hello, i think you meant $ \frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\geq\frac{2(a^4\plus{}b^4\plus{}c^4)}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c}$
Sonnhard.

[quote="PHONGVAN"][quote="PHONGVAN"]Let $ a,b,c > 0$ .Prove that:
$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq\frac {2(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}$[/quote]
No one likes it. :([/quote]
we have $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\ge 3(\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c})$ since vasC 's ineq
and  $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\ge \frac {3(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2}$ since Holder inequality and EV theorem for case $ n\equal{}3$
Add these inequality we have done :?:

[quote="Allnames"] $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\ge \frac {3(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2}$ since Holder inequality and EV theorem for case $ n \equal{} 3$
Add these inequality we have done :?:[/quote]
Can you post your solution ?   :?:

[quote="Allnames"] $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\ge \frac {3(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2}$ since Holder inequality and EV theorem for case $ n \equal{} 3$
Add these inequality we have done :?:[/quote]
This inequality does not holds. Try $ a\equal{}\frac{5}{2},b\equal{}\frac{3}{2},c\equal{}1$.

[quote="PHONGVAN"]Let $ a,b,c > 0$ .Prove that:
$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq\frac {2(a^4 \plus{} b^4 \plus{} c^4)}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}$[/quote]
Because $ 2\left(\frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\right)\geq7(a^2\plus{}b^2\plus{}c^2)\minus{}5(ab\plus{}ac\plus{}bc).$ :wink:


calculus
integration
geometry
logarithms
calculus computations

Answer the following questions.

(1) By setting $ x\plus{}\sqrt{x^2\minus{}1}\equal{}t$, find the indefinite integral $ \int \sqrt{x^2\minus{}1}\ dx$.

(2) Given two points $ P(p,\ q)\ (p>1,\ q>0)$ and $ A(1,\ 0)$ on the curve $ x^2\minus{}y^2\equal{}1$. Find the area $ S$ of the figure bounded by two lines $ OA,\ OP$ and the curve in terms of $ p$.

(3) Let $ S\equal{}\frac{\theta}{2}$. Express $ p,\ q$ in terms of $ \theta$.

$ x \plus{} \sqrt {x^{2} \minus{} 1} \equal{} t$

therefore $ x \minus{} \sqrt {x^{2} \minus{} 1} \equal{} \frac {1}{t}$

adding both we get $ 2x \equal{} t \plus{} \frac {1}{t}$

therefore $ dx \equal{} \frac {t^2 \minus{} 1 }{2t^2}dt$

subtracting the first two equations we get $ 2\sqrt {x^2 \minus{} 1} \equal{} t \minus{} \frac {1}{t}$

therefore the integral becomes $ \int \frac { ( t^2 \minus{} 1)^2 }{4t^3}$

therefore the ans is  $ \frac {t^2}{8} \minus{} \frac {2\log (|t|)}{4} \minus{} \frac {1}{2t^2} \plus{} C$

Hope this is correct .

I think in question 2, S should be expressed in terms of p and q.

By the way, I have got

\[ S\equal{} \frac{1}{4}\ln\left| \frac{p\plus{}q}{p\minus{}q} \right|\]


[quote="AstroPhys"][hide]
$ t+\frac{1}{12}\cdot \frac{t}{\frac{3}{2}}= 5$
$ \frac{19t}{18}=5$
$ t = \frac{90}{19}\approx 4.73684211$
$ d=3t$
$ d = \frac{3 \cdot 90}{19}\approx 3(4.73684211)$
$ \boxed{ d = \frac{270}{19}\approx 14.2105263}$.
[/hide][/quote]
You're on the right track, but you made a little mistake. You divided the entire time by $ \tfrac{3}{2}$, but it doesn't divide evenly, and you get a non-integral amount of stops.

[hide="um.. i think it would be..."]
\[ 14\frac{1}{4}\]miles
[/hide]
but i could be wrong

Post your solution if you think the answer is different.

Supposed they take a rest x time ( x dense in the set of integer), for 0<m<90 we can write an equation :
90x+5x+m=5.60=300
95x+m = 300=>300-m=95x.
In other words, 300-m is a multiple of 95.300=15 mod 95 The value of m statisfying the condition above is 15, x=3.
So, they plan to hike : 3.(90.3+15)/60=3.285/60=285/20 miles=14,25 miles like the answer of Jeffboot.
I only want to desribe how to obtain the answer like that. God bless you all.

qruni you put a comma instead of a . for the number 14.25

[hide="that other guy said i should post my solution so here it is"]
in one and a half hour of walking they take 5 minuets of rest and they travel 4.5 miles.

miles: 4.5

time: 90 minutes

rest: 5 minuets

i multiplied averything by 3 to get the time closest to 5 hours so now it would be:

miles: 13.5

time: 270 minutes (4.5 hours)

rest: 15 minuets

after this, to get the miles they would go in the leftover 15 minuets we multiply 3 by .25 we get .75 miles. so now we add and get 14.25 miles. :lol:  :D  :roll:  :)  :first: [/hide]

I've heard in some places like Germany they use commas instead of full stop for fractional decimals

e.g 0,5 instead of 0.5

@ Jeff - $ 14.25$ miles is the correct answer.

@ Identity - That's true, but please refrain from posting irrelevant posts.

[quote="i_like_pie"]A group of friends are about to go hiking. They plan to hike at 3 mph, and make 5 minute stops every 90 minutes to rest. If they want to hike for a total of 5 hours, how many miles should they plan to hike?[/quote]
[hide="My Solution"]
So in 5 hours, they will make 3 rest stops.
Therefore, they will hike for 4 hours and 45 minutes, which is $ 4\frac {3}{4}$ hours.
Since they go at 3mph, the answer is $ (3)(4\frac {3}{4})$ Which is $ 14\frac {1}{4}$ miles.
[/hide]


geometry
inradius
circumcircle
geometry proposed

Let a triangle $ABC$ and arbitary points $K,\ L$ at $AB,\ AC$ respectively .From a point $R\in KL$ we drow the projections $PD,\ RE,\ RZ$ of $R$  to the segments $AB,\ BC,\ CA$ respectively .Find the positions of $K,\ L$ such that while $R$ varies on $KL$ the $2(RD+RE+RZ)=ct$

A result quite well know (follow me) locus of $M$ such that
$d(M,BC)+d(M,CA)+d(MAB)=k$ is a line perpendicular $OI$ and through $M_{0}$ satify the equation $\frac{1}{R}\vec{OI}\cdot\vec{IM_{0}}+3r=k$ with $r$ is inradius,$R$ circumradius !
thus $K,L$ are intersection of that line with $AB,AC$


function
real analysis
real analysis unsolved

Let X and Y be sets and T be a sigma-algebra of subsets of Y. Let f be a function mapping X to Y. Show that  {F:F subset of Y, f^-1[F] element of the family of sigma algebras} is a sigma-algebra of subsets of Y.

Here $T$ should be a $\sigma$-algebra of subsets of $X$. Use the same identities as in the other [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145273]similar thread[/url].


quadratics
number theory solved
number theory

prove that there are infinitely prime in the form 8k+5  :lol:

As I remember,The solution is
Assume there are finite primes in form 8k+5.
Denote p1,p2,...,pn are all the prime with pn id the greatest prime with form 8k+5.
Let n=(p1*p2*...*pn)^2+3.
And ...I forget.
Sorry

Actually, you can do this:

Assume, as you said, that $p_1,\ldots,p_n$ are all the primes of the form $8k+5$. Take $N=p_1\cdot\ldots\cdot p_n$. Now look at $(6N)^2+1$. $6N$ is of the form $8k+6$, so $(6N)^2+1$ is of the form $8k+5$, meaning that not all of its prime factors are of the form $8k+1$. However, $(6N)^2+1$ can only have prime factors of the form $4k+1$, so all its prime factors are $8k+5$, but different from $p_1,\ldots,p_n$, which is a contradiction.


analytic geometry
geometry
graphing lines
slope
3D geometry
sphere
inequalities

Hi,

I was actually going to repost on AMC A day post but I decided to start a new thread since I want to have this little bit different.

1. Instead of you sending me the answers, post your answer in the spoiler WITH FULL EXPLANATION.
That way, I can see whether you're cheating or not.

Points rule are same.
1 point for correct answer
0 point for unanswered answer
-1 point for incorrect answer

Since it's spoiling text, you SHOULD NOT change your answer. Of course, you can do it but it would make you a liar. Furthermore, this AMC B day's idea is to see many problems, not getting them all right by cheating.

Day 1

Let Q be a set of N different integers from the set {1,2,3....100} such that sum of two different elements is never divisible by 10. The greatest possible value of N is:

A) 42
B) 45
C) 49
D) 50
E) 55

Make sure ----> ANSWER IN SPOILER :)

Couldn't we just copy off someone else's answer if we wanted a full explanation.

[hide="Answer"]A. {0,1,11,21,...,91,2,12,22,...,92,3,13,23,...,93,4,14,24,...,94,5} is one such set. We would build this by noticing that we can't have two numbers that both end in 0, or two numbers that end in 5. Then we see that if we have a number ending in 1, we can't have a number ending in 9. We also can't have a number ending in 2 if we have a number ending in 8. And so on for 3 and 7, and 4 and 6.[/hide]

Seriously, I'm not going to say "you copied this from him/her and you're exclude." That's not what whole thing's about.

You see the problem and you solve it. And let's say you didn't get it but made an approach to it. You'll get wrong or maybe not answer it but still, the idea that you tried to solve the problem is what that matters.

And if anyone copies by word by word,  it wouldn't be hard for me to catch it.  :lol:

Honesty!

[hide]
For any number that is divisble by 10, the remainder x is 0leqxleq9. If you have a number x = 4 (mod 10) and y = 6 (mod 10), then x+y = 0 (mod 10), which we don't want to have. That means that the set can't have numbers that are 1 (mod 10) and 9 (mod 10), or 2 (mod 10) and 8 (mod 10) and so on. The most mods you can have that fit the decription abouve if 5. There are 100 number, 5 numbers in set Q for every 10 number, making set Q have 50 numbers, or (d).[/hide]

My attempt:
[hide]The following pairs of units digits will cause the sum to end in 0 and thus be divisible by 10:
0 0
1 9
2 8
3 7
4 6
5 5
This means that there can be at most one number that ends with 0 in Q, since if there are two such numbers they will add up to a multiple of ten. Similarly, there can be at most one number that ends with 5. Also, in the table above, only one of the units digits from each row can be present. There are 4 rows not counting the special cases 0 0 and 5 5 that we already dealt with. There are 10 numbers with any particular units digit in {1, 2, 3,...100}, so that's 4*10=40 numbers. Add the two numbers ending in 0 and 5, so the answer is A) 42 numbers in Q.[/hide]

[hide="My solution"]We can consider the list (mod 10) for the purposes of addition; the set becomes one with 10 1's, 10 2's, 10 3's, ... and 10 0's.  The set can never have both a 1 and a 9, a 2 and an 8, a 3 and a 7, a 4 and a 6, two 5's, or two 0's.  To maximize the number of elements, we can just take all the 1's (or all the 9's), all the 2's (or all the 8's), all the 3's (you get the idea), all the 4's, one 5, and one 0.  This maximizes the number of elements, so the answer is 10+10+10+10+1+1=42, so the answer is (a).[/hide]

Here be my solution:

[hide]Realize that to accomplish this, we can't include two numbers whos sum ends in 0. So, we can include all numbers ending 1 - 4.

That's 1 - 4, 11 - 14, 21 - 24, ... , 91 - 94. However, we can also include one 5 and one 10, since 10 + A, if A doesn't end in 0, isn't divisible by 10. 5 because 5 + B is never divisible by 10 if B doesn't end in 5. Thus, we have 4 * 10 + 2 = 42.

(A), would be the answer.[/hide]

My answer:
[hide]Answer: A. Solution: Just considering the numbers mod 10, it is clear that you can't have 2 numbers mod 10 that add up to each other.  So if you have a number x mod 10, you can't have any others which are 10-x mod 10.  Taking all the numbers 1, 2, 3, 4 mod 10 gives you 40 total.  Now you can't have any 6, 7, 8, 9 mod 10.  But you can have exactly 1 number mod 5 and 1 number 0 mod 10 since they will be added to another number and will no longer work.  This makes 42 total.[/hide]

Day 1's answer:

[hide="Correct_answer"]The answer is A. The reason is that we can first have:

1-11-21-31-41-51-61-71-81-91
2-12-22....
3-13-23...
4-14....

Which adds up to 40. And we can have numbers 0 and 5 or any two multiples of 5 that has different last number.

Adding them up, it's 42.[/hide]

Day 2 will be posted soon.

Day 2 Problem

(Very easy)

The number of digits in $4^8*5^{17}$ is:

A) 8
B) 10
C) 11
D) 17
E) 23

ANSWER IN SPOILER!

YAY. Fun problem. I have a feeling it was somewhere....

[hide]This is 2^16 * 5^16 * 5 = 16 0's.

Thus, 17 digits including the 5.[/hide]

Day 2 Answer

[hide="Correct_answer"]The correct answer is D. The reason is:

Notice that we can form 10 by 2*5, using that we can rewrite the orginal expression $4^8*5^{17}$ as $2^{16}*5^{17}$ which is equal to $10^{16}*5^1$. $10^{16}$ is total of 16 zeros, or digits and 5 is one digit.

Therefore, 16+1 or 17, which is D.
[/hide]

Day 3

This isn't actually an AMC question (neither the ones above) but I found it quite interesting.

Given that $\frac {3}{2} < x < \frac {5}{2}$, find the value of:

$\sqrt {x^2-2x+1} + \sqrt {x^2-6x+9}$

A) 1
B) 2
C) 2x-4
D) 4-2x
E) None of A,B,C, or D

You have:
[hide]
$\displaystyle \sqrt {x^2-2x+1} + \sqrt {x^2-6x+9} = \\ \displaystyle \sqrt{(x-1)^2} + \sqrt{(x-3)^2} = \\ (x-1) + (x-3) = 2x - 4$.

I'm tired, so I'm expecting this to be a trick question...[/hide]

Just curious, where are you getting these problems from? AMC10 or 12 or what?
1.[hide]
The probability of getting a red wrapper and a mint is the product of the chance of each of those happening.
So .8*x=.2
x=.25
so the answer is [b]C[/b], or 1/4.[/hide]
2.[hide]
It must be like this:
ABCDEFGH, A=H, B=G, C=F, and D=E.  Since there is 50% chance that will not happen for each of those equations, the answer is (1/2)^4=[b]1/16[/b].[/hide]
3.[hide]
a^2-a-1 must be either 1, or -1.  However, it can only -1 if a is even, because otherwise it would be equal to -1, not 1. (For example (-1)^3 is -1 because 3 is odd.)
If it equals 1:
a^2-a-2=
(a-2)(a+1)
a=2,-1
If it equals -1:
a^2-a
a(a-1)
a=0, 1
However 1 is not in the domain because it isn't even.
So the answer is [b]B[/b], there are 3 answers: 2, -1, or 0.
Edit: after thinking about this one, I concluded that I missed a choice.  If the exponent is 0, then a is -2, meaning there are actually [b]4[/b] choices.
[/hide][/b]

Hehe..

That is hard question to answer. :D

I get the questions from many competitions, including that ones you probably don't even know.

Just do some researches and do some stuffs..

Day 22

[hide="Correct_answer"]

Number one,

This one was rather obvious. Since there are 0.2 chance of being red and mint and the probability we have now is 0.8, since the wrapper is red. So, this is just:

0.2/0.8 = 0.25 or 1/4 C.

Number two,

The total combination is, since there are two choices per each, 2^8 = 256. To be COINODROME, you may change the first four but leave the last four so it's going to be COINODROME. Then, 2^4 = 16 and it's out of 256, meaning:

16/256 = 1/16

Number three,

There are several ways we can have 1 as result.

Case 1: 1^1

To do that, the $\alpha$ must be -1. Then outside equals ^1 and inside equals  1-(-1)-1 = 2-1 = 1 so the result is 1^1 as we expected. So, one solution is -1.

Case 2 : something^0

To do that, the outside must equal 0. That's when $\alpha$ = -2.

Case 3 : -1^2

Using the outside one, we get $\alpha$ = 0. This works so the another root is 0.

Case 4 : 1^some power

Different from the case 1, another case is 1^some power. To get the bottom part to the 1, we can set up the equation as:

$\alpha^2 - \alpha - 1 = 1$

And solve from there, we'll get 2.

Since that's only case we have, it's 4.

Notice that AMC gave a solution that's more than correct answer to trick you. :) [/hide]

I couldn't get really tough geometry problems. But I did get three so here it goes:

1. Sides AB, BC, CD, and DA of convex quadrilateral ABCD have lengths 3,4,12, and 13, respectively; and <CBA is a right angle. The area of the quadrilateral is:

A) 32
B) 36
C) 39
D) 42
E) 48

2. A square and a regular hexagon are drawn with the same side length. If the area of the square is $\sqrt 3$, what is the area of the hexagon?

(no choice given)

3. Four of the eight vertices of the cube are vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.

A) $\sqrt 2$
B) $\sqrt 3$
C) $\sqrt \frac {3}{2}$
D) $\frac {2}{\sqrt 3}$
E) 2

[hide="23.1"]
Since B is a right angle, AC is the hypotenuse of a 3-4-5, so it's 5., Then notice that AC=5, CD=12, AD=13 forms a 5-12-13 Right triangle. From there it's just (bh/2 for ABC)+(bh/2 for ACD). 3*4/2+5*12/2=6+30=36. (B)
[/hide]

[hide="23.3"]
Let the sides of the cube be 1. Then the area of the cube is 4*12=6. A regular tetrahedron can be made from four of the cube's vertices, by making them all of its vertices a distance of one face diagonal apart, or $\sqrt{2}$ Using the fact that there are four 30-60-90 triangles making up this tetrahedron, we can easily calculate the area of each triangle (face) to be $\sqrt{3}/2$ and then multiplying this by 4 gives us the total surface area, $2\sqrt{3}$.
$6/2\sqrt{3}=3/\sqrt{3}=\sqrt{3}$ (B)
[/hide]

[hide="23.2"]
If the area of the square is $\sqrt{3}$ then the length of each side is $3^{1/4}$. As given in the problem the length of each side of the hexagon is equal to that of the square. A regular hexagon can be cut into 6 equilateral triangles.  Again, using the 30-60-90 relation we can calculate the area of one triangle to be $3^{1/4}*3^{3/4}/4=3/4$. Multiply this by 6 and you get 9/2.  The area of the hexagon is $9/2$.
[/hide]

Edit: I can't multiply....

23.1
[hide]3*4/2 + 12*5/2 = 6 + 30 = 36 (B)[/hide]

23.2
[hide]area of hexagon =  6/4 s2 :sqrt: 3
s2 = :sqrt: 3

6/4 * 3 = 9/2[/hide]

23.3
[hide]let s = side of cube
surface area of cube = 6s2
side of tetrahedron = s :sqrt: 2
surface area of tetrahedron = 4/4(s :sqrt: 2)2 :sqrt:3
= 2 :sqrt: 3 * s2
6/2 :sqrt: 3 = :sqrt: 3 (B)[/hide]

1.[hide]Splitting the quad into 2 triangles, which both happen to be right triangles, then finding the area of those 2 right triangles = 6+30 = 36.[/hide]

2. [hide]Set the side=s.  Square area = $s^2=\sqrt{3}$.  Hexagon area = $\frac{1}{2}Perimeter*apothem=.5*6s*\frac{\sqrt{3}s}{2}=\frac{3\sqrt{3}}{2}s^2.$  Since $s^2=\sqrt{3}$, total area comes to 9/2=4.5.[/hide]

3.[hide]This one was fairly tricky.  SA of square=$6s^2$  Side of the tetrahedron=$\sqrt{2}s$.  Then A of one face of tetrahedron=$\frac{\sqrt{3}}{2}s^2$.  Total SA=4*that=$2\sqrt{3}s^2$.  Ratio is therefore 6/that=$\frac{3}{\sqrt{3}}=\sqrt{3}$[/hide]

You were right, these problems weren't very difficult.
1.[hide]
The quadrilateral can be split into two right triangles.  Since one is a 3-4-5, its area is 6.  The other is a 5-12-13, so the area of that is 30.  Combining them we get that the area is 30+6=36, so the answer is [b]B[/b].
[/hide]
2.[hide]
Since a hexagon is made of 6 equilateral triangles, the formula for its area is 3*(s^2* sqrt(3))/2.  S squared is root 3, so the area is [b]4.5[/b].[/hide]
3.[hide]
If it is a unit cube the surface area of the cube is 6.  In the tetrahedon, the lengh of each edge will be the same as the lengh of a diagonal of the cube, which is sqrt(2).  Since the surface area of the tetrahedron is 4 regular triangles with side length sqrt(2), we find that the surface area is 2*sqrt(3).  So the ratio is 6/(2*sqrt(3), which is equal to 3/sqrt(3), which is equal to sqrt(3), so the answer is [b]B.[/b][/hide]

Day 23

[hide="Correct_answer"]

Number one,

We can find the area of ABC using Pythagorean Theorem, yielding 3*4/2 = 6. Another triangle then has side 12,5,13 and by Heron's Formula, we get 30. 30+6 = 36 and that's the area of the quadrilateral.

Number two,

The formula for the hexagon's area is:

$\frac {3 s^2 \sqrt 3}{2}$

Since $s^2$ is given, just plug in and get 9/2.

Number three,

Suppose the side length equals 1. Then the surface area of the square is 6. Since the area of the tetrahedron is determined by the four of the equilateral triangle with side length of the cube's face diagonal, we can find the tetrahedron's area by:

$4 * \frac {(\sqrt 2)^2 * \sqrt 3}{4}$

And thus, the surface area of the tetrahedron is $2 \sqrt 3$. Then the ratio is:

$6/2 \sqrt 3 = 3/\sqrt 3 = \sqrt 3$

[/hide]

Today's only AIME. I can't think of any good questions now but since it's weekend, I'll have some better ones. But they're still good! :D

1. For how many $\xi$ is lcm($6^6,8^8,\xi$) = $12^{12}$

2. Find positive integer $\varphi$, $\delta$ such that $\varphi * \delta! = (((3!)!)!)/3!$ and $\delta$ is a large as possible.

3. Find the smallest positive integer $\kappa$ such that $10^{\kappa} = \Omega * \gamma$, where $\Omega$ and $\gamma$ are positive integers, then at least one of $\Omega$ and $\gamma$ must contain the digit 0.

I changed the variable into some fancy Greek letter because I think they're cool. :D

[quote="Silverfalcon"]

3. Find the smallest positive integer $\kappa$ such that $10^{\kappa} = \Omega * \omega$, where $\Omega$ and $\omega$ are positive integers, then at least one of $\Omega$ and $\omega$ must contain the digit 0.[/quote]

You used two omegas (capital and lowercase) in one problem--That's not cool. :)

Fine.

I just used them as variable. If they have other values, just ignore it for this case.
 ;)

24.1
[hide]66 = 26 * 36
88 = 224
1212 = 224 * 312
therefore to get 312 needs to be a factor in the lcm of /xi
the only other prime factors that /xi can have are powers of 2 ranging from 0 to 24
thus there are 25 possible /xi[/hide]

24.2
[hide]((3!)!)!/3! = 720!/3!
the highest factorial possible is 719
thus the desired expression is 719! * 120
answer = 120[/hide]

24.3
[hide]i think im misinterpreting it, but the lowest possible values of the two numbers satisfying the conditions are 10 and 1, yielding k = 1[/hide]

1.[hide]x must have 12 3's and any number of n's 0-24, therefore 25 possible values.[/hide]

2.[hide](6!)!/3! is the same as (6!-1)!*6*5*4, So those are the two variables.[/hide]

3.[hide]I might be misinterpretting this one.  If k=1, the other two values are 1 and 10.[/hide]

Announcement:

I'm really really sorry to tell all of you this but I'll be really busy from now on so I can't do AMC B day anymore.

I'm very very sorry to tell you this but I'll tell the answers to the yesterday's question.

Just answers

[hide]
1. 25
2. 719
3. 8 [/hide]

If I solve LOTS of problems, I might post a 15 question problem thread for AMC B day but sorry, I can't do this anymore since it's close to AMC!

Sincerely,

Silverfalcon


Let L, M is Fréchet spaces and $ u: L\rightarrow M$ is a linear homomorphism. Proof u(L) is complete. Help me. Thank

That's wrong. Take $ L \equal{} L^2(0,1),\,M \equal{} L^1(0,1)$ and $ u$ the canonical embedding for example. This one is even continuous.
Did you mean to write "linear hom[b]e[/b]omorphism"?


limit
logarithms
calculus
derivative
calculus computations

Evaluate $ \lim_{x\rightarrow 0}\frac{\ln (\cosh x)}{\sqrt[m]{\cosh x}\minus{}\sqrt[n]{\cosh x}}$

$ L\;\equal{}\;\lim_{x\to 0}\;\dfrac{\ln\,\left(1\plus{}\dfrac{x^2}{2}\plus{}\textbf O(x^4)\right)}{\left(1\plus{}\dfrac{x^2}{2}\plus{}\textbf O(x^4)\right)^{\dfrac{1}{m}}\;\minus{}\;\left(1\plus{}\dfrac{x^2}{2}\plus{}\textbf O(x^4)\right)^{\dfrac{1}{n}}}$

$ \equal{}\;\lim_{x\to 0}\;\dfrac{\dfrac{x^2}{2}\plus{}\textbf O(x^4)}{\left(1\plus{}\dfrac{x^2}{2m}\plus{}\textbf O(x^4)\right)\;\minus{}\;\left(1\plus{}\dfrac{x^2}{2n}\plus{}\textbf O(x^4)\right)}$

$ \equal{}\;\lim_{x\to 0}\;\dfrac{\dfrac{x^2}{2}\plus{}\textbf O(x^4)}{x^2\;\left(\dfrac{1}{2m}\;\minus{}\;\dfrac{1}{2n}\right)\plus{}\textbf O(x^4)}$

$ \equal{}\;\lim_{x\to 0}\;\dfrac{\dfrac{1}{2}\plus{}\textbf O(x^2)}{\dfrac{1}{2}\left(\dfrac{1}{m}\;\minus{}\;\dfrac{1}{n}\right)\plus{}\textbf O(x^2)}$

$ \equal{}\;\dfrac{mn}{n\minus{}m}$

How did you find the expansion of $ \cosh x$

Generally its a well known expansion, and quite easy to derive either from its exponential definition and using the exponential's series, or just computing the taylor series directly, which is also easy since the n-th derivative of cosh x is either sinh x or cosh x :)


[hide="Hyper hint"] Let the sum be $S$ (this is a useful tool for a lot of summations).  What is $S-\frac{S}{2}$?  [/hide]

It was!!!

And yeah, I like questions like that.

Can you always use this for any recursion like, say that

$a_{n}=a_{n-1}+2a_{n-2}$, and $a_{1}=1$ and $a_{2}=1$

Find the sum of $\frac{a_{1}}{2^{1}}+\frac{a_{2}}{2^{2}}+...$.

You can also find $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{F_{n}}{2^{n}}$ , where $F_{n}$ is the n-th fibonacci number with $F_{1}=1, f_{2}=1$, by decomposing $F_{n}$ into $F_{n-1}+F_{n-2}$.  Then all you have to do is have $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{F_{n}}{2^{n}}$ as terms in a linear equation.  Then solve to get $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{F_{n}}{2^{n}}=2$.

As for mathg...'s problem, you cannot always use the method.  Your sum is equal to [EDIT:approaching] $\infty$.  If you use my method above, you will get $x=0$ where $x\neq 0$.  Contradiction.

Why wouldn't it be possible to say that it's an infinite geometric series who's common ratio approaches $\frac{\phi}{2}$?

It cannot be an infinite geometric series when it's common ratio is [i]approaching[/i] a certain value.  It's a geometric series when you have a fixed common ratio.
${\lim_{n\rightarrow\infty}\frac{\frac{F_{n+1}}{2^{n+1}}}{\frac{F_{n}}{2^{n}}}=\frac{1}{2}\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_{n}}}=\frac{\phi}{2}$

Well you could do out the first few terms and then approximate it with an infinite geometric series, though I have no idea why you'd want to do that...

If you [i]really[/i] want to use a geometric series, you could always rewrite it as:

$\frac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{4^{n}}$

$=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{3-\sqrt{5}}-\frac{1-\sqrt{5}}{3+\sqrt{5}}\right)=\frac{1}{\sqrt{5}}(2\sqrt{5})=2$

[quote="seamusoboyle"]Dont forget to prove it converges... though this is trivial convergence of sequences and sum is incorperated into a lot of higher level maths. So ive been retrained to care about convergence.[/quote]

see A M Y 's post...it is obvious both are ratios less than 1...

Ah yeah i know its trivial, but just to point it out for people who might not know or just forget.


I can't seem to get the answer on this problem.

6(x+4)

I am haveing trouble with it and if someone would be kind enough to answer it and give the explination I would be grateful.

um..

is this a joke or did you forget to give the question :)

What is the question? 6(x+4) isn't a question.

Yes it is or so says my math book.  It may not be I'll ask my teacher tomarow.

Thanks anyway.

No I don't want that though you got the right answer rep.  The answer is 6x+24.

Whatever they wanted me to do you gave me the right answer.

Thanks I am going to puzzle over it some more though.

Perhaps there's the equation "x=blah" somewhere on thne text?

No.  The problem is simply stated as 6(x+4).

If this is a text-book question, there's usually a sequence of questions that you are supposed to answer in the same way, and the way you're supposed to answer them is usually at the beginning of the problems.  Somewhere, it has to say what you want.

You know, not all questions end up with a number...

You'll never get a number, you could substitute a number for x and get a number, or even graph it as y=6(x+4) and play around with it but 6(x+4) itself can never be a number.

Well its always a number but yeah :D never a single one.

BHorseMath - assuming that 6(x+4) wasn't on a page by itself, what was before it?

It must be something like, distribute, or solve if x is equal to 8 or something. Unless, this is a trick question and you really wanted us to just post for no reason.

Before this problem it had ...........Oh use distributive property to remove the parentheses.

I could of helped you. She never comes to me always to yall first. She thinks Im stupid I guess but Im not believe it or not. I know alot of this math Im just not into math at the moment thanks. Ive had yet another tragic day I hate b-days!!!!!!

I didn't come to you aalindsey because you wouldn't talk to me.

I appriciate any help that you people have given me.


MATHCOUNTS email Gauss calculus AMC USA(J)MO USAMO

Are you guys going to someday apply to IMSA?

I am!!!!

Btw I'm taking the SAT in January. Have u taken the SAT suma? If so did you find it easy? Hard?

I haven't taken the SAT yet.  I'm not sure if I should apply this year b/c don't you have to have a year of high school math and science.  Math isn't a problem, but science sort of is.  I'm tested out of science at school but we just do more advanced projects that still pertain to 8th grade curriculum.

I'm not sure what you have to do about the science part....but if that's truly required wouldn't that prevent 8th graders from getting accepted... :maybe:

I don't know.. hopefully it'll be ok with the science activities I do.  I'm going to apply this year but if I get in I'll probably postpone attending until 9th grade b/c I don't want to skip a grade.

Wow I was just going to make a post about this, but you beat me to it :p. I am going to submit an application. I'm not really confident that I'll make it though.

I thought there was only a reading and math component to the SAT? Do we have to take the subject tests too for the science?

I think you only have to take the SAT.  (btw I've decided to apply this year and I'm taking it in March)  Also sunehra you live in Illinois?

maybe was he a mathcounts participant too?

I'm back in the forum! lol
I was in disneyworld for vacation :D

Well I got accepted last year...and I was in 8th grade.  Kennedy provided high school physics course in 8th grade, so that's how I fulfilled the science part.

And for the SAT you just take the normal test (English, math, writing).  They look at the math part closely, but the other parts are also looked at.  You guys shouldn't have problems with the math section, it's extremely easy, just don't make stupid mistakes. =P

And they also look at your math, sciecne awards, as well as extracurricular awards...I finished all of it in one day including the essays.  It's not hard, it just takes time I guess.

Yeah I just got an sat powerpack.  Here's one of the questions it says is typical:

$ a^2 \equal{} \sqrt{b}$  If a = -8 what is b?
a) -8
b) $ 4\sqrt{2}$
c) 8
d) 64
e) 4096

It's obviously E.  I can't believe it's multiple choice too! :o

Do you think there is a chance of an 8th grader getting in without the year of high school science ManHatTaNatiVe94?

@ guass : I'm pretty sure the proficiency tests are what you take if you get accepted.

.... :D

[quote="suma_milli"]Yeah I just got an sat powerpack.  Here's one of the questions it says is typical:

$ a^2 \equal{} \sqrt {b}$  If a = -8 what is b?
a) -8
b) $ 4\sqrt {2}$
c) 8
d) 64
e) 4096

It's obviously E.  I can't believe it's multiple choice too! :o

Do you think there is a chance of an 8th grader getting in without the year of high school science ManHatTaNatiVe94?

@ guass : I'm pretty sure the proficiency tests are what you take if you get accepted.

.... :D[/quote]
physics/chemistry=HS science.

And that's what we're doing in science now.

 :D  :D  :D

Does newton's laws and the other basic physics count as high school? B/c that's something we've done. :P

I think I'll ask my mom to call IMSA later and ask about the high school science thing.

yeah, i believe HS studies newton's laws?

And why do all of us in this thread have  :D other than ManHatTanNatIve?

Sunehra doesn't either.

Maybe it's because all of the posts except for two belong to us. :P

[quote="mathcountsloser"]The SAT's / ACT's works like this for talented children:

4-7th grade = competition
8-9th = something to show off to the teachers
10+ = a waste of time

For the average children:

4-7th grade = idk what SAT and ACT are
8-9th = I'll take it later
10+ = Gotta prepare.[/quote]

I think that for the first part you probably mean children who are exposed to standardized testing and for the second part you probably mean children who are not. It's not that the first group is talented and the second group is average.

Kids who are bent on the college preparation track often take the SAT/ACT early for practice and because standardized testing is necessary for entrance into elite boarding schools. Again, the SAT/ACT may be a relatively weak predictor of intelligence, but taking the exams is not a waste of time.

People who complain about standardized testing are often just whining about the exams. I think that those who can do well on it should not complain much, and those who can't do well on it should study instead of complaining.

Just my two cents there.

[quote="JamesZhou13"]on the principal evaluation form, it says:
Is this candidate enrolled in honors courses and/or AP courses?
Will this candidate run out of science courses prior to the senior year at your area high school?
Will this candidate run out of mathematics courses prior to the senior year at your area high
school?

so will it be better for your application IF you will run out of math and science courses?[/quote]

generally yes, because it shows you "need" IMSA, but this depends on if you're good enough to meet IMSA's desires.

but seriously, don't fill out your application based on how it looks. fill out your application honestly.

FYI: it's usually a bad idea to post sensitive questions about your application on a public forum if your username clearly tells who you are. This question you're posting makes it seem like you're very desperate, and an admissions counselor who could be prowling the AoPS forum could see your question and realize that he/she might have to take your application with a grain of salt because you're filling it out for the sake of looking good and not necessarily honestly.

*Pay?

Yes, there's a $50 (?) application fee required to submit an application to be considered.

[quote="Aryth"][quote="mathcountsloser"]Darn, I wanted to apply for IMSA and when I get accepted, say no.[/quote]

Why would you do that? Also, why do you assume you'd get accepted?[/quote]

Yeah, I'm confused as to why they'd accept a "mathcounts loser" :maybe:

and someone who spells signature incorrectly in his "signiture". . .

and someone who thinks they're good at math because they know pigeon hole OMG SO PRO.

oh god i trolled

[quote="mathcountsloser"]
 Prove that if the length of the line segment is x, then the distance between any 2 points is less than or equal to x/6.[/quote]

cough fail please

[quote="resurrection"]and someone who spells signature incorrectly in his "signiture". . .

and someone who thinks they're good at math because they know pigeon hole OMG SO PRO.

[b]oh god i trolled[/b][/quote]

I think you're doing it wrong.

i tried but it didn't work. also i applied to imsa as well. i got rejected.

hmmmmmmmmmmmmmmmmmmmmm

Let's all be nice.

Also, I suggest everyone listen to my infense song. Here, i'll put it in a hide tag.
[hide]infense infense infense infense
so many fenses, so many infenses,
infense infense infense infense
so many fenses so many infenses

why so much infension
for this i should get pension
this song a brilliant invention
a geneva convention

infense infense infense infense
so many fenses, so many infenses,
infense infense infense infense
so many fenses so many infenses

please, please, pay attention
to this amazing infension
noted is your dissension
you of the third dimension

infense infense infense infense
so many fenses, so many infenses,
infense infense infense infense
so many fenses so many infenses

nothing can surpass prevention
nothing of least pretension
fear my comprehension
get starved by condescension

infense infense infense infense
so many fenses, so many infenses,
infense infense infense infense
so many fenses so many infenses

finally, at least, my reinvention
beware of your hypertension
you shall float in suspension
calling your telephone extension

infense infense infense infense
so many fenses, so many infenses,
infense infense infense infense
so many fenses so many infenses

center of attention, rightless ascension,
misapprehension, striped gentian,
american baptist convention
honored you are with a mention

infense infense infense infense
so many fenses, so many infenses,
infense infense infense infense
so many fenses so many infenses[/hide]


[color=green]How many distinct points does the 9-point circle have when the triangle is
(a) isosceles
(b) equilateral
(c) right-angled isosceles

Thanks very much.....[/color]

What do you mean by distinct points?
Do you mean how many distinct points in the 9 points?

I don't know what it means, actually this is the last assignment for this geometry course, but the instructor hasn't covered this topic yet, the other students are quite frustrated.


geometry
rectangle
analytic geometry
linear algebra
matrix
ratio
trigonometry

In rectangle $ ABCD$, points $ F$ and $ G$ lie on $ \overline{AB}$ so that $ AF \equal{} FG \equal{} GB$ and $ E$ is the midpoint of $ \overline{DC}$. Also, $ \overline{AC}$ intersects $ \overline{EF}$ at $ H$ and $ \overline{EG}$ at $ J$. The area of the rectangle $ ABCD$ is $ 70$. Find the area of triangle $ EHJ$.
[asy]
size(180);
pair A, B, C, D, E, F, G, H, J;
A = origin;
real length = 6;
real width = 3.5;
B = length*dir(0);
C = (length, width);
D = width*dir(90);
F = length/3*dir(0);
G = 2*length/3*dir(0);
E = (length/2, width);
H = extension(A, C, E, F);
J = extension(A, C, E, G);

draw(A--B--C--D--cycle);
draw(G--E--F);
draw(A--C);

label("$A$", A, dir(180));
label("$D$", D, dir(180));
label("$B$", B, dir(0));
label("$C$", C, dir(0));
label("$F$", F, dir(270));
label("$E$", E, dir(90));
label("$G$", G, dir(270));
label("$H$", H, dir(140));
label("$J$", J, dir(340));
[/asy]
$ \displaystyle \textbf{(A)} \ \frac {5}{2} \qquad \textbf{(B)} \ \frac {35}{12} \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ \frac {7}{2} \qquad \textbf{(E)} \ \frac {35}{8}$

[hide]

Let's take a coordinate geometry approach. For ease of numbers, I let  and , so then . Then label the points , , , , , , . We wish to find the coordinates of  and . To do this, we find the lines  (),  (), and  (), and find the intersections of  and the other two to be  and . Using the Shoelace Theorem on , we find the area to be , so the answer is C.[/hide]

Not that nice of a solution, but it got what we needed and was fairly quick.

[quote="agolsme"]What is the shoelace theorem?[/quote]

You take the determinant of the matrix that includes all the points of the vertices of a polygon in counterclockwise order.

[quote="The Master"][quote="Iversonfan2005"]hint
[hide]So therefore, the area of $\triangle{AGJ}=\frac{40}{3}$ and the area of $\triangle{EJC}=\frac{15}{2}$ and we know the area of $\triangle{FGE}=\frac{35}{3}$[/hide][/quote]I don't get it. How do you come up with those conclusions from what you previously stated?[/quote]

Yeah, I realize that that way really isn't the best way to do it, I think assigning coordinates is a lot better as shown by elennop

So basically it's what they teach in Geometry as the "Surveyer's Method", right?

I think another way to do it would be to find the heights to H and J from DC using similar triangles.  H turns out to be 3/5 BC and J turns out to be 3/7 BC.  Then the desired area can be found by subtracting Area(EHC)-Area(EJC) which is easy since it is in terms of CD*BC, and we know CD*BC=70.

[hide="Similarity"]Note that triangles AJG and CJE are similar (side ratio of 4/3), as are triangles AHF and CHE (side ratio of 2/3).

Thus, $ JE = \frac 37 GE$ and $ HE = \frac 35 FE,$ and so
\[ \begin{eqnarray*}
 [EHJ] &=& \frac35 \cdot \frac 37 \cdot [EFG] \\
&=& \frac{9}{35} \cdot \frac{70}{6} = 3.
 \]
(We used the fact the area of a triangle is $ \frac12 ab \sin C.$)
[/hide]

[quote="Narcissa"]umm wow, i did it with another formula (heron's formula), which only works if (1) you don't make careless mistakes at all or (2) you get a calculator. which you won't be getting starting now! so do u think the 2008 test will have these types of problems??

i dunno how to do the "hidden content" thing so ill just say that I found teh coordinates as someone above me did, then I found the length of each of side, and then used the sqrtS(s-A)(s-B)(s-c) formula, and it came out to be 3. but honestly, there's no way i would have that much time / patience / skill to do it without a calculator! do they expect us to??[/quote]

Well once you found the coordinates, a way to avoid messy calculations is to use the shoelace theorem instead of heron's.

[quote="Narcissa"]umm wow, i did it with another formula (heron's formula), which only works if (1) you don't make careless mistakes at all or (2) you get a calculator. which you won't be getting starting now! so do u think the 2008 test will have these types of problems??

i dunno how to do the "hidden content" thing so ill just say that I found teh coordinates as someone above me did, then I found the length of each of side, and then used the sqrtS(s-A)(s-B)(s-c) formula, and it came out to be 3. but honestly, there's no way i would have that much time / patience / skill to do it without a calculator! do they expect us to??[/quote]
Coordinate bashing with shoelace prevents you from having to deal with side of radical length, instead of Heron's which makes you add and multiply them.

the similar triangle argument is more straightforward in this case, i think.

Let $ AB \equal{} x$, $ AD \equal{} \frac {70}{x}$. Then $ [EFG] \equal{} \frac {35}{3}$
Let DF intersect AC at I. By Menelaus on $ \triangle CID$,
$ \frac {1}{2} \cdot \frac {DI}{IF} \cdot \frac {FH}{HE} \equal{} 1$
Note that $ \triangle CID \sim \triangle DIA \sim \triangle AIF$, so $ \frac {DI}{IF} \equal{} \frac {CD}{AF} \equal{} 3$.
Then $ \frac {FH}{HE} \equal{} \frac {2}{3}$.
By Menelaus on $ \triangle AJG$, $ \frac {1}{2} \cdot \frac {GJ}{JE} \cdot \frac {EH}{HF} \equal{} 1$.
From this we find $ \frac {GJ}{JE} \equal{} \frac {4}{3}$
So $ [EHJ] \equal{} \frac {EH}{EF} \cdot \frac {EJ}{EG} \cdot [EFG] \equal{} \frac {3}{5} \cdot \frac {3}{7} \cdot \frac {35}{3} \equal{} 3$.
This took a lot of work, but almost no thinking for me. Of course, the previous similar triangles solution is best. Is it worth coming up with a solution like this if you see it right away?

A diagram is supposed to be included with this problem in the contest. Here it is:
[asy]
size(200);
pair A, B, C, D, E, F, G, H, J;
A = origin;
real length = 6;
real width = 4;
B = length*dir(0);
C = (length, width);
D = width*dir(90);
F = length/3*dir(0);
G = 2*length/3*dir(0);
E = (length/2, width);
H = extension(A, C, E, F);
J = extension(A, C, E, G);

draw(A--B--C--D--cycle);
draw(G--E--F);
draw(A--C);

label("$A$", A, dir(180));
label("$D$", D, dir(180));
label("$B$", B, dir(0));
label("$C$", C, dir(0));
label("$F$", F, dir(270));
label("$E$", E, dir(90));
label("$G$", G, dir(270));
label("$H$", H, dir(140));
label("$J$", J, dir(340));
[/asy]

[quote=Iversonfan2005]hint
[hide]We see that $\triangle{AGJ}$ is similar to $\triangle{EJC}$ and it's side/altitude is in a $4: 3$ ratio. So therefore, the area of $\triangle{AGJ}=\frac{40}{3}$ and the area of $\triangle{EJC}=\frac{15}{2}$ and we know the area of $\triangle{FGE}=\frac{35}{3}$ From there, it's just a matter of finding out the area that's been overcounted...[/hide][/quote]

I don't see how you find those areas from the information in the given problem. Since something happened to Post #3 (everything is blank), I'll [hide=sketch the analytic sol]WLOG AD = 7, AB = 10. Then, we are able to find the equations of the lines EG and EF and AC. Then, we can get the coordinates of the desired triangle using intersections to then use Shoelace to kill it.[/hide] it out.

EH:HF = EC:AF =3:2, so EH:EF = 3:5; As the same, EJ: EG= 3:7
S(EHJ)/S（EFG）=(EH X EJ)/(EF X EG)= 9/35
S(EFG)=1/6 X 70
so S(EFG)= 1/6 X 70 X 9/35= 3

[quote=Beeinthecar.EQ]Anyone try the analytical geometry method? It seems to be the easiest way out of this problem.[/quote]

wouldn't similar triangles be the easiest method?

AHF~CHE, AJG~CJE, then do some altitude and area finding

[hide=Lotus ratio lemma!] So first of all we Join $\overline{CG} ,\overline{AE},\overline{HG}$\\

mark $[ABCD]=A$\\

notice $[CGB]=\frac{A}{6}\implies [CAG]=\frac{A}{3}$\\

Also , $\triangle{HEC}\sim \triangle{HFA}$\\

so $HC:HA=3:2\implies [HGC]:[HAG]=3:2$\\

so $[HAG]=\frac{2A}{15}\implies [HAF]=[HFG]=\frac{A}{15}$\\

also $HE:HF=3:2\implies [HEG]=[HEA]=\frac{A}{10}$\\

so we get $JE:JG=[AEH]:[AHG]=3:4$\\

so from Lotus ratio lemma we get $AH:HJ=7:3$\\

So $[EHJ]=\frac{3A}{70}\implies [EHJ]=\frac{3\cdot 70}{70} \implies [EHJ]=\boxed{3 (C) }$[/hide]

[hide=The lemma][url]https://artofproblemsolving.com/community/c6h509770p25873315[/url][/hide]


geometry
parallelogram
analytic geometry
ratio
transformation
Hard Problems
hard

The parallelogram bounded by the lines $ y \equal{} ax \plus{} c,y \equal{} ax \plus{} d,y \equal{} bx \plus{} c$ and $ y \equal{} bx \plus{} d$ has area $ 18$. The parallelogram bounded by the lines $ y \equal{} ax \plus{} c,y \equal{} ax \minus{} d,y \equal{} bx \plus{} c,$ and $ y \equal{} bx \minus{} d$ has area $ 72.$ Given that $ a,b,c,$ and $ d$ are positive integers, what is the smallest possible value of $ a \plus{} b \plus{} c \plus{} d$?

$ \textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

The complete solution was already posted somewhere... multiple times by different people

Can anybody help to do this? I really wanna know.

[url=https://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12B_Problems/Problem_20]Isn't there some sort of wiki with these problems on it?[/url]

It looks like the wiki solution is wrong, the official answer key says that it is D, while the wiki says C.
I used shoestring to find the area and also got D, but it's not that elegant. Is there a better solution elsewhere?

So you might as well just look at the wiki solution like other people said, but this is how I got it (pretty much the same as the wiki, but since people asked... :lol: )

[hide]

You know that 2 of the opposite corners of the parallelogram lie on the y-axis at (c, 0) and (d, 0) because of the y-intercepts.  The parallelogram can be split into 2 equal trangles.  Each has a base of c - d, and a height of (d - c) / ( a - b), which is taken from either one of the x-coordinates of the intersection points (corners of the parallelogram) that do not lie on they-axis.  So, (c - d)^2 / (a - b) = 18.

The same can be done for the 2nd parallelogram that has an area of 72 by using c + d as the base of the two triangles and finding the heights through the same method.  Therefore, (c + d)^2 / (a - b) = 72.

Combining the two equations gives you c = 3d.

Sub c = 3d back into the first equation, simplify, and you'll get 2d^2 = 9 (a - b).  Therefore, the lowest possible value for a - b is 2, as they're all positive integers.  From this equation, you can see that the lowest possible values for a, b, and d are 3, 1, and 3, respectively, and c = 9 (from the previous equation).

Adding all of the values up gives you 16, or choice (D).

[/hide]

[quote=w squared parabola]My solution is sort of complex, but here it is...
[hide]
The two parallelograms have vertices at $(0,d)$, $(0,c)$ and $(0,-d)$, so the diagonal of the smaller parallelogram is $d-c$ and the diagonal of the larger parallelogram is $d+c$. Since the ratio of areas is $\frac{18}{72}=2^{2}$, so that means $2(d-c)=c+d$, solve we have $c=3d$. Now if we look at the parallelograms, the diagonals divided them into 2 congruent triangles. So the area of 2 triangles in parallelogram with area of $72$ is $36$. The height of the triangle (with diagonal as a base) is $\frac{18}{c}$. Note this is also the x value for the intersection point of line $bx+c$ and $ax-d$. Plug everything in, we have $\frac{18b}{c} + 4c=\frac{18a}{c}$. Simplify that into $18(b-a)=-4c^{2}$ and into $9(b-a)=-2c^{2}$. Now if everything need to be smallest, $c^{2}=9$, $c=3$. $b=1$, $a=3$, and $d=3c=9$. The answer is therefore $1+3+3+9=16$ or $\boxed{D}$.

Note $a$ and $b$ are interchangeable.

[/hide][/quote]


Given the real numbers $a_1,a_2,\cdots ,a_n$ all of which are not always $0,$ such that $a_1\leq a_2\leq \cdots \leq a_n$  and $a_1+a_2+\cdots +a_n=0.$
Prove that $a_1+2a_2+\cdots +na_n>0.$

Why? It's very easy...

Notice that $S_k=a_k+a_{k+1}+...+a_n \ge 0$

Then $\sum_{k=1,n}S_k \ge 0$

The equal can't hold, because if that $S_k=0$ for all $k=1,n$, then a_k=0 for all $k=1,n$.

[quote="hungkhtn"]Notice that $S_k=a_k+a_{k+1}+...+a_n \ge 0$

Then $\sum_{k=1,n}S_k \ge 0$[/quote]

Why is that inequality so "trivial"? :)

Indeed, it is very easy. Note that $1 <2 <3 <... <n.$ From Chebyshef, the inequality is obvious. ;)

[quote="Arne"]I know, but I don't see why hungkhtn introduced the partial sums $S_k$.[/quote]

   If $a_k >0$ then obviously $S_k >0$
  Consider : $a_k <0$ .So $a_1+...+a_{k-1} <0$
 But $a_1+a_2+...+a_n =0$ , this infers $S_k >0$


function
logarithms
group theory
abstract algebra
calculus
integration
algebra

A very beautiful problem:
  Suppose that $a_n$ is a decreasing sequence of positive numbers which converges to 0. Consider the family of functions $ f_u(x)=\sum_{u\leq n\leq ux}{a_n}$ for $x>1$. Prove that $ na_n$ converges if and only if the function $ f_u(x)$ has a finite limit at infinity for all $x>1$.

You mean the limit when $x$ is fixed and $u\to\infty$?

There was a paper in AMM about this one, but I don't remember the issue

it is related to $\ln$

Yes, it is not an easy problem. But it doesn't surprise me anymore, after seeing the oral problems this year: anticommutation, finite subgroups of $GL(Q)$,..., nice but difficult problems.

I'll change it a bit, because I find it cleaner to speak of integrals from $u$ to $xu$ of some measurable function $f$ (which can be taken to be $a_n$ on $[n,n+1)$, for instance) decreasing to $0$ at $\infty$ than of sums of the form $\sum_{u\le n\le xu}a_n$.

We know that for all $x>1,\ \int^{xu}_u fdt\to \ell_x$ as $u\to\infty$. It's pretty clear now that $\ell_{xy}=\ell_x+\ell_y$, and $\ell:(1,\infty)\to[0,\infty)$ is an increasing function. The solution to this Cauchy-like functional equation is well-known to be of the form $\alpha\ln x$ for some $\alpha\ge 0$. We may assume for simplicity that $\alpha=1$ (we can deal with the case $\alpha=0$ separately).

We thus know that $\int^{xu}_ufdt\to \ln x,\ \forall x>1$. Fix some $x>1$ and an $\varepsilon_x>0$. For large $u$'s, we have $(x-1)uf(u)\ge \int^{xu}_ufdt\le \ln x-\varepsilon_x$. This means that for large $u$, we have $uf(u)\ge \frac{\ln x-\varepsilon_x}{x-1}$. By making $x\to 1$ and choosing $\varepsilon_x$ s.t. $\frac{\varepsilon_x}{x-1}\to 0$, we find that $\liminf_{u\to\infty}uf(u)\ge 1$. We can do something similar for the $\limsup$ to conclude that $uf(u)\to 1$.

If $\alpha=0$ (see above; this means that $\int^{xu}_ufdt\to 0,\ \forall x>1$), it's even easier to conclude that $uf(u)\to 0$.
I hope it's correct.


search
factorial
number theory proposed
number theory

Let $a \in \mathbb{N}$ and $ Z_1 (a) $ be the Number of 1-Digits, $ Z_2 (a) $ the Number of 2-Digits, ..., $Z_9 (a)$ the Number of 9-Digits the decimal number $a$ consists of.
Let $p,q,r,s,t,u,v,w,x,y \in \mathbb{N}$ and $\mathbb{F}$ be the Set of all positive Integers $a$ with

$ Z_1(a) = 3^{222p} + 22 \cdot 633^{qrs} $
$ Z_2(a) = 239 \cdot 912 ^ {rs + 23} $
$ Z_3(a) = 944 ^ t $
$ Z_4(a) = 111 ^ t + 3 \cdot u $
$ Z_5(a) = Z_4(a) + 912 \cdot Z_3(a) $
$ Z_6(a) = v $
$ Z_7(a) = 2^w $
$ Z_8(a) = 18 ^ {227 \cdot wxyz} $
$ Z_9(a) = 3 \cdot Z_6(a) + Z_2(a) $

Now let $\mathbb{G} := \{ n\in \mathbb{N} |  n > 6 \land n! \in \mathbb{F} \}$

Decide wether $\mathbb{G}$ is infinite or, if it is finite, give all Elements.

I do not understand what does the expression $\mathbb{G}: = \{n \in \mathbb{N} | n>6 \land n! \in \mathbb{F} \}$ mean, can you explain it to me?

It's the set of all positive integers greater than $6$ such that their factorials fulfill all nine equations

Not exactly. Actually it is the set of all integers $n$ with $n!$ satisfying all the 9 equations. $n$ itself does not have to satisfy them.

Very nice problem! :)
Have you already submitted it to the problem selection committee of the DeMO?
Otherwise, I would like to do so ...

[quote="Schoppenhauer"]Let $a \in \mathbb{N}$ and $Z_1 (a)$ be the Number of 1-Digits, $Z_2 (a)$ the Number of 2-Digits, ..., $Z_9 (a)$ the Number of 9-Digits the decimal number $a$ consists of.
Let $p,q,r,s,t,u,v,w,x,y \in \mathbb{N}$ and $\mathbb{F}$ be the Set of all positive Integers $a$ with

$Z_1(a) = 3^{222p} + 22 \cdot 633^{qrs}$
$Z_2(a) = 239 \cdot 912 ^ {rs + 23}$
$Z_3(a) = 944 ^ t$
$Z_4(a) = 111 ^ t + 3 \cdot u$
$Z_5(a) = Z_4(a) + 912 \cdot Z_3(a)$
$Z_6(a) = v$
$Z_7(a) = 2^w$
$Z_8(a) = 18 ^ {227 \cdot wxyz}$
$Z_9(a) = 3 \cdot Z_6(a) + Z_2(a)$

Now let $\mathbb{G} : = \{ n\in \mathbb{N} | n > 6 \land n! \in \mathbb{F} \}$

Decide wether $\mathbb{G}$ is infinite or, if it is finite, give all Elements.[/quote]

:)

Wait,can you explain $Z_i(a)$ please,I have some trouble to understand.


Find all rationals a, b, c, d such that

(a + b:sqrt:5)4 + (c + d:sqrt:5)4 = 2 + :sqrt:5.

Quite simple ... but nice :)

Nice, yes, Quite simple ... mmmmm

I must admit having spent some time darkening a few pages of stupid equations  :evil: until I remember about conjugate  :mrgreen: !!!!!!

[hide]

The conjugate of x = a+b:sqrt:n is c(x) = a-b:sqrt:n

Just like cxe numbers, c(x+y) = c(x)+c(y) AND c(x*y) = c(x)*c(y) and so on.

So

(a + b:sqrt:5):^4: + (c + d:sqrt:5):^4: = 2 + :sqrt:5.

is equivalent to

(a - b:sqrt:5):^4: + (c - d:sqrt:5):^4: = 2 - :sqrt:5 < 0

So there is NO solution to the equation   [/hide]

Hehehe ... :twisted:
When seeing something like a + b:sqrt:n don't forget about conjugates !
Have also a look at "Arne's problem nr. 4". It uses the same technique.

one of you mind explaining the solution??...im not sure i follow his work...

MithsApprentice:

This works in general:

[hide]If (a+b:rt:5)^n = p+q:rt:5 then (a-b:rt:5)^n = p-q:rt:5. This is because all the bits without a square root of 5 must have an even power of the b:rt:5 term and thus whether its plus or minus won't matter. For the :sqrt:5 terms, it has an odd power and thus positive and negative will be interchanged.

Of course it works with other numbers instead of 5.. (but not squares and stuff like that).

[/hide]


Two circles $ S_{1}$ and $ S_{2}$ touch each other externally at $ K$; they also touch a circle $ S$ internally at $ A_{1}$ and $ A_{2}$ respectively. Let $ P$ be one point of intersection of $ S$ with the common tangent to $ S_{1}$ and $ S_{2}$ at $ K$. The line $ PA_{1}$ meets $ S_{1}$ again at $ B_{1}$ and $ PA_{2}$ meets $ S_{2}$ again at $ B_{2}$. Prove that $ B_{1}B_{2}$ is a common tangent to $ S_{1}$ and $ S_{2}$.

Such an easy problem  :wink:

Inversion with center $P$ and power equal to the power of $P$ to $S_1$ and $S_2$ takes $S_1,S_2$ into themselves and takes the circle $S$ into a line $\tau$ passing though the inverses $B_1,B_2$ of $A_1,A_2,$ respectively. Due to conformity, if $ S$ is tangent to $S_1,S_2$ through $A_1,A_2,$ then $\tau \equiv B_1B_2$ is tangent to $S_1,S_2$ through $B_1,B_2.$


modular arithmetic
number theory
relatively prime
Diophantine Equations

Let $a, b, c$ be positive integers such that $a$ and $b$ are relatively prime and $c$ is relatively prime either to $a$ or $b$. Prove that there exist infinitely many triples $(x, y, z)$ of distinct positive integers such that \[x^{a}+y^{b}= z^{c}.\]

Suppose $ (a,c)\equal{}1$. We'll look for solutions with $ y\equal{}y_0^a$ and $ x\equal{}y_0^b$, so that $ x^a\equal{}y_0^{ab}\equal{}y^b$. Then we're solving $ 2x^a\equal{}z^c$. Now let $ x\equal{}2^uX$ and $ z\equal{}2^mZ$, so that $ X,Z$ are odd. Then $ 2^{1\plus{}au}X^b\equal{}2^{mc}Z^c$ and we must have $ 1\plus{}au\equal{}mc$ and $ X^a\equal{}Z^c$. Because $ (a,c)\equal{}1$, $ 1\plus{}au\equal{}mc$ has infinitely many solutions $ (u,m)$ in positive integers: just take $ m$ to be $ ka\plus{}c'$, where $ cc'\equiv1($ mod $ a)$, for any $ k>0$. This $ c'$ exists, as $ (a,c)\equal{}1$. Also $ X^a\equal{}Z^c$ is obviously solvable: take $ X\equal{}r^c$ and $ Y\equal{}r^a$, for some $ r>0$ odd. In this case we obtain $ x\equal{}2^ur^c$. Because $ x$ must be of the form $ y_0^b$, we need $ r$ to be a power of a positive integer with exponent multiple of $ b$ and $ u$ to be multiple of $ b$. For $ r$ there's no problem, take $ r\equal{}s^b$, for some $ s$ odd. For $ u$ it's a bit more complicate. Remember $ u$ has the form $ ka\plus{}c'$. Now the condition $ (a,b)$ helps. It ensures the existence of a positive integer $ 1\le a'<b$ such that $ aa'\equiv1($ mod $ b)$. We may take $ k\equal{}lb\minus{}a'c'$ for any $ l$ great enough to ensure that $ k>0$. Then $ ka\plus{}c'$ will be a multiple of $ b$, as needed.
We can even neglect $ X,Z$ (that is, set them equal to $ 1$) and obtain solutions with $ x,y,z$ all powers of $ 2$.

[quote="freemind"]Remember $ u$ has the form $ ka+c'$.[/quote]I thought that was $m$.  I don't think you can get a solution with all powers of two if $(a,c) = 1$ but $(b,c) \neq 1$.

We can suppose $a$ and $c$ are relatively prime without loss of generality.
Let $s$,$t$,$k$ be positive integers such that $k = s^{c}-t^{b}$.
Since $a$ and $bc$ are relatively prime, there exist infinity many positive integers $w$ such that
$w \equiv -1 \pmod{a},w \equiv 0 \pmod{bc}$ by Chinese Remainder Theorem.

We take that $x = k^{\frac{w+1}{a}}$, $y = k^{\frac{w}{b}} \cdot t$, $z = k^{\frac{w}{c}} \cdot s$, then they satisfy the equation $x^{a}+y^{b} = z^{c}(=k^w \cdot s^{c})$.
So,we have done.


Let $ a,b,c,d > 0$,prove that

$ {\frac {{c}^{2} \minus{} ab}{\left( a \plus{} b \right) \left( b \plus{} c \right) }} \plus{} {\frac {{d}^{2} \minus{} bc}{\left( b \plus{} c \right) \left( c \plus{} d \right) }} \plus{} {\frac {{a}^{2 } \minus{} cd}{\left( c \plus{} d \right) \left( d \plus{} a \right) }} \plus{} {\frac {{b}^{2} \minus{} da}{ \left( d \plus{} a \right) \left( a \plus{} b \right) }} \geq 0.$

[quote="xzlbq"]Let $ a,b,c,d > 0$,prove that

$ {\frac {{c}^{2} \minus{} ab}{\left( a \plus{} b \right) \left( b \plus{} c \right) }} \plus{} {\frac {{d}^{2} \minus{} bc}{\left( b \plus{} c \right) \left( c \plus{} d \right) }} \plus{} {\frac {{a}^{2 } \minus{} cd}{\left( c \plus{} d \right) \left( d \plus{} a \right) }} \plus{} {\frac {{b}^{2} \minus{} da}{\left( d \plus{} a \right) \left( a \plus{} b \right) }} \geq 0.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-21-b-q-liu.html


[quote="shobber"]$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
\[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1)  \][/quote]
Did you post to wrong place?
1. Obviously (10 second, Cauchy)  ;)
2. Chinese TST without solution??  :huh:

[quote="langtupanda"]
2. Chinese TST without solution??  :huh:[/quote]
What's wrong with that?

[quote="langtupanda"]Nobody solved it, so how it became the test? At least one who gave the test.[/quote]
I still do not know what you are talking about.

[quote="shobber"]$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
\[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1)  \][/quote]
Let $xy=\frac{1}{c},$ $xz=\frac{1}{b}$ and $yz=\frac{1}{a}.$
Then $x=\sqrt{\frac{a}{bc}},$ $y=\sqrt{\frac{b}{ac}}$ and $z=\sqrt{\frac{c}{ab}}.$
Hence, $a+b+c=1$ and
[quote="langtupanda"]
 Obviously (10 second, Cauchy)  ;)
[/quote] :)

[quote="langtupanda"]Nobody solved it, so how it became the test? At least one who gave the test.[/quote]
[quote]Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it.[/quote]

shobber's just posting the problems here and probably he hasn't thought about it. So what's the problem with this?

[quote="langtupanda"] 2. Chinese TST without solution??  :huh:[/quote]

Why do you think that there is no solution, actually there is. But Shobber is just sending the problems, maybe he doesn't need the solutions. Maybe he is just with the problems from TST. :-).

Davron Latipov

ahhh maybe you are confusing about sending this problem in the Unsolved  section, there is no problem, if it is unsolved for the user he can send it as unsolved, since it is unsolved for him ...  :-)

Davron Latipov

Sorry shobber :blush: I just mean you should post this problem to Solved Section.
This is my solution:
 From x + y + z = x.y.z => x.y.z >= 3.[3^(1/2)] ( Cauchy)
           and 1/(xy) + 1/(yz) + 1/(zx) = 1
                => 1/x.[(y+z)/(yz)] = [(yz - 1)/(yz)]
                => (yz - 1) = 1/x.(y + z)
So we have the inequation similar to this one:
x^6.(y + z) + y^6.(z + x) +z^6.(x + y)
Cauchy 6 numbers and using x.y.z >= 3.[3^(1/2)] so we can have the minimum

[quote="shobber"][quote="langtupanda"]Sorry shobber :blush: I just mean you should post this problem to Unsolved Section.
[/quote]
Never mind. I just like to post the Contest problems in the Unsolved section.

From your solution I guess that you are from Vietnam since you call the inequality Cauchy but we call it AM-GM.  :)[/quote]
Cool! you are right, but what is AM-GM?

[quote="langtupanda"][quote="silouan"]AM-Gm is the inequality you called Cauchy  :rotfl:
These names ........
BTW , nice solution Arqady .[/quote]
I mean its real name, not its initials[/quote]
Arithmetic Mean-Geometric Mean Inequality, which states that the arithmetic mean of $n$ non-negative real numbers is greater or equal than the geometirc mean of those $n$ numbers. Equality holds when all the numbers are equal.

Same solution as AwesomeToad. This, for me, was probably one of the easiest China TST problems I have ever seen.

Alternatively, one could use Muirhead to solve this problem.

As $x^7(yz-1) + y^7(zx-1)+z^7(xy-1) = x^6(xyz-x) + y^6(xyz-y) + z^6(xyz-z) = x^6(y+z) + y^6(z+x) + z^6(x+y),$ we have from the fact that
$(6, 1, 0)$ majorizes $(3, 2, 2)$ and Muirhead that:

$$x^7(yz-1) + y^7(zx-1) + z^7(xy-1) \ge 2x^2y^2z^2(x+y+z) = 2(x+y+z)^3. \qquad (1)$$

If we set $s = x+y+z,$ then we know that $s \ge 3 \cdot s^{\frac13} \Rightarrow s \ge 3 \sqrt 3.$ Hence $(1)$ yields that the sum in question is at least $162 \sqrt 3.$ Since this value is obtained at $x = y = z = \sqrt 3,$ we have shown that the answer is $\boxed{162 \sqrt 3}.$

$\square$

The original problem is a special case of the following generalization where
$$p=2,\lambda =1$$

[b]Generalization 1[/b]
Let $x,y,z$ be positive reals such that $x+y+z=\lambda xyz$. Then prove that

$$\sum_{cyc}{\left(x^{3p+1}\left(\lambda yz-1\right)\right)}\geq \dfrac{2\left(3\sqrt{3}\right)^{p+1}}{\sqrt{\lambda ^{3p+1}}}$$

[b]Generalization 2[/b]
Let $x_{1},x_{2},\cdots,x_{n},\lambda,p,n$ be positive reals ($n\geq 3$) such that $\sum_{cyc}{x_{1}}=\lambda \prod{x_{1}}$. Then prove that

$$\sum_{cyc}{\left(x^{pn+1}\left(\lambda \dfrac{\prod{x_{i}}}{x_{1}}-1\right)\right)}\geq \dfrac{\left(n-1\right)\sqrt[n-1]{n^{n\left(p+1\right)}}}{\sqrt[n-1]{\lambda ^{pn+1}}}$$


logarithms
number theory
relatively prime

Let $ S$ be the set of all positive real numbers $ x$ such that $ 1997^x \equal{} 1996^{2x}$. Then:

(A) $ S$ is the empty set.
(B)  $ S$ contains exactly one number.
(C) $ S$ contains exactly two numbers.
(D) $ S$ contains more than 2, but a finite amount of numbers.
(E) $ S$ contains infinitely many numbers.

Hints only please.

[hide="Maybe?"]Try representing it with logs?[/hide]

[hide="I hope this does not give it away..."]
Take logarithms of both sides (any base) then look for solutions[/hide]

[hide="Solution"]$ 1997^x \equal{} 1996^{2x}$

$ \ln 1997^{x} \equal{} \ln 1996^{2x}$

$ x \ln 1997 \equal{} x \ln 1996^2$

$ \ln 1997 \equal{} \ln 1996^2$

$ 1997 \neq 1996^2$

Obviously, there is no solution, EXCEPT for $ \boxed{x \equal{} 0}$, so the answer is $ \boxed{B}$.[/hide]

[quote="RoFlLoLcOpT"][quote="isabella2296"]
Hints only please.[/quote][/quote]Well, he got the wrong answer anyways, so that's what he gets for not following the directions :P

lol but I was close! Didnt pay attention to "[u]positive[/u] reals"

Taking log is not necessary. You can simply work from

$ (1997)^x\equal{}(1996^2)^x$

[quote="chrischris"]Taking log is not necessary. You can simply work from

$ (1997)^x \equal{} (1996^2)^x$[/quote]

thats  an  intelligent  solution . good work

Once you get $ (1997)^x = (1996^2)^x$, you notice that if $ a^n=b^n$, then $ |a| = |b|$. But that's not happening here so the answer is[hide][hide="Awww you clicked..Do you really want to see the answer?"][hide="Just one more click..almost there!"]$ \boxed{A}$[/hide][/hide][/hide]

[quote="PowerOfPi"][hide="Another Solution I Found"]For positive powers of distinct numbers to be equal, they must have common factors. We can notice that 1996 and 1997 are relatively prime, so there are no positive values of $ x$.[/hide][/quote]
[hide="Something wrong."]That shows there are no integer solutions, but $ x$ is real. Many powers won't be integers, so it makes no sense to talk about their factors.[/hide]


for $a,b,c>0$ prove:
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\frac{1}{3\sqrt[3]{abc}}\geq \frac{1}{abc+ab+a}+\frac{1}{abc+bc+b}+\frac{1}{abc+ac+c}$
any ideas? :o

[quote="tondor"]any ideas? :o[/quote]
Yep,
it's from current Polish MO (http://www.om.edu.pl/zadania/om/om58_1.pdf).

[quote="arqady"][quote="tfr"]
Yep,
it's from current Polish MO (http://www.om.edu.pl/zadania/om/om58_1.pdf).[/quote]
tfr, 10 pazdzierinka 2006 - 6 listopada 2006, what is this? Translate please. Thank you.[/quote]
[i]10 pazdzierinka 2006 - 6 listopada 2006[/i] means that this series of the competition starts at 10.10 and ends at 6.11. :)

[b]edit[/b]: this is the second place where tondor tried to cheat (the first was [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116639[/url]). I repeat what I said in that topic:
[quote="Rzeszut"]@tondor: it seems that you are a cheater! Your two posts up to now are this post and [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=662486#662486[/url], where you asked for the solution of another problem from this competition. The location [i]Spain[/i] is false, because you email [email]tondor@poczta.fm[/email] is registered on a Polish server.[/quote]


vector
algebra
polynomial
linear algebra
matrix
abstract algebra
group theory

Let $ G$ be the group of invertible $ 3\times 3$ matrices over the field $ \mathbb{Z}/2\mathbb{Z}.$

1. What is the order of the group $ G?$

2. Find explicit examples of elements of order 2, 3, 4, and 7.

3. Prove that 1, 2, 3, 4, and 7 are the only possible orders of elements. In particular, there are no elements of order 6 and no elements of order greater than 7.

4. Is this a simple group?

[hide]1. The first column can be any $ 3\times1$ column vector over $ \mathbb Z/2\mathbb Z$ except the zero vector, so there are 7 choices for the first column. The second column can be any column vector except a scalar multiple of the first, so there are 6 choices. And the third coluimn can be any column vector except a linear combination of the first two, which gives 4 choices. Hence $ |GL(3,2)| \equal{} 7\cdot6\cdot4 \equal{} 168.$

In general, the order of $ GL(n,q)$ (the general linear group of $ n\times n$ invertible matrices over a field with $ q$ elements) is $ (q^n \minus{} 1)(q^n \minus{} q)(q^n \minus{} q^2)\cdots(q^n \minus{} q^{n \minus{} 1}).$[/hide]

No, there are many different nonabelian groups of that order. That's not a valid argument.

Sorry, the result I was thinking of is that there is only one (up to isomorphism) simple group of order 168. :oops:

You are probably interested in the minimal polynomial of the elements and their rational canonical form.

For instance, let's prove there are exactly two conjugacy classes of elements of order 7 and they are interchanged by the outer automorphism:

An element of order 7 satisfies the polynomial x^7-1, so its minimal polynomial divides it (mod 2).  Since x^7-1 = (x-1)(x^3-x-1)(x^3-x^2-1) mod 2, the minimal polynomial of an element of order 7 is either x^3-x-1 or x^3-x^2-1, and so an element of order 7 is conjugate to the companion matrix of one of those two polynomials.  However, taking the inverse transpose of an element with minimal polynomial f(x) of degree n gives an element whose minimal polynomial is x^n*f(x), the reversal of the polynomial f.  In particular, applying the inverse transpose to the companion matrix of x^3-x-1 gives a matrix with minimal polynomial x^3+x^2-1.

As another example, there are no elements of order 21 since (x^21-1)/(x^7-1)/(x^3-1)*(x-1) has no irreducible factors of degree 1, 2, or 3.

[quote="JackSchmidt"]In particular, applying the inverse transpose to the companion matrix of x^3-x-1 gives a matrix with minimal polynomial x^3+x^2-1.[/quote]
Ok, so $ \begin{bmatrix}0&1&0\\0&0&1\\1&1&0\end{bmatrix}^6\equal{}\begin{bmatrix}0&1&0\\0&0&1\\1&1&0\end{bmatrix}^{\minus{}1}\equal{}\begin{bmatrix}1&0&1\\1&0&0\\0&1&0\end{bmatrix}$

[quote="SD5225272"]Jack, what's the best way to prove this group is simple? The best I can come up with is showing it's perfect and noting there are no non-abelian simple groups, save $ A_5$, of order $ \le84$.

Steve[/quote]

I don't know about the best way, but a very different way that is easy if one happens to have already applied my earlier strategy is to note that normal subgroups are unions of conjugacy classes.  The conjugacy classes are described by rational canonical form and have sizes 1, 21, 24, 24, 42, 56.  The only sub(multi)set of those number containing 1 that sums to a divisor of 168 is the whole sub(multi)set {1,21,24,24,42,56}.

The class sizes can be computed fairly directly: everything commutes with a scalar, so min poly x-a has size 1.  Centralizing something with a minpoly of (x+1)^2 in jordan canonical form requires a matrix like [a,b,c;0,a,0;0,d,e], and to be invertible requires ae<>0; but over a field with 2 elements that means a=e=1 and (b,c,d) in 2^3 gives the 8 elements of the centralizer, so a class size of 168/8=21.  A matrix with minpoly xxx+x+1 (or its reverse) acts irreducibly, so Schur's lemma implies its centralizer has order 2^3-1=7, and so the class has size 168/7=24.  A minpoly of (x+1)^3 in JCF has centralizer all [a,b,c;0,a,b;0,0,a], but to be invertible a≠0, and over the field of 2 elements that means a=1 and (b,c) in 2^2, so the centralizer has size 4, and the class has length 168/4=42.  A minpoly of (xx+x+1)(x+1) has centralizer the direct product of the centralizer in GL(2,2) and GL(1,2) of matrices with minpoly xx+x+1 and x+1, so has order 3*1=3, and the class has length 168/3=56.

Now me personally I would want to have these calculations done anyways to understand different features of the group, but I can understand it is a lot of tedium if you just wanted to show GL(3,2) was simple.  Iwasawa's Criterion (Isaacs's FGT, 8C, Lemma 8.30, p252) is a very nice method of proving matrix groups (have solvable residuals whose inner automorphism groups) are simple.  Isaacs does this for PSL(n,q) in theorem 8.33.  Grove's Classical Groups and Geometric Algebra does this for classical groups (orthogonal, unitary, symplectic) over fairly arbitrary fields. I believe Carter's Simple Groups of Lie Type's Theorem 11.1.1 is just Iwasawa's criterion specialized to groups with a (B,N) pair; this is where he proves finite, untwisted, simple groups of lie type are in fact simple.

At any rate, here is a table of the classes of GL(3,2):

Length, Order, Minpoly
1, 1, x+1
21, 2, (x+1)^2
24, 7, xxx+xx+1
24, 7, xxx+x+1
42, 4, (x+1)^3
56, 3, (xx+x+1)(x+1)

[quote="RareFind"][quote]A matrix with minpoly xxx+x+1 (or its reverse) acts irreducibly, so Schur's lemma implies its centralizer has order 2^3-1=7, and so the class has size 168/7=24.[/quote]

I have two questions:
1) How do you know this matrix acts irreducibly? Is it because it's minimal polynomial is the same as its characteristic polynomial?
2) What version of Schur's lemma are you using? I have never seen it used this way! In particular, I have no idea where the 2^3-1 comes from.

Sorry if these questions are trivial.[/quote]

For 1, it is because the minimal polynomial is irreducible and of the same dimension as the whole space. In terms of k[x]-modules, we get k[x]/(xxx+xx+1) is simple because (xxx+xx+1) is a maximal left ideal.

For 2, we do the same as in 1, but we get even more in terms of rings (commutative rings are so weird), we get k[x]/(xxx+xx+1) is a field because (xxx+xx+1) is a maximal ideal.  Since k here has order 2, and k[x]/(xxx+xx+1) has dimension 3 with a basis {1,x,xx}, the field k[x]/(xxx+xx+1) has order 2^3.

Now Schur's lemma tells us that the centralizer is a field (indeed a splitting field, maybe this is the part that is new; over finite fields everything is easy) and so the group of its units has order 2^3-1.

If the splitting field thing is not clear, let me know and I'll write it up clearly after lunch.

[quote]
For 1, it is because the minimal polynomial is irreducible and of the same dimension as the whole space. In terms of k[x]-modules, we get k[x]/(xxx+xx+1) is simple because (xxx+xx+1) is a maximal left ideal.[/quote]

OK, I get this.  Since the minimal polynomial is irreducible, the corresponding k[x]-module is simple.  I'm guessing this is the only time the k[x]-module is simple?

[quote]
For 2, we do the same as in 1, but we get even more in terms of rings (commutative rings are so weird), we get k[x]/(xxx+xx+1) is a field because (xxx+xx+1) is a maximal ideal.  Since k here has order 2, and k[x]/(xxx+xx+1) has dimension 3 with a basis {1,x,xx}, the field k[x]/(xxx+xx+1) has order 2^3.

Now Schur's lemma tells us that the centralizer is a field (indeed a splitting field, maybe this is the part that is new; over finite fields everything is easy) and so the group of its units has order 2^3-1.

If the splitting field thing is not clear, let me know and I'll write it up clearly after lunch.[/quote]

This is really the sticking point for me.  The version of Schur's lemma I know is that if V is a simple R-module, then Hom_R(V, V) is a division ring (and it seems maybe you're using that a finite division ring is a field?).  But I don't understand what that has to do with the centralizer.  I checked around online and all I could find is that Schur's lemma implies the centralizer should be scalar matrices.  But wouldn't those be with entries in the F_2 field?

I really appreciate your help.

I'm no algebraist and am out of my element here. I do believe that mastery of examples is an important part of learning and understanding, so I thank all of you (especially Jack) for a lucid discussion.

One amusing thing that was pointed out to me: let $ A\in G$ have order $ 7,$ let $ \langle A\rangle$ be the cyclic subgroup of $ G$ generated by $ A,$ and consider the set of matrices $ F\equal{}\langle A\rangle\cup\{0\}.$ Then $ F$ is closed under both addition and multiplication, and is hence a field of order $ 8$.


function
analytic geometry
geometry
geometric transformation
reflection
graphing lines
slope

I know that if you are given any graph $f(x)$ on the coordinate system, the graph of $|f(x)|$ would just reflect all points with a negative y-coordinate in the original function across the x-axis; but what if you took $f(|x|)$? how would you manipulate the graph $f(x)$?  I would like an explanation also, if possible.  Thanks!

wait, so if the numbers are the absolute value of x then it is symmetric about the y-axis, and vice versa.

But, x can go below 0, and the graph is like a v.

If I'm wrong, please correct me.

[quote="banblink182"]The graph of what you have described is the reflection of the image in quadrant one onto quadrant two across the y axis.

This is because f(x)=f(-x), which is an even function - a function symmetric about the y axis.

Tell me if this isn't clear enough, but I hope it helps.[/tex][/list][/quote]

so if there isnt any part of the graph in the 1st quadrant, the graph doesnt change?

After reflecting the image in quadrant 1 into quadrant 2, it could end up not being a function right? because there might already be a line there  under or over the reflected portion?  Could you explain it some more please? Thanks.

It's a line with slope 1 and y-intercept 0.

Use this line, and anything in quadrant III move as a reflection to quadrant II.

I hope this helps.

Yes, I'm sorry, I did make a little mistake. Anything in quadrants I and IV is reflected across the Y-Axis onto the II and III quadrants respectively.

The graph of abs(f(x)) is different from f(x) in that all the points that fall below y=0 is reflected across the X-Axis.

The graph of f(abs(x)) is different from f(x) in that the region to the right of x=0 is the mirror image of the region to the left of it, since f(3)=f(-3), and f(a)=f(-a).

So when you reflect the images in quadrants I and IV into II and III, the original image in quadrants II and III is not there anymore?

Also, why are the points reflected from the positive side to the negative side?  Doesnt f(x)=f(-x)?  Why isnt it reflected the other way across the y-axis?

Correct.

The quadrant II and the quadrant III are affected by the av, not I nor IV.


Find function f, such that $ f(x^2\plus{}y^2)\equal{}f(x)\plus{}f(y),$ for all $ x,y\in\mathbb{R}$

[hide]$ x\equal{}y\equal{}0$ gives $ f(0)\equal{}2f(0) \iff f(0)\equal{}0$. Then $ y\equal{}0$ gives $ f(x^2)\equal{}f(x)$.
Now plugging in $ y\equal{}x$ gives us $ f(2x^2)\equal{}2f(x)\equal{}2f(x^2)$, or $ f(2a)\equal{}2f(a)$. Then, plugging in $ x\equal{}2a, y\equal{}0$, we have $ f(4a^2)\equal{}f(2a)$, or $ 4f(a^2)\equal{}2f(a)$. Since $ f(a^2)\equal{}f(a)$, we must have $ f(a)\equal{}0$ for all real $ a$.[/hide]

Fun problem!

[hide="Solution"]
Let $ x\equal{}y\equal{}a^k$ to get $ f(2a^{2k})\equal{}2f(a^k)$ in general.

Now let $ x\equal{}a^k$ and $ y\equal{}0$ to get $ f(a^{2k})\equal{}f(a^k)\plus{}f(0)$, or $ f(a^{2k})\equal{}f(a^k)$. The only way that can be true is if $ f(x)$ is a constant function, so let $ f(x)\equal{}c$.

Then $ c\equal{}2c$, so $ c\equal{}0$.

Therefore, $ \boxed{f(x)\equal{}0}$.
[/hide]


induction
function
combinatorics unsolved
combinatorics

Consider a n*n table and on each square have a non-negative real such that the sum of each row (or each column) equals 1.
Prove that: There exits n squares satisfied each of them have a positive real and no two-squares of them lie on the same row (or the same column).

Put a + if a positive number occurs. On every k rows , $1<=k<=n$, we must have k columns which each intersects at least one of these rows in a +. Otherwise, the total sum of the numbers would be k (summed up by the rows ) and would also be at most $k-1$ summed up by the columns, contradiction. Now just use marriages lemma and get that for every row we can choose a column which has + at their intersection and this is a bijection.
I couldn't find other aproaches, maybe it could work by induction but it's more difficult.

:( Sorry for my little knowledge :Can you explain more clearly your sol with Marriage 's Theorem.Or can anybody help me with that theorem?(I've read a book: it writes a problem whose sol  use Marriage 's theorem but that book isn't showed the theorem and proof! :D ).

Ok. It is not that hard.
Marriages Theorem (Lemma). Suppose we have 2 sets A and B, and we associate to every element in A, some elements in B. We can build an injective function from A to B basing on these associations, iff from every k elements in A ($1\leq k\leq |A|$) there are at least k elements in B which are associated with them.
Alternate version of the lemma (from which it's name comes). There are n boys in a village and some girls. For every k boys ($1\leq k\leq n$) there are at least k girls that they would like to marry (a boy can like many girls). Now we have that we can make n pairs such that every boy marries a girl.
In this problem. the rows are the boys. If there is a + at the intersection with a column, than that row likes that column. Now we must show that for every k rows there are at least k columns which intersect them in +'s. Suppose it's at most (k-1), then we shall count the sum of the reals in this $k\cdot (k-1)$ rectangle. But on every line the sum is 1, so we have sum$=k$. On the other hand, on every column (and they are not complete), the sum of the number is at most 1, so at most $k-1$, contradiction. So we have at least k such columns no matter of k or of what rows we chose.
That's all.
I also couldn't understand this lemma from the start, but I got when it was explain to me with the alternate version in RoMop.


Let point $A$ and circle $O$ be given such that point $A$ is outside the circle. Find the locus of points $X$ such that the length $XA$ is equal to the length of the tangent from $X$ to $O$.

[hide="hint"]see $A$ as a circle with zero radius[/hide]


The diameters of two circles are $ 8$ inches and $ 12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is:
$ \textbf{(A)}\ \frac{2}{3} \qquad\textbf{(B)}\ \frac{4}{9} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{1}{2} \qquad\textbf{(E)}\ \text{none of these}$

Since the area of a circle varies with the square of the diameter or radius, a diameter ratio of $ \frac{8}{12}\equal{}\frac{2}{3}$ implies an area ratio of $ \frac{4}{9}$, or choice $ \fbox{B}$.


geometry AMC AIME MATHCOUNTS USA(J)MO USAMO blogs

OK... i am starting a team tournament.
every week I will post new problems... but this is how it will work.
each team will have 4 positions to fill.
1. Captain
2. Co-Captain
3. (unnamed)
4. (unnamed)

Their are 10 problems each week, 8 of which are split into pairs of two in any of these categories.
Number theory
algebra
geometry
high school
wild card

and there will be a bonus and sub-bonus.

I will explain more later.
Post to join.
join or create a new team and name.
Do NOT bicker over positions yet.

I join and create team 1337. I am not sure if I want to be captain. Pm me if you want to be captain.

you... didn't hear me, did you?
DON'T TALK ABOUT POSITIONS!!!
I haven't explained enough.

I would like to join pythag's team, if it's all right with him.

thats gonna be manslaughter...
teams are:

Team 1337:
pythag011
AIME15
isabella2296

post any complaints about teams... but don't complain about unfairness.

If Urc, Poincare, and Aylsero all get approval, I would like a team with us four. Team GUASS (xD). Oh, and don't forget the xD in the parentheses IS part of our team name :D

Other than that, I don't want to be with eggylv999 or biblioman. (the spammers)

I'll join Team 1337 if it's OK with them.

I am in. I would also like to join 1337. aleph0, who will?

okay, team list:
parenthesis () mean that their are too many for that team and they must get permission from current team to join (only 4 per team!!!). I will place them in order of request.

brackets [] mean that these people have reserved spots (current team may want to PM them). I will post updates if people are reserved but don't want to be on the team.

angle brackets mean they [b]CANNOT[/b] join that team.

[b]Team 1337:[/b]
pythag011
AIME15
isabella2296
(aleph0)
(Bugi)

[b]Team GUASS (xD):[/b]
gauss1181
[Urc]
[Poincare]
[Aylsero]
<eggylv999>
<biblioman>

more explanation on rules (don't argue about positions yet!!!):
each question is worth 2 points.
sub-bonus is worth 3, and bonus is worth 5.
I will post the 8 questions and the bonuses, and i will PM each member of each team with a pair.
[b]DO NOT[/b] discuss questions with teammates or tell teammates what questions you got. this is cheating.
captain gets bonus.
co-captain gets sub-bonus.
Oh yeah, dont think your captain is getting the same pair of questions as the other captains.
that goes for all positions.
each round will require a certain amount of points for moving on.
The final round will be like the masters round of mathcounts. but you can work together.
I will explain more when it comes along.
I am REALLY hoping for at least 10 teams.
SO... advertise using your sig.
please?

I'm fine with Bugi, but not so with aleph.

Maybe aleph can be on my team if AIME doesn't want him?
 :D  avatars rock!!!!

ughhhh.... confusing....

team list:

Team 1337:
pythag011
AIME15
isabella2296
Bugi

Team GUASS (xD):
gauss1181
[Urc]
[Poincare]
[Aylsero]
(Aleph0)
<eggylv999>
<biblioman>

Team 1337, you may start bickering about positions, but keep the rules in mind.

[img]http://home.tele2.fr/s2long/Sonic/Avatars/AvatarSonic-001.jpg[/img]

sonic is awesome!!!
(Sorry, random outburst.)

[quote="isabella2296"]You could probably start a tournament in the AMC forum.[/quote]

There have been such tournaments in AMC forum before. (I think also in Pre-Olympiad once).

jeez....
HERE IS WHATS GOING ON!!!
regular problems are MC level.
Sub-bonuses and bonuses... a little tougher.
A LITTLE!!!!

[quote="ernie"]Nominate:

Captain: xpmath
Co-Captain: FantasyLover
Important Person Under the Captains: ernie
Participant: high_flyer10

Sorry, high_flyer10.[/quote]
jeez, talk about the international guide to making people depressed...
3 and 4 are virtually the same, and EVERYONE is important.
[size=75]ill give you participant...[/size]

When is the next problem/problems coming?

Yeah, this tourney is moving awfully slow...

Has this thing started yet?

If not I make team RandomPeople1234

XPMATH AND ERNIE

LETS SWITCH HIGH FLIER WITH BOGTRO

no offense high flier

TEAM NOMINATIONS

Captain: xpmath
Co-Captain: FantasyLover/BOGTRO
3rd Participant: FantasyLover/BOGTRO
4th Participant: ernie

I'm tired of this. Call me when it starts.

I agree. This tournament should begin. It has taken too long already.  :mad:

[quote="AIME15"]asterday, have you even made the problems?[/quote]

Asterday, if you haven't, I will make some problems for you. (Not co-captain problems, because I am co-captain, but other ones... I am pretty good at making problems.)

Yes, would you mind PMing me 2 number theory problems and 2 geometry problems(MC LEVEL). If you don't mind.
I'm still confused about team positions, thats part of the reason why i haven't started. I WANT FINAL DECISIONS NOW! PLEASE POST A TEAM LIST.

bump...

is this tourney even starting?

when are you gonna post the first round?


calculus
integration
logarithms
trigonometry
limit
calculus computations

I think this is not the right place to post this problem, so if anyone can suggest me or move this to appropriate place.

Solve:

$\int^\frac{\pi}{2}_{0}{\ln(\sin x)}dx$

You can only use the fact that $\int{dx}= x$

[quote="joml88"][quote="xxxyyyy"]I think this is not the right place to post this problem, so if anyone can suggest me or move this to appropriate place.[/quote]

Announcements are most effective when [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]read[/url] ;)[/quote]

my mistake... dont know how I missed the sticky post... I tried searching for the forum where to port. Apologies once again.

[quote="xxxyyyy"][quote="Elemennop"]$\int^\frac{\pi}{2}_{0}{\ln(\sin x)}dx = \int^\frac{\pi}{2}_{0}{\ln(\cos x)}dx$ is the only tricky part. Trivial after that.[/quote]

If its not too much trouble to you, can you give complete and detailed solution.[/quote]

I posted the solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116550]here[/url].

[quote="RIP"]By the way, here's another simple method.
\[\int\limits_{0}^{\frac{\pi}2}\ln\sin x\,dx=\frac12\int\limits_{0}^{\pi}\ln\sin x\,dx=\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n-1}\ln\sin\frac{\pi k}n \]
Further
\[\prod_{k=1}^{n-1}\sin\frac{\pi k}n=\prod_{k=1}^{n-1}|\sin\frac{\pi k}n|=\frac1{2^{n-1}}\prod_{k=1}^{n-1}|1-e^{\frac{2\pi ik}n}|= \]

\[=\frac1{2^{n-1}}\Bigl|(1+x+x^{2}+\ldots+x^{n-1})\bigl|_{x=1}\Bigr|=\frac n{2^{n-1}}\]
Finally $\int\limits_{0}^{\frac{\pi}2}\ln\sin x\,dx=-\frac{\pi}2\ln2$[/quote]

Nicely done, RIP. Indeed, thank you for this for I have been rather wanting to read [u]any[/u] other solution to this particular integral. Here is yet another solution:

Recall that $\sin \pi z = \pi z\sideset{}{^\prime}\prod_{k=-\infty}^{\infty}( 1+\frac{z}{k})$ converges uniformly for all $z\in\mathbb{C}$ and hence we have that ${\log( \sin \pi z) =\log\pi z+\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}\log( 1+\frac{z}{k}})$. Now write

\[{J: =\int_{0}^{\frac{\pi}{2}}\log\sin x\,dx=\pi\int_{0}^{\frac1{2}}\log( \sin \pi z)\,dz =\pi\int_{0}^{\frac1{2}}\log\pi z\, dz+\pi\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}\int_{0}^{\frac1{2}}\log( 1+\frac{z}{k}})dz \]
\[= \pi[ z\log \pi z-z ]_{0}^{\frac1{2}}+\pi\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}k\int_{1}^{1+\frac{1}{2k}}\log u\, du =\frac{\pi}{2}(\log\frac{\pi}{2}-1)+\pi\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}k[u\log u-u]_{1}^{1+\frac1{2k}}\]
\[=\frac{\pi}{2}(\log\frac{\pi}{2}-1)+\pi\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}k[( 1+\frac1{2k})\log( 1+\frac1{2k})-( 1+\frac1{2k})+1]\]
\[=\frac{\pi}{2}(\log\frac{\pi}{2}-1)+\frac{\pi}{2}\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}[(2k+1)\log( 1+\frac1{2k})-1]\]
\[=\frac{\pi}{2}(\log\frac{\pi}{2}-1)+\frac{\pi}{2}\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}\log( 1+\frac1{2k})+\frac{\pi}{2}\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}[2k\log( 1+\frac1{2k})-1]\]
\[=\frac{\pi}{2}(\log\frac{\pi}{2}-1)+\sin\frac{\pi}{2}+\frac{\pi}{2}\sideset{}{^\prime}\sum_{k=-\infty}^{\infty}[2k\log( 1+\frac1{2k})-1]\]

I'm takin' a break now...


For real numbers $ a$ and $ b$, define $ a \diamond b \equal{} \sqrt{a^2 \plus{} b^2}$. What is the value of \[(5\diamond 12)\diamond ((\minus{}12) \diamond (\minus{}5))?\]
$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ \frac{17}{2}\qquad
\textbf{(C)}\ 13\qquad
\textbf{(D)}\ 13\sqrt{2}\qquad
\textbf{(E)}\ 26$

$ 5 \diamond 12 \equal{} 13$

$ \minus{}12 \diamond \minus{}5 \equal{} 13$

$ 13 \diamond 13 \equal{} \boxed{ \textbf{(D)}\ 13\sqrt{2}}$


probability
geometry
algorithm
IMO
IMO 2005

What is the score of the top 6 students from ex-USSR (Russia + formerly Soviet republics)
Does Russia place #1 if one of its students is replaced by the perfect scorer from Moldova?

[b]Edit:[/b]  If the two lowest scores of Russia (28 and 30) are replaced by 42 and 42 from Moldova and Ukraine,
one gets 6 gold medals and a total score of 236, one higher than China.

However, one could argue that this is not necessarily the team that would have been selected and we are choosing after the competition.
It would be interesting to take the results of all students from the ex-Soviet territories, assume some randomness in selecting the top students, and see what is the probability that Russia is #1.

[quote="hello"]Fleeting guest! You are correct! The dominence of USSR mathematics at work!![/quote]

Come on, be at least semi-relevant.

Another question (more speculative) is whether any provinces in China could send a top 10 IMO team, or what would happen if China sent additional teams from its students ranked 7-12, 13-18, etc.

cool. some of you guys are talking about "virtual score" :lol:   well, why don't we talk about virtual score of china?
taiwan is recognized as part of china by most of the world.(yet, i don't wanna talk about the historical argument for now)
china has 4 perfect scorers while taiwan has 2.  :lol:  then you guys do the maths. does former USSR still beat china?

by the way, hongkong did pretty well,too.

china ranked no.1 8 times and  no country has beaten china twice since 1995. (didn't check the stats before 1995)

well, there are many reasons why the chinese mathematicians aren't as good as those in europe and north america, such as low quality education in the chinese universities, but it's definitely not because they're not clever enough.

[quote="chika"]well, there are many reasons why the chinese mathematicians aren't as good as those in europe and north america, such as low quality education in the chinese universities, but it's definitely not because they're not clever enough.[/quote]
Big difference between doing IMO well and becoming an outstanding mathematician.

My record shows that:
1988 USSR: 217 China: 201
1989 China: 237 USSR: 217
1990 China: 230 USSR: 193
1991 USSR: 241 China: 231

[quote="fleeting_guest"]What is the score of the top 6 students from ex-USSR (Russia + formerly Soviet republics)[/quote]

42(BLR)+42(MDA)+42(RUS)+42(UKR)+41(RUS)+38(UKR)=247. However, less than China plus their best friends Taiwan.

I dont understand what u mean Fedor? Chto ti imeyesh vidu Fedor? Ya xochu chtobi oni ne ponili chto ya napisal, ti bil na imo?

Hello, I do not mean nothing special. I just sum up best 6 six results of post-Soviet contestants and get 247. For China nad Taiwan, it is $6\times 42$.

All questions in Russian are welcome either in Russian subforum or by e-mail.

[quote="Valentin Vornicu"][quote="Fedor Petrov"]However, less then China plus their best friends Taiwan.[/quote] :rotfl:  :rotfl:  :rotfl: this was a good one Fedor![/quote]

We ate a number of meals with the Taiwan deputy. Once we sat down at a table with him and two observers (red-tags, very recognisable) and I thought they were Taiwanese observers. Upon closer inspection... One (maybe both) was actually PRC! They seemed to be chatting very happily.

Seriously, I do believe that they ARE generally friendly towards each other, except for the politicians.

What about Belarus, LEVIN (BLR4) HAS 42 AND BELARUS was in USSR!

Well done, Levin and all Belarussian team!

[quote="manuel"]Why are russians better mathematicians in the future than chinese or asian?
i mean, maybe im wront but this is what i've seen a bit. the strong research professors in math are usually russians or eastern european.
is this because chinese and aisan people are more techinical in solving problems and dedicate more time to algebra and more technical areas and the russians are more creative in their approach and they focus more in combinatorics ?[/quote]

I don't think so. The mainland China didn't realise to develop its science until 1976. Many scientists and mathematicians are still very young. Also, many young men filled with talent have gone abroad especially USA.

maybe i'm a bit ignorant but, what i've seen in the strong math departments of the world's university is eastern european names and russian names..
few asian..
though obviuosly there are some excellent asian mathematicians..
but i think the difference come when the asians see math more as a science and tend to be more technical (you said 'scientists and mathematicians') and there research might be complicated and good but they might lack of the aesthetic sense...
in the other hand, russians and eastern europeans.. are known for their beautiful, brilliant , artistic and direct results in math..
you can see this for example china is known for their algebra strength (a more linear way of thinking) and russia is known for their strenght in combinatorics (ingenious ideas)

[quote="Fedor Petrov"][quote="manuel"]Why are russians better mathematicians in the future than chinese or asian?
i mean, maybe im wront but this is what i've seen a bit. the strong research professors in math are usually russians or eastern european.
is this because chinese and aisan people are more techinical in solving problems and dedicate more time to algebra and more technical areas and the russians are more creative in their approach and they focus more in combinatorics ?[/quote]

I think, it is rather because strong research professors are little bit older than IMO participants, and some 20 years ago, or more, Asians did not dominate on IMO. So, just wait few years, and the percentage of Asian, especially Chinese, researchers, must increase much.[/quote]

The Russian mathematicians who are dominant generally did not get there through the IMO, but by surviving a much more extreme Darwinian selection at the Moscow seminars of Gelfand, Manin, and some others.   The same pattern is true with Chinese students, the ones who become amazing are generally students of people like Chern and Yau and others at that level.  Chinese IMO champions are doing well, but not amazingly well, in research mathematics.

true, fleeting-guest..
i think the russian logic and way of thinking enables students to do well in olympiads (maybe not all perfect scorers) but also be prepared for future research and harder problem solving.
the chinese way of training students it is basically based in TRAINING as a methodic, algorithmic and automatic job. and math is not just this. it is learning math. and sometimes too much olympiad training can be dangerous to matheamtics. for more on this topic visit this topic and read my first post..
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=44859&postorder=asc&start=40[/url]

[quote="manuel"]i think the russian logic and way of thinking enables students to do well in olympiads (maybe not all perfect scorers) but also be prepared for future research and harder problem solving.
the chinese way of training students it is basically based in TRAINING as a methodic, algorithmic and automatic job. and math is not just this. [/quote]

This idea of "Chinese automaton training" is probably exaggerated, and based on some (partially true, but also exaggerated) cultural stereotypes.   The Chinese IMO people that I met were as creative as IMO champions from other countries.  The Chinese system is much different from the Russian one at university and PhD level (well, for the Ph.D, the algorithm for the best students is usually to study outside China), and I think it is there, not the high school olympiad training, that the important differences develop.  There is nothing really comparable to Gelfand's seminars in China (or anywhere else), or Manin's, although of course, such a system requires a very large number of highly-prepared students of which very few survive to the end.

[quote="fleeting_guest"][quote="shobber"]It is better to create problems than solving problems. Math olympiad only trains students to solve problems.[/quote]

There is not such a major difference between creating and solving.  People who are very good at one activity are usually active in the other.  Most chess masters are quite interested in inventing and solving problems, even if it is not their main activity, and all chess problemists are skilled chess players.  Similar statements are true about math olympiads.[/quote]

Quite agree! :D
I also think solving problems is a type of creating problems(such as lemmas!)


Given any eight real numbers $ a,b,c,d,e,f,g,h$ . Prove that at least one of the 6 numbers $ ac\plus{}bd;ae\plus{}bf;ag\plus{}bh;ce\plus{}df;cg\plus{}dh;eg\plus{}fh$ is not negative.

try using scalar products of vectors in 2 dimension