this is a problem i did in 4th grade, so dont scream if its too easy... :lol:

There was a man who was in town for a business trip and needed to stay at a hotel for 7 days. He came up to the desk clerk and asked him how much it would be, and he said "That'll be 100 dollars a day". when he went to search his pockets for his wallet, he realized he had left it at home and was flat broke except for a solid gold chain. He said to the manager: "I dont have money, but i do have this solid gold chain with exactly 7 links. Will that pay for the room?" Luckily, the man said that the gold chain was indeed worth 100 dollars a link, nd that he could use it to pay for each night. But the hotel manager didnt eant it in a whole piece, for he might end up staying for less time, so he said, " Take it to the jeweler next door, and he will cut the chain so you can pay each day instead of all at once." Now, the chain was very valuble and quite fragile, so he didnt want to ruin it by cutting it too much.

What is the least amount of cuts he needed, why, and how did he end up paying for the room?

Let the chain look like this:

1-2-3-4-5-6-7

--------------------

I came up with three cuts.

Make cuts at 2, 4, and at the 6 link positions.

[This is to think that we are to just cut ONE link with ONE each cut.]

Then you are left with uncut links at positions 1, 3, 5, and 7.

All seven pieces are separated so the hotel room can be paid one day at a time with either an uncut link or a cut link, depending on what you have on hand.

there's actually a shorter way to do it. its kinda a logic prb, but yea, that was my first guess, but its wrong :(  :(  :(

[hide]

It should be enough, if he cuts only the third link?

Then he would have three pieces of the chain - pieces of 1, 2 and 4 links.

On the first day he gives 1 link to the manager.
On the second day he gives 2 links and gets 1 link back - manager has now 2 links for 2 days.
On the third day he gives 1 link - manager has 3 links.
On the fourth day he gives 4 links and gets 3 links back - manager has 4 links.
On the fifth day he gives 1 link.
On the sixth day he gives 2 links and gets 1 link back - manager has 6 links.
On the seventh day he gives 1 link[/hide]

good job! i took so much time figuring it out


geometry
perimeter
articles
summer program
Mathcamp
trigonometry
algorithm

If you know how to do this, please post it.

I am not sure if this has been discussed before on AoPS, but here is an interesting forumla to find any binary/hexadecimal digit of $\pi$ without needing to know the preceding terms:

$\pi = \sum_{k=0}^\infty \frac{1}{16^k} \left(\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right) $

This formula is basically used to answer if the digits of $\pi$ are random or not, but as described before, I would say the best approximation of $\pi$ would be to use the polygon method. If you want, here are two sites which discuss the topic further:

[b]Sources:[/b]

http://plus.maths.org/issue21/news/pi/index-gifd.html
http://mathforum.org/library/drmath/view/52589.html
http://mathworld.wolfram.com/PiFormulas.html

[quote="K81o7"]I guess you could take a unit circle, inscribe a regular polygon with any number of sides into it, find the perimeter, double the number of sides, repeat.  So on and so on.  Or you could inscribe one and circumscribe one and find a geometric mean between their perimeters, then double the number of sides, so on.  Essentially the same thing.[/quote]

In fact, this is the way that Archimedes found the first 3 decimal places of pi. He (essentially) used a polygon of 96 sides to get this approximation (remember that this was before the days of calculators!). So you see, while in theory this method can give you digits of pi, it converges very slowly.

Take a look at this article for more information on the history of finding digits of pi: http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Pi_through_the_ages.html

Or here is an even more interesting chronology of the digits of pi: http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Pi_chronology.html

This discussion is probably better suited for Other Problem Solving Topics.

[color=olive]bumped from intermediate[/color]

[quote="ComplexZeta"]that gives 30 or 40 digits[/quote]
you mean .14159265358979323846264338327950288?

It was really fun at mathcamp seeing people memorizing more than hundreds of digits

here's an easy way but takes forever to converge:

$\frac{\pi}{4}=Arctan(1)=1-\frac{1}3+\frac{1}5-\frac{1}7+...$

here's a better way:

$\frac{\pi}{4}=Arctan(\frac{1}2)-Arctan(\frac{1}3)=...$

[quote="ComplexZeta"]Some of the Ramanujan identities converge to $\pi$ incredibly fast. I think there's one that gives 30 or 40 digits after just three terms! You might try looking at some of those.[/quote]

Before the computer age, the best formulas for computing the digits of pi were ones involving the inverse tangent of simple fractions. For example Machin's Forumla: $\displaystyle \frac{\pi}{4} = 4\tan^{-1}(1/5) - \tan^{-1}(1/239)$

The Ramanujan Formula for pi is: $\displaystyle \frac{1}{\pi} = \frac{\sqrt{8}}{9801} \sum_{n=0}^\infty \frac{(4n)!(1103 + 26390n)}{(n!)^4396^{4n}}$

Its not practical to compute by hand because it involves some hard sub-calculations, but it does in deed converge very fast. I think it gives something like 8 decimal places for each term.

There are now algorithms for computing digits of pi that are even more efficient, but many of them are based on ideas behind Ramanujan's Formula.

[quote="gaus202"]In fact, this is the way that Archimedes found the first 3 decimal places of pi. He (essentially) used a polygon of 96 sides to get this approximation (remember that this was before the days of calculators!).[/quote]
It's not just that it was before calculators. These methods do allow for hand calculation if you're sufficiently patient. What stopped Archimedes from going any further was that he didn't have an effective place-value number system to use.

Archimedes reported his answer as $3\,\frac{10}{71}\, <\pi < 3\,\frac17.$

(Well, he didn't use the symbol $\pi$, but that was the substance of it.)

Using geometric ideas little different than those of Archimedes, Al-Kashi published about 20 decimal places of $\pi$ in the 1400's. (He gave it both as a decimal and as a hexagesimal.) What advantage did Al-Kashi have over Archimedes? Effective arithmetic. Don't sell short the advantages and impact of the base-10 system and its associated algorithms.

[quote="gauss202"]
The Ramanujan Formula for pi is: $\displaystyle \frac{1}{\pi} = \frac{\sqrt{8}}{9801} \sum_{n=0}^\infty \frac{(4n)!(1103 + 26390n)}{(n!)^4396^{4n}}$[/quote]

Wow. Now that's complicated. How does one derive a formula like this?

When Ramanujan wrote this formula in one of his notebooks he provided no proof or explaination for where it came from. (His notebooks were filled with strange and curious formulas like this one).

People now believe that he somehow derived it from the theory of modular forms. They're not exactly sure how though.

If you're [i]really[/i] interested you can read a part of an article that talks about the derivation of this formula [url=http://www.cecm.sfu.ca/organics/papers/borwein/paper/html/node15.html]here[/url]. Be warned though, it is quite advanced.

By the way, this formula has a tangential, but curious, relationship to the fact that $e^{\pi \sqrt{163}}$ is [i]very[/i] nearly an integer.

If you want to learn more about modular forms you have to study complex analysis first. But for students here who are already in college and have done so, a great book to learn about modular forms is the classic book [u]A Course of Modern Analysis[/u] by Whittaker and Watson.


How many different combinations of coins could a person have if she has exactly 21 cents?

Clearly we could only have nickels, dimes and pennies.

Possibilities listed from dimes down to pennies:
two dimes:  DDP  That's it
one dime:  DNNP/DN 6P/ D 11P
zero dimes: NNNNP/NNN 6P/NN 11P/N 16P/21P

So, I get 9.

Or we could also note that since it's 21 cents, every combination includes at least 1 penny. So all we have to do is find the number of combinations to make 20 cents.

20 pennies
15 pennies, 1 nickel
10 pennies, 2 nickels
10 pennies, 1 dime
5 pennies, 3 nickels
5 pennies, 1 dime, 1 nickel
4 nickels
2 nickels, 1 dime
2 dimes

We find there are $ \boxed{9}$ ways.

Or you could also use mewto's handy dandy little formula.

$ 3 \plus{} \frac{\text{floor}(\frac{21}{5}) \plus{} \text{ floor}(\frac{21}{5} \minus{} 1)}{2}$

That gives 9, but only works for 10-49 cents.

Hmmm...

The numerator's supposed to be this.

$ \text{floor}(\frac{21}{5}) \plus{} \text{ floor}(\frac{21}{5} \minus{} 1)$

Actually, I believe in the numerator you multiply, not add, and we see that this formula does not work. I believe any time it works is simply a coincidence.


I don't know if this problem or result is known but this is a problem I came up with.  My approach was kind of long and messy so hopefully someone has a simpler solution:
For all acute triangles with angles A,B, and C with area K prve that the area of the orthic triangle K'=2KcosAcosBcosC

Let $AD, BE, CF$ are the altitudes of the triangle $ABC$. Using well-known fact that triangle $BFD$ is similar to $ABC$ with similiarity coefficient(maybe it is a wrong word) $\cos{B}$ we obtain: $DF=b\cos{B}$, in similar way $DE=c\cos{C}$.Now notice that $\measuredangle{FDE}=\pi-2A$, so : $K'=bc\sin{A}\cdot{\cos{A}\cdot{\cos{B}\cdot{\cos{C}}}}$. But $K=\frac{bc\cdot\sin{A}}{2}\Rightarrow2Kcos{A}\cdot{\cos{B}\cdot{\cos{C}}}=bc\sin{A}\cdot{\cos{A}\cdot{\cos{B}\cdot{\cos{C}}}}=K'.$
QED

Good Job Armo, your solution is far shorter than mine :D


LaTeX
inequalities unsolved
inequalities

For $ a,b,c  \geq 0$ prove that
     $ \sum_{cyc} \frac {a^4}{a^3\plus{}b^3} \geq \sum_{cyc} \frac {a}{ 2}$

  Can you use S.O.S to solve

$ (a \plus{} b \plus{} c)^3 \geq 9(ab \plus{} bc \plus{} ca)$

For $ a,b,c \geq 0$ Prove that
                  $ \sum_{cyc} \frac {a^4}{a^3\plus{}b^3} \geq \sum_{cyc} \frac {a^2}{a\plus{}b}$

Can you help by S.O.S

Just put $ \$$ before and  \ $ \$$ after your statement.

$ \sqrt{a^2\plus{}b}\plus{}\sqrt{b^2\plus{}c}\plus{}\sqrt{c^2\plus{}a} \geq 2 $

[quote="thanhak5"]For $ a,b,c \geq 0$ Prove that
                  $ \sum_{cyc} \frac {a^4}{a^3 \plus{} b^3} \geq \sum_{cyc} \frac {a^2}{a \plus{} b}$

Can you help by S.O.S[/quote]
Try $ a\equal{}0,9,b\equal{}1,c\equal{}0$.
By the way, the following is true for all $ a,b,c$ are the sides of an acute triangle
\[ \sum \frac{a^3}{a^2\plus{}b^2} \ge \sum \frac{a^2}{a\plus{}b}\]
See here: http://www.mathlinks.ro/viewtopic.php?t=193004 :)


Given $ a, b, c, d$  be the real nonnegative numbers satisfying $ a\plus{}b\plus{}c\plus{}d\equal{}3$. Prove that:
\[ \frac{ab}{3b\plus{}c\plus{}d\plus{}3}\plus{}\frac{bc}{3c\plus{}d\plus{}a\plus{}3}\plus{}\frac{cd}{3d\plus{}a\plus{}b\plus{}3}\plus{}\frac{da}{3a\plus{}b\plus{}c\plus{}3} \le \frac{1}{3}\]
When does equality occurs ?  :)

[quote="nguoivn"]Given $ a, b, c, d$  be the real nonnegative numbers satisfying $ a \plus{} b \plus{} c \plus{} d \equal{} 3$. Prove that:
\[ \frac {ab}{3b \plus{} c \plus{} d \plus{} 3} \plus{} \frac {bc}{3c \plus{} d \plus{} a \plus{} 3} \plus{} \frac {cd}{3d \plus{} a \plus{} b \plus{} 3} \plus{} \frac {da}{3a \plus{} b \plus{} c \plus{} 3} \le \frac {1}{3}\]
When does equality occurs ?  :)[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-28-t-q-anh.html


algebra
polynomial
trigonometry
algebra unsolved

Polynomial $ P(x)$ of degree $ n$ has exactly $ n$ real roots. Prove that polynomial $ P(x\plus{}ai)\plus{}P(x\minus{}ai)$ also has $ n$ real roots, $ a$ is an arbitrary real number.

Ok. Since no one interested in it, I will post 2 known to me solutions.

Solution 1 (by Prof. A.A.Soloviev)
We may assume $ a>0$. Let $ P(x)\equal{}\prod_{i\equal{}1}^{n}(x\minus{}x_k)$, $ x_k\in \mathbb R$. Clearly, we have $ P(x\plus{}ai)\plus{}P(x\minus{}ai)\equal{}2Re\ P(x\plus{}ai)$. Suppose $ x\minus{}x_k\plus{}ai\equal{}r_ke^{\varphi_k}$. Then \[ 2Re\ P(x\plus{}ai) \equal{} 2Re\ r_1...r_ne^{i(\varphi_1\plus{}...\plus{}\varphi_n)}\equal{}2r_1...r_n\cos(\varphi_1\plus{}...\plus{}\varphi_n).\]
Note that $ \varphi_i$ decreses from $ \pi$ to $ 0$ while $ x$ moves from $ \minus{}\infty$ to $ \plus{}\infty$, so $ \varphi_1\plus{}...\plus{}\varphi_n$ decreases from $ \pi n$ to $ 0$. It follows that $ \cos(\varphi_1\plus{}...\plus{}\varphi_n)$ takes exactly $ n$ zeros.

Solution 2 (by Prof. S.M.Voronin)
Suppose $ a>0$ and $ |\tau|\equal{}1$ ($ \tau \in \mathbb C$). We will prove general fact that $ P(x\plus{}ai)\minus{}\tau P(x\minus{}ai)$ has exactly $ n$ real zeros. Indeed, suppose we got non-real root $ x\equal{}u\plus{}iv$, $ v>0$ (or $ v<0$, it doesn't matter). We have $ \prod_{i\equal{}1}^n \frac{(u\minus{}x_k)\plus{}(v\plus{}a)i}{(u\minus{}x_k)\plus{}(v\minus{}a)i}\equal{}\tau$. Clearly $ \left|\frac{(u\minus{}x_k)\plus{}(v\plus{}a)i}{(u\minus{}x_k)\plus{}(v\minus{}a)i}\right|>1$, so the product is greater than 1 by modulus, but $ |\tau|\equal{}1$.


[quote="nostriel"]Does anybody know some resources(links, books) about Darboux property?[/quote]
Generally, a real - valued function $f$ defined on interval $[a,b]$ has the Darboux property $\iff$ for any $x,y\in [a,b]$ function $f$ reaches any value from the interval $[f(x),f(y)]$, i.e. for any $t\in [f(x),f(y)]$ there exists such $\lambda\in [x,y]$ that $f(\lambda)= t$.
The [url=http://planetmath.org/encyclopedia/IntermediateValueTheorem.html]intermediate value theorem[/url] states that any continuous function has the Darboux property. The [url=http://planetmath.org/encyclopedia/DarbouxsTheorem.html]Darboux theorem[/url] states that the deriviative of a differentiable function has the Darboux property.
See [url]http://eom.springer.de/D/d120050.htm[/url].

[quote="perfect_radio"][quote="Rzeszut"][quote="nostriel"]Does anybody know some resources(links, books) about Darboux property?[/quote]
Generally, a real - valued function $f$ defined on interval $[a,b]$ has the Darboux property $\iff$ for any $x,y\in [a,b]$ function $f$ reaches any value from the interval $[f(x),f(y)]$, i.e. for any $t\in [f(x),f(y)]$ there exists such $\lambda\in [a,b]$ that $f(\lambda)= t$.
[/quote]

Make that $\lambda \in [x,y]$.[/quote]
Yes you're right, but what I said is true since $[x,y]\subset [a,b]$. I'll correct it.


Consider a function $ f$ from real numbers to it self such that
 \[ f(xy)^2\equal{}f(x^2)f(y^2)\]
for all $ x,y$ satisfying the condition$ x^2y^3>2008$
Prove that  $ f(xy)^2\equal{}f(x^2)f(y^2)$ for all $ x,y$

[u]Solution[/u]Let us consider the function
$ t(x,y)\equal{}\frac{2008}{x^2y^3}$ for all positive real $ x,y$
Observe that $ (\frac{x}{t(x,y)})^2(yt(x,y))^3\equal{}2008$
hence,for any $ z>t(x,y)\equal{}2008$ we get $ (\frac{x}{z})^2(yz)^3>2008$,then $ (f(xy))^2\equal{}f(\frac{x^2}{z^2})f(y^2z^2)$
for all $ z>t(x,y)\equal{}2008$
For any positive real numbers $ x,y$.We choose z such that $ z>max [t(x,y),t(y,x),t(x,x),t(y,y)]$
then we get
$ for z>t(x,x)\rightarrow (f(x^2))^2\equal{}f(\frac{x^2}{z^2})f(x^2z^2)$
$ for z>t(y,y)\rightarrow  (f(y^2))^2\equal{}f(\frac{y^2}{z^2})f(y^2z^2)$
$ for z>t(x,y)\rightarrow   (f(xy))^2\equal{}f(\frac{x^2}{z^2})f(y^2z^2)$
$ for z>t(y,x)\rightarrow    (f(xy))^2\equal{}f(\frac{y^2}{z^2})f(x^2z^2)$
$ \therefore f(xy)^4\equal{}f(xy)^2f(xy)^2\equal{}f(\frac{x^2}{z^2})f(y^2z^2)f(\frac{y^2}{z^2})f(x^2z^2)
\equal{}[f(\frac{x^2}{z^2})f(x^2z^2)][f(\frac{y^2}{z^2})f(y^2z^2)]
\equal{}(f(x^2))^2(f(y^2))^2$ i.e $ (f(xy))^2\equal{}f(x^2)(f(y^2)$
which is obvious that this is true for all  positive real numbers$ x,y$,we are done


calculus
integration
logarithms
complex analysis
complex analysis unsolved

Let $f(z)$ be analytic in $|z| \leq 1$. Prove that $\int_0^1 f(x)dx =\frac{1}{2\pi i} \int_{|z|=1}f(z) lnz dz$.

$\ln z$ isn't exactly defined there, but I see what you mean. The integral on the right doesn't change if you take different liftings of the identity map under the covering map $\zeta\mapsto e^\zeta$. You do, however, need to put a $\frac 1{2\pi i}$ in front of the integral on the right, I think.

The integral on the right can then be written as $\frac 1{2\pi i}\int_0^{2\pi}f(e^{i\theta})e^{i\theta}\theta\ \mbox d\theta$. We can check equality in the cases $f(z)=z^n,\ n\ge 0$, and the conclusion then follows from the fact that on the closed unit disk, $f$ can be approximated uniformly with polynomials.

There is also another approach with contour integration. Just take the branch of logarithm that has a jump on $[0,1]$ and consider the contour starting near $0$, going slightly above the interval $[0,1]$ to the point on the unit circumference slightly above $1$, then going around almost the whole unit circumference to a point slightly below $1$ and then, finally, back to $0$ slightly below the interval $[0,1]$. The integral over that contour must be $0$. But this integral consists of the integral over the unit circumference and two integrals over $[0,1]$ that almost cancel each  other up to the jump $-\int_0^1 2\pi i f$. Hence the result.


linear algebra
matrix
function
induction
combinatorial geometry
combinatorics proposed
combinatorics

Given a $ {n}\times{n}$ table with non-negative real entries such that the sums of entries in each column and row are equal, a player plays the following game: The step of the game consist of choosing n sells no two of wich share a column  or row, and subtracting the same number from each of the entries of the n cells, provided that the resulting table has all non-negative entries. Prove that the player can change all entries to zero.

this problem is equivalent to showing that for every positive entry, there exists a transversal through it with all positive entries. anyone have a clue how to show this? i have thought about it for more than an hour and the solution simply escapes me.

Do you really need to show that much?  It seems like it would be enough to show that as long as there is a nonzero entry in the matrix, there is a nonzero transversal (you just subtract off from that transversal until some entry in the transversal becomes 0).

um, i think what i said is exactly what you said!  :)

...that is, if by "non-zero" transversal you mean a transversal with no 0 entries. because once a transversal has a 0 entry you cannot subtract anything from it except 0. if that happens for all transversals through a positive entry, there's no way to reduce that entry to 0. this is why i said "equivalent" and not merely sufficient.

This problem is essentially the Birkhoff-von Neumann theorem: The set of doubly stochastic matrices is the convex hull of the set of permutation matrices.

[url]http://planetmath.org/encyclopedia/ProofOfBirkoffVonNeumannTheorem.html[/url]

i thought of using the marriage lemma but abandoned it for lack of seeing how. but this looks promising... the marriage lemma appears in this proof.

Here's the solution I found.

[u]Solution.[/u] Suppose that we have a matrix $ (a_{ij})\in\mathcal{M}_n(\mathbb{R}_{\geq 0})$ satisfying the given equal summations property. Consider the function $ f: \mathbb{N}^2\rightarrow \mathbb{Z}_n$ such that $ f(a,b) \equal{} a \plus{} b\bmod n$, where $ (i,j)$ represents entry $ a_{ij}$.  Let $ S_m \equal{} \{a_{ab}|f(a,b) \equal{} m\}$, $ m\in\mathbb{Z}_n$.  Then from each element of $ S_m$, substract $ \min S_m$.  Now, let $ g(a,b) \equal{} a\minus{}b\bmod n$, and $ T_m \equal{} \{a_{ab}|g(a,b) \equal{} m\}$, $ m\in\mathbb{Z}_n$.  After performing the transformations using $ f$ and each $ S_m$, perform the same tranformation using $ g$ and each $ T_m$.  Then all the entries must be $ 0$, which is easy to prove. $ \Box$

Hopefully, I do not have any mistakes in my solution.

[quote="boxedexe"]Here's the solution I found.

[u]Solution.[/u] Suppose that we have a matrix $ (a_{ij})\in\mathcal{M}_n(\mathbb{R}_{\geq 0})$ satisfying the given equal summations property. Consider the function $ f: \mathbb{N}^2\rightarrow \mathbb{Z}_n$ such that $ f(a,b) \equal{} a \plus{} b\bmod n$, where $ (i,j)$ represents entry $ a_{ij}$.  Let $ S_m \equal{} \{a_{ab}|f(a,b) \equal{} m\}$, $ m\in\mathbb{Z}_n$.  Then from each element of $ S_m$, substract $ \min S_m$.  Now, let $ g(a,b) \equal{} a \minus{} b\bmod n$, and $ T_m \equal{} \{a_{ab}|g(a,b) \equal{} m\}$, $ m\in\mathbb{Z}_n$.  After performing the transformations using $ f$ and each $ S_m$, perform the same tranformation using $ g$ and each $ T_m$.  Then all the entries must be $ 0$, which is easy to prove. $ \Box$

Hopefully, I do not have any mistakes in my solution.[/quote]

unfortunately that doesn't work. there are other transversals aside from these trivial ones, so all we have to do to construct a counterexample is pick one of these other transversals, put the value 1 for each of its entries, and make everything else 0. then none of the 1's can be reduced because the unique $ S_i$ and the unique $ T_j$ passing through it both contain 0's.

yeah, Hall's Theorem = the Marriage Lemma

the following statement holds (I think this is what [b]nsato[/b] said):

[b]let $ Q$ be a nonnegative real matrix in which every row and every column sums to $ 1$,Prove that $ Q$ can be expressed $ Q \equal{} c_1P_1 \plus{}\ldots\plus{} c_mP_m$,where $ c_1,\ldots,c_m$ are nonnegative real numbers summing to $ 1$ and $ P_1,\ldots,P_m$ are permutation matrices.

(a permutation matrix is a $ 0,1$-matrix having exactly one $ 1$ in each row and column.)[/b]

this statement (as the link of nsato shows) can be proved using induction and hall's theorem,and the problem here follows immediately...


\[
\begin{gathered}
  \left( {\frac{{\sum {\left( {a + b} \right)^{\frac{2}
{3}} } }}
{3}} \right)^{\frac{3}
{2}}  \leqslant \left( {\frac{{\sum {\left| {a + b} \right|} }}
{3}} \right) = 1 \hfill \\
  \sum {\left( {a + b} \right)^{\frac{2}
{3}} }  \leqslant 3 \hfill \\
\end{gathered}
\]

Nice solution, siuhochung.

Here is the back ground of making this problem.

$x^{\frac{2}{3}}\leqq \frac{2}{3}|x|+\frac{1}{3}$


Today I read about the java Scanner class and can think of a good number of times that this would really simplify my code. However, I don't want to slow down my code, for obvious reasons. Do ya'll use Scanner s? Is there any performance hit in using Scanners over readLine()s+parse each character afterwards?

I think that Scanner slows down your code. Here is a comparison on random data, 100,000 lines, 50 characters per line, each character is randomly selected from the string " abcde" (note that the first character is white-space). Here are the methods I used to read in and parse the data (I didn't do anything else, i.e. I didn't print the data or store it anywhere):

[code]
FILE *fin = fopen("test.in","r");
char buffer[51];
while(fscanf(fin,"%s ",&buffer)!=EOF);
[/code]

Exec time in ms across 5 trials:
590 560 600 580 610

[code]
ifstream fin("test.in");
string s;
do fin >> s; while(!fin.eof());
[/code]

Exec time in ms across 5 trials:
510 530 540 540 540

It is, however, my experience that fscanf is much faster than using an ifstream if some information about the input format is known.

[code]
BufferedReader infile = new BufferedReader(new FileReader("test.in"));
String str; String[] s;
str = infile.readLine();
do {
  s = str.split("\\s");
  str = infile.readLine();
} while(str != null);
[/code]

Exec time in ms across 5 trials:
1522 1402 1442 1412 1472

[code]
BufferedReader infile = new BufferedReader(new FileReader("test.in"));
String str; StringTokenzer s;
str = infile.readLine();
do {
  s = new StringTokenizer(str);
  while(s.hasMoreTokens()) s.nextToken();
  str = infile.readLine();
} while(str != null);
[/code]

Exec time in ms across 5 trials:
641 621 631 651 641

[code]
Scanner sc = new Scanner(new File("test.in"));
while(sc.hasNextInt()) sc.nextInt();
[/code]

Exec time in ms across 5 trials:
Unknown (unfortunately, I don't have JAVA 1.5).

I did some tests myself and figured out that Scanner runs about 110 ms slower. For my analysis of a 1000 line input file containing 1000 floats and 1000 ints alternating, storing each, and summing the result, Scanner took an average of .34s, while split+parse{Int,Double} took an average of .22 s. The only difference was between the Scanner and the long way; the overhead was the same otherwise.

However, if the difference in performance isn't critical, simplifying your code can't be a bad idea. I would go for it, even if it's a little slower. (Unless you need to process big input files.)

As bus said, it really depends on what you're wanting to do. If you're after something that runs fast, don't use scanner. If you want something that can be coded fast, use scanner. Its pretty much as close you can get to printf in c++ as you can, which you can't really beat.
In the ICPC competition our team uses Scanner just because its simpler rather than having to split strings, parse things, etc.


How do you prove that the converse of Fermat's Little Theorem is not true?

$ 2^{341}\equiv 2\pmod{341}.$

$ 341 \equal{} 31\cdot 11$ is not prime.

What converse?

Probably Lehmer's theorem.

http://www.jstor.org/pss/2304732
http://mathworld.wolfram.com/FermatsLittleTheoremConverse.html


trigonometry
complex numbers
calculus
calculus computations

In mechanical vibrations, you model an unforced damped system as $ m\ddot{x} \plus{} c\dot{x} \plus{} kx \equal{} 0$.  Say the characteristic equation has two imaginary solutions $ \lambda_{1,2} \equal{} \alpha\pm\beta i$, you get a solution of the form
\[ x(t) \equal{} c_1e^{(\alpha \plus{} \beta i)t)} \plus{} c_2e^{(\alpha \plus{} \beta i)t)}\]
which simplifies to
\[ x(t) \equal{} (c_1 \plus{} c_2)e^{\alpha t}\cos(\beta t) \plus{} i(c_1 \minus{} c_2)e^{\alpha t}\sin(\beta t)\]
But in differential equation theory, you just simply drop the imaginary term in the general solution.  I understand we are concerned with just real solutions, but why are we just able to drop the imaginary term?

The key is that $ c_1$ and $ c_2$ are complex- which means that $ c_1\plus{}c_2$ and $ ic_1\minus{}ic_2$ can be any complex numbers $ k_1$ and $ k_2$. The solutions are real on the real line if and only if buth $ k_1$ and $ k_2$ are real.


Let a, b, c be three integers such that $a^3+b^3+c^3=\left(b+c\right)\left(c+a\right)\left(a+b\right)$. Prove that at least one of the integers a, b, c is 0.

  Darij

Your problem make me remember the equation from HongKong MO
Find all integer a,b,c such that $\frac{(a+b)(b+c)(a+c)} {2}+(a+b+c)^3=1-abc$
But your problem is travial and easier than this problem.Now I'll give you my solution
 We have $(a+b)(b+c)(a+c)=a^3+b^3+c^3$ hence $2(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2)$ and $4(a^3+b^3+c^3)=(a+b+c)^3$.So that $a^2+b^2+c^2=2(ab+bc+ac) (1)$.Oherwise $(a+b)(b+c)(a+c)=a^3+b^3+c^3$ hence $(a+b+c)(2(ab+bc+ac)-a^2-b^2-c^2)=4abc$.From $(1)$ we're easy get $4abc=0$ So we have the result. :)  :)
I think you should solve my problem.It isn't hard and very attractive :)  :)

[quote="nhat"]$2(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2)$[/quote]

Hmm... are you sure this is correct? ;) No, the problem is not that easy, although it is quite easy. Well, to be more precise, it uses quite a similar trick as the equation

$\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{2}+\left(a+b+c\right)^3=1-abc$

from the HongKong MO: Just write it in the form (b + c + 2a) (c + a + 2b) (a + b + 2c) = 2. My equation $a^3+b^3+c^3=\left(b+c\right)\left(c+a\right)\left(a+b\right)$ is even a bit more difficult (both sides, not just the left hand one, have to be written as products of expressions containing a, b, c).

  darij

I 've just proved it when a,bc, are non-negative real :$abc\geq 0$.Since the hypothesis we have :$2(a^3+b^3+c^3-abc)=(a+b+c)(a^2+b^2+c^2)\leq \frac{1}{3}(a^3  +b^3+c^3)$(by Chebysev's ineq)
$<\to 15abc\leq 5(a^3+b^3+c^3)\leq 6abc<\to abc\leq 0$
Hence $abc=0$,
Yes,as you saw ,I did not prove the problem!!

Can you tell me about your identity for writting Lhs as a product,Darij??


trigonometry
inequalities
geometry solved
geometry

Let ABC be a triangle. Prove that:

a(2a 2 -b 2 -c 2) + b(2b 2 -a 2 -c 2 )+ c(2c 2 -a 2 -b 2 )  \geq 0

It works for any positive numbers a, b, c:
From Schur?s ineq a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)>= 0, so a^3+b^3+c^3+3abc >= ab^2+ac^2+bc^2+ba^2+ca^2+cb^2 and because a^3+b^3+c^3>=3abc we get the ineq we want.


geometry
circumcircle
search
trigonometry
angle bisector
geometry proposed

$ D,E$ are points on $ AC,AB$ of $ \triangle ABC$ such $ BE \equal{} CD$. $ X$ is intersection of circumcircle of $ \triangle ADB,AEC$. Let $ O$ be circumcenter of $ \triangle ABC$. Show that $ DE\perp OX$.

remark.
When $ BE \equal{} CD \equal{} BC$, This problem become http://www.mathlinks.ro/viewtopic.php?search_id=110272116&t=147540

[quote="Erken"]The problem is very easy:
Solution:
We need to prove that $ DE\perp OX$,or which is equivalent to $ OD^2 \minus{} OE^2 \equal{} XD^2 \minus{} XE^2$.
It is easy to understand that $ OD^2 \minus{} OE^2 \equal{} x(c \minus{} b)$,where $ AB \equal{} c,AC \equal{} b,BE \equal{} CD \equal{} x$.
It is also easy to see,that $ \triangle XBE \equal{} \triangle XDC$.Hence $ XD \equal{} XB$ and
$ AX$ is an angle bisector of $ \angle ABC$.           $ (1)$
So let $ \angle ABC \equal{} 2\alpha$,$ \omega_B(O_B,R_B),\omega_C(O_C,R_C)$ be the circumcirlcles of the triangles $ \triangle AEC,\triangle ADB$ respectively.
Hence from $ (1)$:
$ XB \equal{} 2R_C\sin{\alpha}$.And $ 2R_C \equal{} \frac {BD}{\sin{2\alpha}}$,so $ XB \equal{} \frac {BD}{2\cos{\alpha}}$.Similar $ XC \equal{} \frac {CE}{2\cos{\alpha}}$.
$ BD^2 \minus{} CE^2 \equal{} (c^2 \plus{} (b \minus{} x)^2 \minus{} 2c(b \minus{} x)\cos{2\alpha}) \minus{} (b^2 \plus{} (c \minus{} x)^2 \minus{} 2b(c \minus{} x)\cos{2\alpha})$.From this we obtain that
$ XD^2 \minus{} XE^2 \equal{} XB^2 \minus{} XC^2 \equal{} x(c \minus{} b) \equal{} OD^2 \minus{} OE^2$.Thus we are done.
Thank you for nice problem.[/quote]
After $ (1),$ we an also say that $ (XD)^{2} \minus{} (XE)^{2} \equal{} (XB)^{2} \minus{} (XE)^{2} \equal{} 2(BE)\cdot (MN),$ $ ,(2)$ where $ M$ is the midpoint of the segment $ BE$ and $ N$ the orthogonal projection of $ X,$ on $ AB.$

From $ (2)$ $ \Longrightarrow$ $ (XD)^{2} \minus{} (XE)^{2} \equal{} x(c \minus{} b)$ because of it is easy to show that $ 2(MN) \equal{} c \minus{} b.$

Kostas Vittas.

i saw this problem in my teacher's notebook and he solve for us!! :idea:


A scientist ran a program to give him 8 random numbers. He noticed a pattern. The numbers were:
$ \minus{}1,0,\minus{}6,\minus{}15,\minus{}22,\minus{}19,8,83$
what is the next number?
(i will post the answer... eventually)

Well, first you add 1, then you subtract 6, then you subtract 9, then you subtract 7, then add 3, then add 27, then add 75. Is there any pattern in that?

I have a feeling it's very easy, but we are looking for to difficult pattern...

just saying, but i think it might be some function with a leading coeffecient of ax^n where n>2. (i think most of u'll already know that)  :)

well, there is a [i]super slow way[/i]. well, looking at the behavior of the numbers, it looks kinda like a x^3 function. u could put $ f(1) \equal{} a(1)^3 \plus{} b(1)^2 \plus{} c(1) \plus{} d \equal{} \minus{} 1$ and so on with the other numbers by putting $ f(n) \equal{} an^3 \plus{} bn^2 \plus{} cn \plus{} d$ where the domain of f(x) is the natural numbers. then u simplify them and make four equations to solve for the four variables. if there is no solution, then u do it for a x^4 function, a x^5, a x^6,....

yes, it is [i]very slow[/i].  :|

I'll give you a hint.

[hide]it is just addition and subtraction.
as simple as that.
and... Wickedestjr started it right... just go a little deeper.
to toadoncart: :yoda: overcomplicating it, you are.[/hide]

i still don't think you will figure it out. :huh:

w00t i think i got it

difference of the adjacent terms are +1, -6, -9, -7, +3, +27, +75

and difference of that is -7, -3, +2, +10, +24, +48

and difference of that is +4, +5, +8, +14, +24

and difference of that is +1, +3, +6, +10

yay triangular numbers

so next 4th difference is +15
next 3rd difference is +24+15=39
next 2nd difference is +39+48=+87
next 1st difference is +75+87=+162

so the answer is 83+162=245?

answer : 245

...
has anybody ever solved a problem while you are trying to post the solution???
good job, FantasyLover!

That is what I was planning on doing, but I thought that it would have been a waste of time. Even if I had gotten that pattern of triangular numbers, I wouldn't of have known what to do next.


projective geometry
geometry unsolved
geometry

let a, b, c are the sides of triangle ABC and D is the mid-point of the line BC. show that |AD| = t satisfies

t = $\frac{2b^{2}+2c^{2}-a^{2}}{4}$

Please post it in High-School topic.
We have: $\vec{AD}=\frac{1}{2}(\vec{AB}+\vec{AC})\implies AD^{2}=\frac{1}{4}(AB^{2}+AC^{2}+2\vec{AB}\cdot\vec{AC})=\frac{2b^{2}+2c^{2}-a^{2}}{4}$

then where should i post?
post in high school basics, intermediate topics, pre-olmypiad or other problem solving topics?

yrhc@eht, please edit your problem.

This problem can be solved in just using Pappus thorem.

sorry, what is Stewart thoerem and Pappus thorem?

[url=http://www.ies.co.jp/math/java/geo/pappus/pappus.html]Pappus's theorem[/url]

If your problem is proving this theorem, then our proof doesn't make sense. :)

[quote="yrhc@eht"]sorry, what is Stewart thoerem?[/quote]

If there is a point $D$ on the line $BC$ of the $\triangle ABC$,we have that $AB^{2}*DC+AC^{2}*BD-AD^{2}*BC=BC*DC*BD$.


function
integration
limit
calculus
real analysis
trigonometry
real analysis unsolved

Got this question from my class that we need to present next week, wonder if anyone could help, thx.

Let $g: R \rightarrow R$ is a continuous function. Suppose that there is some $p>0$ such that $g(x)=g(x+p)$ for all $x\in R$ and define $\eta(g)= \frac{1}{p} \int_{0}^{p} g(t) dt$

Fix $a<b$

(1). Define $G: R\rightarrow R$by $G(x)= \int_{0}^{x} g(t) dt$
    Show that if $\eta(g)=0$ then $G(x+p)=G(x)$ for all $x\in R$
    Conclude that if $\eta(g)=0$ then $G$ is bounded

(2). Suppose that $\phi : [a,b] \rightarrow R$ is a step function.
   Show that if $\eta(g)=0$ then $\lim_{m\to\infty} \int_{a}^{b} \phi(t) g(mt) dt=0$

(3). Let $f: [a,b] \rightarrow R$ be an integrable function. Show that if $\eta(g)=0$ then $\lim_{m\to\infty} \int_{a}^{b} f(t)g(mt) dt =0$

(4). Let $f: [a,b] \rightarrow R$ be an integrable. Show that $\lim_{m\to\infty} \int_{a}^{b} f(t) g(mt) dt = \eta(g)\int_{a}^{b} f(t) dt$

(1) is straightforward.

For (2), it suffices to prove the theorem when $\phi$ is a characteristic function of a single integral. (The general case involves a linear combination of these special cases, and the integral is linear.) Make the change of variable $u=mt,$ express what you have using $G,$ and use the fact that $G$ is bounded.

For (3), make use of this theorem: If $f$ is integrable on $[a,b]$ and $\epsilon>0,$ then there exists a step function $\phi$ such that $\int_a^b|f-\phi|<\epsilon.$

For (4), write $g(x)=\eta(g)+h(x).$  Note that $\eta(h)=0$ and we can write $G(x)=\eta(g)x+H(x).$ Use part (3) as it applies to $h.$

(Note concerning #3: you didn't mention whether your theory of integration was the Riemann theory or the Lebesgue theory. But that theorem exists in either theory.)

Shouldn't that expression have an $n$ in it?

[color=blue][Yes. Fixed now.][/color]

well.. what expression you talked about?
I got this from my class, so...

by the way, what's to do with the integration by "kent"

I guess it's should deal with the Riemann integra in (3), referes to kent's previous post, does it make difference, how would you approach?

Thx

The theorem we want - the approximation in integral of an integral function by a step function - is easier to prove in the Riemann theory than the Lebesgue theory, being very close to the definition of the integral in the Riemann case.

Hey, Kent,
When you said it's straghtforward for the first part of question, can you exhibt in some way to show it? Thx

There would be several ways to do that. Here's one:

Let $w(x)=G(x+p)-G(x).$ Then $w'(x)=g(x+p)-g(x)=0.$

Hence $w(x)$ is a constant.

$w(0)=\int_0^pg(t)\,dt=p\,\eta(g).$

If $\eta(g)=0,$ then $w(x)=w(0)=0$ for all $x$ and hence $G(x+p)=G(x)$ for all $x.$

What do u mean by " For (3), make use of this theorem: If $f$ is integrable on $[a,b]$ and $\epsilon>0,$ then there exists a step function $\phi$ such that $\int_a^b|f-\phi|<\epsilon.$ " ?

Is this related to Riemann Integra, could u explain more? Thx

How will u show the part 2, anyone can say it more?

Suppose $f$ is the characteristic function of $[a,b].$ That is, suppose $f$ is one on $[a,b]$ and zero off of that interval. Just write out what you're trying to prove here and do the appropriate change of variables. What happens?

Once you prove that one case, we can go back and deal with a finite linear combination of such functions.


What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

[hide]Let $n$ be the integer to find. If $d$ is an odd divisor of $n$, then the even divisors of $n$ will be given by the $2d$, $4d$, $8d$, etc. until $2^kd$ where $k$ is the largest power of 2 dividing $n$. Then the number of positive even divisors must be $k$ times the number of positive odd divisors $\Rightarrow k=2\Rightarrow n=4m$, where $m$ is odd.

The number of positive divisors of $m$ will be the number of odd positive divisors of $n$, which is 6. If the prime factorization of $m$ is given by $p_1^{e_1}p_2^{e_2}\cdots$, then $(e_1+1)(e_2+1)\cdots=6$. Then the possibilities for $m$ are given by $p^5$ and $p^2q$ where $p$ and $q$ are distinct primes. The smallest of these numbers is $3^2*5=45\Rightarrow \min n=\boxed{180}$.[/hide]

[hide]Letting the desired number be $N$, we first examine the odd divisors. Notice that if the prime factorization of $N$ were $N=m^2n$, it would have 6 divisors. The smallest odd combination that works is $3^2\cdot5$. Now, we need there to be twice as many even divisors as odd divisors, which can be done by multiplying by $4$. (Do you see why?) Our final result is thus $4\cdot9\cdot5=\boxed{180}$.[/hide]

[hide="Answer"]Removing all factors of $2$, the remaining prime factors have exponents which multiply out so that $(e_1+1)(e_2+1)=6$. The two possibilities here are that $e_1=5$ or $e_1=2$, $e_1=1$. Taking the smallest possible values, $e_1=2$ for $p_1=3$ and $e_2=1$ for $p_2=5$ give smaller values than $e_1=5$ for $p_1=3$. There are eighteen divisors total, so we multiply the six by three, meaning that for $p_3=2$, we have $e_3=2$. Therefore, our answer is $2^2\cdot 3^2\cdot 5=\boxed{180}$.[/hide]

[quote="shadysaysurspammed"]What exactly is he $(e1+1)(e2+1)$....stuff?[/quote]

$p_1^{e_1}p_2^{e_2}...$ is the prime factorization of the number.  If we want to find the number of factors of the number, we would have it of the form $p_1^{a_1}p_2^{a_2}...$ and $0\le a_1 \le e_1$, so there are $(e_1+1)$ possible values for $a_1$.  Use this for all of the factors to get $(e_1+1)(e_2+1)...$.  In other words, in constructing a factor of the number, there are $e_1+1$ possible exponents for $p_1$, $e_2+1$ for $p_2$, and so on.

[hide]You obviously want 2 and 3 to be divisors so then 6 is a divisor too. For each odd divisor there is an even divisor that is two time s that odd number. So then we it to be 4*1*3^2*5=120.  [/hide]

[hide=easier thought process]take the 2 smallest odd primes, 3,5. note that 3^2*5^1 gives 6 odd divisors, and for it to have 12 even divisors, it needs to have 18 total divisors, meaning we must multiply the quantity by 2^2, since that will multiply the number of divisors by 3. it is easy to see that 3^5*2^2 is far larger than 3^2*5^1*2^2, so the answer is 180.[/hide]

Play around with the numbers and the factors of $18,$ and you can get that $n = 5 \cdot 3^2 \cdot 2^2 = \boxed{180}.$

Quick morning problem.

This number must have a total of $18$ positive integer factors. There are not that many such numbers, so we can list them out. Such a number will either be in the form $x^{17}, x^8 \cdot y, x^5 \cdot y^2$, or $x^2 \cdot y^2 \cdot z$. We will start from the last case as it should yield the smallest possible solutions. We can have $x = 2, y = 3, z = 5$ as our smallest possible outcome. This results in $180$, which indeed has six positive odd integer divisors and twelve possible even integer divisors. After doing a quick check over the other options, we can see that the next-smallest option would be $2^5 \cdot 3^2$ which is much larger than $180$. Thus, $\boxed{180}$ is the answer.


I am studying The transform $T=3F+F^{-1}$ which in $F$ be Fourier transform in solving partial differential equations, it Ý shown in the following attached file. But I have some problems with the 0.1-0.5 problems, please counsel me!


inequalities
calculus
function
inequalities theorems

[quote="kdt"]What school math subject would these inequalities, not to mention everything in this Olympiad section, fit under?
By subjects I mean like Algebra 1 or Geometry.
Clearly (or at least I HOPE not) this is not Algebra1 material. I would fit this under uber-advanced Pre-Calculus. What about you guys :)?[/quote]

Much calculus are involved in inequalities!

Calculus is very much involved if, like myself, you often try to use Jensen's inequality. Of course, there are many non-calculus inequalities such as AM-GM and Cauchy. Smoothing, a general technique which is very similar to Jensen, does not require calculus either.

Jensen's itself is not calculus, but the criteria we apply to determine convexity of functions (if $f$ is twice differentiable, then $f''(x)>0\Longrightarrow f$ is convex) uses calculus.


You guys know that game where you try to make 5 in a row on a board  by taking turns putting peices on it?

Is there an online version of it where you can play other people?
Can someone post a link please?

Yes I know that game. It's called "connect 5". You can search it in google and find some good links. One of them is:
[url]http://en.t45ol.com/play/159/connect-5.html[/url]


I don't know if anyone has brought this up, but this game is addicting and becomes really, really hard. There are 24 levels in total. Links is http://onemorelevel.com/game/spin_the_black_circle

Hah. I've played this before... (just recently)
Not sure which level I got up to, though
Funn game  :lol:

I got to this spiral thing with a huge number of spikes.

ummm, all i see and a black screen with a monkey flashing saying like bubble games or something

stopped at 12 (I think the same as 1=2)

Really irritating level, and I had NEARLY won when I screwed up
takes about a minute to get close...

addicting got to lvl 18 and quit cause my other class started

Got to level 19. I do not know how to get past the long string of flames. :(

EDIT: I got to lvl 20!

Wait until you get to level 23. That becomes really tricky. And 22 is a nice level also.

[quote="1=2"][quote="andersonw"]Currently on this level with a big rotating plus sign in the middle.
Yay, I'm ahead of 1=2 and klebian![/quote]

Beat that. :)[/quote]

Haha, I'm on level 21 now! :)

i can't get past level 1... :D
i suck @ games like these...

Fun game.  Can't get past the last level.

How do you beat #16?
That's the one with the big U and spikes all around it.
I can't do it!!!!

EDIT:
I did it!!!!!!!

I beat the game. It's fun. Current total time: 6:15:35.


Let $a$ and $b$ be positive integers such that $a | b^{2}, b^{2}| a^{3}, a^{3}| b^{4}, b^{4}| a^{5},\hdots$.  Prove that $a = b$.

Masoud Zargar

This is a nice problem  :)

[hide]Clearly, $a$ and $b$ have the same prime divisors, say $p_{1}, p_{2}, \cdots p_{n}$. Let $a = \prod_{i=1}^{n}p_{i}^{\alpha_{i}}$ and $b = \prod_{i=1}^{n}p_{i}^{\beta_{i}}$. It's given that $a^{2k-1}| b^{2k}$ and $b^{2k}| a^{2k+1}$ for all positive integers $k$, which is equivalent with $(2k-1)\alpha_{i}\leq 2k\beta_{i}$ and $2k\beta_{i}\leq (2k+1)\alpha_{i}$ for all $i\in\{1,2,\cdots,n\}$. Therefore, $1-\frac{1}{2k}\leq \frac{\beta_{i}}{\alpha_{i}}\leq 1+\frac{1}{2k}$. Let $k\rightarrow\infty$ to get that $1\leq \frac{\beta_{i}}{\alpha_{i}}\leq 1$, so $\alpha_{i}=\beta_{i}$, which proves the statement. [/hide]

[hide="Without UPF"] $n | m \implies n \le m$

So we have

$b^{2k}< a^{2k-1}$
$a^{2k}< b^{2k+1}$

Let $r = \frac{a}{b}$.  Then

$r^{2k}< a \forall k$
$r^{2k}> \frac{1}{b}\forall k$

$\lim_{k \to \infty}r^{2k}= 0, 1, \infty$.  Clearly only $1$ works, which implies $r = 1$.  QED. [/hide]

I used unique factorization, since I was reading an article on unique factorization followed by problems without solutions.


trigonometry
geometry
analytic geometry
geometric transformation
reflection
ratio
similar triangles

In the isosceles triangle $ ABC$ the angle $ BAC$ is a right angle. Point $ D$ lies on the side $ BC$ and satisfies $ BD \equal{} 2 \cdot CD$. Point $ E$ is the foot of the perpendicular of the point $ B$ on the line $ AD$. Find the angle $ CED$.

[hide="Trig Solution"]

There is definitely a straightforward 2-line solution since I cannot solve geometry problems without trigonometry bashing. From the problem's condition, $ BAC$ is 45-45-90 triangle. So, let $ AB \equal{} 1, AC \equal{} 1,$ and $ BC \equal{} \sqrt{2}$. Then, $ BD \equal{} \frac{2 \sqrt{2}}{3}$ and $ CD \equal{} \frac{\sqrt{2}}{3}$. Let $ \angle BAD \equal{} \alpha$. Then, $ \angle DAC \equal{} 90 \minus{} \alpha$, $ \angle ABE \equal{} 90 \minus{} \alpha,$ $ \angle DBE \equal{} \alpha \minus{} 45,$ $ \angle BDE \equal{} \alpha \plus{} 45,$ $ \angle CDE \equal{} 135 \minus{} \alpha$ (the numbers are in degree measures). Using Law of Sine on $ \triangle ACD$ and $ \triangle ABD$, we can solve for $ AD$ and get $ \frac{2 \cdot CD}{\sin \alpha} \equal{} \frac{CD}{\sin(90 \minus{} \alpha)}$. So, $ 2 \equal{} \frac{\sin \alpha}{\cos \alpha} \equal{} \tan \alpha$.

Using this, we can apply Law of Sine again on $ \triangle ABE$ to get $ AE \equal{} \frac{1}{\sqrt{5}}$ and $ BE \equal{} \frac{2}{\sqrt{5}}, DE \equal{} \frac{2}{3 \sqrt{5}}$ using Pythagorean Theorem. Now comes the annoying part. Let $ \angle ACE \equal{} \beta$ and $ \angle DCE \equal{} 45 \minus{} \beta$. Using Law of Sine on $ \triangle ACE$ and $ \triangle CED$, we get:

$ \frac{\frac{1}{\sqrt{5}}}{\sin \beta} \equal{} \frac{CE}{\sin(90 \minus{} \alpha)}$ and $ \frac{\frac{2}{3 \sqrt{5}}}{\sin(45 \minus{} \beta)} \equal{} \frac{CE}{\sin(135 \minus{} \alpha)}$. Solving for $ CE$ and setting them equal, we still have messy equation. So, expanding the sine part for all of them, we get

$ \sin(135 \minus{} \alpha) \equal{} \frac{3 \sqrt{2}}{2 \sqrt{5}}$ and $ \sin(45 \minus{} \beta) \equal{} \frac{\sqrt{2}}{2} (\cos \beta \minus{} \sin \beta)$.

So, simplifying as much as possible, we get $ 3 \sin \beta \equal{} \cos \beta$. Now, applying Pythagorean Identity, we see that $ \sin \beta \equal{} \frac{1}{\sqrt{10}}, \cos \beta \equal{} \frac{3}{\sqrt{10}}$. Putting these back to find $ \sin(45 \minus{} \beta)$ and calculating, we get $ CE \equal{} \frac{\sqrt{10}}{5}$.

Now, applying Law of Sine AGAIN on either of the small triangle ($ \triangle ACE$ or $ \triangle CED$), $ \sin \angle CED \equal{} \frac{\sqrt{2}}{2}$.

So, $ \angle CED \equal{} 45^\circ$ or $ 135^\circ$. From the diagram, it is obvious that $ 45^\circ$ is the right answer. But to be more rigorous, if $ \angle CED \equal{} 135^\circ$ in my diagram, then with $ \triangle CED$, $ \beta$ must be 0 or otherwise, triangle's total sum will surpass 180 by adding only two angles. But since $ \beta$ is not 0, $ \angle CED \equal{} 45^\circ$.

$ \therefore \ \angle CED \equal{} 45^\circ \ \blacksquare$ QED [/hide]

WLOG, let AB=AC=1, then $ BD \equal{} 2\sqrt {2}/3$ and $ DC \equal{} \sqrt {2}/3$. By Stewart's, $ AD \equal{} \sqrt {5}/3$. But the area of $ \triangle ABD$ is 1/3 (BD is 2/3 the length of BC, the total area is 1/2). So $ BE \equal{} 2\sqrt {5}/5$ using area. So by Pythag $ AE \equal{} \sqrt {5}/5$ and $ ED \equal{} 2\sqrt {5}/15$. Since DE/CD=CD/AD, $ \triangle CED$ is similar to $ \triangle ADC$, so CED=45.

The two line proof probably involves showing the similarity easily.

[hide]
Let T be the intersection of BE with the circumcircle. $ \angle BTA\equal{} \angle BCA \equal{} 45^\circ \Rightarrow AE\equal{}ET$.
$ \frac {BD}{DC} \equal{} \frac {BE}{ET}\equal{}\frac {BE}{AE} \Rightarrow \tan \mu \equal{} \frac {1}{2}$
$ \tan (45^\circ \minus{} \mu) \equal{} \frac {\tan 45^\circ \minus{} \tan \mu}{1\plus{}\tan 45^\circ  \tan \mu}\equal{}\frac {1}{3}\equal{}\frac {TC}{BT}$
$ \frac {TC}{BE}\equal{}\frac {TC}{\frac {2}{3} BT}\equal{}\frac {3}{2} \cdot \frac {1}{3} \equal{} \frac {1}{2} \Rightarrow CT \equal{} AE \equal{} ET$
[/hide]

You could also solve it using coordinates (I just used the isosceles triangle at $ (0,3)$, $ (\minus{}3,0)$, and $ (3,0)$ and then applied a scaling argument)  :D

[hide="Solution"]
Let the midpoint of $ BC$ be $ M$, let $ AM$ and $ BE$ meet at $ H$, and let $ BE$ meet $ AC$ at $ N$.  Since $ AB = AC$, we have that $ AM\perp BD$ and we already know that $ BE\perp AD$, so $ H$ is the orthocenter of $ \triangle ABD$, so $ DH\perp AB$.  Yet, $ AC\perp AB$, so $ DH\parallel AC$.  Thus,\[ \frac {HM}{AH} = \frac {MD}{DC} = \frac {MC - DC}{DC} = \frac {\frac {BC}{2} - \frac {BC}{3}}{\frac {BC}{3}} = \frac {\frac {BC}{6}}{\frac {BC}{3}} = \frac {1}{2}\]It follows that since $ H$ is on median $ AM$, we have that $ H$ is the centroid of $ \triangle ABC$.  Thus, $ N$ is the midpoint of $ AC$.  Now, notice that $ \angle AEN = 90 = \angle BAN$, so $ \triangle AEN\sim \triangle BAN$, which means that $ \frac {EN}{AN} = \frac {AN}{BN}$, so $ BN\cdot EN = AN^2 = CN^2$.  This gives that $ \frac {EN}{NC} = \frac {NC}{BN}$, so $ \triangle ENC\sim \triangle CNB$, so $ \angle NCE = \angle NBC$.  Furthermore, from similar triangles $ AEN$ and $ BAN$, we have that $ \angle EAN = \angle ABN$, so\[ \angle AEC = 180 - \angle EAC - \angle ECA = 180 - \angle ABN - \angle CBN = 180 - 45 = 135\]from which we conclude that $ \angle DEC = 180 - \angle AEC = \boxed {45}$. [/hide]

As in the previous solution, the goal is to prove that, if $M$ is midpoint of $BC$ and $N=BE\cap AM$, then $AN=2\cdot MN$. Take $P$ the reflection of $A$ across $BC$, so $ABPC$ is a square and the perpendicular from $B$ to $AD$ will divide $AP$ into the same ratio as $AD$ does for $BC$: $\frac{AN}{NP}=\frac{CD}{BD}=\frac{1}{2}$. Consequently, $AN=2\cdot NM$ and, the continuation is as above.

Best regards,
sunken rock

[color=darkred][b][u]Extension[/u].[/b] Let $\triangle ABC$ so that $AB\perp AC$ . Consider

$\left\{\begin{array}{ccc}
D\in (BC) & ; & \frac {DB}{DC}=k>\frac {c^2}{b^2}\\\\
E\in (AD) & ; & BE\perp AD\end{array}\right|$ . Prove that $\cot\widehat{CED}=k\cdot\frac bc-\frac cb$  .[/color]

[u]See[/u] [b]PP14[/b] from [url=http://www.artofproblemsolving.com/Forum/blog.php?u=9882&b=60549][color=darkred][b]here[/b][/color][/url].


Princeton college calculus topology email vector integration

Just curious, who from AoPS will be attending Princeton?

Does everyone else plan to take 215 1st semester?

Do you think I should take 215 if I've never taken a class in multivariable calculus or differential equations?  I have the requisite test scores to place into 215, but is it normal to take 215 without having taken multi or diff?  Do those topics come later in the sequence?

[quote="shbmsc90"]yes, definitely. i haven't taken a class in mv calc and diff eqns (or lin alg) at all either, but i'll end up taking it with you. it's doable, i hear. hard (like every other class at princeton), but doable. i guess it's more of an intro class in proofs for some people.[/quote]
Yeah, I took 215 and I can second that. 215 is "from first principles" (that is, you don't technically need to know much other than arithmetic to do it). It helps a lot to have BC calculus or equivalent, but anything beyond that is pretty much unnecessary.

I was leaning towards taking 217 in addition to 215 just because Conway's teaching it this fall and, um, it's Conway.  Any anecdotes about his classes Xevarion?

You are opening Pandora's box here! I had 215 with Conway my first semester. Here are some tidbits about him:

- He probably won't use a textbook. As far as I know, he doesn't prepare his lectures. He just does everything from memory.

- He shouts the important stuff to make sure you realize it is important.

- Sometimes he digresses into totally unrelated stuff (in 215, most of the second half of the semester was spent on topology and on the Bernoulli numbers).

- He doesn't remember to give homework every week. Sometimes he gives the same problems more than once for homework.

- He's always sitting around in the math common room. You can go talk to him any time just by going there and walking over to him. He's one of the most accessible profs in the department. HOWEVER, he often doesn't respond to email (he might check it though, I'm not sure), and he doesn't often check his mailbox either.

:D

217 also isn't a horribly difficult class, especially if you've worked with matrices and vectors before. And Conway is fun.

Thanks for the replies.  I'm not so much concerned about the difficulty of the class, because if there's a textbook that goes along with the course I can usually figure stuff out on my own that I'm having trouble with.  However, I am worried about skipping through important topics like multi and diff.  I took a semester of linear and a semester of combinatorics this year, so I'm pretty comfortable with proofs, although I'm sure there is a lot more I have to learn.  What would a typical math sequence look like that starts with 215?  It would make me more comfortable topic-wise if I saw when I would cover multi and diff.

215 - intro to analysis
217 - honors linear algebra
218 - multi
(that gets you into the math dept)
301 or 303 or something - ODEs
330 - fourier analysis and PDEs
there are also engineering PDE classes.

edit: even if you've already had linear algebra, you should still take 217, unless your lin alg class was really hardcore. 217 is really more like a second course in lin alg. Also, you can take some of these classes simultaneously. Here's what I've done in math:

freshman fall: 215
freshman spring: 217, 330, 332 (intro to measure theory and integration theory)
sophomore fall: 390 (probability theory), 333 (special topics in analysis), 322 (algebra)
sophomore spring: 325 (topology), '44' (a reading course on geometric measure theory), ORF527 (stochastic calculus)

NB: this is not a typical math sequence. A typical first four semesters might be
ff 215
fs 217
sf 218, 322
ss 330, 3xx

or if you want to take a lot of math

ff 215, 217
fs 218, 322 or 330
sf 3xx, 3xx
ss 3xx, 3xx

It's not really advisable to take 3 math classes at once. You need to get your distribution requirements done, and you want to make sure you understand the material well, which means spending a fair amount of time on the homework and on studying. Especially this semester (soph spring), I really didn't get as much out of 325 and ORF527 as I should have, because I spent too much time on the reading course.

[quote="yunorman"]Xaverion,

Did you ever take 218? I know other people at princeton who haven't taken one of 215, 217 or 218 and they still get into the math dept. Is that standard?

If I wish to skip one of 217 or 218, would that be fine for me getting into the math dept? I heard that the requirements aren't so strict at princeton.

Do you know if that's the case in other depts too? For example to get the engineering and management systems cert. you must take an introductory stats course, but do you really need to need to take the course?[/quote]
1) I did not take 218 and I do not plan to. Almost everyone takes at least two of the three 21x classes, but you can get away with less. But you do have to take a later course that's "equivalent" (I took 332, real analysis, to replace 218). You could probably use 332 to replace 217 or 215 also; I'm not sure if they'll like it if you try to use that to replace all 3, but if you are skipping all three of those then you can take some other 300-leve classes before the end of soph year too.

2) Yes, they're very lenient. Talk to (email) Professor Gunning if you're not sure (you can find his contact info on the math dept web page http://www.math.princeton.edu )

3) Depends on the department. Talk to the undergrad rep of the department in question. Smaller departments tend to be more lenient.


calculus
integration
function
real analysis
trigonometry
limit
real analysis unsolved

Was looking up for a definition for an improper riemann integral.. I couldnt really find a proper one to answer the following questions:

1. Is it true? An improper riemann integral exists implies that the limit of the integral is the same, whichever way you approach it.

2. royden says that an improper reimann integral may exist without the function being lebesgue integrable. and he gave the example, $(\sin x)/x$ over the interval $[0,\infty)$. Pardon my stupidity, but what is the improper riemann integral of this function?

3. can someone provide me an example, where $f$ is lebesgue integrable over $E$, but the improper riemann integral does not exist?

thanks.

1. "Whichever way you approach it" is a concern for those cases in which more than one limit must be defined in order to come up with an improper integral. Some examples:

$\int_{0}^{1}\frac1{\sqrt{x}}\,dx$ is improper for only one reason. Hence, unambiguously, we can define it as

$\lim_{a\to0^{+}}\int_{a}^{1}\frac1{\sqrt{x}}\,dx$

Since this limit does exist, this is a convergent improper integral. (In this case, it's also an instance of a Lebesgue integrable function - for nonnengative functions, the convergence of the improper integral will be the same as Lebesque integrability.)

Where we run into trouble is with integrals that are improper for more than one reason. An instance of that would be

$\int_{-1}^{2}\frac1x\,dx,$ which we must interpret as

$\lim_{a\to0^{-}}\int_{-1}^{a}\frac1x\,dx+\lim_{b\to0^{+}}\int_{b}^{2}\frac1x\,dx$

Since each of those limits, separately, diverges, we must by convention declare the integral divergent.

One could extract a number out of this - called the Cauchy principal value - by balancing the two limits against each other. But we must stress that the principal value is not the same thing as the improper integral. I would add that this example is of a function which is not Lebesgue integrable.

2. Let's prove that the improper integral $\int_{0}^{\infty}\frac{\sin x}x\,dx$ converges.

We will integrate by parts. We will also consider $\frac{\sin x}x$ to be a continuous function at zero whose value at zero is $1.$ In other words, remove the removable singularity. (And as I proceed with the calculation, I will also remove some other removable singularities at zero.)

$\lim_{b\to\infty}\int_{0}^{b}\frac{\sin x}x\,dx = \lim_{b\to\infty}\left[\left.\frac{1-\cos x}x\right|_{0}^{b}+\int_{0}^{b}\frac{1-\cos x}{x^{2}}\,dx\right]$

$=0+\lim_{b\to\infty}\int_{0}^{b}\frac{1-\cos x}{x^{2}}\,dx$

The latter is the integral of a nonnegative function; as such, as a function of $b$ it is nondecreasing. We can prove that it is convergent by proving that it is bounded above.

But $\int_{0}^{1}\frac{1-\cos x}{x^{2}}\,dx$ is the integral of a continuous function (note that I am again removing a removable singularity) and thus Riemann integrable, while

$\int_{1}^{\infty}\frac{1-\cos x}{x^{2}}\, dx\le \int_{1}^{\infty}\frac2{x^{2}}\,dx,$ which converges.

Note: it is possible to compute an exact value, namely that

$\int_{0}^{\infty}\frac{\sin x}x\,dx = \frac{\pi}2,$

but that computation is not necessary to make the point about the improper integral converging.

On the other hand,

$\int_{0}^{\infty}\left|\frac{\sin x}x\right|\,dx \ge \sum_{n=1}^{\infty}\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{(n+1)\pi}=\sum_{n=1}^{\infty}\frac2{(n+1)\pi},$

which diverges, so the function is not absolutely integrable and hence not Lebesgue integrable.

3. Nothing special about being improper; you do know that the Lebesgue integral handles functions which the Riemann integral cannot touch. Try this example on the interval $(0,1):$

$f(x)=\begin{cases}\frac1{\sqrt{x}}&x\text{ irrational}\\ 0&x\text{ rational}\end{cases}.$

Then $\int_{0}^{1}f(x)\,dx=2$ exists as a Lebesgue integral; since $\int_{a}^{1}f(x)\,dx$ does not exist as a Riemann integral for any $0<a<1,$ we cannot even define the improper integral.

Thanks a lot!

About point 3. Hmmm. I was thinking of something else.

Is it possible to have $f$ integrable over $[0,\infty)$ and $f$ being Riemann integrable on $[0,a)$ for all $a$, but yet the improper Riemann integral of $f$ over $[0,\infty)$ does not exist?

[quote="mao_zai"]Is it possible to have $f$ integrable over $[0,\infty)$ and $f$ being Riemann integrable on $[0,a)$ for all $a$, but yet the improper Riemann integral of $f$ over $[0,\infty)$ does not exist?[/quote]
Of course. That's like asking if it is possible for a series to diverge.

Try this: $\lim_{a\to\infty}\int_{0}^{a}\sin x\,dx$

i want $f$ to be Lebesgue integrable over $[0,\infty)$.

You're not going to get that example. If the Riemann integral exists over $[0,a],$ then the Lebesgue integral exists over $[0,a].$ The integral is Lebesgue integrable iff it it absolutely convergent; just as for series, absolute convergence implies convergence.

Theorem: Suppose $f$ is Lebesgue integrable on $[0,\infty).$ Then

$\lim_{a\to\infty}\int_{0}^{a}f(x)\,dx=\int_{0}^{\infty}f(x)\,dx$

Proof: take any sequence $a_{n}\to\infty.$ Let $f_{n}(x)=\begin{cases}f(x)&x\in[0,a_{n}]\\0&x>a_{n}\end{cases}.$

Then $\lim_{n\to\infty}f_{n}(x)=f(x)$ pointwise everywhere, and $|f_{n}(x)|\le|f(x)|$ everywhere, so by the Dominated Convergence Theorem, we can interchange the limit and the integral.

:blush:  i'm sorry for all the confusion. basically, i cannot formulate the confusion i'm having myself.

ok. consider the following non-negative measurable function $f=\frac{1}{x^{2}+1}$ defined over $(-\infty,\infty)$.

Define the following sequences of non-negative and non-decreasing functions:
$f_{n}(x)=\begin{cases}f(x) & x\in [-n,n]\\ 0 & x\notin[-n,n]\end{cases}$, and,
$g_{n}(x)=\begin{cases}f(x) & x\in [-n,\frac{2n}{1-n^{2}}]\\ 0 & x\notin[-n,n]\end{cases}$.

Now, both converges to $f$, but $\int f_{n}=2 \tan^{-1}n\to \pi$, and $\int g_{n}=3 \tan^{-1}n \to \frac{3}{2}\pi$. But this would mean that $\int f$ has two different values by Monotone Convergence Theorem. Is there something wrong?

Your calculation for $g_{n}$ doesn't make sense. For one thing, for $n>1,$ $\frac{2n}{1-n^{2}}$ is negative and tends to zero. Hence, $g_{n}$ does not tend to $f$ at all, but rather to a function which is equal to $f(x)$ for $x<0$ and to zero otherwise.

But the integral of that should be $\frac{\pi}2,$ not $\frac{3\pi}2.$ The latter number appears to be a mistake.


modular arithmetic
number theory
prime numbers
number theory proposed

Find all prime numbers $ p$ for which $ 6p^3\plus{}1$ and $ 8p^3\plus{}1$ are prime numbers.

[quote="arqady"]Find all prime numbers $ p$ for which $ 6p^3 \plus{} 1$ and $ 8p^3 \plus{} 1$ are prime numbers.[/quote]

Are you sure the problem is correct? $ 8p^3\plus{}1\equal{}(2p)^3\plus{}1^3\equal{}(2p\plus{}1)(4p^2\minus{}2p\plus{}1)$, and both factors are bigger than one, so $ 8p^3\plus{}1$ is not prime.

[quote="arqady"]Find all prime numbers $ p$ for which $ 6p^3 \plus{} 1$ and $ 8p^3 \plus{} 1$ are prime numbers.[/quote]
If $ p\neq 7$ then $ p^3 \equiv \begin{cases} \minus{}1 &\pmod{7}\\ 1 &\pmod{7} \end{cases}$. In the first case $ 8p^3\plus{}1$ is divisible by $ 7$, in the second case $ 6p^3 \plus{} 1$ .
Both numbers are $ >7$ and therefore no pirmes.
So we get $ p\equal{}7$, but $ 8\cdot 7^3 \plus{} 1$ is divisible by $ 3$.
$ \Rightarrow$ There is no $ p$ for wich $ 6p^3 \plus{} 1$ and $ 8p^3 \plus{} 1$ are prime numbers.

[quote="caffeineboy"][quote="arqady"]Find all prime numbers $ p$ for which $ 6p^3 \plus{} 1$ and $ 8p^3 \plus{} 1$ are prime numbers.[/quote]

Are you sure the problem is correct? $ 8p^3 \plus{} 1 \equal{} (2p)^3 \plus{} 1^3 \equal{} (2p \plus{} 1)(4p^2 \minus{} 2p \plus{} 1)$, and both factors are bigger than one, so $ 8p^3 \plus{} 1$ is not prime.[/quote]
It's correct,caffeineboy. The answer is: Does not exist. What about the following problem?
Find all prime numbers $ p$ for which $ 10p^5 \plus{} 1$ and $ 12p^5 \plus{} 1$ are prime numbers.

[quote="arqady"]Find all prime numbers $ p$ for which $ 10p^5 \plus{} 1$ and $ 12p^5 \plus{} 1$ are prime numbers.[/quote]
One of the numbers is divisible by $ 11$ if $ p \neq 11$ and both numbers are $ >11$.
We get $ p\equal{}11$, but $ 3 | 10\cdot 11^5 \plus{}1$.
Again there are no solutions.


The distance from a point $P(x, y)$ on a parabola to the focus is called the focal radius of the point. Show that the length of the focal radius to the point $(a,b)$ on the parabola with the equation $y^{2}= 4px$ is $|a+p|$.

Any hints on how to get started on this one?

Thanks.

Heres my take, hope this helps:

[hide]I dont know latex, so bear with me..
The equation is y^2=4px, which means that p is the distance from the vertex to the focus. Since y is squared in this equation, it is a sideways parabola, and since 4 is positive it opens to the right, which means the focus is p units to the right of the vertex.

So you have a parabola with vertex (0,0) and focus (p,0). Solving for y in the equation, y = 2 sqrt(px), so you have to find the distance from (p,0) to (x, 2 sqrt(px)). Use the distance formula, and it becomes p + x. I guess theres an absolute value sign there because distances cant be negative.[/hide]

ohhh solve for y..

yeah I'd already done what you did in your first paragraph, but I solved for x, not y :blush:

Ok thanks for your help!


Assuming non-zero denominators, $\frac{k}{x-2}+\frac{w}{(x-2)^{2}}+\frac{bx+f}{x^{2}+1}= \frac{3x^{2}+3x+7}{(x-2)^{2}(x^{2}+1)}$ for all values of x. Find the value of $k+w+p+f$.

According to the solution:
When we simplify it, we get
$k(x-2)(x^{2}+1)+w(x^{2}+1)+(px+f)(x-2)^{2}=3x^{2}+3x+7.$

And then we let
$x=2; x=0; x=1; x=-1$
so we get a system of equation of f, k and p and find that w=5.

I understood that $x=2$ as for one of the factors to be $0$ but why letting $x=1$ and $x=-1$ how do you know which numbers to let $x=...$?

Thanks!

Well they could have picked any four numbers, since the two sides must be equal for ALL x. It's just that the system is easier to deal with for smaller $x$.

so basically, any four numbers would give a system and that system would help solve for k, w, b, and f. If there are three variables, then we would should get three equations to make up that system, and so forth...?


A particular right square-based pyramid has a volume of
63,960 cubic meters and a height of 30 meters. What
is the number of meters in the length of the lateral
height ($ \overline{AB}$) of the pyramid? Express your answer
to the nearest whole number.

[asy]draw((-8,0)--(0,20)--(8,0)--cycle);
draw((8,0)--(11,4.5)--(0,20));
draw((9.5,2.3)--(0,20));
draw((9,3.3)--(9.7,4.3)--(10.6,3.1));
dot((9.5,2.3));
label("$B$",(9.5,2.3),SE);
label("$A$",(0,20),N);
dot((0,20));[/asy]

first- u find the square root of 63960, which approximately equals 252.903 (yes i do have a life and i am lazy, i used a calc)

find half of the side, approx. = 126.452

then treat the measurements as a right triangle

a^2 (2 lazy 2 latexify) = height = 30^2 = 900
b^2 = half the side of square base = 126.452^2, approx. = 15990.108
c^2 = lateral height = the square root of (900 + 15990)

ans = 130

(wow, that was long) :sleeping:

[hide="try again"]Okay, so we have $ \frac{1}{3}\text{Bh}\equal{}63,960$, and after we plug stuff in, we can find that B=6396. Now, since I don't want to type, let each side of the square be x, where $ x\equal{}\sqrt{6396}$. Now, we can create a right triangle with half the side length and the height. Letting what we want be y, we have $ \frac{x^2}{4}\plus{}900\equal{}y^2$, and we can find $ y^2\equal{}2499$, so it's approximately 50, our answer.[/hide]


For a triangle $ ABC$, let $ AD\bot BC$ ( $ D\in BC$), $ DE\bot CA$ ( $ E\in CA$), $ AF\bot BE$ ( $ F\in DE$).

When $ \frac {BD}{DC} \equal{} \frac {2}{3}$  $ \Longrightarrow$  find $ \frac {FE}{DF}$

I got two solutions, which one is analytic and the other is by vector. But I want only synthetic ones, (not trigonometric).
May I ask you a favor ? Thanks for reading my post.

I think there is some data you missed. I built the figure in Geogebra leaving $ B$, $ D$, $ C$ as fixed and moving $ A$ the ratio is not constant.

Sorry, but it is constant. Please check it in detail~

We claim that there exists the equality: BD/DC = EF/DF.
Take P - the foot of altitude of B on AC in ABC.
The triangles ADC and BPC are similar and have the omologous sides perpendicular. AS BE and AF are perpendicular too, we get that E and F divide PC and DE in the same ratio, i.e. PE/CE = EF/FD, but PE/CE = BD/CD, hence our claim is true.

Best regards,
sunken rock

"Because BE and AF are perpendicular too, we get that E and F divide PC and DE in the same ratio"
$ \Longrightarrow$ I can't understand  :what?:

You may take F' on (DE)| EF'/F'D = PE/CE, and get same result: PC and DE are both opposite to equal angles in the right angled triangles BPC and ADE and, for PE/EF' = BP/AE we conclude triangles AEF' and BPE are similar, hence <EAF' = <PBE, the quadrialteral PABQ is cyclic (Q - intersection AF' with BE) and <AQB = 90. Since AF and BE are perpendicular too, we get F and F' are the same.

Best regards,
sunken rock


Let be $ a,b,c,d\in (0,\infty)$ such that $ a^6\plus{}b^6\plus{}c^6\plus{}d^6\equal{}1$ . Prove that :

$ (a\plus{}b\plus{}c\plus{}d)^2\le \frac{b^2c^2d^2}{a^{10}}\plus{}\frac{c^2d^2a^2}{b^{10}}\plus{}\frac{d^2a^2b^2}{c^{10}}\plus{}\frac{a^2b^2c^2}{d^{10}}$

I hope it holds  :D

[quote="alex2008"]Let be $ a,b,c,d\in (0,\infty)$ such that $ a^6 \plus{} b^6 \plus{} c^6 \plus{} d^6 \equal{} 1$ . Prove that :

$ (a \plus{} b \plus{} c \plus{} d)^2\le \frac {b^2c^2d^2}{a^{10}} \plus{} \frac {c^2d^2a^2}{b^{10}} \plus{} \frac {d^2a^2b^2}{c^{10}} \plus{} \frac {a^2b^2c^2}{d^{10}}$

I hope it holds  :D[/quote]
Yes, this true.

Does this holds with condition $ a^5\plus{}b^5\plus{}c^5\plus{}d^5\equal{}1$ or $ a^7\plus{}b^7\plus{}c^7\plus{}d^7\equal{}1$ ?

[quote="alex2008"]Does this holds with condition $ a^5 \plus{} b^5 \plus{} c^5 \plus{} d^5 \equal{} 1$ or $ a^7 \plus{} b^7 \plus{} c^7 \plus{} d^7 \equal{} 1$ ?[/quote]
Well. It holds for $ a^r \plus{} b^r \plus{} c^r \plus{} d^r \equal{} 4^{\frac {6 \minus{} r}{6}}, r \ge 1$, with equality when $ a \equal{} b \equal{} c \equal{} d$. And thus not when $ a^7 \plus{} b^7 \plus{} c^7 \plus{} d^7 \equal{} 1$.. But it holds for $ a^5 \plus{} b^5 \plus{} c^5 \plus{} d^5 \equal{} 1$, but there is never equality..


geometry
circumcircle
cyclic quadrilateral

could someone pls prove that if
 Angle  ABD and Angle ACD are equal for the segment AD
Prove that ABCD is a cyclic Quadrilateral

Well that's one of the basic "definitions" of a cyclic quadrilateral (or one of the facts that follow immediately)... but a proof I guess goes:

Draw the circumcircle of $\triangle ABD$. But since $\angle ABD = \angle ACD$ and intercept the same arc, $C$ must lie on the circumcircle. Thus the quadrilateral is cyclic.

Let $C$ be an arbitrary point outside of the circumcircle $(O)$ of the triangle $\triangle ABD$ and $C'$ the intersection of the line $OC$ with the circumcircle. The triangles $\triangle DOA, \triangle AOB, \triangle BOD$ are all isosceles, because their sides $OA = OB = OD$ are radii of the circle $(O)$.

$\angle ABD = \angle ABO + \angle DBO = \angle BAO + \angle BDO =$

$= (\angle BAD - \angle OAD) + (\angle BDA - \angle OBA) =$

$= (\angle BAD + \angle BDA) - (\angle OAD + \angle OBA) =$

$= (180^o - \angle ABD) - (180^o - \angle AOD) = \angle AOD - \angle ABD$

$2 \angle ABD = \angle AOD,\ \ \angle ABD = \frac{\angle AOB}{2}$

In exactly the same way, it can be shown that the angle $\angle AC'D = \frac{\angle AOD}{2}$, hence, $\angle ABD = \angle AC'D$. Since the point $C$ is outside of the circle $(O)$, we have $OC > OC', AC > AC', DC > DC'$ in the triangles $\triangle AOC, \triangle DOC$ and consequently,

$\angle ACD = \angle ACO + \angle DCO < \angle AC'O + \angle DC'O = \angle AC'D = \angle ABD$

Thus the angle $\angle ACD < \angle ABD$ for any point $C$ outside of the circle $(O)$. In a similar way, it can be shown that the angle $\angle ACD > \angle ABD$ for any point inside of the circle $(O)$. Therefore, if these 2 angles are equal, the point $C$ must be on this circle.


I have a TI-89. How do I program it? Thanks in advance!

Go to the app "Program Editor."

Have fun playing!

Where's "Program Editor"? I can't find it.  :blush:

EDIT: Oh wow, I'm blind. Found it. But, um, how do I use it? (I'm completely hopeless...)

Ech.... now is the time to consult the manual. :lol:

It's pretty easy. It's good if you learn some basic java/C stuff though.

ti basic is very very easy
just learn how to use inputs lables outputs and the store button
u can make tons of stuff
i suck and i made a program to prime factor things

However, I think that you program differently on a TI-84 and a TI-89.

Anyway, this [url=http://www.ticalc.org/pub/]link[/url] is for lazy bums only.

It's pretty much the same. I've done both TI-89 and TI-83+/84+.
Some differences is that the defining of variables is different
When you do a loop like for:
for(blah
thing to be looped
EndFor

so, can you make formula programs and use them on MC?!!

only on the team round (maybe target? I forgot the rules)

I'm pretty sure you can use it on Target too. If they allow graphing calculators, they must know that people are going to program them...

According to JBoyd, programmed calculators are allowed.

They are to be used in Target Round and Team Round.

but then people could just make a program that if you put in say volume of sphere it would spit out $ \frac{4\pi r^3}{3}$ and that wouldn't be fair would it? cause i know that many people can't afford a programmable calculator!

ernie... that problem would be wayy to easy for a target or team round question

Do you mean poincare?
Then you have to type it in manually. It's the same amount of work you know...
To make that, it'll be
PROGRAM: JKESKEGJ
:Prompt A
:Disp $ 4/3\piA^2$

It's only 2 lines of code... it doesn't save any time. In fact programming formulas before math contests is useless.

No, i mean the spherical thing was just an example, it could be the shoelace algorithm or even on those rates problems you could just make a program that solves them for you! that's really unfair.


linear algebra
matrix
linear algebra unsolved

Suppose $ A$ and $ B$ are two equally-sized symmetric square matrices with real entries.

Suppose that there exist two matrices $ X$,$ Y$ of $ M_{n}(R)$ such that $ rank(AX \plus{} BY) \equal{} k$

Prove that $ rank(A^2 \plus{} B^2)\geq k$

$ A \equal{} B \equal{}
\left(
\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}
\right)$, $ X\equal{}
\left(
\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}
\right)$,  $ Y\equal{}
\left(
\begin{array}{cc}
1 & 1\\
0 & 0
\end{array}
\right)$

That doesn't seem to be a counterexample - everything, including $ AX\plus{}BY,$ has rank $ 1.$ But what about this?

$ A\equal{}B\equal{}\begin{pmatrix}0&1\\0&0\end{pmatrix},$ $ X\equal{}Y\equal{}I.$

Err... I think you missed the word "symmetric" (or Moubinool added it at a later time???).

$ A^2,B^2,A^2\plus{}B^2$ are non negative symmetric matrices.
There exists $ E$ a subspace of $ \mathbb{R}^n$ of dimension $ k$ s.t. $ \forall{x}\in{E}\minus{}\{0\},x^T(AX\plus{}BY)\not\equal{}0$. Assume that $ rank(A^2\plus{}B^2)<k$; then $ \exists{x}\in{E}\minus{}\{0\}$ s.t. $ x^T(A^2\plus{}B^2)x\equal{}0$ or $ \parallel{}Ax\parallel{}^2\plus{}\parallel{}Bx\parallel{}^2\equal{}0$ what is a contradiction.


Hi to all !

You know, the russian problem is good about combinatorics..

_______________________

On the blackboard, following numbers are written. [tex] 1000,1001,...,2999[/tex].

We hands numbers between two, like [tex]a,b[/tex] are changed in [tex] \frac{1}{2} min(a,b)[/tex]. We repeat this operation exactly 1999 times, then only one number (let us call it [tex] c [/tex]) leave. Then show that [tex] c<1 [/tex].

_______________________

   :D   BEAUTIFUL  ....  CHALLENGE TO IT !

Consider binary tree which contains history of our operations. The root of this tree is c, it has two children a and b (c=min(a,b)/2), these children both has two children too, and so on, until we reach initial numbers (1000,...,2999).
Consider leafs 1000,1001,...,2047. Let m be arbitrary one of them and $c=x_0\to x_1\to x_2 \to\dots\to x_k=m$ is path from root c to m. If $k\geq 11$ then since $x_{i-1}\leq x_i/2$ we obtain $c<1$. So suppose $k\leq 10$ for all these leafs, but it is impossible since tree of height 10 contains at most 1024 leafs.

Are you english native speaker, olympiad-math?


function
calculus
algebra
domain
topology
real analysis
real analysis unsolved

I am using Apostol's calculus text, and in it he resorts to a wordy bisection argument to prove that continuous functions are bounded. I was wondering why we couldn't simply assume the existence of an unbounded point in the interval and then immediately invoke the definition of continuity to obtain a contradiction.

Let $f: [a,b] \to \mathbb{R}$ continuous function in $[a,b]$ .

You can assume that for all $M>0$ there exist $x \in [a,b]$ such that $|f(x)|>M$ .

Then for $M=n$ you have a sequence $x_{n}\in [a,b]$ s.t. $|f(x)|>n$ .

Use Bolzano - Weierstrass theorem  to get the contradiction .

[quote]I was wondering why we couldn't simply assume the existence of an unbounded point in the interval and then immediately invoke the definition of continuity to obtain a contradiction.[/quote]
Because there's no such thing as an "unbounded point."

Let's look at it this way: consider the functon $f(x)=\frac1x$ on the domain $(0,1).$

Of course, that is an unbounded function. It's also continuous at every point of its domain. I'm sure you want to say that it's unbounded "at" zero; but there's no such thing as being unbounded "at" a point. Not only that, there is no such point as zero - or at there very least, there is no such point as long as we look only within the domain of the function.

A continuous function can be unbounded. That alone dooms your idea. The theorem isn't that "contunuous functions are bounded" but something much more specific: A continuous function [b]on a closed bounded set[/b] is continuous and assumes its maximum. Any proof of this must deal very specifically with the fact that the hypothesized domain is a closed bounded set (or, if we're slightly less ambitious about what we're proving) a closed bounded interval.

This theorem is one of the earliest appearances in elementary analysis of the notion of a compact set. (Another such early appearance is the Bolzano-Weierstrass Theorem, one version of which says that a bounded sequence has a convergent subsequence.) Successive bisection arguments are quite at home as a way of proving properties that depend on compactness.

--

Note added: "Ramanujan" has invoked Bolzano-Weierstrass; that's yet another sign that this is a theorem about compactness. Back up a step behind that - how would you prove Bolzano-Weierstrass?

I'm sorry for not being more specific...it was assumed that that the theorem refered to the function being continuous on a closed interval and bounded on that closed interval. The book is not at the level to be concerned with compactness or metric space or anything that typically belongs to the abstract analysis. The proofs typically only involve the definition of continuity and least upper bounds. As far as unbounded point, I did not mean to assert a formal definition.

Anyways, supposing our function is continuous on [a,b] , and further supposing there is some c in [a,b] such that f is not bounded at c...we recall that f is continuous at c and that there is some d such that for every x

|x-c| < d   , then |f(x) - f(c)| < 1

and we see that in this way f(x) would be bounded in this neighborhood of c.

Does this fail because functions cannot be unbounded at distinct points but only on intervals? Is the manner in which f(x) = 1/x is unbounded on (0,1) the only manner in which such functions can be unbounded? That is, when we consider f(x)= 1/x on (-1, 0) and (0, 1), we see that it isn't that the function is unbounded inside the interval (-1, 1), but only in neighborhoods of the endpoints of the separate intervals (-1,0) and (0, 1)  because this function is not defined at x=0? Can I conclude that it doesn't make sense to consider an "unbounded point" because a function would not even be defined at such a point?

With all of this in mind, is seems almost absurd to then assert that a function would be unbounded on a closed interval yet somehow not be unbounded at any particular point in that interval.....

[quote]Does this fail because functions cannot be unbounded at distinct points but only on intervals?[/quote]
That's the key point right there. If a real-valued function is defined at a point, then it it bounded "at" that point, because its value is one specific real number.

You're trying to see "unbounded" as an affirmative, as something that happens, somewhere. But it's not that; it's a negative concept, the lack of a bound. Here's a function $f(x).$ Is $1000$ an upper bound? No, because I can find some $x$ such that $f(x)>1000.$ Well, then, is $10^{6}$ an upper bound? No, because I can find some different point $y$ such that $f(y)>10^{6}.$ And so on. And that's the way it goes; if the function is truly unbounded, I can respond to each such challenge, but with different points each time.

Thank you for your clarification of what you meant. It turns out that I understood you the first time; this is indeed what I thought you were trying to say. I was specifically warning you against this approach to the proof; it goes nowhere.

I also didn't mean that you would find the concept of a compact set familiar. In my mind the best way to creep up on that concept by first proving a number of theorems, highly important in themselves, that will be seen in retrospect as manifestations of the same underlying concept. This includes the Bolzano-Weierstrass theorem, the Extreme Value Theorem (the theorem we are talking about here) and the theorem that a continuous function on a closed bounded interval is uniformly continuous. In a similar vein, we will state and prove the Intermediate Value Theorem long before we define the concept of connectedness.

The Extreme Value Theorem says that a continuous real-valued function on a closed bounded interval is bounded and assumes its maximum. Somewhere down the road, a generalization of this will read "the continuous image of a compact set is compact." In a number of ways, I find the more specific theorem - the Extreme Value Theorem - to be more fundamental. Similarly, I find the Intermediate Value Theorem to be more fundamental than its eventual generalization, "the continuous image of a connected set is connected."

Thank you for the feedback. My only remaining question is how this notion even became an issue. If we assume the function is continuous on a closed interval but then become worried of the possibility that it may be unbounded on this interval, it seems that something like a ghost is being feared. We say, "we are worried that such functions may be unbounded, we must prove otherwise". That is, we worry that for such a function, there does not exist a C such that C>f(x) for all x on the interval. It seems then we are asserting the existence of some point c in the interval such that f(c) is greater than any number one could name. And this elusive point is what I mean by "unbounded" point. But we say there is no such point, that it is altogether meaningless to assert that there can even be such an "unbounded" point. But then it seems that we are saying "of necessity there can be no point c in the interval such that f(c) is greater than any number one can name."  And this seems to be equivalent to "for all x in the interval, f(x) is LESS than some number"  But isn't this exactly what we wanted to show? Does it seem that we have known it all along without any kind of fancy proof?

[quote]It seems then we are asserting the existence of some point c in the interval such that f(c) is greater than any number one could name.[/quote]
You're still falling into the same trap. For an unbounded function (like $\frac1x$ on the interval $(0,1)$ ), there's isn't one magical point $c$ where, once and for all, $f(c)$ is bigger than anything. Rather it is a process: for each alleged possible upper bound $M,$ we can find an $x$ such that $f(x)>M,$ The $x$ depends on the $M.$ Different $M,$ different $x.$

Now we need to prove that this can't happen for continuous functions on closed bounded intervals. To you this seems obvious - but it can only be deemed "obvious" on the basis of your experience in drawing examples of graphs of continuous functions. That experience is quite insufficient. Have you ever drawn a continuous function with infinitely many local maxima and minima (within a finite interval)? Have you ever drawn a continuous function that has no tangent lines at any point? Strange as those things may sound, there are examples of such continuous functions.

Because there are such strange things, you cannot rely on your intuition or experience to tell you what is obvious. The Extreme Value Theorem is neither obvious nor easy. It is in fact quite a deep theorem of analysis.


function
trigonometry
algebra
polynomial
real analysis
real analysis unsolved

We have:
\[\sum_{k=1}^{n-1}\dfrac 1{\sin^2(\frac{k\pi}n)}=\sum_{k=1}^{n-1}\bigg(1+\cot^2\bigg(\frac{k\pi}n\bigg)\bigg)=\frac{n^2-1}3\]
since $ \cot\frac{k\pi}n \;\text{with}\; k=1,2,\cdots n-1$ are the roots of the polynomial:
\[\binom{n}{1}x^{n-1}-\binom{n}{3}x^{n-3}+\cdots.\]


algebra
polynomial
function
difference of squares
special factorizations

[quote="pkothari13"]Given a polynomial P whose degree is 5, P has the following characteristic:  if z is a complex number such that $z^5 + 2004z = 1$, then $P(z^2)=0$.  Compute $\frac{P(1)}{P(-1)}.$

I figured there was some way I could solve it using methods found in AoPS II, but I came up with nothing.[/quote]

we can change the first equation to $z^5 + 2004z - 1= 0$.  Since $P(z^2)=0$ when the first equation is satisfied, $P$ has roots that are sqaures of the roots of the first polynomial, and here's a way to find a polynomial with such property:

we can rewrite first polynomial as $\prod x-r_i$, then if we multiply it by $\prod x+r_i$, it will become the difference of squares and we get $\prod x^2 - r^2_i$, now substitute in $y=x^2$ then we're done

The second polynomial can be found by plugging in $-x$ into the first polynomial, so that it becomes $\prod -x-r_i$, or $\prod -(x+r_i)$.  since there are five roots, we have one more negative sign and we have to multiply it by $-1$.

call the first polynomial $f(x)$, then $f(-x) = -z^5 - 2004z - 1$, multiply it by $-1$ we get $g(x) = z^5 + 2004z + 1$, multiply it with $f(x)$ (notice it is a difference of two squares) and substituting $y = x^2$ we get $P(x) = z^5 + 4008z^3 + 2004^2z - 1$.  substituting $1$ and $-1$ and dividing we get $\frac{2004^2 + 4008}{-(2004^2 + 4008 +2)}$ as a very nasty answer.   Did I do something wrong...??

Wow... nice :surf:  That's the right answer, by the way.  Thanks for the explanation!

[quote="scorpius119"][hide]If $z$ is a root of the equation $z^5+2004z=1$, square both sides to get $z^{10}+4008z^6+2004^2z^2=1$.

From here it follows that $P(x)=x^5+4008x^3+2004^2x-1$. Then just plug in 1 and -1 for $x$ and divide.[/hide][/quote]

How can you just square both sides?? How do you know that will give you your new function?

Oh right, I wasn't exactly clear on that. It works because the function $x^\frac 52 +2004x^\frac 12-1$ is zero when $x=z^2$. To get a polynomial, the powers need to be integers, so squaring does this.


One-half of a road construction project was completed by 6 workers in 12 days. Working at the same rate, what is the smallest number of workers needed to finish the rest of the project in exactly four days?

With the same number of workers, it would take $ 12$ days to finish the remaining job. Thus, we need to triple the amount of workers to $ 6\cdot3\equal{}\boxed{18}$ to reduce that time to $ 4$ days.

Or we could notice that they are inversely proportional.  We have $ 72\equal{}4x$, so $ x\equal{}18$.

...which is exactly the same logic I used, only I explained it. :wink:


Να βρεθεί ο αριθμός $ A\equal{}\overline{aaabb}$  όπου είναι τετράγωνο φυσικού, περιττός και διαιρείται με το $ 9$.

[quote="Spribo"]Απαιτείται κατ'αρχάς 3a+2b=0mod9. Έπειτα ο Α λήγει είτα σε 1 είτε 5 είτε 9 (περιττός και τετράγωνο φυσικού). Άρα πρέπει είτε 3a=7mod9 είτε 3a=8mod9 είτε 3a=0mod9. Με απλές δοκιμές προκύπτει ότι ο Α δεν υπάρχει.
 (Έαν έχω λάθος διορθώστε με!)[/quote]
Θα προτείνω και εγω μια λύση , παρακαλώ να με διορθώσετε:
Σύμφωνα με την υπόθεση του προβλήματος προκύπτει ότι
\[ 3a \plus{} 2b \equal{} 9k \Leftrightarrow k \equal{} \frac{{3a \plus{} 2b}}
{9} \equal{} \frac{a}
{3} \plus{} \frac{{2b}}
{9}\]
Όμως \[ \left( {9,2} \right) \equal{} 1\].Έτσι έπεται ότι b=9.
Έτσι α=3 ή α=6 ή α=9.
Γνωρίζουμε οτι τοτράγωνο περιττου είναι της μορφής 8π+1. Με δοκιμές προκύπτει ότι δεν υπάρχει τέτοιος αριθμός

[quote="Spribo"]Στην πραγματικότητα, από τη σχέση σου έπεται ότι [b]όλος[/b] ο αριθμός σου είναι ακέραιος, όχι τα χωριστά τμήματά του.[/quote]
Ναι αλλά για να είναι ακέραιος δεν πρέπει να είναι και οι όροι του αθροίσματος ακέραιοι???

Είναι συνθήκη ικανή, αλλά όχι αναγκαία. Για παράδειγμα, 1/3 + 2/3=1.

Αυτό που λες, ισχύει αν και μόνο αν οι παρονομαστές των κλασμάτων σου είναι πρώτοι μεταξύ τους, όπως εδώ. Πρόκειται για ειδική περίπτωση.

[quote="Papalakis"]Να βρεθεί ο αριθμός $ A \equal{} \overline{aaabb}$  όπου είναι τετράγωνο φυσικού, περιττός και διαιρείται με το $ 9$.[/quote]
Έχω την εντύπωση ότι ακόμα και αν δεν είχαμε την condition $ 9|A$ πάλι δεν θα υπηρχε λύση...
$ A \equiv 1 (mod8) \Leftrightarrow 10.000a\plus{}1.000a\plus{}100a\plus{}10b\plus{}b \equiv 1(mod 8) \Leftrightarrow 4a\plus{}3b \equiv 1 (mod8)$
Αλλά εύκολα συμπερένουμε ότι $ b \in \{1,5,9\}$.Παίρνοντας περιπτωσεις καταληγουμε σε ατοπο χωρίς το $ 9|A$.....

Όμως, ούτε εδώ δεν είναι σχετικά πρώτοι. Εδώ τυχαίνει να βγαίνεις σωστός ο τρόπος. Όμως, εάν είχες (2a+b)/3 ακέραιο, δεν μπορούσες να υποθέσεις ότι 3|a και 3|b.
(Το προηγούμενο αν και μόνο αν ξέχασέ το).