algebra
difference of squares
special factorizations

Prove that ${\underbrace{999\dots 99}_{n\ \ 9\text s}}^{2}=\underbrace{99\dots 9}_{n-1\ \ 9\text s}8 \underbrace{00\dots 0}_{n-1\ \ 0\text s}1$.

Just use the binomial expansion of $(1000000000\ldots)^{2}-1$, and do a difference of squares on it.

My solution:
[hide]The series of 9's can be written as 10^(n)-1
Therefore the square is (10^(n)-1)^2
which expands to 10^(2n)-2*10^(n)+1
2*10^(n) is greater than 1 so the first term will cause the string of nines, the second will cause the 8 and the last will cause the one at the end.[/hide]

That was basically my idea too.
Nice solution.

[quote]2*10^(n) is greater than 1 so the first term will cause the string of nines, the second will cause the 8 and the last will cause the one at the end.[/quote]

How does the first term cause the string on nines?
-or the second cause the 8?

I guess I should go into more detail about how the 9's and 8 come about. When you subtract a number from a power of 10 the result is a number starting with a string of 9's. When the number subtracted starts with a 2 there is am 8 after the string of 9's. This is true because if you add the subtracted number back in it is added to the 8 and the one is carried all the way to the left.

Think of how 10-2=8
100-2=98
1000-2=998
1000-20=980

[hide]subtract 1 from both sides, and it becomes difference of squares[/hide]

very nice solutions guys! but the "difference of squares" solution was pretty clever. :lol:
[hide="my solution"]the idea is almost the same as the other posts in this thread.
$(10^{n}-1)^{2}=10^{2n}-2\cdot 10^{n}+1=10^{n}(10^{n}-2)+1=1000\dots 00\times 999\dots 998+1=999\dots 998000\dots 001$
[/hide]


so, as I live here in Dallas and will be at AMSP, I decided to do something cool. Since harry potter releases while we're at camp, how about this: instead of reserving a book to be mailed to you, give me the money (face value) and I'll have my parents reserve books at the local bookstore and bring them to us on that saturday the 21st so you don't have to wait to get the books.

I don't know about this but it just kinda popped into my head and I thought it would be a nice idea.

some parents you got there...

but yeah i would go for it

sorry guys - the parents nixed the plan.

Pshaw. Just get your parents to get you a copy and tell us all what happens  :rotfl: .

I'll make a shirt

Like those "dumbledore dies on page [insert page #]" shirts
and then I will be assassinated

if you have a barnes and noble gift card even u can mail it to your dorm...

dr. t won't let us for mathpath however...

jorian

I have confirmation from
junggi
13375P34K
jli

any more? please, I want my discount so let me know ASAP


Sorry for reviving this, but could someone post a solution for this problem please?

[quote="4everwise"]Anumber of schools took part in a tennis tournament. No two players from the same school played against each other. Every two players from different schools played exactly one match against each other. A match between two boys or between two girls was called a [i]single[/i] and that between a boy and a girl was called a [i]mixed single[/i]. The total number of boys differed from the total number of girls by at most 1. The total number of singles differed from the total number of mixed singles by at most 1. At most how many schools were represented by an odd number of players?[/quote]

Credits to Kalva website
[hide]Answer: 3.

Let there be $n$ schools. Suppose the $i^{th}$ school sends $B_i$ boys and $G_i$ girls. Let $B = \sum B_i$ and $G = \sum G_i$. We are given that $|B - G| = 1.$

The number of same sex matches is $1/2 \sum B_i(B - B_i) + 1/2 \sum G_i(G - G_i) = (B^2 - \sum B_i^2 + G^2 - \sum G_i^2)$. The number of opposite sex matches is $\sum B_i(G - G_i) = BG - \sum B_iG_i$. Thus we are given that $B^2 - \sum B_i^2 + G^2 - \sum G_i^2 - 2BG + 2 \sum B_iG_i = 0$ or $\pm2$. Hence $(B - G)^2 - \sum(B_i - G_i)^2 = 0$ or $\pm2$. But $(B - G)^2 = 1$, so $\sum(B_i - G_i)^2 = -1, 1$ or $3$. It cannot be negative, so it must be $1$ or $3$. Hence $B_i = G_i$ except for $1$ or $3$ values of $i$, where $|B_i - G_i| = 1$. Thus the largest number of schools that can have $B_i + G_i$ odd is $3$. [/hide]

This solution uses a slightly differently worded problem, one that says the number of boys and girls differed by 1 (not at most 1).  But it doesn't make a difference (for difference 0, the largest value is 2).


Let $ {a,b,c,m,n}$ be positive real numbers. Prove that the following inequality holds:

${ \frac {a^{m + n} + b^{m + n}}{a^{m}b^{n} + b^mc^n} + \frac {b^{m + n} + c^{m + n}}{b^{m}c^{n} + c^ma^n} + \frac {c^{m + n} + a^{m + n}}{c^{m}a^{n} + a^mb^n}}\geq3 .$


Prove that    a/b2+1  +  b/c2+1  +  c/a2+1  >= 3/2

when a+b+c=3 with a, b, c >=0.

Can someone please solve the above problem?
thanks

[quote="fragos72003"]Prove that    $\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} \geq \frac{3}{2}$  when $a+b+c=3$ with $a, b, c \geq 0$.

Can someone please solve the above problem?
thanks[/quote]

It was solved before at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13943


Solve for x in the equation: \[2x^{2}-5\sqrt{x^{2}-3x-1}= 6x+5\]

[quote="shobber"]Rewrite the equation: $2x^{2}-6x-2-\boxed{2}\sqrt{x^{2}-3x-1}=3$. Let $t= \sqrt{x^{2}-3x-1}$, then it becomes: $2t^{2}-5t-3=0$. Hence $t=3$. Now we get $x^{2}-3x-1=9$, which gives: $x=5$, $x=-2$.[/quote]

Where did you get that 2?
I think you meant 5.

P.S. Please hide your answer next time.

[quote="lotrgreengrapes7926"][quote="buzzer11"]would subbing -5 for -2 also work? I dont think it would...[/quote]
No, a quadratic has only 2 roots, and 5, -2 are the correct 2.[/quote]
The most number of roots a polynomial=0 of degree $n$ can have is $n$ roots.

[quote="lotrgreengrapes7926"][quote="buzzer11"]would subbing -5 for -2 also work? I dont think it would...[/quote]
No, a quadratic has only 2 roots, and 5, -2 are the correct 2.[/quote]

Sorry I meant that i dont think that you get the right answer if we:

[quote="Totally Zealous"]Where did you get that 2?
I think you meant 5. [/quote]


geometry
trapezoid
function
trigonometry
ratio
Pythagorean Theorem
AMC

An isosceles trapezoid is circumscribed around a circle.  The longer base of the trapezoid is $ 16$, and one of the base angles is $ \arcsin(0.8)$.  Find the area of the trapezoid.

(A) 72
(B) 75
(C) 80
(D) 90
(E) not uniquely determined

I don't get how the solution (below - sorry that it's a bit blurry) uses the fact that $ \arcsin(0.8)$.  Furthermore, is this in radians or degrees?  Isn't there more than one value that satisfies it?

Thanks a lot!

$ \arcsin(0.8)\equal{}x \implies \sin(x)\equal{}\frac{4}{5}$

so $ \arcsin(\theta)\equal{}x \implies \sin(x)\equal{}\theta$

[quote="undefined117"]If you look at the arcsin graph, you'll see that it's not a function (it doesn't pass the vertical-line test). To make it a function, it's convention to limit the range to $ \left[ \minus{} \cfrac{\pi}{2},\cfrac{\pi}{2}\right]$.

This works similarly for arccos, as well as the sine and cosine graphs.[/quote]

Why would whether or not arcsin is a function matter?  And why do we try to make it a function?

My understanding is that we try to make it a function so we don't get two equivalent answers for one $ x$-value.

For example, if the range was not restricted, $ \arcsin\cfrac{1}{2}$ would return $ \cfrac{\pi}{6},\cfrac{5\pi}{6},\cfrac{13\pi}{6},\cdots$

However, $ \cfrac{\pi}{6}\equal{}\cfrac{13\pi}{6}\equal{}\cfrac{(1\plus{}12k)\pi}{6}$. Also, $ \sin\cfrac{\pi}{6}\equal{}\sin\cfrac{5\pi}{6}\equal{}\sin\cfrac{(1\plus{}12k)\pi}{6}\equal{}\sin\cfrac{(5\plus{}12k)\pi}{6}\equal{}\cfrac{1}{2}$. It's pointless to consider all those other solutions, so we make it easier on ourselves by limiting the range.

If you'd really wanted to, you could've limited the range to $ [\tfrac{\pi}{2},\tfrac{3\pi}{2}]$, as long as each $ x$-value returns only one $ y$-value (be one-to-one). Notice then, $ \arcsin\cfrac{1}{2}\equal{}\cfrac{5\pi}{6}$, which is actually equivalent to $ \cfrac{\pi}{6}$ anyway.

This problem can be solved by common tangents, and recognizing that the arcsin of the base angle = .8 means the ratio of the altitude to the legs is .8...
Using the Pythagorean theorem we eventually get 80 (c)

Realize that you dont need to solve for $ \theta$ to solve the problem, just use the fact that its arcsin is .8


Prove that any rational number is the sum of the cubes of 3 rational numbers. :!:

hello, see here
http://www.mathlinks.ro/viewtopic.php?t=150907
The hint comes from Darij.
Sonnhard.


function
vector
LaTeX
advanced fields
advanced fields unsolved

Count all functions from {0,1}^N -> {0,1} that:

1. f(000...) = 0
2. f(x) != f(x XOR ei)
3. f(x XOR y) = f(x) XOR f(y)

ei is a vector with only one "1" on i-th place.

It's easy to show that f(ei) = 1, for all i, but problem arises for vectors with infinite number of "1".

Existence should be proven with Zorn lemma.

Can someone post a solution to this? I don't see a way to apply the lemma here :(.

[quote="JBL"]Please use a bit of [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ) to make your posts more understandable -- did you mean $ f: \{0, 1\}^N \to \{0, 1\}$ (e.g., for a positive integer $ N$) or $ f: \{0, 1\}^{\mathbb N} \to \{0, 1\}$?[/quote]

I'm sorry. I meant $ f: \{0, 1\}^{\mathbb N} \to \{0, 1\}$ -- vectors are infinite.


advanced fields
advanced fields unsolved

Show that a Banach space X is reflexive iff it dual space X' is reflexive

If $ X$ is reflexive, then $ (X^*)^{**} \equal{} (X^{**})^* \equal{} X^*$ so $ X^*$ is reflexive (although formally I guess you have to write down what the canonical embeddings do).

If $ X^*$ is reflexive, then we just proved that $ X^{**}$ is reflexive. Then $ X$, as a closed subspace of $ X^{**}$ is also reflexive (closed subspaces of reflexive spaces are reflexive).


geometry
3D geometry
inequalities
ratio
probability
circumcircle
inradius

Since people may be wondering about the answers to some problems, why don't we put all of the detailed solutions on this thread. Some rules
-let's try to post in the general order of questions
-try to explain in as much detail as possible
-If anyone has any questions, lets post in another thread.
-Try to quote the question.

Here's questions 1-4:
1. A space diagonal of a polyhedron is a line that connects two vertices of the polyhedron that do not lie on the same face. How many space diagonals does a cube have?

(A) 0 (B) 2 (C) 4 (D) 8 (E) 28

Answer: Let us call the cube ABCDEFGH, ABCD being the face closer to us with A: top left corner, B: top right corner, C: bottom right corner, D:bottom left corner; EFGH being the face farther away from us with E: top left corner, F: top right corner, G: bottom right corner, H:bottom left corner. The space diagnolls have to be on opposite faces, left/right positions, and up/down positions. So the space diagnols are AG,BH, EC, and FD. There are 4, so the answer is: (C) 4.

2. Find the units digit of the product of the first 2004 odd primes.

(A) 1 (B) 3 (C) 5 (D) 7 (E) 9

Answer: The first odd prime numbers are 3,5,7,11,17..... The product of the first 3 prime #'s is 105, first 4 primes is 735, and so on. Therefore, from this pattern, we determine that the units dgit of the first 2004 odd primes will be 5. Another way to look at it: 5 times any number of odd #'s (such a 2003 more) will always have a unit digit of 5. Therefore the answer is: (C) 5.

3. The area of the triangle bounded by the lines y=x, y=-x, and y=6 is

(A) 12 (B) 12 :sqrt:2  (C) 24 (D) 24 :sqrt:2  (E) 36

Answer: The three intersection points of these lines are (0,0),(-6,6), and (6,6). This forms a triangle of height 6 (from a point (x,6) to (x,0)) and length (base) of 12 ((6,6)-(-6,6)=(12,0)). The area of a triangle  is bh/2, so the area of this triangle is 12*6/2=36. Therefore the answer is: (E) 36.

4. How many odd positive integers between 1000 and 2000, inclusive, have all digits different?

(A) 224 (B) 280 (C) 504 (D) 512 (E) 729

Answer: This is a combanatorics/counting problem. There are 4 digits:_ _ _ _. We know that the thousands digit can only be 1 (2000 isn't odd). There are [b]4 [/b]possibilities for the units digit: one of the odd numbers 3,5,7,9 since 1 has already been used and we need all the digits different. The tens digit can have only [b]8[/b] possibities from the 10 digits; 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place. The hundreds digit has [b]7 [/b]possibilities: 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place, and yet another digit can't be used since it has been used in the tens digit. We already know there is [b]1[/b] possibility for the thousands digit: the number 1. Therefore the total number of possibilities for this number are [u]1[/u]  *  [u]7[/u]  *  [u]8 [/u] *  [u]4[/u]  = 224. The answer is therefore:  (A) 224.

Again: lets keep questions in order and POST YOUR ANSWERS!

Don't we already have a discussion thread??
[b]5.[/b] For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18  (B) 19  (C) 20  (D) 21  (E) 22

Solution: By Triangle Inequality, 11+15>x, or 26>x. By Triangle Inequality again, 11+x>15, x>15-11=4. Therefore 26>x>4. The total number of possiblity for x is 26-4-1=21. However, we included 11 and 15 in the 21 values of x. Thus 21-19=2, or B.

[b]6.[/b] How many right triangles with integer side lengths are such that their side lengths are in a nonconstant geometric sequence?

(A) 0  (B) 1  (C) 2  (D) 4  (E) Infinitely many

 Solution: Call the sides a b and c. Since they are in geom. progression they can be labeled like this: a, ar, and ar^2 respectively. For this to be a right triangle a^2+(ar)^2=(ar^2)^2 must be true.

Factor an a^2 from the left sides and divide by a^2 on both sides and we are left with 1+r^2=r^4. To find r we use quad form. and take the square root of it(when using quad form we find r^2 so we need to take the sq. root). You can find this and figure out it is irrational. But the sides must be integers. With integer sides and an irrational ratio there are no solutions.

Yay number 5!

5. For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18 (B) 19 (C) 20 (D) 21 (E) 22

First we look for x such that 11 + x > 15 and x <= 15.  Thus x > 4 from our first inequality, so the minimum x is 5.  Next we look for 11+15 > x.  So 26 > x, and the maximum x is 25.  There are 25-5+1 = 21 numbers between 5 and 25 inclusive BUT WAIT!  The problem said DISTINCT side lengths!  Thus 11 and 15 don't work for x, so we subtract 2.  21 - 2 = 19, thus our answer is B.

[quote="nr1337"]BUT WAIT!  The problem said DISTINCT side lengths![/quote]

<Evil cackle>  :twisted:

Only 6 people got this right!!!  :(

7.  7. A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is D/Q, with D and Q relatively prime positive integers, find D + Q.

(A) 3 (B) 17 (C) 331 (D) 346 (E) 349

First note that:
Even+Even=even,  Even+Odd=odd, and Odd+odd=even and that we have 6 odds and 5 evens in our set of numbers.

We are going to try to find the number of times that it makes it odd over the total number of times.  You can choose 6 from a group of 11 in \[{11\choose 6}=462\]

Now to find how many ways it will be even we break it into cases.

1st case- 6 evens... this doesn't create an odd

2nd-5 evens and 1 odd...this creates odd numbers.  There is one way to get 5 evens and 6 to get an odd making 6 total ways.

3rd-4 evens 2 odd... none

4th-3 even 3 odds... this works.  10 ways to choose 3 odds and 20 for the evens, 200 ways

5th-2 evens 4 odds...none.

6th-1 even 5 odds... works.  5 ways for the evens and 6 for the odds 30.

So we have $\frac{200+30+6}{462}=\frac{118}{231}$.

 D+Q=349 (E)

[b]7.[/b] A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is [tex]\frac{D}{Q}[/tex], with D and Q relatively prime positive integers, find D + Q.

(A) 3  (B) 17  (C) 331  (D) 346  (E) 349

Solution: There are 6 odd numbers and 5 even numbers. If the sum of the six numbers chosen is odd, then we have 3 cases:
1)5 odd  and 1 even
2) 3 odd and 3 even
3) 1 odd and 5 even.
In case 1), the total number of ways to get the result is (6C5)(5C1)
In case 2), the total number of ways to get the result is (6C3)(5C3)
In case 3), the total number of ways to get the result is (6C1)(5C5)
Thus, the total number of way of getting the sum of 6 numbers odd is
(6C5)(5C1)+(6C3)(5C3)+(6C1)(5C5)=236
The total number of ways to draw 6 marbles simultaneously is 11C6
The probability is 236/11C6=118/231. 118+231=349, the answer is E.

[b]8.[/b] Find the number of integers [tex]k[/tex] such that [tex]k^3 + k^2 + k + 7[/tex] is divisible by [tex]k+1[/tex].

(A) 2  (B) 4  (C) 6  (D) 8  (E) Infinitely many

Solution:$k^3+k^2+k+7$ must be divisible by $k+1$, so divide it to get $\displaystyle k^2+1+\frac{6}{k+1}$. If it's divisible, that will have to be an integer, meaning $k+1|6$. From this, we see $\pm(k+1)\in \{1,2,3,6\}$ So there are 8 possible values, 1, 2, 5, 0, -2, -3, -4, -7 for k, and the answer is d.

[b]9.[/b] Circle A is inscribed in an isosceles right triangle and circle B is circumscribed about the same triangle. The ratio of the area of circle A to the area of circle B is

(A) [tex]\frac{1}{2}[/tex] (B) [tex]3 - \sqrt{2}[/tex]  (C) [tex]3 + \sqrt{2}[/tex]  (D) [tex]3 - 2\sqrt{2}[/tex]  (E) [tex]3 + 2\sqrt{2}[/tex]

Solution: Without Loss of generality let the length of the leg of right triangle be 1. Thus the length of the hypotenuse is $\sqrt2$.Since the center of the circumscribed circle of a right triangle is the midpoint of the hypotenuse of the right triangle. The length of the circumradius is $\frac{\sqrt2}{2}$. Now note that the area of the triangle is $1(1)/2=\frac{1}{2}=rs$, where r is the inradius, and s is the semiperimeter. The semiperimeter=$\frac{1+1+\sqrt2}{2}$. Thus the inradius$=\frac{1}{2}/ \frac{2+\sqrt2}{2}=\frac{1}{2+\sqrt2}$. Inradius: Circumradius=$\frac{1}{2+\sqrt2}:\frac{\sqrt2}{2}=\frac{\sqrt2}{2+\sqrt2}=\sqrt 2-1$. Ratio of area=ratio of radius^2
=$(\sqrt 2-1)^2=3-2\sqrt2$, or D

Thx to all solution posters.

This is what a getting started problem solver needs.

10. A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn randomly and without replacement. Given that the probability that the value of the coins drawn is at least 50 cents is m/n, with m and n relatively prime positive integers, find m + n.

(A) 961 (B) 1015 (C) 1051 (D) 1056 (E) None of these

The total number of ways to choose 6 is . We are left with finding how many ways to get 50 or over.

We can have
dddddd
dddddn
dddddp or
ddddnn
There is one way to get the first case.

For the second case you can chose the 5 d's in  ways and the n in 4 ways(since order doesn't matter). So we have 6*4=24 ways for this case.

For case 3, the d's can be chosen in 6 again. The p can be chosen in 2 which means there are 12 for this case.

For the last case, the d's can be picked in  ways and the n's in ${4\choose 2}=6$ ways. This makes 90 ways.

So the answer is 127/924. m+n= 1051 (C)

11.  11. Given that $i^2=-1$, for how many integers n is $(n+i)^4$  an integer?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Expanding the binomial we get
\[n^4+4n^3i+6n^2i^2+4ni^3+i^4=n^4+4n^3i-6n^2-4ni+1\]

Taking away the parts that we know are integers we are left with:
\[4n^3i-4ni=4ni(n+1)(n-1)\]
This is only an integer when it is 0. so $n\in \{-1,0,1\}$.  (D)

[quote="Pork_Chop8"]
5. I missed the distinct part. Great trick, evil problem writer. :lol: [/quote]

Hey, I resent that...  ;)

Only 6 people out of 29 got that one right.

Also, I think this thread was intended have the solutions in order of the questions...so you may want to edit your post, or move it to the other discussion thread where solutions are not being discussed in order.

[b]12.[/b] A quadrilateral circumscribed about a circle has three of its side lengths 132, 162, and 201. What is the positive difference between the minimum and maximum lengths for the fourth side?

(A) 60  (B) 78  (C) 93  (D) 138  (E) 231

Solution:
Lemma: If a circle can be inscribed in a quadrilateral ABCD, then AB+CD=AD+BC,  so AD=AB+CD-BC.
Proof: Let the circle be tangent at AB, BC, CD, AD at W, X, Y, Z respectively.
Thus AW=AZ, BW=BX, CY=CX, DY=DZ. Add all the equations up yields
AB+CD=BC+AD as desired.
Therefore, let AD be the fourth side, AD is at minimum when AB+CD=132+162=294, and BC=201. AD=294-201=93.
AD is at maximum when AB+CD=201+162=363, BC=132. AD=363-132=231.
Thus 231-93=138, or D.

[b]13.[/b] For how many primes [tex]p[/tex] is [tex]p^4 + p^5[/tex] the square of an integer?

(A) 0  (B) 1  (C) 2  (D) 3  (E) Infinitely many

Solution:   First we factor \[p^4(p+1)\].
So we need the square root of this to be an integer.  $p^4$ will come out of the square so we are left with finding were $\sqrt{p+1}$ is an integer.  In other words we want \[p+1=a^2\] where a is an integer.

Subtract by one on both sides and factor to get \[p=(a+1)(a-1)\]  Wait a minute a prime is only divisible by one and itself.  So this could only work when a-1=1.  this makes p=3 and 3 is prime so we only have one solution, or B

[b]14.[/b] If [tex]x[/tex], [tex]y[/tex], and [tex]z[/tex] are positive numbers satisfying [tex]x + \frac{1}{y} = 4[/tex], [tex]y + \frac{1}{z} = 1[/tex], and [tex]z + \frac{1}{x} = \frac{7}{3}[/tex], then find [tex]xyz[/tex].

(A) [tex]\frac{2}{3}[/tex]  (B) [tex]1[/tex]  (C) [tex]\frac{4}{3}[/tex]  (D) [tex]2[/tex]  (E) [tex]\frac{7}{3}[/tex]

Solution:  $(x+\frac{1}{y})(y+\frac{1}{z}) (z+\frac{1}{x})$
$=xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}$
$=\frac{28}{3}$

Since
$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
$=(x+\frac{1}{y})+ (y+\frac{1}{z})+(z+\frac{1}{x})$
$=\frac{22}{3}$
Sub that in, and let $xyz=k$
$k+\frac{1}{k}+\frac{22}{3}=\frac{28}{3}$
$3k^2-6k+3=0$
$(k-1)^2=0$
$k=1$, or B.

[b]15.[/b] If [tex]\frac{\pi}{2} < \alpha < \frac{3\pi}{4}[/tex], [tex]\cos{\alpha} + \sin{\alpha} = \frac{1}{3}[/tex], and [tex]\left|\cos{2\alpha} + \sin{2\alpha}\right|= \frac{n + \sqrt{p}}{q}[/tex], where [tex]n[/tex], [tex]p[/tex], and [tex]q[/tex] are relatively prime positive integers and [tex]p[/tex] is not divisible by the square of any prime, find [tex]n+p+q[/tex].

(A) 34  (B) 35  (C) 36  (D) 37  (E) 38

Solution:  $(\sin{\alpha}+\cos{\alpha})^2$
$=(\sin^2{\alpha}+\cos^2{\alpha})+2\sin{\alpha}\cos{\alpha}$
$=1+\sin{2\alpha}$
$=\frac{1}{9}$
Solve for $\sin{2\alpha}$
to get $\sin{2\alpha}=-\frac{8}{9}$
$\cos{2\alpha}=\pm \sqrt{1-(\frac{8}{9})^2}$
$=\pm \frac{\sqrt{17}}{9}$
Because of our restrictions, $\cos{2\alpha}$ is negative
$=-\frac{\sqrt{17}}{9}$
absolute value of $\cos{2\alpha}+\sin{2\alpha}=\frac{\sqrt{17}+8}{9}$
17+8+9=34, or A.

[b]16.[/b] If [tex]r_1[/tex], [tex]r_2[/tex], and [tex]r_3[/tex] are the solutions to the equation [tex]x^3 - 4x^2 - 4x + 17 = 0[/tex], find [tex]r_1^3 + r_2^3 + r_3^3[/tex].

(A) -68  (B) -4  (C) 0  (D) 61  (E) 68

Solution:First off, rename $ r_1=a; r_2=b; r_3=c $(so that you don't get all the r's confused).  Since they are solutions to the equation, the equation $x^3-4x^2-4x+17=(x-a)(x-b)(x-c)$.  Expanding the right side gives $x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=0$.  Therefore, $a+b+c=4; abc=-17$; $ab+bc+ac=-4$.  Now take $a+b+c=4$ and multiply both sides by (a+b+c), giving $a^2+b^2+c^2+ab+ac+bc=16$.  Replace $ab+ac+bc$ with -8, swing the -8 to the other side, then multiply both sides by $(a+b+c)$ again.  Now you have (with a little bit of simplification):
 $a^3+b^3+c^3+a(ab+ac)+b(ab+bc)+c(ac+bc)=96$
Change the equation we already had to ab+ac=-4-bc and put that into the equation we have, then repeat for the other two with the other variables.  It now looks like this:
$a^3+b^3+c^3+a(-4-bc)+b(-4-ac)+c(-4-ab)=96$  which is equivalent to this:
$a^3+b^3+c^3-4(a+b+c)-3abc=96$   Substituting from our equations for a, b, and c, gives us $a^3+b^3+c^3-16+51=96$; therefore, $a^3+b^3+c^3=61$ and the answer is (D).

[b]17.[/b] Find [tex]\sum_{k=0}^{49} (-1)^k  {99 \choose 2k}[/tex].

(A) [tex]0[/tex]  (B) [tex]-2^{49}[/tex]  (C) [tex]2^{49}[/tex]  (D) [tex]-2^{50}[/tex]  (E) [tex]2^{50}[/tex]

Solution:Note that
$(1-i)^{99}=1-{99 \choose 1}i-{99 \choose 2}+ {99 \choose 3}i+{99 \choose 4}.....$
$=\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}+i\sum_{k=0}^{49} (-1)^k {99 \choose {2k+1}}$
And $1=-(\sqrt2)(\cos\frac{3\pi}{4}), -1=-(\sqrt2)(\sin\frac{3\pi}{4})$
By DeMorvie's Theorem
$(1-i)^{99}=-((\sqrt2)(\cos\frac{3\pi}{4})+i(\sqrt2)(\sin\frac{3\pi}{4}))^{99}
=-(\sqrt2)^{99}(\cos\frac{(99)(3)\pi}{4}+i(\sin\frac{(99)(3)\pi}{4}))$
Thus
$\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}$
$=-(\sqrt2)^{99}(\cos\frac{297\pi}{4})$
$=-2^{49}(\sqrt2)(\cos\frac{\pi}{4})$
$=-2^{49}$, or B.

That took A LOT of Copy and Paste. :)

18. A rising number is a positive integer, each digit of which is larger than each of the digits to its left, and no digit of which is 0. For example, 12689 is a rising number. When all five-digit rising numbers are listed from smallest to largest, the 97th number in the list does not contain the digit

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

The smallest is 12345.  With 4 as the second number we have 5 that work(i.e.12345,12346,12347,12348,12349).
If we keep the first 3 the same and keep increasing the second number by 1 we will get 4, 3, 2 ,and 1 that work for each one.  Therefore there are 5+4+3+2+1=15 rising numbers of the form 123_ _ .  Continuing in this way we find that there are 4+3+2+1=10 rising numbers of the form 124_ _.  6 with a 5 in the third, 3 with a 6 and 1 with a 7.  This makes 35 of the form 12_ _ _.  For 13_ _ _ there are 4+3+2+1  + 3+2+1 + 2+1 + 1= 20.   For 14_ _ _ we have 10. For 15_ _ _  there are 4. For 16_ _ _ there is 1.  This makes a grand total of 70.
For 2 as the first number we do the same thing.  We find that 24589 is the 96 so 24678 is the 97.   This doesn't contain a 5 so (B)

[quote="kool_dudy"]Joml, I understood your solution.  Thx.

I got one question though.  How do you know that 24589 is the 96th number in that pattern?[/quote]
Get as close as you can with the method he used up to 70, and then list the rest.

[b]19.[/b] [tex]\bigtriangleup ABC[/tex] has all side lengths distinct. Two altitudes of this triangle have lengths 4 and 12. If the length of the third altitude is also an integer, find its maximum value.

(A) 4  (B) 5  (C) 6  (D) 7  (E) None of these

Solution: Let the length of AB be a, BC be 1, AC be c. Let the altitude of length 4 and 12 be opposite to AB and BC respectively. Let the length of the third altitude be h. The area of ABC=base*altitude/2 $=\frac{4a}{2}=\frac{12}{2}=\frac{hc}{2}$. Multiply by 2 yields 4a=12=hc, thus a=3
Since AB, BC, AC must satisfy triangle inequality, c+1>3, c>2. Since hc=12,
h<6. The maximum integral value of h is 5, or (B).

[b]20.[/b] It is given that [tex](\log_z{3})(\log_x{3})+ (\log_y{3})(\log_x{3}) + (\log_y{3})(\log_z{3}) = 6(\log_z{3})(\log_x{3})(\log_y{3})[/tex] and [tex]x<y<z[/tex]. If [tex]x[/tex], [tex]y[/tex], and [tex]z[/tex] are in a geometric sequence with common ratio 2, and [tex]x + y + z = \frac{m}{n}[/tex], with m and n relatively prime positive integers, find [tex]m + n[/tex].

(A) 61  (B) 62  (C) 63  (D) 64  (E) 65

Solution: Note that $(\log_m{n})(\log_n{m})=1, \log_m{n}=\frac{1}{\log_n{m}}$. Divide both side of the original equation by $(\log_z{3})(\log_x{3})(\log_y{3})$ we have $\frac{1}{\log_y{3}}+\frac{1}{\log_z{3}}+\frac{1}{\log_x{3}}=6$, or $\log_3{y}+\log_3{z}+\log_3{x}=6$,
$\log_3{xyz}=6$, which yields $xyz=3^6=729$. Since x, y, z are in geometric squence of common ratio 2, y=2x, z=2y=4x. xyz=8x^3=729, solve for x yields $x=\frac{9}{2}$. $x+2x+4x=7x=\frac{63}{2}$. m+n=63+2=65, or (E).

Where can I find the solutions for the rest of the test? Please, I need this.
I appreciate it

22. Find the number of distinct pairs of integers (x, y) such that 0<x<y and sqrt{1984} =sqrt{x} +sqrt{y}

(A) 0 (B) 1 (C) 3 (D) 4 (E) 7

you have to factor out 1984 which is 2^6*31
root(1984) simplifies to 8root(31)

x has to be 31 and that multiplied by a perfect square for root(X) and root(y) to have a common factor of root 31

when x is 31*1, the other number is (8-root(1))squared*31 which is 49*31 and is 1519
(31,1519) is another solution

then you find the next perfect square

when x is 31*4, the other number is (8-root(4))squared*31 which is 36*31 and is 1116
(124,1116) is another solution

when x is 31*9, the other number is (8-root(9))squared*31 which is 25*31 and is 775
(279,775) is another solution

you can do this for the next perfect square until you realize that x and y will be the same in that

therefore:
(279,775),(124,1116), and (31,1519)  are the 3 ordered pairs that work

the answer is (C). or 3

p.s. sorry for first response, i felt i had to correct myself

:(  sorry, i didnt know calculators werent allowed, look back up for the non- calculator solution  :P


Let ABCDE a convex pentagon with CD=DE  and /angle BCD = /angle DEA = 90^o. Let F be the point on side AB such that AF/FB=AE/BC. Show that /angle FCE = /angle FDE and /angle FEC = /angle BDC.

Sorry if I made double posting.

[quote="raf92"]Let ABCDE a convex pentagon with CD=DE  and /angle BCD = /angle DEA = 90^o. Let F be the point on side AB such that AF/FB=AE/BC. Show that /angle FCE = /angle FDE and /angle FEC = /angle BDC.[/quote]
Are you sure that F lies on AB ?.

Assume that $ \angle FCE \equal{} \angle FDE$ $\Longrightarrow$ quadrilateral $ FCDE$ is cyclic. Then it follows that $ \angle FEC \equal{} \angle FDC,$ but we have $ \angle FEC \equal{} \angle BDC$ $\Longrightarrow$ $ \angle BDC \equal{} \angle FDC$ $\Longrightarrow$ $ F \equiv{} B,$ which is absurd.

P.S. raf92, may you write the reference of this problem?

Yes, luis, I noticed some strange things in the question. I just copied from "Nots On Euclidean Geometry". Anyway, thank you for your remarks.

[quote="raf92"]I just copied from "Nots On Euclidean Geometry".[/quote]
I just checked [i]''Nots on Euclidean Geometry"[/i], and the enunciation is indeed incorrect. The problem is proposed in section 4.3, when it's supposed to be proved by Pascal theorem :what?:


analytic geometry
inequalities
geometry unsolved
geometry

A convex hexagon is given. Let $ s$ be the sum of the lengths of the three segments connecting the midpoints of its opposite sides. Prove that there is a point in the hexagon such that the sum of its distances to the lines containing the sides of the hexagon does not exceed $ s.$

[i]Author: N. Sedrakyan[/i]

If the lines connecting the midpoints of the opposite side-segments of the given hexagon, are concurrent at one point, then their concurrency point has the property as the problem states.

A configuration of this particular case, may be constructed by three similar isosceles triangles, as the [b][size=100]Kyriazis problem[/size][/b] states, in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131804]A difficult concurrency[/url].

In the general case of three distinct points $ X,\ Y,\ Z,$ of intersection of the above three midlines, taken per two of them, these points have the property as the problem states too.

In the schema [b][size=100]t=215641[/size][/b], we can see a ''constructive'' solution without words, which may be used also in the particular case.

It is easy to show that all the points inwardly to the triangle $ \bigtriangleup XYZ,$ including the points of its side-segments, have the property as the problem states, by the argument of $ (d_{3} \plus{} d_{6}) < M'Q' < MQ,$ where $ MQ'\parallel MQ.$

Kostas Vittas.

I've got to say that I am not entirely satisfied with the following solution; at the end, I resort to some rather messy computations to solve a fact that seems solvable by purely geometric means  (I know a solution to the problem which asks to prove that if a hexagon has parallel opposite sides then the lines that connect midpoints of opposite sides are concurrent, which is a particular case of the more general result I proved). I would be really interested if anyone could point a geometric approach to the final part. Till then, here is my solution:

We first work with the line $MQ$; the exact same considerations can be then carried out for the other similar lines. Suppose that $AB$ and $ED$ intersect at $X$ (possibly an infinite point) and let $H_{MQ}$ be the semiplane determine by $MQ$ that contains $X$ (if $X$ is an infinite point we choose the semiplane arbitrarely). Let $L$ be a point that is both inside of the hexagon and of $H_{MQ}$. We assert that the sum of distances from $L$ to the lines $AB$ and $ED$ is at most equal to $MQ$: let $L_1$ and $L_2$ be the intersections of the parallels from $L$ to the lines $ED$ and $AB$, respectively (see figure 1.1). Because of the semiplane in which $L$ lies (and the fact that it lies inside the hexagon), the order of the points on line $MQ$ is $M-L_2-L_1-Q$. This means that $MQ\geq ML_2+QL_1$. But $QL_1$ is at not less than the distance from $L$ to $ED$ (because this distance is equal to the distance from $L_1$ to $ED$ and in a right triangle, the longest side is the hypotenuse). Similarly, we have that the distance from $L$ to $AB$ is a t most equal to $ML_2$ and we have thus proved that $MQ$ is at least equal to the sum of distances from $L$ to the sides $AB$ and $ED$. As we will use them further, we will define (in a similar way to the one we defined $H_{MQ}$) $H_{NR}$ and $H_{PS}$.

It is rather easy to see that each of the semiplanes defined contains three vertices. We may suppose that no vertex belongs to all three semiplanes, because otherwise the problem would immediately result from the fact proved in the preceeding paragraph. We may therefore suppose that $A,F,E\in H_{MQ}$, $E,D,C\in H_{NR}$ and $A,B,C\in H_{PS}$ (the other possible case, $B,C,D\in H_{MQ}$ and so on is proved similarly; in any other case we may find a vertex that belongs to all three semiplanes). We will prove that the three semiplanes have a point in common, from which the problem results by invoking the fact proved earlier.

First, we will prove that we may consider that at least two pairs from $(AB,DE)$, $(BC,EF)$ and $(CD,FA)$ consist of parallel lines. Suppose that this is not the case, and let's suppose that neither $AB$ and $DE$, nor $BC$ and $FE$ are parallel. To prove that the three semiplanes have a point in common, it is enough to prove that if $MQ\cap NR=\{X\}$ and $MQ\cap PS=\{Y\}$ then the order of the points on $MQ$ is $M-X-Y-Q$ (we have used the fact that $A,F,E\in H_{MQ}$). Let $B'$ be in the interior of the heaxagon so that $BB'||MQ$ and at least one of the pairs of sides $(AB',ED)$ and $(CB',FE)$ consists of parallel sides, while the semiplane defined by the other still "points" in the same direction as the semiplane defined by the corresponding line in $ABCDEF$ (or, in other words, $H_{M'Q'}$, $H_{N'R'}$ and $H_{P'S'}$ have no vertex in common; see figure 1.2). If $B'$ happens to be in the interior of the penatgon $ACDEF$, we replace it with a point $B'$ so that $BB'||MQ$ and $B'\in AC$. We shall study this case, in which $B'\in AC$, first. We have that $A,F,E\in H_{M'Q'}$ and $C,D,E,\in H_{N'R'}$.

Let $\{Z\}=ED\cap AB$. Then $A$ lies between $Z$ and $C$. Because $ACDEF$ is convex, $A$ also lies between $C$ and $U$, the intersection of $EF$ and $AC$ (see figure 1.3). Because $B'\in AC$, this means that $A,F,B'\in H_{N'R}$, a contradiction.

We may therefore suppose that $B'$ does not lie in the interior of $ACDEF$ and so $AB'CDEF$ is a convex hexagon. Because of the definition of $B'$ we have that $M'\in MQ$ and therefore $Y$ remains unchanged. Because $\overrightarrow{NN'}=1/2\overrightarrow{BB'}$ and $\overrightarrow{BB'}$ and $\overrightarrow{MQ}$ have the same \textit{direction}, $\overrightarrow{NN'}$ and $\overrightarrow{MQ}$ also have the same direction. Therefore, as both $X$ and $X'$ are in the interior of $MQ$, we have that $X\in (MX')$. This means that if $Y\in{MX'}$, then $Y\in {MX}$. As such, if we manage to prove the problem for $AB'CDEF$ the problem for $ABCDEF$ immediately follows. We may therefore suppose that two of the pairs $(AB,DE)$, $(BC,EF)$ and $(CD,FA)$ consist of parallel lines. Without loss of generality, let $(AB,ED)$ and $(BC,EF)$ be those.

Although now we may theoretically choose $H_{MQ}$ as we wish, the transformations we made in the previous paragraphs force us to take it as the one containing $A,F$ and $E$ (similarly for $H_{NR}$). Let $\{T\}=AB\cap CD$, $\{V\}=CD\cap FE$ and $\{W\}=AB\cap EF$ (see figure 1.4). To make computations easier, we shall make an affine transformation that sends $\Delta TWV$ into an isosceles right triangle with sidelenght $1$ and right angle at $W$ (we are allowed to perform such a tranformation maintains parallelism and and midpoints).

Let now the coordinates of $W$ be $(0,0)$, of $V=(1,0)$, $T=(0,1)$, $A=(0,x)$, $F=(x_1,0)$, $B=(0,x_2)$ and $E=(x_3,0)$ (see figure 1.5). Because of the parallelisms we have (and the fact that $A,B,C\in H_{PS}$) we have that $C=(1-x_2,x_2)$, $D=(x_3,1-x_3)$ and $x\geq x_1$. As before, let $\{X\}=MQ\cap NR$ and $\{Y\}=PS\cap NR$. We shall now find the coordinates of $X$ and $Y$. Let the coordinates of $X$ be $a(\frac{x_1+x_3}{2},0)+(1-a)(\frac{1-x_2}{2},x_2)=b(0,\frac{x+x+2}{2})+(1-b)(x_3,\frac{1-x_3}{2})$ and those of $Y$ be $c(\frac{x_1+x_3}{2},0)+(1-c)(\frac{1-x_2}{2},x_2)=d(\frac{x_1}{2},\frac{x}{2})+(1-d)(\frac{1+x_3-x_2}{2},\frac{1+x_2-x_3}{2})$. All we have to prove is that $c\geq a$. By solving the systems of equations we get that \[c\left(\frac{x_1+x_3+x_2-1}{2}\cdot\frac{1+x_2-x-x_1}{2}+x_2\cdot\frac{x_3+1-x_1-x_2}{2}\right)=\]

\[\frac{x_3}{2}\cdot\frac{1+x_2-x_3-x}{2}+\frac{x_3+x_2-1}{2}\cdot\frac{1+x_3-x_1-x_2}{2}\] and
\[a\left(x_2x_3-\frac{x_1+x_2+x_3-1}{2}\cdot\frac{x+x_2+x_3-1}{2}\right)=\]

\[\frac{1-x_2}{2}\cdot\frac{x+x_2+x_3-1}{2}+x_3\cdot\frac{x_2-x}{2}.\]

By using the fact that the hexagon is convex we get the following inequalities: $x_2\geq x(\geq x_1)$, $x_3 \geq x_1$ and $x_1+x_2\geq 1$. From these inequalities it results that all the brackets written above are positive (actually, nonnegative; however, we may suppose that in the inequalities written earlier, equality does not hold, as this case follows by continuity from the other). As \[\frac{x_1+x_2+x_3-1}{2}\cdot\frac{1+x_2-x-x_3}{2}+x_2\cdot\frac{x_3+1-x_1-x_2}{2}=\]

\[x_2x_3-\frac{x_1+x_2+x_3-1}{2}\cdot\frac{x+x_2+x_3-1}{2}\] all we have to prove is that \[\frac{x_3}{2}\cdot\frac{1+x_2-x-x_3}{2}+\frac{x_3+x_2-1}{2}\cdot\frac{1+x_3-x_1-x_2}{2}\geq\]

\[\frac{1-x_2}{2}\cdot\frac{x+x_2+x_3-1}{2}+x_3\cdot\frac{x_2-x}{2}\] which is actually equivalent to $(x-x_1)(x_2+x_3-1)\geq 0$ which follows from the inequalities written at the beginning of this paragraph. The problem is now solved, as we have proved that the three semiplanes $H_{MQ}$, $H_{NR}$ and $H_{PS}$ have a point in common, which then has the property requested in the problem.


linear algebra
matrix
vector
induction
linear algebra unsolved

Let $ M\equal{} \begin{pmatrix}
m_1 & m_2 & m_3 & \cdots &m_{n\minus{}1}& m_n\\
m_2 & m_2 & m_3 &  \cdots &m_{n\minus{}1}&  m_n\\
m_3 & m_3 & m_3&   \cdots & \vdots &\vdots \\
\vdots  & \vdots  &\vdots  & \vdots &\vdots  & \vdots \\
m_{n\minus{}1} & m_{n\minus{}1} & \cdots & \cdots &m_{n\minus{}1} & m_n\\
m_n & m_n & \cdots & \cdots & m_n & m_n\\
\end{pmatrix}$ with $ 0<m_1 < m_2 <\cdots <m_n$ and denote by $ N\equal{}(n_{i,j})\equal{}M^{\minus{}1}$.
Show that $ \sum_{1\leqslant i,j\leqslant n}n_{i,j} \equal{}\frac{1}{m_n}$

If $ j$ is the column vector of all ones, we want to calculate $ j^TM^{\minus{}1}j$.

Since $ M\cdot\begin{pmatrix}0\\0\\ \vdots\\ 0\\1\end{pmatrix}\equal{}m_n\cdot j$, we have that $ M^{\minus{}1}j\equal{}\frac1{m_n}\begin{pmatrix}0\\0\\ \vdots\\ 0\\1\end{pmatrix}$ if $ M$ is invertible, and $ j^TM^{\minus{}1}j\equal{}\frac1{m_n}j^T\cdot \begin{pmatrix}0\\0\\ \vdots \\0\\1\end{pmatrix}\equal{}\frac1{m_n}$.

It remains to show that $ M$ is actually invertible whenever the $ m_k$ are not equal and $ m_n$ is not zero. To do this, subtract the last row from each other row; this leaves the last entry as the only nonzero entry in the last column, and the upper left $ (n\minus{}1)\times(n\minus{}1)$ block as a smaller version of $ M$ with translated entries; they are still unequal, and the last is not zero. By induction, we can continue this process and row reduce to a lower triangular matrix with nonzero diagonal entries, making $ M$ invertible.


An altitude of a convex quadrilateral is a line through the midpoint of a side perpendicular to the opposite side. Show that the four altitudes are concurrent iff the quadrilateral is cyclic.

This is well known. The lines concur at the anticenter of quadrilateral $ABCD$.

[quote="mecrazywong"]The lines concur at the anticenter of quadrilateral $ABCD$.[/quote]

More information here:

[url]http://www.cut-the-knot.org/Curriculum/Geometry/Brahmagupta2.shtml[/url]
[url]http://www.pballew.net/cycquad.html[/url]

etcetera

Or you can complex bash this :)

Once you've proved the cyclic part, you can use that to disprove noncyclic quads.

Dear Mathlinkers,
we can speak instead altitude, maltitude...
see

http://jl.ayme.pagesperso-orange.fr/Docs/A%20propos%20de%20l'anticentre%20d'un%20quadrilatere%20cyclique.pdf

Sincerely
Jean-Louis

The midpoints of a quadrilateral form a parallelogram (even better known). Let $X$ be the center of the parallelogram. Rotate $ABCD$ 180 degrees through $X$.
If $ABCD$ is cyclic, let $O$ be the circumcenter. The lines from $O$ to the midpoints are the perpendicular bisectors of the respective sides; the rotation takes the midpoint of each side to the midpoint of the opposite side, and the images of the lines from $O$ to the midpoints are precisely the altitudes of $ABCD$, which meet in the image of $O$. Likewise if the altitudes are concurrent, the rotation takes the altitudes to the perpendicular bisectors of the respective sides, and if these are concurrent then $ABCD$ is cyclic.

Just consider the homothety with center in the center of the parallelogram formed by the midpoints of the sides, and radius $-1$. Notice that the altitudes are taken in the perpendicular bisectors of the sides of the quadrilateral. Thus, they concur if and only if this lines meet at an unic point, i.e, ABCD is cyclic.


[quote="ashpotter"]I used to be able to make a picture apear on my desktop that i could drag all over my screen (i would get cool pokemon images and make them battle on my screen) How do you do it? You can also do it with a web page I think, any suggestons[/quote]

what are you using, XP? Linux?

[quote="ashpotter"](i would get cool pokemon images and make them battle on my screen)[/quote]

[quote="7h3.D3m0n.117"]what are you using[/quote]

I'd guess absinthe.

[quote="ashpotter"]I'm using an XP[/quote]

hmm I have XP and i don't recall being able to uh drag images but maybe you need a program?

I once had a cat on my desktop which I could drag and drop anywhere. I had windows 98 then and it was a simple application I think. I only had it for one evening, but liked it.

I'll see if I can help you.


vector
linear algebra
matrix
Gauss
algebra
system of equations
linear algebra unsolved

Assume that you have the following system of equations:
P=AP+C
where P is the vector of variables, C is a non-negative irreducible matrix
and C is a vector of constant values. Moreover, matrix C has a spectral
radius less than unity.
Is it possible to know which element of P has the highest modulus without
computing the inverse of A?

Thank you,
Jad

I rewrite your question in this way:
$A$ is a (n,n) nonnegative  irreducible matrix and $\rho(A)<1$.

It's sufficient to find an approximation of $p$.
Then $p=(I+A+A^{2}+....)c$; in general there exists an unique $A$'s eigenvalue $\lambda$ associated to an eigenspace of basis ${{u}}$ (dimension 1) such that $\lambda={\rho(A)}$ and if $\mu$ is another $A$'s eigenvalue then $\mid{\mu}\mid<\rho(A)$.
Moreover there exists $E$ a $A$-stable subspace such that $\mathbb{R}^{n}=u\oplus{E}$. According to this decomposition $c=\alpha{u}+v$.
For big $q$ $A^{q}c\approx\lambda^{q}\alpha{u}$ and for big $k$:
$p=(I+A+...+A^{k-1})c+(A^{k}+...)c \approx(I+A+...+A^{k-1})c+\frac{1}{1-\lambda}A^{k}c$.
  Remark 1:  Choice of $k$: let $\nu$ be one $A$'s eigenvalue such that $\lambda$ is the unique $A$'s eigenvalue which is greater than  $\nu$; If $\frac{\lambda}{\nu}$ is near 1 (far 1) then $k$ is big (is little).
  Remark 2: Numerically this calculation may be not interesting: we obtain $(I-A)^{-1}c$ in $\sim\frac{2}{3}n^{3}$ operations; in another hand the calculation of $p$ by the approximation formula is in $\sim{kn^{2}}$ operations.
Thus this calculation presents an interest only if $\frac{k}{n}$ is little.

Thank you loup blanc,

But What I need is to know which is the element of P that have the highest value even if its value is not known. In other work, I am searching for
arg(max(Pj) ).

Remark 1: My approximation formula is valid only if $c\not\in{E}$; if $c\in{E}$ then don't write the last term.
Remark 2: The calculation of a $\lambda$'s approximation has a negligible cost: $\lambda\approx\frac{\parallel{A^{k}c}\parallel}{\parallel{A^{k-1}c}\parallel}$.
Remark 3:  The condition $\mid\frac{\lambda}{\nu}\mid\gg1$ isn't sufficient so that we can conclude with my method if the $\mid{p_{j}}\mid$ aren't spaced enough.
Example: \[A=\frac{1}{200}\begin{pmatrix}10&56&64\\58&61&75\\86&17&62\end{pmatrix},Spectre(A)=\{\approx0.8,\approx-0.2,\approx0.066\},\mid\frac{\lambda}{\nu}\mid\approx4\]
Let $c=[0.91000000000001,-1.36,0.345]$; $p\approx[1,-1,1]$ and $\mid{p_{1}}\mid$ is the biggest modulus ($p_{1}-p_{3}\approx4.10^{-14}$). I use my formula with the $\lambda$'s calculation as in Remark 2. $p_{1}$ is in first position only when $k\ge22$.Thus here the best method is: calculate $(id-A)^{-1}$;there don't exist any difficulty because $cond(A)\approx9$ ($A$ is well conditioned; during the Gauss method only one digit is lost).
Conclusion: I don't see any method to solve this problem without the calculation of a solution's approximation.


Prove that a trigonometric polynomial of degree N has a maximum of 2N roots in an interval (0,2$ \pi$).

[quote="Albanian Eagle"]What do you mean with trigonometric polynomial??[/quote]

$ \sum_{n\equal{}0}^{N}{a_n sin nx \plus{} b_n cos nx}$, N is the degree.

I have already found a solution. All elementary proofs are still welcome though.

Let $ y \equal{} e^{ix}$, then $ sin(nx) \equal{} \frac {y^n \minus{} y^{ \minus{} n}}{2i},cos(nx) \equal{} \frac {y^n \plus{} y^{ \minus{} n}}{2}$, therefore $ \sum_{n \equal{} 0}^N(a_nsin(nx) \plus{} b_ncos(nx)) \equal{} 0$ equavalent to $ P(y) \equal{} 0$ were $ P(y)$ is polinom with degree $ 2N$.

Im not very into e^{something complex} but I guess its impossible for $ {y\plus{}y^{\minus{}1}}\over{2}$ to be equal to $ sin nx$ for all $ n\in N$! Assuming you meant $ sin x$, I have no idea why would $ P(y)$ be a polynomial because of that $ y^{\minus{}1}$...

[quote="t0rajir0u"]Rust means $ y \equal{} e^{inx}$.  Rust's argument tells you directly what the connection between trigonometric polynomials and the usual polynomials are.[/quote]

Keep in mind that $ n$ isn't constant.

[quote="t0rajir0u"]We are asked to prove that the result is true for all $ n$.  $ y$ is defined for all $ n$.  Where is the problem?[/quote]

Not $ n$ but $ N$.

Oh, sorry, I didn't see how Rust used his variables  :oops:  His revision is correct as is.


im posting this because i have a feeling people on this site took the exam or are taking it.  i just want to know who ot into the problem to give me advice on what topics in math and science that are going to be on it. also how did you study and what sources did you use to get a high score

I attended the program from 2005 to 2007, and the simple answer is: you don't prepare for the exam. The material they cover in the classes is so advanced in comparison to the typical high school curriculum to the point that, if you have to prepare for the exam, you probably won't gain much from the program even if you were accepted.
But that's probably not the answer you wanted. From what I remember, there were algebra, geometry, and elementary combinatorics on the math part, and geology, physics, chemistry, and biology on the science part.


[quote="Farenhajt"]If this is discussed elsewhere, just redirect me.

Why doesn't the latex search work? Or maybe it's working, but I don't know how to use it anymore? Thanks.[/quote]

I don't think you can search by $\LaTeX$, because doesn't AoPS render it into a picture? That's what I think at least.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=latex+search&t=99159[/url]


Find the greatest positive constant $p$ such that
\[ \sqrt{\frac{a^2+pbc}{b^2+c^2}}+\sqrt{\frac{b^2+pca}{c^2+a^2}}+\sqrt{\frac{c^2+pab}{a^2+b^2}}\geq3\sqrt{\frac{1+p}2}, \]
for all positive real numbers $a,b,c$.


[color=blue][i]Μου κάνει εντύπωση που αυτή την άσκηση την έχουν για την τάξη VII.[/i][/color]

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

 Δίνεται τετράγωνο $ ABCD$  και σημεία $ E,Z$ των πλευρών $ AB,BC$ αντίστοιχα , ώστε $ AE\equal{}BZ$.

Αν οι ευθείες $ AC,BD$ τέμνονται στο  $ O$ ,  οι ευθείες  $ DE,AZ$ τέμνονται στο  $ T$

και οι ευθείες   $ DA,CE$ τέμνονται στο  $ K$ , να αποδειχθεί ότι τα σημεία   $ K,T,O$ βρίσκονται στην ίδια ευθεία.

[u] Μπάμπης[/u]

[quote="stergiu"][color=red][b]ΑΣΚΗΣΗ[/b][/color]
Δίνεται τετράγωνο $ ABCD$  και σημεία $ E,Z$ των πλευρών $ AB,BC$ αντίστοιχα , ώστε $ AE \equal{} BZ$.

Αν οι ευθείες $ AC,BD$ τέμνονται στο  $ O$ ,  οι ευθείες  $ DE,AZ$ τέμνονται στο  $ T$ και οι ευθείες   $ DA,CE$ τέμνονται στο  $ K$ , να αποδειχθεί ότι τα σημεία   $ K,T,O$ βρίσκονται στην ίδια ευθεία.[/quote]
[hide]Από ίσα ορθογώνια τρίγωνα $ \bigtriangleup ADE,$ $ \bigtriangleup BAZ,$ έχουμε $ \angle ADE \equal{} \angle BAZ$ $ ,(1)$

Από $ (1)$ στο ορθογώνιο τρίγωνο $ \bigtriangleup ADE,$ προκύπτει ότι $ AT\perp DE$ $ ,(2)$

Από $ (2)$ και $ AO\perp DO$ συμπεραίνουμε ότι το τετράπλευρο $ ATOD$ είναι εγγράψιμο και άρα έχουμε $ \angle OTZ \equal{} \angle ADO \equal{} 45^{o}$ $ ,(3)$

Από $ (2)$ επίσης, συμπεραίνουμε ότι τα ορθογώνια $ \bigtriangleup ADE,$ $ \bigtriangleup TDA,$ είναι όμοια και άρα $ \frac {TA}{TD} \equal{} \frac {AE}{AD}$ $ ,(4)$

Έστω $ F,$ σημείο μεταξύ των $ D,$ $ T,$ έτσι ώστε $ TF \equal{} TA.$

Επειδή $ \angle TAF \equal{} 45^{o}$ $ ($ από ορθογώνιο ισοσκελές τρίγωνο $ \bigtriangleup TAF$ $ ),$ αρκεί να αποδειχθεί ότι $ AF\parallel KT.$

Αλλά $ \frac {TF}{TD} \equal{} \frac {TA}{TD} \equal{} \frac {AE}{AD} \equal{} \frac {AE}{CD} \equal{} \frac {KA}{KD}$  $ \Longrightarrow$ $ \frac {TF}{TD} \equal{} \frac {KA}{KD}$ $ ,(5)$

Από $ (5)$ $ \Longrightarrow$ $ AF\parallel KT$ $ \Longrightarrow$ $ \angle KTA \equal{} \angle TAF \equal{} 45^{o} \equal{} \angle OTZ$ $ \Longrightarrow$ $ K,$ $ T,$ $ O,$ συνευθειακά και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.[/hide]

Για μια διαφορετική λύση του παραπάνω δείτε και το τελευταίο post από εδώ

http://www.mathlinks.ro/viewtopic.php?t=126554&start=20

(Όσο για το ότι είναι για VII ταξη αυτή η άσκηση δεν έχω τι να πω . Τα αποτελέσματα των Ρουμάνων στις ΙΜΟ τα λένε όλα
Όταν εμείς μαθαίνουμε για την παραλληλία και τις εντός εναλλάξ γωνίες αυτή μαθαίνουν Simpson  :P )

Θα δώσω και εγώ μια λύση σε αυτήν την πολύ ωραία άσκηση (πιστέυω πως υπάρχουν πάρα πολλοί τρόποι ακόμα με τους οποίους μπορεί να λυθεί :wink: )

Η λύση μου βασίζεται στο ότι οι ριζικοί άξονες 3 κύκλων συντρέχουν!

Οπως και στη λύση του κ. Βηττα έχουμε $ ATOD$ εγγράψιμο $ \iff \angle{OTZ}\equal{}45\equal{}\angle{OBZ}$ που συνεπάγεται με τη σειρά του ότι $ OTBZ$ εγγράψιμο. Μάλιστα επειδή $ EBZ\equal{}ETZ\equal{}90$ και το $ E$ βρίσκεται σε αυτόν τον κύκλο.

Εστω τώρα $ L$ η τομή των $ EC, DZ$. Οπως και με το $ T$ προκύπτει $ \angle{ZLC}\equal{}90\iff$ και το $ L$ βρίσκεται στον κύκλο $ OTEBZ$. Επίσης έπεται ότι $ AELD$ εγγράψιμο αφού $ A\equal{}L\equal{}90$.

Αρα οι ευθείες $ AD, EL, TO$ αποτελούν ριζικούς άξονες των 3 κύκλων που προανέφερα και άρα συντρέχουν (στο $ K$).

$ QED$ :)

[quote="stergiu"][quote="bilot4"]Θα δώσω και εγώ μια λύση σε αυτήν την πολύ ωραία άσκηση (πιστέυω πως υπάρχουν πάρα πολλοί τρόποι ακόμα με τους οποίους μπορεί να λυθεί :wink: )

Η λύση μου βασίζεται στο ότι οι ριζικοί άξονες 3 κύκλων συντρέχουν!

Οπως και στη λύση του κ. Βηττα έχουμε $ ATOD$ εγγράψιμο $ \iff \angle{OTZ} \equal{} 45 \equal{} \angle{OBZ}$ που συνεπάγεται με τη σειρά του ότι $ OTBZ$ εγγράψιμο. Μάλιστα επειδή $ EBZ \equal{} ETZ \equal{} 90$ και το $ E$ βρίσκεται σε αυτόν τον κύκλο.

Εστω τώρα $ L$ η τομή των $ EC, DZ$. Οπως και με το $ T$ προκύπτει $ \angle{ZLC} \equal{} 90\iff$ και το $ L$ βρίσκεται στον κύκλο $ OTEBZ$. Επίσης έπεται ότι $ AELD$ εγγράψιμο αφού $ A \equal{} L \equal{} 90$.

Αρα οι ευθείες $ AD, EL, TO$ αποτελούν ριζικούς άξονες των 3 κύκλων που προανέφερα και άρα συντρέχουν (στο $ K$).

$ QED$ :)[/quote]

Δημήτρη,

 χαίρομαι που κατάφερες μια τέτοια λύση :) . Όταν και γω την πρωτοκοίταξα , είδα τα εγγράψιμα , αλλά προσπάθησα να αποδείξω ότι τα τρία σημεία βρίσκονται στον ίδιο ριζικό άξονα. Δεν τα κατάφερα αμέσως και εγκατέλειψα την προσπάθεια. Φαντάσου ότι όλα αυτά τα κάνω  την ώρα που κάποιος μαθητής προσπαθεί να λύση καμιά άσκηση και έτσι ξεκλέβω λίγο χρόνο για τις ολυμπιάδες.
 Η επόμενη προσπάθεια ήταν να πάω μετρικά με Μενέλαο , αλλά οι υπολογισμοί ήταν σχετικά δυσχερείς από πρώτη άποψη .
   Απόψε πρόλαβα και έλυσα δυο άλλες ασκήσεις(πιο εύκολες) από το ίδιο περιοδικό, τη μία από τις οποίες σας την έστειλα ήδη.
 Καλό βράδυ !
 Μπάμπης[/quote]
Αυτή η άσκηση λύνεται πολύ εύκολα με αναλυτική-γεωμετρία. Βέβαια η λύση χάνει την ομορφιά της Ευκλείδειας αλλά έτσι γίνεται σχολικού και όχι διαγωνιστικού επιπέδου


inequalities
trigonometry
inequalities unsolved

$ \sum_{cyc} \frac{a \sqrt{a^2\plus{}b^2\plus{}c^2}}{ \sqrt{a^2\plus{}b^2} \cdot \sqrt{a^2\plus{}c^2}} \le \frac{3 \sqrt{3}}{2}$

It's obviously trues because:
$ (ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c) \leq\ \frac{9}{8}(a\plus{}b)(b\plus{}c)(c\plus{}a)$
 :wink:

Alternatively you can substitute $ x \equal{} a^{2} \plus{} b^{2}$, $ y \equal{} b^{2} \plus{} c^{2}$, $ z \equal{} c^{2} \plus{} a^{2}$. As $ x$, $ y$, $ z$ are the sides of a triangle with semiperimeter $ s$ you get
$ \sum_{cyc}\frac {a\sqrt {a^{2} \plus{} b^{2} \plus{} c^{2}}}{\sqrt {a^{2} \plus{} b^{2}}\cdot\sqrt {a^{2} \plus{} c^{2}}} \equal{} \sum_{cyc}\frac {\sqrt {s \minus{} y}\cdot\sqrt {s}}{\sqrt {xz}} \equal{} \sum_{cyc}\cos\frac {\alpha}{2}\leq 3\cos\frac {\pi}{3} \equal{} \frac {3\sqrt {3}}{2}$ (Jensen inequality).

[quote="nguoivn"]It's obviously trues because:
$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \leq\ \frac {9}{8}(a \plus{} b)(b \plus{} c)(c \plus{} a)$
 :wink:[/quote]
Nice! (We also have to use $ \sqrt{x\plus{}y\plus{}z}\frac{\sqrt{3}}{\sqrt{2}} \ge \sqrt{x\plus{}y} \plus{} \sqrt{y\plus{}z} \plus{} \sqrt{z\plus{}x}$ which is just cauchy swartz)


I want to start a new team in my school....if there's only about 8 - 10 members, do the top 2 scorers in the local exam still go to nationals or what?

I assume you are referring to USNCO... only 2 per school are allowed to take the national (although I have heard that exceptions have been made in the past...)


geometry
incenter
trigonometry
geometry proposed

It's given a triangle $ ABC$ with incenter $ I$ and radius of its incircle $ r$. Prove that $ \sin\frac{\alpha}{2}.\sin\frac{\gamma}{2}\equal{}\sin\frac{\beta}{2} \Leftrightarrow AB\plus{}BC\equal{}3AC.$

hello, using the half-angle formulas we get
$ \sqrt{\frac{(s\minus{}b)(s\minus{}c)}{bc}}\sqrt{\frac{(s\minus{}a)(s\minus{}b)}{ab}}\equal{}\sqrt{\frac{(s\minus{}a)(s\minus{}c)}{ac}}$
this is equivalent with(after squaring)
$ (s\minus{}b)^2\equal{}b^2$
$ s\minus{}b\equal{}b$
$ a\plus{}b\plus{}c\equal{}4b$
$ a\plus{}c\equal{}3b$
Sonnhard.

And how you get that $ \sin\frac {\alpha}{2} \equal{} \sqrt{\frac {(s \minus{} b)(s \minus{} c)}{bc}}$?


Την παλεύω από χθες αλλά δεν έχω καταφέρει να την αποδείξω τελείως..

Να δείξετε ότι:

$ \alpha^{\beta}+\beta^{\alpha}>1$ για κάθε $ \alpha,\beta>0$

eliyi xronu... :P
profanos aksizi na dume mono tin periptosi $ 1>a>b$ (afu mporume na ipo8esume oti a>b)
thetontas a=1/x b=1/y pernume oti $ y>x>1$  +i anisotita ginete
$ (1/x)^{1/y}+(1/y)^{1/x}>1$ pu grafete tlk
$ (1-x^{1/y})*(1-y^{1/x})<1$ pu isxii
(xano pu8ena?:s)

orea 8a itan kai i lisi me paragogus, ala me kolai ligo...(an exi kanis as tin postari)
(btw congratz gia ta pedia tis imo  :) )

Τώρα που τελείωσε η εξεταστική και βγάζουμε όλη τη μέρα στην παραλία δεν υπάρχει αρκετός χρόνος και κάτι άλλο... :lol:

Πάντως πως ξέρουμε ότι ισχύει $ (1-\chi^{\frac{1}{\psi}})*(1-\psi^{\frac{1}{\chi}})<1$??

Κι εγώ ψάχνω μία λύση με την χρήση παραγώγων γιατί την άσκηση αυτή την άσκηση αυτή την έχω βρει σε βιβλίο ανάλυσης...

$ (1-\chi^{\frac{1}{\psi}})*(1-\psi^{\frac{1}{\chi}})<1$
gt ka8enas paragontas ine $ <1$(afu to $ \chi^{\frac{1}{\psi}}$ ine 8etiko) kai ara 8a ine kai to ginomeno  tus $ <1$  :P

Εφόσον $ \chi,\psi>1$ τότε έχουμε ότι $ 1-\chi^{\frac{1}{\psi}}<0$ γεγονός που δεν μας εγγυάται ότι το γινομενο είναι μικρότερο του 1.... :wink:

Πάντως μου έστειλε μία λύση o socrates εδώ που είχε μπει η ίδια άσκηση πιο παλιά

http://www.mathlinks.ro./viewtopic.php?t=52020&sid=f579cd35e8304ce323a8a802d5c25ce0

Ο Ράλλης στην άρχη έκανε λόγο για $ 0<b<a<1$. Μόνο τότε αξίζει να μελετηθεί. Αν $ a,b>1$ τότε είναι προφανές ότι ισχύει !


conics parabola trigonometry ellipse algebra function domain

Well it's the beignning of the school year and I've forgotten alot about ways to do these and I was wondering if anyone cna help me with these problems.

The problem gives a parametric equation like x=2t-5, y=4t-7, -inf<t<infi, asking for the initial and ending points, then asks for a Cartesian equation that contains the curve. I know how to do probelms like that, just subsiting T into the Y from the x, exp: t=(x+5)/2, y= 4(x+5/2)-7, but what about the following ones, I forget what to do with cos and sins.

x=-sqroot(t), y=t, t> or =0

x=4cos(t), y=2sin(t), 0<t<2pie

x=sin(t), y=cos(2t), -infi<t<inf

Thanks for the help

[quote="Logg"]x=-sqroot(t), y=t, t> or =0

x=4cos(t), y=2sin(t), 0<t<2pie

x=sin(t), y=cos(2t), -infi<t<inf
[/quote]

[hide="First one"] Not sure what your asking here, since its obvious that $ y\equal{}x^{2}$, which is a parabola[/hide]
[hide="Second one"]We have $ x^{2}\plus{}4y^{2}\equal{}16\cos^{2}{t}\plus{}16\sin^{2}{t}\equal{}16$ which is an ellipse[/hide]
[hide="Last one"]We have $ y\equal{}\cos{2t}\equal{}1\minus{}2\sin^{2}{t}\equal{}1\minus{}2x^{2}$, which is simply a parabola[/hide]

[b]Edit:[/b] The gentleman above got to the post first. My final answers are [b]similar[/b].
However (directed to both of you), [b]pay attention to the domain and range![/b]

[hide="Solution for 1"]
$ x =-\sqrt{t}\implies x =-\sqrt{y}$.
The above is a Cartesian equation in itself.

If you want it in terms of $ y$.
Squaring, $ y = x^{2}$.
However, since square root is always defined positive, $ x =-\sqrt{t}=-(\text{some non-negative number})$ means that $ x$ must be non-positive.

Hence, $ \boxed{y = x^{2}\;\forall x\geq 0}$
[/hide]

[hide="Hint for 2"]
The identity: $ \sin^{2}\theta+\cos^{2}\theta = 1$
[/hide]

[hide="Solution for 2"]
$ x = 4\cos t\iff\cos t =\frac{x}{4}$
$ y = 2\sin t\iff\sin t =\frac{y}{2}$

$ \cos^{2}t+\sin^{2}t = 1\iff\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{2}\right)^{2}= 1$

${ \iff\frac{x^{2}}{16}+\frac{y^{2}}{4}= 1\iff x^{2}+4y^{2}= 16}$

Since $ t$ is an angle, sine and cosine of t repeat periodically, and the given interval will create the entire ellipse, except for the point at $ t = 0\equiv t = 2\pi$, which is $ (4,0)$

Our revised formula is $ \boxed{x^{2}+4y^{2}= 16,\; x\neq 4}$
[/hide]

[hide="Hint for 3"]
The identity: $ \cos 2\theta =\cos^{2}\theta-\sin^{2}\theta$
[/hide]

[hide="Solution for 3"]
$ y =\cos 2t =\cos^{2}t-\sin^{2}t$
$ x =\sin t\implies 2\sin^{2}t = 2x^{2}$

Adding the above, we have:
$ y+2x^{2}=\cos^{2}t-\sin^{2}t+2\sin^{2}t =\cos^{2}t+\sin^{2}t = 1$

However, since $ -1\leq\sin 2t\leq 1\;\forall t$

Our final Cartesian equation is:

$ \boxed{y = 1-2x^{2},\; |x|\leq 1 }$
[/hide]


trigonometry
limit
ratio
logarithms
complex numbers
geometric series
calculus

Let $ \theta _n$ be the minimum root of $ \sum_{k\equal{}1}^n \cos k\theta \equal{}0$ for $ n\equal{}1,\ 2,\ \cdots n$.
Find $ \lim_{n\to\infty} (1\plus{}\theta _n)^n$.

What do you mean by minimum? Absolutely or? Cause for $ n=1$, we have $ \cos \th = 0$ and that has infinitely many roots...

Probably minimum positive?  (Every single one of those equations will have infinitely many roots if they have any: for any root $ r$, $ r \plus{} 2\pi$ is a root as well.)

[hide="A possible first step"]We can sum that series explicitly (it's the real part of a geometric series with ratio $ e^{i\theta}$), and this tells us that our root is the root of the equation $ \sin (t/2) \equal{} \sin (nt \plus{} t/2)$, if that's helpful.[/hide]

Here's a very stupid and very probably a wrong answer. Fist we suppose there actually is a limit. Then, for n "big enough" we have:

$ (1\plus{}\theta_n)^n \equal{} L$

or

$ \theta_n \equal{} \sqrt[n]{L}\minus{}1$

Now, since this is a solution to the equation JBL wrote, we need to have

$ \sin\frac{\sqrt[n]{L}\minus{}1}{2}\equal{}\sin\left(\frac{\sqrt[n]{L}\minus{}1}{2}\plus{}n \left(\sqrt[n]{L}\minus{}1\right)\right)$

$ \left(\sin\frac{\sqrt[n]{L}\minus{}1}{2}\right)\equal{}\sin\frac{\sqrt[n]{L}\minus{}1}{2}\cos n \left(\sqrt[n]{L}\minus{}1\right)\plus{}\cos\frac{\sqrt[n]{L}\minus{}1}{2}\sin n \left(\sqrt[n]{L}\minus{}1\right)$

Now, letting $ n \rightarrow \infty$ we have

$ \lim_{n\rightarrow \infty}\sin n \left(\sqrt[n]{L}\minus{}1\right)\equal{}0$

$ \sin \ln L\equal{}0$

$ L \equal{} e^{k \pi}$

To be honest, even if this is a good attempt of a solution, I don't know what to to with this! :)

[quote="milin"]What do you mean by minimum? Absolutely or? Cause for $ n = 1$, we have $ \cos \th = 0$ and that has infinitely many roots...[/quote]

Sorry I forgot to write the critical condition. :blush:

Let $ \theta _n$ be the minimum [color=red]positive[/color] root of $ \sum_{k = 1}^n \cos k\theta = 0$ for $ n = 1,\ 2,\ \cdots n$.
Find $ \lim_{n\to\infty} (1 + \theta _n)^n$.

kunny


floor function
function
calculus
derivative
inequalities
symmetry

For $n\ge2$, find the integer part of
\[(\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n\]

[hide]
it depends on whether $n$ is odd or even.
If odd: $5 - 6 + 7 = 6$
if even: $5 + 6 + 7 = 18$

since only $(\sqrt[n]{5})^n$, $(\sqrt[n]{6}^{n})$, and $\sqrt[n]{7}^{n}$ will be integers because the numbers 5,6,7 are relatively prime.
[/hide]

I think he means $\lfloor (\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n \rfloor$.

I think it's [hide]always 5[/hide], but I haven't proven it yet.

Oh, I think I got it. Seems a bit too easy...
[hide]
Since $\sqrt[n]{x}$ is concave, we have $\sqrt[n]{5}+\sqrt[n]{7} < 2\sqrt[n]{6} \Rightarrow \sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7} < \sqrt[n]{6} \Rightarrow (\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n < 6$

And $0 < \sqrt[n]{7}-\sqrt[n]{6} \Rightarrow \sqrt[n]{5} < \sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7} \Rightarrow 5 < (\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n$

So $5 < (\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n < 6 \Rightarrow \lfloor (\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n \rfloor = 5$.[/hide]

This isn't so bad.  [hide]You can conjecture that it'll always be 5 by just imagining large n.

We pretty clearly have that it's always greater than 5 (because $7^{\frac{1}{n}}-6^{\frac{1}{n}}>0$.  We just need to prove that it's always less than six.  This is equivalent to showing that $6^{\frac{1}{n}}>5^{\frac{1}{n}}-6^{\frac{1}{n}}+7^{\frac{1}{n}}$.

I'm sure there's a better way to do this next part.

To do this, it suffices to show that the function $f(x)=(x+1)^{\frac{1}{n}}-x^{\frac{1}{n}}$ has a positive derivative for every $x>0$.  We have $f'(x)=\frac{1}{n}\frac{1}{(x+1)^\frac{n-1}{n}}-\frac{1}{n}\frac{1}{(x)^\frac{n-1}{n}}$ which is obviously greater than zero, because $x+1>x$.

Sorry this is a little sketchy in some places, and I used obviously to mean if you take out a pencil for two seconds it'll be obvious.[/hide]

Edit: Oh, wait, it's already been solved.  Damn.

are these from the problem solving classes?  if so would someone be a doll and post some more of the problems?

Whose problem solving classes? Not Andrica's right?

fred youre not there?  are you doing something else or just staying home?

I've been taking classes at a college and working. Steve, what about you?

[quote="paladin8"]
Since $\sqrt[n]{x}$ is concave, we have $\sqrt[n]{5}+\sqrt[n]{7} < 2\sqrt[n]{6} \Rightarrow \sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7} < \sqrt[n]{6} \Rightarrow (\sqrt[n]{5}-\sqrt[n]{6}+\sqrt[n]{7})^n < 6$
[/quote]

why does it matter if $\sqrt[n]{x}$ is concave??

[quote="pkothari13"][quote="Magnara"]Jensens says that if f is concave then f(a)+f(b)<=2f((a+b)/2)[/quote]
ohhh. but it was convex, say $x^n$, where n is positive, then would it be $f(a)+f(b)\geq 2f(\frac{a+b}{2})?$[/quote]

Yeah, or more generally, given any $w_i$ for $i=1,2,...,k$ such that $w_1+w_2+\cdots+w_k = 1$, and a convex function $f$ on an interval $I$, we have

$w_1f(a_1)+w_2f(a_2)+\cdots+w_kf(a_k) \ge f(w_1a_1+w_2a_2+\cdots+w_ka_k)$

for all $a_j \in I$.

[quote="pkothari13"]so in this case, $w_1 = w_2 = \frac{1}{2},$ right?

i dont mean to go too much off topic, but are there are any other problems like this where one would see jensen's to be useful?[/quote]

Yes, $w_1 = w_2 = \frac{1}{2}$.

Jensen's is one of the most useful inequalities. Try proving the famous Nesbitt's inequality with it:

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ with positive reals $a,b,c$.

fred, im not doing anything really just being a bum.

[quote="pkothari13"][quote="paladin8"]$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ with positive reals $a,b,c$.[/quote]
man, i cant even think of a function that will somehow get that symmetry in it!  :(  can i have a hint?[/quote]

[hide="hint"]
Let $S = a+b+c$. Try rewriting the problem so you get a convex function. Treat $S$ as a constant.[/hide]

You can prove it using rearrangement.

a >= b >= c
1/b+c >= 1/a+c >= 1/a+b

a/b+c + b/a+c + c/a+b >= a/a+c + b/a+b + c/b+c
a/b+c + b/a+c + c/a+b >= a/a+b + b/b+c + c/a+c

2(a/b+c + b/a+c + c/a+b) >= a+c/a+c + b+c/b+c + a+b/a+b = 1 + 1 + 1 = 3
a/b+c + b/a+c + c/a+b >= 3/2.

That's by far my favorite proof for Nesbitt's.  :D

An alternate solution via Jensen's:

[hide]Let $a+b+c=1$ (since the inequality is homogeneous of degree zero).  The function $f(x)=1/x$ is convex on $[0,\infty)$, so that Jensen's gives

$\sum \frac {a}{b+c}\geq \frac {1}{\sum a(b+c)}=\frac {1}{2\sum ab}\geq \frac {3}{2}$ since $(a+b+c)^2\geq 3(ab+bc+ca)$

Using a similar method leads to a quick proof of the inequality from IMO 2000.
[/hide]

[quote="paladin8"]Hmm, I personally like the Cauchy one :P (there are so many different proofs)

$\sum \frac{a}{b+c} = \sum \frac{a^2}{ab+bc} \ge \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$.[/quote]

paladin8, what was your solution using jensen's?

Set $S=a+b+c$.  Since $x/(S-x)=$ is convex (the algebra is trivial, or write it as $S/(S-x)-1$), we see that

$\sum \frac {a}{S-a}\geq 3\left(\frac {\frac {a+b+c}{3}}{S-\frac {a+b+c}{3}}\right)=\frac {3}{2}$


I can't see any LaTeX on any of the forum. Here's a screenshot of my problem:

I checked my settings and Javascript and Java are enabled. Anyone know what the problem is?

Highlight the text from [i]inequality[/i] to [i]show that[/i] then right-click and choose View Selection Source. Does the highlighted source text you get include
[code]inequality
AMP

. Show that[/code]
This would tell you if the browser is downloading the whole page and not showing the image or only downloading part of the page or even messing with the code. All of this is shouldn't happen but it clearly does.

I am wondering if it is hiding the image - what happens if you click on [i]Solution[/i] in azjps post? Are there other pages with missing images and if so, is the common factor hidden text where you have to click a link to see it?

What happens if you turn Javascript off (Tools, Options, Content, Enable Javascript); can you see the images then?

(There's a similar report at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180739]Can't see LaTex[/url])

Woah ... Did I seriously write that I never had done any inequalities before?   :what?:

Seriously though, it might be possible that you are clicking on the hide/show tag before the page finishes loading, which often causes the images/latex to stop from loading (happens to me quite often). If so, then you can just refresh the page and wait until everything is done loading.


Tha mporouse kapoios na lusei tin parakatw askisi!! i dikia mou lish einai kapws xronovora kai me arketes prakseis! An kapoios skeftei kapoia alli lish as apantisei! Tha ithela poli pantws na dw kapoia diaforetiki!
Loipon einai!

Na luthei i parakatw eksiswsi stous [u]pragmatikous[/u] arithmous

(x^8)(y^4)+2(y^8)+4(x^4)-6(x^4)(y^4)= 0 στο R

[quote]Egregium Theorema[/quote]

[quote="maniopas"]Η παράσταση είναι ισοδύναμη με:
$ y^4(x{}^4){}^2 + x^4(4 - 6y^2) + 2y^8 = 0$, την οποία μπορούμε να θεωρήσουμε τριώνυμο του $ x^4$
Η διακρίνουσά της πρέπει είναι μη αρνητική για να έχει λύσεις, άρα
$ D > = 0 < = > 0 < = 16 - 48y^4 + 28{}^y{}^8 = f(y^4)$
Η διακρίνουσα της f είναι Δ = $ 2 ^ 9 > 0$ άρα ... (χρησιμοποιώντας απλή θεωρία τριωνύμου βρίσκουμε το $ y^4$ για να είναι $ D = f(y^4) > = 0$)
Και χρησιμοποιώντας πάλι θεωρία τριωνύμου βρίσκουμε τις ρίζες $ x_1, x_2$ της πάνω σχέσης συναρτήσει του y.[/quote]

Κατ' αρχάς μπορούμε να θεωρήσουμε το $ x^4=a\geq0, y^4=b\geq0$ για να απαλλαγούμε από μεγάλες δυνάμεις.

Κατά δεύτερον η παράσταση γίνεται $ y^4(x{}^4){}^2 + x^4(4 - 6y^4) + 2y^8 = 0$ και

κατά τρίτον μπορείς να εξηγήσεις πώς έβγαλες τη διακρίνουσα$ =16 - 48y^4 + 28y^8$...εμένα μου βγαίνει τελείως διαφορετική...Πάντως και εγώ με τριώνυμο προσπάθησα να την προσεγγίσω.

maniopa καλή η λύση που εδώσες αλλά δεν μου φαίνεται και τόσο σωστή:
Νομίζω πως έχεις κάνει λάθος στην παραγοντοποίηση:

Είναι

(x^8)(y^4)+2(y^8)+4(x^4)-6(x^4)(y^4)= 0 <=> (y^4)(x^4)^2+(x^4)[4-6(y^4)]+2(y^8)= 0

[quote]Egregium Theorema[/quote]

Ίσως να την έδωσα όντως λάθος (έκανα όλες τις πράξεις κατευθείαν με το μυαλό και κάπου πρέπει να κόλησα). Πάντος αυτή είναι η σκέψη.

πολύ καλό Αρχιμήδης 6!!! Θα το κοιτάξω για σιγουριά αλλά ωφείλω να ομολογήσω οτι είναι αρκετά έξυπνο!!!

Egregium Theorema[quote][/quote]


Suppose a test consists of 25 question and awards 6 points for each correct answer, 1.5 points for each unanswered question, and 0 points for each incorrect answer, how many distinct scores are possible?

[hide="solution"]
we have $ a,b,c \in \mathbb{N}_0$;
$ 6a \plus{} \frac {3}{2}b \plus{} 0c \equal{} S$ and $ a \plus{} b \plus{} c \equal{} 25$. We want the number of values of $ S$, so since $ 6 \equal{} 4(\frac {3}{2})$ an equivalent question is;
How many solutions to $ 4a \plus{} b$ given $ a \plus{} b\le 25$

Letting $ 0\le b < 4$ then the epression $ 4a \plus{} b$ is like the division alogithm.
and we can make values from $ 0$ to the largest value $ 4\cdot 25 \plus{} 0 \equal{} 100$
However we cannot make $ 4\cdot 23 \plus{} 3, 4\cdot 24 \plus{} 2/3$, $ (\because 23 \plus{} 3,24 \plus{} 2,24 \plus{} 3 > 25)$

Hence the number of possible scores is $ 101 \minus{} 3 \equal{} \boxed{98}$

[/hide]

I think you are correct... but could you also find the score that has the most different ways of attaining it?  :)

[hide]
let $ 6a\plus{}1.5b\plus{}0\cdot c\equal{}S$ be the number with the largest number of ways of attaining it.

if $ (a',b',c')$ is another solution then
$ 6a \plus{} 1.5b \equal{} 6a'\plus{}1.5b' \Longrightarrow 4(a\minus{}a') \plus{} (b\minus{}b') \equal{} 0$

Hence $ b\equiv b' \mod 4$, and if $ (a,b)\not \equal{} (a',b')$ then $ |b\minus{}b'| \ge 4$

So assume $ (a,b,c)$ is one of the solutions to $ S$ such that $ b$ is max.
to find the next solution we take $ (a,b,c) \mapsto (a\plus{}1,b\minus{}4, c\plus{}3)$

We can continue this until we reach $ (a\plus{}6,b\minus{}24,c\plus{}18)$
taking it one further gives $ (a\plus{}7,b\minus{}28,c\plus{}21)$ which is impossible because $ 0\le b\le 25$

So $ S$ has at most $ 7$ solutions if $ b\ge 24$. So we have $ (1,24,0), (0,25,0), (0,24,1)$
$ \therefore S \equal{} 42, 37.5, 36$ which all have $ 7$ solution triples.

[/hide]


trigonometry search inequalities algebra polynomial function

Could u guys post some useful trigonometric identities in this topic?
 :D

$\sin(a\pm b)=\sin a \cos b \pm \sin b \cos a$
$\cos(a \pm b)=\cos a \cos b \mp \sin a \sin b$.
 :D

I think you can find millions of trig identities if you search in Google.

$\sin a+\sin b=2 \sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$
   ...there are very much trigonometric identities, and they are very useful

$\sin^2\alpha+\cos^2\alpha=1$
   This is very useful.

There are double-angle, triple-angle and half-angle formulas, but you can derive all of these from sum and difference (the ones kimby_102 posted).

Also...

Sum and difference for tangents:
$\tan (a+b)=\frac{\tan a\pm \tan b}{1\mp \tan a\tan b}$ ;)

[quote="t0rajir0u"]$\tan \alpha \tan \beta \tan \gamma = \tan \alpha + \tan \beta + \tan \gamma$

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma = 1$ (if memory serves...)

:)[/quote]
These properties are true iff $\alpha+\beta+\gamma=\pi$, right?

Yes.Another identity:
$\tan(a+b+c)=\frac{\tan a+\tan b+\tan c-\tan a\tan b\tan c}{1-\tan a\tan b-\tan a\tan c-\tan b\tan c}$

For double-angle,triple-angle,..., we can use the chebyshev polynomials to dinf the cosine:

$cos(na)=T_n(cos a)$

where:
$T_0(x)=1, T_1(x)=x$
$T_n(x)=2x T_{n-1}(x)-T_{n-2}(x),n \geq 2$

[quote="frt"]
[quote="t0rajir0u"]$\tan \alpha \tan \beta \tan \gamma = \tan \alpha + \tan \beta + \tan \gamma$

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma = 1$ (if memory serves...)

:)[/quote]
These properties are true iff $\alpha+\beta+\gamma=\pi$, right?[/quote]

Yes.  Well, the first one can also be shown to be true if $\alpha + \beta + \gamma = n\pi, n \in \mathbb{Z}$.

These two are nice:

$sin^4\alpha+cos^4\alpha=\frac{3}{4}+\frac{1}{4}cos4\alpha$
$sin^6\alpha+cos^6\alpha=\frac{5}{8}+\frac{3}{8}cos4\alpha$

These can be useful
$tan(a)+tan(b)=\frac{sin(a+b)}{cos(a)cos(b)}$
$cot(a)+cot(b)=\frac{sin(a+b)}{sin(a)sin(b)}$
$sin^2(a)=\frac{1-cos(2a)}{2}$
$cos^2(a)=\frac{1+cos(2a)}{2}$
$sin(a)+sin(b)=2sin(\frac{a+b}{2})cos(\frac{a-b}{2})$
$sin(a)-sin(b)=2cos(\frac{a+b}{2})sin(\frac{a-b}{2})$ $(a>b)$

How can we prove this :?:

$cos\frac{\pi}{2n+1}cos\frac{2\pi}{2n+1}cos\frac{3\pi}{2n+1}...cos\frac{n\pi}{2n+1}=\frac{1}{2^{n}}$

$sen\frac{\pi}{2n+1}sen\frac{2\pi}{2n+1}sen\frac{3\pi}{2n+1}... sen\frac{n\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^{n}}$

for $n \in \mathbb Z$

Multiply with $1$ where $1 = \frac{\sin{\frac{\pi}{2n+1}}}{\sin{\frac{\pi}{2n+1}}}$

You can also use roots of unity to prove those

Try to use search function on the forum. This topic should be usefull [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16103[/url]


eyy this is sean g from aloys, howd everyone find the amoc?
I thought it was pretty easy compared to the few past papers we got sent and got 4 of the qs out pretty well within 2 and a half hours. Q5 was gayness though, someone post a solution!
expected score: 25ish unless there was a stupid mistake

Well done Sean. =D I reckon you did pretty well.

I got three questions out but I reckon most people did... :huh:

Hoping for 20-ish. Maybe scab marks for q3 which I didn't get.

Hum, I apparently copied everyone else and solved the first three, wrote multiple pages on problem 4 to no avail, and ran away from 5. I don't suppose anyone's willing to provide proofs for the last two questions? =3


Could you please help me with this inequality.
x-1      2x+5
----  -  --------  > -2
3              4

Thank You,
Bob

While you're on this site, you should learn $\LaTeX$. It's pretty cool, and makes things look better! :)

$\frac{x-1}{3}-\frac{2x+5}{4}>-2$

Getting rid of denominators, $4x-4-6x-15>-24$.
$5>2x$
$\frac{5}{2}>x$

this does not look like it belongs in Pre-Olympiad.

well it is "pre" olympiad, would you say it's not? :-P

lol it's his first post, give him some time to find his way around the site.


Let $a,b,c$ be positive real numbers which satisfy the condition $ab+bc+ca=3$.Prove that the following inequality occurs:
\[\frac{a}{b+a+a^{3}}+\frac{b}{c+b+b^{3}}+\frac{c}{a+c+c^{3}}\leq 1 \]

Nobody likes it? :(  I have a solution for this inequality with cauchy :wink:

ok..I will give a hint:
In the right,rewrite $1$ as $\frac{a^{2}+b^{2}+c^{2}+ab+bc+ca+3}{(a+b+c)^{2}}$ :)
I think it is easier from here :)

Strange and nice inequality and solution, too!
From your hint, we have

$\sum \frac a{b+a+a^{3}}\leq \sum \frac{1+ab+c^{2}}{(a+b+c)^{2}}=1$.

[quote="Beat"]why
$\sum \frac a{b+a+a^{3}}\leq \sum \frac{1+ab+c^{2}}{(a+b+c)^{2}}=1$[/quote]
By the Cauchy Schwarz Inequality, we get
$\sum \frac a{b+a+a^{3}}\leq \sum \frac{ab+c^{2}+1}{(a+b+c)^{2}}$

Or even more explicitely,\[(b+a+a^{3})(ba+c^{2}+1)\geq \left(b\sqrt{a}+c\sqrt{a}+a\sqrt{a}\right)^{2}=\sqrt{a}^{2}\left(b+c+a\right)^{2}\]which means $\frac{a}{b+a+a^{3}}\leq\frac{ba+c^{2}+1}{(a+b+c)^{2}}$


probability
function
combinatorics proposed
combinatorics

There are $ 64$ players,in a tennis tournament where if a contestant lose a match, he´s eliminated.The players are p1,p2,... and p64,where p1 is better than p2,p2 is better than p3,... and p64 is the worst one.They play with each other in a random way.knowing that the better palyer always wins his match,what is the probability of p2 be one of the finalists?

[hide="Solution"]More generally, let's find the probability that $ p_2$ makes it to the round where there are $ 2^{k\minus{}1}$ players remaining. This is simply the probability that $ p_2$ does not play against $ p_1$ in all of the preceding rounds, which is $ P_{k,1,8} \equal{} \frac{62}{63}\times\frac{30}{31}\times\cdots\times\frac{2^k\minus{}2}{2^k\minus{}1}$ or, more succinctly, $ P_{k,1,8} \equal{} \prod_{i\equal{}k}^8 \frac{2^i\minus{}2}{2^i\minus{}1}$ which can generalize via the number of players and will telescope somewhat when writing $ 2^i\minus{}2 \equal{} 2(2^{i\minus{}1}\minus{}1)$ and $ 2^i\minus{}1 \equal{} (2^{i/2}\minus{}1)\cdot(2^{i/2}\plus{}1)$ (the latter is only useful when $ i$ is even). (We used the notation for $ P$ specified in the next paragraph.)

It is not hard to generalize to an arbitrary player: if it is possible for $ p_{n\plus{}1}$ to make it to the round with $ 2^{k\minus{}1}$ players remaining in a single-elimination tournament with $ 2^j$ players in total, he will do so with probability $ P_{k,n,j}\equal{}\prod_{i\equal{}j}^{k} \frac{2^i\minus{}1\minus{}n}{2^i\minus{}1}$. We can make provisions for double-elimination tournaments and for byes as well, although in the former case, there might be too much casework due to the variable tournament structure (for an arbitrary player) and in the latter case, the specific function of the byes must be specified.

By the way, in the specific case where $ n\equal{}1$, $ j\equal{}8$, and $ k\equal{}2$, we find that $ P \equal{} \frac{992}{1953}$.[/hide]

[hide="Solution"]The only way player 2 is in the final is if he's placed in opposite halves of the draw as player 1, and so the probability is 1/2. Is that right?[/hide]

In that case, you must specify the problem more clearly, or perhaps check your work. If by "final round" you mean the round where $ 2$ players remain, we know that $ p_2$ must have not played against $ p_1$ in the semifinals, quarterfinals, round of 16, etc. In the first round, there are $ 63$ potential players to play against, and $ 62$ of them are not $ p_1$. In the second round, there are $ 31$ potential opponents, and $ 30$ of them are not $ p_1$. Continuing this way, in the quarterfinals, there are $ 3$ possible opponents, and $ 2$ of them are not $ p_1$. In order to reach the final round, we simply calculate $ \frac {62}{63}\times\frac {30}{31}\times\cdots\times\frac {2}{3}$, which equals $ \frac {992}{1953}$ after simplifying. Or did I misinterpret the problem/make a computational error?

EDIT: I saw Math Champion's solution which appeared before I was able to post my response. That's another interpretation of this problem: that a bracket structure is generated before the tournament. I assumed that the players played each other in a "random way" after each round, because that seems more random.  :P In that case, we can fix the half of the draw that $ p_1$ appears in and we know that $ p_2$ must appear in the $ 64/2 \equal{} 32$ spots in the opposite half of $ p_1$, out of a total of $ 64 \minus{} 1 \equal{} 63$ spots; in other words, the probability is $ \frac {32}{63}$.

oops, yeah you're right vworld. It's not $ 1/2$ :blush:


Tak cesky IMO team 2006 je (v abed. poradi):

Jarda Hancl
Zbynek Konecny
Jakub Oprsal
Vojta Riha
Pavel Salom
Jan Uhlik

PS.: vite jejich prezdivky na AoPS? vetsina lidi to netaji..

Mirek

Myslim, ze Jarda sem nechodi, aj ked konto mozno nejake ma.
Len pre zaujimavost slovensky imo team (v poradi po vyberku)

Ondrej Budac (ja)
Jano Mikulas
Jaro Knebl
Istvan Estelyi
Miso Takacs
Samo Hapak

Prvni den za nama... ja jsem to uplne skopal, snEk s Pavlem se tvari ze maji dve ulohy, jednicku mame asi vsici... zadani snad budou jeste dnes viset na AoPSu. Tak drzte palce at to zitra dotahnem!

Slovinska tipkovnica ... :(


Find all functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ satisfying
\[ f(f(x) \plus{} y) \equal{} f(x^2 \minus{} y) \plus{} 4f(x)y
\]
for all $ x,y \in \mathbb{R}$.

You do not actually need to prove that $f(0) = 0$. The $f(0)$ term will ultimately cancel out if you start from step 3.

5) Why do you need to show this fact?

You do not need to show that $f(0) = 0$; the $f(0)$ term will cancel out ultimately if you start from step 3.

5) Why do you need to show this fact, which does not apply to the function $f(x) = 0$?

my solution is as follows...

set $y: =x^2$, then

\[ f(f(x)+x^2)=f(0)+4x^2f(x)  \]

now, set $y: =-f(x)$, then

\[ f(0)=f(x^2+f(x))-4f(x)^2  \]

from this we conclude that

\[ f(x)^2=x^2f(x)  \]

at this point everything seems to point that $f(x)=0$ or $f(x)=x^2$... but in "Problems and Solutions from Around the World 1999-2000" they prove that $f$ is not a combination of the two functions mentioned above, and this is what i don't understand from it...

can anybody explain me why we have to prove $f$ is not a combination if we have arrived to $f(x)\bigl(f(x)-x^2\bigl)=0$?  :?

thanks in advance  :D

Hi [b]campos[/b],

What you and the others have proved so far is the following: If we define the set $S = \{x \mid f(x) = 0\}$, then $\mathbb{R}\setminus S = \{x\mid f(x) = x^2, x \neq 0\}$. You still haven't proved that $S = \{0\}$ provided $f\not\equiv 0$.

To prove the latter, we use the fact that $f(x) \geq 0$, $\forall x \in \mathbb{R}$. Take any $x_0$ such that $f(x_0)> 0$ and arbitrary $y > 0$. Then

$f(f(x_0) + y) = f(x_0^2 - y) + 4f(x_0)y \geq 4f(x_0)y > 0\qquad \forall y > 0.$

Thus $f(x) > 0$ for all $x > f(x_0)$ and hence $f(x) = x^2$ for all $x > f(x_0)$.

Now in the given functional equation take $x = 0$ to get $f(y) = f(-y)$ and therefore

$f(y) = f(\bar{x}^2 + y)$

for any $\bar{x}\in S$. Thus $\bar{x}^2 + \bar{x} \in S$  for all $\bar{x}\in S$. If you assume that there is $\bar{x}\in S$ such that $\bar{x} \neq 0$, then you get that $S$ is unbounded, which is a contradiction.

Regards!

At this point you proved that the solutions are $f(x)=0$ ; $f(x)=x^2$ & $f(x)=g(x)$ where $g(x)=0$ if $x\in A$ and $g(x)=x^2$ if $x\in R-A$
And you have to find the subset $A$
that's why in point 5) i try to prove that the subset $A$ contains just one element $0$
Hope it's clear

To jivko777
What about if $y=\bar{x}^2-y$ To complete your proof you have to prove that $f(\frac{\bar{x}^2}{2})=0$

Be more careful [b]AYMANE[/b]!

Why do I have to prove it?

Regards!

Because your proof is based on  $f(y)=f(\bar{x}^2-y)$ so if $y=\bar{x}^2-y$
And your proof not gives a contradiction but proves that $S=R$ and thus $f(x)=0$ for all reals

Hi [b]AYMANE[/b],

What I prove is that if there exsits $x_0$ such that $f(x_0)> 0$, (and therefore $f(x_0) = x_0^2$) then $f(x) >0$ for all $x > f(x_0)$ and thus $f(x) = x^2$ for all $x > f(x_0)$. Please note that if we assume $f(x)\not\equiv 0$, then such $x_0$ necessarily exists, since we have proved that $f(x)\geq 0$ for all $x \in \mathbb{R}$.

From the first part of my proof it follows that the set $S$ must be bounded. Then I prove that

$f(y) = f(\bar{x}^2 + y)$ for all $y\in \mathbb{R}$ and all $\bar{x}\in S$

and now I have the freedom to choose $y = \bar{x}$ and if I further assume that $\bar{x}\neq 0$, then this would imply (by induction or whatever) that the set $S$ must be unbounded (to see that consider the function $g(x) = x^2 + x$).

Hope it's clear now?

Regards!

Your idea is good but i persist that is incomplete!!!
1) if $f(x_{0})+y=x_{0}^2-y$ then $f(x_{0})=0$
2) See my post above ( I idited it )

Hi [b]AYMANE[/b],

If you assume $f(x_0)\neq 0$ (that's what I have done) then this implies $f(x_0)>0$ and then this implies $f(x_0) = x_0^2$. Then I take arbitrary $y > 0$ and get

$f(f(x_0) + y) > 0$ by a small chain of inequalities.

If $f(x_0) + y = x_0^2 - y$, then this would mean that $y = 0$ and does not contradict to what I have written in any way. Do you agree now??

Regards!

I dont know how to explain you:
Your proof is based on $f(y)=f(\bar{x}^2-y)$ and so you have to include the case $y=\frac{\bar{x}^2}{2}$ so you have to proove that in this case $f(\frac{\bar{x}^2}{2})=0$
Till now you proved that $S=R-[\frac{\bar{x}^2}{2}]$
Can someone help us :!:

Please read carefully [b]AYMANE[/b].

In the equation

$f(f(x) + y) = f(x^2 - y) + 4f(x)y\qquad \mbox{(1)}$

let $x: =0$. Then you get $f(y) = f(- y)$ for all $y\in \mathbb{R}$. Now suppose there is $\bar{x}\in S$ such that $\bar{x}\neq 0$. Then, after setting $x: =\bar{x}$ in (1), you get

$f(-y) = f(y) = f(\bar{x}^2 - y)$ for all $y\in\mathbb{R}$.

Now change $y$ with $-y$ in the last equation to get

$f(y) = f(\bar{x}^2 + y)$ for all $y \in \mathbb{R}$.

And now you have the freedom to choose $y: =\bar{x}$ in the very last equation. So you get

$f(\bar{x}) = f(\bar{x}^2 + \bar{x})$,

which means that $\bar{x}^2 + \bar{x} \in S$ whenever $\bar{x}\in S$. Since we have assumed that $\bar{x}\neq 0$, this means that the set $S$ must be unbounded, which is the desired contradiction.

Does it seem OK now?

Regards.

$1) f(f(0))=f(0)$
$2) f(x^2)=f(f(x))$
$3) f(f(x)+x^2)=4f(x)x^2+f(0)$
$4) f(0)=f(x^2+f(x))-4f^2(x)$
from $3,4$follows that $f(x)(f(x)-x^2)=0$
then suppose that $f(a)=0,f(b)=b^2$ where $a!=b!=0$ plugging $x->a,y->y$ we will get condraction)
thus the answer is $f(x)=0 and f(x)=x^2$

let $y=0$,
$f(x^2)=f(f(x))$
let $x=0$,$f(f(0))=0$
let $y=x^2$
$f(x^2+f(x))=4x^2f(x)+f(0)$
let $y=-f(x)$
$f(0)=f(x^2+f(x))-4f^2(x)$
hence $f(0)=f(f(0))-4f^2(0),f(0)=0$
then it's trivial $f(x^2+f(x))=4f^2(x)=4x^2f(x)$
obtaining $f(x)=0 or x^2$
if there exists $c\ne 0,f(c)=0$
then $f(y)=f(c^2-y)$
for arbitrary $y$
since $y^2=(c^2-y)^2,0=c^2-y,y=0$
have a finite number of solutions,hence except for some number,$f(x)\equiv 0$
obtaining $f(x)\equiv 0$.
hence all solutions are $f(x)=0,f(x)=x^2$
QED

[quote=Dilshodbek]My solution 1).y=0 f (f (x)) =f (x^2) we assume that f (0) =a 2).x=0  f (y+a)=f (-y) + 4ay  3).y=-a  f (0) =f (a) -4a^2 thence a=o f (0) =0   5).y=x^2 f (f (x)+x^2)=f (0) +4x^2f (x)  we achieve f (f (x)+x^2) =4x^2f (x) 6). y=-x^2  f (x^2 +f (x)) =4f(x)^2   therefore we have f(x)=x^2 or 0[/quote]

Common error (see all the previous posts in the thread) :
From $f(x)^2=x^2f(x)$ you can conclude :
$\forall x$, either $f(x)=0$, either $f(x)=x^2$

But you can not conclude (as you did) :
Either $f(x)=0$ $\forall x$, either $f(x)=x^2$ $\forall x$

Look for exemple at the function $g(x)=\frac 14(x+|x|)^2$
It matches also $g(x)^2=x^2g(x)$

In fact, your proof is not finished (look at previous posts for a correct end of proof).

Let $P(x,y)$ be the assertion $f(f(x) + y) = f(x^2 - y) + 4f(x)y.$ $P(x,\tfrac{x^2-f(x)}{2})$ implies $f(x)\in \{0,x^2\}.$ Indeed if $f(v)=v^2$ and $f(u)=0$ then, $P(u,v)$ gives $f(u^2-v)=v^2\neq 0$ hence $2v=u^2$, not always true. So $f\equiv x^2$ or $f\equiv 0$.


What is the remainder of $ 2^{87} \plus{} 3$ when divided by 7?

First find a pattern of powers of two in modulo 7. Then add 3 to it.

Use Euler's totient function,which simplifies $ 2^{87}$ to $ 2^3$.

Like hunter said, you could find a pattern and stuff, or from Fermat's Little Theorem, $ 2^6$ is 1 (mod 7), so it's 11 (mod 7)=4.

Or, $ 2^3\equiv 1$ (mod 7), so $ 2^{87}\equal{}(2^{3})^{29}\equiv 1$ (mod 7), so just add 3 to get four

You really don't need to use totient function or even Fermat's in this problem if you know your powers of 2.

Well, I just did 7-3 and got 4.  I know this isn't the way to do it but hey, it worked. :D

find a pattern in the expontents of 2, do the modualr stuff. add 3

find remainder

simple.easy.to the point. lol


cmr: cần ít nhất là 2^k so tự nhiên x1,x2,x3,...,x(2^k);
để với mọi số n tự nhiên thì n=(x1)^k + (x2)^k + (x3)^k +...(x(2^k))^k;[code] :rotfl:  :D  :ninja:  :D [/code]

Waring's Problem

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

http://en.wikipedia.org/wiki/Waring's_problem


Two cars travel in opposite directions. The slower cars travels at 56mph and the faster at 60mph. In how many hrs will the cars be 464 miles apart?

Ans. 4 hrs

A passenger train and a frieght train start at the same time from stations that are 405 miles apart. And move towards eachother. The rate of the passenger is twice that of the frieght, in 3 hrs the trains pass eachother. Find the rate of each train.

Passenger- 90
Frieght- 45

Mr stone started from home on a trip that averaged 48 mph. How fast must his son travel in order to overtake him in 3 hrs and if carl started 30 minutes after his father

Carl must travel 40 mph.

A cargo plane left an airport at noon and flew towards new york at an average rate of 400 mph. At 2 pm a passenger  plane left the same airport for new york and flew the same route at 560 mph. How many miles did the passenger plane fly before it overtook the cargo plane?

10 miles.

Mr. Fields spent 6 hrs hiking along a trail. He walked out at the rate of 4 mph, and walked back at the rate of 2 mph. How far along did the trail did he go?

18 miles.

Ronald rode away from home in a car at the rate of 32 mph. He walked back at the rate of 4 mph. The round trip took him 2 and 1/4 hours. How long did Ronald ride?

36 miles.

If its wrong please give an explanation.  :)

We're here to help you with math, not make sure you get 100% on your homework.  We want to help you understand the problems, not give you answers.

I second this. Try to UNDERSTAND everything, and it won't matter if you got it right or wrong. In fact, colleges would ignore a 0.0 GPA in middle school. :wink:


3.  The incircle X of the acute angled triangle ABC is tangent to BC at K. Let AD be an altitude of triangle ABC and M be the midpoint of AD. If an is the other point of X and KM then prove than X and circumcircle of triangle BCN are tangent at N.

If you call $U$ the A-excenter of $ABC$ then $U, M, K$ are collinear.
 Suppose the tangent at $N$ cuts $BC$ at a point $J$. Then $IJ$ bisects the segment $NK$ at a point $Q$ (here $I$ is the center of $(X)$). We have $B, Q, I, C, U $ lie on a circle with diameter $IU$. Hence $JN^2 = JK^2 = JQ . JI = JB . JC$, which implies that $JN$ touches the circumcircle of $BNC$. And the result follows.


How many non-empty subsets $ S$ of $ \{1,2,3,\ldots,15\}$ have the
following two properties?
\begin{enumerate}
\item[(1)] No two consecutive integers belong to $ S$.
\item[(2)] If $ S$ contains $ k$ elements, then $ S$ contains no
number less than $ k$.
\end{enumerate}

I can't read this, I think the problem forgot some \$ signs in there.

Um, just get rid of the $ \text{\LaTeX}$ then  :P (too lazy to fix it and too lazy to do problem)

How many non-empty subsets $ S$ of $ \{1,2,3,\ldots,15\}$ have the
following two properties?

No two consecutive integers belong to $ S$.

If $ S$ contains $ k$ elements, then $ S$ contains no
number less than $ k$.

its 405, or thats the Alcumus answer anyway ... i dont get it though

Neither do I. Can anyone simplify it so I can understand it?


[quote="Arvind_sn"]Phil Bob looked at the clock. The hands pointed in the same direction. How much time does Pablo have to wait if he wants to see the clock's hands point in different directions?[/quote]

I hope you mean they form a 180 degree angle...because then the problem would be sort of pointless...

[hide]I assumed it was 12:00, but it doesn't matter. Each minute the minute hand moves 360/60=6 degrees, and the hour moves $\frac{360}{12\times 60}=\frac{1}{2}$ degrees per minute. So if $x$ is the number of minutes passed, $6x-\frac{x}{2}=180$ since we want a 180 degree angle. Solving, $\boxed{x=\frac{360}{11}}$. [/hide]


Hey i wasn't at the last math circle and so don't know if there is one this Monday (the 12th) Anyone know if its happening?
Thx.

afaik the waterloo one is no more this week..so the toronto one is prolly gone too


Prove that the two-variable polynomial

$P(x, y) = x^3 + y^3 - 3xy + 2$

Has no positive roots (that is, no roots where $x, y$ are both positive).

Hint for t0rajir0u: Intermediate Topics ;)

[quote="lkryptonicl"]what is AM-GM?[/quote]

The Inequality of Arithmetic-Geometric Means states that, for a set of positive reals $a_1, a_2, ... a_n$,

$\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 ... a_n}$

[quote="surge"][quote="hydro"][hide]

since $x,y$ are positive, by AM-GM we get $x^3+y^3+1\geq 3xy$[/hide][/quote]

Are there any steps to getting that answer?  :huh:[/quote]

That's actually how I came up with the question  :rotfl:

$x^3 + y^3 + 1 \ge 3xy$
$x^3 + y^3 - 3xy + 1 \ge 0$
$x^3 + y^3 - 3xy + 2 \ge 1$

Hence it can never equal $0$.  :)

How is:

$x^3 + y^3 + 1 \ge 3xy$

in the form:

$\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 ... a_n}$?

My brain's dead today.

Let $a_1=x^3$, $a_2=y^3$ and $a_3=1$.

Then we have $\frac{x^3+y^3+1}{3}\ge \sqrt[3]{x^3\cdot y^3\cdot 1}\Longleftrightarrow x^3+y^3+1\ge 3xy$ ;)


When written in base 12, a positive integer has one terminal zero. When written in base five, this same positive integer has two terminal zeroes. In how many other positive integral bases greater than one must the representation of this same positive itneger have at least one terminal zero?

[hide]Well, since it has two terminal zeros in base 5, it has one in base 25, and it must have at least one for all factors of 12, so that's 2, 3, 4, and 6 also.  So I'm getting 5...[/hide]

[hide="I think there is more"]

Since it ends in 0 in base 12, the number must be divisible by $3$ and by $2^2$.  Since it's divisible by $5^2$ we are looking for all the divisors of the number $2^2*3*5^2$, except for 1,5, and 12.  So that's 3*2*3 - 3 = 15.

EDIT: Small mistake in last calculation fixed.

[/hide]


If $a, b$ and $c$ are positive integers less than 10, then $(10a+b)(10a+c)$ equals $100a(a+1) + bc$ if:
\[ \text{(A)}\ b+c=10 \qquad \text{(B)}\ b=c \qquad \text{(C)}\ a + b = 10 \qquad \text{(D)}\ a = b \qquad \text{(E)}\ a + b + c = 10 \]

[hide]
Let's expand

$100a^2 + 10ac + 10 ab + bc = 100a^2 + 100a + bc$

Cross out identical terms on both sides to get

$10 ac + 10 ab = 100a$
$10a(c + b) = 100a$
$a(c + b) = 10 a$
$c + b = 10$

$(A)$
[/hide]

[hide]FOIL and distribute, then combine like terms.  The answer is $b+c=10$ which is $a$[/hide]


Is there a way to produce a shorter line than if I use the hyphen on the keyboard? For example, I want to write "45-45-90 triangle".

Is this what you want? $45\raisebox{0.5ex}{\tiny-}45\raisebox{0.5ex}{\tiny-}90$
If so, then add to your document
[code]\newcommand{\sdash}{\raisebox{0.5ex}{\tiny -}}[/code]so you can write [code]45\sdash45\sdash90[/code]
If that's too small then you could use [code]\newcommand{\sdash}{\raisebox{0.2ex}{\footnotesize -}}[/code] to get $45\raisebox{0.2ex}{\footnotesize-}45\raisebox{0.2ex}{\footnotesize-}90$
Play around to get exactly what you need.


geometry
geometric transformation
reflection

Given the points $A(1,7)$, $B(-2,4)$, $C(3,10)$, and $D(-1,5)$. Find the shortest distance possible for a path that starts at Point A; then goes to a line determined by any two of the points B, C, and D; and then goes to the remaining fourth point. Express your answer as a decimal rounded to 4 significant digits.

[hide] If $A$ wants to go to point $X$ and touch the line $YZ$ along the way, just reflect $X$ across $YZ$ to get $X'$. The minimum distance is $AX'$. Do this for all three possibilities and find the smallest one.[/hide]


1. In trapezoid $ ABCD$, $ AD\parallel BC$. $ K$ on $ AB$ and $ L$ on $ CD$, so that $ \angle ABL\equal{}\angle CDK$. Prove that $ \angle ALB\equal{}\angle CKD$.

2. Find all primes $ p$ that make
$ 5^p\plus{}4p^4$ is a perfect square

3. Prove that $ 2(1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)\ge(1\plus{}a)(1\plus{}b)(1\plus{}c)(1\plus{}abc)$

for 2
i dont like it but anyway it s here:(
\[ 5^p\equal{}(n\minus{}2p^2)(n\plus{}2p^2)\]
so if $ (n\minus{}2p^2)$ is not 1 it must be a multiple of 5 and $ (n\plus{}2p^2)$ is also multiple of 5 so their difference must be
a multiple of 5 from there 5 becomes our first solution if we try it.And also for the condition $ (n\minus{}2p^2)\equal{}1$ we have
$ 5^p\equal{}4p^2\plus{}1$ but it is possible because right side gets bigger so rapidly it is enough to try 2,3 which are not solution.So we only have p=5.

[quote="Dr Sonnhard Graubner"]hello, is there any condition for the inequality in 3.?
Sonnhard.[/quote]
I think $ a,b,c\in \mathbb{R^\plus{}}$. because the source where I got the problem, didn't give the condition of $ a,b,c$

@mehmetcantu: should we prove that for $ p>3,4p^2\plus{}1>5^p$?
Thanks mehmetcantu :)

How about number 1?


The Statement:
 "Suppose that f and g are 1-1 on R. If f, and gof are continuous on R, then g is continuous on R"
 is false

    I cannot find a counterexample

    Any help would be appreciated!

Here's a cheap one. Let $ f(x)\equal{}e^x.$ Let $ g(x)\equal{}\begin{cases}x&x>0\\x\minus{}1&x\le 0\end{cases}.$