geometry
perpendicular bisector
geometry solved

let PA and PB be tangents to the circle at A,B.  show that any chord XY passes thru M, the midpoint of AB, has PM as the bisector of angle XPY

Let X'Y' be a chord of the circle which also passes through M s.t. XX' || YY' ||AB. It's clear that X' is the symmetric of X with respect to PM and the same can be said for Y' and Y. It should be enough to prove that XY' and X'Y intersect at P.

Let P' be the intrsection between XY' and X'Y. It's clear that P is on PM because the entire figure is symmetric with respect to PM. Let Z be the intersection of P'X and AB. PX/PY'=XX'/YY' (*). ZM/YY'=XZ/XY' and ZM/XX'=ZY'/XY'. By dividing one of these by the other one we get ZX/ZY'=XX'/YY' (**). From (*) and (**) we get PX/PY'=ZX/ZY' => Z is on the polar of P' with respect to the circle => Z is on the line joining the points of tangency between the circle and the tangents from P', but this line is parallel to AB because P' is on the perpendicular bisector of AB, so Z is, at the same time, on AB and on a line parallel to AB, which means that the 2 lines are the same, so the polar of P' with respect to the circle is AB, so P'=P. I guess this ends it.

This is very similar to the 1'st prb from the Greek selection test posted by Arne on Artofproblemsolving ("Advanced Problems" section).

this problem seems to be an old problem  :D i have seen it or similar ones in many math olympiad books. and i remembered that there was also a similar problem in the homework of plain geometry in my junior middle school years.  :D


I have 300 unit cubes.  I attach them all by gluing faces of cubes together.  What is the largest surface area I can possibly create for my new solid?

[hide]The rectangular prism with the largest suface are always has two of its sides equal 1.

That would mean that with 300 cubes, you have 2 sides that are 1, and 4 sides that are 300.

300+300+300+300+1+1 = 1202

[/hide]

[hide]The largest surface area would be all the cubes in a straight line.  All the cubes have at least 4 faces so 300*4=1200.  The two cubes at the end each have one more face, so 1200+2=1202[/hide]

An infinite amount, given you can cut the unit cubes into infinitely small pieces.

Otherwise, I would say 1202.


geometry
incenter
circumcircle
geometry proposed

In triangle $ABC$, $AB > AC$. Through $A$, draw the tangent $l$ of the circumcircle of $ABC$. Draw a circle $w$ with center $A$ and radius $AC$. $w$ intersects $AB$ at $D$, and $w$ intersects line $l$ at $E, F$, $E$ is closer to $C$. Prove that $DE$ passes through the incenter of $ABC$, and $DF$ passes through one of the excenters of $ABC$.

It's just an angle chase. Since $C,D$ are symmetric wrt $AI$, in order to prove that $DE$ passes through $I$, it suffices to prove that $\angle ADE=\angle ACI$ (and we can do something similar for in order to show that $DF$ passes through the excenter of $ABC$ opposite to $A$).


[quote="saxy_mcc"]I can't compile the document because my AoPS user name has an underscore in it.  It compiles fine if I delete the one character, so I was wondering if it's ok to send with my correct username, but knowing that it doesn't compile, or is it better to do something else? :?[/quote]

Do you get an error message that underscores can be used only in math mode? TeX has a few special characters, such as \, _, $, and %, that are used for commands. It is a common mistake to forget this when writing ordinary text and try to use a special character as a character rather than a command. To use a special character as a character, you have to use a command. For example, the command to make an underscore character is \_. Thus, to write saxy_mcc in LaTeX, I type saxy\_mcc, as in:

$\text{saxy\_mcc}$
Hmm, the underscore looks ugly in that particular font and size. It looks better at a smaller size.
$\text{\footnotesize saxy\_mcc}$

Erin Schram
USAMTS grader

$\text{saxy\_\ \!mcc}$ looks a little better. It's a crude solution, but it works pretty well, as far as I can tell.

If your username contains any of the following characters:
 # % $ & _ ~ ^ { }
then it must be preceded by a backlash (\).  So if your username were math_genius, then the line in the solutions.tex file would be

\username{math\_genius}

Don't worry about making it look pretty.  Definitely do *not* put your username into math mode.


[quote="MCrawford"]What percent of 44 is 55?[/quote]
[hide]$\dfrac{x}{100} \times 44= 55 \rightarrow x= \dfrac{55}{44} \times 100=125%
$[/hide]

jensen, your latex is messed up.  You might want to edit your post and fix it.

In my perspective, you can go about this problem in a simple way. You must remember that is/of = percent/100 with the percent being what you do not know.  So if it says 55 is what percent of 44 keep this in mind.

55--is
-------- = percent/100
44--of

[hide=" :D "]
$\frac{55}{44} = \frac{x}{100}$
when you cross multiply you get
$5500=44x x=\frac{5500}{44} x=125$
[/hide]
click on mr. green.


1.Let $ z$ be a complex number such that $ |z| \equal{} 1$.

   Find the minimum and maximum value of

   $ |1 \plus{} z| \plus{} |1 \minus{} z \plus{} z^2|$.

2.If $ a>0$ and $ |z\plus{}\frac{1}{z}|\equal{}a$.

   Then find minimum and maximum value of $ |z|$.

    What does the equation $ |z\plus{}\frac{1}{z}|\equal{}a$ denote geometrically

maximum value is 13/4 and minimum is 3^1/2 .[/b]

[quote="pankajsinha"]1.Let $ z$ be a complex number such that $ |z| \equal{} 1$.

   Find the minimum and maximum value of

   $ |1 \plus{} z| \plus{} |1 \minus{} z \plus{} z^2|$.

2.If $ a > 0$ and $ |z \plus{} \frac {1}{z}| \equal{} a$.

   Then find minimum and maximum value of $ |z|$.

    What does the equation $ |z \plus{} \frac {1}{z}| \equal{} a$ denote geometrically[/quote]

[b]Question 1 :[/b]
Let $ z\equal{}e^{i 2\theta}$
$ |1\plus{}z|^2\equal{}(1\plus{}z)(1\plus{}\frac{1}{z})\equal{}2\plus{}2\cos 2\theta\equal{}4\cos^2 \theta$

Similarly $ |1\plus{}z^3|^2\equal{}4 \cos^2 3 \theta$

So $ |1 \plus{} z| \plus{} |1 \minus{} z \plus{} z^2|\equal{}|1\plus{}z|\plus{}\frac{|1\plus{}z^3|}{|1\plus{}z|}$
$ \equal{}|2 \cos \theta|\plus{}|\frac{\cos 3 \theta}{\cos \theta}|$
$ \equal{}|2 \cos \theta|\plus{}|4 \cos^2 \theta \minus{}3|$

WLOG, $ y\equal{}\cos \theta \ge 0$
[b]Case 1:[/b] $ y \ge \frac{\sqrt{3}}{2}$ , then $ \sqrt{3} \le 4y^2\plus{}2y\minus{}3 \le 3$
[b]Case 2:[/b]$ y \le \frac{\sqrt{3}}{2}$ , then $ \sqrt{3} \equal{}\frac{13}{4}\minus{}(\sqrt{3}\minus{}\frac{1}{2})^2 \le 2y\minus{}4y^2\plus{}3\equal{}\frac{13}{4}\minus{}(2y\minus{}\frac{1}{2})^2 \le \frac{13}{4}$

Combining two cases , max. value $ \equal{}\frac{13}{4}$ , min. value $ \equal{}\sqrt{3}$

[quote="A K Bhattacharyya"]Could you prove the max. and min. part for me?

As for the second part,

it's not hard to see, that 1/z is actually z-conjugate in this case. Hence |z+1/z|=a denotes a straight line in this case.
[/quote]

Is $ |z|\equal{}1$ is a condition of Question 2 ?

[quote="stephencheng"]

Is $ |z| \equal{} 1$ is a condition of Question 2 ?[/quote]

No.


induction
linear algebra
linear algebra unsolved

Let $f_1(x)=f(x)=\frac{a_1x+b_1}{c_1x+d_1}\text{ } (a_1d_1\neq b_1c_1)$
And let
\[ f_2(x)=(f\circ f)(x)=\frac{a_2x+b_2}{c_2x+d_2}  \]
\[ f_3(x)=(f\circ f\circ f)(x)=\frac{a_3x+b_3}{c_3x+d_3}  \]
\[ \vdots  \]
\[ f_n(x)=(f\circ f\circ\cdots\circ f)(x)=\frac{a_nx+b_n}{c_nx+d_n}, n\in\mathbb N  \]
Let $\mathbf A=\left(\begin{array}{cc}a_1&b_1\\c_1&d_1\end{array}\right)$ and $\mathbf{A}_n=\left(\begin{array}{cc}a_n&b_n\\c_n&d_n\end{array}\right)$. Prove that $\forall n\in\mathbb N: \mathbf{A}_n=\mathbf A^n$.

[quote="alekk"]just induction ..[/quote]
but how to use the induction?

Actually, that should be $A^n=\lambda_nA_n$ for some nonzero constants $\lambda_n$.

Show that composition of linear fractional transformations is multiplication of their matrices. The group of such transformations is thus $PGL_2(F)$.

[quote="ifai"]Let $f_1(x)=f(x)=\frac{a_1x+b_1}{c_1x+d_1}\text{ } (a_1d_1\neq b_1c_1)$
and let .........
\[ f_n(x)=(f\circ f\circ\cdots\circ f)(x)=\frac{a_nx+b_n}{c_nx+d_n}, n\in\mathbb N ....... \][/quote]
Only [b] a question[/b] :  suppose that $\{ a_1,b_1,c_1,d_1\}$ are given real numbers such that $\begin{array}{|cc|} a_1 &b_1\\ c_1 &d_1 \end{array}\ne 0 .$  [b] Which is the largest set $D\equiv D(a_1,b_1,c_1,d_1)$ having the property that for any $x_0 \in D$ the numbers  $f_n(x_0)$ exist for all $n$  ?? [/b]


There's probably a typo, otherwise it's obvious ...


[quote="bobbym"]Hi;

For a=sqrt(2) and b = 0 and c = -1 the inequality does not hold. Are there more constraints?[/quote]

if a,b,c>=0?

$ a^2\plus{}b^2\plus{}c^3\equal{}c^3\minus{}c^2\plus{}3$

$ \frac{77}{27} \leq a^2\plus{}b^2\plus{}c^3 \Longleftrightarrow$

$ (3c\plus{}1)(3c\minus{}2)^2\geq0$ which is obviously true. Equality holds $ a^2\plus{}b^2\equal{}\frac{23}{9}$ and $ c\equal{}\frac{2}{3}$

$ a^2\plus{}b^2\plus{}c^3\leq3^{\frac{3}{2}} \Longleftrightarrow$

$ c^2(c\minus{}1)\leq3^{\frac{3}{2}}\minus{}3$

Since $ a,b,c \geq0$ , $ c \leq \sqrt{3}$

$ c<1 \Longrightarrow$

$ c^2(c\minus{}1)<0<3^{\frac{3}{2}}\minus{}3$

$ c \geq 1 \Longrightarrow$

$ c^2(c\minus{}1)\leq3(\sqrt{3}\minus{}1)\equal{}3^{\frac{3}{2}}\minus{}3$. Equality holds $ a\equal{}b\equal{}0$ and $ c\equal{}\sqrt{3}$.

I think this is an easy (maybe very easy) question for olympiad section.  :wink:


geometry
3D geometry
inequalities
pigeonhole principle
factorial
quadratics
summer program

Hi,

I'm currently working through some British maths olympiad problems, and I was wondering if anyone could make any suggestions as to the following, which I'm a bit stuck on..... I hope they're not too easy to be in the advanced section.

1.
The repeat of a positive integer is obtained by writing it twice in a row (e.g. 254254 is the repeat of 254). Is there a positive integer who's repeat is a perfect square?

2.
How large a subset A of {1,2....2002} can we have so that no element of A is a factor of any other element of A?
(I'm pretty sure this is 1001, but I can't come up with a conclusive proof.

3.
On a 9x9 board, either a +1 or a -1 is written on each square. An operation is defined as: associate with each square the product of the numbers in all the squares sharing a common edge with it, not including the square itself. Simultaneously replace the number in each square by the corresponding product. Is it true that whatever the starting position, all the squares will eventually contain +1?

4.
Can 19^19 be written as the sum of a cube (positive or negative) and a fourth power?

5.
Let a,b,c,d,e,f be positive real numbers. Prove that:
(a :^3: +d :^3: )(b :^3: +e :^3: )(d :^3: +f :^3: )  :ge:  (abc+def) :^3:
(This looks like a generalisation of Holder's inequality, but I can't see how to get it? My inequality keeps coming out the wrong way round!)

6.
Find a positive integer m>1 such that the sum of the first m squares is itself a perfect square, and a positive integer n>1 such that the arithmetic mean of the squares of the first n positive integers is itself a perfect square.
(Easy by brute force, but is there an elegant way?)

7.
Let n :ge: 3 be an odd integer, and let a1, a2, a3....an be fixed positive integers. For each of the n! permutations  :pi: =( :pi: 1, :pi: 2, .... :pi: n) of (1,2,....n) define f( :pi: )=a1 :pi: 1 + a2  :pi: 2 .....+an :pi: n
Prove that there exist distinct permutations A, B of (1,2,....n) such that
f(A)-f(B) is divisible by n!
(It's clearly the pigeonhole principle, but I've got one too many pigeonholes!)

8.
Two players play a game on a 10x10 board with all it's squares initially coloured white. The players take it in turn to either
A. Change a white square to a black square or
B. Choose a row or column containing strictly more white squares than black squares, and change the colour of every square in that row or column.
Show that the second player will win the game if she plays correctly
(I tried considering the number of white squares mod 2, which the second player can always force to be zero, but the first player might be able to do so too!)

9. Find all pairs of integers a, b such that a :^2: +4b and b :^2: +4a are both perfect squares.

Thanks in advance for any hints. These problems are starting to really get to me!

Paul

1) Let N be a n-digit number. A repeat will be N(10^n+1). First note that 10^n+1 can never be a perfect square. If it is, it=(10k :pm: 1)^2=
k(100k :pm: 20)+1. So k(100k :pm: 20) must be a perfect power of 10 because the tenth digit is always 2, or 8. So (10^n+1)=N(k^2)
First k can never be 1 since N is a n-digit number. and k can only be odd.
Since N is a n-digit number, k must be 3, and k^2=9. But 10^n+1 can never be divisible by 9. So N(10^n+1) can never be a perfect square.

#7 looks like IMO2001 #4. Why did they use the same problem for BMO?

Beta -- what if (10n + 1) is equal to a perfect square times some smaller number?  Then N would not have to divide (10n + 1).

Quick check -- on number 7, that exclamation point: is it for emphasis or factorial?

A small start on number 9:  [hide]WLOG assume a :ge: b.  Then

a :^2: + 4b :le: a :^2: + 4a < (a + 2) :^2:

Since a :^2: + 4b has the same parity as a, we see that b :le: 0.[/hide]

I assume that in #5, the problem actually reads:

(a :^3: +d :^3: )(b :^3: +e :^3: )(c :^3: +f :^3: ) :ge: (abc+def) :^3:

(As written it's silly false - just let c be big and everthing else small.)

Hint: [hide]Let a :^3:  + d :^3:  = 1, b :^3:  + e :^3:  = 1; and c :^3:  + f :^3:  = 1. This is standard operating procedure for a homogeneous inequality like this one.[/hide]

Further hint: [hide]After making that reduction, try using the AM-GM with three factors on abc.[/hide]

In answer to various questions, yep, it should be a c not a d, the exclamation mark means factorial, and it's from BMO training papers, rather than the actual competition (where I expect they are somewhat more original).

Thanks very much for all the help,

Paul

2) Write the numbers as [tex]x_i = 2^{a_i}\times b_i[/tex] where each b_i is odd. Notice that there are only 1001 possibilities for b_i. If we had 1002 numbers from the set {1,2,...,2002} then we would have b_i = b_j for some i<>j by pigeonhole. If a_i<a_j then x_i divides x_j. If a_i>a_j then x_j divides x_i. Finally, a_i cant equal a_j since then we would have x_i = x_j, which isn't true. This shows that we can have at most 1001 numbers. On the other hand, the set {1001,...,2002} has 1001 numbers and works perfectly fine.

Zach meant to say "the set {1002,...,2002} has 1001 numbers and works perfectly fine."

Problem #4

Hint: [hide] Try it mod 13 [/hide]

Full solution: [hide]

The cubes mod 13 are 0,1,5,8,12 and the fourth powers are 0,1,3,9.

The possible sums are thus 0,1,2,3,4,5,6,8,9,10,11,and 12; everything but 7.

19^19 :equiv: 6^19 :equiv: 6^7 :equiv: 7, so 19^19 cannot be expressed as a sum of a cube and a fourth power.[/hide]

jmerry -- could you give some of the process that got you to those solutions?  I saw or understand the basic approaches that you used for both of them but would not have ever carried my calculations far enough to find the particular values you looked at.  Did you just get lucky, start checking mods working from 4 on up, or does something about 19^19 get you started thinking immediately about mod 13?

Answering on jmerry's behalf:

On problem #1, you had already seen that showing that 10^n + 1 isn't a square isn't enough; you'd also have to show it was square-free. So to disprove it, we're looking for a prime p such that p:^2: divides 10^n + 1. That would mean that 10^n  :equiv: -1 (mod p:^2:) as well as 10^n  :equiv: -1 (mod p). The second condition rules out p = 2, 3, or 5, since 10^n is always 0 or 1 mod those primes. This leaves p = 7 as our first chance. By Euler's Theorem, the order of any element has to divide p(p-1), so 10^(2n) = 1 would force 2n to be a divisor of 7(7-1) = 42. From that, we're led to try p = 7, n = 21.

On problem #4, we're trying to prove that it's impossible, and that suggests modular arithmetic. But we have both a cube and a fourth power, and we want the sum of those to be restricted. In what mod are cubes and fourth powers highly restricted? If they are both restricted, then 12th powers should be restricted. Where are 12th powers restricted? By Fermat's Theorem, all 12th powers are 1 if you're working mod 13. That suggests that mod 13 is worth trying - with no guarantee of success.

[quote="beta"]what's Euler's Theorem?[/quote]

It's a generalization of Fermat's little theorem.

http://mathworld.wolfram.com/EulersTotientTheorem.html

6)Sum of first m squares is[tex] \frac{m(m+1)(2m+1)}{6}.  [/tex]
First, m, m+1, 2m+1 are all relatively prime
Because any perfect squares can never be 2(mod 3)
it's impossible that
m/2, (m+1)/3, and 2m+1 are all  perfect squares
(m+1)/6, m, 2m+1 are all perfect squares
(2m+1)/3, m/2, m+1 are all perfect squares

because difference of 2 positive integer squares can never be 1
m/3, (m+1)/2, (2m+1) are not all perfect squares
and 2 can't divide 2m+1
that leaves only case
m/6, m+1, 2m+1 are all perfect squares.
m=24 we have all perfect squares.[/tex]

For problem 4, another reason why you want to try mod 13 is because if p is 2 mod 3, then cubing mod p is 1 to 1. So, you want to try the primes that are 1 mod 3, because cubing is 3 to 1. In that case, you would try 7, then 13. And it turned out that 13 already works.

For problem 9, consider the four cases:

a, b > 0
a, b < 0
a > 0, b < 0
a = 0.

That should get you started.

and for the second part of 6):
[tex]2n^2+3n+1=6a^2[/tex]
[tex]2n^2+3n+1-6a^2=0[/tex]
by quadratic formula, we want 1+48a^2 to be a perfect square.
And you can cheat with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12182&highlight=pell]Pell's Equation[/url]
to find that n=337 works, is there a way to do this without Pell's Equation?[/url]

Interesting, #1 was on the Mathcamp 2003 Qualifying Quiz. So that's where they get their problems...

Are you sure these are BMO problems? I don't recall seeing some of these in BMO questions, but in other sources..

Any number 7 is identical to the one in IMO. The question that got me a bronze, hehe :

[hide]

Suppose that f(A)-f(B) is not 0 (mod n!) for any permutations. Since we have n! possible permutations, we must have f(permutation) taking the values 0, 1, ..., n!-1 (mod n!), exactly once each. So the sum of f(each permutation) is congruent to (n!-1)n!/2 (mod n!), which is NOT 0 mod n!, since n>2 and n!-1 is odd.

However, when we sum up f(each permutation) we get each number from 1 thru n occuring with a_j (n-1)! times. So the sum of f(each permutation) is (sum a)(n-1)!(n(n+1)/2), which is divisible by n!. Contradiction.

[/hide]

[quote]Are you sure these are BMO problems? I don't recall seeing some of these in BMO questions, but in other sources.. [/quote]

They're taken from the BMO Senior Mentoring scheme, which is a correspondance course for those who did well in the first few rounds of the BMO. Unfortunately, I'm not one of those people - I'm a maths teacher, and I'm trying to get people up to that level, so I thought I should work my way through as many of these problems as I can over the summer, and try to distill what tools contestants need. Nearly finished working through them now, thanks to the very useful hints on this board!

Which does pose an interesting question - how much of this stuff can be taught? I didn't used to think you could teach it, but now I'm not so sure - there are certainly lots of tricks which you really need, which you just don't get taught in school. The BMO certainly advertises itself as not requiring anything more knowledge than is on the school syllabus. Whilst this is true to an extent, people who go into it without the AM-GM, or the power of a point have to reinvent the wheel in the middle of the contest!


the USACO training pages now have an Java grader. It's still in the development stages. The more people try submitting programs in Java the quicker they'll work out any problems. Of course, C/C++ is still fully supported, as is (I think) Pascal.

I use Java.  It's really annoying though cause their compiler sucks.


Find the maximum value of real number $k$ such that \[\frac{(b-c)^{2}(b+c)}{a}+\frac{(c-a)^{2}(c+a)}{b}+\frac{(a-b)^{2}(a+b)}{c}\geq k(a^{2}+b^{2}+c^{2}-ab-bc-ca\; \forall a,b,c\in (0,\infty).\]


is there a number n, such that there is a number kn(k>1), where kn contains digits ONLY found in n.

if so, what is the smallest possible value of n?

Yeah, but I think there's another solution.

If you are asking whether or not this is MATHCOUNTS level, then I think that it might be one of the last 5 of a chapter sprint.... as for the answer, I'm with alanchou

It has more than one solution.
Example, K is two and N is 7368421.
$ 7368421\times 2 \equal{} 14736842$

[quote="abacadaea"]is there a number n, such that there is a number kn(k>1), where kn contains digits ONLY found in n.

if so, what is the [u]smallest possible value[/u] of n?[/quote]

The smallest one is....

10*10=100

Do the solution and the multiple have to have the same number of each digit? :huh:

 :10:

um, possible, not positive :wink:

I still think that this is mathcounts level, assuming $ 0$ is the answer

21978*4=87912, I think that's the smallest non-trivial solution

[quote="ZhangPeijin"]n is 1 and K is 11.
Ha.
Smallest positive.[/quote]

That is correct. +1 for you!

When a question asks something about some smallest positive integer and the quickest method is guess and check, make sure you start at 1 (though most of the time, it's not 1). I'd say this is for Chapter Sprint Round maybe?


geometry
rectangle
search
combinatorics unsolved
combinatorics

A $ 6\times 6$ square is dissected in to 9 rectangles by lines parallel to its sides such that all these rectangles have integer sides. Prove that there are always [b]two[/b] congruent rectangles.

[hide="Sorry, this solution is kind of incomplete. I assumed there were only four lines."]Very nice.

If the sides are cut into $ l_1, l_2, l_3$ and $ w_1, w_2, w_3$, you will form rectangles of $ l_i \times w_j$ for $ i,j$ in $ (1,2,3)$.

Then,
$ l_1 \plus{} l_2 \plus{} l_3 \equal{} 6$
$ w_1 \plus{} w_2 \plus{} w_3 \equal{} 6$

Assume no congruent rectangles exist. We then know that no two of $ l_1, l_2, l_3$ or $ w_1, w_2, w_3$ can be equal. However, this implies $ (l_1,l_2,l_3)$ and $ (w_1,w_2,w_3)$ are permutations of $ (1,2,3)$.

But, you can always pick $ l_i, l_j, w_k, w_l$ such that $ l_i \equal{} w_k$ and $ l_j \equal{} w_l$, giving two congruent rectangles of different orientation. Hence there will always be two congruent rectangles.[/hide]

[quote="makar"]http://www.mathlinks.ro/viewtopic.php?search_id=820928092&t=122541[/quote]I can't believe the solution was that simple.  :blush:

The area of $6\times 6$ squares is 36 square unit. By contrary, let $A_1, A_2, \ldots, A_9$ be 9 distinct rectangles with integers sides. \\Therefore,  Area$(A_1 + \cdots + A_9) = 36.$\\Choosing the minimum possible areas for the rectangles, we have $A_1$ as $1 \times 1$, $A_2$ as $1\times 3,\ldots, A_6$ as $1\times 6$ and $A_7$ as $2\times 2$, $A_8$ as $2\times 3$ and $A_9$ as $2\times 4$  rectangles. \\Thus the minimum area is $1 +2 +3+4 +5+6+2+6+8 = 37$(and the total area is 36) which is a contradiction. \\Thus at least two rectangles are equal.

Consider all possible combinations of non-congruent rectangles...
You will observe that the least possible area=$1$ is a $(1,1)$ square. $Area=2$ is of a rectangle (1,2).proceed this way till $area=8$.
The least possible sum of areas of these incongruent rectangles comes out to be $39>36$.
CONTRADICTION!!!!


ratio
geometry
trigonometry
geometry unsolved

Let $D,E$ be points on side $AB$ of triangle $ABC$ such that $\measuredangle ACD=\measuredangle DCE=\measuredangle ECB=\varphi$. Find ratio ${CD \over CE}$ if sides $AC$, $BC$ and angle $\varphi$ are given.

$\angle BCA=3\phi$

$0<3\phi<180^o$

The area $(ABC)$ of the triangle $ABC$ can be expressed on two different ways:

$(ABC) = (BCE)+(ECA)$
$(ABC) = (BCD)+(DCA)$

$\Rightarrow$

$(BCE)+(ECA) = (BCD)+(DCA) \Rightarrow$

$\frac{1}{2}BC\cdot CE\cdot\sin\phi+\frac{1}{2}AC\cdot CE\cdot\sin 2\phi = \frac{1}{2}BC\cdot CD\cdot\sin 2\phi+\frac{1}{2}AC\cdot CD\cdot\sin\phi \Rightarrow$

$\frac{1}{2}CE(BC\cdot\sin\phi+AC\sin 2\phi) = \frac{1}{2}CD(BC\sin 2\phi+AC\sin\phi) \Rightarrow$

$CE(BC+2AC\cos\phi)\sin\phi = CD(2BC\cos\phi+AC)\sin\phi \Rightarrow$

$CE(BC+2AC\cos\phi) = CD(2BC\cos\phi+AC) \Rightarrow$

$\frac{CD}{CE} = \frac{BC+2AC\cos\phi}{AC+2BC\cos\phi}$

[b]Lemma.[/b] $\triangle ABC\ : \ M\in BC\Longrightarrow \frac{MB}{MC}=\frac bc\cdot \frac{\sin\widehat {MAB}}{\sin \widehat {MAC}}\ .$

[b]The proof of the proposed problem.[/b] I shall apply the above lemma thus :

$1.\blacktriangleright$ For the triangle $ABC$ and the point $D\ : \ \frac{DA}{DB}=\frac ba\cdot \frac{\sin \phi}{\sin 2\phi}\Longrightarrow$
$\frac{DA}{b}=\frac{DB}{2a\cos \phi}=\frac{c}{b+2a\cos\phi }\Longrightarrow \boxed {\ AD=\frac{bc}{b+2a\cos \phi}\ }\ .$

$2.\blacktriangleright$ For the triangle $ABC$ and the point $E\ : \ \frac{EA}{EB}=\frac ba\cdot \frac{\sin 2\phi}{\sin \phi}\Longrightarrow$
$\frac{EA}{2b\cos \phi}=\frac{EB}{a}=\frac{c}{a+2b\cos\phi}\Longrightarrow \boxed {\ AE=\frac{2bc\cos\phi}{a+2b\cos\phi}\ }\ .$

$3.\blacktriangleright$ For the triangle $CDE$ and the point $A\ : \ \frac{AD}{AE}=\frac{CD}{CE}\cdot \frac{\sin \phi}{\sin 2\phi}\Longrightarrow$
$\frac{CD}{CE}=2\cos\phi\cdot\frac{AD}{AE}=2\cos\phi\cdot\frac{bc}{b+2a\cos\phi}\cdot\frac{a+2b\cos\phi}{2bc\cos\phi}\Longrightarrow\boxed {\ \frac{CD}{CE}=\frac{a+2b\cos\phi}{b+2a\cos\phi}\ }\ .$

[b]Remark.[/b] $\frac{DA}{EB}=\frac{\frac{DA}{DE}}{\frac{EB}{ED}}=\frac{\frac{b}{CE}}{\frac{a}{CD}}=\frac ba\cdot \frac{CD}{CE}\Longrightarrow \boxed {\ \frac{DA}{EB}=\frac ba\cdot \frac{a+2b\cos\phi}{b+2a\cos\phi}\ }\ .$

We must know only $\lambda=\frac ab\ .$ In this case, $\boxed {\ \frac{CD}{CE}=\frac{\lambda+2\cos\phi}{1+2\lambda\cos\phi}\ }\ .$


modular arithmetic
induction
number theory
prime numbers

Let $p$ and $q$ be primes with $5 \leq p < q.$ Show that $p^2 \cdot q^2 - (p^2 + q^2) + 1$ is a multiple of $576$.

[quote="orl"]Let $p$ and $q$ be primes with $5 \leq p < q.$ Show that $p^2 \cdot q^2 - (p^2 + q^2)$ is a multiple of $576$.[/quote]
Am I missing something obvious? As  the expression on left an odd number which can not be a multiple of 576.. Now if you added 1 to the left side then, of course, it will be trivial to show...:)

Question corrected. Now you can prove it.  :)

Yes, and you don't even need   $p>q$  .. a weeker condition eg $5 \le p \le q $ would suffice...

I guess you should give a proof and not only reason about how to weaken the conditions that the problem still can be solved.  ;)

$\displaystyle p^2 \cdot q^2 - (p^2 + q^2) + 1=(1-p^2)(1-q^2)$

Consider any prime $p\ge5$:
$p\ge5\rightarrow p\equiv 1\pmod 3$ or $p\equiv 2\pmod 3$
Then $p^2\equiv 1\pmod 3$ and $(1-p^2)\equiv 0\pmod 3$.

$p\ge5\rightarrow p\equiv1\pmod 8$ or $p\equiv3\pmod 8$ or $p\equiv5\pmod 8$ or $p\equiv7\pmod 8$
Then $p^2\equiv 1\pmod 8$ and $(1-p^2)\equiv 0\pmod 8$.

Since $(1-p^2)\equiv 0\pmod 3$ and $(1-p^2)\equiv 0\pmod 8$, then $(1-p^2)\equiv 0\pmod{24}$.

The same applies for $(1-q^2)$ for $q\ge5$ since it is also a prime.
Since $24\mid (1-p^2)$ and $24\mid (1-q^2)$ then $576\mid (1-p^2)(1-q^2)$.

How do you know the $\displaystyle\ p^2$ must be congruent to 1 or 2 mod 3, and all those other nice congruence relationships you used?

Since p >= 5 and is a prime, it can't be divisible by 3. Hence it must be congruent to 1 or 2 (mod 3). Squaring 1 (mod 3) gives 1 (mod 3), and squaring 2 (mod 3) gives 4 (mod 3) which is again equivalent to 1 (mod 3).
Similarly, p can't be divisible by 2, so it must be congruent to 1,3,5 or 7 (mod 8). Squaring these returns 1, 9, 25, 49 (mod 8) respectively, each of which is equivalent to 1 (mod 8).

[quote]
How do you know the \displaystyle\ p^2 must be congruent to 1 or 2 mod 3, and all those other nice congruence relationships you used?[/quote]
[hide]..Also  any prime greater than 3 can be written as $6k \pm 1$  so $(p^2-1)$ is of the form
$36 k^2 \pm 12k =  12k(3k \pm 1) $
now either  $k$ is even or $3k \pm 1$ is even so $k(3k \pm 1)$ is even

Hence $p^2 -1 $ is divisible by $24$ and so is $q^2 -1$ and the expression is divisble by $24^2 = 576$

Another way to look at it:
$(p^2-1)=(p-1)(p+1)$ since (p-1) and (p+1) are two consecutive even integers, one of them is divisible by 4, and hence the product is divisible by  $8$
Also in three consecutive integers $(p-1), p, (p+1)$ one of them is divisible by 3, and it isn't $p$ (because p is prime)
... So ...$(p-1)(p+1)$ has to be divisible by $3$.

So $(p-1)(p+1)$ is divisible by $8 \cdot 3 = 24 $

[hide="Also "](In other words  for  any odd number, $p^2 \equiv 1 \pmod 8 $  and
 for any  number not divisible by 3 $p^2 \equiv 1 \pmod 3$

(Or any  odd number not divisible by 3  - that includes all prime numbers >3 too :) $p^2 \equiv 1 \pmod {24}$[/hide][/hide]


geometry
geometric transformation
homothety
ratio
geometry proposed

The points $ A,B,C,D,E,F$ lie on a circle and the lines $ AD,BE,CF$ are concurrent. Let $ P,Q,R$ denote the midpoints of $ AD,BE,CF$ respectively. If chords $ AG,AH$ are drawn  parallel to $ BE,CF$  respectively, prove that the triangles $ PQR$ and $ DGH$ are similar.

$ P,Q,R$ lie on a circle with diameter $ OM (M$ is the intersection of $ AD,BE,CF$) then $ \angle QPR\equal{}\angle FME\equal{}180^o\minus{}\angle GAH\equal{}\angle GDR$.
But $ \angle RQP\equal{}\angle RMP\equal{}\angle HAD\equal{}\angle HGD \Rightarrow QED$

Well, really nice solution. I have done it using some homothetic mappings. Please tell if it is correct.
Consider a homothety mapping D to P about A ratio 1/2. D maps to P. M maps to A. Be maps to AG, CF maps to AH. The concurrency property is also verified. And G maps to X where X is midpoint of AG. H maps to Y where Y is midpoint of AH. Also, R maps to Y and Q maps to X as AG , Be and AH, CF are mapped. So we get that both triangles map to PXY. All mappings are homothetic. (ie not that DGH and PQR are homothetic, but they are mapped by interlinked homotheties, retaining angles-isometries. ). SO they are similar.


Alcumus probability MATHCOUNTS videos Support

Get the book.  I feel that combining Alcumus with the book is better than either alone, as Alcumus gives you problems that you should be able to solve, and the book goes in-depth with examples, and lots and lots of problems.  On Alcumus, you should definitely have pencil and paper, and if you come across some giant calculation, I think it's ok to use a calculator.  :)

Make sure you buy the solutions manual along with the book; it's like a whole 'nother textbook.

I hear the class is also pretty good :) even though I've never taken it.

i don't have introduction to counting and probability, so what i say might be a little biased...

anyway, coming from a person who doesn't have the book, i think that alcumus is a great way of teaching the material
if you come across problems you don't get, you can always watch the movies
buying the book will probably help you learn the material in a little more detail, but i would do some alcumus problems first while you're waiting for the book to arrive, and while you're reading the book too, because it will give you more practice with the information
i totally recommend buying the book though, as i have the art of problem solving volume 1 and 2, and they are wonderful!

as for solving the problems, i agree that paper and pencil is best, but if you're practicing for speed (say for mathcounts sprint), then try to do it mentally
calculators i don't suggest, but if it's a computation that involves big numbers, sure, go ahead

I personally think Alcumus is a better reviewing tool than a learning tool, but that's probably because I only used it as a reviewing tool... :P Of course, there are a lot of challenges for anyone. Plus, as AIME15 said, the book goes into more depth that any of the videos will.

[quote="MrMagician"]I was wondering what the rules for doing the problems in Alcumus were. How are we supposed to solve the problems? Mentally? With a Calculator? Pencil and Paper? Sorry I am new to the site. Should I even start Alcumus? Or wait until I purchase a book? I was thinking of getting the the Introductory Probability and Counting book to start off. Should I finish that book first and then try Alcumus? Any other advice is greatly appreciated :)[/quote]

I'll echo the others in recommending that you get the book and work with Alcumus alongside working through the book.  (And feel free to start with Alcumus while waiting for the book to arrive.)

I also recommend trying to solve every problem with pencil and paper.  On rare occasion, you'll hit one that requires some larger multiplications, and it's OK to use a calculator for those, but for the most part, you should be able to do these with pencil and paper alone.

Alcumus is designed to supplement both the class and the book -- students in the Spring class will have an Alcumus schedule to complete, and Alcumus suggests specific sections and chapters to go with particular problems.

Yeah, I use Alcumus in addition to other math books, including but not limited to Aops. I use pencil and paper but I've seen a couple where you're expected to do calculations that I would use a calculator for (and I am known for not using calculators. Ever.)


要开学了，联赛也要来了，但我总觉得我平时上学的时候时间不够用！

高一高二还是高三啊？ :maybe:
如果没有精力去搞数学竞赛，还是不要太花时间去做数学竞赛题了。
毕竟平时的成绩还是最重要的。
无聊的时候，做数学题消遣一下了～～

[quote="lonelyhunter"]要开学了，联赛也要来了，但我总觉得我平时上学的时候时间不够用！[/quote]
same here


inequalities
function
vector
calculus
derivative
inequalities unsolved

hi
im studying the topic of majorization at the moment and it raised a couple questions but i only want to post one here:
i thought i understood the concept of schur-convex functions a little bit but when i tried to proof the GM-AM inequality i came to a shocking result, i wanna share with u. please tell me where their is the error in my logic.

let $\phi(x)=\prod_i x_i^2$ for all vectors $x=(x_1;\dots;x_n)$
$\phi(x)$ is schur-convex, since its symetrical, and every $f(x_i)=\phi(x)$ for all $x_j$ with $j\ne i$ fixed is convex.
[i]Proof:[/i]
$a=\prod_{j\ne i} x_j\\
f(x_i)=a^2x_i^2\\
f'(x_i)=2a^2x_i\\
f''(x_i)=2a^2\ge 0$

since then $(a;b)\succ (\frac{a+b}2;\frac{a+b}2)$ for all $a,b\in \mathbb{R}^+$ and $\phi(x) $ is schur-convex,
$\phi(a;b)\ge \phi(\frac{a+b}2;\frac{a+b}2)\\
%Error. "Rightleftarrow" is a bad command.
a^2b^2\ge \left(\frac{a+b}2\right)^2\left(\frac{a+b}2\right)^2\\
%Error. "Rightleftarrow" is a bad command.
\sqrt{ab}\ge \frac{a+b}2$

and this is obviously almost the direct opposite of the GM-AM inequality.
great THX to the one who can show me my error and explain how it is done correctly.
peeta

[quote="darij grinberg"]Hmm, I don't know exactly what a Schur-convex function is, but convexity in every variable is [b]not[/b] enough to claim that, if $\left(x_1;\;x_2;\;...;\;x_n\right)\succ\left(y_1;\;y_2;\;...;\;y_n\right)$, then $f\left(x_1;\;x_2;\;...;\;x_n\right)\geq f\left(y_1;\;y_2;\;...;\;y_n\right)$. You can claim this if the function f is really convex on $\mathbb{R}^n$ (not just convex in every variable), or if some strange condition on partial derivatives is fulfilled (I don't remember that condition, sorry).

  Darij[/quote]
The conditions incldue sylvester's criterion and hessian test.
Anyway, what's schur-convex??

first thanks to you

but now there is rasied a new question: how is convexity of a function on $\mathbb R^n$ defined and how can one proof its convexity?

peeta

ps:
i have a book from the library of the university here. its called "Inequalities: Theory of Majorization and its Applications" by Marshall and Olkin. in it schur-convexity of a function is defined as the following.

A real-valued function $\phi$ defined on a set $\mathbb A \subset \mathbb R^n$ is said to be  [i]Schur-convex[/i] on $\mathbb A$ if
\[x\succ y \textnormal{  on  }\mathbb A \Rightarrow \phi(x)\ge \phi(y)\]

oh, we simply call it convex here.
Anyway, you can look up the topics in Inequalities theorems and formulaes.


[b]Prove that $ x^4\minus{}18x^2\plus{}4dx\plus{}9 \equal{} 0$ has 4 real roots , if $ d^4\le 1728.$[/b]

[quote="iammathlover"]is this high school basic? I don't think so.[/quote]

Yes. But i think a difficult one!!!


linear algebra
matrix
induction
linear algebra unsolved

$ A,B,C\in M_{n}(k)$ s.t $ A^2C^2\equal{}B^2C^2\equal{}B^2A^2\equal{}O_{k}$.Prove that
$ det(I_{k}\plus{}A^{2m}\plus{}B^{2n}\plus{}C^{2p} )\geq0$,for any natural m,n,p.

Here we are in $ \mathcal{M}_k(\mathbb{R})$.
It's wellknown that if $ UW\equal{}0$ then $ U,W$ are simultaneously triangularizable; but if $ UW\equal{}VW\equal{}VU\equal{}0$, then I think that $ U,V,W$ are simultaneously triangularizable but I haven't any proof. We assume that this result is true; then $ A^2,B^2,C^2$ are simult.triang..Let $ (\lambda_i),(\mu_i),(\nu_i)$ be the eigenvalues of $ A,B,C$; the following condition is satisfied: $ \forall{i}$  the set $ \{\lambda_i,\mu_i,\nu_i\}$ admits at most only one non zero element. $ A^{2m}\plus{}B^{2n}\plus{}C^{2p}$ is a real matrix which has the following spectrum: $ (\lambda_i^{2m}\plus{}\mu_i^{2n}\plus{}\nu_i^{2p})$. Then these eigenvalues are real or conjugate by pair.
$ det(I_k\plus{}A^{2m}\plus{}B^{2n}\plus{}C^{2p})\equal{}(1\plus{}\lambda_1^{2m}\plus{}\mu_1^{2n}\plus{}\nu_1^{2p})\cdots$.
It remains to prove that the negative eigenvalues $ \lambda_i^{2m}\plus{}\mu_i^{2n}\plus{}\nu_i^{2p}$ are of even order. We assume for example that $ \lambda_i^{2m}<0,\mu_i^{2n}\equal{}\nu_i^{2p}\equal{}0$. Then $ \lambda_i$ isn't a real number. If $ \lambda_i$ is of order $ r$ then its conjugate is non zero and also of order $ r$. Therefore $ u\equal{}\lambda_i^{2m}\plus{}\mu_i^{2n}\plus{}\nu_i^{2p}$ is at least of order $ 2r$; it isn't finished! Now we have a look on the $ j$ s.t. $ u\equal{} \mu_j^{2n}$ and on the $ k$ s.t. $ u\equal{}\nu_k^{2p}$. The conclusion is similar. We are done .  QED

loup blanc,
Since you consider matrices in $ \mathcal{M}_k(\mathbb{R})$, when you wrote
[quote="loup blanc"]if $ UW \equal{} VW \equal{} VU \equal{} 0$, then I think that $ U,V,W$ are simultaneously triangularizable[/quote]
did you mean "simultaneously triangularizable as matrices in $ \mathcal{M}_k(\mathbb{C})$", and not necessarily in $ \mathcal{M}_k(\mathbb{R})$?

Let's say that $ m,n,p > 0$. Then, $ \left( I \plus{} B^{2n} \right) \left( I \plus{} A^{2m} \right) \left( I \plus{} C^{2p} \right) \equal{} I \plus{} A^{2m} \plus{} B^{2n} \plus{} C^{2p}$. Taking the det of this, we get the conclusion. If $ mnp \equal{} 0$, then a similar argument works.

However, it may be that posabodgan didn't mean [quote]$ A^2C ^2 \equal{} B^2 C^2 \equal{} B^2 A^2 \equal{} 0,$[/quote] but something cyclic instead.

1) To JC_math; yes, $ U,V,W$ would be simultaneously triangularizable over $ \mathbb{C}$. The eigenvalues are complex numbers.
2) To perfect_radio; how do you prove that: $ B^{2n}A^{2m}\equal{}B^{2n}C^{2p}\equal{}A^{2m}C^{2p}\equal{}0$ ?
However your result is correct if $ A^2,B^2,C^2$ are simultaneously triangularizable and then your proof is much more simple than mine.

$ B^{2n} A^{2m} \equal{} B^{2n \minus{} 2} \left( B^2 A^2 \right) A^{2m \minus{} 2} \equal{} 0$ (now it's apparent why I considered two cases).

However, I'm still wondering whether the original problem meant $ A^2 C^2 \equal{} B^2 C^2 \equal{} B^2 A^2 \equal{} 0$ or $ A^2 C^2 \equal{} C^2 B^2 \equal{} B^2 A^2 \equal{} 0$...

Well played perfect_radio; in my mind $ A^2B^2\equal{}...\equal{}0$ or $ B^2A^2\equal{}...\equal{}0$ hasn't any importance but in fact this choice is very important. Your proof works iff, in the hypothesis, one of $ A^2,B^2,C^2$ is 2 times in first position. Here it's $ B^2$.
Have you a proof if the hypothesis is, as you write: $ A^2C^2\equal{}B^2A^2\equal{}C^2B^2\equal{}0$ ?

The problem is corectly.I have found it in Gazeta Matematica 12/1999

[quote="loup blanc"]Here we are in $ \mathcal{M}_k(\mathbb{R})$.
It's wellknown that if $ UW \equal{} 0$ then $ U,W$ are simultaneously triangularizable[/quote]

Proof: by induction on $ k$


let x, y, z positive numbers
prove that
$ \sqrt{\frac{2x}{x\plus{}y}}\plus{}\sqrt{\frac{2y}{y\plus{}z}}\plus{}\sqrt{\frac{2z}{z\plus{}x}}\le 3$


a,b,c are natural numbers

ab divides c(c^2-c+1)

c^2+1 divides a+b

=>{a,b}={c^2-c+1,c}[/hide][/url]

obviously for $ c = 1,2$ the problem is true therefore  suppose that $ c > 2$ we have:
$ ab|c(c^2 - c + 1)\Rightarrow{ab\leq{c(c^2 - c + 1)}}\Rightarrow(a + b)^2 - (a - b)^2\leq{4(c(c^2 - c + 1))} (I)$
and
$ c^2 + 1|a + b\Rightarrow{(c^2 + 1)^2\leq(a + b)^2 (II)}$
let $ t = |a - b|$ from $ (I)$ and $ (II)$ we have :
$ (c^2 + 1)^2 - 4(c(c^2 - c + 1))\leq{t^2}\Rightarrow(c - 1)^2\leq{t}$
WLOG $ a > b$ and suppose that $ c(c^2 - c + 1) = yab$ and $ a + b = x(c^2 + 1)$
$ c(c^2 - c + 1) = yab = yb(x(c^2 + 1) - b)\Rightarrow {yb^2 + 1 = (c^2 + 1)(ybx - c + 1)}$ therefore $ yb^2 + 1\geq{c^2 + 1}\Rightarrow\sqrt {y}b\geq{c}$
also   $ c(c^2 - c + 1) = yb(b + t)\geq\sqrt {y}c(c - 1)^2\Rightarrow\sqrt {y}\leq1 + \frac {c}{(c - 1)^2}$
and $ c > 2$ therefore $ \sqrt {y} < 2\Rightarrow{y = 1}$
also we have $ c(c^2 - c + 1) = yb(b + t)\geq{b(b + (c - 1)^2)}$ therefore $ b\leq{c}$
and also $ b^2 + 1 = yb^2 + 1\geq{c^2 + 1}\Rightarrow{b\geq{c}}$ therefore $ b = c$ and $ a = c^2 - c + 1$


Can someone show me this please?

Given compass points that are the vertices of an equilateral triangle, give
a [b]compass only[/b] construction for the centre of the triangle.

Let $ A$, $ B$, and $ C$ be the vertices, and $ \ell$ the sidelength of the equilateral triangle.  At $ A$, draw a circle of radius $ \ell$.  At $ B$, draw a circle of radius $ \ell$.  These two circles meet at $ C$ and another point $ C'$.  The points $ A'$ and $ B'$ are constructed similarly.  Now, the circles at $ A'$, $ B'$, and $ C'$, all with radius $ \ell$, are concurrent at the center of the triangle.

Do you mind just telling me quickly why this works?

Thanks for the response.


Can any member explain just what is exactly a"[b] Triangulation[/b] "[b] in communication technology? How does it apply to any given area?

                                         Respectfully,
                                          A.Eddington[/b]

well triangulation in general is just figuring out the distance between you and an point and then bashing out the distance.

so in communications this may refer to for example, how far a radio tower would work if its a given height


SO, when's everyone taking it?
Preliminary round is set for Feb 6th at my school

my round is february 9th. Did you know they changed the number of multiple choice from 30 to 20. I have a bad feeling this will mean that the multiple choice will be much harder.

oh. I've been practicing with 2004 and before, which all had 30 questions. Thats why assumed the change went into effect this year. Can you tell me if you noticed any shift in difficulty in last years MC compared to previous years?

AAAAAHHHH physics olympiad on thursday. I'm freaking out!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Wow, i get really bored on days when I have to stay home from school. I'm reduced to upping my post count with weird random posts that contribute nothing to this forum.

Hey,

I do believe the shift from 30 to 20 problems does make things a tad harder...

Anyway, I took the MC today.  If anyone's reading this, a small reminder that we shouldn't discuss the problems here, until everything's been sent in yet.

G'luck, =)

Is the deadline a postmark deadline or an arrival deadline?

I'm asking because we're doing the physics olympiad for the first time in a very long time at my school, and we don't know any of the rules...it would stink to be disqualified by a technicality.

i'm not sure... instructions would probably be included with the exam packets, though.  Ask your teacher to read it over?

My teacher said the postmark deadline is Monday... that's when I'm taking it.  :P

Good luck, everybody!

can we discuss the scores we got online, cuz im curious as to how everyone did...

I myself did horrible and I'm saying no more than that until I'm sure that I can divulge my rather low score over the internet. I don't even know if im gonna make semis. Argh....

i did bad too!
but looking at the problems, i think everyone did horribly! ;)

I did even worse than last year ... and I knew less than one semester of first-year physics last year ...

i really hope they lower the cutoff this year... :(

wow, i cant even explain how relieved I am that I wasnt the only one who found that test hard...

ok, im posting my score: 95/140  :blush:

and this is one of those rare cases when I use smilies, since that is the only way to actually convey how i feel.

I dont know my score yet, but I know I def. misread the first FR, and didnt finish the last one.  I guess we can't discuss the solutions just yet (tuesday? because the postmark dead. is monday), but I am very confident that I did worse than last year.
It's okay though, the semi cutoffs are relatively low(and, judging from what people thought this year, the cutoffs might be lower this year).

Psychicalchemist, 95 isn't anything to be embarassed about.  These tests dont determine how smart you are,  :) Physics is "phun" ! woo.

plus, a 95 is pretty darn good too.

Crap, that's no good... I think I got around 130, but I'm a senior.

(Lost close to all 50 on the last question thanks to an impossibly stupid error on the very first part.)

If you got 130, then I'm sure you made it, since last year Chor got slightly more than half and made the team.  Not only that, but this year's was a lot harder.

I probably missed them... but, guys, where can I find the questions from Usapho that you are talking about?

thanks in advance.

How many can you miss on the first test and still qualify as semifinalist?

Letters were mailed yesterday via FedEx priority mail.  I made it, and so did 2 others from my school.

Do you only get a letter if you made it?

[quote="blendecho"]Yes.

And I made it. Totally shocked, but eh, I'm not complaining.[/quote]

Lol, guess that means I didn't make it. Nor did 2 people who I was hoping would make it. Eh, oh well, still 2 more years...

So in light of this, time to study for USAMO!  ;)

Anyone have advice on how to prep for the team? I really, really want to make travelling team but I don't think I stand a chance against the people who were on the team/travelling team last year...

probably, more problems?  They'll send us a big fat syllabus in a few weeks, just be comfortable with those things, I guess.

So basically lots of IPhO problems. Alright. Those are fun anyways.

There's gonna be a pretty good list available in about 4 days on the website:-p


I wonder if this is true without any conditions?

$C' = A_0 B_0 \cap A_1 B_1$, $C = A_0 A_1 \cap B_0 B_1$, and $X = A_0 B_1 \cap A_1 B_0$.

 :arrow: $NX \perp CC'$.

[quote="M4RI0"]I wonder if this is true without any conditions?

$C' = A_{0}B_{0}\cap A_{1}B_{1}$, $C = A_{0}A_{1}\cap B_{0}B_{1}$, and $X = A_{0}B_{1}\cap A_{1}B_{0}$.

 :arrow: $NX \perp CC'$.[/quote]

Without any conditions, this cannot be true, because the conditions you gave are invariant under projective transformations while the assertion is not.

But what was your original problem?

  darij


Determine all solutions (x,y,z) of positive integers such that (x+1) y+1 + 1 = (x+2) z+1 .

Ok, let's end the problem so as to move it into the solved problem section (the most difficult part has been done...) :

First, since I do not understand what w means in the above solution, I rewrite the last sentence :
We have
2 n  \geq 2 d  = (2b+1) k  + 1 > (2b+1) k  - 1 = 2 n-d b n  \geq  b n , thus b = 1, and the equation becomes
3 m  - 2 n  = 1.  (*)

Recall that m,n > 1.

Since n  \geq  2. Then 3 m = 1 mod[4] which forces m to be even, say m = 2m'.
Then (3 m' +1)(3 m' -1) = 2 n , which leads to :
3 m' +1 = 2 u and 3 m' -1 = 2 v , for some positive integers u,v such that u+v = n.
Thus 2 u = 2 v  + 2.
But, the only power of 2 which differ by 2 are 2 and 4, and then u = 2, v= 1 and n = 3. That gives m = 2.
And then we get x = 1, y = 2 and z = 3.

Pierre.

[quote="StRyKeR"]hehe we COULD just cite Catalan's conjecture, but that would not suffice as a proof  :D http://mathworld.wolfram.com/CatalansConjecture.html[/quote]

should we call it catalan's theorem now? :D


integration
logarithms
calculus
calculus computations

solve the equation:

$ \frac{dy}{dx}\equal{}\minus{}\frac{e^{\minus{}y}\plus{}1}{x}$

\[ \begin{gathered}
   \Leftrightarrow \int {\frac{{dy}}
{{1 \plus{} e^{ \minus{} y} }} \equal{}  \minus{} \int {\frac{{dx}}
{x}} }  \hfill \\
   \Leftrightarrow \ln \left( {e^y  \plus{} 1} \right) \equal{}  \minus{} \ln \left| x \right| \plus{} \ln C \hfill \\
   \Leftrightarrow e^y  \plus{} 1 \equal{} \frac{C}
{{\left| x \right|}} \hfill \\
   \Leftrightarrow y \equal{} \ln \left( {\frac{C}
{{\left| x \right|}} \minus{} 1} \right) .\hfill \\
\end{gathered}
\]

why did the negative $ y$ exponent become positive?


How many ordered pairs (x,y) satisfy BOTH conditions below?
Condition 1: x=1 or y=2
Condition 2: x=0 or y=1

Seperate the conditions first.
1a. x=1
1b. y=2
2a. x=0
2b. y=1

At least one of the following must be met:
1a+2a
1a+2b
1b+2a
1b+2b

The first and fourth cannot happen, so only 1a+2b and 1b+2a are possibilities. These both give exactly one ordered pair each as a solution, so the answer is 2.

In case you wanted to know, the ordered pairs are (1,1) and (0,2).


A jar contains 10 red, 7 blue, and 5 yellow marbles. Blue marbles are then added in order to change tho probability of randomly selecting a blue marble from the jar to "greater that 1/2." What is the least number of blue marbles that must be added?

The total number of non-blue marbles is $ 15$. This must be half the total number of marbles after the blue marbles have been added. That means there must be a total of $ 30$ marbles, and since there are already $ 22$ marbles, we have to add $ 8$ blue marbles to make the probability $ \frac{1}{2}$. Thus, we must add $ \boxed{9}$ to make it greater than $ \frac{1}{2}$


pigeonhole principle
geometry
congruent triangles
combinatorics proposed
combinatorics

Each of the vertics of a regular nonagon has been coloured either red or blue. Prove that there exist $ 2$ congruent monochromatic triangles; that is triangles whose vertices are all the same colour. :D

[hide="is it really that easy or am i missing something?"]There are 8 patterns of colours: RRR,RRB,RBR,BRR,RBB,BRB,BBR,BBB. Consider nine triangles: $ A_1A_2A_3,A_2A_3A_4,A_3A_4A_5,A_4A_5A_6,A_5A_6A_7,A_6A_7A_8,A_7A_8A_9,A_8A_9A_1,A_9A_1A_2$. They are all congruent, and by pigeonhole principle two of them are monochromatic. So we are done.[/hide]

follow-up question: does the result still hold if there is another colour, say green?

You are missing something.
[quote="Johan Gunardi"]by pigeonhole principle two of them are monochromatic.[/quote]  No: by pigeonhole principle two of them have the same coloring, but none of them need be monochromatic.  (Just consider the coloring RBRBRBRBR.)

@mathVNpro: please give your posts more informative titles.  For example, this one could have been called, "Nonagon in two colors."

by monochromatic, do you mean that ALL VERTICES in a triangle are of the same color, or do you mean that two corresponding triangles have the same configuration? moreover, if it were the former, do the monochromatic triangles need to be of the same color?

The question is quite explicit that they should be monochromatic, and it says nothing about whether they should be the same color.  Therefore, it's probably safe to assume that you should find two congruent triangles, both of which are monochromatic (but not necessarily the same color).  Although I am having trouble finding an example with 5 vertices of one color that doesn't give two congruent triangles, so the stronger statement might be true as well.

JBL is right---the stronger statement is true.

Five vertices determine $ {5\choose 3} \equal{} 10$ triangles, but there are only 7 non-congurent triangles that can be formed from three vertices of a regular nonagon.

Sorry, I completely misunderstood the problem. I have solved this, but my solution has a massive number of caseworks. :blush:

edit: very nice, otsu!

This is a problem from a combinatorics of [b]Titu Andrescu[/b], so I think that there should be no wrong in the problem statement. And the solution of this problem is very simple, but not very simple like the solution which is provided by [b]otsu[/b]. Can you please solve it in detail? :D .
Thank you very much.

Hint for the general problem given by JBL:

Since the number of distinct triangles formed by vertices of the nonagon is $ 7$ (we can count it directly), it is enough to prove that in every way of coloring, we always obtain the number of monochromatic triangles are greater then $ 7$.

Let $ r,b,g$ respectively be the number of triangles colored Red, Blue and Green. We have $ r\plus{}b\plus{}g\equal{}9$. Therefore, it is enough to prove that:

$ min_{r\plus{}b\plus{}g\equal{}9} \left( {r \choose 3} \plus{} {b \choose 3} \plus{} {g \choose 3} \right) \ge 7 (1)$.

In order not to consider too many cases, we will assume that $ r \le b \le g$. Now we will just need to consider $ (r,b) \in \{(1;1),(1;2),(1;3),(1;4),(2;2),(2;3),(3;3)\}$. It easy to test that $ (r,b) \in \{(1;1),(1;2),(1;3),(1;4),(2;2)\}$ makes $ (1)$ right. In case $ (r,b)\equal{}(2,3)$ or $ (r,b)\equal{}(3,3)$, we will perform a great cases considering to see that there are triangles satisfied what we wanted.
(It is lengthy but not so much cases to consider :lol: )


Let $ ABC$ be triangle and denote by $ (I)$ its incircle and $ M$ the midpoint of $ BC$. Let $ \{N,P\} \equal{} AM \cap (I)$. Draw the parallel lines with $ BC$ through $ N$ and $ P$ and let them intersect $ (I)$ at $ S,T$ respectively. Let $ AS \cap BC \equal{} X$ and $ AT \cap BC \equal{} Y$. Prove that $ MX \equal{} MY$.

Draw the parallel line $ l$ from the vertex $ A$ to $ BC$. The quadrilateral $ F,N,F',P$ is harmonic. The tangent at $ N,P$ are intersecting on the line $ l$. From Pascals theorem we can prove that the intersection points of the lines $ PF',FN$ and $ F'N,PF$ are on the $ l$. We consider the complete quadrilateral $ NF'PFJV$ where $ J,V$ are the he intersection points of the lines $ PF',FN$ and $ F'N,PF$. So from brockard's theorem we know that the $ IS'$ is perpendicular at $ l$ and of course on the parallel line from $ P$. So now if we consider the isosceles trapezium $ NSPT$ the $ S'$ will be the intersection point of its diagonals $ TS$ and $ PN$, because $ S'$ is on the midlebisector of $ TP$ (and of course $ NS$).
So we prove that $ S,S',T$ are collinear.

Let $ D,E,F$ be the tangency points of $ (I)$ with $ BC,CA,AB.$ It is well-known that $ AM,DI$ and $ EF$ concur, i.e. $ ST,AM$ and $ EF$ concur at a point $ K.$

$ \frac {MX}{NS} \equal{} \frac {AM}{AN} \ , \ \frac {MY}{PT} \equal{} \frac {AM}{AP} \Longrightarrow \frac {MX}{MY} \equal{} \frac {NS}{PT} \cdot \frac {AP}{AN} \equal{} \frac {KN}{KP} \cdot \frac {AP}{AN}$

Since $ EF$ is polar of $ A$ WRT $ (I),$ it follows that $ (P,N,K,A) \equal{} \minus{} 1 \Longrightarrow$

$ \frac {MX}{MY} \equal{} \frac {KN}{KP} \cdot \frac {AP}{AN} \equal{} 1 \Longrightarrow MX \equal{} MY.$

notice the fact that AKTL are harmonious division we can conclude that X,T,Y are collinear
then the rest part are trivial

it is one of the geometry exercises given to the students participating in iran3rd round 2009.i will try to post most of them.their topics will be named like this topic(geometry(2),etc).

The proposed problem can be generalized:

Let $ ABC$ be a triangle, $ \Gamma\left(J\right)$ is a circle that tangent to the lines $ AB,AC$ (the point $ J$ lies on the bisector of the angle $ \angle BAC$),
$ M$ is the midpoint of $ BC$ and $ \left\{N,P\right\} \equal{} AM\cap\Gamma$. The parallel lines with $ BC$ through the points $ N,P$ intersect $ \Gamma$ at the points
$ S,T$, respectively. If $ \left\{X\right\} \equal{} AS\cap BC$ and $ \left\{Y\right\} \equal{} AT\cap BC$ then prove that $ MX \equal{} MY$.

The proof is similarly to the Luis' proof.

Best regards, Petrisor Neagoe


91?  Im not sure..anyway ill try and explain...

[hide]3m^2n^2 +m^2 = 6m^2n^2.  6m^2n^2 = 30n^2 +517.

divide by 6 so m^2n^2 = 5n^2+ 86 1/6.

So m^2 = 5+ (86 1/6) / n^2.

So then to find 3m^2 that equaled 273.5/3n^2.  Then when u brought n^2 in the numerator and denominator canceled each other out.  So you got 546/6 which reduced to 91.[/hide]

Im not sure if i solved correctly for n^2.  Someone plz enlighten me.[/hide]

here's how i did it, a little differently.

[hide]

move everything to one side, so m^2+3n^2m^2-30n^2=517.  Factor the left side to (m^2-10)(3n^2+1) +10 = 517.  So

(m^2-10)(3n^2+1) = 507 = 3*13^2.  Play around with the equations a little, and find that  m^2-10=39 and 3n^2+1 = 13.  So n = 2, m^2 = 49.  we want 3*49*4, which is 588.[/hide]

Im not exactly sure what you did, but i'll try to be of some more help tomorrow.  i have to sleep now.

it's really only a minor point, just a simplification: [hide]when we have (m^2-10)(3n^2+1)=3*13^2, we know the second factor on the LHS cannot be a multiple of 3, so it must be 1, 13, or 169. Just a tad bit less casework, methinks.[/hide]

Astral, where did you drag 6m:^2:n:^2: from?

jeffrey wrote:m,n are integers such that m^2+(3m^2)(n^2)=30n^2+517.
Find (3m^2)(n^2)

it took me some playing around, but here goes:

[hide]so expanding the equation thing out one gets:
m^2+3(mn)^2=30n^2+517. so then we look to factor, this was the part that took me the longest, heh. so then subtract 30n^2 from both sides to get:

m^2+3(mn)^2-30n^2=517

then that factors to (3n^2+1)(m^2-10)+10=517

subtraction here and there:

(3n^2+1)(m^2-10)=507

then you want the two factors on the lhs etc so then i prime factorized the rhs to 3x13^2. if you look at the (3n^2+1) part, you know its not a factor of 3, so then it could be 1,13, or 169 according to the prime factorization. so that does limit my search for the integers. righteous. okay, so setting up the equations i ended up with 3n^2+1=13

so weve taken 1 factor of 13 from 3 x 13^2 so then we know that m^2-10=13x3=39

of course, i had to play around a bit, but you dont want to see my failed attempts now do you

so then m^2-10=39, m^2=39, m=7

3n^2+1=13, 3n^2=12, n^2=4, n=2

looking back at what the question asks for:
find (3m^2)(n^2)

we take then (3x7^2)(2^2)=147x4=588 i believe

well im pretty sure my steps were right, but factoring i prbly messed up, but hrm. just hrm

[/hide]

All good, except that m and n needn't be 2 and 7 -- they could also be negative values.

[color=cyan]yeah, thankfully that didnt play a role in a different answer, so i am saved[/color]


modular arithmetic
number theory proposed
number theory

You only need to prove $ x^9\equiv x^5 \pmod{240}$
this follows from $ x^9\equiv x^5 \pmod{3}$
$ x^9\equiv x^5 \pmod{5}$ and
$ x^9\equiv x^5 \pmod{16}$


In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum, $N$. If told the value of $N$, the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if $N= 3194$.

[hide="Answer"]The sum of the numbers is $222(a+b+c)=3194+\text{abc}$. $a+b+c\ge 15$ (since $222(14)=3108$, giving a negative value for $\text{abc}$), so testing possible values, we see that the answer must be $\text{abc}=\boxed{358}$, as this is the answer for $a+b+c=16$, and $3+5+8=16$.[/hide]

[hide="Solution"]Let the chosen number be $ m\equal{}100a \plus{} 10b \plus{} c$ and $ d(m)$ be its digital sum, so
\[ 222d(m) \equal{} 3194 \plus{} m
\]The RHS is a multiple of 222 if and only if $ m\equiv136\pmod222$, so $ m$ is $ 136$, $ 358$, $ 580$, or $ 802$. These leaves $ 15$, $ 16$, $ 17$, and $ 18$ as respective values for $ d(m)$. However, if we add the digits of each possible $ m$, we get 10, 16, 13, and 10, respectively. Only the second choice matches, so $ m\equal{}\boxed{358}$.[/hide]

Well, the solutions above are pefect. However, I am wondering how could the magician determine the number for any given N, is that possible?

In the same manner as chesspro did, replacing $136$ with $n \mod 222$

Can you elaborate on the casework that you did? I'm pretty sure that most people can add all of the numbers, but finding $a, b,$ and $c$ is the difficult part.


I'm sure most of you are acquainted with metal puzzles, also known as metal disentanglement puzzle, tavern puzzle, metal wire puzzle, and so on (see attached image).

Do you guys have any ideas where I could find some cheap ones? (15 dollars + shipping doesn't seem a good deal at the moment.)

Examples: http://www.seriouspuzzles.com/uncpuzall.html

I suppose you could always buy one off of ebay...

You could post on Craigslist... I'm sure there are people out there with disentanglement puzzles that are collecting dust.


geometry
circumcircle
incenter
angle bisector
geometry proposed

Be $ABCD$ a cyclic quadrilateral.If $AC$ cut $BD$ at $P$,
$BC$ cut $AD$ at $E$, $CD$ cut $AB$ at $F$ and from $P$ draw $PM$
orthogonal to $EF$ ($M$ on $EF$). Prove that:
$ \frac {AC}{BD}= \frac {MA+MC}{MB+MD}$

I've observed something, but alas! I mut go to school, so I'll just state the property for now:

$\frac{MA}{PA}=\frac{MB}{PB}=\frac{MC}{PC}=\frac{MD}{PD}$. From that the problem follows easily. The equalities $\frac{MA}{PA}=\frac{MC}{PC},\ \frac{MB}{PB}=\frac{MD}{PD}$ can be deduced from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16824]here[/url], as for the rest of it, I'll think about it later.

Indeed, Grobber's equation is most important here. Once we have established $\frac{MA}{PA}=\frac{MB}{PB}=\frac{MC}{PC}=\frac{MD}{PD}$, by corresponding addition we get $\frac{MA+MC}{PA+PC}=\frac{MB}{PB}$ and $\frac{MB+MD}{PB+PD}=\frac{MB}{PB}$, so that $\frac{MA+MC}{PA+PC}=\frac{MB+MD}{PB+PD}$, or, equivalently, $\frac{MA+MC}{AC}=\frac{MB+MD}{BD}$, what immediately yields Juan Carlos' problem. So the question is to prove $\frac{MA}{PA}=\frac{MB}{PB}=\frac{MC}{PC}=\frac{MD}{PD}$. This would clearly follow from simple inversive geometry if we succeed to show that the points M and P are mutually inverse with respect to the circumcircle of the quadrilateral ABCD.

So it remains to show that the points M and P are mutually inverse with respect to the circumcircle of the quadrilateral ABCD. Well, the orthogonal projection of a point on its polar with respect to a circle k is the inverse of this point in the circle k. Since, by a well-known theorem, the line EF is the polar of the point P with respect to the circumcircle of the quadrilateral ABCD, and the point M is the orthogonal projection of the point P on this line EF, it follows that the point M is the inverse of P in that circumcircle, and we are done.

Juan Carlos has a beautiful proof of this without using poles and polars, based on the fact that M is the Miquel point of the complete quadrilateral formed by the lines AB, BC, CD and DA.

  Darij

Another interesting fact:
$PM$ cut the circumcircle of $ABCD$ at $Q$. Now: $Q$ is the common incenter of $AMC$ and $BMD$ triangles.

Well, that is easy.

I will only prove that the point Q is the incenter of triangle AMC (the fact that Q is the incenter of triangle BMD is analogous). In fact, I call Q' the incenter of triangle AMC, and aim at showing that Q' = Q. What is clear first is that the points Q and Q' both lie on the angle bisector of the angle AMC (in fact, Q lies there because of the Lemma posted by Treegoner in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16824]another thread[/url], and Q' lies on that angle bisector since Q' is the incenter of triangle AMC). Now, we will prove that the point Q' lies on the circumcircle of the quadrilateral ABCD; once this will be proven, it will clearly follow that both points Q and Q' are points of intersection of the angle bisector of the angle AMC with the circumcircle of the quadrilateral ABCD. The angle bisector of the angle AMC intersects the circumcircle of the quadrilateral ABCD at two points, but only one of them lies inside the triangle AMC; and since these two points Q and Q' both lie inside the triangle AMC, this implies that Q and Q' both coincide with this one point of intersection, and thus Q = Q'.

It only remains to show that the point Q' lies on the circumcircle of the quadrilateral ABCD. In fact, since the point Q' is the incenter of triangle AMC, we have < AQ'C = 90 + < AMC / 2 (we don't use directed angles here). On the other hand, the point M is the inverse of the point P in the circumcircle of the quadrilateral ABCD; since the point P lies on the line AC, the point M must lie on the inverse of this line AC with respect to the circumcircle of the quadrilateral ABCD, i. e. on the circle through the points A, C and O, where O is the circumcenter of the quadrilateral ABCD. Thus, the quadrilateral AMCO is cyclic. Hence, < AMC = 180 - < AOC. By the central angle theorem, < AOC = 2 < ADC, and since the quadrilateral ABCD is cyclic, we have < ADC = 180 - < ABC. Thus,

  < AMC = 180 - < AOC = 180 - 2 < ADC = 180 - 2 (180 - < ABC)
             = 180 - 360 + 2 < ABC = 2 < ABC - 180.

Therefore,

  < AQ'C = 90 + < AMC / 2 = 90 + (2 < ABC - 180) / 2 = 90 + < ABC - 90
              = < ABC,

so that the point Q' lies on the circumcircle of the quadrilateral ABCD. Proof complete.

  Darij

We can use the fact that $MP$ is the bisector of $\angle AMC$ (proven in the thread mentioned by Darij). It now suffices to show that $MBOD$ is cyclic. This follows from the fact that $O$ is the orthocenter of $PEF$. The inversion wrt circle $(O)$ turns circle $(AOC)$ into $AC$ and circle $(BOD)$ into $BD$, meaning that it turns their intersection into $P=AC\cap BD$, i.e. $(AOC)\cap (BOD)$ is the inverse of $P$ wrt $(O)$, and this inverse is $M$ (I think it was shown by Darij above).

Also a simple sequel: $AP.PC=BP.PD=OP.PM$, where $O$ is the circumcircle of $ABCD$.

Indeed, this follows from the fact that $OBMD$ is cyclic. We write the power of $P$ wrt to its circumcircle and we get $PO\cdot PM=PB\cdot PD=PA\cdot PC$.

Another proof:
We know $M$ is the Miquel's point of $ABCD$ by the Mandelier's theorem:
$AB.AD=2MA.JK$ ...(1)
$AB.BC=2MB.JK$ ...(2)
$AD.CD=2MD.JK$ ...(3)
$BC.CD=2MC.JK$ ...(4)
Where $J$ and $K$ are the midpoints of $AC$ and $BD$.
Further by Viete's theorem:
$\frac{AB.AD+BC.CD}{AB.BC+AD.CD}=\frac{AC}{BD}$ ...(5)
Combining (1)-(5):
$\frac{AC}{BD}=\frac{MA+MC}{MB+MD}$ QED.

I remember I asked Grobber what the Miquel point was, but now I forget it. One more time, what is Miquel point ?  :) Thank you very much.

Take four lines in general position (no three are concurrent and no two are parallel). They form four triangles. The circumcircles of these four triangles are concurrent in the Miquel point of the quadrilateral.


Let $ n\in N^*$. Prove that
$ \sum^n_{j\equal{}1}\frac{1}{C^j_n}\equal{}\frac{n\plus{}1}{2^n}\sum^{n\minus{}1}_{j\equal{}0}\frac{2^j}{j\plus{}1}$


Can anyone give me a hint to do the following problem?

Let $ N$ be a nonnegative integer valued random variable and $ X_1, X_2, ...$ be an iid sequence of random variables with $ EX_n\equal{}\mu$.
Then $ E[\sum_{n\equal{}1}^{N}X_n|N]\equal{}N\mu$

$ E[X_k|N]\equal{}\mu$ for every k...
$ E[\sum_{k\equal{}1}^n|N]\equal{}n\mu$ by linearity

since we know the value of N, let n=N and we get the desired result?
i dont know if we need even need walds equation?

Sorry there is a correction in the question. The $ N$ is independent of $ X_i$'s

Is my following argument correct?

$ E[\sum_{i \equal{} 1}^{N}X_i|N\equal{}n]\equal{}E[\sum_{i \equal{} 1}^{n}X_n|N\equal{}n]\equal{}E[\sum_{i \equal{} 1}^{n}X_n]\equal{}n\mu$

Thus $ E[\sum_{i \equal{} 1}^{N}X_i|N]\equal{}N\mu$


AMC AIME function logarithms geometry perimeter incenter

Please read every part of the spoiler. Please send me the answers through the private message. If I don't get your message or reply to it, that's because my inbox is full or I'm busy.

[hide="Rules"]
No calculator is permitted and there is no penalty for guessing. You must take this test in 3 hours. Get someone who can time you and take the test under the real AIME rule.

This test is belong to me as source of COP. You're free to print this page and use it to study but you're not allowed to produce without getting permission from me.

[hide="Test"]
1. Function $g(y)$ is given such way that for all $y$,

$6g(1+\frac {1}{y}) + 12g(y+1) = \log_{10} y$

If $g(9) + g(26) + g(126) + g(401) = \frac {m}{n}$ where $m$ and $n$ are relatively prime integers, computer $m+n$.

2. Two 5-digit numbers are called "responsible" if they are:

\[ \text {i. In form of abcde and fghij such that fghij = 2(abcde)}\]
\[ \text {ii. all ten digits, a through j are all distinct.}\]
\[ \text {iii.} a+b+c+d+e+f+g+h+i+j = 45\]

If two "responsible" numbers are small as possible, what is the sum of the three middle digits of $\text {abcde}$ and last two digits on the $\text {fghij}$? That is, $b+c+d+i+j$.

3. The triangle is called Heronian if its perimeter equals the area of the triangle. Heronian triangle COP has sides 29,6, and $x$. If the distance from its incenter to the circumcenter is exprssed as $\frac {\sqrt b}{a}$ where $b$ is not divisible by squares of any prime, find the remainder when $b$ is divided by $a$.

4. Eight boxes are numbered A to H. The number of ways you can put 16 identical balls into the boxes such that none of them is empty is expressed as $\binom {a}{b}$. What is the remainder when $\binom {a}{b}$ is divided by $a+b$? (Just use $\binom {a}{b}$, not $\binom {a}{a-b}$.

5. There are three rooms in Phil's Motel. One room for one person, one room for three people, and another one room for four people. Let's say Peter and his seven friends came to the Phil's Motel. How many ways are there to house Peter and his seven friends in these rooms?

6. Larry and his two friends toss a die four times. From all possible outcomes, the probability that outcome has at least one occurrence of 2 is $\frac {p}{q}$ where $p$ and $q$ are relatively prime integers. Find $|p-q|$.

7. In $\triangle RST$, X is on $\overline {RT}$, dividing $RX:XT = 1:2$. Y is on $\overline {ST}$, dividing $SY:YT = 2:1$. V is on $\overline {XY}$, dividing $XV:VY = 1:2$. It is found that ray VT intersects $\overline {RS}$ at Z. Find:

$128 ({\frac {TV}{VZ} + \frac {RZ}{ZS})}$

8. Let $x,y, \in \mathbb {Z}$ and that:

$x + \sqrt y = \sqrt {22 + \sqrt {384}}$.

If $\frac {x}{y} = \frac {p}{q}$ with $(p,q) = 1$, then what is $p+q$?

9. On a circle with center $\zeta$, two points, $X$ and $Y$ are on a circle and $Z$ is outside the circle but in ray XY. $\overline {XY} = 33.6, \overline {YZ} = 39.2,$ and $\overline {\zeta X} = 21$. If $\overline {\zeta Z} = \frac {j}{k}$ where $j$ and $k$ are relatively prime integers, and such that $\frac {j+k}{j-k} = \frac {o}{p}$, find $o+p.$

10. Given that:

$(\frac {1}{r})(\frac {1}{s})(\frac {1}{t}) = \frac {3}{391}\\
r + \frac {1}{s} = \frac {35}{46}\\
s + \frac {1}{t} = \frac {1064}{23}\\
t + \frac {1}{r} = \frac {529}{102}$.

Then what is the smallest integer that is divisible $rs$ and $12t$?

11.  There are $z$ number of ways to represent 10000000 as product of three factors and $Z$ number of ways to represent 11390625 as product of three factors.  If $|z-Z| = p^q$, find $p+q$.

12. Let $m = 101^4+256$. Find the sum of digits of $m$.

13. If the reciprocal of sum of real roots of the following equation can be written in form of $\frac {r}{s}$ where $(r,s) = 1$, find $r+s$.

$1000x^6-1900x^5-1400x^4-190x^3-130x^2-38x-30 = 0$

14. The set $Z$ contains complex numbers ${\zeta_0,\zeta_1,\zeta_2...}$ such that $n = 0,1,2,3.....$.

$\zeta_n$ is defined this way:

$\zeta_{n+1} =  (\frac {\zeta_n - i}{\zeta_n + i})^{-1}$

If $\zeta_0 = i + \frac {1}{121}$ and 2008th number of the set is in form of $a + bi$, find $ab$.

15. Call the value of $[\log_2 1]+[\log_2 2]+[\log_2 3]+....[\log_2 64]$ as $\alpha$. Let the value of $[\log_2 1]+[\log_2 2]+[\log_2 3]+....[\log_2 52]$ as $\delta$. Find the absolute value of $\delta - \alpha$.
[/hide]
[hide="Reminder"]Please do not post anything directly about answers or post on the forum concerning the answers.

Thank you[/hide][/hide]

Umm, for #2... how can the sum of 10 distinct digits equal 55? Don't they have to sum to 45?

#4: $\binom{a}{b} = \binom{a}{a-b}$ so there are two answers.

#7: XV should read XY.

I think they mean the floor function.

Also, I don't think 14 has an integer answer. Perhaps you mean 2005?

Use "\lfloor f(x) \rfloor" to make $\lfloor f(x) \rfloor$ (the left and right floor function brackets :) ).

Oi, just saw this. And I have to go to delta college this afternoon. darn. Can you post the answers afterwards so I can take it later?

I'm going to give people sometime to take this.

So, the answer would be posted little bit later.

Are we allowed to take it in pieces, as long as the total  time doesn't exceed 3 hours, since I don't have time for 3 hours in a row?

Yes.

It depends on you, however. It's "better" and "realistic" to take on the original AIME rule.

i just started, in #4, how do you know which one is a-b, which one b

Just do $\binom {a}{b}$.

I'm not sure why you asking it. :?

?

lets say the answer is 10C3, b can either be 3 or 7

and for # 14, i thought you say n = 1,2,3,4

how does zeta_0 come in play?

Zn is defined as the inverse of the floor value of something ?

#12  distinct factors?

#12, what are you talking about? It's asking some of digits of $m$.

For #14, it should've been $n=0,1,...$.

For a and b one, I don't see why you have trouble on determining which one. You should find it when you're doing combinatorics.

#11 was what i was referring to

you asked for remainder of aCb  when divided by a+b,   and i believe you know you can have 2 possible b values,
there can be 2 answers...

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3744975_2005_silverfalcon_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS. Answer key is [url=https://artofproblemsolving.com/community/c5h28368p175847]here[/url]


How many square feet are in the area of a rectangular chalkboard whose width is 4 feet and diagonal length is 10 feet? Round to the
nearest tenth of a square foot.

This implies that the diagonal length is $ \sqrt{100\minus{}16}\equal{}\sqrt{84}\equal{}4\sqrt{21}$, so the area is $ 16\sqrt{21} \approx \boxed{73.3}$.

whoa, um, AIME15, you kinda messed up on $ \sqrt{84}$... it's $ 2\sqrt {21}$, not $ 4\sqrt {21}$... so the answer would actually be $ 4*2\sqrt {21} \equal{} 8\sqrt {21}\approx\boxed{36.66}$...  :P

$ The$ $ Origami$ $ Guy$


geometry
circumcircle
inequalities solved
inequalities

Solution:

using the fact that 1=ab+bc+ca
we replace it  and we have that:

sum (a/sqrt(1+a^2))=sum(a/sqrt((a+c)(a+b)))= sum(sqrt(a/(a+c))*sqrt(a/(a+b)))<= 1/2 sum(a/(a+c)+a/(a+b))=3/2.

cheers!!! :D

or we make the substitution:
a=cot(x) b=cot(y) x=cot(z)  x+y+z=PI
and we have to prove
cos(x)+cos(y)+cos(z)<=3/2

which is obvious as cos X + cos Y + cos Z = 1 + r/R where r is the radius of the incircle of the triangle ABC and R is the radius of the circumcircle of the triangle ABC.


AwesomeMath
summer program
AMC
USA(J)MO
USAMO
AMC 12
MATHCOUNTS
\/closed

Hi,

I was looking at AwesomeMath in Texas Austin for a summer camp. But if I don't get in, are there other similar summer camps, in CA, or western US, from third week of June to end of July timeframe?

I'm in eighth grade, and I made it to the USAMO selections. I want to take a summer camp to improve my chances in high school, USAMO, AMC12 etc.

Thanks.

[url]mathpath.org[/url] is in Vermont, which isn't exactly Western USA, but you may want to check it out. I went two years ago and it was a great experience.

It may be a little late to apply (it's due pretty soon I believe), but if you are going to MathCounts nationals, they will give you an extension of like two weeks. Also, AMP is at UT Dallas, not Austin.


[quote="GameBot"]How many digits are required to write $ 2^{15}\times 5^{10}$ as a whole number?[/quote]

$ 2^{15}\times 5^{10}\equal{}2^5\times 2^{10} \times 5^{10}\equal{} 2^5\times 10^{10}$

Here we can say the number would be 2^5 (which we know as 32) followed by 10 zeroes. Thus 2 digits (the 32) followed with 10 digits (1 for each zero), makes 12 digits in total.


Given that a, b and c are positive, prove that

$\sqrt{a^2 + \frac{1}{a^2}} + \sqrt{b^2 + \frac{1}{b^2}} + \sqrt{c^2 + \frac{1}{c^2}} \geq \sqrt{82}$

such that $a+b+c\leq 1$

[b]PLEASE... GIVE ME HINTS ONLY!!![/b] :D

This is a problem in the University entrance exam of Vietnam
So many solutions!!!
Use am-gm or Minxkopxki's ineq.

I tried AM-GM but it only complicated things :?

I will try Minkowski now..

Could someone show me an example that uses Minkowski please? I am having trouble applying it to the problem :(

Applying Minkowski we obtain \[ \sqrt{a^2 + \frac{1}{a^2}} + \sqrt{b^2 + \frac{1}{b^2}} + \sqrt{c^2 + \frac{1}{c^2}}\geq \sqrt{(a+b+c)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=A.\]
So we are to prove $A^2\geq 82$. Indeed, let $a+b+c=s\leq 1$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}=\frac{9}{s}$. Therefore, $A^2\geq s^2+\frac{9^2}{s^2}\geq 1+\frac{81}{1}=82$, since $s\leq 1$.

If $s \leq 1$ I don't think you can write that

$ s^2+\frac{9^2}{s^2}\geq 1+\frac{81}{1}=82 $

Am I wrong?

Inequality $s^2+\frac{9^2}{s^2}\geq 1+\frac{9^2}{1}$ holds for $s\leq 1$.

maybe this one is harder :D :

let a,b,c>0 and a+b+c<=3/2

then prove that:

$ \sum\sqrt{a^2+\frac{1}{b^2}+\frac{1}{c^2}}\geq3\sqrt{33}/2 $

 ;)


[quote="Mathx"]Let $ A \equal{} \{1,2,...,n\}$ and $ B_1,\dots,B_b$ be subsets of $ A$ each with $ k\ge 2$ elements,such that for each pair of elements in $ A$ , the pair appears exactly in $ \lambda$ of $ B_i$s prove there exist a number like r such that each element of A appears in exactly r of the $ B_i$s[/quote]

Let some element $ x\in A$, say $ x\equal{}1$, belong to $ p$ different sets $ B_i$s, say to $ B_1$, $ B_2$, $ \dots$, $ B_p$. Note that each pair $ (1,i)$, $ 2\le i\le n$ appears in exactly $ \lambda$ $ B_i$'s, and each $ B_i$ ($ 1\le i\le p$) contins exactly $ k\minus{}1$ such pairs. Hence $ p(k\minus{}1)\equal{}\lambda(n\minus{}1)$ (both sides are equal to the number of pairs $ \{(1,i)$, $ B_j$ such that $ \{1,i\}\subset B_j\}$). So, we may take $ r\equal{}\lambda(n\minus{}1)/(k\minus{}1)$.


$ x^2 \plus{} 19x \plus{} 92$ is a perfect square for how many $ x \epsilon I$

What is that $ I$?

Anyways, $ \minus{}11$ and $ \minus{}8$ work. They both give $ 4$.

Checking through the interval [-1000000,1000000] (integers), one can see that only -11 and -8 work.

Also, note $ x^2\plus{}19x\plus{}92\equiv x^2\plus{}x\plus{}2 \mod 3$ and residues mod 3 are 0,1. Then only x=1 mod 3 will produce a possible square.

As of now I don't know how to prove those are the only ones...

Completing the square will work wonders :)

Oh wow...

[hide] $ x^2\plus{}19x\plus{}92\equal{}(x\plus{}19/2)^2\plus{}7/4\equal{}\frac{(2x\plus{}19)^2\plus{}7}{4}$
Then the numerator must be a square. Only squares 7 apart are 9,16.

So we have $ (2x\plus{}19)^2\equal{}9\rightarrow 2x\plus{}19\equal{}\pm 3\rightarrow \minus{}11,\minus{}8.$

 [/hide]

That's a clever trick to keep in mind.

$ x^2\plus{}19x\plus{}92 \equal{} (x\plus{}9)^2\plus{}(x\plus{}9)\plus{}2 \equal{} t^2\plus{}t\plus{}2$ with $ t\equal{}x\plus{}9$

$ t^2\plus{}t\plus{}2$ lies strictly between $ t^2$ and $ t^2\plus{}2t\plus{}1$ if $ t \notin (,\minus{}2\minus{}1,0,1)$ Checking for the values of t, we get perfect square when t = 1 or t = -2.

Correspondingly, x = -8 or x = -11

$ I$ is set of integers. You are supposed to know the general notations.

[quote="hsiljak"]The most general notation for integers is $ \mathbb{Z}$ ([i]Zahlen[/i]), though.

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

Note that some even use the $ \mathbb{I}$ for imaginary numbers (not too often, though)...[/quote]
Rajiv, we just use that in India as I+ and I- for +ve and -ve integers only. Zahlen is the most common one you would see in the world , like some countries use log for log to the base 10 but many use lg.


Made it into TJ? Cool!  :D
I look forward to seeing you there next year.

Alot of those guys are graduating or have left Virginia on that thread so I'm interested in hearing about you!
Feel free to introduce yourself although I know there's a similar thread that Thomas Mildorf posted years ago somewhere on this forum.  :wink:

Good job! I didn't  :(
but that's only because I didn't take the TJ test
and [i]that's[/i] because I'm not old enough yet  :D

my sister and all her friends got in though.

Actually, does anyone know incoming sophomores who didn't attend TJ in ninth grade?

I'm hoping Andrew Kim and Adam Hood come. I miss them...

im going to tj :D
i hear u actually learn stuff in that school ;)
thats amazing considering our school well cant wait to go

i cant believe it longfellow students can take two math classes at once and they have a geometry problem solving class >:o

oh wow, i didnt know that, no wonder longfellow owns all math-related thing in VA..

well the classes themselves, as you would imagine, are somewhat far in difficulty from contest math.
Im not saying the classes don't help, but i am saying that just following school math at Longfellow wont get you much further than a 20 on AMC 8. You still need to spend a considerable amount of time at home.

Also, u need to spend time in home to get recommendations for the two math class thing(usually GPS and AL2/Trig).

just in case if ur wondering, GPS stands for Geometry Problem Solving (basically advancded geometry)

anyway, im not going to tj

[quote="batteredbutnotdefeated"][quote="turing"]yes, why not hjoon?[/quote]

because he's not cool enough. Okay, not really, but anyways I'm an 8th grader who's currently in the middle pf the TJHS admissions process. What grade are you in aufha? And Adam Hood is attending TJHSST currently....[/quote]

Note that his post was made in 2008.

Is TJ acceptance completely based on an application test?

@Rit

No, you also have to fill out an informational packet. Your math and science grades are also a factor.

I made it into the 2014 Freshman class.

YAY

I've met 4 people in person who are/have/will be going to TJ.

I've met so many, I don't even want to count.


pigeonhole principle
search
number theory proposed
number theory

a. Does there exist 4 distinct positive integers such that the sum of any 3 of them is prime?
b. Does there exist 5 distinct positive integers such that the sum of any 3 of them is prime?

a) Yes, for example $ \{1,5,7,11\}$, since $ 1\plus{}5\plus{}7\equal{}13$, $ 1\plus{}5\plus{}11\equal{}17$, $ 1\plus{}7\plus{}11\equal{}19$ and $ 5\plus{}7\plus{}11\equal{}23$.
b) No, because if there are three of them that are equal mod 3, they sum to $ 0\bmod{3}$, so that sum can't be a prime. Also, we can't have three that are pairwise distinct mod 3, since $ 0\plus{}1\plus{}2\equiv 0\bmod{3}$. So by pigeonhole we can have at most $ 2\times2\equal{}4$ numbers, as desired.

Again, the general form of b) is: In any $ 2n \minus{} 1$ integers, we can find $ n$ such that their sum is divisible by $ n$.
Search for Erdös-Ginzburg-Ziv if you want.

oh ,sorry I don't see that there is the positive integers are distnic!

I try to upload a file but can't it says "the extension out is not allowed", example is $\{1,3,7,9\}$

a) example: $3,7,9,19$
b) if look at remainders modulo 3. If we have $0,1,2$, we have a number divisible by 3 which can't be prime. So we have only $2$ remainders(if we have only one, then every sum of 3 numbers is multiple of 3). Using pigeonhole's principle we have 3 numbers given the samen remainder modulo 3, so their sum is multiple of 3, so it is not prime($1+1+1=3$ is the only such pair, but they are not distinct)

moreover,do there exist infinite number of pairs of such quadruples?

Related:What is the maximal $n$ such that there are $n$ integers (possibly negative) satisfying that condition(the sum of every three numbers is a prime number)?


Given $ a^2\plus{}b^2\plus{}c^2\equal{}1$

Prove that: $ abc\plus{}2(1\plus{}a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)>\equal{}0$

Anyone have any idea? I've been thinking about this for a little over a day now...

Use the fact that the constraint $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ implies the equality
$ abc \plus{} 2(1 \plus{} a \plus{} b \plus{} c \plus{} ab \plus{} ac \plus{} bc) \equal{} abc \plus{} (a \plus{} b \plus{} c \plus{} 1)^2$.

Now for such triples $ (a,b,c)\in\mathbb{R}^3$ for which $ abc > 0$ the inequality is easy.

If any two of the numbers a, b, c are positive and the 3rd is negative the inequality can be proved in relatively elementar way.

I think the most complicated case is when $ (a,b,c)\in\mathbb{R}_ \minus{} ^3$. On the other hand, I think the solution is not hard. :)

[hide="my thoughts, tell me if I did something incorrect"]
We note that $ ab \plus{} bc \plus{} ca \equal{} \frac {(a \plus{} b \plus{} c)^2 \minus{} a^2 \minus{} b^2 \minus{} c^2}{2}$

So, the expression becomes $ abc \plus{} 2(1 \plus{} a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca)\rightarrow$

$ abc \plus{} 2(1 \plus{} a \plus{} b \plus{} c \plus{} \frac {(a \plus{} b \plus{} c)^2 \minus{} a^2 \minus{} b^2 \minus{} c^2}{2}) \rightarrow$

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

This factors into $ (a \plus{} b \plus{} c \plus{} 1)^2 \plus{} abc$. To prove that $ (a \plus{} b \plus{} c \plus{} 1)^2 \plus{} abc \ge 0$, or more importantly that the magnitude of $ (a \plus{} b \plus{} c \plus{} 1)^2$ is greater than the magnitude of $ abc$, I will break this up into four cases.

Case 1: All $ a,b,c$ are greater than $ 0$.
This is trivial because the LHS is greater than $ 0$.
Case 2: One of $ a,b,c$ is negative.
WLOG, assume $ a$ is negative.  Note, $ |a| \le 1 \rightarrow \minus{} 1\le a\le 1 \rightarrow 0\le a \plus{} 1\le 2$, which can be observed by the given restriction $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.

Now, since $ b,c \ge 0$, using AM-GM, we get $ \frac {b \plus{} c}{2} \ge \sqrt {bc} \rightarrow (b \plus{} c)^2 \ge 4bc$.

Now, since $ \minus{} 1\le a \le 1$, $ 4bc \ge |abc|$.  Also, since $ a \plus{} 1 \ge 0, (a \plus{} b \plus{} c \plus{} 1)^2 \ge (b \plus{} c)^2$. So, combining the two results, we get $ (a \plus{} b \plus{} c \plus{} 1)^2 \ge |abc|$.
Case 3: Two of $ a,b,c$ are negative.
This case is also trivial because the LHS is greater than $ 0$.
Case 4: All three of $ a,b,c$ are negative.
Now...I've been stuck here for quite some time....any ideas?[/hide]

EDIT: Well, I guess you posted when I was double checking my post :P

[quote="Zellex"]
Case 4: All three of $ a,b,c$ are negative.
Now...I've been stuck here for quite some time....any ideas?

EDIT: Well, I guess you posted when I was double checking my post :P[/quote]
Oh it's very easy.
Let $ \minus{} b \equal{} x, \minus{} c \equal{} y \therefore x,y\ are\ postive$
And it becomes case 2 :D


integration
calculus
calculus computations

Find a reduction formula for $ \int x^{n}e^{x^{2}}\ dx$ and show that $ \int_{0}^{1} x^{5}e^{x^{2}}\ dx \equal{} \frac{1}{2}(e\minus{}2)$.

What have you done?  Where are you stuck?  This should be pretty straightforward.

Thanks dallasboy, I'm not stuck at all. I integrated by parts to get the reduction formula and then substituted, first $ n\equal{}5$ and then $ n\equal{}3$ to get $ \frac{1}{2}(e\minus{}2)$. I posted it because I hoped it would be of interest to someone.


integration
logarithms
trigonometry
calculus
complex analysis

I am doing a simple problem to calculate $\int_{0}^\infty \frac{x\ln x}{(1+x^{2})^{3}}\, dx$.

I wonder if someone can show me how to calculate the residues at $z = \pm i$ of $\frac{z(\log z)^{2}}{(1+z^{2})^{3}}$, where $\log z$ uses the branch of $\arg$ with values in $(0,2\pi)$. I mean, obviously I could do it, but I can't think of a reasonably efficient way.

edit: Perhaps I should have done the variable substitution $t=x^{2}$ first, but now it's too late :P. Ok, obviously you were meant to see this since the residue is very easy to determine if you do this first. However, I'd still like some suggestions for doing it the non-thinking way.

One approach worth trying is to expand the numerator and denominator as power series about the point in question and then use long division. That's not all that efficient in this case but you could do it. In the case of something like finding the residue of $\frac1{z^{2}\sin x}$ at zero, it would be my method of choice.

One method, avoiding the rezidues technique, would be this:
Let $I$ be your integral, make the substitution $x=\frac{1}{y}$ to get that:
$I=\int_{0}^{1}\frac{x\ln x}{(1+x^{2})^{3}}dx+\int_{1}^{\infty}\frac{x\ln x}{(1+x^{2})^{3}}dx=\int_{0}^{1}\frac{(y-y^{3})\ln y}{(1+y^{2})^{3}}dy$
$=\sum_{n=0}^{\infty}(-1)^{n}\frac{(n+1)(n+2)}{2}\int_{0}^{1}(y^{2n+1}-y^{2n+3})\ln y dy=-\sum_{n=0}^{\infty}(-1)^{n}\frac{1}{8}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)=\frac{-1}{8}$.
I simplified the calculations as much as possible.


Does anyone know what those things below your sn are? Like i'm a hodge conjecture or something and there's P vs NP and Yang-mills theory and some others.

Have you tried googling?

They are the seven millenium problems. The CMI (Clay Mathematics/al Institute) has set aside $1 million for the solution of each one.

[quote="bleumoose"]heres another question for you...
how many posts does it take to "graduate" to the next problem? and how do the stars work? are they in conjunction with the problems, or are they on their own timetable?[/quote]

It depends on what rank you're going to. Some require as small as 50 more posts, while others differ by up to 2000+.

Yaaaaaay! I finally got to (about) 20 posts and my first millenium problem. :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :blush:

I think birch&swinnonston dyer starts at 3000

w00t. Will be reaching 1000 soon, altough its taken me a really long time to get there.

This is my 20th post
Now I'm P vs NP
Yah !!!!:D  :D  :D

[quote="Scrambled"]well ithink its something like this
20-50 PvsNP
50-100 hodge conjecture
100-200 poincare conjecture
200-500 riemanns hypothesis
500-1000 yang mills theorey
1000-? navier strokes equations
?+ birch&swinnonston dyer[/quote]

Not exactly right. It should be

     0 -      19       New Member                      (Zero Stars)
    20 -     49       P versus NP                       (Half Star)
    50 -    99       Hodge Conjecture              (One Star)
  100 -   249       Poincare Conjecture          (Two Stars)
  250 -   499       Riemann's Hypothesis       (Two and Half stars)
  500 -   999       Yang Mills Theory              (Three Stars)
1000 - 2499       Navier-Stroke's Equation   (Four Stars)
2500 - $\infty$   Birch & Swinnerton Dyer.   (Five Stars)
-----------------   Administrator                       (Six Stars)  :D

Note that most of the differences are just 1 at the end, but others are fifty or more.

Best,

Didn't someone recently say he had solved the Poincare conjecture?  A serious mathematician, I think Russian...it was in a magazine a few months ago.

[quote="Hexahedron"]Is it possible to get $\infty$ posts given that we have a finite lifetime?

I should hit 50 posts soon.....hodge conjecture[/quote]

Of course not, but what I mean is that, given that there are seven problems and those are in cathegories up to that, well, I suppose there won't be a new cathegory at 5000, therefore when you get "Birch & Swinnerton Dyer" cathegory you stay there for life (unless someone else come with a new millenium problem or you become an administrator).

Best,

[quote="Hexahedron"]Is it possible to get $\infty$ posts given that we have a finite lifetime?

[/quote]

um...defiently not, however, it is like that one Zeno Paradox (Zeno was quite a strange guy).

link: http://www.brainteaser-world.com/puzzles/paradoxes/paradoxes-4.htm

[quote="thatin"]i think it is possible to get infinite posts... your account will still exists when you will die and you can make a programm that sents replies here as it was you[/quote]

....err....???!!??

Wow
I looked at those Millenium problems you guys were talking about
there's so much Greek in the official problem esriptions that I'm almost too dizzy to type this :o


S2 = { u1 = ( 2, -2, -3), u2 = (3, -3, -6 ) } included in R^3

Normally, i would be doing the determinant of the matrix of which the vectors create, but, it's the first time when i see an exercise [b]with only 2 vectors[/b], what should i do ?

Your main computational tool should never be determinants: it should be Gaussian elimination.

(With just two vectors it's much simpler than what I'm about to state, but what I'm about to do is geared toward a more general case - for instance, four vectors in $ \mathbb{R}^6$)

Form a matrix with the given vectors as columns (rows would work, too, but there are notational reasons for making this the default.) In our particular case, that would be:
\[ \begin{bmatrix}2 & 3 \\
\minus{} 2 & \minus{} 3 \\
\minus{} 3 & \minus{} 6\end{bmatrix}
\]
Then perform row elimination until you have arrived at a matrix in row echelon form. You don't have to make the pivot elements $ 1$ and you don't have to clean out above the pivot elements - row echelon form suffices.

$ \begin{bmatrix}2 & 3 \\
\minus{} 2 & \minus{} 3 \\
\minus{} 3 & \minus{} 6\end{bmatrix} \longrightarrow \begin{bmatrix}2 & 3 \\
0 & 0 \\
0 & \minus{} \frac32\end{bmatrix} \longrightarrow \begin{bmatrix}2 & 3 \\
0 & \minus{} \frac32 \\
0 & 0\end{bmatrix}$

Now: we see a pivot element in every column. That means this is an independent set. We have a dependent set if and only if there is some column without a pivot element, meaning that there are fewer pivots than vectors.

What would happen if the vectors were $ \{(2, \minus{} 2, \minus{} 4),(3, \minus{} 3, \minus{} 6)\}?$

The professor didn't taught us about pivots and row elimination.

Isn't any other method to solve ?

What is the definition of independent?

(That might be a test question for me on the first test of the term in the Intro. Linear Algebra course: "State the exact definition of a set of vectors being independent.")

I'll add that having determinants before developing row elimination seems to me like an odd and counterproductive order for a course. In the course I was talking about, and in nearly all of the standard textbooks for that course, the course begins with systems of linear equations and row elimination. Determinants are postponed to perhaps the third or fourth chapter.

[url]http://en.wikipedia.org/wiki/Linear_independence#Alternative_method_using_determinants[/url]

Looks that the determinant method it still applies in this case  :)

That's not the definition.

Also, how are you calculating those determinants? For a large matrix, the most efficient way to calculate a determinant is... Gaussian elimination.

Well, we never calculate large matrixes ( larger than 4x4 or 5x5 ).

Have you ever calculated the determinant of a "full" $ 5\times 5$ matrix? That is, one where all or nearly all of the terms are nonzero? Did you really do all 480 multiplications and 119 additions/subtractions that that would entail?

By the expanded-out definition (or by minors, which amounts to the same thing), an $ n\times n$ determinant expands out to have $ n!$ terms, each of which is the product of $ n$ factors. That's where I'm getting the number of multiplications I mentioned above: $ (n \minus{} 1)n!.$ By contrast, if done by Gaussian elimination, computing the determinant requires approximately $ \frac13n^3$ multiplications or divisions.

The set of vectors $ \{v_1, v_2, ..., v_n \}$ is said to be linearly dependent if there is one vector of the set that can be written as a linear combination of the others. If that doesn't happen, then the set is said to be linearly independent. A single vector is linearly independent if it is not the zero vector.

From this definition we can conclude that:

1) Two vectors are dependent iff each is a multiple of the other.

2) The set $ \{v_1, v_2, ..., v_n \}$ is linearly independent iff the only way to construct the zero vector as a linear combination of the vectors of the set is to have all coefficients zero. If at least one is non-zero, then the set is linearly dependent.

3) Any set that includes the zero vector is linearly dependent.


calculus
integration
function
trigonometry
calculus computations

Find the function $f(x)$ which satisfies the following integral equation.

\[f(x)=\int_0^x t(\sin t-\cos t)dt+\int_0^{\frac{\pi}{2}} e^t f(t)dt\]

Integrating gives us

$f(x) = \sin x - x\sin x - \cos x - x\cos x +1 + \int_0^{\frac{\pi}{2}} e^t f(t) \ dt$

Since $\int_0^{\frac{\pi}{2}} e^t f(t) \ dt$ is a constant, we have

$f(x) = \sin x - x\sin x - \cos x - x\cos x +1 + \alpha$

So now we must solve the equation

$\alpha = \int_0^{\frac{\pi}{2}} e^t f(t) \ dt$

$\alpha=\left[\frac{-e^t\left(3\cos t +(2t-1)\sin t -2(\alpha +1)\right)}{2}\right]_0^{\frac{\pi}{2}} = \frac{ 2\alpha \left( e^{\pi/2}-1\right) - (\pi-3)e^{\pi/2}+1}{2}$

$\alpha = \frac{(\pi-3)e^{\pi/2}-1}{2\left(e^{\pi/2}-2\right)}$

So finally, we have

$f(x) = \sin x - x\sin x - \cos x - x\cos x +1 + \frac{(\pi-3)e^{\pi/2}-1}{2\left(e^{\pi/2}-2\right)}$


Let a,b,c,d >0 satisfy: a^2+b^2=(c^2+d^2)^3.
 Prove that: c^3/a +d^3/b>=1
 When does the equality hold?

 Notes: You can try this problem and other problems from SMo at the official site of SMS: sms.maths.nus.edu.sg. Good Luck! : :lol:

It seems to me that I saw this problem somewhere before.
[b]Solution:[/b]
By CBS inequality:
$ \frac {c^3}{a} \plus{} \frac {d^3}{b} \equal{} \frac {c^4}{ac} \plus{} \frac {d^4}{bd}\geq\frac {(c^2 \plus{} d^2)^2}{ac \plus{} bd}\geq\frac {(c^2 \plus{} d^2)^2}{\sqrt {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}} \equal{} 1$.
So we are done,I guess.
Added after seeing [b]arqady[/b]'s reply:
I was too late :roll: