.$ I'm not getting $ \frac34.$

Alternately, you could choose the other two points on the circle rather than inside. That gives a probability of $ \frac12$- only the central angle can be obtuse.

$ P($obtuse$ )\equal{}\frac{1}{2}\plus{}\frac{1}{8}\plus{}\frac{1}{\pi}E(\cos^{ \minus{} 1}r \minus{} r\sqrt {1 \minus{} r^2})\equal{}\frac{3}{4}$.

Correcting my mistake:

$ \frac1{\pi}\int_0^12r\int_r^12\sqrt {1 \minus{} x^2}\,dx\,dr$

$ \equal{} \frac4{\pi}\int_0^1\sqrt {1 \minus{} x^2}\int_0^xr\,dr\,dx \equal{} \frac2{\pi}\int_0^1x^2\sqrt {1 \minus{} x^2}\,dx \equal{} \frac{2}{\pi}\cdot\frac{\pi}{16}\equal{}\frac18.$

OK, now I have $ \frac12\plus{}\frac18\plus{}\frac18\equal{}\frac34$ and I agree.

There is a slightly better way to compute it. The probability that the obtuse angle is at O is $ 1/2$ (fix $ A$ and see where $ B$ can lie). The probability that the obtuse angle is at $ B$ is $ 1/8$ (Kent did it fine fixing $ A$) and for the probability that the obtuse angle is at $ A$, you'd better fix $ B$ (or just note that it is the same as for $ A$). There is no need to integrate those terrible square roots :).


How would you explain the problem solving method detailed below?

\begin{eqnarray*}
\log (x-1) &=& y \\
10^{\log (x-1)} &=& 10^y \\
\end{eqnarray*}

Application of an inverse function
Undoing a base-10 logarithm
Solving for x
Doing the same thing to both sides of an equation

... is that what you're looking for?

The base of the logarithm (10) raised to the exponent (y) equals the stuff in the middle of the logarithm (x-1).  This is a rule that you applied.....is this what you are asking?

Thanks for the replies. Normally you can say "raise both sides to the 10th power" but I wasn't sure what the "opposite" would be.

Ah! I see. I've struggled with that myself. You could try things like "apply the base-10 exponential" or "apply the function f(x) = 10^x". I'm not sure if "exponentiate with base 10" is a correct way of saying it -- if it is, you could use that.

I usually say "raise 10 to both sides" as a short form of "take ten to the power of both sides"
I [i]knew[/i] I wasn't the only one who wondered about this..


The digits of a four-digit positive integer are added. If the digits are all different, what is the greatest possible sum?

We need to maximize the digits, while keeping them all distinct. This is accomplished with $ 6$, $ 7$, $ 8$, and $ 9$.
$ 6\plus{}7\plus{}8\plus{}9\equal{}\boxed{30}$.

[hide="ANSWER"]1.we know the following facts

 I. the # is a 4 digit integer

 II.they want it to be the largest

2.knowing these facts we will make the # as big as possible by putting all the higest # 's into it :9,8,7,6
adding all these digits 9+8+7+6=30

3.[hide="ANSWER"]30[/hide][/hide]

[= i hope my answer is well!


There is a combination of exactly nine pennies, nickels, dimes, and quarters in Lisa's pocket. The coins have a total value of 79 cents. If one coin is randomly selected, what is the probability that it is a dime? Express your answer as a common fraction.

In this problem, we have to do a bit of guess-and-check.  Since $ 3$ quarters and $ 4$ pennies makes $ 7$ coins, we have to produce $ 2$ more--by changing $ 1$ quarter into $ 2$ dimes and $ 1$ nickel.  Thus, there are $ 2$ dimes.  The probability of drawing a dime is, therefore, $ \boxed{\frac{2}{9}}$.

Weird.  I had that question just yesterday.


On the circumference of a circle there are 24 equally spaced points. In how many ways can we choose 8 out of these 24 points so that the length of the minor arc joining any two of the chosen points is not equal to 3 or 8?

please explain to me with steps.. THZ A LOT!!!

[quote="wailam010"] ...so that the length of the minor arc joining any two of the chosen points is not equal to 3 or 8? [/quote]

Can you explain this part of the problem?


Given a convex quadrilateral $ ABCD$ and a point $ P$ in its interior such that $ AP \equal{} CP$, $ \angle ABC \equal{} \angle APD$  and $ \angle CDA \equal{} \angle CPB$. Prove that

$ DA\cdot AB\cdot BP \equal{} BC\cdot CD\cdot DP$.

[quote="jbmorgan"][quote="The QuattoMaster 6000"]This problem is an inverted image of http://www.artofproblemsolving.com/Forum/viewtopic.php?p=131841#p131841[/quote]

Can you explain a bit more? I cannot see clearly how they are related...[/quote]
For notation purposes, call $ B$ in this problem to be $ B'$ and $ D$ to be $ D'$.  Apply an inversion with radius $ AP^2$ to map $ B'$ to $ B$ and $ D'$ to $ D$.  The problem now becomes $ \angle APD' \equal{} \angle AB'C \equal{} \angle PBA \plus{} \angle PBC$, meaning that $ \angle CPB' \equal{} \angle PDB \plus{} \angle PDB$.  Analagously, $ \angle CPD \equal{} \angle PCB' \plus{} \angle PCD'$.  Then, $ ABCD$ has the same properties of $ ABCD$ in the link I posted above.  Now, in the link, I prove that  $ \triangle APD'\sim \triangle ABC$, meaning that $ \frac {PD'}{AP} \equal{} \frac {BC}{AB}$.  Since $ PD' \equal{} \frac {PA^2}{PD}$, we have that $ \frac {PA}{PD} \equal{} \frac {BC}{AB}$.  This means that $ BC\cdot PD \equal{} PA\cdot AB$.  Similarly, $ AD\cdot PB \equal{} PC\cdot CD$.  Multiplying and noticing that $ PC \equal{} AP$ gives us that $ CD\cdot BC\cdot PD \equal{} AD\cdot PB\cdot AB$.

Thanks. That's a nice solution.

Any solution without inversion?  :)


calculus
geometry
geometric transformation
rotation
special factorizations

What's the minimum of the expression $x^{2}+2xy+3y^{2}+6y+4x$, where $x,y\in\mathbb{R}$?

[hide="Cheap"]
Apply a rotational transformation.[/hide]

Alternatively, set first partials equal = 0 and solve. It would be disingenuous to solve this any other way. Sometimes calculus distorts math, but in this case, elementary math (rotation plus completing the square) is the ugly culprit.

A question like this would never appear on a real math contest because users of calculus could solve this in less than 30 secs.

Obviously:

\[x^{2}+2xy+3y^{2}+6y+4x=(x+y+2)^{2}+2(y+\frac{1}{2})^{2}-\frac{9}{2}\]
 :P

trivial inequality...blah blah...and there is an equality case...


Find all the pair of positive integer $ (a,n)$ satisfying each prime divisor of $ a^n\plus{}1$ is also prime divisor of $ a\plus{}1$

if n is even we have $ (a^n\plus{}1,a\plus{}1)\equal{}1 or 2$
if $ (a^n\plus{}1,a\plus{}1)\equal{}1$ contradicts
if $ (a^n\plus{}1,a\plus{}1)\equal{}1$  we have $ a\plus{}1$ and $ a^n\plus{}1$ both power of 2 but $ a^n\plus{}1\equiv2(mod 8)$so $ a\equal{}1$
so n is odd.
if we difine $ \parallel{}a\parallel{}(p)$ is the greatest pewer of p that divide $ a$(p is prime)
for n odd and p (prime) and p is not 2 with condition $ p|a\plus{}1$ we have:
$ \parallel{}a^n\plus{}1\parallel{}(p)\equal{}\parallel{}a\parallel{}(p)\plus{}\parallel{}n\parallel{}(p)$
and for p=2 also w have equation because both side are 1.(n is odd)
so if we multiple this equation for all primes that divide$ (a\plus{}1)$
so we  have $ a^n\plus{}1\equal{}n(a\plus{}1)$
now it is easy to find .

[quote="sumita"]
if we difine $ \parallel{}a\parallel{}(p)$ is the greatest pewer of p that divide $ a$(p is prime)
for n odd and p (prime) and p is not 2 with condition $ p|a \plus{} 1$ we have:
$ \parallel{}a^n \plus{} 1\parallel{}(p) \equal{} \parallel{}a\parallel{}(p) \plus{} \parallel{}n\parallel{}(p)$[/quote]
Hey Sumita , I think you made a mistake.
you only say that $ p|n$, cant say that the greatest power of$ p$ that divide  $ (a^{n\minus{}1} \minus{}...\plus{}1)$ is equal to  the greatest power of $ p$ that divide $ n$.

Zsigmondy's theorem strikes again :wink:

i said with condition $ p|a\plus{}1$
we have$ \parallel{}a^{n\minus{}1}\minus{}.....\plus{}1\parallel{}(p)\equal{}\parallel{}n\parallel{}(p)$


find $ f: N*\minus{}>N*$ such that:
$ f(m\plus{}f(n))\equal{}n\plus{}f(m\plus{}1)$

$ m\equal{}1$ gives $ f(m\plus{}f(1))\equal{}f(m\plus{}1)\plus{}1$
so $ \forall n\in\mathbb{N}^*: \ f(n\plus{}k)\equal{}f(n)\plus{}1$ where $ k\equal{}f(1)\minus{}1>0$
then $ f(m\plus{}1\plus{}kn)\equal{}f(m\plus{}1)\plus{}n$ gives $ \boxed{(1): \ f(m\plus{}f(n))\equal{}f(m\plus{}1\plus{}nk)}$

$ f(n)\equal{}f(m)$ gives $ n\plus{}f(m\plus{}1)\equal{}f(m\plus{}f(n))\equal{}f(m\plus{}f(m))\equal{}m\plus{}f(m\plus{}1)$ gives $ n\equal{}m$ ($ f$ is injectiv)
so $ (1)$ gives $ m\plus{}f(n)\equal{}m\plus{}1\plus{}nk\Rightarrow f(n)\equal{}1\plus{}nk$
and $ \boxed{f(m\plus{}f(n))\equal{}n\plus{}f(m\plus{}1))}$ gives $ 1\plus{}k(m\plus{}kn\plus{}1)\equal{}n\plus{}1\plus{}k(m\plus{}1)\Rightarrow k\equal{}1$

$ \boxed{\forall n\in\mathbb{N}^*: \ f(n)\equal{}n\plus{}1}$

What is the asterisk * meant over $ \mathbb{N}$?

[quote="Fermat's Little Turtle"]It means including $ 0$.[/quote]
It means not including $ 0$ :blush:
thanks pilot!

No...the natural numbers strictly greater than zero is denoted $ \mathbb(N)$
I've come across many ways to represent the whole numbers, such as $ \mathbb{W},\mathbb{N_*},\mathbb{N_0}$

But then again with a superscript I've seen it denote excluding 0.. like $ \mathbb{N^*}$

[quote="Math pro"]find $ f: N* \minus{} > N*$ such that:
$ f(m \plus{} f(n)) \equal{} n \plus{} f(m \plus{} 1)$[/quote]
Very bad titile .  :?:


superior algebra
superior algebra unsolved

Determine the roots of unity in the field of $ p$-adic numbers.

[i]L. Fuchs[/i]

[hide="Hint"]Use the decomposition $ \mathbb Q_p^*\equal{}p^{\mathbb Z}\times \mu_{p\minus{}1}\times (1\plus{}p\mathbb Z_p)$ which is a  group isomorphism. $ \mu_{p\minus{}1}$ denotes the group of $ (p\minus{}1)$-th roots of unity having order $ p\minus{}1$, by Hensel's lemma.[/hide]


function
calculus
derivative
real analysis
real analysis theorems

Suppose we have a function $f$ and for all value in the interval $(a,b)$, $f$ is a real function.

If $f$ is continuos and differentiable, is possible that $f$ have a complex value for its derivative in the interval $(a,b)$.

I think it's impossible.
But is it true?

It is for a problem that i made to evaluate some integral.

thanks.

Look at the definition of derivative.  Since $f$ is always real in that range, you have a limit of real values.  A limit of real values cannot have nonzero imaginary part.

I know the fermat definition of tangent.

I just asking if there is no strange function like this.


quadratics
algebra
polynomial
function
system of equations

[quote="Temperal"]Given for quadratic polynomial function $ f: \mathbb{C}\rightarrow\mathbb{C}$ that $ f(1) \equal{} f( \minus{} 1) \equal{} i$, find $ f^{2008}(i)$.[/quote]

[hide="its a start"]
it's implied that the function f has no x term
so it's $ f(x) \equal{} ax^2 \plus{} b$
now if we plug in 1, it gives $ a \plus{} b \equal{} i$
dunno how to continue
[/hide]

This is just sort of a guess, but here it is anyway. If anyone else gets it or thinks this might be correct let me know.

I guessed that $ f(x) \equal{} x^2 \minus{} x \plus{} i$, because it satisfies $ f(1) \equal{} f(\minus{}1) \equal{} i$

So for $ f(i)$, we get $ i^2 \minus{} i \plus{} i \equal{} \minus{}1 \minus{} i \plus{} i \equal{} \minus{}1$. Now the second iteration will put -1 back into the function, but we already know that $ f(\minus{}1) \equal{} i$, so we have a loop. It isn't hard to see that when the iteration is an even number, the output is $ i$, and when it's an odd number, the output is $ \minus{}1$. Therefore, $ f^{2008}(i) \equal{} i$

[quote="Lulze"]

I guessed that $ f(x) \equal{} x^2 \minus{} x \plus{} i$, because it satisfies $ f(1) \equal{} f( \minus{} 1) \equal{} i$

[/quote]
Uh-oh, I think $ f(\minus{}1)$ is not $ i$ but $ 2\plus{}i$.

[hide="Comments"]
@Lulze: $ f(\minus{}1) \equal{} (1) \minus{} (\minus{}1) \plus{} i \equal{} 2 \plus{} i \neq i$.

As for the original question, consider $ f(x) \equal{} x^2\minus{}1\plus{}i$ and $ f(x) \equal{} 2x^2 \minus{} 2 \plus{} i$. Both satisfy the conditions but give different values for $ [f(i)]^{2008}$.
[/hide]

Hm, it seems that I made a mistake in formulating the problem. I'll need to fix that...

[quote="ejr322"]All I have so far is:

[hide]

ax^2 + bx + c = 0

pluggin in 1 and -1,

a + b +c = a -b + c

This implies there is no "b" term, and that a + c = i.

So we have: ax^2 + c = i.

This imples that we have a polynomial with complex coefficients, since if it were real, our criteria would be false. (this might be a bit of a leap, but i see no other way to solve it, since if c is a complex number, we can get nowhere - there are infinitely many polynomials that we can make based on these criteria eg (x^2 + (-1+i)).The only way this could be true is if it were a zero coefficient, but this would not be a quadratic polynomial.

Making a system of equations, we have:

a + c = i, from our original equations
a - c = i, since it has complex coefficients, and using our original criterion.

We can see there is no c term either. It is implied the coefficient,a, is just i.

Therefore, if i am understanding the question right, f^2008 (i) = f(i) ^ 2008.

This means a(i) ^ 2008. Since a is i, our answer is (i^2)^2008 = (-1) ^ 2008, which, since negative 1 raised to any even power is 1, is equal to 1. [/hide][/quote]

no. u r misunderstanding in your last step. $ f^{2008}(i)\neq f(i)^{2008}$
instead, $ f^{2008}(i)\equal{}f(f(f(f(.....f(i))))))))$ 2008 times :D

If my other assumptions are right, then:

[hide] f(x) = i(x2)

f(i) = i(i^2) = -i
f(-i) = (-i)^2 (i) = -i

....

So its just -i, i think.

$ f(x)\minus{}i$ has two roots, $ 1$ and $ \minus{}1$, so since it is a quadratic, we can write

$ f(x)\equal{}a(x\minus{}1)(x\plus{}1)\plus{}i$ where $ a$ is a constant.

Now this is obviously not uniquely defined.


Show that the equation x2 + y2 + z2 = 3xyz has infinitely many solutions in positive integers.

Pretty easy. The first one on the test. Have fun!

[hide]
Note that $x=y=z=1$ satisfies the equation.

Assume a solution exists, then look for a bigger solution. Hence, we know:
$x^{2}+y^{2}+z^{2}= 3xyz$ (1)
Replace $z$ with $z+k$ for some positive $k$. Hence:
$x^{2}+y^{2}+z^{2}+k^{2}+2kz = 3xyz+3xyk$
$k=3xy-2z$ {subtracting (1) and dividing across by $k$}

Hence, we can create a larger $z$ that satisfies the equation. We can repeat the same process for $x$ and $y$ instead of $z$, ad infinitum, hence an infinite number of $(x,y,z)$ pairs can be created.
[/hide]

Now try solutions to $x^{2}+y^{2}+z^{2}=xyz$

problem already discussed:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=455324#p455324


Fie functia $\ f: \mathbb{C}\rightarrow\mathbb{C}$.Aflati toate functiile $\ f$ astfel incat sa avem $\ f(xy)=f(x)+f(y)$, $\ x,y>1$.

Stiu ca o solutie este $\ f(x)=\log x$, adica $\ f$ e logaritmica.

Cred ca enuntul era urmatorul:
Determinati functiile $ f : (1,\infty)\rightarrow\mathbb{R}_{\plus{}}^{\ast}$ pentru care $ f(xy) \equal{} f(x)\plus{}f(y)$, pentru orice $ x, y > 1$.
[b]Solutie[/b]. Consideram $ f$ indeplinind conditiile ipotezei. Punem $ x \equal{} e^{a}$, $ y \equal{} e^{b}$, $ \forall a, b\in\mathbb{R}_{\plus{}}^{\ast}$. Atunci $ f(e^{a\plus{}b}) \equal{} f(e^{a}\cdot e^{b}) \equal{} f(e^{a})\plus{}f(e^{b})$, deci functia $ g:\mathbb{R}_{\plus{}}^{\ast}\rightarrow\mathbb{R}_{\plus{}}^{\ast}$, $ g(x) \equal{} f(e^{x})$ satisface $ g(a\plus{}b) \equal{} g(a)\plus{}g(b)$, pentru orice $ a, b\in\mathbb{R}_{\plus{}}^{\ast}$, deci ecuatia functionala a lui Cauchy pe $ \mathbb{R}_{\plus{}}^{\ast}$. Rezulta $ f(x) \equal{} g(\mathrm{ln}(x))$, pentru orice solutie $ g$ a ecuatiei lui Cauchy, care reciproc verifica ipoteza.
Pentru solutia uzuala $ g(x) \equal{} kx$, $ k\in\mathbb{R}_{\plus{}}^{\ast}$ o constanta, avem $ f(x) \equal{} k\mathrm{ln}(x) \equal{}\log_{e^{k^{\minus{}1}}}(x)$, adica [b]logaritmica de baza supraunitara[/b].

Daca functia din ecuatia functionala a lui Cauchy este pozitiva, atunci ecuatia are doar solutia $ f(x)\equal{}kx$


AMC
USA(J)MO
USAMO
geometry
perpendicular bisector

How do you describe a solution to a construction problem?

that is a very generic question. try asking a more specific question('how does one construct...') and you might get a better answer.

I am not sure. You might get more response if you put this in  the high school section.

I think you will be okay if you just first describe each step of the construction, leaving out the explanation for each part of the construction, then prove that your construction works at the end.

Usually, most of the construction problems I do will have extraneous lines (for example, constructing a simple set of perpendicular lines) that indicate you constructed it, instead of using a protractor and ruler.

You could also describe how you constructed, similar to a procedure, but I'm pretty sure that most construction problems will just have you...construct.

i mean like how do you describe where you put the compass and the starightedge? Like, how can i describe how to place the compass and stuff?
also, do any competitions have construction problems? Maybe MOP?

When you are writing up a solution to a USAMO level construction problem, certain basics will not need to be proven.

For example, one basic construction is constructing a perpendicular bisector. You can just say "constuct the perpendicular bisector of $ AB$." You will not need to describe how you use the intersection of circles and stuff to make the perpendicular bisector. People reading your proofs at Olympiad level would know that that would be trivial for you to prove. You can think of the basic constructions as your "tools" for construction problems.

Other basics you can just use include, but are not limited to
1. constructing the midpoint of a line
2. constructing a line through some point $ C$ that is parallel to some line $ AB$
3. constructing a line segment of length ax+by, where you are given the lengths a and b, and x and y are integers, not necessarily positive
4. constructing a line of length $ \sqrt a$, where you are given the length a.
5. Reflecting anything over a line
6. rotating any figure about a point.
hm there are probably many more, but I am not the person to ask, since I have done 0 olympiad level construction problems. (everything I am saying here is a guess of what I am pretty sure how these proofs work.....but I have had one of my proofs for an easy construction problem graded by a teacher who knew what he was doing, and I followed these guidelines on that proof)
The above six are only a few of the basic constructions.

Basically if you are asking whether you should write "center your compass about point $ A$ and open your compass to a length of......", then the answer is no, you don't have to describe the construction like that. You can just say "construct the circle centered at $ A$ with radius $ r$" where $ r$ is a length you can copy. You don't need to describe where you put your compass and straightedge - everything in construction problems can pretty much be described by some combination of the tools of construction.

whether a construction is considered a "basic" contruction that you can use as a tool might be hard to determine in some cases - just use your best discretion (and think about who your audience is). If you are unfamiliar with construction problems, it should not hurt to practice coming up with a proof for a basic constructions every now and then.

USAMO and IMO will have construction problems for time to time, (probably not rarely, but not every year either, I would think. Maybe one every few years. Then again, how would I know?...)


function
calculus
integration
algebra
floor function
number theory solved
number theory

Let $p$ be prime number.Find the value of :$\displaystyle S(p)=\sum_{k=1}^{(p-1)(p-2)}[\sqrt[3]{kp}]$

This is a nice cute little problem. Here is my solution:

We show that S(p)=(p-1)(p-2)(3p-5)/4.
First observe that S(p)=:Sigma: ([(i+1)3/p]-[i3/p])*i, where [] denotes the floor function with i ranging from i to p-2. Note that this is indeed true because as k ranges from 1 to (p-1)(p-2), kp stays inside interval (i3, (i+1)3) for exactly ([(i+1)3/p]-[i3/p]) values of k (for each i from 1 to p-2), namely k=[i3/p]+1 to k=[(i+1)3/p]. Here, the interval is open on both sides because i3/p is not integral for all i=(1,2,..,p-2). (Recall that p is prime!) Hence for those values of k listed above, [(kp)1/3] assumes the value of i. Summing over all values of i plainly give S(p).

Now S(p)=:Sigma:([(i+1)3/p]-[i3/p])*i=(p-1)*[(p-1)3/p]-:Sigma:[i3/p], where the latter sum is taken over i=1 to i=p-1. (Note that sum S(p) is telescopic.) But while i ranges from 1 to p-1, :Sigma:[i3/p]=:Sigma:[(p-i)3/p]=:Sigma:(p2-3pi+3i2-1) - :Sigma:[i3/p]. So by summing twice the desired sum, we have :Sigma:[i3/p]=(p-1)(p-2)(p+1)/4.

Finally, we have S(p)=(p-2)(p-1)2-(p-1)(p-2)(p+1)/4=(p-1)(p-2)(3p-5)/4. As claimed.


Катюша ...
I know that this is quite well-known russian song and I would like to have words ... Does anybody have them ?   In russian, I mean ... I will be able to understand them  :wink:  ...

Расцветали яблони и груши,
Поплыли туманы над рекой.
Выходила на берег Катюша,
На высокий берег на крутой.

Выходила, песню заводила
Про степного, сизого орла,
Про того, которого любила,
Про того, чьи письма берегла.

Он ты, песня, песенка девичья,
Ты лети за ясным солнцем вслед.
И бойцу на дальнем пограничье
От Катюши передай привет.

Пусть он вспомнит девушку простую,
Пусть услышит, как она поет,
Пусть он землю бережет родную,
А любовь Катюша сбережет.

Расцветали яблони и груши,
Поплыли туманы над рекой.
Выходила на берег Катюша,
На высокий берег на крутой.

This is the version I have on my PC...


Is there a general method to solve recurrence relations that depend on n, like this one:

$f(n)=(1-\frac{1}{2n-2})f(n-1)$

with f(1)=1

[quote="Wumbate"]Is there a general method to solve recurrence relations that depend on n[/quote]

No.  For the example you gave, the obvious thing to do is to write it as a product.

However, methods exist to find explicit forms of certain linear relations such as $a_{n}= \alpha a_{n-1}+\beta a_{n-2}+\gamma a_{n-3}+\cdots.$ For other such strange relations, you may have to play around with the relation. In this case, products would work very well. A well known technique for proving strange relations is experimentation, conjecturing an explicit form, and proving the form to be true by induction.

Google should produce much more results: http://mathworld.wolfram.com/RecurrenceEquation.html


Earlier today, my friend showed me a site in an attempt to prove me wrong that there was only one student from my school accepted into MIT, and suceeded.  The site said there were none.  /endsegway

However, the site was (according to him) also supposed to show applications of rejected students, and there were none when we searched the site, although several people in my class saidthey knew people from my school who had applied.

I was wondering what sites similar to that one exist, and their use/ reliability.

Hmmm?

Colleges don't post the applications of their students. High schools usually don't have access to the full applications of their students.

If there was a site in which students posted applications to colleges from where they were rejected, that would obviously not be reliable. I could imagine someone pumping up their applications and going on a tirade about how unfair the admissions process is, or something like that.

Well, there is collegeconfidential.com, which applies somewhat.

My school has Naviance, which makes scatterplots of [ACT score] vs. [GPA] for a certain college and each dot represents an anonymous person's ACT/GPA from our school and the dot's color shows whether they got accepted/rejected/waitlisted. It's fun to go through it, but it doesn't work for colleges where not enough applied due to privacy reasons.

[quote="alexhhmun"]/endsegway[/quote]

As much as I'd like to end those two-wheel, self-balancing vehicles, I think you mean "segue."

Naviance is probably the site you are thinking about. If a particular high school doesn't upload information to a Naviance account, the information isn't there for that high school.

Aside: Yes, Segway is a brand name for a silly motorized vehicle, while a segue is a transition to another topic in a narrative.

I got my naviance account at school a few weeks ago and that's exactly what I was referring to, but is there anything similar to that?


My professor recently taught us partial derivatives and used it on an implicit function to find the first derivative. Knowing that dy/dx = -Fx/Fy, I thought it was great that I don't have to do the implicit differentiation the long way. However, I was wondering, by using partial derivatives, if there is something similar to the first derivative for the second derivative. Is there one? and if there is, can somebody post the proof. Thanks.

Joel

alright then, then just tell me if there is one.

The only thing I can think of for this is that once you have the first derivative, you just apply the same process again to get the second derivative.  I don't know of a pretty way of getting straight to a second derivative using parital derivatives, however.  (Why don't you ask your professor?)


vector
linear algebra
combinatorics unsolved
combinatorics

Let X be a non-empty set and suppose we are given
some subsets of X. Each of these subsets have an odd
number of elements, and each pair of the subsets
have an even number of elements in common (non-zero).

Find the maximum possible number of elements in X.

Excuse me my silly question. "In commoon" it is $A\cup B$ or $A\cup B$?

Anyway consider the set $X=\{1,2,...,2n\}$ and its subsets $X_i=X\{i\}$. Then $|X_i|=2n-1$, $|X_i\cap X_j|=2n-2$ and $|X_i\cup X_j|=2n$.

So number of elements of $X$ cannot be bounded. :?

I think that Clo wanted to ask the following. Let $X$ be an $n$-element set. Find, in terms of $n$, the maximal possible cardinality of a collection $F \subset 2^X$ of subsets of $X$ such that: 1) every set in $F$ has odd cardinality; and 2) for any two different sets $A,B \in F$, $|A \cap B|$ is a non-zero even number.

Well, assign to each set $A \in F$ its characteristic vector $v_A \in \{0,1\}^n$. Then, the $|F|$ vectors $v_A$ must be linearly independent over $\mathbb{F}_2[x_1,\ldots,x_n]$, giving immediately the bound $|F| \leq n$. It is easy to construct an $n$-element collection with the required properties for $n \geq 6$, and the small cases are easy to check. Thus, the answer to the problem is:

$n$, for $n \geq 6$;
$4$, for $n = 5$;
$2$, for $n=4$;
$1$, for $n=3$;

--Vesselin

Have a look here for this problem and related ones.
http://www.mathlinks.ro/Forum/viewtopic.php?t=13941&highlight=linear+algebra
[Moderator edit: Actually, this exact problem doesn't occur in the above link. Instead, the link considers questions where the sets $ A$ and $ B$ are not required to be different.]

Pierre.


let $ a \leq b \leq 0$ and $ x,y$ is nonegative.prove that:
$ (x^a\plus{}y^a)^\frac{1}{a} \leq (x^b\plus{}y^b)^\frac{1}{b}$

[quote="tuantam1lan"]let $ a \leq b \leq 0$ and $ x,y$ is nonegative.prove that:
$ (x^a \plus{} y^a)^\frac {1}{a} \leq (x^b \plus{} y^b)^\frac {1}{b}$[/quote]

try

$ x \equal{} 2,y \equal{} 1,b \equal{} \minus{} 1,a \equal{} \minus{} 2$

 :lol:


When you think the qualifying quiz will come out?

i'm not sure though i guess feb.
for some reason everything happens then. if not then jan

[quote="ztbb"]We're working on it now, and hope to have it ready before the end of the year.[/quote]

Cool, thanks.


Two cans containg coffee have the same shape and are made out of the
same tin. One can weighs 2 kg and is 12 cm high; the other weighs 1 kg
and is 9.5 cm high. Find the net weight of the coffee in both cans.

[quote="G-UNIT"]i dont think it would be right blurting out the answer until someone gives a genuine attempt to solve the problem.......[/quote]
I tried, but I can't get it.

hmmm, perhaps a hint?

[hide="hint"] Denote the wieght of the coffee in the largen can $x$, and the smaller by $y$.  Let the weights of the cans themselves be $z$ and $t$, repectively.  We then have the equations:

x + z=2

y+t = 1

Now we must find the ratio of x:y.  Remember [b] volumes[/b] [/hide]

i need this question answered: Are they proportional or is only the height changing?

[quote="math92"]this is challenging, this is going to take me a while[/quote]

umm... yeah, someone already did it  :roll:  ;) i'm not saying it's right, but why don't you look at his solution and see if it makes sense or not?

[quote="Hamster1800"]assuming that the cans are similar solids and the coffee in each weighs the same:

$x = \text{weight of coffee}$
$y = \text{what makes } V=yh^3$

working in grams:

$12^3y + x = 2000$
$6\cdot12^2y + \frac x2 = 1000$

$9.5^3y + x = 1000$

$6\cdot12^2y + \frac x2 = 9.5^3y + x$

$6\cdot12^2y-9.5^3y = \frac x2$

$12^3y - 18 \cdot 9.5^2y = x$

$12^3y + 12^3y - 19\cdot9.5^2y = 2000$

$3456y - 1714.75y = 2000$

$1741.25y = 2000$

$y = 1.1486001...$

$1728y + x = 2000$

$1728\cdot1.1486001... + x 2000$

$1984.7810... + x = 2000$

$ x = \boxed{15.21895... \text{ grams}} = \boxed{0.01521895... \text{ kg}}$

i think that's right but it is such a small amount of coffee.....[/quote]

  well you're suppose to find the net weight of the coffee in both cans, but either way thats wrong

well either i'm misreading the question or maybe not but if they have different amounts of coffee wouldn't there be an infinite number of solutions? there are 3 variables and only 2 equations...

Well, I guess it's been long enough, heres the solution:

  Denote the weight of the coffee in the larger can by $x$, and that of the smaller by $y$.  Let the weights of the cans themsleves be $z$ and $t$, respectively.  We then have the equations:

$x$ + $z$ =2

$y$ + $t$ =1

  Since the weights of the contents of the contents of full cans are in the smae ratio as their volumes, that is, as the cubes of their heights,* follows that:

$\frac{x}{y}$ = $\frac{12^3}{9.5^3}$ $\approx$ 2.02 or $x$= 2.02y

  Now the weights of the empty cans are in the same ratio as their total surface areas, which is to say, as the squares of their heights.  Therefore,

$\frac{z}{t}$= $\frac{12^2}{9.5^2}$ $\approx$ 1.60 or $z$= 1.60 $t$

  Substituting the values of $x$ and $z$ into the first equation, we get the following system:

2.02$y$ + 1.60$t$ =2
      $y$ +        $t$ =1

  Solving it, we find:

$y$= $\frac{20}{21}$=0.95,   $t$ = 0.05

  And so

 $x$=1.92,    $z$=0.08

  Thus, the net weight of the coffee in the larger can is 1.92kg, nad that in the smaller is 0.94 kg.

uhh...shouldn't the weight of the empty cans also be in the same ratio as their volumes. You didn't state that the cans were made out of sheets. Besides, they'd need to be nonexistant with a thickness of 0

btw, that surface area equation was the key to the whole thing. It was the 4th equation that was needed to solve 4 variables

This proportion is applicable only when the cans are thin. This in because,
strictly speaking, the outer and inner surfaces of cans are not similar; what
is more, the height of the inside part of a can is, strictly speaking, different
from the height of the can itself.

However you never stated that they were thin. Therefore, as my teammate would say, that problem was faulty.


arithmetic sequence
Diophantine equation
Divisibility Theory
Pell s Equation

Find infinitely many triples $(a, b, c)$ of positive integers such that $a$, $b$, $c$ are in arithmetic progression and such that $ab+1$, $bc+1$, and $ca+1$ are perfect squares.

But those aren't even in arithmetic progression??

My idea:
\[ (a, 4m(4ma\minus{}1), 4n(4na\plus{}1) )\]
We have three variables, in order to tinker:
1. the expressions in an arithmetic progression
2. solve the pell eq. $ 4mn(4ma\minus{}1)(4na\plus{}1) \plus{}1 \equal{} k^2$

This is possible but messy unless mods start messing things up. In that case play around with the +/- 1 in the second two of the three numbers ...

me@home // Hawk Tiger's solution is almost right. It just has a little mistake.
In his solution, (b,c,a) (not (a,b,c)) is arithmetic progreesion. (b=2u-v, c=2u, a=2u+v)

since the pell's equation gives us infinite number of positive integers $u,v$ such that $v^{2}-3u^{2}=1$, you can easily check that : $a=u$ ,$b=4u+2v$,$c=7u+4v$ works, as we have :
$ab+1=(u+v)^{2},$
$bc+1=(5u+3v)^{2},$
$ca+1=(2u+v)^{2}.$

[quote="Hawk Tiger"]Maybe not hard,but I spend 1 hour to solve it..... :oops:
We know that there exist infinitely many solution for the Pell's Equation:
$v^2-3u^2=1$
Let $a=v+2u,b=-v+2u,c=2u$,(we can check that $a,b,c>0$)
then $ab+1=4u^2-v^2+1=u^2$,
$bc+1=4u^2-2uv+1=u^2-2uv+v^2-1+1=(u-v)^2$,
$ac+1=4u^2-2uv+1=u^2+2uv+v^2-1+1=(u+v)^2$[/quote]

I understand the latter half of your solution, mainly once we have a, b and c, we can check to see if they are part of an arithmetic progression and that ab+1, bc+1 and ca+1 are perfect squares. However, I am confused about how your solution guarantees an infinite number of triples that meet these requirements.

Can you clarify the connection between your Pell Equation and your a, b, and c triple?

Thank you

Are these the only triples that satisfy the conditions?

Well a crude Python search up to $10000$ motivated me to consider the Pell Equation $x^2-3y^2=1$.
[hide=code]
[pywindow]L=100 #Don't set to higher because your computer may crash :D
#I used IDLE to compute mine, AoPS can't really handle big code

def isSquare(n):
    if n**0.5==int(n**0.5):
        return True
    return False

def progsq(x,y,n):
    ssum=0
    for i in range(0,n-1):
        for j in range(i+1,n):
            ssum+=isSquare((x+i*y)*(x+j*y)+1)
    return ssum

for i in range(1,L):
    for j in range(1,L):
        if progsq(i,j,3)==3:
            print([i,j])
[/pywindow]
[/hide]

By [b]Lemma 13[/b] [url=https://web.math.pmf.unizg.hr/~duje/pdf/bound.pdf]here[/url], if $a<b<c$ and all of $ab+1.bc+1,ca+1$ are perfect squares, then either $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If $a,b,c$ form an arithmetic progression, then the second possibility cannot occur, and the first (together with the AP condition) then implies that $b$ is even and $y^2 - 3(b/2)^2 = 1$, where $y=b-a=c-b$. Hence all such triples come from the Pell equation.

This implies that there are no arithmetic progressions of length $4$ satisfying a similar condition.

[quote=hamup1]Could someone explain the motivation for the above solution? I understand fully, but where would you get the motivation to use the Pell Equation $v^2-3u^2=1$?[/quote]

I am not entirely sure if this constitutes as motivation, but the goal is to find some parameterization of (a, b, c) in terms of polynomial expressions. (In our case, that would be polynomials in u and v, which are specifically a=u, b=4u+2v, and c=7u+4v). The existence of an infinite number of solutions for (a,b,c) is then analogous to an infinite number of (u,v) pairs.

"Infinite solutions" and "polynomials" triggers Pell's equations, since this imposes favorable constraints that can help manipulate (u, v) to their desired form (that is ab+1, bc+1, ac+1 being squares). At that point you can start searching for linear expressions in u, v for a, b, c and some D for which $u^2-Dv^2=1$. Keep in mind you want things to be as simple as possible. Following these guidelines does eventually get you to a solution.

Hope that helps.

[quote=hamup1]Could someone explain the motivation for the above solution? I understand fully, but where would you get the motivation to use the Pell Equation $v^2-3u^2=1$?[/quote]
Another possible motivation:
Firstly, to utilize the AP condition, we make the standard substitution $(a,b,c) \mapsto (a-v,a,a+v)$, an then we have that
$$\left\{\begin{array}{l}a^2-av+1=A^2\\a^2+av+1=B^2\\a^2-v^2+1=C^2\end{array}\right.$$
Now, the key step is to consider the [hide=transformation]There are many methods to motivate this.
Firstly, looking at the annoying $av$ terms motivates setiing $v=2$, but that does not help... but using this idea we can think of setting $v=2z$. That seems better, but not that useful. But what if we set $a=2u$... Boom!
Another (useful) common trick is to multiply a square expression by $4$, for instance if we have $k^2+k+1=x^2$, then $(2k+1)^2-3=4k^2+4k+4=(2x)^2$.
[/hide] $a \mapsto 2u$. Then some wishful thinking gives
$$\left\{\begin{array}{l}4u^2-2uv+1=(u-v)^2+3u^2-v^2+1\\4u^2+2uv+1=(u+v)^2+3u^2-v^2+1\\4u^2-v^2+1=u^2+3u^2-v^2+1\end{array}\right.$$
Where motivated by the $2uv$, we wrote the first expression in the form $(u-v)^2+\cdots$ and similarly for the second one, and seeing carefully, we find that the annoying (common) [hide=component]$3u^2-v^2+1$[/hide] in the first 2 expressions also appears in the last expression! So, quite naturally, we try and set $3u^2-v^2+1=0$, and this is, fortunately, a form of Pell's equation!

[quote=Hawk Tiger]Maybe not hard,but I spend 1 hour to solve it..... :oops:
We know that there exist infinitely many solution for the Pell's Equation:
$v^2-3u^2=1$
Let $a=v+2u,b=-v+2u,c=2u$,(we can check that $a,b,c>0$)
then $ab+1=4u^2-v^2+1=u^2$,
$bc+1=4u^2-2uv+1=u^2-2uv+v^2-1+1=(u-v)^2$,
$ac+1=4u^2-2uv+1=u^2+2uv+v^2-1+1=(u+v)^2$[/quote]

In your solution I  see that $ 2b \neq a+c $

[quote=inom]I found an expression which gives infinitive sln.
$ (n^2\minus{}1,n^2\plus{}2n,1)$

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

In this case $ 2b \neq a+c $

[quote=Tuhin_Bose]
In your solution I  see that $ 2b \neq a+c $[/quote]
But $2c=a+b$  ;)


Let $ a,b,c$ be non-negative real numbers such that
$ a \plus{} b \leq c \plus{} 1,b \plus{} c\leq a \plus{} 1,c \plus{} a\leq b \plus{} 1$
Prove that:  $ a^2 \plus{} b^2 \plus{} c^2\leq2abc \plus{} 1$

[quote="aakansh92"]Let $ a,b,c$ be non-negative real numbers such that
$ a \plus{} b \leq c \plus{} 1,b \plus{} c\leq a \plus{} 1,c \plus{} a\leq b \plus{} 1$
Prove that:  $ a^2 \plus{} b^2 \plus{} c^2\leq 2abc \plus{} 1$[/quote]
Assume $ a \ge b \ge c$, then from
\[ a\plus{}b \le c\plus{}1 \le b\plus{}1\]
we get
\[ a,b,c \le 1\]
The inequality is equivalent to
\[ (b\plus{}c)^2 \plus{}a^2\minus{}2bc \minus{}2abc \le 1\]
\[ \Leftrightarrow (b\plus{}c)^2\plus{}a^2 \minus{}2(a\plus{}1)bc \le 1\]
\[ \Leftrightarrow \frac{1}{2}(b\plus{}c)^2(1\minus{}a)\plus{}a^2 \plus{}\frac{1}{2}(a\plus{}1)(b\minus{}c)^2\minus{}1\le 0\]
From $ a\plus{}b \le 1\plus{}c$ and $ a\plus{}c \le 1\plus{}b$, we get
\[ \left\vert b\minus{}c \right\vert \le 1\minus{}a\]
and using $ b\plus{}c \le a\plus{}1$, it suffices to prove that
\[ \frac{1}{2}(1\minus{}a)(1\plus{}a)^2 \plus{}a^2 \plus{}\frac{1}{2}(a\plus{}1)(1\minus{}a)^2 \minus{}1 \le 0\]
But it is true since
\[ \frac{1}{2}(1\minus{}a)(1\plus{}a)^2 \plus{}a^2 \plus{}\frac{1}{2}(a\plus{}1)(1\minus{}a)^2 \minus{}1  \equal{}0\]
we have done. :)

:D .You're so clever ;) .
Another solution ;)
Setting :
$ {a + b = 1 + c + x ;\\
c + b = 1 + a + y; \\
c + a = 1 + b + z ;\\
x,y,z \ge 0}$
The ineq equivalent to
$ 18(xy + yz + zx) + 16xyz + 6(x^2 + y^2 + z^2) + 7[xy(x + y) + zy(z + y) + xz(x + z)] + 2(x^3 + y^3 + z^3) \ge 0$
;)

My solution:
The inequality is equivalent to proving that :
$ a^{2}\plus{}b^{2}\plus{}c^{2}\leq 2(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \plus{}1$ because of the well known alternate form of the Schur's Inequality: $ abc \ge (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$.

Now, let $ r\equal{} a\plus{}b\minus{}c$
             $ p\equal{}b\plus{}c\minus{}a$
             $ q\equal{}c\plus{}a\minus{}b$
The inequality can now be equivalently written as follows:
$ (p\plus{}q)^2 \plus{} (q\plus{}r)^2 \plus{} (r\plus{}s)^2 \le 8pqr \plus{} 1$

Now consider $ f(p,q,r) \equal{} (p\plus{}q)^2 \plus{} (q\plus{}r)^2 \plus{}(r\plus{}s)^2 \minus{}8pqr$ . It can be easily checked that $ f$ is convex.
By the first part of the given problem statement, $ 0\le p,q,r \le 1$.
So now checking the maxima cases, the minimum value comes to $ 1$.
Done.


A biased coin with probability p; 0<p<1 of head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p equals to

Hi umesh;

This is how I interpret the problem:

[hide]

The chance that a head comes up on an even toss is:

$ p(1\minus{}p)\plus{}p(1\minus{}p)^3\plus{}p(1\minus{}p)^5\plus{}...\plus{}$

Now sum the series:

$ \sum_{k\equal{}1}^{\infty}p(1\minus{}p)^{2k\minus{}1}\equal{}\minus{}\frac{(p\minus{}1)^2}{(1\minus{}p)(p\minus{}2)}$

Set the sum to 2/5 and solve for p.

$ \minus{}\frac{(p\minus{}1)^2}{(1\minus{}p)(p\minus{}2)}\equal{}\frac{2}{5}$

Solve the equation any way you can, computer, hand method, iteration you get p = 1/3
[/hide]


function
calculus
derivative
trigonometry
limit
geometry unsolved
geometry

$\boxed{\ \begin{array}{c}
\mathrm{Let\ a\ circle}\ w=C(O,1)\ ,\ \mathrm{the\ points}\ \{A,B\}\subset w\ \mathrm{such\ that}\ OA\perp OB\  \mathrm{and\  a\ line}\ d\ \mathrm{for\ which}\ A\in\ d\ \mathrm{and}\ d\perp OA\ .\ \mathrm{For\ a\ point}\ P\in \mathrm{ray}\ [AO\ \mathrm{and}\ (\forall )\ X\in \overarc {AB}\\\\
\mathrm{denote}\  M_x\in PX\cap d.\ \mathrm{Ascertain\ the\ position\ of\ the\  remotest\ point}\  P\ \mathrm{from}\ O\ \mathrm{(if\ it\ exists)}\ \mathrm{such\ that}\  \forall\ X\in \overarc{AB},\ \mathrm{we\ have\ the\ length\ of\ the}\ \overarc {AX}\le AM_x\end{array}\ }$

This problem is a very interesting aspect of trigonometry. Even nobody ?

Let t = <AOX in(0,Pi/2), L = OP, f(t) = AMx - arc(AX) = (1+L)sin t/(L+cos t)-t
if g(t) = (sin t -t cos t)/(t-sin t), as (t-sin t)(g(t)-L) = (L+ cos t)f(t), f(t) and g(t)-L have the same sign.
The result follows from the fact that the minimum of g is g(Pi/2) = 2/(Pi-2)

Why ? What is te monotony of the function ? This is problem.

This is clear : sin t - t cos t and t - sin t are both positive and increasing.

No, here is division, not multiply ! I am sorry.

:coolspeak:

$\left(\forall\right) x\in \left(0,\frac{\pi}{2}\right)$, we have:$u(x)=sinx-xcosx>0\nearrow\ ,\ v(x)=x-sinx>0\nearrow \ ,\ w(x)=\frac{1}{v(x)}>0\searrow.$
$f(x)=\frac{u(x)}{v(x)}=u(x)\cdot\frac{1}{v(x)}=u(x)\uparrow\cdot w(x)\downarrow.\ \ ?\ ?\ ?$

From this don't results that the function f is monotonously.
See http://www.mathlinks.ro/Forum/topic-44733.html
:spider:

Yes, of course! Sorry for this stupid confusion; drawing the graph of g, I thought that g  was obviously decreasing and i was too lasy to check it.
But I think that you have probably a better solution than a laborious study of the sign of a derivative. Haven't you?

It is O.K. The study of the sign of the derivative is only way. Here is a great artifice. I will present later my solution. With friendship, Levi

Without artifice (but tedious) :
dg/dt has the same sign than u(t) = t^2+(cos t -1) (1+t/sin t)
du/dt has the same sign than v(t) = t - sin t (2 +cos t)/(1+2 cos t)
dv/dt = -2 sin^2 t (1-cos t)/(1+2 cost)^2 <=0 hence dg/dt<=0 on (0,Pi/2)

[b]I will ascertain the monotony of the function[/b] $f(x)=\frac{\sin x-x\cos x}{x-\sin x},x\in \left(0,\frac{\pi}{2}\right)$ [b]and[/b] $\mathrm {Im\ f}.$

[b]Definition (same sign).[/b] $A\ s.s.\ B\Longleftrightarrow AB>0\ or\ A=B=0.$ [b]Thus,[/b]

$ f'(x)\ s.s.\ x\sin x(x-\sin x)-(1-\cos x)(\sin x-x\cos x)\ s.s.\ (\sin x)\cdot x^2-(1-\cos x)\cdot x-\sin x(1-\cos x)\ s.s.\ \left[x-(1-\cos x)\cdot \frac{1+\sqrt{5+4\cos x}}{2\sin x}\right] =\ g(x)\ .$

\begin{eqnarray*} g'(x) &s.s.& (2\sin^2 x+\cos x-1)\cdot \sqrt{5+4\cos x}-(1-\cos x)(5+4\cos x)+2\sin^2 x (1-\cos x)\\\\
&s.s.& (1+2\cos x)\cdot \sqrt{5+4\cos x}-(2\cos^2 x+4\cos x+3)\\\\
&s.s.& (1+2\cos x)^2\cdot (5+4\cos x)-(2\cos^2 x+4\cos x+3)^2\\\\
&s.s.& -4(\cos^4x-2\cos ^2x+1)=-4\sin^4x<0\end{eqnarray*}

Observev that $g(0)=0$ and $g'(x)<0\implies g\searrow\implies$ $(\forall )\ \in\left(0,\frac {\pi}2\right)\ ,\ g(x)<g(0)=0$ $\implies f'(x)<0\implies f\searrow\ .$ Prove easily

that $\lim_{x\searrow 0}f(x)=2\ ,\ \lim_{x\searrow \frac {\pi}2}f(x)=\frac 2{\pi -2}$ . In conclusion, $0<x<\frac {\pi}2\implies \frac 2{\pi -2}<f(x)<2\implies \mathrm {Im\ f}=\left(\frac 2{\pi -2},2\right)$ .

[u]See and [b]PP2[/b][/u] from [url=http://www.artofproblemsolving.com/Forum/blog.php?u=9882&b=47595][color=red][b]here[/b][/color][/url].

[quote="levi"][b]I will ascertain the monotony of the function[/b] $f(x)=\frac{sinx-xcosx}{x-sin},x\in \left(0,\frac{\pi}{2}\right)$ [b]and[/b] $Im f.$

[b]Definition (same sign).[/b] $A\ s.s.\ B\Longleftrightarrow AB>0\ or\ A=B=0.$ [b]Thus,[/b]

$\begin{eqnarray*} f'(x) &s.s.& xsinx(x-sinx)-(1-cosx)(sinx-xcosx)\\
&s.s.& (sinx)\cdot x^2-(1-cosx)\cdot x-sinx(1-cosx)\\
&s.s.&\left[x-(1-cosx)\cdot \frac{1+\sqrt{5+4cosx}}{2sinx}\right] =\ g(x)
\end{eqnarray*} .$

[color=red]i did not understand this step.  will you please give more details?[/color]

\begin{eqnarray*} g'(x) &s.s.& (2sin^2 x+cosx-1)\cdot \sqrt{5+4cosx}-(1-cosx)(5+4cosx)+2sin^2 x (1-cosx)\\\
&s.s.& (1+2cosx)\cdot \sqrt{5+4cosx}-(2cos^2 x+4cosx+3)\\\
&s.s.& (1+2cosx)^2\cdot (5+4cosx)-(2cos^2 x+4cosx+3)^2\\\
&s.s.& -4(cos^4x-2cos^2x+1)=-4sin^4x<0.\ \ \ Thus,\ g\searrow\ and\ f\searrow.
\end{eqnarray*}[/quote]

$g'(x) .s.s. 2g'(x)=(2x)'-\left( \frac{1-cosx}{sinx}\right) '\cdot (1+\sqrt {5+4cosx})-\frac{1-cosx}{sinx}$ $\cdot\ (1+\sqrt {5+4cosx})'=$

$2-\frac{1-cosx}{sin^2x}\cdot (1+\sqrt{5+4cosx})+\frac{1-cosx}{sinx}\cdot \frac{2sinx}{\sqrt {5+4cosx}}=$

$2-\frac{1-cosx}{sinx}\cdot \left( \frac{1+\sqrt {5+4cosx}}{sinx}-\frac {2sinx}{\sqrt  {5+4cosx}}\right) =$

$2-\frac{1-cosx}{sinx}\cdot \frac{\sqrt {5+4cosx}+5+4cosx-2sin^2 x}{sinx\cdot \sqrt {5+4cosx}}$

$.s.s.\ 2sin^2 x\cdot \sqrt {5+4cosx}-(1-cosx)\cdot \sqrt {5+4cosx}-(1-cosx)(5+4cosx-2sin^2 x)=$

$(2sin^2+cosx-1)\cdot \sqrt{5+4cosx}+(1-cosx)(2sin^2 x-4cosx-5).$

IS IT O.K. ? I wait for answer !

I didn't have the courage to check your dexterous solution last night, when I posted my series-based solution at http://www.mathlinks.ro/viewtopic.php?t=44733 , but I did check it tonight and agree with it 100% -- congratulations!

Both proofs can be extended to (0, π), but the given function is actually decreasing on (0, 2π)  :lol:

George Baloglou

[quote=Virgil Nicula][b]I will ascertain the monotony of the function[/b] $f(x)=\frac{\sin x-x\cos x}{x-\sin x},x\in \left(0,\frac{\pi}{2}\right)$ [b]and[/b] $\mathrm {Im\ f}.$
[b]Definition (same sign):[/b] $\ A\ s.s.\ B\Longleftrightarrow AB>0\ or\ A=B=0.$ [b]Thus,[/b] $f'(x)\ s.s.\ x\sin x(x-\sin x)-$ $(1-\cos x)(\sin x-x\cos x)\ s.s.\ (\sin x)\cdot x^2-(1-\cos x)\cdot x-\sin x(1-\cos x)\ s.s.\ \left[x-(1-\cos x)\cdot\frac{1+\sqrt{5+4\cos x}}{2\sin x}\right] =\ g(x)\ .$

$g'(x)\ s.s.\ (2\sin^2 x+\cos x-1)\cdot \sqrt{5+4\cos x}-(1-\cos x)(5+4\cos x)+2\sin^2 x (1-\cos x)\ s.s.\ (1+2\cos x)\cdot \sqrt{5+4\cos x}-(2\cos^2 x+4\cos x+3)\ s.s.\ (1+2\cos x)^2\cdot (5+4\cos x)-(2\cos^2 x+4\cos x+3)^2\ s.s.\ -4(\cos^4x-2\cos ^2x+1)=-4\sin^4x<0\ .$ Observe that $g(0)=0$ and $g'(x)<0\implies g\searrow\implies$ $(\forall )\ \in\left(0,\frac {\pi}2\right)\ ,\ g(x)<g(0)=0$ $\implies f'(x)<0\implies f\searrow\ .$ Prove easily that $\lim_{x\searrow 0}f(x)=2\ ,\ \lim_{x\searrow \frac {\pi}2}f(x)=\frac 2{\pi -2}$ . In conclusion, $0<x<\frac {\pi}2\implies \frac 2{\pi -2}<f(x)<2\implies \mathrm {Im\ f}=\left(\frac 2{\pi -2},2\right)$ .

[u]See and [b]PP2[/b][/u] from [url=http://www.artofproblemsolving.com/Forum/blog.php?u=9882&b=47595][color=red][b]here[/b][/color][/url].[/quote]

See and here => https://artofproblemsolving.com/community/c1090h1448274


IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}1$ prove the inequality:

\[ \frac{a^{5}\plus{}b^{5}}{ab(a\plus{}b)}\plus{} \frac {b^{5}\plus{}c^{5}}{bc(b\plus{}c)}\plus{}\frac {c^{5}\plus{}a^{5}}{ca(a\plus{}b)}\geq 3(ab\plus{}bc\plus{}ca)\minus{}2.\]

we know $ a^5 \plus{} b^5 >\equal{} a^2b^2(a\plus{}b)$(prove easy)
so the inequality is $ >\equal{} ab \plus{} bc\plus{} ca$
now we have to prove $ ab \plus{} bc \plus{} ca \plus{}2>\equal{} 3(ab \plus{} bc \plus{} ca)$
or $ 2(a^2 \plus{} b^2 \plus{} c^2) >\equal{} 2(ab \plus{} bc \plus{} ca)$,which is true.

Since $ y \equal{} x^5$ is an increasing and downwards convex function for $ x > 0$, by Jensen's inequality we have

$ \frac {a^5 \plus{} b^5}{2}\geq \left(\frac {a \plus{} b}{2}\right)^5\Longleftrightarrow \frac {a^5 \plus{} b^5}{ab(a \plus{} b)}\geq \frac {1}{16}\cdot \frac {(a \plus{} b)^4}{ab} \equal{} \frac {1}{16}(a \plus{} b)^2\cdot \frac {(a \plus{} b)^2}{ab}$

$ \geq \frac {1}{16}(a \plus{} b)^2\cdot 4\ \because (a \plus{} b)^2\geq 4ab$ for $ a > 0,\ b > 0$. Thus for $ a > 0,\ b > 0,\ c > 0$,

$ \frac {a^5 \plus{} b^5}{ab(a \plus{} b)} \plus{} \frac {b^5 \plus{} c^5}{bc(b \plus{} c)} \plus{} \frac {c^5 \plus{} a^5}{ca(c \plus{} a)}$

$ \geq \frac {1}{4}\{(a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2\}$

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

$ \geq ab \plus{} bc \plus{} ca\ \because a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$

$ \geq 3(ab \plus{} bc \plus{} ca) \minus{} 2$

Then we are to prove $ ab\plus{}bc\plus{}ca\geq 3(ab\plus{}bc\plus{}ca)\minus{}2$ which can be proved by

$ ab\plus{}bc\plus{}ca\geq 3(ab\plus{}bc\plus{}ca)\minus{}2\Longleftrightarrow 1\geq ab \plus{} bc \plus{} ca$

$ \Longleftrightarrow a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca\ Q.E.D.$

[quote="delegat"]IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$ prove the inequality:

$ \frac {a^{5} \plus{} b^{5}}{ab(a \plus{} b)} \plus{} \frac {b^{5} \plus{} c^{5}}{bc(b \plus{} c)} \plus{} \frac {c^{5} \plus{} a^{5}}{ca(a \plus{} c)}\geq 3(ab \plus{} bc \plus{} ca) \minus{} 2$[/quote]
The following inequality is true too:
$ \frac {a^{5} \plus{} b^{5}}{ab(a \plus{} b)} \plus{} \frac {b^{5} \plus{} c^{5}}{bc(b \plus{} c)} \plus{} \frac {c^{5} \plus{} a^{5}}{ca(a \plus{} c)}\geq6\minus{}5(ab \plus{} bc \plus{} ca).$ :wink:

[quote="arqady"][quote="delegat"]IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} + b^{2} + c^{2} = 1$ prove the inequality:

$ \frac {a^5 + b^5}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq 3(ab + bc + ca) - 2$[/quote]
The following inequality is true too:
$ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq6 - 5(ab + bc + ca).$ :wink:[/quote]
it is equivalent to
$ \sum(\frac {a^5 + b^5}}{ab(a + b)} - \frac {1}{2}(a^2 + b^2)} \geq \frac {5}{2}(\sum(a - b)^2)$
$ \sum(a - b)^2(\frac {2a^2 + ab + 2b^2}{2ab} - \frac {5}{2}) \geq 0$
$ \sum\frac {(a - b)^4}{ab} \geq$ 0 which is true

i don't know why the latex looks like that  :(

Nice work[b] HTA[/b]  You have made small mistake in writing Namely you wrote \frac { ...}[b]}[/b] :wink:

[quote="delegat"]IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$ prove the inequality:

$ \frac {a^{5} \plus{} b^{5}}{ab(a \plus{} b)} \plus{} \frac {b^{5} \plus{} c^{5}}{bc(b \plus{} c)} \plus{} \frac {c^{5} \plus{} a^{5}}{ca(a \plus{} b)}\geq 3(ab \plus{} bc \plus{} ca) \minus{} 2$[/quote]

[hide="Solution using Power Mean"]Observe the following

$ ab \le a^{2}\minus{}ab\plus{}b^{2}$ for $ (a\minus{}b)^{2} \ge 0$.

Then $ \frac{a^{5}\plus{}b^{5}}{ab(a\plus{}b)} \ge \frac{a^{5}\plus{}b^{5}}{a^{3}\plus{}b^{3}}$.

By Power Mean Inequality, we have $ \frac{a^{5}\plus{}b^{5}}{2} \ge (\frac{a^{4}\plus{}b^{4}}{2})^{\frac{5}{4}}$, and we also have
$ \frac{a^{3}\plus{}b^{3}}{2} \le (\frac{a^{4}\plus{}b^{4}}{2})^{\frac{3}{4}}$.

So $ \frac{a^{5}\plus{}b^{5}}{a^{3}\plus{}b^{3}} \ge (\frac{a^{4}\plus{}b^{4}}{2})^{\frac{1}{2}}$.

But by Power Mean again, we have $ (\frac{a^{4}\plus{}b^{4}}{2})^{\frac{1}{2}} \ge \frac{a^{2}\plus{}b^{2}}{2}$.

So, by adding all the equivalent forms, we get that we have to prove that

$ a^{2}\plus{}b^{2}\plus{}c^{2} \ge 3(ab\plus{}bc\plus{}ca)\minus{}2$ which is equivalent to $ 3(a^{2}\plus{}b^{2}\plus{}c^{2}) \ge 3(ab\plus{}bc\plus{}ca)$, which is true by multiplying $ \frac{3}{2}$ to the following inequality $ (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2} \ge 0$.[/hide]

[quote="HTA"][quote="arqady"][quote="delegat"]IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} + b^{2} + c^{2} = 1$ prove the inequality:

$ \frac {a^5 + b^5}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq 3(ab + bc + ca) - 2$[/quote]
The following inequality is true too:
$ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq6 - 5(ab + bc + ca).$ :wink:[/quote]
it is equivalent to
${{ \sum(\frac {a^5 + b^5}}{ab(a + b)} - \frac {1}{2}(a^2 + b^2)} \geq \frac {5}{2}(\sum(a - b)^2)$
$ \sum(a - b)^2(\frac {2a^2 + ab + 2b^2}{2ab} - \frac {5}{2}) \geq 0$
$ \sum\frac {(a - b)^4}{ab} \geq$ 0 which is true

i don't know why the latex looks like that  :([/quote]
It can be:
it is equivalent to
${ \sum(\frac {a^5 + b^5}{ab(a + b)} - \frac {1}{2}(a^2 + b^2)}) \geq \frac {5}{2}(\sum(a - b)^2)$ :wink:

By Muirhead inequality we have $a^5+b^5\geq a^4b+b^4a=ab(a^3+b^3)=ab(a+b)(a^2-ab+b^2)\geq a^2b^2(a+b)$ [last by AM-GM], so:
$\sum_{cyc}\frac{a^5+b^5}{ab(a+b)}+2\geq \sum_{cyc}\frac{a^2b^2(a+b)}{ab(a+b)}+2=ab+bc+ca+2=ab+bc+ca+2(a^2+b^2+c^2) \geq 3(ab+bc+ca)$ [last by AM-GM].

This inequality is equivalent with
$ \sum\frac{(a+b)^2}{ab}(a-b)^{2}\geq 0 $
This is SOS form

http://www.artofproblemsolving.com/community/c6h481723p2698355:
IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2}+b^{2}+c^{2}=1$ prove the inequality:
\[ \frac{a^{5}+b^{5}}{ab(a+b)}+ \frac {b^{5}+c^{5}}{bc(b+c)}+\frac {c^{5}+a^{5}}{ca(a+b)}\geq 1-\lambda (ab+bc+ca-1).(\lambda\leq5)\]
[url=http://www.artofproblemsolving.com/community/c6h1200710p5902466]Stronger than B&H 2008[/url]

$\frac{c^5+a^5}{ca(a+b)}$ should be corrected as $\frac{c^5+a^5}{ca(c+a)}$ :)

[quote=arshakus]$a^5+b^5\ge a^4b+b^4a=ab(a+b)(a^2-ab+b^2)$
whriting for all fractions we will get the following inequality which is obvious)
$a^2+b^2+c^2\ge ab+ac+bc$[/quote]

Then

$P(a,b,c)$=LHS(say)

$P(a,b,c)\ge 2(a^2+b^2+c^2)–2(ab+bc+ac)\ge
3 (ab+bc+ac)–2$ will suffice
$\implies 2(a^2+b^2+c^2)+2\ge 4(ab+bc+ac)$

$2.1=2(a^2+b^2+c^2)$

Above reduces
$4(a^2+b^2+c^2)\ge 4(ab+bc+ac)$ obvious true


calculus
integration
logarithms
function
algebra
partial fractions
calculus computations

I think the first one is
$x+\frac 1x +\ln \left| \frac{x-1}{x+1} \right| + C$

Also, I think that in the one you solved, you should have factored $1-x^2$ when you did the partial fractions, although I don't know whether it affects the answer or not.

For the second integral:

We rewrite the integrand as 1/3. 3(X^2)/(X^3). (X^3 +1)/(X^3 -8) and make the substitution Y = X^3.This gives dY/dX =3.(X^2) and this

reduces the problem to integrating with respect to Y the function 1/3.(Y+1)/Y(Y-8) ,which is dealt with by partial fractions.


trigonometry
logarithms
function
calculus
derivative
algebra unsolved
algebra

Some body help me
Prove that:Exist only one $ x$ such that $ x \in [\frac{3 \pi}{2},\frac{5 \pi}{2}]$ and
$ lgx\equal{}sinx$
(there $ lgx$ is the number that $ 10^{lgx}\equal{}x$)
So who can help me

[hide]Clearly, $ \sin x$ and $ \log x$ are one-to-one over $ \left[ \frac {3\pi}{2},\frac {5\pi}{2} \right]$.

Also, $ \sin(\frac {3\pi}{2}) \equal{} \minus{} 1 < \log (\frac {3\pi}{2})$ and $ \sin(\frac {5\pi}{2}) \equal{} 1 > \log (\frac {5\pi}{2})$

Thus, there is one and only one $ x \in \left[ \frac {3\pi}{2},\frac {5\pi}{2} \right]$ such that $ \sin x \equal{} \log x$.[/hide]
This seems too easy to be in the Olympiad section.

Edit: nvm, I wasn't thinking very well.

[quote="Xantos C. Guin"]
Thus, there is one and only one $ x \in \left[ \frac {3\pi}{2},\frac {5\pi}{2} \right]$ such that $ \sin x \equal{} \log x$.[/quote]

Why?

You seem to be saying that if $ f(x)$ and $ g(x)$ are both 1:1 on an interval $ [a,b]$, $ f(a)<g(a)$ and $ f(b)>g(b)$ then $ f(x)\equal{}g(x)$ has a unique solution in that interval. That's not true.

Take $ g_1(x) \equal{} 0$ on the interval and let $ f_1(x)$ be a function with three zeros on the interval with $ f_1(a)$ negative and $ f_1(b)$ positive. Assume further that $ f_1$ is continuously differentiable and $ |f_1'(x)| < 0.9$ throughout the interval. This is clearly possible, say with $ f_1(x) \equal{} \alpha (x\minus{}t_1)(x\minus{}t_2)(x\minus{}t_3)$ with the zeros $ t_1, t_2, t_3$ in $ (a,b)$ and $ \alpha$ small enough. Now $ f(x) \equal{} f_1(x)\plus{}x$ and $ g(x) \equal{} g_1(x)\plus{}x$ intersect three times but both have a strictly positive derivative throughout $ [a,b]$, so they are 1:1.

Here is my solution:
[hide][b]Existence:[/b] Let be $ g(x) \equal{} sin(x) \minus{} log(x)$. $ g$ is trivially continuous in $ x \in I\equal{}\left[ \frac {3\pi}{2},\frac {5\pi}{2} \right]$
$ g(\frac {3\pi}{2}) < 0$ and $ g(\frac {5\pi}{2}) > 0$, so there is at least one $ x\in I$ such as $ sin(x) \equal{} log(x)$ (Bolzano).
[b]Unicity:[/b] $ g'(x) > 0$ $ \forall$ x in the interval (g is monotone). Thus, there can't be two solutions. [/hide]

[quote="Klaus"]Here is my solution:
[hide][b]Existence:[/b] Let be $ g(x) \equal{} sin(x) \minus{} log(x)$. $ g$ is trivially continuous in $ x \in I \equal{} \left[ \frac {3\pi}{2},\frac {5\pi}{2} \right]$
$ g(\frac {3\pi}{2}) < 0$ and $ g(\frac {5\pi}{2}) > 0$, so there is at least one $ x\in I$ such as $ sin(x) \equal{} log(x)$ (Bolzano).
[b]Unicity:[/b] $ g'(x) > 0$ $ \forall$ x in the interval (g is monotone). Thus, there can't be two solutions. [/hide][/quote]

Sorry,this proof is not true because $ g'(x)>0$ for all x is is not true ( you can check it), So it is a difficult problem and Posting it here is suitable

[quote="Allnames"]Some body help me
Prove that:Exist only one $ x$ such that $ x \in [\frac {3 \pi}{2},\frac {5 \pi}{2}]$ and
$ lgx \equal{} sinx$
(there $ lgx$ is the number that $ 10^{lgx} \equal{} x$)
So who can help me[/quote]
Thank every body,i have had its proof
Since $ lgx \equal{} sinx$ we have $ x\equal{}arcsin(lgx)$
Then let $ f(x)\equal{}x\minus{}arcsin(lgx)$ ,$ f`(x) >0$ for all $ x \in [\frac {3 \pi}{2},\frac {5 \pi}{2}]$,so the proof is finished
Sorry because of costing your time


What is your favourite genre of music? :)

mine's a bit contradicting... techno and classical :)

hey!!! no kidding, me too, actually i think it's quite a nice combination, and those remixes of classic music aer just great

psykyk

how about piano rock?

oh don't tell me you've never heard of Something Corporate!

[quote="demonface"]nothing[/quote]

U=(funny) ? :?

I voted for pop/dance but I also like rock a lot!
It depends on the mood!

I'm suprised you didn't include fractal music. (it's not my favorite, but still...)

I'll have to go with the "marching band" genre.

[quote="uglyman621"]what about rap??[/quote]
:?

Irish music forever!

rock pop and classical all work for me...

Yay classical!
I'm a violinist and a pianist and I just love orchestra...I like Chinese pop too...
And I compose ALL the time with classical music...

Rock n Rolleth ruleth  (60s-80s)

      Guns N' Roses, Pink Floyd, Deep Purple , led zep, Blue Oyster Cult, Lynyrd skynyrd, jeff beck, metallica, peter frampton, queen, the who, rainbow, scorpions, thin lizzy  , the beatlesblahblahblah i could go on forever

My favourite guitarist is Buckethead, pure genius ... i seriously cant believe how good he is

I hate nirvana and the other grunge bands that took rock to a bad direction and gave birth to stuff we have today ( falloutboy, funeral for a friend,linkin fart etc, etc)

people saying rap should be up there, the poll asks what kind of MUSIC  :ninja:

Punk Rock, Emo (depends), Pop, Parody, Classical, Metal.

dnb s' good.

BUCKETHEAD, OPETH, IRON MAINDEN! others.

[quote="shadysaysurspammed"]DnB?

gud to knw sumone here listens to buckethead[/quote]

Drum and Bass

I love Rap,R&B.Especially Usher,Craig David,2Pac and most of the Russian rap.

[quote="neelnal"][quote="junggi"]My favorite is rock and classical.
I actually like all kinds of music except pop, hiphop, rap,and heavy metal stuff. I hate those.[/quote]

I agree. Rap is the stupidest stuff I have ever heard.[/quote]
U make me very mad :mad: .At first i can't stand rock,i really hate it,but i never told smth bad about this kind of music.Also maybe rap isn't stupid,maybe you're too stupid to understand it,there are many lyrics in it,of course not in all of them,but in some of them,believe there are many clever things.So just close your mouth and never told bad about the things,you don't know. :mad:  .And one more thing,i tried to understand rock,but i failed,i can not find nothing in it.Rock just a...Everyone knows what i mean :rotfl:

Hey,  I listen to classic....... with a synthesizer


trigonometry
calculus
calculus computations

what is y(2) if y" - 2y' + 3y = 0 and y(0)=1, y'(0)=3

not sure how work this but the answer is e^6
I've only gotten this far  y' = 3y/2 + y"/2

You have an ordinar linear differential equations. Its general solutions is $ae^x\cos \sqrt{2}x+be^x\sin \sqrt{2}x$. Using conditions $y(0)=1$ and $y'(0)=3$ we obtain $a=1$ and $b=\sqrt{2}$. :?

Myth is right. In the general case of $ay'' + by' +cy =0$, we first solve the characteristic equation $am^2+bm+c=0$.  In this case the discriminant is less than zero, and thus the roots are imaginary and are denoted as $\alpha \pm \beta i$.  The solution to the differential equaion is then $e^{\alpha x}\left(C_1\sin(\beta x) + C_2\cos(\beta x)\right)$. If you're confused about this, just use the identity $e^{xi}=i\sin(x)+\cos(x)$.

So $y=e^x\cos(x\sqrt{2})+e^x\sqrt{2}\sin(x\sqrt{2}) \Rightarrow y(2) \not= e^6$.

Probably there is a typo either in the post or in the book and the answer $e^6$ relates to the equation $y''-2y'-3y=0$...

Here's the solution to the equation $y''-2y'-3y=0$
$m^2-2m-3=0$
$m=3$ or $m=-1$
$y_1=e^{3x}$
$y_2=e^{-x}$.
$y=y_h=C_1y_1+C_2y_2$  (This is because the original equation is homogenous and therefore has no particular solution, $y_p$).
$y=C_1e^{3x}+C_2e^{-x}$.
Solving for the initial conditions yields $C_1=1$ and $C_2=0$.
$y(2)=e^{(3)(2)}=e^{6}$


algorithm
real analysis
real analysis solved

$x\in ]0,1]$ prove that there exist  $y_n$ increasing integer sequence $y_n \geq 2$ such that

$x=\sum_{k\geq 0}\frac{1}{y_0y_1...y_k}$

Dear Moubinool I have a book where I found the following problem:
([u]The Algorithm of Ostrogradski[/u])
[b]Prove that every rational number from interval $]0,1[$ can be uniquely represented as a sum
$\frac{1}{a_{1}}+\frac{1}{a_{1}a_{2}}+\cdots+\frac{1}{a_{1}a_{2}{\cdots}a_{n}}$,where
$a_{i}$ is natural,$a_{i+1}{\geq}a_{i},a_{i}{\geq}2$.Moreover every such sum is a rational from $]0,1[$.[/b]
I think this will help.

Wasn't the initial problem pretty clear in the first place? :? I know I'm probably doing something wrong, but it looks like there's a unique natural $y_0\ge 2$ s.t. $y_0x\in[1,2]$, and, moreover, $y_0x-1\le x$ (take $y_0$ to be the smallest natural for which $y_0x\in[1,2]$). After this, we can apply the same procedure to $x_1=y_0x-1$, finding the smallest $y_1$ for which $y_1x_1\in[1,2]$, and so on. We will have, for example, $y_1\ge y_0$ because $x_1\le x_0$.


Find an integer $ n$ such that numbers $ n$, $ 2n\plus{}1$, and $ 4n\plus{}1$ are primes.

I think its $ 3$  :?:

But how we can find it mathematically  :!:

Consider $ n\equal{}0,1,2 (mod \;\; 3)$ we have one of the three numbers is divisible by $ 3$, i.e. $ \equal{}3$, then it's easy to see that $ n\equal{}3$.

[quote="hasan4444"]Is there any other way to solve it without Modular?[/quote]
It is too simple my friend :wink:
Take perpuses
For $ n \equal{} 3k$,contradiction if $ k$ is not $ 1$
For $ n \equal{} 3k \plus{} 1$ contradioction
For $ n \equal{} 3k \plus{} 2$ contradiction
So the only solution is $ n \equal{} 3$
It is the same way as stephencheng but without modular

hasan4444 can you explain your signature, how is it supposed to be read?

Since it's been brought up, he probably means for it to be read as "Future Olympiad Crusher", as in he intends to blow everyone away.

In almost all cases, you should ask such personal questions by Private Message [PM] rather than in the thread itself.

[quote="hasan4444"]Find an integer $ n$ such that numbers $ n$, $ 2n \plus{} 1$, and $ 4n \plus{} 1$ are primes.

I think its $ 3$  :?:

But how we can find it mathematically  :!:[/quote]

Um, the problem says "find [i]an[/i] integer $ n$" (emphasis added).  You don't need to show that $ n\equal{}3$ is the only solution.  Just plugging in $ n\equal{}3$ shows that $ 3,7,13$ are all primes and therefore you have found an $ n$ that satisfies the problem's conditions.

It's funny how both you and the other people posted on this topic misread the problem and overcomplicated it.  :D

But one minute if we say $ 7$ is also a solution so $ 2(7)\plus{}1\equal{}15$ and $ 15$ isn't prime? Or do you mean that he didn't want $ 2n\plus{}1$ do be prime over all  :huh:


number theory
least common multiple
MATHCOUNTS

A church rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells every 25 minutes. If they all ring their bells at noon on the same day, at what time will they next all ring their bells together?

Hmmm... I asked this question.  I put $ \boxed{5: 00}$...  Why am I wrong?

The game only accepts 5, for some reason. 5:00 will be considered wrong.

[quote]I suppose, ernie, that you forgot to put A.M. or P.M. --- or isabella2296 is right.[/quote]
_______________________________
I love Mathcounts and AMC! ~ pi man  :roll:


9.2 Given 19 cards. Is it possible to write a nonzero digit on each card in such a way that you can compose from these cards an unique 19-digits number, which is divisible by 11?
([i]R. Zhenodarov, I. Bogdanov[/i])

suppose there is an unique number build with the cards, color the cards alternately black and white, changing the possition of 2 monochromatic cards, also give a good numbers so we must have 10 cards have the same number and the other 9 the same number.

Let $a$ be the number in the black cards, and $b$ the number in the white cards we must have $10a-9b=0$(mod11), or $a=2b$ (mod 11) so we have the possible pairs: $(2,1), (2,4), (3,6),(4,8), (6,1), (7,3),(8,5), (9,7)$

from here i suggest trial and error with this combinations of numbers to check if some of them works...

Hmm... Too many unfinished solutions last time, Pascual!

We have $n=n_1n_2...n_{19}$ with $11|n$
If we set $m_k$ the number $n$ with $n_k\leftrightarrow n_{k+2}$ and $1\leq k\leq 17$ we have $m_k$ divisible by 11 because
$n-m_k=(n_k-n_{k+2})10^{19-k}+(n_{k+2}-n_k)10^{17-k}=(n_{k+2}-n_k)10^{17-k}\times (1-100)=11\times 9(n_k-n_{k+2})10^{17-k}$
Consequently $\forall k, m_k=n\Rightarrow \forall k, n_k = n_{k+2}$
We have $10$ cards with $n_1$ and $9$ cards with $n_2$
Now we have to prove that with these cards the only number divisible by 11 is $n_1n_2n_1...n_2n_1$
We use the fact that $x_1x_2x_3...x_{2n}x_{2n+1}\equiv x_1-x_2+x_3-...x_{2n}-x_{2n+1} (11)$
For our numbers we have $k$ times $+n_1$, $(10-k)$ times $-n_1$, $(10-k)$ $+n_2$ and $(k-1)$ times $-n_2$ so we have to prove that $k\times n_1+(10-k)n_2 - (10-k)n_1 - (k-1)n_2\equiv (2k+1)n_1-2k\times n_2\equiv 0(11)$ has only the $k=10$ solution (By supposition $k=10$ works since $11|n$ thus we have $n_1\equiv 2n_2 (11)$)
$(2k+1)n_1-2k\times n_2\equiv 0(11)$
$\Leftrightarrow (4k+2)n_2 - 2k\times n_2\equiv 0(11)$
$\Leftrightarrow (2k+2)n_2\equiv 0(11)$
$\Leftrightarrow 2k+2\equiv 0(11)$
$k=10$ is the unique solution since $1\leq k\leq 10$

So  with $10$ cards $n_1$ and $9$ cards $n_2$ and $(n_1,n_2)\in ((2,1), (4,2), (6,3), (8,4), (1,6), (3,7), (5,8), (7,9))$ there is an unique 19-digits number, which is divisible by 11, it's $n_1n_2...n_2n_1$


Is it always possible to connect any four points with a triangle? Prove why or why not.

any 4 points? a triangle only needs 3.

and yes, it is true that any three of four points will always make a triangle. the three points determine a plane, and if you connect them, you get a triangle. The fourth point may or may not be on the plane of the three you picked.

Except when 3 (or 4) of them are collinear, in which case if you pick those three they don't make a triangle.

Is that what you were asking?

well, not exactly. Sorry, let me clarify.

I meant to say on one plane. So four points on one plane.
For instance, four points in the shape of a square may be connected with a isoceles right triangle (with the legs extending past two adjacent sides of the "square" and the hypotenuse passing through the remaining point).

Is there a way to PROVE this conjecture?

[hide] For 4 collinear points: trivial.
For 3 collinear points: still trivial.
Otherwise: Name the 4 points $A, B, C, D$. Draw lines $AB$ and $BC$. (The lines cannot be coincident, because that would imply 3 collinear points.) Construct a line through $D$ parallel to $AB$ intersecting $BC$ at point $E$. Pick any point $X$ on $BC$ not on the same side of $DE$ as point $C$. Draw line $XD$.

The triangle is bound by the lines $AB$, $BC$, and $XD$, containing all 4 points. [/hide]

I totally do not understand this problem...

Do you mean like this:

[img]http://theani.isgreat.org:8080/php/uploads/ani/trainglesinplane.JPG[/img]

[quote="lotrgreengrapes7926"]He means "for any 4 points, you can always draw a triangle so that the 4 points are contained in the sides of the triangle."[/quote]

Yes, that was a much better way to put it. Thank you.  :)

If all four of the points are collinear, simply make the line on which they lie one of the sides of the triangle. If 3 points are collinear, (say A, B, and C, with B between A and C) draw line AC. If D is the third point, draw line DA and DC, or connect D to any 2 points on AC which would keep A, B, and C on the side of the triangle.

Now the easy, collinear cases are done. No three of A, B, C, or D lie on the same line. Therefore, form quadrilateral ABCD. The quadirilateral is either a paralellogram or it is not. If it is not, extend 2 non-parallel sides until they intersect. Thake the point of intersection, and the two opposite points of the quadrilateral, to form the triangle.
if ABCD is a parallelogram, extend AB and AD. Draw BD, and draw a line through C paralell to BD. let the intersections of this line (through C) with line AB and line AD be 2 verticies of the triangle, with A as the 3rd

Let the points be ABCD

draw AB, BC and DC.  theres your triangle.  did i miss something in the question?

[quote="k8reindeer"]what if you AB is parallel to CD? Not a triangle...[/quote]
Extend CD in both directions. Extend AB in both directions. Connect AC and extend that in both directions. Then pick a point on AC that is above A and connect it through B to intersect CD. You now have a triangle with all four of these points on it.

what if the four points are positioned like one of the following?
[img]http://img238.imageshack.us/img238/3964/abcdefaqw0.png[/img]


number theory
greatest common divisor
inequalities

Prove there exist positive integers a,b,c for which $a+b+c=1998$, the gcd is maximized, and $0<a<b\leq c<2a$.
Find those numbers.
Are they unique?

Yes. When we talk about gcd then we're dealing with integers. :)

:)

Let $a=a'd, b=b'd, c=c'd$ where d is the gcd of the three.  Lucky for us, that inequality is changed to $0<a'<b'\leq{c'}<2a$.  The equation becomes $d=\frac{1998}{a'+b'+c'}$.  We then want the smallest possible value of $a'+b'+c'$ that divides 1998.
The smallest divisor is 1.  This obviously doesn't work.  Next smallest is 3.  Also, obviously doesn't work.  Next is 9.  We see that the smallest triple is {2,3,3}, whose sum is 8, and the next smallest is {3,4,4}, whose sum is 10.  That's a shame.  But lucky for us, there are a few pairs for 27, which means that the largest value for the GCD is 37.
Specifically, these pairs are {7,8,12}, {7,9,11}, {7,10,10}, and {8,9,10}.

Perhaps I am confused about what the question is asking? I'll erase it if I am mistaken...

[hide]

a+b+c=2*3^3*37

x*gcd(a,b,c)+y*gcd(a,b,c)+z*gcd(a,b,c)=2*3^3*37

0<x<y<=z<2x

so we need x+y+z to sum to a factor of 2*3^3*37, and to maximize gcd(a,b,c) we want it to sum to the smallest factor possible (2,3,6,9,18..)

it is impossible to find x,y,z to fit the inequality for x=1

x=2 ... (x,y,z)=(2,2,3) not a factor of 1998

x=3 ... (x,y,z)=(3,4-5,4-5) cannot sum to a factor of 1998

x=4 ... (x,y,z)=(4,5-7,5-7) ... (4,7,7) sums to 18 and gives the solution

(a,b,c)=(4*3*37,7*3*37,7*3*37)=(444,777,777) with gcd =111

for x=5 ... (x,y,z) = (5,6-9,6-9) and this yields (x,y,z)=(5,6,7), so one other solution?

[/hide]

Oh.  Um.  I sort of forgot about that factor of 2.


If a farmer can trade four chickens for a pig, three pigs for two sheep, and five sheep for two cows, what is the minimum number of cows he needs to trade for $ 20$ chickens?

c=number of chickens
p=number of pigs
s=number of sheep
o=number of cows
4c=p
3p=2s
5s=2o
20c=5p
6p=4s
5s=2o
since you cannot trade a fraction of an animal, and the initial trading prices have no common factors, the farmer needs to trade $ \boxed2$ cows for 20 chickens


function
inequalities
limit
topology
algebra unsolved
algebra

Given $ f(0;1)\minus{}\minus{}>(0;1)$ function and number $ M$ from the interval $ (0;1)$. Assume that for any numbers $ x,y$ from $ (0;1)$, the following inequality holds: $ |f(x)\minus{}f(y)|<\equal{}M|x\minus{}y|$.
    Show that the equation $ f(x)\equal{}x$ has exactly one root.

hint:
[hide]prove $ f$ is continious and consider a sequence $ (u_n)$ such that $ u_0 \equal{} a \in (0,1)$ and $ u_{n \plus{} 1} \equal{} f(u_n)$, prove $ (u_n)$ converge and then.. and it's clear that f  has at most one fixed point.[/hide]

gamarjoba lasha  :wink:
I want to show you more general theorem:
suppose $ X$ is [b] complete [/b]  metric space and $ f: X \to X$ also $ f(X) \equal{} X$
with the property:
$ d(f(x),f(y))\leq M d(x,y)$
for all $ x,y \in X$ and some $ 0 < M < 1$
then $ f$ have exactly one
motionless point or $ f(x_{0}) \equal{} x_{0}$

suppose we have two such points say $ f(x) \equal{} x$ and $ f(y) \equal{} y$ hence :
$ (1 \minus{} M)d(x,y)\leq 0$ therefore $ x \equal{} y$

Now, Let's show that for all $ x_{0}$  sequance $ x_{n} \equal{} f^{n}(x_{0})$ is Cauchy.
(by $ f^{n}$ I mean composition of $ f$ -$ n$-th times at point $ x_{0}$)
$ d(x_{n},x_{n \plus{} p})\leq M^{n} d(x_{0},x_{p}) \leq M^{n} \sum_{i \equal{} 1}^{p}d(x_{i \minus{} 1},x_{i}) \leq M^{n} d(x_{0},x_{1}) \sum_{i \equal{} 1}^{p} M^{i \minus{} 1} \leq \frac {M^{n} d(x_{0},x_{1})}{1 \minus{} M} \to 0$ as $ n \to \infty$
thus we have $ x \equal{} \lim x_{n}$
note that $ f$ is continuous , hence $ f(x) \equal{} x$

What about $ f(x)\equal{}\frac{x}{2}$? ;)

then by theorem  $ f$  have exactly one
motionless point or there exist $ x_{0}$ such that $ f(x_{0}) \equal{} x_{0}$ in your case $ x_{0} \equal{} 0$ :)

Yes you're right. I just made a mistake while posting the condition. The correct condition would be $ f: [0;1) \minus{} \minus{} > [0;1)$. :)

you can't choose $ f(x) \equal{} x/2$ since  theorem holds only when $ f(X) \equal{} X$ in your case $ X \equal{} [0,1)$  and if $ f \equal{} x/2$ hence :
$ f([0,1)) \equal{} [0,1/2) \neq [0,1)$.
also your problem (original) is true if $ f(0,1) \equal{} (0,1)$

This result is usually referred to as the [url=http://en.wikipedia.org/wiki/Banach_fixed_point_theorem]Banach fixed point theorem[/url] or as the contraction principle.


What does everybody think of this book.  I was thinking of trying to work through this book in order to work up to Linear Algebra Done Right.  I am not especially mathematically mature (currently working slowly through Spivak's Calculus) but am willing to struggle through the problems, so I figure maybe the Linear Algebra Problem Text will give me enough intuition and experience to go onto the Linear Algebra Done Right text, without having to start with the typical less abstract stuff.  Thanks

good as companion book to any standard linear algebra text which has all major theorems stated and proved, but be aware that it is definitely advanced undergraduate level, and I believe it requires some familiarity with abstract algebra to get the full benefit of the author's insight.

I suggest starting out with easier books such as Lang or Apostol, or Lipshetz (Shramm Outline) or any cheap Dover title.

the book has some good reviews on amazon.com.


Given points $ A,B,C$ on a straight line $ \ell$ in this order construct point $ D$ on $ \ell$ satisfying $ AD\cdot BD\equal{}CD^2$.

[geogebra]17fc107814bd5c565e25f5fff381267e566c8ed9[/geogebra]
Power of a point (Circle Power) ?


function
real analysis
real analysis unsolved

I know the proof that there exist continous functions,say on $[-\pi,\pi]$ which Fourier series diverges on a dense set of points.But the proof i know is non-constructive and uses Banach-Steinhaus Theorem.

Do you know any constructive way of showing that such functions exist.

*(Except the example of Fejer).

I know a Lebesgue's example of a continuous function whose Fourier series diverges at a point.  Unfortunately
I don't know whether the divergence may be extented to cover a dense set, sorry.