trigonometry geometry trapezoid algebra polynomial LaTeX ratio

Hi,

I was originally trying to build Mock AMC C but I couldn't find much problems to last 6 so here are just 19 problems that I was thinking to use on Mock AMC C. They're arranged in level of difficulty so I hope you guys enjoy them.

[hide="Problems"]

1. Which of the following numbers is equal to the sum

\[7^7+7^7+7^7+7^7+7^7+7^7+7^7\]

A. $7^8$
B. $7^9$
C. $49^7$
D. $7^{49}$
E. $49^{49}$

2. $\triangle PQR$ has a right angle at $R$. If $\sin P = \frac{4}{5}$, then $\tan Q$ is

A. $\frac{\sqrt 5}{4}$
B. $\frac{3}{4}$
C. $\frac{4}{3}$
D. $\frac{\sqrt 3}{4}$
E. $\frac{\sqrt 7}{5}$

3. If $x$ is $x \%$ of $y$ and $y$ is $y \%$ of $z$, where $x$ is a positive real number, what is the value of $z$?

A. 100
B. 200
C. 10,000
D. It does not exist
E. It cannot be determined

4. During registration, a student discovers that he may register for a Mathematics class at the hours of 8:00,10:00, or 11:00; English at 9:00,10:00, or 11:00; and Science at 8:00,10:00, or 11:00. In how many different ways can he schedule the three courses?

A. 12
B. 27
C. 15
D. 18
E. 10

5. A square whose area is 16 partitioned into four congruent smaller squares. What is the area of the circle which passes through the centers of the four subsquares?

A. $\pi$
B. $2 \pi$
C. $3 \pi$
D. $4 \pi$
E. $5 \pi$

6. By placing a 2 at both ends of a number, its value is increased by 2317. The sum of the digits of the original number is

A. 5
B. 6
C. 7
D. 8
E. 9

7. The lengths of the two parallel sides of a trapezoid are 10 and 20, and the lengths of the other two sides are both $\sqrt{89}$. Find the length of either diagonal.

A. 15
B. $6 \sqrt 3$
C. 17
D. $8 \sqrt 2$
E. $\sqrt{2001}$

8. There are 100 people in a room. 95 of them speak Russians, 80 speak Mandarian, and 90 speaks Spanish. At least how many speak all 3 languages?

A. 30
B. 35
C. 65
D. 75
E. 80

9. If $f(x) = x^{2001} (x-1)^{2001}$, then $f(9) - f(3)f(4)$ equals

A. 9
B. 0
C. 3
D. 4
E. 2

10. An alloy consists of two metals (copper and tin) in proportion 1:2; another alloy consists of the same metals in proportion 2:3. Let $C$ be the number of pounds of alloy 1, and $T$ be the number of pounds of alloy 2. How many pounds of each alloy should be taken to get 1,320 pounds of an alloy containing the metals in proportion 17:27?

A. alloy 1: 1050, alloy 2: 270
B. alloy 1: 810, alloy 2: 510
C. alloy 1: 660, alloy 2: 660
D. alloy 1: 600, alloy 2: 720
E. alloy 1: 270, alloy 2: 1050

11. Two circles, one of radius 8 and one of radius 18, are tangent at point $P$. Two lines, $l$ and $m$, also are tangent to two circles and intersect in $E$. Line $l$ is tangent to the small circle at $B$ and the large circle at $A$. Line $m$ is tangent to the small circle at $D$ and the large circle at $C$. Find the distance from $E$ to a center of the small circle (one with radius 8).

A. 16.2
B. 18.5
C. 20.8
D. 22.6
E. 24.4

12. The sum of the lengths of the twelve edges of a rectangular box is 140, and the distance from one corner of the box to the farthest corner is 21. The total surface area of the box is

A. 812
B. 800
C. 798
D. 784
E. 776

13. For how many positive integers $n$, is $3^n + 81$ the square of an integer

A. 0
B. 1
C. 2
D. More than 3 but not infinitely many
E. Infinitely many

14. Starting with the numbers 1,2,3,...,500, a new sequence is formed by deleting numbers so that the sum of any two numbers of the new sequence is never a multiple of 7. What is the maximum length of the new sequence?

A. 213
B. 216
C. 217
D. 284
E. 287

15. For each positive integer $n$, let

\[S_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\]

Then for $n \geq 7$, $n S_n - (S_1+S_2+ \cdots + S_{n-1}) = $

A. $n$
B. $n+1$
C. $n-1$
D. $n+2$
E. $n-2$

16. Suppose $ABCD$ is a quadrilateral, inscribed in a circle. Then among the following identities:

\[\bullet \sin A = \sin C\\
\bullet \sin A + \sin C = 0\\
\bullet \cos B + \cos D = 0\\
\bullet \cos B = \cos D\]

how many are always true?

A. 0
B. 1
C. 2
D. 3
E. 4

17. During a recent period of time, eleven days had some rain. A morning rain was always followed by a clear afternoon. An afternoon rain was always preceded by a clear morning. A total of nine mornings and twelve afternoons were clear. How many days had no rain at all?

A. 6
B. 4
C. 3
D. 5
E. 7

18. Assume that all the roots of the polynomial $x^8+a_7 x^7 + a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2+ a_1 x + 420$ are positive integers. The maximum number of distinct roots of such as polynomial is

A. 3
B. 5
C. 6
D. 7
E. 8

19. Let $a_0 = 1,000,000$. For $n \geq 1$, define

\[a_n = \frac{a_{n-1}}{4} + 48\]

Then $a_{100}$ is closest to

A. 48
B. 60
C. 61
D. 63
E. 64 [/hide]

Use answer keys to check your answers. I do not have solution so you can ask or post them in this thread I guess. This is not mock AMC or anything like that but just a set of problems I gathered.  :P

[hide="Answer"]

1. A
2. B
3. A
4. E
5. B
6. D
7. C
8. C
9. B
10. E
11. B
12. D
13. B
14. C
15. A
16. C
17. D
18. B
19. E
[/hide]

Thanks! always nice to have new problems
I'll prob do the set after I wake up tmr, :D

[hide="1"]$7^7+7^7+7^7+7^7+7^7+7^7+7^7=7(7^7)=7^8\Rightarrow \boxed{A}$[/hide]

[hide="2"]$\sin P=\frac{4}{5}=\frac{opp}{hyp}$

Therefore, we have a 3-4-5 right triangle. $Q$ is the opposite angle; $\tan Q=\frac{opp}{adj}=\frac{3}{4}\Rightarrow \boxed{B}$.[/hide]

[hide="3"]$x=\frac{x}{100}\cdot y \Rightarrow 100x=xy \Rightarrow y=100$
$y=\frac{y}{100}\cdot z \Rightarrow 100=\frac{100}{100}\cdot z \Rightarrow z=100 \Rightarrow \boxed{A}$[/hide]

[hide="4"]If he takes the Math class at 8:00, he can take the English class at 9:00, leaving him with two options for Science; if he takes the English class at 10:00, he has one option for Science; if he takes the English class at 11:00, he has one option for Science. So, if he takes the Math class at 8:00, he has four options.

If he takes the Math class at 10:00, he can take the English class at 9:00, leaving him with two options for Science; if he takes the English class at 11:00, he has one option for Science. So, if he takes the Math class at 10:00, he has three options.

If he takes the Math class at 11:00, he can take the English class at 9:00, leaving him with two options for Science; if he takes the English class at 10:00, he has one option for Science. So, if he takes the Math class at 11:00, he has three options.

$4+3+3=10 \Rightarrow \boxed{E}$[/hide]

[hide="5"]The area of the square is $16$, so $s=4$. Each smaller square must be $\frac{1}{4}$ of the larger square, giving $s=2$. The center of each subsquare is $\sqrt{2}$ units from the center of the original square, so the radius of the given circle is $\sqrt{2}$, and the area of the circle is $\pi (\sqrt{2})^2=2\pi \Rightarrow \boxed{B}$.[/hide]

[hide="6"]By placing a $2$ at the end, the units digit was increased by $7$, so the original number was a $5$. This $5$ is now in the tens place, and that was increased by $1$, plus the $1$ carried over by having $5+7=12$, so the original digit there was a $3$. Therefore, the original number was $35$, and $3+5=8 \Rightarrow \boxed{D}$.[/hide]

[hide="7"]Since the lengths of the non-parallel sides are equal, it must be an isosceles trapezoid. Therefore, there is an extra length of $5$ on each side of the larger base, so we have right triangles with hypotenuse $\sqrt{89}$ and one leg $5$. Therefore, the height of the trapezoid is $\sqrt{(\sqrt{89})^2-5^2}=\sqrt{64}=8$. $\sqrt{8^2+15^2}=\sqrt{289}=17 \Rightarrow \boxed{C}$.[/hide]

[hide="8"]$95+80+90=265$. The average number of languages spoken per person is 2.65, so we must have at least $65$ people who speak all three languages (which would mean that the other $35$ speak two). $\boxed{C}$.[/hide]

No time to do the rest at the moment; I might have more time a little later.

Can you make a more printer friendly version... if it's not too much of a nuisance.. maybe a LaTeX document?

Because this is really helpful. I've already done all the AMC problems I can probably find.

19.

[hide]$a_{100}=\frac{\frac{\frac{\frac{\frac{a_0}{4}+48}{4}+48}{4}+48} \ddots}{4}+48$

Which ends up as $a_{100}=\frac{a_0}{4^{100}}+\frac{48}{4^{99}}+\frac{48}{4^{98}}+ \ldots +\frac{48}{4}+48$

the first term is very close to 0, and the rest of the series can be approximated as an infinite series $ 48\sum^{\infty}_{n=0}(\frac{1}{4^n})=64$

this approximation for $a_{100}$ has error much, much, much, much smaller than $\frac{1}{2}$, so we can rest assured that the closest integer to $a_{100}$ is 64.[/hide]

18.

[hide]by viete, the product of the roots must be 420.

$420=2\times 2 \times 5 \times 3 \times 7 \times 1$

is the prime factorization (with the 1 at the end).  there's 5 distinct primes in there.  no matter how else you divide up 420, there's not going to be more than 5 distinct factors.

A certain polynomial of that form with 5 distinct integer roots: $(x-1)(x-2)(x-2)(x-5)(x-3)(x-7)(x-1)(x-1)$[/hide]

14.

[hide]
Consider this: The ways that could lead the sum of two numbers to be a multiple of 7:

1mod7+6mod7 ---------(1)
2mod7+5mod7 ---------(2)
3mod7+4mod7 ---------(3)
0mod7+0mod7 ---------(4)

In the set, there are 71 of 4mod7 through 0mod7, with 72 of 1,2,3mod7. Since we dont want any two numbers to add to 0mod7, we should only have, in maximum, one side of (1),(2),(3),(4). Since there are 72's of 1mod7, 2mod7, and 3mod7, and we can keep 1 of 0mod7, then as a maximum we can have $72 \cdot 3 + 1 = 217$ numbers on the new list. C[/hide]

16.
[hide]
Remember that if the quadrilateral is concyclic, Angles A and C add up to 180, and Angles B and D add up to 180. Then $\sin{A}=\sin{C}$, and $\cos{B} = -\cos{D}$, which would satisfy condition #1 and #3. So 2 of the conditions are always true. (C)[/hide]

15.
[hide]
\begin{eqnarray*}
nS_n-(S_1+\cdots+S_{n-1}) &=& S_n+(S_n-S_{n-1})+(S_n-S_{n-2})+\cdots+(S_n-S_2)+(S_n-S_1)\\
&=& \left(1+\frac 12+\cdots+\frac 1n\right)+\left(\frac 12+\frac 13+\cdots+\frac 1n\right)+\cdots + \left(\frac 1{n-1}+\frac 1n\right)+\left(\frac 1n\right)\\
&=& 1+2\left(\frac 12\right)+3\left(\frac 13\right)+\cdots+n\left(\frac 1n\right)\\
&=& n
\end{eqnarray*}[/hide]

[hide="11"]Designate $M$ and $N$ the centers of the smaller and larger circles, respectively. Drawing the segment $EM$, we have that the distance is equal to $8+x$, and it is the hypotenuse of right triangle $BME$, with short leg $8$. Continuing $EM$ to $N$, we have that $EN=34+x=EM+26$, forming right triangle $ANE$ and similar to right triangle $BME$, with short leg $18$. We now have the ratios $\frac{18}{34+x}=\frac{8}{8+x}\Rightarrow 144+18x=272+8x \Rightarrow 10x=128 \Rightarrow x=12.8$. Therefore, $EM=8+12.8=20.8\Rightarrow \boxed{C}$.[/hide]

[hide="12"]There are three distinct edge lengths: $x$, $y$, and $z$. $4x+4y+4z=140 \Rightarrow x+y+z=35$, and the total surface area of the box is $2(xy+xz+yz)$. $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$, so we just need to find $x^2+y^2+z^2$ and subtract that. Conveniently enough, we are given the length between two opposite corners. $21=\sqrt{(\sqrt{x^2+y^2})^2+z^2} \Rightarrow 441=x^2+y^2+z^2$, Therefore, $(x+y+z)^2=1225=441+2(xy+xz+yz) \Rightarrow 2(xy+xz+yz)=784 \Rightarrow \boxed{D}$.[/hide]

[hide="13"]The only possible number is when $m^2=3^4(1+3^{n-4})$, and $(1+3^{n-4})$ is a perfect square, which is true only for $n=5$. Therefore, the answer is $\boxed{B}$.[/hide]

[hide="14"]We can take numbers in the pattern $7x+1$, $7x+4$, $7x+5$, and none will be able to add to a multiple of $7$, where $x$ is an integer. This gives a total of 215 numbers before we exceed $500$, and we can also use one multiple of $7$, so our answer is $\boxed{B}$.[/hide]

[hide="17."]
You first notice that a single day cannot rain twice.
Thus, each 11 rainy days must have one rain and one clear.

Also, the number of total clears is 9+12=21. A total clear day has 2 individual clears. Since the 11 rainy days each need a clear, they need 11 of them, so the number of total clear days is 21-11=10/2=5.

Then we check if this case can be constructed. Let 0's be clear and 1's be rainy. The first row is the morning and the second row is the afternoon, the columns are the individual days:

0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1
1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0

The answer is D, 5.
[/hide]


i want to get some test questions from you.such as AIME and   AMC.i want to get the intact test questions !can you give me some?

There are some questions in the "Questions" Column of this page.

http://www.unl.edu/amc/e-exams/e5-amc10/archive10.html

you could try going to the AIME or AMC website.  They probably have sample problems.
Go [url=http://www.unl.edu/amc/index.html]here[/url]


For any finite set $A \subseteq \mathbb{R}$, define $f(A) = \max A - \min A$. Take $S = \{1,2,\,\cdots,n\}$. Find $\sum_{A \subseteq S} f(A)$.

[hide]
$k$ is the maximum of any subset that contains only elements in $\{1,2,\ldots,k-1\}$ and $k$ itself. So $k$ is the maximum in $2^{k-1}$ subsets. Similarly, $k$ is the minimum in $2^{n-k}$ subsets. We are thus looking for the summation

$\sum_{i=1}^n i(2^{i-1}-2^{n-i})$,

which I believe we can inductively prove to be $2^n(n-3)+(n+3)$.[/hide]


《[b]High-School Mathematics[/b]》(China Tianjin)No.3(1992)
Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1x_2>0,x_1y_1>z_1^2, $ and $ x_2y_2>z_2^2,$ prove that \[(x_1+x_2)(y_1+y_2)\ge(z_1+z_2)^2+4\sqrt{(x_1y_1-z_1^2)(x_2y_2-z_2^2)}.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=500408&p=2811799#p2811799

Hungarian  mathematics olympic
Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1x_2>0,x_1y_1>z_1^2, $ and $ x_2y_2>z_2^2,$ prove that \[(x_1+x_2)(y_1+y_2)\ge(z_1+z_2)^2.\]

[quote="Rushil"]Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1>0,x_2>0,x_1y_1>z_1^2$, and $x_2y_2>z_2^2$, prove that: \[ {8\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le{1\over x_1y_1-z_1^2}+{1\over x_2y_2-z_2^2}. \] Give necessary and sufficient conditions for equality.[/quote]
To strengthen it, I got
Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1x_2>0,x_1y_1>z_1^2, $ and $ x_2y_2>z_2^2,$ prove that \[ {4\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le\frac{1}{\sqrt{(x_1y_1-z_1^2)(x_2y_2-z_2^2)}}\]
$\Leftrightarrow (x_1+x_2)(y_1+y_2)\ge(z_1+z_2)^2+4\sqrt{(x_1y_1-z_1^2)(x_2y_2-z_2^2)}.$( you can see It's proof:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=500408]here[/url])

See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366742

[quote=Rushil]Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1>0,x_2>0,x_1y_1>z_1^2$, and $x_2y_2>z_2^2$, prove that: \[ {8\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le{1\over x_1y_1-z_1^2}+{1\over x_2y_2-z_2^2}. \] Give necessary and sufficient conditions for equality.[/quote]
Let $x_1, x_2,y _1,y_2$ be real numbers such that $x_1^2 + x_2^2 \le 1$. [url=https://artofproblemsolving.com/community/c1068820h2085828p15037588]Prove the inequality[/url]$$(x_1y_1 + x_2y_2 - 1)^2 \ge (x_1^2 + x_2^2 - 1)(y_1^2 + y_2^2 -1)$$
[b]Austrian - Polish 1998[/b]


function
real analysis
real analysis unsolved

first, prove the particular case:

$ tg{1}\in{R\minus{}Q}$

Now, prove the general one:

if $ x\in{Q^{*}} \equal{}> tg{x}\in{R\minus{}Q}$

[quote="hpe"]This is in Ivan Niven's book, Irrational numbers, Cambridge University Press, 2005, on page 17-18.[/quote]

I believe you, but i don`t have this book:)

come ooon:(
nobody knows? I think here is some genius who can solve it:D just show yourself

i can't understand what you ask...
tg=tangent?
 :maybe:

Would make sense. $ \tan(0)$ certainly is not irrational :)

Yeah well, I did not know how to prove that.
See [url=http://www.philjol.info/index.php/JRSCE/article/view/96/101]this link right here[/url], though.

[quote="Tyl"]I think it's this:

Prove: $ \tan(1) \in \mathbb{R} \setminus \mathbb{Q} \equal{} \mathbb{I}$ (tangent of one is irrational).

However, what is $ Q^{*}$?[/quote]

lol people..of course $ tg$ is $ tangent$ and of course $ Q^{*}$ is $ Q\minus{}{0}$
however, your link doesn`t provide a solution for my problem

[quote]Every trigonometric function for any ''non zero'' rational values of the arguments is irrational....
[/quote]

this is the answer...

[quote="Zamfirmihai"][quote="Tyl"]I think it's this:

Prove: $ \tan(1) \in \mathbb{R} \setminus \mathbb{Q} \equal{} \mathbb{I}$ (tangent of one is irrational).

However, what is $ Q^{*}$?[/quote]

lol people..of course $ tg$ is $ tangent$ and of course $ Q^{*}$ is $ Q \minus{} {0}$
however, your link doesn`t provide a solution for my problem[/quote]

Your thread said you wanted help. You ask for help on a problem that apparently is not clear. We ask for clarification and get this? :)
See the second part of that link. It says something stronger than what Christiano89 said.

I think, that the link that Tyl suggests answer to your question...  :lol:
What you want more? :huh:


trigonometry
logarithms
function
modular arithmetic
real analysis
real analysis solved

Prove or disprove:

$\sum_{n=1}^\infty (-1)^n\frac{\sin nx}{\ln n}$ converges.

But it is easy to show taht it converges using Dirichlet.First of all $\frac{1}{\ln{n}}$ tends to zero monotonically and $\sum_{k=1}^{n}{(-1)^{n}\sin{nx}}=0$ for $x=\pi{m}, {m}\in{\mathbb{Z}}$ and  $\sum_{k=1}^{n}{(-1)^{n}\sin{nx}}{\leq}{\frac{2}{\sin{x}}}$ in other cases so for every $x$ the partial sums $\sum_{k=1}^{n}{(-1)^{n}\sin{nx}}$ are bounded(as a sequence).And applying Dirichlet we can deduce that the given series converges.The better question is to study it's convergence as a functional series.I mean does it cinverge uniformly or not.But it seems to converge uniformly in segments not containing points $\pi{m}, {m}\in{\mathbb{Z}}$ and converge not uniformly in the intervals containing such points.Unfortunately now I don't have time to post the full reply with it's proof.

[quote="LevonNurbekian"]The better question is to study it's convergence as a functional series.I mean does it cinverge uniformly or not[/quote]
It does not converge uniformly. Worse, it does not converge in $L^2[0,2\pi].$ The Fourier series $a_0+\sum_{n=1}^{\infty} a_n\cos(nx)+b_n\sin(nx)$ converges in $L^2$ if and only if $\sum_{n=1}^{\infty}(a^2+b^2)<\infty.$

Since $\sum_{n=2}^{\infty}\frac1{\ln^2n}=\infty$, this series does not converge in $L^2$ Furthermore the function it converges to cannot belong to $L^p$ for any $p>1.$

But Levon is right. His argument shows that the series converges for $x\not\equiv 0\pmod{\pi}$, and for $x\equiv 0\pmod{\pi}$ the sum is trivially zero. It is well-behaved near zero. The local bad behavior it must have (the unbounded limit function) must be in the neighborhood of $\pi.$

I should add that a summation by parts argument - what Levon called Dirichlet's test - does show that the series converges uniformly on any interval of the form $[-\pi+\delta,\pi-\delta].$ Hence the limit is continuous except at points equivalent to $\pi$ mod $2\pi.$ But ii is unbounded near those exceptional points.


linear algebra
matrix
vector
analytic geometry
linear algebra unsolved

Let $ A \equal{} \left(a_{ij}\right)$ be an $ n\times n$ matrix over a field $ \mathbb F$, and suppose that its permanent $ \text{Per}\left(A\right)$ is nonzero (over $ \mathbb F$). Then, for any vector $ b \equal{} \left(b_1, b_2,\dots ,b_n\right)\in \mathbb F^n$ and for any family of subsets $ S_1, S_2,\dots ,S_n$ of $ \mathbb F$, each of cardinality $ 2$, there is a vector $ X\in S_1\times S_2\times\dots\times S_n$ such that for every $ i$, the $ i$-th coordinate of $ AX$ differs from $ b_i$.

This follows directly from the Combinatorial Nullstellensatz Theorem. Consider the vector $AX$, where is is the vector with entries $x_1,x_2,\ldots,x_n$. If we assume that one entry of this vector is necessarily one of the $b_i$ we get $P(x_1,x_2,\ldots,x_n)=\prod_{i=1}^n (u_iX-b_i)=0$ whenever $x_i\in S_i$, where $(u_i)$ are the row vectors of $A$. Hence by Nullstellensatz, $P=\sum_{i=1}^n g_i(x)(x-s_{i,1})(x-s_{i,2})$ where $S_i=\{s_{i,1}, s_{i,2}\}$ and $deg(g_i)\leq n-2$. This is impossible since the coefficient of the monomial $x_1x_2\ldots x_n$ in $P$ is $Per(A)\neq 0$


I have a problem i'm wondering about:

If $ f(x) \equal{} xln|x|$ for x ≠ 0 , is it possible to define f(0) in such a way that f is continous on the whole real line?

I can't see why it could be continous on (-infinity,0)

thanks

if x<0 the limit of f as x approach 0 from left is 0. So you can expand definition of f to be continuous in R if f(0)=0=limf(x) as x->0


Why in My a profile my E-mail address is not displayed? :maybe:

You can set your options here: http://www.mathlinks.ro/Forum/usercp.php?mode=prefs&sub=privacy

thank you  :) , in next time  I will be more attentive .


A circle divided into $ n$ congruent arcs by points, which are numerotated by $ 1,2,..,n$ in a consecutive way, represents a game-field for two players, which make moves by turn. At his move, the player draws a line through two 'free' points of same parity, without intersecting the already drawn lines.The player that makes the last move wins. Who wins the game if both players use correct strategies for playing?

Can the lines intersect only on the interior of the circle?

Yes, the lines intersect only on the interior of the circle.
Sorry for missing that condition.
Now the problem is OK.


How many different numbers can be formed by the product of two or more of the numbers 3, 4, 4, 5, 5, 6, 7, 7, 7?

I have worked for over an hour on this problem and have gotten nothing.  I figure the total number of different combinations is 2^9 - 10 = 502 (the 10 represents all the sets with only 1 element and the empty set) where order is unimportant.  But from here I'm stuck.  I figure the next step would be to count the number of duplicate products (i.e. 3*4 is counted twice) but I can't think of any good way to do this.

Any help would be much appreciated.

Note:  This is a problem from the first week of an upper division combinatorics class.  The only things I can really use to count this are the addition and multiplication principles (in case there might be other methods).

Note that any good number has the form $3^a \cdot 4^b \cdot 5^c \cdot 6^d \cdot 7^e$  so that it is determined by the choice of $(a,b,c,d,e)$ according to the conditions :
$a,d \in \{ 0,1 \},b,c \in \{0,1,2 \}, e \in \{0,1,2,3 \}$ (1) and $a+b+c+d+e \geq 2$ (2).

Conversely, any number of this form is a good number and, fortunately, if $(a,b,c,d,e) \neq (a',b',c',d',e')$ then the associated numbers are distincts.

It follows that the number of good numbers is the number of $(a,b,c,d,e)$ satisfying the above conditions.
There are $2 \times 3 \times 3 \times 2 \times 4 = 144$ 5-uples satisfying (1). Only one of them staifies $a+b+c+d+e=0$ and 5 of them satisfy $a+b+c+d+e=1$. Thus, the desired number is $144-6=138$.

Pierre.


geometry
parallelogram
geometric transformation
rotation
homothety
similar triangles
geometry proposed

On the sides $AB,BC,CD,DA$ of a quadrilateral $ABCD$ we construct alternately to outside and inside similar triangles with vertices $Y,W,X,Z$ respectively .Show that $YWXZ$ is a parallelogram .

Denote $\phi = \angle AYB,\ k = \frac{BY}{AY}.$ Spiral similarity $\mathcal S_{Y}$ with center Y, rotation angle $\phi$ and similarity coefficient k carries the vertex A into B. Likewise spiral similarity $\mathcal S_{W}$ with center W, rotation angle $-\phi$ and similarity coefficient 1/k carries B into C, spiral similarity $\mathcal S_{X}$ with center X, rotation angle $\phi$ and similarity coefficient k carries C into D, and spiral similarity  $\mathcal S_{Z}$ with center Z, rotation angle $-\phi$ and similarity coefficient 1/k carries D into A. The similarity transformation $S_{Z}\circ S_{X}\circ S_{W}\circ S_{Y}$ is a translation, because the sum of all rotation angles is zero and the product of all similarity coefficients one. Since this translation has the fixed point A, it is an identity transformation. On the other hand, $S_{W}\circ S_{Y}= \mathcal T_{1},\ S_{Z}\circ S_{X}= \mathcal T_{2}$ are also translations, because in both cases, the sum of rotation angles is zero and product of similarity coefficients one. Since $\mathcal T_{2}\circ \mathcal T_{1}= I$ is identity transformation, translations $\mathcal T_{1},\ \mathcal T_{2}$ are by segments of the same length with opposite directions. Spiral similarity $\mathcal S_{Y}$ leaves Y in place, hence translation $\mathcal T_{1}$ carries Y into the same point Y' as spiral similarity $S_{W},$ so that the triangles $\triangle YWY' \sim \triangle BWC$ are spirally similar. Likewise, spiral similarity $\mathcal S_{X}$ leaves X in place, hence translation $\mathcal T_{2}$ carries X into the same point X' as spiral similarity $S_{Z},$ so that the triangles $\triangle XZX' \sim \triangle DZA$ are also spirally similar. Since the triangles $\triangle BWC \sim \triangle DZA$ are directly similar, the triangles $\triangle YWY' \sim \triangle XZX'$ are directly similar as well. From $\mathcal T_{2}\circ \mathcal T_{1}= I$ and $C = \mathcal T_{1}(A),\ A = \mathcal T_{2}(C),$ we have $YY' = AC =-CA =-XX'$ and $YY' \parallel AC \parallel XX',$ so that the last 2 triangles are centrally similar (congruent) with similarity coefficient -1 and similarity center $O \equiv YX \cap WZ \cap Y'X',$ the common midpoint of YX, WZ, Y'X'. Since the diagonals YX, WZ of YWXZ cut each other in half at O, it is a parallelogram.

Thank you very much very nice solution .
My solution was by complex numbers .

$B$ is the center of similitude of triangles $\triangle BYA$ and $\triangle BWC.$ Such a similitude is the product of the rotation $ (B,\angle ABC)$ and the direct homothecy $ \nabla(B,\frac {_{YW}}{^{AC}} ).$ This takes $ A$ into $ C$ and $ Y$ into $ W$ $\Longrightarrow$ $\frac {YW}{AC} \equal{} \frac {BY}{BA} \equal{} k$

Similarly, we'll have: $ \frac {ZX}{AC} \equal{} \frac {XD}{CD} \equal{} k$  $\Longrightarrow$ $ YW \equal{} XZ$

[i]Mutatis mutandis[/i], $ YZ \equal{} WX$ $\Longrightarrow$ Quadrilateral $ XWYZ$ is a parallelogram.

Is spiral similarity of triangles BYA and BWC with center B, which takes $ A \rightarrow C, Y \rightarrow W$ (as you say), really product of the rotation $ (B,\angle ABC)$ and a direct homothety $ \nabla(B,\frac {BY}{BA})$ ?

I apologize for such nonsense, I'm pretty drowsy right now.  :oops:

What I'm trying to explain is that $\triangle BYW$ and $\triangle BAC$ are similar, then $ \frac {_{YW}}{^{AC}} \equal{} \frac {_{BY}}{^{BA}} \equal{} k$ due to SAS criterion. This similitude with coefficient $\frac{_{BA}}{^{BC}} $ takes $A$ into $C$ and $ Y$ into $W.$

Based on the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=258422]Prove APRQ is a parallelogram[/url], we have that the angle formed by the line segments $ YW,\ AC,$ is equal to $ \angle \omega$ and that $ YW \equal{} u\cdot AC,$ where $ u \equal{} \frac {YB}{AB} \equal{} \frac {WB}{BC}.$

Similarly, we have also that the angle formed by the line segments $ XZ,\ AC,$ is also equal to $ \angle \omega$ and that $ XZ \equal{} u\cdot AC,$ where $ \frac {XD}{CD} \equal{} \frac {DZ}{AD} \equal{} u \equal{} \frac{YB}{AB} \equal{} \frac{WB}{BC}.$

Hence, we conclude that $ YW\parallel \equal{} XZ$ and the proof is completed.

Kostas Vittas.


Is this true? If true, How to prove? Let $ x, y, z$ be three directed lines. Then $ \measuredangle(x,z)\equal{}\measuredangle(x,y)\plus{}\angle(y,z)$.

How do you define directed angles between directed lines? Different possible definitions come to my mind (and some are probably wrong, as I am not an expert in this kind of angles). And do you mean the same by $ \measuredangle$ and $ \angle$ ?

  darij

sorry. I mean $ \angle(x,z) \equal{} \angle(x,y) \plus{} \angle(y,z)$ and $ x,y,z$ are directed line

This is not hold since $ \angle( x,y)\plus{}\angle(y,x)\equal{}0\neq \angle (x,x)$.


search
function
inequalities unsolved
inequalities

Let $ x,y,z$ real numbers such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$. Find the minimum value of
$ \frac {x }{ x^2 \plus{} 1} \plus{} \frac {y}{ y^2 \plus{} 1} \plus{} \frac {z}{ z^2 \plus{} 1}$

[quote="lambruscokid"]http://www.mathlinks.ro/Forum/viewtopic.php?p=1297645#1297645
Please try to use the search function when you know the source of the problem :)

Daniel[/quote]

I don't really understand why $ x\plus{}y\plus{}z\equal{}0$.

That was a wrong solution, a correct one was posted lower.


Ring Theory
vector
algebra
polynomial
linear algebra
matrix
Galois Theory

Question: Let $ M_n(\mathbb{Q})$ be the ring of nxn matrices over the rationals. Let K be a subring of $ M_n(\mathbb{Q})$ such that K is a field and K contains $ \mathbb{Q}$. Prove that the degree $ [K: \mathbb{Q}]$ is finite and $ [K: \mathbb{Q}]$ divides n.

Well, as a vector space, $ M_n(\mathbb{Q}) \approx \mathbb{Q}^{n^2}$ has dimension $ n^2$ over $ \mathbb{Q}$, so since K sits (up to isomorphism) between  $ \mathbb{Q}^n$ and $ \mathbb{Q}$, we have
$ [\mathbb{Q}^n: \mathbb{Q}] \equal{} [\mathbb{Q}^n: K][K: \mathbb{Q}]$ and so $ [K: \mathbb{Q}]$ divides $ n^2$.

But I don't know how to show it must divide $ n$. Help is appreciated!

Hint: show that this problem is equivalent to the following: if $ M\in M_n(\mathbb{Q})$ and the minimal polynomial $ p(x)$ of $ M$ is irreducible, then $ \deg{p}$ divides $ n$.

I'm having difficulty drawing the parallel between field extensions over $ \mathbb{Q}$ and minimal polynomials, and minimal polynomials in the context of matrices.

I'm wondering if we can draw a field diagram from bottom to top
$ \widetilde{Q} \, - \, K \, - \, \widetilde{Q}(A)$
where $ \widetilde{Q} = \{qI: q\in \mathbb{Q}\}$, where I the identity matrix, and where $ A\in M_n(\mathbb{Q})$ has an irreducible minimal polynomial over Q. But I can't justify why K sits below $ \widetilde{Q}(A)$; ie, I don't know how to use that $ K\subseteq M_n\mathbb{Q}$.

Am I on the right track?

It's the right idea, but you'll have an easier time trying to find a matrix $ A$ such that $ K \equal{} \tilde{\mathbb{Q}}(A)$.  The existence of such an element, in fact, has nothing to do with matrices.

Once you have such an $ A$, try to write down the properties that $ A$ must have to guarantee that $ K$ is a field.

Thinking about K over $ \widetilde{Q}$ as a general field extension, I can't see what's forcing K to be a simple extension. I can only tell that $ K$ must be the result of adjoining $ \widetilde{Q}$ with <= n elements. What other information do we have at hand to use?

Ahah! So how is this:
1) Since $ \widetilde{\mathbb{Q}} \subseteq K \subseteq M_n(\mathbb{Q})\approx \widetilde{\mathbb{Q}}^{n^2}$, we have
$ [\widetilde{\mathbb{Q}}^{n^2}: K][K: \widetilde{\mathbb{Q}}] \equal{} [\widetilde{\mathbb{Q}}^{n^2}: \mathbb{Q}] \equal{} n^2$,
so  $ [K: \widetilde{\mathbb{Q}}]$ is finite.

2) Therefore $ K\approx \widetilde{\mathbb{Q}}(A)$ for some $ A\in M_n(\mathbb{Q})$.

When I thought about why this has to divide n, I came up with a stronger statement, so I suspect my reasoning is flawed:

3) Now, $ \widetilde{\mathbb{Q}}(A) \approx \widetilde{\mathbb{Q}}/<m(x)>$, where $ m(x)$ is the minimal polynomial for $ A$ over $ \widetilde{\mathbb{Q}}$. Since $ \widetilde{\mathbb{Q}}(A)$ is a field, $ m(x)$ is irreducible. On the other hand, the minimal polynomial for $ A$ has to divide the characteristic polynomial p(x) of A, which will be of the form $ (x\minus{}\lambda_1)^{t_1}\cdots(x\minus{}\lambda_k)^{t_k}$ with $ \lambda_i\in \mathbb{Q}$. So the only way for m(x) to be irreducible over $ \mathbb{Q}$ is for it to be linear: $ m(x) \equal{} x \minus{} \lambda_1$, say. Then $ [K: \widetilde{\mathbb{Q}}]\equal{}1$, which trivially divides $ n$.

Is this too strong?

@jamesc:
What you have said is wrong, because the eigenvalues of a matrix with rational entries may not be rational - they may not even be real.

A way that could work would be rational canonical form.

[quote="jamesc"][quote="LydianRain"]All finite extensions of $ \mathbb{Q}$ are simple.  More generally, any finite separable extension is simple; this is sometimes called the [url=http://en.wikipedia.org/wiki/Primitive_element_theorem]primitive element theorem[/url].[/quote]
Thanks for your help so far. Could you suggest where I can steer my approach?[/quote]
Soarer provided the last piece.  You should read up on the rational canonical form, particularly on how it relates to the minimal and characteristic polynomial.

By the way, it helps to have some examples in mind when doing this kind of problem.  For example, you could have known that $ K$ can have degree larger than one, with some standard example like $ M \equal{} \left(\begin{array}{cc} 0 & 1 \\ \minus{}1 & 0 \end{array}\right)$, $ K \equal{} \mathbb{Q}(M) \cong \mathbb{Q}(i)$.


Hi folks,

In order to attract attention to this forum, in which people can put up results as they will be available on IMO 2004 :) we must visit this forum a lot.

So please whenever you feel like it, please click on the link. :)

When are you planning to post partial results ?  any news on cut-offs ?


Let x, y, z positive integers such that:
$ \frac{1}{x}-\frac{1}{y}=\frac{1}{z}$, let h= gcd(x,y,z), prove that hxyz and h(y-x) are perfect squares

$ x=ah,y=bh,z=ch,gcd(a,b,c)=1$
we have $ (b-a)c=ab$
so $ (gcd(a,b))^{2}|an=(b-a)c$
then $ gcd(a,b)|c$
so $ gcd(a,b)|gcd(a,b,c)=1\to\ gcd(a,b)=1$
and we have $ b-a|ab$ so $ b-a=1$ because $ gcd(ab,b-a)=1$
then $ \exists n\in N^{*}: \ (a,b,c)=(n,n+1,n(n+1))$
so $ xyzh=abch^{4}=(n(n+1)h^{2})^{2}$
and $ h(y-x)=h^{2}(b-a)=h^{2}$


Hello,

I happen to be in Beijing for one week and I heard there are some cool libraries here where I can buy some cheap (well, a thirtieth of what I'd pay in France) math copies of good books, in English of course (for instance I heard of such a library near the Academy of Science).

Anybody to go with me on Monday morning or Tuesday afternoon ? Any suggestion of good libraries ? - Warning: I don't speak Chinese -

Thank you :)

Julien

Where are you now? In which part of Beijing? Do you ride a bike?

If you have use MSN, then send me your adress. We can talk over it.

Hi Shobber,

Right now I am right in the center of Beijing (Metro stop Chongwenmen); I don't have a bike, well, I've just arrived today from Nanchang.
My Msn email is julien_santini@hotmail.com.

See you !

[quote="José"]Be careful! Never put your adress like that! Just put abc, where a=julien_santini b=@ and c=hotmail.com, It's safer.[/quote]

Whaddya mean by that?

Anyway, I can't help you julien, but I hope you have a nice time in beijing. Latch on to shobby! He is quite dependable.

[quote="José"]Be careful! Never put your adress like that! Just put abc, where a=julien_santini b=@ and c=hotmail.com, It's safer.[/quote]
José is right, on Wikipedia they remove that immediately for your own sake.
There are evil bots on the web scanning pages automatically for adresses. :mad:  :o

[quote="MithsApprentice"]There aren't too many bookstores in Beijing that sell books in foreign languages - period. Much less math books. You'll probably want to start looking at the Bookworm. Look up directions online.[/quote]
There is a small and not-so-famous bookstore in Beijing that only sells math books. It also sells a lot of Math in English, and the price is, indeed, cheaper than the price in Europe and America.

yeah,these books in china are more cheaper than in other counties.and they are so valuable.

[quote="MithsApprentice"]There aren't too many bookstores in Beijing that sell books in foreign languages - period. Much less math books. You'll probably want to start looking at the Bookworm. Look up directions online.[/quote]

Ok, this is a bit late of a reply, but I just wanted to point out that that's DEFINITELY not true.

The largest foreign language bookstore that has an EXCELLENT selection of undergraduate/graduate textbooks is the Foreign Language Bookstore on Wangfujing Street. They have quite a large collection of those yellow graduate texts in mathematics, as well as a variety of other titles.

I think the way it works is that Chinese publishers buy printing rights from the authors/publishers in the U.S., and they print them exclusively for sale in mainland China. As a result, a good selection of titles are available for a VERY low price. (I just bought Topology by Munkres for RMB 60, which is approx. USD 7.50.)

Other resources are Wangfujing Bookstore, Xidan Bookstore, and Book City, as well as the aforementioned Bookworm. (Shameless plug: I'd just like to mention that if you go to a foreign joint like Jenny Lou's/Subway/Starbucks/etc., you can pick up a copy of [i]that's Beijing[/i] for free, and they'll have listings of bookstores in Beijing, along with the Chinese address for taxi purposes.)

FYI - the Foreign Language Bookstore also has an excellent selection of undergrad/graduate textbooks in other subjects. I saw a copy of the Feynman lectures there and would have gotten them if I didn't already own them. So go and enjoy!

Hey there,

No problem for my email adress, this is not my principal one ...

Otherwise ... Shobber was kind enough to take me there, and I bought two books of immediate use. I'l go back there and buy more before I visit my family in France next May (I'll have to transit by Beijing anyhow).

Thank you all for the references and the help !!!


inequalities
function
inequalities proposed

I prefer the following
\[\frac1{a^{2}+bc}+\frac1{b^{2}+ca}+\frac1{c^{2}+ab}\leq\frac32\Bigl(\frac{a+b+c}{ab+bc+ca}\Bigl)^{2}.\]

pvthuan, I prefer Vasc's to yours.because when $c\to 0,b\to \infty$ the equality of Vasc's inequality is also attained.,But yours don't have this property.And Mr.Vasc's ineq is stronger than yours.

The following is also nice
\[\frac1{a^{2}+ab+b^{2}}+\frac1{b^{2}+bc+c^{2}}+\frac1{c^{2}+ca+a^{2}}\leq \Bigl(\frac{a+b+c}{ab+bc+ca}\Bigl)^{2}.\]
This is complimentary to a known one.

For all things there exist first time .This is the first time I solve a Vasc's inequality in the forum if I am not wrong .

The inequality is equivalent to
$\frac{(ab+bc+ca)(2a^{2}+2b^{2}+2c^{2}+ab+bc+ca)}{(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab)}\leq\frac{(a+b+c)^{2}}{(ab+bc+ca)^{2}}$

whic is equivalent to
$3(a^{4}b^{4}+b^{4}c^{4}+c^{4}a^{4})+2abc(a^{4}b+b^{4}a+b^{4}c+c^{4}b+c^{4}a+a^{4}c)+4abc(a^{5}+b^{5}+c^{5})\geq 2abc(a^{3}b^{2}+b^{3}a^{2}+a^{3}c^{2}+c^{3}a^{2})+6a^{2}b^{2}c^{2}(ab+bc+ca)+a^{2}b^{2}c^{2}(a^{2}+b^{2}+c^{2})$

which is clearly true by Muirhead or whatever you want

***Notice that
$\frac{1}{a^{2}+b^{2}+c^{2}}+\frac{2}{ab+bc+ca}\leq\frac{1}{a^{2}+2bc}+\frac{1}{b^{2}+2ca}+\frac{1}{c^{2}+2ab}\leq\frac{(a+b+c)^{2}}{(ab+bc+ca)^{2}}$

nice chain Vasc  :)

[quote="pvthuan"]The following is also nice
\[\frac1{a^{2}+ab+b^{2}}+\frac1{b^{2}+bc+c^{2}}+\frac1{c^{2}+ca+a^{2}}\leq \Bigl(\frac{a+b+c}{ab+bc+ca}\Bigl)^{2}. \]
This is complimentary to a known one.[/quote]
It is nice indeed, but it is not true (see $a\rightarrow \infty$).

Can someone prove the initial inequality by SOS method ?

[quote="Vasc"]If $a,b,c$ are positive numbers, then
$\frac 1{a^{2}+2bc}+\frac 1{b^{2}+2ca}+\frac 1{c^{2}+2ab}\leq (\frac{a+b+c}{ab+bc+ca})^{2}$.[/quote]

I prove this by rewritng it in the form
\[\sum_{cycl}\left( \frac{1}{ab+bc+ca}+\frac{1}{a^{2}+2bc}\right) (a-b)(a-c) \ge 0 \]
The rest should be easy.

[quote="pvthuan"]I prefer the following
\[\frac1{a^{2}+bc}+\frac1{b^{2}+ca}+\frac1{c^{2}+ab}\leq\frac32\Bigl(\frac{a+b+c}{ab+bc+ca}\Bigl)^{2}. \]
[/quote]

This is the one I used to derive the inequality at
http://www.mathlinks.ro/Forum/viewtopic.php?t=116853
as mentioned by Yimin Ge.

I prove this partly (only partly - since Cauchy is also invoked) in the same way as my proof for Vasc's inequality above.

Also related to this topics is
\[\frac a{b^{2}+bc+c^{2}}+\frac b{c^{2}+ca+a^{2}}+\frac c{a^{2}+ab+b^{2}}\geq\frac{a+b+c}{ab+bc+ca}.\]
This is true and easy.

[quote="ductrung"]I prove this by rewritng it in the form
\[\sum_{cycl}\left( \frac{1}{ab+bc+ca}+\frac{1}{a^{2}+2bc}\right) (a-b)(a-c) \ge 0 \]
The rest should be easy.[/quote]
Is the rest easy indeed?
Does SOS method fails for this inequality?

[quote="silouan"]If the function
$f(x)= \frac{1}{xb+bc+cx}+\frac{1}{x^{2}+2bc}$
has $f''(x)>0$

then it is true by Vornicu-Schur inequality . But is it ? :maybe:[/quote]
No, silouan.
Let's $f(x)=\frac{1}{x+2b}.$ Then $f''(x)>0,$ but the following inequality
$\sum_{cyc}\frac{(a-b)(a-c)}{a+2b}\geq0$ is wrong:$a=3,$ $b=1$ and $c\rightarrow0^{+}.$ :wink:
[quote="Vasc"][quote="ductrung"]I prove this by rewritng it in the form
\[\sum_{cycl}\left( \frac{1}{ab+bc+ca}+\frac{1}{a^{2}+2bc}\right) (a-b)(a-c) \ge 0 \]
The rest should be easy.[/quote]
Is the rest easy indeed?
[/quote]
Yes, it's easy by arqady's :D  method:
Let's $a\geq b\geq c.$
Hence, $\left( \frac{1}{ab+bc+ca}+\frac{1}{c^{2}+2ab}\right) (c-a)(c-b) \ge 0$ and
$\left( \frac{1}{ab+bc+ca}+\frac{1}{a^{2}+2bc}\right) (a-b)(a-c)+$
$+\left( \frac{1}{ab+bc+ca}+\frac{1}{b^{2}+2ac}\right) (b-a)(b-c)=$
$=\frac{(a-b)^{2}}{ab+ac+bc}+(a-b)\left(\frac{a-c}{a^{2}+2bc}-\frac{b-c}{b^{2}+2ac}\right)=$
$=(a-b)^{2}\left(\frac{1}{ab+ac+bc}-\frac{2c^{2}-3(a+b)c+ab}{(a^{2}+2bc)(b^{2}+2ac)}\right)\geq0.$ :wink:

[quote="Vasc"][quote="ductrung"]I prove this by rewritng it in the form
\[\sum_{cycl}\left( \frac{1}{ab+bc+ca}+\frac{1}{a^{2}+2bc}\right) (a-b)(a-c) \ge 0 \]
The rest should be easy.[/quote]
Does SOS method fails for this inequality?[/quote]
$\sum_{cyc}(a-b)(a-c)S_{a}=\frac{1}{2}\cdot\sum_{cyc}(a-b)^{2}(S_{a}+S_{b}-S_{c}),$
but I think, it doesn't work here.

My proof is to write the inequality as
$\frac{(a-b)^{2}+(b-c)^{2}+(a-b)(b-c)}{ab+bc+ca}+\sum\frac{(a-b)(a-c)}{a^{2}+2bc}\geq 0$.
Assuming that $a \geq b \geq c$, it suffices to show that
$(a-b)^{2}+(b-c)^{2}+(a-b)(b-c)-\frac{(ab+bc+ca)(a-b)(b-c)}{b^{2}+2ca}\geq 0$,
which is equivalent to
$(b-c)^{2}+\frac{c(a-b)^{2}(2a+2b-c)}{b^{2}+2ca}\geq 0$. :lol:

[quote="arqady"]
Yes, it's easy by arqady's :D  method:
Let's $a\geq b\geq c.$
Hence, $\left( \frac{1}{ab+bc+ca}+\frac{1}{c^{2}+2ab}\right) (c-a)(c-b) \ge 0$ and
$\left( \frac{1}{ab+bc+ca}+\frac{1}{a^{2}+2bc}\right) (a-b)(a-c)+$
$+\left( \frac{1}{ab+bc+ca}+\frac{1}{b^{2}+2ac}\right) (b-a)(b-c)=$
$=\frac{(a-b)^{2}}{ab+ac+bc}+(a-b)\left(\frac{a-c}{a^{2}+2bc}-\frac{b-c}{b^{2}+2ac}\right)=$
$=(a-b)^{2}\left(\frac{1}{ab+ac+bc}-\frac{2c^{2}-3(a+b)c+ab}{(a^{2}+2bc)(b^{2}+2ac)}\right)\geq0.$ :wink:[/quote]

Yes, arqady, that's exactly my proof too.
I did try SOS but it involes significantly more computation than the above proof.

[quote="silouan"]The inequality is equivalent to
$\frac{(ab+bc+ca)(2a^{2}+2b^{2}+2c^{2}+ab+bc+ca)}{(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab)}\leq\frac{(a+b+c)^{2}}{(ab+bc+ca)^{2}}$
whic is equivalent to
$3(a^{4}b^{4}+b^{4}c^{4}+c^{4}a^{4})+2abc(a^{4}b+b^{4}a+b^{4}c+c^{4}b+c^{4}a+a^{4}c)+4abc(a^{5}+b^{5}+c^{5})\geq 2abc(a^{3}b^{2}+b^{3}a^{2}+a^{3}c^{2}+c^{3}a^{2})+6a^{2}b^{2}c^{2}(ab+bc+ca)+a^{2}b^{2}c^{2}(a^{2}+b^{2}+c^{2})$which is clearly true by Muirhead or whatever you want [/quote]
silouan  said that Muirhead succeeds, but arqady said SOS fails.I didn't have any computing.But I think Muirhead can be proved by SOS,so one of silouan and arqady may have a mistake.I guess it may be silouan.

To Vasc. Nice! :lol:
[quote="Hawk Tiger"]... but arqady said SOS fails.[/quote]
I am sorry, Hawk Tiger, you are wrong. I don't say it. I say that
[quote="arqady"]
$\sum_{cyc}(a-b)(a-c)S_{a}=\frac{1}{2}\cdot\sum_{cyc}(a-b)^{2}(S_{a}+S_{b}-S_{c}),$
but I think, SOS doesn't work [b]here[/b].[/quote] :wink:

[quote=Vasc]If $a,b,c$ are positive numbers, then
$\frac 1{a^{2}+2bc}+\frac 1{b^{2}+2ca}+\frac 1{c^{2}+2ab}\leq (\frac{a+b+c}{ab+bc+ca})^{2}$.[/quote]

[img]http://s8.sinaimg.cn/middle/006ptkjAzy76seRcIILb7&690[/img]


$a,b,c >0$.Prove that:
           $\sum_{cyclic} \sqrt[3] {\frac{a^3+b^3}{2}} \leq a+b+c+\frac{(a-b)^2+(b-c)^2+(c-a)^2}{a+b+c}$
Question:Find the best contanst $k$ such that:
           $\sum_{cyclic} \sqrt[3] {\frac{a^3+b^3}{2}} \leq a+b+c+k\frac{(a-b)^2+(b-c)^2+(c-a)^2}{a+b+c}$ holds for all $a,b,c >0$

Yes, it's equivalent to $(4,1,0)\geq(3,1,1).$

[quote="libra_gold"]:D I have tested it by computer and the last inequalities is true.[/quote]

And what kind of software did you use?

See here for a similar one
http://www.mathlinks.ro/Forum/viewtopic.php?t=44070

[quote="bodan"]Just an idea: Using Mildorf's lemma for $k=3$ we obtain \[ \sum_{\mathrm{cyclic}}\sqrt[3]{\frac{a^3+b^3}{2}}\leq\sum_{\mathrm{cyclic}}\frac{a^2+b^2}{a+b}  \] so the inequality is equivalent to \[ \sum_{\mathrm{cyclic}}\frac{2a^2+2b^2-(a+b)^2}{2(a+b)}\leq\sum_{\mathrm{cyclic}}\frac{(a-b)^2}{a+b+c}\;\;\Leftrightarrow  \]\[ \sum_{\mathrm{cyclic}}\left(1+\frac{c}{a+b}\right)(a-b)^2\leq\sum_{\mathrm{cyclic}}2(a-b)^2\;\;\Leftrightarrow\;\;\sum_{\mathrm{cyclic}}(a-b)^2\frac{a+b-c}{a+b}\geq0.  \]
Maybe somebody could prove this inequality?[/quote]

Thank you, Nhat.

Moreover, the above result can be prove easily by SOS method. Notice that
\[ a^2\frac{a+c-b}{a+c}+b^2\frac{b+c-a}{b+c}\ge 0.  \]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G_{1},G_{2},...,G_{m}$ be subgroups of the group $ (\mathbb{Q}, \plus{} )$ and $ G_{i}\neq Q ,\forall i \equal{} \overline{1,m}$. Prove that $ G_{1}\cup G_{2}\cup ... G_{m}\neq \mathbb{Q}$.

That's cogent.
Have you found a counterexample?

very obvious: take some finite group which is not cyclic and all its cyclic subgroups.

Q is actually $ \mathbb {Q}$, my mistake.

ok then it's true. here is a proof, probably it's involved ;).

we may assume $ G_i \neq \{0\}$ for every $ i$. the product of some fixed non-zero elements in $ G_i$ lies in every $ G_i$, and multiplication with its inverse does not change the whole thing. thus, we may assume $ \mathbb{Z} \subseteq G_i$ for every $ i$, and we have to show the claim for proper subgroups $ G_i$ of $ \mathbb{Q}/\mathbb{Z}$. recall that $ \mathbb{Q}/\mathbb{Z}$ is the direct sum of the infinite prüfer $ p$-groups $ \mathbb{Z}/p^\infty$ and that every proper subgroup of $ \mathbb{Z}/p^\infty$ is finite - in particular the claim holds for these groups. so we only have to show that there is some prime $ p$, such that $ G_i \cap \mathbb{Z}/p^\infty \neq G_i$ for every $ i$. but the converse would mean that for every prime there is some $ i$ such that $ G_i \subseteq \mathbb{Z}/p^\infty$. this $ p$ is unique for $ i$ and we would map one-to-one infinitely many primes to finitely many indices, a contradiction.


hello everybody,
can anyne give some pdf notes on inequality(e.g. kiran kedlaya's note, mildorf's note)?
i need it urgently........
please help me...............

Exactly here http://www.animath.fr/cours/inegalites.pdf  :)


geometry
circumcircle
perpendicular bisector
geometry solved

How to contruct the point M inside triangle ABC so that:
$\widehat{MAB}=\widehat{MBC}=\widehat{MCA}$

Easy one.... first of all, our condition means that \[ \angle AMB= 180-\angle ABC  \] \[ \angle BMC= 180-\angle BCA  \] \[ \angle CMA=180- \angle CAB  \] As angles $\angle A,\angle B, \angle C$ are given, all we have to do is to construct three Thales arcs (Thales arc of two points $X,Y$ and angle $\alpha$ is a geometrical place of points $P$ such that $\angle XPY=\alpha$.) From our "angle equations" it's easy to see which arcs are to be constructed. Our point $M$ will be the intersection point of these arcs.

[quote="Virgil Nicula"]In exterior of the triangle $ABC$ you construct the equilateral triangles $BMC$, $CNA$, $APB$. Then the circumcircles of the last triangles are concurrently in the point $P$.[/quote]

This is not true. He asked for the point M satisfying < MAB = < MBC = < MCA, not < AMB = < BMC = < CMA.

In fact, the point M satisfying < MAB = < MBC = < MCA is called [url=http://mathworld.wolfram.com/BrocardPoints.html]the [b]first Brocard point[/b][/url] of triangle ABC and can be constructed by the method given in post #2. Here is a slightly different way to formulate this method:

By the tangent-chordal angle theorem, we have < MBC = < MAB if and only if the circle through the points A, M, B touches the line BC at the point B. In other words, < MBC = < MAB holds if and only if the point M lies on the circle passing through the points A and B and touching the line BC at the point B. Denote this circle by $k_a$; this circle $k_a$ can be easily constructed (in fact, since it passes through the points A and B, its center must lie on the perpendicular bisector of the segment AB, and since it touches the line BC at the point B, its center must lie on the perpendicular to the line BC at the point B; thus, its center can be located as the point of intersection of these two lines, and knowing the center, we can readily construct the circle). Similarly, define the circle $k_b$ passing through the points B and C and touching the line CA at the point C, and define the circle $k_c$ passing through the points C and A and touching the line AB at the point A. In the above, we showed that < MBC = < MAB holds if and only if the point M lies on the circle $k_a$; by the same argument, we can establish that < MCA = < MBC holds if and only if the point M lies on the circle $k_b$, and that < MAB = < MCA holds if and only if the point M lies on the circle $k_c$. Now, if we denote by M the point of intersection of the circles $k_a$ and $k_b$ different from B, then this point M satisfies < MBC = < MAB and < MCA = < MBC, so that < MAB = < MBC = < MCA. Hence, we have constructed our required point M. Moreover, we get as a bonus the fact that this point M lies on the circle $k_c$ as well (because of < MAB = < MCA).

  Darij

I am sorry ! My point is the Fermat's point. I understood incorrectly the enunciation.


Prove that for any set of 18 consecutive positive integers, one cannot partition the set into two nonempty subsets, such that the products of the numbers in each subset are equal.

Just a first thought -- [hide]the numbers have to be of the form

17n, 17n + 1, . . . 17n + 17, with the largest and smallest numbers being in the two disjoint subsets.  Then you might be able to say something about not being able to form a rational number n/(n + 1) from among the other 16 . . . that part didn't seem so easy though, so this may not be the way to go.[/hide]

I think you can do this one by [hide]thinking mod 19 and using Wilson's theorem.[/hide]


MATHCOUNTS
LaTeX
geometry
quadratics
conics
AwesomeMath
summer program

Okay, I got this game from the college confidential forum...

You say a statement about the person below you and the person below you says whether it's true or false and starts a new statement.

For example:
User 1 TPBM loves math.
User 2 True. TPBM is worried about college
User 3 False. TPBM...and so on.
tpbm= the person below me

I'll start of course:

TPBM is a vegetarian.

True, except for in marginal cases.

TPBM is taller than 100 cm.

True.

TPBM lives in the Northern hemisphere.

${ \text{F}_\text{alse}}$

TPBM is cool and ZZzZZZZZzzzZZZZzZZZZZZZ...

[size=25]ZZzzZ... (get it!)[/size]

  :bomb:      => :furious:
 :yankchain:

True..?

TPBM is sweating like crazy right now.

AH HAHAHAHAHAHA nah False.

TPBM has a proof of the Riemann Hypothesis.

FALSE!!!

TBPM knows who my avatar is...

FALSE.  :huuh:

TPBM is going into the 8th grade

FALSE

TPBM made it in the top 10 in MATHCOUNTS Chapter Tournament

False...  :(

TPBM knows who MY avatar is

FLASE.(wat is it?)

TPBM knows what all of these ranks on AoPS for # posts are name after

cant decide if my classes are boring yet, cuz today was the first day, and we didnt really do any learning, just going over rules.

Idk, true?

TPBM is not from a private school.

True.

TPBM is wearing headphones at the moment.

False.  Unless I'm TZF's alternate account....

True. But I think I have a medulla oblongata. :P

TPBM has been on TV.

False. But I could've been.

TPBM wants to get groovy with it

Depends on what it refers to.........

TPBM has a life.

True, it decreases very slowly.

TPBM has ADD

False.

TPBM talks like a retard, incessantly using the words "like, totally, cool, man, yo, sweet" and "awesome" in their sentences in order to conform to the admirable mould of North American society.

False

TPBM understood what Prime Factorization said.

"The below person me"???
SFFT???

Yo Prime Factorization, like, I'd say it's totally not admirable and, like, just pretty much in the USA, man.


limit
number theory unsolved
number theory

Posted before.
$ \lim_{n\to\infty}{a_n}\equal{}0$

Exactly $ a_n\equal{}\sqrt{\frac{3}{n}}\minus{}\frac{3\sqrt 3}{10}\frac{ln n}{n\sqrt n }\plus{}o(\frac{ln n }{n\sqrt n }).$


geometry
inradius
incenter
circumcircle
area of a triangle
Pythagorean Theorem
perpendicular bisector

[quote="bpms"][quote="ragnarok23"][hide]Once again, use Heron's to find the area of the triangle. \[\sqrt{8(3)(3)(2)=12}\] . Since the area of a triangle is also (inradius)(semiperimeter), we have \[12=8x\rightarrow x=\frac{3}{2}.\] The area of the circle is $\left(\frac{3}{2}\right)^{2}\cdot \pi=\boxed{\frac{9}{4}\pi}$[/hide][/quote]
Here's a suggestion. Don't use Heron's on an isosceles triangle. Draw the altitude to the base ;) and then use pythagorean. No variables needed.[/quote]
Oh crap, I should've seen that.

Extension: Find the area of the circumcircle. :D

[quote="nebula42"][quote="bpms"][hide="hint"]$R=\frac{abc}{[ABC]}$ where a b and c are the sides[/hide][/quote]
Actually, that's not correct.  :wink:  The correct version of that "formula" is
[hide] \[4R=\frac{abc}{[ABC]}.\] [/hide]

(By the way, if you want to show that a 'formula' you have made is incorrect, try plugging in some numbers or some situation that is easy to calculate. For example, in this case, you could use an equilateral triangle, and then you can see that the 'formula' is actually incorrect.)[/quote]
I don't know where that came from. I was being careless.

[quote="sonny"][hide]What do the A,B and C represent? By the way is the answer $625pi/64$? I solved it with another way.[/hide][/quote]
In general $[A_{1}A_{2}...A_{N}]$ means the area of triangle Of the n-gon.

[quote="sonny"]So is my answer correct?[/quote]
I got the same thing. How did you do it without le formule?
Why does le formule work?

Well what you can do is draw an altitude to base 6 whose length would be 4 using pythagorean theorem. Now, draw a line from the center of the circle to base 6. This divides the altitude into x and 4-x where 4-x is the radius. Now draw a radius from another verticy of the triangle.  That creates a right triangle with legs 3 and x and hyp. 4-x. Use pythagorean theorem and you get that x equals 7/8 and therefore 4-x(radius)=3 1/8.

[quote="ProtestanT"][quote="bpms"][quote="bfrock2000"]but how do i find the area of the circle after i find the altitude?[/quote]
The area is $\frac{1}{2}bh$[/quote]
That's the area of the triangle. How will we find the area of the circle after the altitude is found?[/quote]
Once you know the area of the triangle you can use the other formula for the area of the of a triangle, $rs$, where r=inradius (what we are trying to find) and s=semiperimeter.

[quote="bfrock2000"]the area of a circle inscribed in a triangle with sie lengths 5 ,5  and 6[/quote]

Here is it for both inscribed and circumscribed, no formulas, except for maybe some easy ones.

[hide="Incircle"]We will assume that $AB=5$, $AC=5$, and $BC=6$.

Let us call the incenter I, and the inradius $r$.

Now we shall draw the segments AI, BI, and CI.

So, we know that the circle is tangent to all three sides of the triangle, so, we can say that the altitude to each side is $r$.

Hence the area of ABI is $\frac{5r}{2}$, the area of ACI is $\frac{5r}{2}$, and the area of BCI is $3r$.

So this means that the area of ABC is $8r$, adding the areas of the three smaller triangles.

Also, we know that this is an isosceles triangle, with a base of 6, and using Pythagorean Theorem, a height of 4, so the area is 12.

Hence $8r=12$, and $r=\frac{3}{2}$.

Hence the area of the incircle is $\pi r^{2}=\frac{9 \pi}{4}$.[/hide]

[hide="Circumcircle"]Let the circumcenter be H.  Since the circumcenter is the intersection of the perpendicular bisectors of the sides, and we know that ABC is isosceles, then AD, the perpendicular bisector is also an angle bisector and median.  Hence, we know that the circumcenter is on AD.

We know AD=4.  Then, we can assume that AH=R, which is the value of the circumradius.  Then this means that DH=4-R.

Also, note that since it is the circumcenter, AH=BH=CH.

And finally, BDH is a right triangle, with hypotenuse of length $R$, and legs of length $4-R$ and $3$.

So, by Pythagorean Theorem

$(4-R)^{2}+3^{2}=R^{2}$.

So, solving for $R$, we get $R=\frac{25}{8}$.

So, the area of the circumcircle is $\pi R^{2}=\frac{625 \pi}{64}$

These are correct, I believe.[/hide]

[quote="bfrock2000"]the area of a circle inscribed in a triangle with sie lengths 5 ,5  and 6[/quote]

[hide]its 2 3-4-5 triangles. area=12[/hide]

[quote="ProtestanT"][quote="bpms"][quote="bfrock2000"]but how do i find the area of the circle after i find the altitude?[/quote]
The area is $\frac{1}{2}bh$[/quote]
That's the area of the triangle. How will we find the area of the circle after the altitude is found?[/quote]

So now that you know that the area is 12 you can draw three lines to the center of the circle from each verticy. This divides the large triangle into three trianlges with base 6,5 and 5. The altitude of each of the triangles is x which is the radius of the circle. You can find the areas of each of those triangles by multiplying the base by the altitude(radius) and dividing by 2. That gives areas 5/2x+5/2x+3x=12. Solve for that and you get 3/2.

[quote="nebula42"][quote="bpms"][hide="hint"]$R=\frac{abc}{[ABC]}$ where a b and c are the sides[/hide][/quote]
Actually, that's not correct.  :wink:  The correct version of that "formula" is
[hide] \[4R=\frac{abc}{[ABC]}.\] [/hide]

(By the way, if you want to show that a 'formula' you have made is incorrect, try plugging in some numbers or some situation that is easy to calculate. For example, in this case, you could use an equilateral triangle, and then you can see that the 'formula' is actually incorrect.)[/quote]
Another way to check if a theorem or something is correct is to use extremes.


function
number theory
relatively prime
number theory proposed

This is very challenging, but also very beautiful: for any sufficiently large prime $ p$, there exists $ g<p$ a primitive root mod $ p$ which is relatively prime to $ p-1$.

Let $\chi$ be a Dirichlet character mod $p$ of degree $p-1$, and let $q_1, q_2, \dots q_k$ be distinct prime divisors of $p-1$. Consider
$$ f(n) = \prod_{1 \leq j \leq k} (1-\chi(n)^{\frac{p-1}{q_j}}).$$
Note that $f$ is non-zero if and only if the $n$ is a primitive root modulo $p$. Consider also
$$ g(n) = \sum_{d |(n,p-1)} \mu(d). $$
Then $g(n)$ is nonzero (equal to 1) if and only if $n$ is coprime with p-1.

The statement of the problem can be turned equivalently to $\sum_{n=1}^{p-1} f(n) g(n) \neq 0$. To show this, first write
$$ f(n) = \prod_{1 \leq j \leq k} (1-\chi(n)^{\frac{p-1}{q_j}}) = \chi_0(n) \pm \chi_1(n) \pm \chi_2(n) \pm \cdots \pm \chi_{m}(n),$$ where $m = 2^k-1$, $\chi_0 = 1$ and $\chi_1 \dots \chi_m$ are non-principal characters mod $p$. Thus
\begin{align*}
\sum_{n=1}^{p-1} f(n) g(n) &= \sum_{n=1}^{p-1} (\chi_0(n) \pm \chi_1(n) \pm \chi_2(n) \pm \cdots \pm \chi_{m}(n)) \sum_{d |(n,p-1)} \mu(d) \\
&= \phi(p-1) + \sum_{l=1}^{2^k-1} \pm \left(\sum_{d: d|p-1}  \mu(d) \cdot \sum_{1 \leq n \leq p-1: d|n} \chi_l(n) \right) \\
&= \phi(p-1) + \sum_{l=1}^{2^k-1} \pm \left(\sum_{d: d|p-1} \mu(d) \chi_l(d) \cdot \sum_{1 \leq n_1 \leq \frac{p-1}{d}} \chi_l(n_1) \ \right)
\end{align*}
The famous Polya-Vinogradov bound states that $| \sum_{1 \leq n_1 \leq \frac{p-1}{d}} \chi_l(n_1) | \leq \sqrt{p} \cdot \ln{p}.$ Therefore the RHS above is greater than
$$\phi(p-1) - 2^{k} \cdot d(p-1) \cdot \sqrt{p}\ln{p},$$
where $d(p-1)$ is the number of divisors of $p-1$. This is positive because $\phi(p-1) \geq p^{1-\epsilon}$, while $2^k \leq d(p-1) \leq p^\epsilon.$ The result follows.


does anyone know if there are going to be any new Introductory level books? and do you know when they and the intermediate books are going to be released, and what subjects do they cover?

It says [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]here[/url].

[quote]Introduction to Algebra, Intermediate Algebra, and  Intermediate Counting & Probability will be available in early 2007. [/quote]

um... sorry to bring this up again  :maybe:

but could i plz get an estimate as to when the intermediate counting and probability book is gonna come out?

i have a math contest in april, so i was wondering if the book will come out before then, otherwise i'm going to order the introductory one

thx in advance for any help  :)

[quote="princevegetam"]but could i plz get an estimate as to when the intermediate counting and probability book is gonna come out?[/quote]It probably won't come out until this summer (possibly late spring if we're lucky).


This belongs in the FTW forum.

I don't play on FTW much, but it is a good way to improve speed. I like it, but I don't have the time to go on it.

bleh it's pretty addicting like louwiethechowie said.

[quote="louiethechewy123"]I usually don't play on FTW often, but when I start, it's very hard to stop. Maybe that's why I don't go on it much...[/quote]
I don't play FTW anymore. Addicted to Alcumus :)
But I don't own Intro to C&P so I don't know a lot of the problems but I try at least.

[quote="myyellowducky82"]

[quote="louiethechewy123"]I usually don't play on FTW often, but when I start, it's very hard to stop. Maybe that's why I don't go on it much...  [/quote]

I don't play FTW anymore. Addicted to Alcumus
But I don't own Intro to C&P so I don't know a lot of the problems but I try at least.
[/quote]

I do own Intro C&P, and in like May I had a really addictive Alcumus stage where I did like 50 problems a day... but then I decided to save the rest of the problems until more come out.  :)

FOR THE BILLIONTH TIME!!! DONT! SPELL! MY! USERNAME! WRONG!!!!!!
that means you, cgyao15. Copy and Paste If you need to.

Yeah at first, I thought it wouldn't be that great so I tried it out...now I try to do it everyday to help build my speed in solving math problems. The countdown round mode is especially interesting. :P  Btw what's the Alcumus thing you're talking about?

Alcumus is a resource AoPS has created (free) that teaches you counting and probability. It has video lessons and problems. There is a link to it on the left side of the AoPS website that says "Alcumus".

If you want more information, go to the Alcumus forum (Sorry, IDK how do make a link)

here's the link:
[url]http://www.artofproblemsolving.com/Alcumus/Landing.php[/url]

Dude I *severely dislike* FTW, only because im terrible at it.

I would imagine its a great way to improve your mathcounts CD skillz, but im terrible at it (my rating is like 1193 :blush:). Thats why my logic for mathcounts is "ace the written part so you dont even need countdown."

that's true, but you should still practice just in case.

Hey Thunder I'm pretty bad too. I only win one or two of the rounds I play...but that's why i do it, to get better. Btw guys whenever I happen to win a round my rating goes up like from 1000 to like 1020. Is that supposed to happen? Isn't it supposed to decrease? :huh:

Higher rating = higher skill level, although many people have inflated ratings. Like me!

For the Win is really addicting- once you go on and start playing it's almost impossible to stop...

I agree. FTW is addicting...but when I lose 2 games in a row I get really irritated and leave.  :oops:

For the Win [u]is[/u] addicting.  Normally, its either:

(1) I gain a lot of rating
(2) I lose a lot of rating

Recently it's been mostly (2)  :( .  But still I love FTW because it has helped me so much in MATHCOUNTS.

(If anyone wants to know, I play on FTW as professormom, NOT as professordad.  I abandoned professordad on FTW.  But professordad is used on AoPS for discreet math  :D and for posting on the forum.)

ftw seriously sharpened my mental acuity (but they have blunted in recent weeks  :P ) and Its a great resource and game. However too many people play it for rating and too many people abuse chat. That paired with a bunch of my own mental preparation for some contests next year and my need for mindless good for nothing fun during the summer doesnt leave me much time to use FTW to improve myself because i value knowledge of formulas (which i get from the forums occasionally) over the speed practiced on FTW.

On school days I would use this as often as possible as a replacement for the amount of time wasted on the failure (in my eyes) of homework.

My rating: 4.25 it would be a 5 if people stop getting so competitive and worried over this and just relax and try their best without cheating. I havent played this for some time because i am content with my 1337 rating and dont care to raise or lower it.

Yes I do like for the win, although i am new therefore i do lose a lot!!!   :oops:  How did [i]you guys[/i] get better in math?

aparently i study Aops books and read other textbooks and i check the forums :P

[quote="Dillinger"]Lol. FTW is addicting.[/quote]

it is if you are good at it

I go crazy when I lose 3cds in a row with 3points.  :mad:


trigonometry
inequalities unsolved
inequalities

Let $a,b,c$ be three sides of triangle $ABC$ . Prove that :
 $\sqrt{a+b+c}.(\frac{\tan \frac{A}{2}}{\sqrt{b}}+\frac{\tan \frac{B}{2}}{\sqrt{c}}+\frac{\tan \frac{C}{2}}{\sqrt{a}}) \geq 3$

This ineq is the same with a problem in VMEO 11 . When solving VMEO , once I infered this ineq from the original problem  :)
  If you want to solve VMEO , don't change the ineq into this form .


Yes sorry! The inequality is:

a 2 (2a+3)+ b 2 (2b+3)+c 2 (2c+3) \geq 6(ab+bc+ca)-3

cute!!
 \sum (a^3 +b^3 + 1)  \geq   \sum 3ab  by AM-GM
and 3 \sum a 2   \geq   3\sum  ab
add these :D  :D


AMC
AIME
AMC 12
USA(J)MO
USAMO
ARML
AoPS Books

[color=cyan]
How prepared would I have to be to expect a qualifying score for the AMC-12? AIME? Are only AoPS 1 and 2 enough for either? Suggestions?

 :o [/color]

people say that aops 1&2 should be sufficient for amc12 and aime. however for usamo aops3 (aka zeitz book) might be required.

however for usamo aops3 (aka zeitz book) might be required.[quote][/quote]

..and probably more than zeitz's book, since the book doesn't cover all the usamo topics (nor all the usamo topics to the needed extent).

But for AMC12 and AIME, yeah, both the AoPS books should include most of the theory needed to get near-perfect scores on both tests. Applying the learned theory is another matter though...

Fierytycoon

[quote]Applying the learned theory is another matter though... [/quote]

Yep.  Particularly for the AIME, you'll want plenty of practice.  Fortunately, there are now around 20 years of past AIMEs to cut your teeth on.  You can also turn to ARML and Mandelbrot for some practice (the harder questions on these two mimic the AIME fairly well, and the proof portions give you training-wheel guidance for the USAMO).


The lengths of five of six line segments are $ 3x\plus{}1$, $ 2\minus{}2x$, $ 5x\minus{}1$, $ 4x\minus{}3$, and $ 3x\plus{}2$. Find the length of the sixth segment in terms of $ x$ if the mean of all six segments is $ 3x\minus{}2$.

[quote="GameBot"]The lengths of five of six line segments are $ 3x \plus{} 1$, $ 2 \minus{} 2x$, $ 5x \minus{} 1$, $ 4x \minus{} 3$, and $ 3x \plus{} 2$. Find the length of the sixth segment in terms of $ x$ if the mean of all six segments is $ 3x \minus{} 2$.[/quote]

Adding yields 13x+1.  The total is 18x-12.  Thus we still need to add 5x-13.


geometric sequence
real analysis
real analysis solved

[b]Find the range of the value for n- th term of the infinite geometric series whose sum is 1.[/b]

kunny

Sorry for my poor English, Myth.

I have edited. How about this one?

kunny

Two notational or conventional understandings in my answer:
(1) The index of summation starts with $n=0.$
(2) $1+0+0+0+\cdots$ is considered a geometric series.

With those understandings, the range of allowed values for the $n$th term is:

For $n=0,$ the interval $(0,2).$
For $n$ even, $n>0$, the interval $[0,2).$

For $n$ odd, the interval $\left(-2,\frac{n^n}{(n+1)^{n+1}}\right].$

If you reject the convention in (2), then the answers are the same except for the deletion of $0$ from the sets for $n>0.$

If you restrict this to positive term series, then for $n=0$ we get the interval $(0,1)$ and for $n>0$ we get the interval $\left(0,\frac{n^n}{(n+1)^{n+1}}\right].$

To:Kent Merryfield

We restrict $n$ to be natural number.

Best regards

kunny

[quote]We restrict $n$ to be natural number.[/quote]

So your index of summation starts with 1? In that case just shift my answer. Change odd to even, change even to odd, turn $n=0$ into $n=1$ and change $\frac{n^n}{(n+1)^{n+1}}$ into $\frac{(n-1)^{n-1}}{n^n}.$

Oh! I see. :)

Here is the solution.

Let the n-th term of geometric sequence be $a_n$,the answer is

For $n=1,\ 0<a_n<2$
For $n=2,4,6,\cdots\ -2<a_n\leqq \frac{(n-1)^{n-1}}{n^n}$
For $n=1,3,5,\cdots\ 0\leqq a_n<2$


I was going to edit one of my post because I found a mistake on it. I clicked "Edit your post" in my post "While playing cards with my friends " and I got a message "Sorry, but you cannot edit this post anymore. Ask a moderator to do it if it is really necessary." But below on the site I was going to edit my post was said "You [b]can[/b] edit your posts in this forum.". A contradiction. It is a good property to edit your own posts so that I can correct my mistakes, so please give that property back.

Yeah, it says you cannot edit this post [b]anymore[/b]. You can only edit your posts in the first 36 hours from posting. After that, if there is an important modification to make, please ask a moderator of the forum the post is in to do it.

But why can't I edit my posts on my own? I found a mistake in my posts, but I can't fix it. It is so small, only one letter, so it is little boring to ask moderator to edit my post. And honestly, what do we lose if we can edit our own posts forever?


Let $z=e^{i\frac{2\pi}{15}}$.  Note that $\cos \frac{2\pi}{15}=\frac{z^{2}+1}{2z}$.
Note that
\[\frac{z^{2}+1}{2z}+\frac{z^{4}+1}{2z^{2}}+\frac{z^{8}+1}{2z^{4}}+\frac{z^{14}+1}{2z^{7}}=\frac{1}{2}, \]
as required.
I have already answered another one related to secant in another post of yours, namely http://www.mathlinks.ro/Forum/viewtopic.php?t=121127. If you post these in the high school section, then I believe you would receive answers since people may find them interesting.

Masoud Zargar


If ${ \Omega=\mathbb{C}}-{\left[{-1,1}\right]}$, and ${ f: \Omega\rightarrow\mathbb{C}}$, $ f(z)=\log{(z+1/z-1)}$, evaluate

$ \int_{C}{f(z)dz}$ with $ C=\left\{z\in{\mathbb{C}}: \left|z\right|=2\right\}$

[quote="Kalle"]I don't really get it. Is the intention that $ f$ should be continuous on $ \Omega$? But then $ f$ would be analytic in $ \Omega$, and that makes no sense because its growth behavior at $ \infty$ is not like an analytic function.[/quote]

Maybe he means
$ \log\frac{z\plus{}1}{z\minus{}1}$

Yes, $ \log\frac{z\plus{}1}{z\minus{}1}$
Sorry for the confusion

[quote="111222333444555"]Yes, $ \log\frac {z \plus{} 1}{z \minus{} 1}$
Sorry for the confusion[/quote]

In that case, try it with the residue theorem using the OUTSIDE of your curve.


geometry
3D geometry
AMC
AIME
USA(J)MO
USAMO

Well, scores are in.  Post your most recent scores here, even if they are not from the last SAT test date.
Mine:

CR:          710
Math:      800
Writing:   670

I only got 7/12 on the essay.  I was expecting a higher Writing, and lower CR.  But maybe my score looks better this way...

Good job, I missed a question (the cube question) on the math and got a 770 on math :( , and my other two were 560 :( But oh well...

Uhh, from a while ago (last may?)

Math: 800
Writing: 690
CR: 660

[quote="Iversonfan2005"][quote="236factorial"]Geez everyone's getting 800 on math, even I just got it today.[/quote]

Except me :([/quote]

Well, that's just one (maybe two?) question(s). I got that couple of days ago too.  :)

[quote="daveed"][quote="256factorial"]
Geez everyone's getting 800 on math, even I just got it today.
[/quote]

well, it IS AoPS, after all[/quote]

That's my point  :P  of course, the people who get below 700 or 650 on math won't even post here...  :D

i've gotten a 720 (missed 3 questions) and a 710 (missed 4 questions) on math. i don't know what happens.  :(
i guess i'll go join the stupid crowd with bleumoose.  :P

Well, score reports were out today, I got 1 wrong on Math and 18 wrong on both Verbal and Writing :(  :o

I got 740 but that was when I was 11 ohh well what you gonna do.

I got a 740 when I was 13.
I got an 800 in January 2004, but I just recently got a 760. I'm sending the 800 to all my schools, so the 760 hurts nothing but my honor.   :wacko:

well, the sat was hardly intended to determine high math ability anyway, just competence

[quote="PenguinIntegral"]My opinion on the math section is that it is a test of not making mistakes, and not a good indicator of real math ability at the top levels.

The problem for me is is the verbal...:([/quote]

Well, then again, we are on AOPS, and I doubt that anyone on this sight has a real problem with getting a $700+$ on Math, but for the majority of students in America, the Math section is fairly difficult since the National average is a $520$. In fact, students tend to do better on Writing and Verbal... But then again, I don't know how accurate the national average really is since a lot of people sleep through the test while others try to bubble in pretty pictures on the answer sheet :P

I concede that the SAT math section is not indicative of one's full mathematical ability.  The rest of the world (outside the mathy people) overrates the SAT math score's importance.  On the other hand, it seems to me that a lot of mathy people tend to underrate the merit of the SAT test.  There is some element of not making mistakes.  But that is a part of any reasonable test that you could administer to the public to test mathematical ability!  For example, some people might suggest that the AMC or AIME should be used in place of the SAT math section.  But there would still be people complaining that [i]should[/i] have gotten a higher score [i]if[/i] they didn't make stupid mistakes.  The SAT math section is meant to be tricky.  It's not a test of whether or not you can not make stupid mistakes but rather a test of how well you can interpret problems and reason.

And, to add to that, paying close attention IS pretty important in the real world.  Everybody knows that NASA employees aren't dunces, but when they made that whole metric-british unit mixup with the Mars probe a few years back...

[quote="probability1.01"]Well guess what I got a 2370 on the new SAT, but that was way back when I was 10...[/quote]

Unless I'm [i]quite[/i] mistaken, you weren't anything close to 10 when they started administering it. -.-

The only thing that that math section indicates is how much sleep you got the night before. Seriously... It measures little of true math ability, and I'm sure all of you who posted here have the potential for an 800 right now.  :roll:

I guess that's why we have test like AMC, AIME, and USAMO to distinguish from the top 0.99 percentile and the 0.01 percentile. :idea:  ;)

[quote="joml88"]Ok, so if I went and picked an "average" student from my school and made sure they got adequate rest, they could get an 800 right?

I'm sure everyone here [i]is[/i] capable of getting an 800.  But that's because the people here are the top math students not because the test is a huge joke.  The test is not perfect.  In many cases it doesn't distinguish between a good math student and a great math student.  However, it gives a decent estimate of where one ranks mathematically among other people taking the test.[/quote]
It does give a decent estimate, but I would prefer, after ranking in one of the higher precentiles, for more testing. A harder test, maybe giving orally by a trained examiner, would test your math skills more than the SAT and provide the collages with a list of people who are good at math, not people  who didn't make mistakes in the math section.
Im not saying competence in math isn't important, but once you reach a certain level, like multiplying 5 digit numbers, it should be done by a calculator anyway.

[quote="PenguinIntegral"]
It does give a decent estimate, but I would prefer, after ranking in one of the higher precentiles, for more testing. A harder test, maybe giving orally by a trained examiner, would test your math skills more than the SAT and provide the collages with a list of people who are good at math, not people  who didn't make mistakes in the math section.
Im not saying competence in math isn't important, but once you reach a certain level, like multiplying 5 digit numbers, it should be done by a calculator anyway.[/quote]

Yeah...you're right.  It's not like the SAT already costs a ton of money considering it's a test that everyone who has any aspirations of going to college has to take...

The less possibility a test offers for careless mistakes, the more that test costs to administer generally.  For example, very rarely do you hear someone complain about mistakes on the USAMO (and if you do hear someone complain...  :huh: ).  Anyway, the cost of the USAMO must be quite high per person due to the difficulty in writing USAMO problems and grading the solutions.

You may not be satisfied with the SAT but rest assured, you have other opportunities to prove yourself.  No, really!  Colleges do look at more than your SAT score.  And most intelligent people know that there is more to life than IQ and SAT scores.


Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that $ \sum_{cyc}\frac{a^{k}}{a+b}\ge\frac{3}{2}$ for any positive integer $ k \ge 2$.

To obtain the  given lower bound we will use these inequalities,where n is a positive integer larger than 2.

1.$ \frac{a^{n}}_{b+c}$+$ \frac{b^{n}}_{c+a}$+$ \frac{c^{n}}_{a+b}$ $ \geq$ $ \frac{a^{n-1}+b^{n-1}+c^{n-1}}_{2}$

2.$ {3^{n-2}({{a^{n-1}+b^{n-1}+c^{n-1})}}}$  $ \geq$ $ (a+b+c)^{n-1}$.
This second inequality can be proved by Tchebychef's Inequality.
Using the fact that abc=1 and applying the arithmetic-geometric mean inequality to the right hand side in (2) we get $ (a+b+c)^{n-1}$ $ \geq$ $ 3^{n-1}$
and therefore $ (a^{n-1}+b^{n-1}+c^{n-1})$  $ \geq$ $ 3$..........(3).

To prove the inequality (1):

Write the left side $ L$ as $ \frac{a^{n-1}.a}_{b+c}$  + $ \frac{b^{n-1}.b}_{c+a}$ + $ \frac{c^{n-1}.c}_{a+b}$

Suppose that a $ \geq$ b $ \geq$ c   then

$ a^{n-1}$   $ \geq$   $ b^{n-1}$  $ \geq$   $ c^{n-1}$ and

$ \frac{a}{b+c}$   $ \geq$   $ \frac{b}{c+a}$   $ \geq$   $ \frac{c}{a+b}$ .

Tchebychef's Inequality gives

$ \frac{L}{3}$ $ \geq$ ${ \frac{1}{3}}$.( $ \frac{a}{b+c}$ +$ \frac{b}{c+a}$+$ \frac{c}{a+b}$). ${ \frac{1}{3}}$. $ (a^{n-1}+b^{n-1}+c^{n-1})$

Now applying Nesbitt's Inequality to the first parenthesis on the right of this inequality we finally reach inequality (1)


I always read AoPS during digital art class, and you guys inspired my latest project.

[img]http://gosuapm.com/goodpixx.jpg[/img]

Billy: Billy
chicken: Mr Chen
Pig walking around the river: Walk around the river
boy soprano on the horse: boy soprano on the horse
thing in the egg: future georgia ARML B team member/AoPS poster/etc

Lol.

Poor Billy, his shadow ran away from him.  :lol:

Very cool picture.

BTW: Is any of that picture actually taken on Sola Fide Farm?

Nope, it was all constructed from 17+ different images I found online.

mr chen = now a ranger/ Go Around the tree

muhahahahahahahaha I'm glad I came upon this.

how did you know my name? wtf lol

is that picture cicatriz? or just a picture found on the internet.

Oh come on Cicatrix, everyone knows fir trees don't grow in Georgia! :lol: Nor have we any horses, and the pigs are presently in the freezer. As for turkeys, well, that's one skinny turkey there ;), and we've never grown ugly brown turkeys like that one. But we have grown great big fat white-plumed turkeys :).

Oh yes. I was the youngest person in Georgia ever to get their egg-candling license ;)...

Oddly enough, I once [i]did[/i] ride horses.  (And that photo of me works uncannily well for horseback posture) :lol:

[quote="anirudh"]WHAT? I THOUGHT NOW A RANGER AND GO AROUND THE TREE WERE COUSINS :maybe:[/quote]

they could be but some people like me suspect now a ranger has mult. personalities. i am prob. wrong after all i think the gov. is scheming against me.

schweet pic
u need another egg for ME!!! muahahaha


Again I ask you, can you please have the guides show up on the forum so that team members can get in touch even ahead with their guides :P :D

[quote="Sasha"]Other countries don't get guides? :?[/quote]All countries will be getting guides (I belive it's an official IMO regulation but I am not sure about that). It's probably just that niño doesn't yet know the names of all the guides :)

[quote="Albanian Eagle"]do you know the name of the Albanian guide???? :?:[/quote]

Manuel Perez e-mail manueru_p@hotmail.com

I'm almost shure that he's your guide, does anyone else want to know?

by the way, i'm Ireland's guide

[quote="Cristos"]by the way, i'm Ireland's guide[/quote] :welcomeani: finally a guide on the site :P

[quote="Vahe Musoyan"]Do you know Armenia guide's name?[/quote]

no, I don't know, sorry.

[quote="niño"]Guides in imo:

FRANCE: Alexandre Ramos y Peon (ARP 1986)
TUNISIA: Adrian Ramos Peon
NEW ZEALAND: Octavio Arizmendi Echegaray (IMO 2003)
PUERTO RICO: Gerardo Arizmendi Echegaray
IRELAND: Cristos Ruiz Toscano (IMO2004 & IBERO 2003)
KAZAKHSTAN: Carlos Vargas Obieta (IMO 2004 & IMO 2003 & IBERO 2003 & CENTROAMERICAN 2002)
URUGUAY: Mariano Gonzalez Tokman
ARGENTINA: Samuel Estala (IMO 2000)
GREECE: Catia Baron Aznar
CANADA: Monica Carrillo
HOLLAND: Alicia Prieto Langarica
DENMARK: Alonso Prieto Langarica
USA: Nicolas Flores Castillo
MEXICO: Rodrigo Cervantes
HONG KONG: Mirna
LICHSTENTEIN: David Plata
BELGIUM: Ana Laura Garcia
ECUATOR: Fermin Reveles
SLOVAKIA: Abraham Solis
PORTUGAL: Daniel Gonzalez Tokman
CROATIA: Carla Soriano

for more information, please contact Br. Alexandre Ramos, aramos@cimat.mx[/quote]

niño is wrong, I was in IBERO 2004  :lol:

This is all I know by now (albania's guide was changed)

Albania		Lizet Romero Barrera
Argentina		Samuel Estala  Arias
Australia, 		Claudia Verónica Ponce Alfaro
Austria,		Marco David Blanco Sandoval
Bangladesh	Ana Paola Lopez Díaz
Belgium, 		Ana Laura García González
Belarus		Kenya Verónica Espinosa Hurtado
Bolivia, 		Alejandra Donaji Herrera Reyes
Bosnia and Herzegovina, 	Luis Manuel Huerta Aparicio
Brazil, 		Palmas Pérez Santiago Alonso
Bulgaria,  		Laura Mireya Gutiérrez Quintal
Canada, 		Monica Itzuri Delgado Carrillo
People´s Republic of China, 	Yichao Zhu Wu
Colombia, 	Fátima Renée Suárez Baeza
Costa Rica, 	Teresita de Jesus Morales Leal
Croatia, 		Carla Soriano Garza
Cuba, 		Andrés Lara Maldonado
Cyprus, 		José Luis Perera Pomol
Czech Republic,	Gerardo Alfaro Cárdenas
Denmark, 	Alonso Prieto Langarica
Ecuador, 		Fermín Omar Reveles Gurrola
El Salvador,	Abril Paulina Arellano Morales
Estonia,  		Maria Nohemi Bravo Solis
Finland, 		Mariana Carnalla Cortés
France, 		Alexandre Ramos  Peón
Georgia, 		Mildret Yalile Rosado Rivero
Germany		Verónica Arévalo Schoedl
Greece, 		Catia Baron Aznar
Guatemala, 	Hector Jesus Rojas Santana
Hong Kong 	Mirna Lissette Gómez Morales
Hungary, 		Fabiola Gamboa López
Iceland,		José Alejandro Andalón Estrada
India, 		Ariel Hernández Gómez
Indonesia,	Oscar Alberto Quijano Xacur
Islamic Republi of Iran, 	Darío Herrera Sierra
Ireland, 		Cristos Alberto Ruiz Toscano
Israel,		Edgar Alfredo Duéñez Guzmán
Italy, 		Ana Lucía García Pulido
Japan, 		Martha Wario López
Kazakhstan, 	Carlos Vargas Obieta
Republic of Korea	Joyce Allyne Vicario Solís
Kuwait		Alissar Sleiman A.
Kyrgyzstan, 	Natalia Jonard Pérez
Latvia, 		Avila Jáuregui Jorge Antonio
Lichtenstein	David Plata
Lithuania,  	Laura Elena Ochoa Gómez
Luxembourg, 	Mery Pina Hinojosa
Malasia		Raymundo Aguillón Álvarez
Former Yogoslav Republic of Macedonia, 		Omar Betanzos Méndez
Mexico, 		José Rodrigo Cervantes Polanco
Moldova, 		Daniela Olvera
Mongolia		José Felipe Golib Dzib
Morocco		Héctor Fabián Campos Mora
Mozambique	María Luisa Lara Couoh
The Netherlands, 	Alicia Prieto Langarica
New Zealand, 	Octavio Arizmendi Echegaray
Norway, 		Iván Leonardo Pérez Cabrera
Pakistan, 		Adolfo Vázquez Ceballos
Paraguay, 	Orlando Ochoa Castillo
Peru, 		Geomara Asunción Martínez Perera
Philippines, 	Silvia del Rocío López Maldonado
Poland, 		Pawel Cesar Sanjuan Szklarz
Portugal, 		Daniel Matías González Tokman.
Puerto Rico,	Gerardo Arizmendi Echegaray
Romania, 		María Alejandra Lira Pantoja
Russian Federation,		Leonid Serkin
Serbia and Montenegro	Rodrigo Orlando Balam Chacón
Singapore	Ángel Águila Valdez
Slovakia,		Abraham Solís García Inda
South Africa,	Cesar Millan Pacheco
Spain, 		Laura Cecilia Avila Jáuregui
Sri Lanka 		Gloria Guadalupe Yver Cambranis
Sweden, 		Karla Sarahí Aguilera Gómez
Switzerland	Jorge Agustín Albarrán Morales
Taiwan		José Muñoz Porras
Tajikistan, 	Emanuel Pérez Gomez
Thailand,		Grisel Rio Suárez
Trinidad and Tobago, 	Claudia Dinorah Rivera Burgos
Tunisia		Adrián Ramos Peón
Turkey, 		Yasam Dogu
Turkmenistan, 	Itza Tlaloc Quetzalcoatl Curiel Cabral
Ukraine, 		Bernardo Altamirano Salmerón
United Kingdom, 	Sandra González Monge
United States, 	Nicolas Virgilio Flores Castillo
Uruguay, 		Mariano Raúl González Tokman
Uzbekistan,	Ekaterina Jimenez Sabinina
Venezuela, 	Sergio Guzmán Sánchez
Vietnam,		Roger Abraham Medina Brito

[quote="Cristos"]This is all I know by now (albania's guide was changed)

Albania		Lizet Romero Barrera
[/quote]

Is he or is she?

[quote="manuel"]that's a girl name in spanish
good luck  :D[/quote]

THanks!

[quote="Dave Vather"][quote="Cristos"]

Peru, 		Geomara Asunción Martínez Perera [/quote]
I heard that Peru´s Guide was Adrián Chi, are you sure, Cristos?[/quote]

Almost, I took it from nacho's list (the last one).

[quote="manuel"]that's a girl name in spanish
good luck  :D[/quote]

right, is a girl.

[quote="Stijn"]Thank you for the list. I did some Googling and got in contact with Belgium's guide  :) .[/quote]How about telling him to stop by here? :)

[quote="Dave Vather"][quote="portoseb"]I already found Finland's guide. She's 17  :D .
Actually I only found her picture, not her e-mail :( , but I guess that's enough for now :D .[/quote]
I think you are wrong and she was 17 in 2003[/quote]

You're right, since the MSN Groups page where I read that information could have been old. In fact it was old.

[quote="Mónica"][b][color=blue]Hi! I'm Canada's guide.[/color][/b][/quote]Welcome on the site :welcomeani: I'm sure Billzhao and the rest of the team will answer your question soon :)

[quote="Mónica"][b][color=blue]Hi! I'm Canada's guide. Can someone tell me when are they arriving Merida?  ;) [/color][/b][/quote]

Monica, I sent you a private message.


Lakeside 6th grade student Darryl Wu just just won 2008 MATHCOUNS National Champion.  Team WA won 2nd place.

Darryl Wu, 11, is the youngest student to ever become Mathcounts National Champion.

Congratulations!

Happyface do you even live in washington???

:mad: If it weren't for this stupid genius kid, (yes, i do see the oxymoron) Washington Middle School would be winning everything, and if it weren't for his dumb lakeside scholarship, he would be at Washington and all of us WMS kids wouldn't have our glory taken away!

Grrrrrrrr!

It is great that Lakeside has Darryl, but you forgot Odle took team first place (and two members in last year's WA MATHCOUNTS National team).  So with or without Darryl, WMS still has to compete against Odle Middle School.

Lakeside provides financial aid based on needs but no merit scholarship, and of course no dumb scholarship.  Had it not been Lakeside, then Darryl would go to Odle (naturally since Stevenson Elementary is Odle's feeder school).

If Odle can do it without Darryl, Washington Middle should also be able to win the top awards on their own.

:( Darryl's not stupid! He's awesome!! He's super awesome!!

apparently i heard some sad news from romeo that the 10th place kid from Washington beat Darryl in countdown...therefore darryl only got 2nd at state :(
But he did get first written though :D

Is Countdown official in Washington state ?

[color=darkblue][i]i [/i]thought that the tenth place guy winning was inspirational. he had to win through 30+ questions to get to first
BTW, that thenth place guy is Albert Chu from the abovementioned Odle Middle
and their team won first place again at state this year
and two people from that school are going to Nats[/color]

I find it amusing that

Darryl Wu wins Nationals, 3rd written, 1st countdown
.
Darryl Wu "fails" states, 1st written, 2nd countdown

I want to cd that albert chu guy

Maybe Albert lucked out because Darryl was being too calm at countdown? I mean, I watched the webcast and I kinda find Darryl taking his time calmly, but sometimes it doesn't necessarily work for uber-fast people like Bobby Shen (if he didn't careless so much) :maybe:

[quote="ghug"]Derryl wasn't there, and for that matter, neither was Kathleen Zhou from Odle.

Anybody know why?[/quote]
what the heck? :huh:  :maybe: Maybe YOU didn't see him or meet him or something? Or are you being sarcastic?

I was at the middle school competition yesterday, he didn't place, and he didn't go up with any of the Lakeside teams that did; unless he just did really, REALLY, badly, he wasn't there.

Well then I guess you were talking about some other middle school math contest. Darryl doesn't do EVERY one of them, and neither does his school, I believe. Not every beastly person or his/her school does every single math contest available in his/her state.

You think Derryl Wu wouldn't show up at Blaine, it's the second biggest contest in the state (behind mathcounts), and lakeside was there.

Well then maybe I guess they didn't let him participate b/c he was too good?
I've heard that it happened sometimes for some beastly people in certain states.

apperantly there was a mu alpha theta thing that day

1. Albert's a ninja, but everyone already knew that.

2. While he's very good at the countdown round, that was true of about half of the people there.  His main advantage came from the fact that he was aware of the strategy required to win (buzzing in a little before the answer pops into his head).  Other people tended not to use this, possibly because they were a little stressed.

3. Yes, Darryl was not at Blaine, because I saw him that day in a place pretty far away from Blaine.

4. Darryl isn't only focused on math competitions...he's busy with all of his skillz, not just one.

5. I have to go tease Kevin for winning 7th grade division now...If I ever meet him again.

I think i know you lastwarrior......... yeah... that one stupid kid in my english class.  haha.  hi.  Darryl Wu was at State Science olympiad in Vancouver, WA.  so yeah. otherside of WA from blaine.......

Blaine? I've done a LOT of math contests, and I've never even heard of it. :stink:

I just wanted to say that Darryl didn't need a scholarship to attend Lakeside--he lives in one of the best communities in Bellevue.

Odle is a math power house, at least when I went to school there. =D

Blaine, ya know, Washington State Math Championships!

And Albert went to Exerter. But his dad still works here in Washington...

RIP Darryl Wu

He did something that no other sixth grader has done, and went on to win ARML 2014. Huge inspiration

What was the cause of Darryl Wu's death?

RIP Darryl Wu. I don't want to sound inhuman or insensitive, but what happened to Darryl?

Let us mourn the death of Darryl Wu, for all passings are sad. He was an exceptional student, and he will be forever remembered for his many accomplishments. He did feats that no other students have done, and he will be missed.

It's really fishy that he died at such a young age.
Does anyone know the cause of his death?
It's sad that many great mathematicians died at a young age.

Well, I believe it was reported that "no foul play was involved".

Could it be some health issues?
Or maybe food poisoning?

It was most likely just a heart attack. Happens a lot. It is really sad that it had to happen to a good person like Darryl Wu.

[quote=jack74][quote=mathwizarddude]Why and how would a teenager get a heart attack?

Maybe Darryl was not getting enough exercise or sleep?[/quote]

Heart attacks are very unfortunate and they can happen to anyone.

[quote=amalbjacob2]If you tire yourself out too much, a heart attack resulting is highly likely.[/quote]

Yeah, I think I heard of that. They really suck.[/quote]

rip jack74 :(

5 years. Washington states is too bad now……….. we got 7th mathcounts


Prove that for all real $a,$ $b$ and $c$ holds
$6\sum_{cyc}a^{2}(a-b)(a-c)+\sum_{cyc}(a-b)^{2}(a^{2}+4ab+b^{2})\geq0.$
The equivalent form:
Prove that $\sum_{cyc}(4a^{4}-2a^{3}b-2a^{3}c-3a^{2}b^{2}+3a^{2}bc)\geq0.$

I use $SOS$.
and the inequality is equivalent to:
$\sum_{cyc}\left(4(a^{2}+b^{2}+ab)-3c^{2}\right)(a-b)^{2}\ge 0$
Wlog:$a \ge b \ge c$
we have:$S_{c}\ge 0$,$S_{b}\ge 0$,$S_{a}+S_{b}\ge 0$

By the way,very nice inequality.And the equality hold when $a=b=c$ or $a=b=-2c$(or change orders):)

[quote="zhaobin"]
Wlog:$a \ge b \ge c$
we have:$S_{c}\ge 0$,$S_{b}\ge 0$,$S_{a}+S_{b}\ge 0$
[/quote]
It is work mere for $a\geq b\geq c\geq0.$ :wink:
For example, why $S_{c}\geq0$ $?$

Sorry,I am wrong.
but we can prove:
$S_{b}\ge 0$,$S_{b}+S_{a}\ge 0$,$S_{b}+S_{c}\ge 0$

Wlog $a \ge b \ge c$ we have $a^{2}\ge b^{2}$ or $c^{2}\ge b^{2}$
So
$S_{b}=4(a^{2}+c^{2}+ac)-3b^{2}\ge 0$

and
$S_{b}+S_{a}=(a+2c)^{2}+(b+2c)^{2}\ge 0$
$S_{b}+S_{c}=(b+2a)^{2}+(c+2a)^{2}\ge 0$
 :P So your inequality still been proven,
What about your solution :)

My proof irrelevant since, your solution better and very nice.
My congratulations, zhaobin! :lol:

[quote="zhaobin"]Sorry,I am wrong.

Wlog $a \ge b \ge c$ we have $a^{2}\ge b^{2}$ or $c^{2}\geg b^{2}$ [/quote]

Sorry for my stupid but I don't understand the statement above.
I think that with a,b,c be real number it cannot happen.
Thanks if anyone explains it to me.


[quote="Soarer"]Is this true?
$ [6,0,0]\plus{}[2,2,2]\ge [5,1,0]\plus{}[3,2,1]$
where the notation represents symmetric sum of three variables, i.e. $ [a,b,c] \equal{}\sum_{\sigma\in S_{3}}x_{\sigma (1)}^{a}x_{\sigma (2)}^{b}x_{\sigma (3)}^{c}$[/quote]
Have you meant to prove it for all real variables? ( If the variables are non-negative numbers then it's obviously true. )

for non-negative numbers. Sorry I should have checked it in more detail.

err...isn't this just obvious...if you switch sign(s) of the variables, the RHS is the same, but the LHS just stays the same or becomes less.

[quote="Altheman"]...if you switch sign(s) of the variables, the RHS is the same, but the LHS just stays the same or becomes less.[/quote]
Certainly!


Solve>

\[ f(x)\equal{}(x\minus{}y)f((x\minus{}y\plus{}1)x)\plus{}f(y)\]

[quote="cckek"]Solve>
\[ f(x) \equal{} (x \minus{} y)f((x \minus{} y \plus{} 1)x) \plus{} f(y)
\]
[/quote]
Let $ x \equal{} 0$ then
$ f(0) \equal{} \minus{} yf(0) \plus{} f(y)$
So $ f(y) \equal{} (y \plus{} 1)f(0)$
$ f(0) \equal{} a$ then
$ f(x) \equal{} a(x \plus{} 1)$
Replace to the function we have :
$ a(x \minus{} y) \equal{} a(x \minus{} y)[(x \minus{} y \plus{} 1)x \plus{} 1]$
So $ a \equal{} 0$
It mean that $ f(x)\equiv 0$


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I'm interrested,too,but since you are not looking for moderators right now...... :(