analytic geometry
function
geometry
rectangle
3D geometry
trigonometry
counting

This is a self-taking Mock AMC. I just made this so that many people in this site and I myself can have some extra mock AMC. I don't have an answer key so if you have a right answer, post here in the spoiler.

Many questions are came from old AHSMEs or even Mock AMC's. And other contests.

1. A lattice point is a point on a coordinate plane with integer values for x and y. How many lattice points lie on a circle centered at the origin with radius 25?
A) 40
B) 20
C) 10
D) 5
E) None of above

2. Find the unit digits of the sum:

$\sum^{2005}_{k=0} k^5$
A) 1
B) 3
C) 5
D) 7
E) 9

3. How many three element subsets of the set:

{88,95,99,132,166,173}

have the property that the sum of the three elements is even?
A) 6
B) 8
C) 10
D) 12
E) 24

4. If x and y are positive integesr such that $3x+4y=100$, then $x+y=$?
A) 10
B) 12
C) 14
D) 16
E) 18

5. Let a,a',b,b' be real numbers with $a,a' \neq 0$. The solution to $ax+b=0$ is less than the solution to $a'x+b' = 0$ if and only if:
A) a'b<ab'
B) ab'<a'b
C) ab<a'b'
D) b/a<b'/a'
E) b'/a'<b/a

6. In an h-meter race, Sunny is exactly d meters ahead of Windy when Sunny sportingly starts d meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finished the second race?
A) d/h
B) 0
C) $d^2/h$
D) $h^2/d$
E) $d^2/(h-d)$

7. Let $f$ be a function which $f(x/3) = x^2+x+1$. Find the sum of all values of $z$ for which $f(3z) = 7$.
A) -1/3
B) -1/9
C) 0
D) 5/9
E) 5/3

8. In an arithmetic sequence of complex numbers the first term is $3+4i$ and the second term is 7. Find the sum of the first 50 terms.
A) $5050-4700i$
B) $5000-7100i$
C) $5000+7100i$
D) $5050+7100i$
E) None of these

9. Two nonadjacent vertices of a rectangle are (4,3) and (-4,-3) and the coordinates of the other two vertices are integers. The number of such rectangles is:
A) 1
B) 2
C) 3
D) 4
E) 5

10. Inscribed in a circle is a quadrilateral having sides of lengths 25,39,52,and 60 taken consecutively. The diameter of the circle has length of:
A) 62
B) 63
C) 65
D) 66
E) 69

11. A wooden cube with edge length n units (where n is an integer > 2) is painted black all over. By slice parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit dege length. If the number of smaller of cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is n?
A) 5
B) 6
C) 7
D) 8
E) None of these

12. An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(a) the selection of four red marbles
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color

What is the smallest number of marbles satisfying the given condition?
A) 19
B) 21
C) 46
D) 69
E) More than 69

13. In their base 10 representation, the integer $a$ consists of a sequence of 1985 eights and the integer $b$ contains a sequence of 1985 fives. What is the sum of digits of the base 10 representation of the integer $9ab$?
A) 15880
B) 17856
C) 17865
D) 17874
E) 19851

14. Let $D_n$ denote the number of diagonals plus the number of sides in a convex polygon with $n$ sides. Then:

$\frac {1}{D_4} + \frac {1}{D_5} + \frac {1}{D_6}+...\frac {1}{D_2004} = \frac {m}{n}$ where m and n are relatively prime integers. Find $m-n$.
A) -334
B) -330
C) -332
D) -328
E) None of the above

15. In a triangle ABC, angle C is a right angle and CB>CA. Point D is located on segment BC so that angle CAD is twice angle DAB. If AC/AD = 2/3, then CD/BD = m/n, where m and n are relatively prime integers. Find m+n.
A) 10
B) 14
C) 18
D) 22
E) 26

16. Evaluate:

$\frac {\sin 10 + \sin 20}{\cos 10 + \cos 20}$ where numbers are given in degrees.
A) $\tan 10 + \tan 20$
B) $\tan 30$
C) $\frac {1}{2} (\tan 10 + \tan 20)$
D) $\tan 15$
E) $\frac {1}{4} \tan 60$

17. Find the units digit in the decimal expansion of:

$(15 + \sqrt {220})^{19} + (15 + \sqrt {220})^{82}$.
A) 0
B) 2
C) 5
D) 9
E) None of the above

18. If John has 12 quarters to distribute among Akiba, Bobby, Crl, David, and Emanuel, in how many ways can he distribute the money given that Akia gets at least 50 cents?

19. If $x = \frac {-1 + i \sqrt 3}{2}$ and $y = \frac {-1 - i \sqrt 3}{2}$ where $i^2 = -1$, then which of the following is not correct?

A) $x^5+y^5 = -1$
B) $x^7+y^7 = -1$
C) $x^9+y^9 = -1$
D) $x^{11} + y^{11} = -1$
E) $x^{13} + y^{13} = -1$

20. A and C lie on a circle with radius $\sqrt {50}$. The point B is  inside the circle such that <ABC = 90, AB = 6, BC = 2. Find OB.

A) 5
B) :sqrt: 26
C) :sqrt: 31
D) 11/2
E) 6

21. In triangle ABC, BC = 8 and AC = 10. If D is the midpoint of BC, and E is on AC such that EC = 3. Let EB and AD intersect at F. If DE:AF = m/n, where m and n are relatively prime integers, find m+n.
A) 28
B) 57
C) 100
D) 239
E) 87

22. How many elements in the 2002nd row of Pascal's Triangle (the one that begins with 1,2001,....) leaves remainder of 1 when divided by 3?
A) 23
B) 26
C) 28
D) 29
E) 667

23. TWo of the altitudes of the scalene triangle ABC have length 4 and 12. If the length of the third altitude is also integer, what is the biggest it can be?
A) 4
B) 5
C) 6
D) 7
E) None

24. Eight congruent equilateral triangles, each of different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like other)
A) 210
B) 560
C) 840
D) 1260
E) 1680

25. If the base 8 representation of a perfect square is ab3c, where a is not 0, then c is:

A) 0
B) 1
C) 3
D) 4
E) Not uniquely determined

Please note that this is problem thread AND solution thread. So, post your solution so people can see about it. Also, even you're not sure about answer, please post here.

Thanks and Enjoy the problems!

P.S. I have to try this too! :lol:

Number one: :D

[hide="My_answer"]Since the x,y has to fit in equation:

$x^2+y^2=25^2$

The possible pairs are (24,7), (15,20) and reverse. So, total 4. But there are four quadrants so it's 4*4 = 16. AND there are (25,0), (-25,0), (0,25), (0,-25). So, totally the numbers of those points are 16+4 = 20 [b]B[/b][/hide]

Number 10

I'm not so sure about this so I'll give it a shot.

[hide="Um"]I think this quadrilateral is called cyclic but I forgot its properties. Anyway, I thought I have to use some Pythagorean Theorem to do it so here's how I got it:

$25^2+60^2 = 4225$
$39^2+52^2 = 4225$

So that means 4225 is a diameter's square so:

$4225 = d^2\\
d = 65$

I hope that's right.[/hide]

Is there any better way or if I'm wrong, correct way?

[quote="Silverfalcon"][hide]18. We can start with giving Aki 2 quarters. Then we have 5 people and 10 quarters.

$\binom {14}{10} = 1001$

I remembered the formula I saw somewhere although I can't remember why it was like that. It works though. :D [/hide][/quote]

Basically, if you line it up, you have the question: How many ways can I arrange 4 breakers/dividers, and 10 objects which I am dividing. That's just C(14,10).

#15)

[hide]say AC=2, AD=3. $CD=\sqrt{5}$ by Pythagoras. Let AE be the angle bisector of DAC, E is on BC. by angle bisector theorem, ED=\frac{2\sqrt{5}}{5},

Let $\angle CAE=\theta$.
$\tan \theta=\frac{\sqrt{5}}{5}$, $\tan 2\theta=\frac{\sqrt{5}}{2}$

$\tan(3\theta)=\frac{\tan\theta+\tan 2\theta}{1-\tan\theta(\tan 2\theta}$

so $\tan 3\theta=BC/2=\frac{7\sqrt{5}}{5}$

$BC=\frac{14\sqrt{5}}{5}$, $BD=\frac{9\sqrt{5}}{5}$
BD/CD= 9/5. 9+5=14[/hide]

[quote="eryaman"]I don't understand how there can be one single answer to number 4.  Aren't there many different possibilities?[/quote]

Not only that, in fact none of the answers can work!

4(x+y)-x=100, so 4(x+y)>100, x+y>25! Therefore none of the choices can be correct!

[quote="Silverfalcon"]Number 7

[hide="My_ANSWER"]I decide to set x=9z. Then:

$f(x/3) = f(9z/3) = f(3z) = (9z)^2+9z+1 = 81z^2+9z+1$

Then it factors into $(9z+1)(9z+1)$ which leads me that z=-1/9 and that's only possible value for z so it's -1/9.

[/hide]

I thought there was different answer on Harold Reiter's Page though.. ;)[/quote]

[hide="Mistakes in your answer"]
Firstly.. $(9z+1)(9z+1) = 81z^2 + 18z + 1\text{, not } 81z^2+9z+1$..

Secondly.. you forgot that you wanted to solve $f(3z) = 7\text{, not }f(3z)=0$..

And thirdly, two wrongs make a right and you somehow got the right answer :D:D[/hide]

It should have been 13x+4y=100

I must've typed the question wrong.

The correct answer then would be D.

13(4)+4(12) = 100 = 100 :)

Hehe...

I knew I was wrong because the quesiton had another way of solving it using Sum and Product Formula (only sum part this case) but for my equation, that formula didn't work.

Somehow, it worked though. :-D

Right way would be:

$81z^2+9z-6 = (9z+3)(9z-2)$ and this also gives me a right answer.

Hehe... I got the right answer by wrong method... :lol:

1. A lattice point is a point on a coordinate plane with integer values for x and y. How many lattice points lie on a circle centered at the origin with radius 25?
A) 40
B) 20
C) 10
D) 5
E) None of above

[hide="answer"]
In quadrant I (from 0 :le: x < :pi:/2), there are (25, 0), (24, 7), (15, 10), (10, 15), and (7, 24). That times 4 is 20.

B
[/hide]

2. Find the unit digits of the sum:

$\sum^{2005}_{k=0} k^5$
A) 1
B) 3
C) 5
D) 7
E) 9

[hide="answer"]
Well, the units digit of the sum of $\sum^{10}_{k=1} k^5$ is obviously equal to $\sum^{20}_{k=11} k^5$ which is obviously equal to  $\sum^{30}_{k=21} k^5$ . And since the units digit of $\sum^{10}_{k=1} k^5$ is an integer, the units digit of 200( $\sum^{10}_{k=1} k^5$ ) is obviously 0. So, since the units digit of 0:^5: is 0, all we really have to find is $\sum^{2005}_{k=2001} k^5$, which equals $\sum^{5}_{k=1} k^5$ . The units digit of the units digit of 1 + 3[b]2[/b] + 24[b]3[/b] + 102[b]4[/b] + 312[b]5[/b], which is 5.

C
[/hide]

3. How many three element subsets of the set:

{88,95,99,132,166,173}

have the property that the sum of the three elements is even?
A) 6
B) 8
C) 10
D) 12
E) 24

[hide="answer"]
To get an even number, you can have:
E E E - of which there is one distinct way, or six total ways
E O O - which there are three distinct ways, or 18 total ways
So there are 24 ways in total.

E
[/hide]

4. If x and y are positive integesr such that $13x+4y=100$, then $x+y=$?
A) 10
B) 12
C) 14
D) 16
E) 18

[hide="answer"]
Only if x = 4 then the two criteria are met. So, y must be 12.
12 + 4 = 16
D
[/hide]

1 to 4

Just wanted to point a non-plug and chug method for this one:

4. If $x$ and $y$ are positive integesr such that $13x+4y=100$, then $x+y=$?

[hide="Solution"]
Immediatedly we think to look at it modulo 4 since two of the numbers are divisible by 4.  Doing this we see that $13x\equiv 0 \pmod{4}.$  So $x$ can be a multiply of 4.  But now $x=0$ doesn't work (x has to be positive) and x=8 makes the expression too big to start so it must be x=4. That makes y=12.  Thus x+y=16.[/hide]

If you haven't used mods very much this might not seem like a very natural approach.  Plug and chug certainly works without excruciating pain in this case but when you get harder diophantine equations (it just means an equation where you are looking for integer solutions) you will almost have to use modular arithmetic.

This probably isn't a good explanation, but here goes:

Say you have 3 + 2 = 5
We know 5 = 0 (mod 5), and since 3 + 2 = 5, we can say 3 + 2 = 0 (mod 5), which is true.
Thus, when considering a certain mod x, the left hand side and the right hand side of the original equation must be equal to the same y (mod x)

We have 13x + 4y = 100

I considered mod 13, which wasn't the best idea, but if you do that you get
13x or 0 (mod 13) + 4y (mod 13) = 100 (mod 13)

13x is always 0 mod 13, so just ignore it for now

So 4y (mod 13) = 100 or 9 (mod 13)

4y = 9 (mod 13)

I forgot my other steps so I'll just leave it at this...

The smoothest thing to do at that point is to notice that 9 = -4 (mod 13), so 4y = -4 (mod 13) so y = -1 = 12 (mod 13).  (Notice that we divide -- you can only ONLY ONLY ever do this when the thing you're dividing by is relatively prime to the mod.  Luckily, 13 is prime.)

quick question.

On number 14, what does d(2)004 mean, the one that is the last term of the series.  ?????

The main thing to notice in number 4 is that 13 and 4 are relatively prime.

Um.. Number 18 doesn't have any answer choices and on number 14.. are you sure the answer isn't -335?? becuase you have -334 on there, but I'm not sure if its a typo or what.  This is what i did on 14:

[hide]So we do Summation(4,2004) of 1/(n+n(n-3)/2), which is 2/n(n-1).  So we can turn this into 2[Summation(4,2004) of 1/n(n-1)].  The summation part is just the telescoping/recursive, whatever you call it series.. the one where you can cancel out different terms.  So we have 1/3-1/4+1/4-1/5+1/5.... +1/2003-1/2004.  So it becomes 1/3-1/2004, which is 667/2004.  HOwever, we still ahve to multiply in the 2, so we get 667/1002.  667-1002=-335.  [/hide]

So go figure.. maybe i did something wrong.  Some help on 14 please??

-335 is a correct answer.

For #18, I forgot.

I'd like to get the answers for these ASAP since AMC is coming and if you can answer these 25 questions VERY EASILY, you're ready for most of AMC questions (-10 or -12).

I think exception of use of trig and log, this is harder than regular AMC and mock AMC's since it involves some hardest question on mock AMC.

[quote]6. In an h-meter race, Sunny is exactly d meters ahead of Windy when Sunny sportingly starts d meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finished the second race?
[/quote]

Let's call r as the rate of Sunny. Then, (h+d)/r is the time for Sunny to finish race with d meters more. To find Windy's rate, we set proportion:

Windy's rate: r = (h-d): h

Then we get (h-d)r/h as Windy's rate. Since the time for d+ race is (h+d)/r, then:

$\frac {(h-d)r}{h} * \frac {(h+d)}{r}$

Which equals h - d^2/h so Windy's ahead by d^2/h.

Few answers to the problems.

#13

Since it's in form of 9*a*b, we can try in simpler case:

9*8*5 = 360
9*88*55 = 43560
9*888*555 = 4435560

So, the sum of digits equals 9n then 1985 = 9*1985 = 17865

There was a constructive method that Mr. Crawford showed although I didn't quite get it.

If anyone can explain here, I'll be thankful.

#16

I'm not sure but can I do this?

$\frac {\sin 10}{\cos 10} + \frac {\sin 20}{\cos 20} = \tan 10 + \tan 20$?

Can anyone gives some helps for the rest?

Thanks!

Silverfalcon....I'm disappointed.  You should know that $\frac{a+b}{c+d}\neq \frac ac+\frac bd.$

[hide="Hint"]
There are certain formulas that when applied to this problem make it really easy.  Do not look them up.  Rather try deriving them right now to solve the problem.  Here's  a hint:

Play around with $\sin(a+b)$, $\sin(a-b)$, $\cos(a+b)$ and $\cos(a-b)$.

[hide="Another Hint"]
You want to turn the sum of the sines on the top of the expression into a product and the same with the cosines on the bottom.  Try finding a way to convert sums to products using   $\sin(a+b)$, $\sin(a-b)$, $\cos(a+b)$ and $\cos(a-b)$.[/hide][/hide]

That's what I thought too.

Except I wasn't sure for "TRIG" part.

No wonder it came out so nice and easy..

Hmmm...

[hide]
19.
both x and y are cube roots of 1.  i think it's from one of the chapters in AOPS2.
just looking at the answers, x^9+y^9 should equal 2, so that's wrong.  C.
[/hide]

#11. [hide]
here's my solution (i don't know if it's right)

number of smaller of cubes with just one face painted black= 6(n-2)^2
number of smaller cubes completely free of paint =  (n-2)^3

put them equal to each other and get n=8. [b]D[/b] [/hide]

#12. [hide]
again, i don't know if this is right.

let the number of red marbles = r.
from (a) and (b), you know that the probability of choosing 1 out of all the white marbles is equal to the prob. of choosing 1 red marble out of (r-3) marbles.. so there are (r-3) white marbles.
continue with blue and green and add them all up to get total = 4r - 6
since r is a whole number, 4r - 6 is an even number. The only even answer is [b]C.[/b] 46 [/hide]

To make this mock AMC more useful, I'll now post the keys to the answers that has posted up here.

[hide="KEYS"]

1. B
2. C
3. Treething posted E. I don't think that's right, however.
4. D
5.
6. C
7. B
8.
9.
10.
11. D
12. C
13. C
14. E
15. B
16.
17.
18. 1001 (NO CHOICE. I forgot)
19. C
20.
21.
22.
23. B
24.
25.
[/hide]

Answers needed:

5,8,9,10,16,17,20,21,22,24,25.

#3, I get 10. Sets that are even:

{88,95,173}
{132,95,173}
{166,95,173}
{88,99,173}
{132,99,173}
{166,99,173}
{88,95,99}
{132,95,99}
{166,95,99}
{88,132,166}.

Which one's right?

And thanks again for the answers. More answers to fill in the spaces are appreciated!

Okay here are the formulas I was talking about for that trig problem:

[hide="Formulas"]
$\sin A +\sin B =2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

$\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$[/hide]

[quote="Silverfalcon"]
#3, I get 10. Sets that are even:

{88,95,173}
{132,95,173}
{166,95,173}
{88,99,173}
{132,99,173}
{166,99,173}
{88,95,99}
{132,95,99}
{166,95,99}
{88,132,166}.

Which one's right?

And thanks again for the answers. More answers to fill in the spaces are appreciated![/quote]

[hide]I think TreeThing was counting ordered sets. If order doesn't matter, you either have EEE or EOO, giving us
\[1+\binom{3}{2}\binom{3}{1}=1+9=10\]
So I think you're right. C for #3.[/hide]

#5
[hide]\[\frac{-b}{a}<\frac{-b'}{a'}\Rightarrow\frac{b}{a}>\frac{b'}{a'}\]
Answer is E.[/hide]

#8

[hide]$d=4-4i$. Let the sum we want to find be $S$. We find the real part and the imaginary part separately.
\[Re(S)=3+7+\cdots+(3+4\cdot49)=5050\]
\[Im(S)=4+0+\cdots+(4-4\cdot49)=-4700\]
Therefore,
\[S=5050-4700i\]
A's the answer.[/hide]

Thanks to joml88 for the formula. I think I can figure out this.

See if this is right.

\[\frac {\sin 10 + \sin 20}{\cos 10 + \cos 20}\\
= \frac {2 \cdot \sin \frac {(10+20)}{2} \cdot \cos \frac {(10-20)}{2}}{2 \cdot \cos \frac {(10+20)}{2} \cdot \cos \frac {(10-20)}{2}}\\
= \frac {\sin 15}{\cos 15} = \tan 15\]

\[\text {So, our final answer is D.}\]

The question about the expansion:

Find the units digit of:

\[\displaystyle (15 + \sqrt {220})^{19} + (15 + \sqrt {220})^{82} \]

I noticed that this question is extremely hard since it was #30 on one of old AHSMEs on 80's but how do you do it?

Do you expand it? Use conjugates? Would this help?

\[(15+\sqrt {220})^{19} \cdot [1+(15+\sqrt {220})^{63}]\]

Applying binomial theorem is the key to this problem but in what way?

P.S. Rereading all these problems again, I noticed why there aren't many replies -----> This is a set of DIFFICULT PROBLEMS!

Thanks for this great mock test! Not having the time to look at the whole thing, I worked only on the ones we needed answers for. Of course I was triple tasking when I did them, so my sols might be flawed. Let me know if I made any mistakes, or if there are problems I missed! And I think you should recheck problem 21.

9
[hide]The other two vertices will be some pair of opposite lattice points on the circle with center at the origin and radius 5. There are 5 such pairs: $\pm(3,4),\pm(5,0),\pm(0,5),\pm(-3,4),\pm(-4,3)$.[/hide]
E

10
[hide]Noticing that 25=5*5 and 60=5*12, we see 5 and 12 and think 5-12-13 triangle. If these two formed a right angle, the hypotenuse would be 5*13=65. This suspicion is confirmed, because 39=3*13 and 52=4*13 implies that the hypotenuse would also be 5*13=65. So those two pairs of sides meet at right angles, and the diameter is 65.
C[/hide]

16
[hide]by sum to product formulas, $\sin10+\sin20=2\sin15\cos5$, and likewise $\cos10+\cos20=2\cos15\cos5$. The answer is $\tan15$.
D[/hide]

17
[hide]Define $S_n=(15+\sqrt{220})^n+(15-\sqrt{220})^n=x^n+y^n$. It is easy to check that $x$ and $y$ are the roots of $a^2-30a+5=0$, so the $S_n$ satisfy the recursion $S_{n+1}=30S_n-5S_{n-1}$. Since $S_1=30$ and $S_2=890$, all of the terms are therefore divisible by 10. In particular, $S_{19}+S_{82}$ ends in 0. Since $0<y<1$, $y^n$ is small and positive, so $S_{19}+S_{82}$ is just a small amount greater than $(15+\sqrt{220})^{19}+(15+\sqrt{220})^{82}$. So this number is $\ldots9.9\ldots$, and its units digit is 9.
D[/hide]

20
[hide]Let B be the point $(x,y)$. Then $(x+2)^2+y^2=x^2+(y+6)^2=50$. Subtracting the two equations gives $x=3y+8$, and then subbing this back in gives $(x,y)=(5,-1)\text{ or }(-7.-5)$. The second one lies outside the circle, so B is (5,-1) and the answer is $\sqrt{26}$.
B[/hide]

21
I think there's an error in this problem.

22
[hide]The number 2001 is $2202010_3$. There is a theorem (I forgot the name) that says that for the number $n=a_6a_5a_4a_3a_2a_1a_0$ in base 3, \[\binom{2001}{n}\equiv\binom{2}{a_6}\binom{2}{a_5}\binom{0}{a_4}\binom{2}{a_3}\binom{0}{a_2}\binom{1}{a_1}\binom{0}{a_0}\bmod3.\] We just have to make that product equal $1$. There are 2 ways to do this.
First Way: each of the 7 binomial coefficients comes out to 1. Then $a_6,a_5,a_3$ are either 0 or 2, $a_0,a_2,a_4$ are each 0, and $a_1$ is either 0 or 1. There are therefore 16 ways in this case.
Second Way: Two of the three "2" terms equal 2, all other terms equal 1. In this case, $a_0,a_2,a_4$ are still 0 and $a_1$ is either 0 or 1, but two of $a_6,a_5,a_3$ equal 1 and the other is 0 or 2. There are $2^2\cdot\tbinom{3}{2}=12$ ways for this. Answer: 16+12=28.
C[/hide]

24
[hide]The idea is to fix as much as possible until the octahedron is completely determined. First, rotate so color 1 is on top. There are 4 cases:
CASE I: color 2 is adjacent to color 1. In this case, rotate it so 1 is still on top and 2 is in front. This completely determines the octahedron. The other 6 colors can arrange themselves in 6! ways.
CASE II: color 2 is not adjacent to 1 but not opposite it. In this case, rotate it so the vertex that colors 1 and 2 share is in front (and color 1 is still on top). This fixes the octahedron, so there are again 6! ways.
CASE III: colors 1 and 2 are opposite each other and color 3 is adjacent to color 1. Rotate so 1 is top, 2 is bottom, 3 is front. There are 5! ways.
CASE IV: 1 and 2 opposite, 3 adjacent to 2. Again there are 5! ways.
Answer: $6!+6!+5!+5!=2\cdot5!\cdot(6+1)=1680$.
E[/hide]

25
[hide]A perfect square can be 0,1,4, or 9 modulo 16 (verify!), which means that it can be 0, 1, 4, 9, 16, 17, 20, 25, 32, 33, 36, 41, 48, 49, 52, 57 modulo 64. (There are actually less than this, but I didn't want to check 64 numbers :)) This gives us the last two digits base 8. Of these, the only one with an "eights" digit of 3 is 25, which converts to $31_8$. The answer is 1.
B[/hide]

[quote="Silverfalcon"]This is a set of DIFFICULT PROBLEMS![/quote]

Indeed! I say this is a little harder than the hardest AMC would be. But it's still very good review! (Luckily, I've seen one or two of these problems before  :D  )

Many thanks to Silverfalcon for this mock AMC!

Also, zabelman, your #17 solution is irresistably good. Puts me to shame.  :blush:

[quote="litoxe2718281828"]Also, zabelman, your #17 solution is irresistably good.[/quote]
Thank you! It just goes to show that simple conjugates can go a long way!  :P

[quote="litoxe2718281828"]Puts me to shame.  :blush:[/quote]
Why should it do this? There's no shame in struggling with a difficult problem. That's what problem solving's all about! And indeed, that problem [i]is[/i] difficult.

Wow, new solution!

 :) Thanks for the great replies!


We have $ n$ colors named $ A_1,...,A_n$,we use them to color $ K_m$ . $ a_1\ge a_2 \ge ...\ge a_n\ge 1$ are given positive integers.
Find the smallest value $ f(a_1,a_2,...,a_n)$ such that however we colored $ K_m$,there always exists $ i$ $ a\le i\le n$ s.t. we can find $ a_i$ edges which are colored with$ A_i$,and they are "independent" from each other.
(edge $ A$ is independent from edge $ B$ means they have no common vertices)

Sorry,I had two typoes
1.$ m\equal{}f(a_1,...a_n)$
2.$ 1\le i \le n$


geometry
circumcircle
power of a point
radical axis
geometry proposed

Five distinct points $A, B, C, D$ and $E$ lie on a line with $|AB|=|BC|=|CD|= |DE|$.
The point $F$ lies outside the line.
Let$G$ be the circumcentre of the triangle $ADF$ and $H$ the circumcentre of the triangle $BEf$.
Show that the lines $GH$ and $FC$ are perpendicular.

Let $P$ be the intersection of $CF$ with the circumcenter of $\triangle ADF$.  So because $A,D,F,P$ are concyclic, we have $AC\cdot CD = FC\cdot CP$.  However, we have $BC = CD$ and $CE = AC$ so $FC\cdot CP = BC\cdot CE$ and hence $P$ lies on the circumcircle of $\triangle BEF$.  Thus, $FP$ is the radical axis of the two circles and hence is perpendicular to $GH$.


Let ABCDE is a pentagon such that : $ AB\equal{}BC\equal{}CD\equal{}DE\equal{}EA  ,  <A \geq <B \geq <C \geq <D \geq <E$
prove that ABCDE is regular.


induction
number theory unsolved
number theory


vector
abstract algebra
superior algebra
superior algebra unsolved

I need  in such question:

We know that Z2={0,1} is a field.
Let G be Abelsk group.
We prove to give G structure of one vector space over Z2={0,1} in such way:
If a belong Z2 and g belong G, thus we define ag in G
ag=0 (identity in G) if a=0 and ag=g if a=1 .

1. Show that this not working.
2. Analyze exact what is wrong?
3. It must be that g+g=1g+1g=(1+1)g=0g=0 for all g in G for this work, but why?
4. Giv one concrete example at one group (with more then 2 elements!) G which in default way is one vector space over Z2={0,1}.
Thanks for all help.
I need very simple explain.

Best regards
Tanja

[quote="Tanja"]
Let G be Abelsk group.
[/quote]
I suppose you mean abelian :)

Well what is not working is what you stay in 3), that for all $g\in G$ must be $g+g=0$ but this is not true in all the groups (for example in $\mathbb{R},\mathbb{Z}$ and many others). You can find a simple example in $\mathbb{Z}$ when you definition for a scalar moltiplication is fallacios: let me write  $X_{\mathbb{Z}}$ for an integer $X$ and $Y_{\mathbb{Z}_2}$ for an element of $\mathbb{Z}_2$ and simillary I'll write $+_{\mathbb{Z}}$ for sum in $\mathbb{Z}$ and ${+_{\mathbb{Z}_2}}$ for sum in $\mathbb{Z}_2$ (usually this notation is too long so it's not used, but I think it can be usefull for the begginers...) so we have $0_{\mathbb{Z}}\neq 2_{\mathbb{Z}} = 1_{\mathbb{Z}}+_{\mathbb{Z}}1_{\mathbb{Z}}=1_{\mathbb{Z}_2}\cdot 1_{\mathbb{Z}}+_{\mathbb{Z}}1_{\mathbb{Z}_2}\cdot 1_{\mathbb{Z}}=(1_{\mathbb{Z}_2}+_{\mathbb{Z}_2}1_{\mathbb{Z}_2})\cdot 1_{\mathbb{Z}}=0_{\mathbb{Z}}\cdot 1_{\mathbb{Z}}=0_{\mathbb{Z}}$ well you see that this is wrong. And I said before why. To find a concrete example of a group in which it works we must find one in which $g+g=0$ for all $g\in G$ (formally all elements of $G$ has order two) it can be proved that such a group have $2^n$ elements (for some natural $n$, but we don't need to prove it here), you can simply consider the Klein's group $\mathbb{Z}_2\times\mathbb{Z}_2$ or $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ or  $\mathbb{Z}_2^n$ for any $n\in \mathbb{N}$. Well generally if $F$ is a field  you should know that $K^n$ is a vector space and if $V$ is a $n$-dimensional vector space over $K$ then $V$ is necessarly isomorphic to $K^n$.

Hope I was clear enought...
P.S are you from Odessa?

P.S are you from Odessa?[/quote]
Jes, I live in Sweden, but I came från Odessa for 7 years ago.
I cannot good English, but I study mathematic Fraleign "Abstrakt Algebra"
and I have very difficulties.
Thank You very much. :lol:

P.S.
Can You give me Russian useful mathematikal lanks?

Best regards
Tanja
 :?:

Try not to write in Russian if you're not in a national forum.
Check your private messages however, I've send you one :)


geometry
floor function
trigonometry
AMC

Consider the system: $ |z^{2}+az+1|=1$ and $ |z|=1$ where $ a$ and $ z$ are complex numbers. Let $ S$ be the set of all points in the complex plane such that: if $ a$ is in $ S$, then there is a solution for $ z$. If $ A$ is the area of $ S$ graphed in the complex plane, find $ \lfloor 10A\rfloor$.

[hide="hint"]
$ z=\cos{\theta}+i\sin{\theta}\implies z^{2}-2z\cos{\theta}+1=0$.[/hide]

There is a slightly faster solution; factor out a $ z$.

[quote="cancer"][quote="K81o7"]
$ z=\cos{\theta}+i\sin{\theta}\implies z^{2}-2z\cos{\theta}+1=0$[/quote]

Umm... how to continue?[/quote]
[hide]
$ z^{2}-2x\cos{\theta}+1=0$
$ z^{2}+az+1=z(2\cos{\theta}+a)$
$ \implies |2\cos{\theta}+a|=1$
This gives that there is a solution for $ z$ as long as $ a$ is within $ 2$ of a root of unity (horizontally).  Then the region in question is formed by taking the unit circle and "dragging" in 2 units to the right, and also two units to the left, and taking everything it covers.  The area in question is then $ \pi+8$, and so our final value is $ 80+31=111$[/hide]

Or, as Altheman said, you can factor out a $ z$ as follows

[hide]
$ |z^{2}+az+1|=|z+a+\frac{1}{z}|$.  But $ \frac{1}{z}=\overline{z}$.  Thus, $ z+\frac{1}{z}=2\cos{\theta}$.  And proceed as before.[/hide]

Yea once you have that $ a=2\cos \theta+e^{i\phi}$, you can see that it is a circle that is shifted horizontally left and right two units.

This was one of my favorite problems of that test. I think it is pretty unique as far as these problems go. The solution is also very short and elegant.


Hi!

Let a, b and c three complex numbers.
Show that :
$|a + b| + |a + c| + |b + c| \leq |a| + |b| + |c| + |a + b + c|$

Good luck! :D

Come on!!! This has already been posted many times before.It's Hlawka's inequality. ;)

heh...  :?

I see this inequality has been mentioned here :
http://www.mathlinks.ro/Forum/viewtopic.php?p=9422#p9422
http://www.mathlinks.ro/Forum/viewtopic.php?p=269015#p269015
http://www.mathlinks.ro/Forum/viewtopic.php?p=276019#p276019

But, I can't find a proper proof...
 :what?:

Not real hard... arrange vectors a, b, c so that they and a+b+c form a quadrilateral.  Two of the given LHS lengths are the diagonals of the quadrilateral, so let's say they are |a+c| and |c+b|.  Then $|c|+|a+b+c| \ge |a+c| + |c+b|$ and $|a| + |b| \ge |a+b|$.

Wow.. nicely done! :)
Thanks a lot dude!

[quote="probability1.01"] Two of the given LHS lengths are the diagonals of the quadrilateral, so let's say they are |a+c| and |c+b|. [/quote]

this step is a little "fuzzy" for me, to say the least

could you show how your procedure works for $\overrightarrow{a}=1\overrightarrow{i}+2\overrightarrow{j}$, $\overrightarrow{b}=3\overrightarrow{i}+2\overrightarrow{j}$ and $\overrightarrow{c}=4\overrightarrow{i}+5\overrightarrow{j}$ ?

Hmmm... actually what I said was wrong (even if the quadrilateral thing is fully justifiable).  I'm not sure what I was thinking at the time...


I just thought of this quick dumb question which is giving my friends trouble for some reason :rotfl:  :rotfl:

Suppose we drop a ball from $h$ metres above the ground.Now suppose we give it some initial velocity by dragging it down from up and leave it as soon as our hand comes down to $h$ meters above the ground.In which case will the ball reach the ground faster?

who were those "friends" shady??? :rotfl: this was dummmmbbb!!!

[quote="thatin"]h=Vo+gt<=> t=(h-Vo)/g
Vo=0 => to=h/g
Vo>0 => t= h/g -Vo/g= to-Vo/g => t<to
 ok?[/quote]

Your dimension is wrong!


calculus
integration
function
calculus computations

let
$ f(x,y) \equal{} y^{ \minus{} 2}$ for $ 0 < x < y < 1$
and $ f(x,y) \equal{} \minus{} x^{ \minus{} 2}$ for $ 0 < y< x < 1$ and $ f(x,y) \equal{} 0$ otherwise
let $ S$ be the unit square $ [0,1]\times[0,1]$

Show that $ f$ is not integrable on $ S$, but $ f$ is integrable on $ [0,1]$ as a function $ x$ or $ y$.

Also, Show that $ \int_0^1\int_0^1 f(x,y)dxdy$ and $ \int_0^1\int_0^1 f(x,y)dydx$ both exist and are unequal.

I haven't really read the problem or considered the functions but it looks like at the bottom you are trying to show a counterexample to Fubini's Theorem.  This occurs when there is a bad discontinuity in the region in question so there must be a bad discontinuity in the function.

With a piecewise continuous function like this, the only thing "not integrable" can mean is "not absolutely integrable": $ \int_S |f|\equal{}\infty$. By Tonelli's theorem, you can test that by integrating in either order. $ \int_0^1\int_0^1 |f(x,y)|\,dx\,dy\equal{}\int_0^1\left(\int_0^y y^{\minus{}2}\,dx\plus{}\int_y^1 x^{\minus{}2}\,dx\right)\,dy\equal{}\int_0^1 y^{\minus{}1}\plus{}y^{\minus{}1}\minus{}1\,dy\equal{}\infty$

As for the rest- just do the calculations.


i think they subtract the amount left from the standard level to get the AMOUNT DECAYED which helps in calculating age

[quote="rituraj007"]i think they subtract the amount left from the [b]standard level[/b] to get the AMOUNT DECAYED which helps in calculating age[/quote]
that seems to be the only solution, which i also thought of.
But is there really a "standard level" for all organisms?  :huh:

[u]it is[/u] (approximately) depending on location of tree

[quote="hash_include"]A possibly silly question, related to a 2006 JEE chemistry comprehension question :

The theory behind carbon dating technique of finding the age of a fossil, is that the C14 content in an organism (say plant) is constant while it is living, yet the C14 amount decays (according to first order kinetics) after it dies. Since half life of C14 is known, it gives a possibility for scientists to find out the age of the plant, i.e. the time passed since it died.
But my question is: while finding the age, do the scientists know the amount of C14 that HAS BEEN decayed, or the amount of C14 YET TO BE decayed?

My logical assumption was that they know the amount yet to be decayed. But the answer turned out to be the former.

Any idea?  :ninja:[/quote]

http://www.fiitjee.com/down/sol/chemistry06.pdf

Well logically speaking. The scientists have only the fossil in hand... So they analyze it and find the amoung of carbon14 in it. At the same time (if available, dinosaurs are dead but still...) they get a fresh sample and find the amount of carbon in it.

Only then can they predict anything about its age.

Edit: Goto the FIITJEE link and read the last few lines. The proportion of C14:C12 in living matter is 1:10^12 and ofcourse C12 does not decay.


How many non-negative integers have a square less than 30?

We see that $ \sqrt{30}\ge5$. Thus there are $ 6$, including $ 0$ non-negative intergers with a square less that 30.


Suppose it is now the month of November. What month will
it be 100 calendar months from now?

i though i've seen this one but i'll give it a go.  Since 100 leaves a remainder of 4 when divided by 12.  December,Jan,Feb,March
March right?
I hope so.

i've already posted this topic.
[hide]march[/hide]


I recently heard about the butterfly effect which i think says that the flap of a butterfly's wing can cause a huge wave in the ocean.  I don't see how this is possible.  My mother says its true.  There's also a move called the Butterfly Effect but i don't know if it has to do with this topic. And its rated R so i can't watch it. :( Can anyone explain the Butterfly Effect in detail?

Will maybe help you:

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

http://www.cmp.caltech.edu/~mcc/chaos_new/Lorenz.html (see the Java thing)

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

I see it simply like this:

If you balance a particle on the top of a sphere, in real life it will roll off in some direction. But to determine that direction depends on knowing the EXACT initial conditions, for even a small change (like a butterfly flapping its wings) could change the outcome of the result.

Small change in input leads to huge change in output.

The film uses part of this idea...the main character keeps travelling back in time to correct things from his past only to find that the implications in the future are greater than he could ever imagine.


topology
advanced fields
advanced fields solved

Let $A$ be the convex set of $\mathbb{R}^2$ including the origin $o=(0,0).$
(1) Show that $A$ is arcwise connected.
(2) Find the fundamental group $\pi_1 (A,o).$

A convex set is connected by segments, not just by arcs, as it easily follows from the definition of convexity.
A convex set is contractible, hence $\pi_1(A,o)$ is trivial.


calculus
integration
trigonometry
calculus computations

integrate $ sin2x/(sin^4x\plus{}cos^4x)$ with respect to x

$ \int \frac{\sin 2x dx}{\sin^4 x \plus{} \cos ^4 x}\equal{} \int \frac{2\sin x \cos x dx}{\sin^4 x \plus{} (1\minus{}\sin^2 x)^2}$

$ \equal{}\int \frac{2udu}{(u^2)^2\plus{}(1\minus{}u^2)^2}\equal{}\int \frac{dv}{v^2\plus{}(1\minus{}v)^2}\equal{}\int \frac{\frac{1}{2}dv}{v^2\minus{}v\plus{}\frac{1}{2}}$

$ \equal{}\int \frac{2dv}{(2v\minus{}1)^2\plus{}1}\equal{}\int \frac{dw}{w^2\plus{}1} \equal{} \arctan w\plus{}C$

$ \equal{}\arctan (2v\minus{}1) \plus{}C \equal{} \arctan (2u^2\minus{}1) \plus{}C \equal{} \arctan (2 \sin^2 x \minus{}1)\plus{}C\equal{} \arctan (\cos 2x)\plus{}C$

hello, we have $ \frac{\sin(2x)}{\sin(x)^4\plus{}\cos(x)^4}\equal{}\frac{2\sin(2x)}{2\minus{}sin(2x)^2}\equal{}\frac{2\sin(2x)}{1\plus{}\cos^2(2x)}$. Now we have $ \minus{}\int \frac{\minus{}2\sin(2x)}{1\plus{}\cos(2x)^2}\,dx\equal{}\minus{}\int\frac{dt}{1\plus{}t^2}\equal{}\minus{}\arctan(\cos(2x))\plus{}C$, when we set $ t\equal{}\cos(2x)$.
Sonnhard.

hello kenn4000, when i differentiate your result with respect to $ x$ i get
$ (\arctan(\cos(2x))+C)^'=-\frac{2\sin(2x)}{1+\cos(2x)^2}$.
Sonnhard.

oooops

$ \arctan (\minus{}\cos 2x)$ is what i had, but i stupidly treated arctan as even  :blush:


calculus
integration
algebra
polynomial
probability and stats

A famous result of Ito in 1951 gives the following formula for n times iterated Ito integrals:

$ n! \int \cdots ( \int ( \int dB_{u_1})dB_{u_2})\cdots dB_{u_n}=t^{\frac{n}{2}}h_n(\frac{B_t}{\sqrt t})$
where $ h_n$ is the hermite polynomial of degree n.

a) Verify each of these n Ito integrals has an integrand which satisfies the requirements,
(i) $ (t,w)\to f(t,w)$ is $ B\times F$ measurable for $ B$ borel algebra,
(ii) $ f(t,w)$ is $ F_t$ adapted
(iii) $ E(\int_0^T f(t,w)^2dt)<\infty$

b) prove this holds for n=1,2,3

c) use b) to prove $ B_t^3-3tB_t$ is a martingale

so, b) is easy since its just computations, but I cant see how I can use it to prove c), besides the fact that it is the iterated integral for n=3.  Also, I just kind of hand waved my way through part a), so I would appreciate if someone could show me if something in there isnt trivial.

Thanks

A stochastic integral with respect to wiener process is a martingale, so the result is clear.

Concerning your solution for a/, expose it, I 'll try to tell if I see something wrong in it


Let $ a,b,c \geq 0$ such that $ a\plus{}b\plus{}c\equal{}2$. Prove that
  $ 1.(a^2\plus{}b^2\plus{}c^2)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2) \leq 2$
  $ 2.(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(ab\plus{}bc\plus{}ca) \leq 1$
  $ 3.(a^2\plus{}b^2\plus{}c^2)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(ab\plus{}bc\plus{}ca) \leq 2$

(1) and (2) can be solved easily by pqr.
(3) was created from: $ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca) \leq\ 2$ and $ \sum\ a^2b^2 \leq\ 1$


LaTeX
combinatorics unsolved
combinatorics

I'm having trouble with this problem:

[color=brown]Prove or disprove: $H$ is an induced subgraph of $G$ if and only if there exists a set $W$ such that $W\subseteq V(G)$ and $H=G-W$.[/color]

I was going to try and put what I have so far, but it takes forever to do this LaTex code stuff.

If you have any tips I would really appreciate it.

[quote]I was going to try and put what I have so far, but it takes forever to do this LaTex code stuff. [/quote]
Give it a try anyway - it's good for you. Use the preview button frequently. Split it into several smaller posts if you have to. The very act of having to try to organize it to present it to us might well move you further toward understanding.


1. A cart is 52g  and is moving at 5m/s in the x direction on the ground. A mass of 2kg comes falling down at 4m/s and lands in the cart. What is the final velocity of the cart in the x direction after the mass lands on it?

2. Superman is flying in the air and catches a mass of 4 kg which is travelling at 9.8m/s. Superman is 50kg and is intially traveling at 6m/s. He catches the mass and keeps flying. What is his velocity?

Just conservation of momentum and/or energy.  Sorry, I'm too lazy to actually solve them.


[color=green]Could anybody prove the following result?

Let ABC be a triangle, let a, b, c be the perpendicular bisectors of BC, AC, AB respectly and let M[size=75]a[/size], M[size=75]b[/size], M[size=75]c[/size] be the midpoints of  BC, AC, AB respectly. let's take A', B' C' in a, b, c respectly ( A and A' are not in the same semiplane with respect to BC, in the same way to B' and C') Then AA', BB' and CC' are concurrent if and only if M[size=75]a[/size]A'/BC=M[size=75]b[/size]B'/AC=M[size=75]c[/size]C'/BC.

José Carlos[/color]

[quote="hucht"]Let ABC be a triangle, let a, b, c be the perpendicular bisectors of BC, AC, AB respectly and let M[size=75]a[/size], M[size=75]b[/size], M[size=75]c[/size] be the midpoints of  BC, AC, AB respectly. let's take A', B' C' in a, b, c respectly ( A and A' are not in the same semiplane with respect to BC, in the same way to B' and C') Then AA', BB' and CC' are concurrent if and only if M[size=75]a[/size]A'/BC=M[size=75]b[/size]B'/AC=M[size=75]c[/size]C'/BC.[/quote]

At first, I guess the equation should be $\frac{M_aA^{\prime}}{BC}=\frac{M_bB^{\prime}}{CA}=\frac{M_cC^{\prime}}{AB}$.

Then, the "only if" part is wrong (e. g., if you take A', B', C' to be the vertices of the tangential triangle of triangle ABC, then the lines AA', BB', CC' concur at the symmedian point of triangle ABC, but the equation $\frac{M_aA^{\prime}}{BC}=\frac{M_bB^{\prime}}{CA}=\frac{M_cC^{\prime}}{AB}$ is, generally, not satisfied).

The "if" part is well-known: If $\frac{M_aA^{\prime}}{BC}=\frac{M_bB^{\prime}}{CA}=\frac{M_cC^{\prime}}{AB}$, then the triangles BA'C, CB'A, AC'B are directly similar isosceles triangles erected on the sides BC, CA, AB of triangle ABC (in fact, they are isosceles since the points A', B', C' lie on the perpendicular bisectors of the segments BC, CA, AB, and they are similar because of the equation $\frac{M_aA^{\prime}}{BC}=\frac{M_bB^{\prime}}{CA}=\frac{M_cC^{\prime}}{AB}$). Hence, by the Kiepert theorem ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=186725#p186725]http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 post #2[/url]), the lines AA', BB', CC' are concurrent.

  Darij


Call a positive integer [u]nice[/u] if

n = x2 + y3 + z7

for certain positive integers x, y and z.
Do there exist infinitely many numbers which are [u]not[/u] nice?

It is a wonderful problem I think :)
The solution is simple and is mainly combinatorial.

Here's my solution.  [hide]Yes, there exist infinitely many non-nice integers.  To see why, let N be a (large) positive integer.  In the range 1..N, there are at most N1/2 squares, N1/3 cubes, and N1/7 7th powers.  Thus in 1..N, there are at most N1/2N1/3N1/7 = N41/42 nice integers.  That means that in 1..N, there are at least N - N41/42 non-nice integers.  Now take the limit as N goes to infinity.      [/hide]

I solved it the same way.  Clever problem.  Someone had to beat me to posting the solution, though...  :(

Very nice problem (no pun intended there :-D)! And good job on the solution Ravi and Set!


I need some help with these 2 questions the first one has to do with force as a vector:

1) Three forces, each of which is perpendicular to the other two, act on an object. If the magnitudes of these forces are 6N, 15N and 10N, respectively, find the magnitude and direction of the resultant. (State the angles that the resultant makes with the two largwer forces.)

Th second question has to do with algebraic vectors:

2) a) Find the length of the medium AM in the triangle ABC, for the points: A( 2 , 3/2 , -4 ), B( 3, -4 , 2 ), and C( 1 , 3 , -7 ).

    b) Find the distance from A to the centroid of the triangle.

Any Help as to how to get started on these would be very appreciated, thanks. PEACE :)

This really doesn't belong in the FTW problem discussion forum...


Hay, I am Sarthak Kalani of class XI studying in jaipur. I too have given NSEA 2nd level round. I have joined today only and want to make friends. Please respond

Hi sarthak! I am Rak from Tamil nadu class XII I think u shall post this in the Welcome india stickie

Yes do that  :D . And dont worry we are all ur friends and dont mind the stupid mod of this forum  :P   :rotfl:

hi  i am ur 'stupid mod' :D
[url=http://www.mathlinks.ro/index.php?f=332]Here[/url] is the link to advanced physics forum again moderated by me...do post ur problems  :D

Can I be guises further in the forum? :huh:

I meant to be guided futher. Well, according to me, i am getting about 70 marks in 2nd level astronomy olympiad. May I have the privelege of knowing your marks?Please also tell me shall I be selected??

There's a separate thread fer that, mpd do ur duty!  :D

actually I am new to this site and since my tping speed is not so fast so please inform me if I commit a mistake anywhere.

[quote="madness"]Well there was no offense, see my signature, i just qualified u as human  :D   :rotfl:   :P[/quote]

I would feel extremely delighted if I woud have been called "STUPID" bcz:
S:-SMART
T:-TALENTED
U:-UNIQUE
P:-POLITE
I:-INTELLIGENT
D:-DILLIGENT

actually he referred to me as stupid :rotfl:

Why dont you merge this with the Intro India thread??

Hey Sarthak

And yeah, mod do your duty da... Merge this with the intro to india thread

actually there is no use doing that because the all but the first post can only be moved so once a thread is made it can either get delted or all except the first  post moved...so
SARTHAK  re-introduce urself again in the introduction STICKY :)
i am locking this


Find all f: N to N such that:
  f(f(n)+f(m))=n+m for all m,n in N

same as this shortlist problem : [url]http://www.kalva.demon.co.uk/short/soln/sh8819.html[/url]


inequalities
LaTeX
inequalities unsolved

Prove, that if R_i >=1 the foloving inequality holds:

$ \sum^{n}_{i \equal{} 1}{\frac {{R}_{i}}{R_{i} \plus{} 1}}\leq\sqrt [n]{\prod^{n}_{i \equal{} 1}{R_{i}}}\cdot(\sum^{n}_{i\equal{}1}\frac{1}{R_{i}\plus{}1})$

I don't belive that is true

take R_i=1 and n>=2

Yes i made a mistake writing it,... i am new with this latex, cant handle it well yet

Denote $ \prod R_i\equal{}g^n$, $ \sum \frac1{1\plus{}R_i}\equal{}ns$, then $ \sum \frac{R_i}{1\plus{}R_i}\equal{}n(1\minus{}s)$, so we need to prove $ 1\minus{}s\le gs$, i.e. $ \frac1{1\plus{}g}\le s$. Putting $ R_i\equal{}e^{x_i}$ and $ f(x)\equal{}\frac1{1\plus{}e^x}$, we rewrite our inequality as \[ \frac1n \sum f(x_i)\ge f\left(\frac1n \sum x_i\right).\]
This is just Jensen inequality, so we need to check that $ f$ is convex on $ (0,\infty)$. We have $ f'(x)\equal{}\minus{}1/(e^x\plus{}e^{\minus{}x}\plus{}2)$, so $ f'$ increases on $ (0,\plus{}\infty)$.


geometry
geometric transformation
reflection
power of a point
radical axis
projective geometry
geometry unsolved

For circle $ (A)$ and point $ O$ outside it, $ PB$ is tangent of circle $ (A)$ at $ B$ and $ OP$=$ PB$.
$ C$ is the reflection of $ B$ in $ PA$. When $ B$ moves on circle $ (A)$,

Prove

(a) the locus of $ P$ is line ( let's call it $ l$ ) and $ l$ is perpendicular to $ OA$
(b) $ BC$ always meets $ OA$ at a fixed point $ E$
(c) If $ D$ is $ l \cap OA$,  $ D$ and $ E$ are inverse in circle $ (A)$

I solved only (a) at most analytically  :(
Help me, please. I hope sythetic proofs. :blush:

Maybe easy?  :D
(a) $ DA^2\minus{}DO^2\equal{}PA^2\minus{}PO^2\equal{}PA^2\minus{}PB^2\equal{}R^2$, so D is a fixed point.
(b) PA meets BC at Q, $ AE.AD\equal{}AQ.AP\equal{}AB^2\equal{}R^2$, so E is also fixed point.
(c) It is obvious because $ AE.AD\equal{}R^2$

Thank you
Well, I don't know what  you proved out of (a),(b),(c).
It seems that you don't prove any. please look at my post in detail.

and one more question please
I see the locus of $ Q$ in gsp in following figure.
Is it well-known ?

[quote="77ant"]Thank you
Well, I don't know what  you proved out of (a),(b),(c).
[b]It seems that you don't prove any. please look at my post in detail.[/b]

Ah, one more question please
I see the locus of Q in gsp in following figure.
Is it well-known ?[/quote]
??? Can you explain about your post :?:  The point Q in your last post is a new point, isn't it ? In my post, Q is the intersection of AP and BC.

$ BC\cap OA$=$ E$
$ BO\cap AP$=$ Q$

1) Given a point $ O$ and a circle $ (A)$, let a point $ B$ be on a circle $ (A)$
2) At $ B$, draw the tangent of circle $ (A)$
3) On the tangent line, take a point $ P$ such that $ PB$=$ OP$
4) Take a point $ C$ of the reflection of $ B$

Everything else is the same like the above

Please read my post in detail.
May I ask you a favor ?

I think mr.danh don't solve particular (not detail) but his idea is true.
His idea is
Let $ D$ lie on $ AO$ and $ PD\perp AO$ and prove $ D$ is fixed point

[quote="mr.danh"] (a) $ DA^2 \minus{} DO^2 \equal{} PA^2 \minus{} PO^2 \equal{} PA^2 \minus{} PB^2 \equal{} R^2$, so D is a fixed point.[/quote]
Also, we can say that the point $ P$ lies on the radical axis of the circles $ (A)$ $ ($ with radius $ R \equal{} OA$ $ )$ and the circle $ (O)$ $ ($ with radius $ R' \equal{} 0$ $ ).$

So, the point $ P$ lies on a line perpendicular to $ OA,$ which is the line segment connecting the centers of the circles $ (A)$ and $ (O).$

From the above relation $ (DA)^{2} \minus{} (DO)^{2} \equal{} R^{2}$ $ \Longrightarrow$ $ (DA \plus{} DO)\cdot (DA \minus{} DO) \equal{} R^{2}$ $ \Longrightarrow$ $ \boxed{DA \equal{} \frac {(OA)^{2} \plus{} R^{2}}{2\cdot (OA)}}$ $ ,(1)$

Hence, $ D$ is a fixed point between $ O,\ A.$

[quote="mr.danh"](b) PA meets BC at Q, $ AE.AD \equal{} AQ.AP \equal{} AB^2 \equal{} R^2$, so E is also fixed point. [/quote]
Also we can say that because of the point $ P$ lies on the line segment $ PD\perp OA,$ then, the line segment $ BC$ as the polar of $ P$ with respect to the circle $ (A),$ intersects the line segment $ OA$ at one point so be it $ E,$ which is a fixed point as well $ ($ the point $ E\equiv DA\cap BC,$ is the harmonic conjugate of $ D,$ with respect to the points $ Z,\ Z',$ as the intersection points of the circle $ (A),$ from the line segment $ DA\equiv OA$ $ ).$

Kostas Vittas.


A circle radius r is inscribed in ABCD.  It touches AB at P and CD at Q.  AP = 19, PB=26, CQ=37, QD=23.  Find r.

I a little lost as to how to finish this problem.  Using the fact that tangents are equal, you know the side lengeths of the quadrilateral.  Ptolemy's theroem won't work.  Now is there a formula for the radius?  If so I assume it would be similar to the heron's formula.

Let the circle touch BC and DA at R and S respectively. Define the four angles
A=<SOA=<AOP
B=<POB=<BOR
C=<ROC=<COQ
D=<ZOD=<DOS,
and notice that A+B+C+D=180. Since we also know that tan(A)=a/r=19/r, etc., we want to get some tangents in there. So, we do the following:
$0 = \tan(A+B+C+D) \\
= \frac{\tan(A+B)+\tan(C+D)}{1-\tan(A+B)\tan(C+D)} \\
= \tan(A+B) + \tan(C+D) \\
= \frac{\frac ar+\frac br}{1-\frac ar\cdot\frac br} + \frac{\frac cr+\frac dr}{1-\frac cr\cdot\frac dr} \\
= \frac{r^3(a+b+c+d)-r(abc+bcd+cda+dab)}{\text{something}} \\
=r^2(a+b+c+d)-(abc+bcd+cda+dab),$
so $r^2=\frac{abc+bcd+cda+dab}{a+b+c+d}=647$.


I'm not trying to flame you but...

These are generally bad questions to ask... for instance, my sequence could be:
"First a 4, then 32432424654, then pi, then all 6's..."

Thanks

Is this an exponential series? And PhilB, sometimes these can be logical. [i]Sometmes[/i]

(Yagaron, you were the one that flamed me the other time, right?)

Well, all I can see is that $\frac{1}{7}$  is $.142857$ repeating, so is it taking some numbers in the number, and dividing by the others to give a repeating decimal, and so on?

yes. but in the last number we have a division by zero.

so. there is no next number.

you don't need to know how to divide the numbers. you just need to know that you need to know that you divides the first algarism by some arrangement of the anothers. a multiplication or division. does not matter. the result is division by zero.

it's just a trick


geometry
angle bisector
perpendicular bisector

Can someone please help me with these two problems?

1. Construct a circle tangent to two given lines that passes through a certain point.

2. Construct a circle tangent to a line and two points.

Also, please do not just give a link to the Problem of Apollonius unless that link explicitly gives the construction for one of the above two (or both) - I am having trouble finding links with regards to the Problem of Apollonius that explicitedly gives the construction for these two.

Thanks.

[quote]Get the perpendiclar lines on B,C repectively [/quote]

the two lines through B,C perpendicular to which other two lines??? To WY (see my picture) an YZ, or to PA and PB?
Also, I had thought for sure that one of the steps would be to construct the angle bisector of $ \angle WYZ$ since the center O of the circle we want to construct clearly must be on this angle bisector.

Ok, I made a not so great geogebra picture, so just move the points around yourself if you want the circle to look like its tangent to WY and YZ and goes through point P. (also, there is some extra stuff you can ignore off to the right that I used to simply construct the angle bisector - I don't know how to hide objects)

 [geogebra]94980c9ce3a647fb9135b9d28b5847ca3bb63796[/geogebra]

Now, hopefully it's easier to talk about this problem

[quote]You mean :
Use ruler and compass to construct ?
Problem (1) : If any, you may use a compass to drop two equal lengths from the given point , say A. B is marked on one line when setting equal length ,C is marked on another one line when setting equal length. Get the perpendiclar lines on B,C repectively . The intersection of two perpendiculars ,say O which is the center of the circle .OA=OB is the radius of the circle.

Probelm (2) : The construct way is more elegant than that of problem(1) , but is more complicated to state without diagrams , waiting for somebody help you.[/quote]

Can you please explain in more detail, using the points labeled as shown in my geogebra picture?

Ok, here is the picture for the second one.

 [geogebra]c304e2fde01581f1cc1aa27cedceb9f5c7e966f8[/geogebra]

Here are my ideas for this one:

 [geogebra]64922089a4723c23578207046c9aea894d82824e[/geogebra]

[hide]I get the feeling we should try to somehow construct the third point D on the line, using the fact that the center is on the perpendicular bisector of EF drawn above, and maybe using the fact that we have similar right triangles in the diagram above[/hide]


I will prove that for all positive real numbers a and b the inequality
a3/(a2 + b2) \geq a - b/2
is true. Indeed, the claimed inequality is equivalent to
2a3 \geq (2a - b)(a2 + b2)
or
2a3 \geq 2a3 + 2ab2 - a2b - b3
or
a2b + b3 \geq 2ab2
which is obviously true by the AM - GM inequality.
So, using the above established inequality we get
x3/(x2 + y2) \geq x - y/2
y3/(y2 + z2) \geq y - z/2
z3/(z2 + x2) \geq z - x/2
and the result now easily follows from adding these three inequalities.

Your "problem" was solved .. I moved it to algebra! :D

For $x,y,z>0$, prove or disprove that: when $0\leq p\leq6$,
$\frac{x^{p+1}}{x^{p}+y^{p}}+\frac{y^{p+1}}{y^{p}+z^{p}}+\frac{z^{p+1}}{z^{p}+x^{p}}\geq\frac{x+y+z}{2}$;
when $-6\leq p\leq0$, the inequality inversely holds.

Not prove or disprove. (prove only  :) ). This famous problem has been proved for n=3,4 (by Hungkhtn) and 5,6 (by Phan Thanh Nam) and discussed in Vietnamese forum. The solution is too long, so I can't post it.

Moreover, if it true for $n\in R$, we can prove that it is still true for all $r\le n$.


Each of eight houses on a street is painted brown, yellow or white. Each house is painted only one color and each color is used on at least one house. No two colors are used to paint the same number of houses. In how many ways could the eight houses on the street be painted?

its solution is pretty easy if u think in a different manner.
take like that if u have to fill three boxes with eight balls of different colors then without repetition of any color u can fill the boxes in 8*7*6=336 ways.
it is same with the question

Does this solution make any sense?  It was the one given in alcumus, but I don't understand why you would multiply 3! for the ways to assign brown, yellow, and white to the letter A,B,and C.  Wouldn't it be easier to use the colors in the first place? Also does anyone check this forum?  :P

Given the restrictions there are 2 possible color distributions for the houses:
1 house color A, 2 houses color B, and 5 houses color C
OR
1 house color A, 3 houses color B, and 4 houses color C.
In the first case, there are $ \frac{8!}{1!\cdot2!\cdot5!}$ ways of assigning letters A, B, and C to the eight houses. In the second case, there are $ \frac{8!}{1!\cdot3!\cdot4!}$ ways of assigning letters A, B, and C to the eights houses. In both cases, there are 3! ways of assigning the colors brown, yellow, and white to the letters A, B, and C. This gives a total of $ \left(\frac{8!}{1!\cdot2!\cdot5!}\plus{}\frac{8!}{1!\cdot3!\cdot4!}\right)\cdot3!\equal{}\boxed{2688}$ ways of painting the houses.

You multiply by 3 cause that's the number of ways to arrange the 3 colors.

[quote=PersonPsychopath]Isn't this possible to do with complementary counting?[/quote]

This is a bump but such action is taken for positive reason. How would you do complementary counting in the problem? You would have to count the number ways there are to paint the same number of houses one color then use PIE to try to not overcount and then you have to find the number of ways for which there are $0$ houses that have one color and then bash PIE because there can be more than one color that does not have any houses and it follows that you have to use PIE for the third time in the complementary bash sol because you have have $0$ houses of a color and also have two colors that have the same number of houses. So can you post the sol using complementary counting, or if anybody else has sol, can you post?

how is this level 25?

complementary counting is possible but it's long winded and unnecessary

based off of the restrictions, there are only 2 ways to do it. 1 2 5 or 1 3 4. amount of ways to arrange is $\frac{8!}{1! \cdot 2! \cdot 5!}$ and $\frac{8!}{1! \cdot 3! \cdot 4!}$ respectively. [hide]adding gives $448$, but each can be arranged in $3!=6$ ways, so it is $448 \cdot 6 = \boxed{2688}$.
[/hide]

ALMOST forgot about ordering
phew that was close

[quote=Adul]@above you realize the answer is revealed[/quote]

Obviously, this is the GameBot thread.


If you mean find the zeroes, here's the method:

$ y \equal{} x \minus{} \frac{1}{x}$, quadratic x3


Prove that the product of the lengths of the internal bisectors of the angles of a triangle is less than the product of the length of the sides

Let la, lb, lc be three length of bisector angle in a triangle.

Now we try to prove this one

la.lb.lc <= sqrt(27/64).a.b.c < a.b.c  (*)

We use these formula :
la = (2b.c.cos(A/2))/(b+c)
lb = (2a.c.cos(B/2))/(a+c)
lc = (2b.a.cos(C/2))/(b+a)

(*) is now equivalent to this one

8abc.cos(A/2).cos(B/2).cos(C/2) <=  sqrt(27/64).(a+b)(b+c)(c+a)  (**)

And (**) is proved if we use 2 lemmas

Lemma 1 : For all non negative x,y,z we have
   8xyz<=(x+y)(y+z)(x+z)
Lemma 2:  For all triangle ABC we have
   cos(A/2).cos(B/2).cos(C/2) <= sqrt (27/64)

The equality case occur iff ABC is an equilateral triangle.


Hi Folks: This one should be easy for you math experts out there ...

Attached is a simple spreadsheet. Blue cells are input cells (each column has two of them for input variables). The yellow cells are the one I solved for. Notice the third column for Mark Up (I am unable to find the right fomula) that will produce the right answer (150) using the two input variables of 33.3333% and 100.00. What is the right formula to solve this problem in cell D5 using D4 and D8 only? Thank you so much!

Or worded differently, how do I solve the following:

Cost = 100.00 (kinown variable)
Mark up = 33.3333% (known variable)
Sales Price = x (answer should be 150.00)

Check out a search engine:
[url]http://www.google.com/search?hl=en&q=Markup+Sales+Price+cost+Gross+Profit&btnG=Search[/url]

[hide="If you really must"]You have D8 labeled as Gross Profit, not markup.  The percentages may need to be relabeled Gross Margin:
\[ \text{SalesPrice}\equal{}\frac{\text{Cost}}{1\minus{}\text{Gross Profit Margin}}\]

The Markup is a percentage relating to the Cost, not the Sales Price, yielding instead:

\[ \text{SalesPrice}\equal{}(1\plus{}\text{Markup})\times\text{Cost}\][/hide]


I finally finished writing those solutions, they probably took me 6 hours to just write the solutions, and solving took even longer! How do you guys feel about your solutions? Are they good? i think mine are okay, but they might be too long and wordy, i have to say, this was quite a good experience to do some proof writing, and i hope that i got my solutions in on time, i sent them in before 9:30 Central Time, so that should be good right, maybe next time i should not procrastinate so much, but i know i will, i have so much to do with school, i have little time for extra things

I finished as well, with about 4 hours to spare :P I've been thinking about the one problem that I still hadn't solved for the whole day, in all my classes and everything. I finally got it done. I heard about the USAMTS by mail, too, on Sep 27th, so I did all of the problems in about a week. It'll be nice to have more time to do the next round of problems.

It's about 8:00PM AoPS time right now (pacific), so we have about 4hrs left, don't we? I hope we do! :?

I solve everything except number 3 in one week, then that one took me a WHOLE month.....

I almost went crazy, every time I saw a blank space on a paper I start writing my works on there to try to solve it.....

I like that problem, tho

I solved everything saturday, sunday, and today (still finishing #5) except #1 I had after about the first week :blush:

It's funny how some of us can find a problem very easy while others of us find it nearly impossible. I did 5 within an hour of reading the problem but what took me the longest was 4. But I guess we have to hold off on discussing which actual elements of the problems made them hard or easy to us. I guess it's just the way problem-solving works. You either get some genius idea of how to solve a problem, or you don't, in which case you have to think about it for much longer before finding a solution.

I don't think we should be discussing anything about the problems until after the Math Jam on Wednesday, just in case we might be giving something away to somebody who hasn't submitted his/her solutions yet.

[quote="sonar"]Sent mine an hour ago  :lol:

I look forward to the mass solution posting on Wedneday  :lol:[/quote]
So do I  :D

Just sent mine with 3.5 hours to spare. Feeling paranoid at the moment, so I'm sure something will happen that will result in me getting no points. Urgh.

[quote="4everwise"]Chess64, if you aren't done yet, be careful. It took me an HOUR to get my fax in.

Now I'm in trouble cause I'm only halfway done with my homework. Panic! Panic! Panic!  :blush:[/quote]

I'm emailing mine :)

I only got 4 and predicting a 15ish score...... I was very lazy in doing the problems.

How many times you check your solution (grammar, answer, etc.) after your done??

I check like 5 times (I feel my english improved just by doing this... :rotfl:)

Also, how many pages are your solutions total??

I believe it would be prudent to stop talking about the problems until Wednesday's math jam.

Okay.

I want to see Zach's solution for problem #3.

[quote="chess64"][quote="ZennyK"][quote="white_horse_king88"]I want to see Zach's solution to all of them... I'm rather selfish on that note.  :blush:

Then I could figure out how to improve my solution for next time...

Plus, does anybody remember last year's USAMTS shirt? Krazy awesome solution by Zach.[/quote]
I know, that was insanely cool. I think he found The Book solution! ;)[/quote]

What USAMTS shirt? :?[/quote]
They gave t-shirts to everyone who participated (or was it just winners?) last year. It featured Zachary Abel's solution to Problem 4, Round 3, which I believe is posted on the USAMTS website under past problems/solutions.

[quote="white_horse_king88"]I just realized I misinterpreted 2 and got it wrong. That's a 15 max - I think a point off another. That's just sad...[/quote]
What was the answer to that?  I got home late and missed most of the math jam :-(

My solutions were handwritten and 12 pages long.

[quote="white_horse_king88"]The answer is: $\dfrac{1}{11} + \dfrac{1}{9} + \dfrac{1}{7} + \dfrac{1}{5} + \dfrac{1}{3} + 1$.

 :cursing:[/quote]
Ooh, so THAT's the elegant way of expressing the solution - I approached it from a completely different angle, so I just got the big messy fraction, and thought "oh shoot, I must have done something wronng, because the answer is really messy."

I got my answer as
$\sum^6_{k=1} \left[ \dfrac{1}{2k-1} \right]$
So expressing it that way was plain to see.

[quote="white_horse_king88"]The answer is: $\dfrac{1}{11} + \dfrac{1}{9} + \dfrac{1}{7} + \dfrac{1}{5} + \dfrac{1}{3} + 1$.

 :cursing:[/quote]
It IS?! Then I actually got something right! YES!!!

[quote="white_horse_king88"]I just realized I misinterpreted 2 and got it wrong. That's a 15 max - I think a point off another. That's just sad...[/quote]
Sorry to hear that. However . . .

If you didn't trivialize the problem by misinterpreting it, then you shouldn't lose many points. I completely misinterpreted a problem last year and got the wrong answer, but the grader still gave me a 4. Apparently I made the problem harder by misreading it. So yeah, unless you are certain you totally trivialized the problem in your misinterpretation, you shouldn't lose hope yet.

Sweet, I got it right! How do you find that really simple answer, though? :what?: I did it the longer way, although it still wasn't very complicated...

I had essentially no time to do the solutions, or even just get answers, due to massive amounts of homework, two instruments to practice, and sports, so I spent 8.5 hours writing solutions last Sunday.  They didn't turn out that good (I got 2 and 4 wrong, and my 5 was just ugly), so I expect I'll receive a 20 at the highest.   :(


inequalities
function
inequalities proposed

[quote="Vasc"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ (a) \ \ \ \ \ a^\frac 1{a} \plus{} b^\frac 1{b} \plus{} c^\frac 1{c} \le 3$;

[/quote]

[hide="click me"]

We first prove that if $ c \geq 0$ then $ c^c \geq c$.

If $ c \geq 1$ it's quite obvius that $ c^c \geq c$, so consider $ 0< c < 1$.

Let $ c \equal{} \frac{1}{b} \ \ \ b > 1$.

We have to prove $ \frac{1}{b^{1/b}} \geq \frac{1}{b}$:

$ 1 \geq \frac{b^{1/b}}{b}$

$ 1 \geq b^{(1/b)\minus{}1}$

$ 1 \geq b^{(1\minus{}b)/b}$

$ 1 \geq b^{1\minus{}b}$

that is true because $ 1\minus{}b < 0$. We proved that $ c^c \geq c$, hence $ c \geq c^{\frac{1}{c}}$ for all positive $ c$.

Finally:

$ a\plus{}b\plus{}c \equal{} 3$ and $ a \geq a^{\frac{1}{a}} \ \ b \geq b^{\frac{1}{b}} \ \  c \geq c^{\frac{1}{c}}$

so

$ \boxed{a^{\frac{1}{a}} \plus{} b^{\frac{1}{b}} \plus{} c^{\frac{1}{c}} \leq 3}$
[/hide]

Right?

My 1st reply! :P

[quote="Haile"][quote="Honey_S"][quote="Haile"]

Finally:

$ a \plus{} b \plus{} c \equal{} 3$ and $ a \geq a^{\frac {1}{a}} \ \ b \geq b^{\frac {1}{b}} \ \ c \geq c^{\frac {1}{c}}$
[/quote]

Wrong![/quote]

Could you please be more specific? I cannot find mistakes.[/quote]

Try $ a < 1$.

[quote="Honey_S"][/quote][quote="Haile"][quote="Honey_S"][quote="Haile"]

Finally:

$ a + b + c = 3$ and $ a \geq a^{\frac {1}{a}} \ \ b \geq b^{\frac {1}{b}} \ \ c \geq c^{\frac {1}{c}}$
[/quote]

Wrong![/quote]

Could you please be more specific? I cannot find mistakes.[/quote][quote="Honey_S"]

Try $ a < 1$.[/quote]

$ a = 0.5$

$ a^{1/a} = 0.5^{1/0.5} = 0.5^{2} = 0.25$

It holds $ \forall a \in \mathbb{R^+}$, as I proved in the 1st reply...

[quote="Haile"]

$ a^{1/a} \equal{} 0.5^{1/0.5} \equal{} 0.5^{2} \equal{} 0.25$

It holds $ \forall a \in \mathbb{R^ \plus{} }$, as I proved in the 1st reply...[/quote]

oh yes, so sorry my mistake.

[quote="Vasc"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ (a) \ \ \ \ \ a^\frac 1{a} \plus{} b^\frac 1{b} \plus{} c^\frac 1{c} \le 3$;

$ (b) \ \ \ a^\frac 3{2a} \plus{} b^\frac 3{2b} \plus{} c^\frac 3{2c} \le 3$.[/quote]
Very nice inequality, VasC. :) Here is my solution (I write it in the simplest)
Assume that $ a \ge b \ge c$, then we will consider 2 cases
Case 1. If $ b \le \frac {3}{4}$, then by consider the function $ f(x) \equal{} x^{\frac {3}{2x}}$, we can easily show that
\[ a^{\frac {3}{2a}} \le e^{\frac {3}{2e}}, \quad b^{\frac {3}{2b}} \le \frac {9}{16}, \quad c^{\frac {3}{2c}} \le \frac {9}{16}
\]
Therefore
\[ a^\frac 3{2a} \plus{} b^\frac 3{2b} \plus{} c^\frac 3{2c} \le e^{\frac {3}{2e}} \plus{} \frac {9}{8} < 3.
\]
Case 2. If $ b \ge \frac {3}{4}$, then we can show that $ f(x)$ is concave for all $ x \in \left[\frac {3}{4},3\right]$, therefore
\[ f(a) \plus{} f(b) \le 2f(t) \equal{} 2t^{\frac {3}{2t}}
\]
where $ t \equal{} \frac {a \plus{} b}{2} \le \frac {3}{2}$.
Now, using Bernoulli's Inequality (with note that $ \frac {3}{2t} \ge 1$), we have
\[ \left( \frac {1}{t}\right)^{\frac {3}{2t}} \ge 1 \plus{} \frac {3}{2t}\left( \frac {1}{t} \minus{} 1\right) \equal{} \frac {2t^2 \minus{} 3t \plus{} 3}{2t^2}
\]
Hence
\[ t^{\frac {3}{2t}} \le \frac {2t^2}{2t^2 \minus{} 3t \plus{} 3}
\]
Similarly,
\[ c^{\frac {3}{2c}} \le \frac {2c^2}{2c^2 \minus{} 3c \plus{} 3} \equal{} \frac {2(3 \minus{} 2t)^2}{2(3 \minus{} 2t)^2 \minus{} 3(3 \minus{} 2t) \plus{} 3} \equal{} \frac {(3 \minus{} 2t)^2}{4t^2 \minus{} 9t \plus{} 6}
\]
We have to prove that
\[ \frac {4t^2}{2t^2 \minus{} 3t \plus{} 3} \plus{} \frac {(3 \minus{} 2t)^2}{4t^2 \minus{} 9t \plus{} 6} \le 3
\]
which can be easily simplied to
\[ (2t \minus{} 3)(t \minus{} 1)^2 \le 0
\]
which is true and we are done. :)

What is your proof, VasC? Thanks :)

Nice your solution, Can_Hang. :lol:
[quote="can_hang2007"] What is your proof, VasC? Thanks :)[/quote]
I have proved it for $ n$ numbers (with $ k\equal{}\frac n{n\minus{}1}$, as you stated in the topic for 2 numbers) by RCF-Theorem. In addition, for the case $ a_2\equal{}a_3\equal{}...\equal{}a_n$, I used only Bernoulli's Inequality.


In triangle $ABC$, $AB=AC$ and $\angle{BAC}=100$ degrees. $D$ is on $AC$ such that $\angle{ABD}=\angle{CBD}$. Prove that $AD+DB=BC$.

[color=darkblue]Denote the second intersection $E$ between the line $BC$ and the circumcircle of the triangle $ABD$.
Obtain $DA=DE$,  $BD=BE$ and $ED=EC$. Therefore, $BC=BE+EC=$ $BD+ED=BD+DA\ .$[/color]


$ f: \mathbb{R^{ \plus{} }}\mapsto\mathbb{R^{ \plus{} }}$, $ f$ is continuous and decreasing  function
$ f(x \plus{} y) \plus{} f(f(x) \plus{} f(y)) \equal{} f(f(x \plus{} f(y))) \plus{} f(y \plus{} f(x)))$  $ \implies$
prove that $ f(f(x))\equal{}x$

Prove what?

I'm guessing you should use substitution?

[quote="cnyd"]$ f: \mathbb{R^{ \plus{} }}\mapsto\mathbb{R^{ \plus{} }}$, $ f$ is continuous and decreasing  function
$ f(x \plus{} y) \plus{} f(f(x) \plus{} f(y)) \equal{} f(f(x \plus{} f(y))) \plus{} f(y \plus{} f(x)))$  $ \implies$
prove that $ f(f(x)) \equal{} x$[/quote]

According to me, there is no solution to this equation and so it's meaningless to show that $ f(f(x)) \equal{} x$ :

Since $ f(x) > 0$ and continuous and decreasing, $ \exists a > 0$ such that $ f(a) \equal{} a$

Setting then $ y \equal{} a$ in the original equation, we get $ f(x \plus{} a) \plus{} f(f(x) \plus{} a) \equal{} f(f(x \plus{} a)) \plus{} f(a \plus{} f(x))$ and so :

$ f(f(x \plus{} a)) \equal{} f(x \plus{} a)$ $ \forall x > 0$

But, since $ f(x)$ is decreasing, $ f(x)$ is injective and so $ f(f(x \plus{} a)) \equal{} f(x \plus{} a)$ $ \implies$ $ f(x \plus{} a) \equal{} x \plus{} a$ $ \forall x > 0$ and so $ f(x)$ is increasing $ \forall x > a$, hence a contradiction with "$ f(x)$ decreasing".

I suppose $ f$ has to be increasing^^  :P

the number of '(' $ \neq$ the number of ')' - correct it, cnyd


The lengths, in centimeters, of the sides of a right triangle are
integers, and the sum of the shortest and longest sides is 18 cm.
How many centimeters are in the perimeter of the triangle?

There is only one triple which satisfies this: $ (5,12,13)$.

Thus, the answer is $ 5\plus{}12\plus{}13$, or $ \boxed{30}$.


I need to find the minimum value among the numbers,
a= 3V7 - 3V6
b= 3V8 - 3V7
c= 3V9 - 3V8
d= 3V10 - 3V9
e= 3V11 - 3V10

where 3V means the cube root(sorry about the lack of TeX)
I can't think of a strategy to find the minimum value, the fact that a+b+c+d+e = 3V11 - 3V6
may have some importance but I can't see it.
Any help is appreciated.

Forget about it, it's apparent when you consider the gradient of the curve y=x^3 and observe the points (3V6,6), (3V7,7) ,etc.
the answer is 3V11 - 3V10

LaTeX: \sqrt[n]{m} gives $\sqrt[m]{n}$.  And yes, it's geometrically obvious because of the concavity of the curve.


$ \lim_{n \to \infty} \left [ C_{2n}^{n}.\sqrt{n}.4^{\minus{}n} \right ]$ , $ n\in N$.

By Stirling's formul $ \frac{1}{\sqrt \pi }$.

[quote="Rust"]By Stirling's formul $ \frac {1}{\sqrt \pi }$.[/quote]

Can you post full solution?


inequalities
function
calculus
inequalities proposed

$a,$ $b$ and $c$ are non-negative numbers. Prove that
$a^{5}+b^{5}+c^{5}+6(ab+ac+bc)abc\geq7(a^{2}+b^{2}+c^{2})abc.$

Wlog assume $abc=1$. Then the inequality is equivalent to $\sum\frac{(a^{3}-1)(a^{3}-k)}{a}\ge0$ for $k=6$.
Once again, we can apply the "Right-Convex Function Theorem" ( http://www.mathlinks.ro/Forum/viewtopic.php?p=226013#p226013 ). With some, but not an awful lot, of calculus, it turns out that the inequality holds iff $k\le 3x+2+\frac1x\approx6.7290315$ where $x\approx1.324717957$ is the root of $x^{3}-x-1=0$. :D


Let n be an integer greater than 1 and let $ x_1 ,x_2 , \ldots ,x_n$  be nonnegative real numbers such that$ x_1\plus{}x_2\plus{}...\plus{}x_n \equal{} \sqrt{2}$. Find the minimum value of

\[ \frac{{1 }}{{1 \plus{} x_1^2 }} \plus{} \frac{{1 }}{{1 \plus{} x_2^2 }} \plus{} ... \plus{} \frac{{1}}{{1 \plus{} x_n^2 }}
\]

I have one proof which uses AC-Theorem of VASC, but I think this inequality can solve without AC-Theorem.

[quote="Altheman"]Hello, what is the source of this problem?[/quote]

Uhm, I discovered it when I solve one problem from Mathematical Reflections. And I saw the similar problem from VASC's book where $ x_1\plus{}x_2\plus{}\ldots\plus{}x_n\equal{}s$

[quote="rofler"]Well, we have a concave function and we want to find a minimum, so Jensens just pwned it :P[/quote]

I don't think so, because F is minimal for $ x_1\equal{}...\equal{}x_k\equal{}\frac{1}{k}$ and $ x_{k\plus{}1}\equal{}...\equal{}x_n\equal{}0$ where $ k \in \{1,2,...,n\}$

This is equivalent to a problem in the current issue...


What is the value of $ 2000^2 \minus{} 1999^2$?


Since Canmoo has become such a popular item at the IMO, let me say a few words about its history.

It all started in 2005 (when I was on the Canadian team as a student). During our training in Banff International Research Station, we decided to get a mascot. After some searching, we found a very cute stuffed moose at a local gift shop, and decided to adopt it as our team mascot. I proposed the name "Canmoo." And hence Canmoo was born :)

[img]http://web.mit.edu/yufeiz/www/imo2006/canmoo.jpg[/img]

We used to also have can for the moose (haha, "Can"-moo). But apparently it has been retired over the years.

[img]http://web.mit.edu/yufeiz/www/imo2006/halifaxpics/can1.jpg[/img][img]http://web.mit.edu/yufeiz/www/imo2006/halifaxpics/can2.jpg[/img][img]http://web.mit.edu/yufeiz/www/imo2006/halifaxpics/can3.jpg[/img]

It surprised me how quickly Canmoo has gained popularity at the IMO. It has become arguably one of the most well-known mascots at the IMO. Lots of people have taken pictures with it -- girls wanted to take a picture with Canmoo because it's cute and adorable, and guys wanted to take a picture with it to show off as a trophy for stealing it. Oh yeah, our mascot also gets stolen a lot too (mostly by other teams, but this is true even at our own training camps!). For instance, in 2006, I saw the following picture as the desktop background on one of the computers.

[img]http://web.mit.edu/yufeiz/www/imo2006/imopics/rec9.jpg[/img]
Note the red circle on Canmoo and a label pointing to it saying "Main Objective."

Of course, everyone likes to steal Canmoo just for fun. It always (we'd hope) gets returned at the rightful owners at the end. One such attempt was even documented by the IMO 2006 documentary "Hard Problems" (its accuracy debatable ... thanks Richard ...).

Of course, we, the Canadian team, love Canmoo. And I hope that you will too! :)

[img]http://i32.tinypic.com/i3t6kg.jpg[/img]
Canmoo at IMO 2005

[img]http://i26.tinypic.com/2642bgz.jpg[/img]
Canmoo at IMO 2006

[img]http://i31.tinypic.com/34o3n78.jpg[/img]
Canmoo at IMO 2007

[img]http://i27.tinypic.com/16111yo.jpg[/img]
Canmoo at IMO 2008

thank you!
very informative. (it brings me so much memories lol)
So Canada is the first country that bring mascot to IMO?

ps. Is the same Canmoo in everyyear or they are different? I remember in 2008 there was two, One snatched by Greeks and other by Americans... How many "official canmoo" are there?

Great post!   Has anyone stolen Canmoo this year? or at least have they tried to steal it?
 :D

[quote="chichi"]thank you!
very informative. (it brings me so much memories lol)
So Canada is the first country that bring mascot to IMO?[/quote]

No, Canada is certainly not the first country to bring a mascot to the IMO. Nonetheless, Canmoo has become, for some reason, one of the most popular targets to steal.  :P

[quote="chichi"]ps. Is the same Canmoo in everyyear or they are different? I remember in 2008 there was two, One snatched by Greeks and other by Americans... How many "official canmoo" are there?[/quote]
No, it was the same one; the moose was returned by Greeks at the closing ceremony and stolen by Americans after it :rotfl: and only returned in the morning right before departure :D
It definitely is the most popular one to steal. :moose:


geometry parallelogram circumcircle trigonometry Stanford college

In triangle ABC, O and R are the orthocentre and circumcentre respectively. Prove that, AO is twicw the distance between R and BC.

thank you...... :)

[hide][asy]pair A=(8,7), B=(0,0), C=(10,0), E, O, R, K; R=circumcenter(A, B, C); O=orthocenter(A,B,C); draw(circumcircle(A, B, C)); draw(A--B--C--cycle); draw(R--A..R--B..R--C); dot(A^^B^^C^^O^^R); draw(C--intersectionpoint(3*(O-C)+C--C,A--B)); draw(B--intersectionpoint(3*(O-B)+B--B,A--C)); draw(A--intersectionpoint(3*(O-A)+A--A,B--C)); E=intersectionpoint(4*(R-B)+R--R,circumcircle(A,B,C)); K=midpoint(B--C); draw(K--R--E..A--E--C); dot(A^^B^^C^^O^^R^^E^^K); label("A",A,NE); label("B",B,SW); label("C",C,SE); label("R",R,NW); label("O",O,SW); label("E",E,NE); label("K",K,S);[/asy]Let $ K$ = midpoint of $ BC$. Extend $ BR$ past $ R$ to intersect circle $ R$ at $ E$. We just need to prove that $ AOCE$ is a parallelogram.

Step 1: Prove $ \angle ECA=\angle OAC$
$ \angle BCA=\angle BEA$, so $ \angle OAC = 90-\angle BCA = 90-\angle BEA=\angle EBA = \angle ACE$.

Step 2: Prove $ \angle EAC=\angle OCA$
(Here we need the fact that $ \angle CBO = \angle RBA$. This is not yet intuitive for me, so I'll prove it.) $ \angle EBC = 90-\angle BEC= 90-\angle BAC=\angle ABO\implies \angle ABR=\angle OBC$.
Ok so $ \angle ACO = \angle ABO = \angle EBC = \angle EAC$.

Therefore we have $ AE\parallel OC, AO\parallel EC\implies AO=EC=2\cdot RK$.
[/hide]

hmm this might be obvious but how did you get that $ \angle BCA = \angle BEA$?

this is how I did it

[hide]

like minseon I just proved that $ AOEC$ is a parallelogram

First to show that $ \angle ECO = \angle EAO$

For my proof I labeled the point where $ AO$ hits $ BC$ point $ T$ and the point where $ CO$ hits $ AB$ point $ S$, then we know:

$ \angle BAE = \angle BCE = 90^{\circ}$

$ \angle BAT = \angle BCS = 90^{\circ}-\angle ABC$

Hence $ \angle BAE-\angle BAT = \angle BCE-\angle BCS \implies \angle ECO = \angle EAO$

Then to prove that $ \angle AEC = \angle AOC$

We know $ \angle AEC = 360^{\circ}-\angle BAE-\angle BCE-\angle ABC = 180^{\circ}-\angle ABC = \angle SOT = \angle AOC$

Thus, $ AO = EC$ and since $ \triangle RKB \sim \triangle ECB$ and $ K$ is the midpoint of $ BC$, we have $ AO = 2 \cdot RK$

[/hide]

I think we are getting a bit complicated here.

Problem: Let $ O$ and $ H$ be the circumcenter and orthocenter, respectively.

$ AH=2*d(O, BC)$.

Solution: $ d(O,BC)=R*\sin 90-A=R\cos A$. Let $ E$ be the projection of $ B$ onto $ AC$. Then $ \cos 90-C=\frac{AE}{AH}$ and  $ AE=c\cos A$. Combining those, $ AH=\frac{c\cos A}{\sin C}=2R\cos A$. The result follows.

P.S. It seems like there is a Naperville North invasion in this forum :P.

haha you mean ex-naperville north for both of us  :wink:

minseons going to exeter next year and im going to stanford

[quote="gotztahbeazn"]hmm this might be obvious but how did you get that $ \angle BCA = \angle BEA$?[/quote]$ ABCE$ is cyclic

[quote="hzbrl"]In triangle $ ABC$, $ H$ and $ O$ are the orthocentre and circumcentre respectively. Prove that, $ AH$ is twice the distance between $ O$ and $ BC$[/quote]
[asy]pair A=(1.5,8), B=(0,0), C=(10,0), H, O, M, G; O=circumcenter(A, B, C); H=orthocenter(A,B,C); M=midpoint(B--C); draw(A--B--C--cycle); draw(O--M); dot(A^^B^^C^^H^^O); draw(A--H); draw(A--M); draw(O--H); G=intersectionpoint(A--M,O--H); dot(G); label("A",A,NE); label("B",B,SW); label("C",C,SE); label("O",O,NW); label("H",H,SW); label("G",G,NW); label("M",M,S);[/asy]
Let $ G$ be the centroid
Clearly $ AH\parallel OM$, so $ \triangle GAH \sim \triangle GMO\Rightarrow \frac{AH}{MO}= \frac{GH}{GO}= \frac{AG}{MG}= \frac{2}{1}$

[b]Pontios[/b], you use the fact that  the centroid lies on OH, which I think is derived using AH=2OM.

Hmm... ok then :(
I have a solution similar to [b][color=blue]minsoen's[/color][/b]

Let $ F$ be the symmetric point of $ A$ wrt $ O$

[img]10077[/img]

It is $ FB\perp BA \Rightarrow FB \parallel CH$. Similarly $ FC\parallel CH$, so $ BHCF$ is a parallelogram. So $ M$ is the midpoint of $ HF\Rightarrow$
$ \triangle OFM \sim \triangle AFH \Rightarrow \frac{AH}{OM}= \frac{AF}{OF}= 2$

Using the fact that $ AH=2OM$, we prove that the centroid lies on $ OH$ and we can use the previous proof I posted :D :D


logarithms
complex numbers
algebra unsolved
algebra

If $xyz=1$,then calculate $k$.
$k=(\log_{z} \frac{x}{y}+\log_{x} \frac{y}{z}+\log_{y} \frac{z}{x})(\log_{\frac{x}{y}} z+\log_{\frac{y}{z}} x+\log_{\frac{z}{X}} y)$ :)

I think $k=9$
just use complex numbers
 ;)

dont know about the primary soloution!!
this was our $QUIZ$ problem
and I`ve solved it with complex numbers :D
its not hard at all ;)

[b][color=blue]If $xyz=1$, then calculate
$k=(\log_{z} \frac{x}{y}+\log_{x} \frac{y}{z}+\log_{y} \frac{z}{x})(\log_{\frac{x}{y}} z+\log_{\frac{y}{z}} x+\log_{\frac{z}{x}} y)$
[/color][/b]
[b][u]Solution:[/u][/b]
Let be
$a = \lg x,b = \lg y,c = \lg z$

then
$xyz = 1 \Rightarrow a + b + c = 0$

and we have
$k = \left( {\log _z \frac{x} {y} + \log _x \frac{y} {z} + \log _y \frac{z} {x}} \right)\left( {\log _{\frac{x} {y}} z + \log _{\frac{y} {z}} x + \log _{\frac{z} {x}} y} \right) =$

$= \left( {\frac{{a - b}} {c} + \frac{{b - c}} {a} + \frac{{c - a}} {b}} \right)\left( {\frac{c} {{a - b}} + \frac{a} {{b - c}} + \frac{b} {{c - a}}} \right) =$

$= \frac{{\left( {a + b - c} \right)\left( {a - b + c} \right)\left( {a - b - c} \right) + abc}} {{abc}}\mathop = \limits_{\mathop {a + b + c = 0}\limits^ \uparrow } 9$

[quote="amin_68"]Can you post your solution? :)[/quote]
sure
we have a theory in geometric that the proof is with complex numbers:
sorry guys but I have to write this in our language because I`ve not the mood of writing this in english:
$lemma:$
[b]majmooe tamame $n$ rishe $n$ome vahed,barabare sefr ast
esbate lemma:
farz konid $w^i (i=0,1,...,n-1)$ rishehaye $n$ome vahed bashand pas majmooe risheha barabar ast ba:
$1+w+w^2+...+w^{n-1}=\frac{1-w^n}{1-w}$
amma dashtim: $w^n=1 (w\not=1)$,pas soorate kasre samte raste tasavie fogh sefr khahad bood va lemma sabet shod[/b]
[b]Theory:
farz konid $A_1,A_2,...,A_n$ rooose yek $n$ zeli montazam ba dayereye mohiti be shoae vahed bashand
sabet konid be ezae har noghte $P$ rooye mohite in dayere darim:
$\sum_{i=1}^n\bar{PA_i}^2=2n$[/b]
esbatesham asoone vali man nemidoonam ke to cheghadr aadad mokhtalet baladio hendesat dar che haddie pas vasat esbatesh mikonam:
[b]esbate ghazie:agar aadad mokhtalete motanazer $A_i$ha ra dar safhe mokhtalet ba $a_i$ha va adade motanazere $P$ ra ba $p$ neshan dahim:
$K=\sum_{i=1}^n \bar{PA_i} \to \sum_{i=1}^n |p-a_i|^2$
(namade $\to$ ra neshan dahandeye tanazor dar safheye mokhtalet dar nazar gereftim)
ke khahim dasht:
$K=\sum_{i=1}^n|p-a_i|^2=\sum_{i=1}^n(p-a_i)(\bar{p}-\bar{a_i})=\sum_{i=1}^n(p\bar{p}+a_i\bar{a_i}-(a_i.\bar{p}+\bar{a_i}.p))$
hal az anja ke $p,a_i$ha rooye mohite dayereye be shoae vahed gharar darand agar markaze dayere ra mabdae mokhtasat gharar dahim darim:
$p\bar{p}=|p|^2=1,a_i\bar{a_i}=|a_i|^2=1$
banabarin:
$K=2n-(\sum_{i=1}^na_i.\bar{p}+\sum_{i=1}^n\bar{a_i}.p)=2n-(\bar{p}.\sum_{i=1}^na_i)-(p\sum_{i=1}^n\bar{a_i})$
hamchenin az lemmae esbat shode midanim ke :$\sum_{i=1}^na_i=0$ va dar natije
$\bar{(\sum_{i=1}^na_i)}=\sum_{i=1}^n\bar{a_i}=0$
pas:
$K=2n-(\bar{p}*0)-(p*0)=2n$
va hokm bedin tartib esbat gardid[/b]
hala age az taarife logarithm estefade konim sooratmasale be soorate zir dar miad:
$K=(\lambda+\gamma+\theta)(\Lambda+\Gamma+\Theta)$
hala age $\lambda,\gamma,\theta,\Lambda,\Gamma,\Theta$ ro bordarhaye makane jesmhaye $a,b,c,A,B,C$ dar nazar begirim
oonvaght ba estefade az nokte $xyz=1$ natije migirim ke bordarhaye makane in $6$ moteharrek dar safhe mokhtasate $\mathbb{R}^2$ tashkile yek $6$ zelie montezam ba dayere mohitie be shoae $1$ midan
hala age azin ghaziehe estefade konim bara noghtei mese $\zeta$ rooye in dayere darim:
$\bar{\zeta\lambda}+\bar{\zeta\gamma}+\bar{\zeta\theta}+\bar{\zeta\Lambda}+\bar{\zeta\Gamma}+\bar{\zeta\Theta}=12$
amma agar biaim va safhei jadid be name $\zeta^2$ ro tarif konim oon vaght dar in safhe mokhtalet agar be $Arg$ aadade mokhtalete $\lambda\Gamma,\lambda\Theta,\gamma\Lambda,\gamma\Theta,\theta\Lambda,\theta\Gamma$ tavajoh konim khahim dasht $\lambda\Gamma+\lambda\Theta+\gamma\Lambda+\gamma\Theta+\theta\Lambda+\theta\Gamma=6$
amma $K=\lambda\Lambda+\gamma\Gamma+\theta\Theta+\lambda\Gamma+\lambda\Theta+\gamma\Lambda+\gamma\Theta+\theta\Lambda+\theta\Gamma=1+1+1+6=9$
dar tasavie fogh az nokte $\lambda\Lambda=\gamma\Gamma=\theta\Theta=1$ estefade kardim ke inam be sadegi az tarife logarithm natije mishe

now again sorry because the solution was some kind of long and I hadnt the mood to translate it
 ;)
did you get it amin??

[quote="amin_68"]Thank you BaBak. :)[/quote]
youre welcome
did you get it???
 :)


graph theory
combinatorics unsolved
combinatorics

A graph $G=(V,E)$ is given. If at least $n$ colors are required to paints its vertices so that between any two same colored vertices no edge is connected, then call this graph ''$n-$colored''. Prove that for any $n \in \mathbb{N}$, there is a $n-$colored graph without triangles.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=no+triangle&t=21943

Pierre.

I must draw your attention about the fact (not mentioned in the link above either) that there exists a much stronger result (this would be awkward in a competition, if someone invoked this result).

[color=blue][b]Theorem (Erdos 1959)[/b] For every integer $ n$, there exists a graph $ G$ with girth $ g(G) > n$ and chromatic number $ \chi(G) > n$.[/color]

The girth of a graph is the length of its shortest cycle ($ \infty$ for acyclic graphs), so $ g(G) > n$ prohibits not just triangles, but any cycles of length at most $ n$.

The chromatic number of a graph is the minimum number of colors needed to color its vertices so that adjacent vertices bear different colors.

The proof uses the Probabilistic Method (and it is unknown to me any construction-type proof).

The reference is [R. Diestel - Graph Theory].

I have an idea: Consider the Ramsey number $R(3, t)$. Since there exist two constants $c_1, c_2\in \mathbb{R}^+$ such that $$c_1\dfrac{t^2}{lnt}<R(3, t)<c_2\dfrac{t^2}{lnt}$$ so for any $n\in \mathbb{Z}^+$, we can find a $t$ such that $$R(3, t)>nt$$ which means there exists a graph $G(V, E)$ with $|V|=nt$ such that $G$ has no triangles and independent sets with $t$ vertices. Clearly, this $G$ satisfies the condition since if it can be colored by $n$ colors it must has an independent set with no less than $t$ vertices.

An example of triangle-free graph with high chromatic number is called Mycielski graph.


we have point $O$ and polygon $F$ in the plane (not necessarily convex). assume that its perimeter is $P$,and $D$ is sum of distance from $O$ to vertices of $F$,an also $H$ is sum distance from $O$ to the edges of$F$ prove that :
       $D^2-H^2\geq \frac {P^2}{4}$

It comes from the 1996 short-list.

Pierre.


arithmetic sequence
geometric sequence
number theory proposed
number theory

Prove that infinite arithmetic progression a, a+d, a+2d,...,a+nd,...
contains an infinite geometric progression b, bq,...,bq*k,... (Bosnian TST 2006)
(a,d,b and q are positive integers)
P.S. Has anybody seen this problem before?

When $a$ is in the progression, so is $a(d+1)=a+ad$. This already gives an infinite geometric progression.


[b]1. [/b] Let $ 0\leq x<1$, prove that if $ 0<p\leq \sqrt {\frac {11 \plus{} 5\sqrt5}2}$, then
\[ \left(\frac {1 \minus{} x}{1 \plus{} x}\right)^{1/p}\leq\frac1{1 \plus{} x^2}.
\]
[b]2. [/b] Let $ a,b,c$ be non-negative real numbers. Prove that
\[ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {a \plus{} b \plus{} c}2.
\]
The second is hard but not new and there is a nice solution that is based on the first.

[quote="pvthuan"][b]1. [/b] Let $ 0\leq x < 1$, prove that if $ 0 < p\leq \sqrt {\frac {11 \plus{} 5\sqrt5}2}$, then
\[ \left(\frac {1 \minus{} x}{1 \plus{} x}\right)^{1/p}\leq\frac1{1 \plus{} x^2}.
\]
[b]2. [/b] Let $ a,b,c$ be non-negative real numbers. Prove that
\[ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {a \plus{} b \plus{} c}2.
\]
The second is hard but not new and there is a nice solution that is based on the first.[/quote]
2/ lemma $ ab^3 \plus{} bc^3 \plus{} ca^3 \leq \frac {1}{3}(a^2 \plus{} b^2 \plus{} c^2)^2$

Use BCS we have:
\[ (\frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3})(a^2(a^3 \plus{} b^3) \plus{} b^2(b^3 \plus{} c^3) \plus{} c^2(c^3 \plus{} a^3)) \geq (a^3 \plus{} b^3 \plus{} c^3)^2
\]
we must prove:
\[ 2(a^3 \plus{} b^3 \plus{} c^3)^2 \geq (a \plus{} b \plus{} c)((a^2(a^3 \plus{} b^3) \plus{} b^2(b^3 \plus{} c^3) \plus{} c^2(c^3 \plus{} a^3))
\]
we have:
\[ a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3 \equal{} (ab \plus{} bc \plus{} ca)(ab^2 \plus{} bc^2 \plus{} ca^2) \minus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca) \\
\equal{} \frac {ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c}(ab^3 \plus{} bc^3 \plus{} ca^3 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c)) \minus{} abc(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca) \\
\equal{} \frac {ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c}(ab^3 \plus{} bc^3 \plus{} ca^3 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} abc(a^2 \plus{} b^2 \plus{} c^2) \\
\leq \frac {ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c}(\frac {1}{3}(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} abc(a^2 \plus{} b^2 \plus{} c^2)
\]
so that we have to prove:
\[ 2(a^3 \plus{} b^3 \plus{} c^3)^2 \geq (a \plus{} b \plus{} c)(a^5 \plus{} b^5 \plus{} c^5) \plus{} (ab \plus{} bc \plus{} ca)(\frac {1}{3}(a^2 \plus{} b^2 \plus{} c^2)^2 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)
\]
standardize $ a \plus{} b \plus{} c \equal{} 1$, put $ ab \plus{} bc \plus{} ca \equal{} \frac {1 \minus{} q^2}{3}, abc \equal{} r (0\leq q \leq 1)$ then we have:
\[ \frac {(1 \minus{} q)^2(1 \plus{} 2q)}{27} \geq r \geq \frac {(1 \plus{} q)^2(1 \minus{} 2q)}{27}
\]
the inequality is equivalent:
\[ (54r^2 \plus{} (28q^2 \minus{} 4)r) \minus{} 3(1 \minus{} q^2)r \plus{} \frac {1}{27}(7q^6 \plus{} 108q^4 \minus{} 39q^2 \plus{} 5) \geq 0
\]
we have $ f(r) \equal{} 54r^2 \plus{} (28q^2 \minus{} 4)r$ increasing $ r$ we have:
\[ VT \geq (54(\frac {(1 \plus{} q)^2(1 \minus{} 2q)}{27})^2 \plus{} (28q^2 \minus{} 4)(\frac {(1 \plus{} q)^2(1 \minus{} 2q)}{27})) \minus{} 3(1 \minus{} q^2)\frac {(1 \minus{} q)^2(1 \plus{} 2q)}{27} \plus{} \frac {1}{27}(7q^6 \plus{} 108q^4 \minus{} 39q^2 \plus{} 5) \\
\equal{} \frac {q^2(q^2(15q^2 \plus{} 26q \plus{} 24) \plus{} (1 \minus{} 3q)^2}{27} \geq 0
\]
The inequality is done, it hold by $ a \equal{} b \equal{} c.$

It's my solution on http://toanthpt.net/forums/showthread.php?t=5946 too

[quote="Ly Tieu Long"]$ \frac {q^2(q^2(15q^2 \plus{} 26q \plus{} 24) \plus{} (1 \minus{} 3q)^2}{27} \geq 0$

The inequality is done, it hold by $ a \equal{} b \equal{} c.$

It's my solution on http://toanthpt.net/forums/showthread.php?t=5946 too[/quote]

It is $ \frac {q^2(q^2(15q^2 \minus{} 26q \plus{} 24) \plus{} (1 \minus{} 3q)^2}{27} \geq 0$ :wink:


The school system here in the US is notorious for underteaching geometry.  So not as a surprise my formal education in it was lacking and uninspiring.  My 9th and 10th grade years I absolutely hated geometry problems.  Luckily I started doing more of them over the this summer and the Intermediate Independent Study program here helped a lot.  Right now I think geometry is my favorite subject.  I have a ways to go but I am also a lot better.  The AoPS books are a great start.  I m not really sure how to tell you to get better but mainly just practice, practice, practice.  We could probably help you better if you told us some specifics you aren't getting.

[quote="samus"]So to improve my AMC scores from last year, I got Aops V.1. So, I chugged through the first 8 chapters, and I made enormous strides in algebra and number theory. I took a sample AMC 12, and my score increased by 12!

But Ch. 9-14, I can't grasp. I just don't understand geometry, even though I am a 10th grader doing Calculus BC easily; I usually miss all of the geometry problems on the AMC. Does anyone have any ideas on how to approach geometry? I did [u]all [/u]the problems in Ch. 1-8 in less than 2 weeks, but for over 2 months, I'm still trying to understand these 6 latest chapters. I've been attempting the problems, but even some of the easier "unthreaded" ones I don't understand. Worst of all, I can't memorize the formulas. I've tried deriving them, but I can't do it well. :(

Has anyone been through the same thing? Any advice? I'm worried that if there are 4+ geometry problemss on the AMC's this year, I won't even qualify for the AIME's :?.

Thanks:
Your frustrated friend, samus[/quote]

Eh, I know how you feel. I learned geometry over a summer and needless to say, my knowledge of it in competitions is shaky at best. But AoPS Volume 1 is really helping a lot. There's a ton more that I understand and know how to do now. It's pretty tedious, but I'm learning more, slowly and surely.

Generally, my strategy is: read the explanations and examples carefully; make fullest effort on problems, whether or not I actually solve them; reread explanations and examples; try again at problems; lather, rinse, and repeat. It also helps to draw diagrams the text doesn't include and even to redraw diagrams the text does include. I don't know why that works but it does. It's like the process of drawing something pounds something into your mind. Oh yeah, and also...whatever you do, DO NOT give up, that is mathematical suicide!

I'll toss in my two cents on geometry.  My short-answer geometry technique is essentially to

$1.$ DRAW A LARGE AND ACCURATE DIAGRAM. Preferrably on difficult problems using a compass and straightedge. Even if it takes a few dozen more seconds, this enables me to cleanly label everything I know.

$2.$ Using this diagram, I am usually able to make accurate conjectures and am better able to recognize which ideas are relevant.

$3.$ At this point, I repeatedly bash the figure with theorems. Of course, with every new idea I pick up, this part becomes more effective.

As for practice, that definitely helps. This is the best means to obtaining the crucial intuitive understanding of which theorems will "work" and which ones won't.

I second the recommendation to use compass and straightedge. Draw lots of diagrams. I had a short course in geometric construction in FIFTH grade, and that was enormously helpful for me for years afterward.  (Well, actually it made my geometric intuition so strong that I found school geometry classes boring, but people who work with AoPS problems won't have that misfortune.) Eventually, you will learn to think about geometric figures as having moving parts. Once you can sense what moves, and what doesn't, you'll be able to go to the next level of solving geometric problems.


Can these equations be solved?

a) cos x = x

b) e^x = 3x

Of course, there are such $x$, but I don't think that you can find a closed expression for $x$.
But, I'm not good enough to prove or disprove it.

The existence can be shown by the intermediate value theorem. But their solutions are transcendental, thus lordWings should be right  (and I'm sure this is already proved, but well...).

I obtained the approximate answer by using calculator

a) x = 0.999847741
b) x = 1.512134552

a) I get $a=0.739085133215161$. Did you use degrees instead of radians¿ (then it's indeed $0.999847741531088$)

b) There are two: $0.619061286735945$ and yours.

Sorry, i forgot to set my calculator back to radian mode

I think the existence may be proved by using cosx - x and e^x - 3x are continuous.
But are there any ways to find the actual value, or it cannot be expressed?