Okay... this means that we can write that
a=kc
b=kd

Because we know that a and b are coprime we know that k=1

Therefore, $ gcd(ab,a\plus{}b)\equal{}gcd(k^2cd,k(c\plus{}d))\equal{}k$

And k=1, so gcd(ab,a+b)=1

You haven't proven anything; you assumed $ \gcd(cd, c\plus{}d) \equal{} 1$, which is equivalent to the problem statement.

[hide="One approach"] $ (x \minus{} a)(x \minus{} b)$ has distinct roots $ \bmod a$ and $ \bmod b$. [/hide]

Let $ a\equal{}p_1^{e_1}p_2^{e_2}\cdots p_j^{e_j}$ and $ b\equal{}q_1^{f_1}q_2^{f_2}\cdots q_k^{f_k}$.
Then
\[ \gcd \left(ab,a\plus{}b\right)\equal{}\gcd \left(p_1^{e_1}p_2^{e_2}\cdots p_j^{e_j}q_1^{f_1}q_2^{f_2}\cdots q_k^{f_k}, p_1^{e_1}p_2^{e_2}\cdots p_j^{e_j} \plus{} q_1^{f_1}q_2^{f_2}\cdots q_k^{f_k}\right).\]
The first term in the $ \gcd$ has $ j\plus{}k$ prime factors. It is clear that none of them divide into the second term in the $ \gcd$. Q.E.D.

The hint I gave was given for reasons of generality.  Does your argument generalize to $ \gcd(abc, ab \plus{} bc \plus{} ca, a \plus{} b \plus{} c)$ given that $ \gcd(a, b, c) \equal{} 1$?

Alternately, if you want to solve the problem that way what you're really saying is that $ \gcd(a, a \plus{} b) \equal{} \gcd(b, a \plus{} b) \equal{} 1 \implies \gcd(ab, a \plus{} b) \equal{} 1$.  This is a consequence of a lemma that can be proven without an explicit appeal to prime factorization.

Any prime p dividing ab would have to either divide a or b, but not both, but then it wouldn't divide a+b.

[quote="mathwizarddude"]Prove that if $ a$ and $ b$ are coprime then $ ab$ and $ a \plus{} b$ are coprime for integers $ a$ and $ b$.[/quote]
[hide]By the sake of contradiction, assume that $ ab$ and $ a\plus{}b$ are not coprime. Then let d=gcd(ab,a+b)(d>=2, d is an integer).
Then we have: a(a+b) is also divisible by d, which means: a^2 is divisible by d
Similarly, we also get: b(a+b) is divisible by d, which means b^2 is divisible by d
Therefore, d is the common factor of a^2 and b^2. Then a and b must have at least one common factor>=2(since d>=2)(contradiction since (a,b)=1).
So, we conclude that $ ab$ and a+b must be coprime if $ a$ and $ b$ are coprime(Q.E.D) :) [/hide]


geometry
integration
trigonometry
analytic geometry
ratio
calculus
function

ABCD is a square. AB=10. Find the area of  PQRS.(the dot area)
(I am not sure if it's the right place for this question... :? )
[img]http://i5.photobucket.com/albums/y194/jelly119/square.jpg[/img]

Hmm, that circle in the middle does make it kind of difficult... I actually don't know how to do it, I have no idea how to take off the area outside of the circle but inside of the two quarter-circles.

You could transform it into cartesian co-ords and use integration.

yeah, use polar coordinates...its easy. :)

Could you show me the details, please! :huh:

I'm very interested in this problem.Can you please post a solution?

wow this is a really nice problem... i'd really like to see a solution to this one, i have no clue how to go about approaching it  :blush:

[hide="hint"]
in the figure below, the area of $BSPDQR$ is equal to:
$\int_{0}^{\frac{\pi}{4}}(20\sin{\theta})^2=57.079$
for circle $A$, $r=20\sin{\theta}$
for circle $B$, $r=20\cos{\theta}$
find points $P,Q,R,S$ in polar...[/hide]

You can find cartesian coordinates of S, P, Q, R (hmm that's something else too), and it's pretty easy from there (plus calculator), but hopefully there is a prettier way.

I spent a while finding the cartesian coordinates of the center intersections.  If the center of the square and circle is the origin, the points will be $\frac{5 \pm 5\sqrt 7}{4}$  I used that to create triangles from the vertices of the square and the center of the circle, then I could find the area of the arc outside the triangles, then just find the area of the rectangle/quadrilateral inside, but the angles I'm getting for the triangles are irrational, and either this a bad problem, or I'll be forced to just use the calculator (which I had hoped not to)

A better problem that doesn't require a calculator (perhaps easier, however) is a square whose vertices each have a circle with radius equal to the side length; find the area inside all four circles.

If you give this problem to a computer scientist:
[hide]
Randomly pick a large sample of points inside the square (say, a million). The ratio of the number of points inside the region over the total number of points picked is a fairly good approximation of the ratio of the area of the region over the area of the square.
[/hide]

If you give this problem to a mathematician:
[hide]
calculus
[/hide]

No calculus in this forum! Although I can't move this topic to the calculus forum :(

i don't think that moving this is necessary, because calculus is only one of the ways that this can be solved....so this problem is not neccessarly a calculus problem. :|

but this is just way too messy without calculus(I'm not sure if calculus will make it any easier though)

[quote="seamusoboyle"]Its way messy with calculus!
Maybe mathworld might be able to help?[/quote]
i don't think so...
all you need to do is find the $P,Q,R,S$ in polar,(or cartesian) cooridantes,and then you need to get find the value for problabley four intergrals....that's all.. ;)

It boils down to find two angles (or trig function values of these two angles): $\angle{PBS}$ and $\angle{POQ}$ where O is the center of the square.

These angles are independent of scaling and rotation. To find them, you can put it into two complex planes with different origins.

For example, if we set B as origin, BD as positive x-axis, then $\cos{\angle{PBD}}=\frac{5\sqrt{2}}{8}$ (this actually can be found by law of cosines).

It's all about a triangle POB of sides, 5, 10, $5\sqrt{2}$, you can find the two angles I mentioned above from this triangle.

[quote="seamusoboyle"]How do you use integration with polar co-ords?[/quote]
$\frac{1}{2}\int_{\theta_1}^{\theta_2}r^2 d\theta$

Let a=5, so AB=2a

Let O be the center of the square.

Let X be the intersection of the BD with the arc PS and let Y  be the intersection of the AC with the arc PQ.

Let $\angle{PBO}=\alpha$ and $\angle{POA}=\beta$.

Consider the triangle POB, we have $PB=2a, PO=a, BO=\sqrt{2}a$, by law of cosines, we have

$\cos{\alpha} = \frac{5\sqrt{2}}{8}$ and

$\sin{\beta} = -\cos(\beta+90^\circ) = -\cos{\angle{POB}} = \frac{\sqrt{2}}{4}$

area(PBO) = $\frac{\sqrt{7}a^2}{4}$

area(PBX) = $2\alpha a^2$

area (POY) = $\frac{\beta a^2}{2}$

Putting all together,

area(PQRS) = 4 x area(PXOY) = 4 x {area(PBX) - area(PBO) + are(POY)}

= $4\times (2\alpha - \frac{\sqrt{7}}{4} + \frac{\beta}{2})a^2$

It sounds interesting that someone does not consider integration as included in Calculus?
Maybe someone can avoid using integration to find the area of that region, but I think "taking limit" is unavoidable anyway.


binomial coefficients
algebra
binomial theorem
number theory proposed
number theory

[i]If $A$ is the sum of the digits of $1001^{1002}$, and $B$ is the sum of the digits of $A$, find $B$.[/i]

Masoud Zargar

Hint. Find good bounds, and show its not 1,10.

It is not correct, because $1001^{1002}$ have 3007 digits. Therefore $A<3007*9=27063 \Longrightarrow B\le8+27=35$.
$B=A(mod 9)=1001^{1002} (m0d 9)=1(mod 9).$ Therefore B=1 or 10 or 19 or 28.


Let us say the are 6 "unit shapes" in the diagram, where a "unit shape" is a polygon shown that includes only one polygon inside it, using the lines drawn.

Here is my casework by number of unit shapes:

1:6
2:7
3:2
4:2
5:0
6:1

we count 18 total concave polygons

Yes sorry that should say convex  :blush:

I don't understand can you please explain with more detail?

Thanks!

Basically, @hurdler defined a "unit shape" to be a convex polygon in the figure shown that has only a single polygon inside it. There are $6$ such unit shapes. Next, we count the number of convex polygons that are made up of $2$ unit shapes. There are $7$. Continuing, we count how many convex polygons are made up of $3$ unit shapes, then $4$, $5$, and finally $6$ unit shapes. We finish here because there are a total of $6$ unit shapes. Adding together, we get a total of $18$ convex polygons.


Find all functions $f: \mathbb{N}_{0}\rightarrow \mathbb{N}_{0}$ such that for all $n\in \mathbb{N}_{0}$: \[f(f(n))+f(n)=2n+6.\]

[quote="lyxuansang"]The general problem of K9 is

Find all fucntions $ f:\mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$ such that for all $ n\in\mathbb{N}_{0}$ :

$ f(f(n))\plus{}f(n)\equal{}2n\plus{}3k$
and it on here http://www.mathlinks.ro/Forum/viewtopic.php?t=161084[/quote]

[hide="Bashy Solution Sketch"]Let $f(0) = a$. We have that $0\le a\le 6$ by the definition of the function. By checking each value of $a$, we find that each value of $a$ leads to a contradiction except $a=2$. Therefore $f(0) = 2$. We will now prove that for $x=2n$, $f(x) = x+2$ for all nonnegative integers $n$. The claim is true for $n=0$. Assume the claim is true for $n=y$. This means that $f(f(2y))+f(2y) = 4y+6$. The LHS expands to $f(2y+2) + 2y+2 = 4y+6$, or $f(2y+2) = 2y+4$. Therefore, if the claim is true for $n=y$, it is also true for $n=y+1$, and by induction we have shown the claim is true for all $n$. Now let $f(1) = b$. We have that $0\le b\le 8$ by the definition of the function. By checking each value of $a$ and using the fact we just proved, we can show that $a=3$. Finally, we use a very similar inductive argument to show that $f(2n+1) = 2n+3$ for all nonnegative integers $n$. Therefore we have that $f(n) = n+2$ as desired. [/hide]

This should work:
Plug in $n=0$

$$f(f(0))+f(0)=6$$

We have the cases $f(f(0))=0, f(0)=6 .... f(f(0))=6, f(0)=0$ with a total of 7 cases.
I will state the contradictions for each case except case 5, which is indeed the valid answer:

Case 1: $f(f(0))=0 f(0)=6$ which translates to $f(6)=0$
Plug $n=6$ into the given conditions yields $f(6)=12$ CONTRADICTION

Case 2: $f(f(0))=1, f(0)=5$ which translates to $f(5)=1$
Plug $n=5$ gives $f(1)=15$, plug $n=15$ then gives $f(15)=-7$ CONTRADICTION (range is non-negative)

Case 3: $f(f(0))=2, f(0)=4$ which translates to $f(4)=2$
Follow the same steps as Case 2, we get contradiction for f(12) being negative.

Case 4: $f(f(0))=3, f(0)=3$ which gives $f(3)=3$
Plugging in $n=3$ gives $f(3)=6$ CONTRADICTION as a function has no more than one image per input.

Case 5: no contradiction can be obtained

Case 6: $f(f(0))=1, f(0)=5$ translates to $f(5)=1$
Plugging in n=1, n=5, and n=3 we get a contradiction for $f(13)$ being negative.

Case 7: $f(f(0))=6, f(0)=0$ translates to $f(0)=6$ contradiction (same reason as Case 4)

Therefore, the only valid case is case 5, which gives us:
$f(0)=2, f(2)=4,$ and $ f(4)=6$
Inducting gives $\boxed{f(n)=n+2}$


Let $ n$ be an odd number and $ \omega\equal{}\cos\frac{2\pi}{n}\plus{}i\sin\frac{2\pi}{n}$ .

Prove that: $ a^n\plus{}b^n\equal{}(a\plus{}b)(a\plus{}\omega b)(a\plus{}\omega^2b)...(a\plus{}\omega^{n\minus{}1}b),\ \forall\ a,b\in\mathbb{C}$ .

It suffices to show that $ ({\frac{a}{b}})^n+1=(\frac{a}{b}+1))(\frac{a}{b}+\omega)\cdots(\frac{a}{b}+\omega^{n-1})$, or $ x^n+1=(x+1)(x+\omega)\cdots(x+\omega^{n-1})$.  Substituting -x for x, ${ 1-x^n=(1-x)(\omega-x})\cdots(\omega^{n-1}-x)$, and taking the negative of both sides, $ (x-1)(x-\omega)\cdots(x-\omega^{n-1})=x^n-1$.  But this is clearly true because the powers of $ \omega$ are the nth roots of unity.


number theory
greatest common divisor
superior algebra
superior algebra solved

In a ring A we have xn=x for all x\in A where n=2a+2b with a, b\in N*. We also know that n-1 isn't divisible with 2k-1 for any natural k>=2. Show that x2=x for all x\in A.

What a beautiful problem!

2^k - 1 doesn't divide 2^a + 2^b - 1 for all k ==> 2^k - 1 doesn't divide 2^(a+1) + 2^(b+1) - 2. If gcd(a+1,b+1)=d>1 then 2^d-1 | 2^(a+1)-1 , 2^d-1 | 2^(b+1)-1, so 2^d - 1 | 2^a + 2^b - 1, a contradiction.

Therefore gcd(a+1,b+1)=1. Let M = {k in N| x^(2^k) = x , for all x in A}. We prove that a+1 , b+1 are in M and that u,v in M implies gcd(u,v) in M. This will settle our matter.

1+1=0 in A, so (x+1)^(2^a+2^b) = x^(2^a+2^b) + x^(2^a) + x^(2^b) + 1 = x + 1 ==> x^(2^a) + x^(2^b) = 0. Now, notice that y^2=0 in A implies y=0. Also [x^(2^(a-1)) - x^(2^(b-1))]^2 = 0 so x^(2^(a-1)) = x^(2^(b-1)), and so on. Finally, we get that if for example a>b x^(2^(a-b)) = x in A. Also let us notice that a is not equal to b or otherwise gcd(a+1,b+1)=a+1>=2>1. Ok.

x^(2^a + 2^b) = x ===> x^[2^b(2^(a-b)+1] = x ===> [x^(2^(a-b)+1]^(2^b) = x ==> x^(2^(b+1))= x. Since x^(2^a) = x^(2^b), we get that x^(2^(a+1)) = x , so a+1 , b+1 are in M.

Now for u, v in M x^(2^u) = x^(2^v) and as above we obtain x^(2^|u-v|) = x so |u-v| is in M.  Now since we know that if a>b gcd(a,b) = gcd(a-b,b) and we also notice (a-b) + (b) < a+b we can obtain the gcd by repeated substractions. Hence we get that 1=gcd(a+1,b+1)  is in M, qed.

I did almost the same thing, except for the proof that a+1 and b+1 are in M:

from x2a=x2b=>
x=x2a+2b=(x2a)2=x2a+1, so a+1\in M. In the same manner b+1\in M. It's no big difference.

I agree, the problem is simply beautiful :).


Hey guys I just hope people here could help me. I've done all the problems I need to do, but this really counts for a lot and I need every correct answer I can get. If you find any faults, please tell me. I really appreciate it guys.

I wrote out all the questions, and at the bottom of each I wrote my answers. Please guide me guys. I need to ace this.

1- Bacteria divides into 4 new cells every 20 minutes. If the bacteria reproduces for x hours, find the
exponential equation that represents the increase in population as a function of time.

I found y= 12^x

2- Simplify:
A- (2/3)^2X+3 = (27/8)^3X-2
I found x=0,27

B- 3^x+3= 13574
I found x= 5.6617

3- Find the inverse of the following function: f (x)= log base5(3-x)
I found y= -5^x + 3

4- For the following function f(x)= 3 (1/2)^4x-1 + 5, State whether the statements are true or false. Correct the false ones.

A. It is decreasing
B. The ordinate is 8
C. The equation of the asymptote is y=5
D. It is always positive
E. It has no abcissa at the origin

I wrote true for A, C, D and E, and I wrote the ordinate is 11 for B

5- Find the equation of the following log function with an asymptote of x= -5.625 and passing through the point (10,3)
I found y= log2.5 (x+5.625)

6- Martha brought a bond for 1500.00$ at an annual rate of 4.8%, compounded semi annually. Find the rule that
describes the value as a function of the number of years. Using the rule, find the value in 10 years.
I found the rule was y= 1500. 1.024^2x, so in 10 years
y= 1500 . 1.024^20= 1901,47
$

7- The following rule is used for calculating the number of years (y) it takes for a bond to reach a certain amount (x)
y= 96.64logx - 460.34
Calculate how much time it would take to reach 65000 and 85000.

I put in 65000 and 85000 for x, and got 4.78 and 16.04.

8. Find the value of the following expression:
8(log4 32 + log2 8) - (log2 16)^2

I found 28

Thanks in advance guys!

1) Say you start off with 1 bacterium.  Then after 20 minutes there will be 4.  After 20 more minutes, those 4 will divide into 4 more for a total of 16 bacteria.  After another 20 minutes, those 16 split into 4, for a total of 64 bacteria.  Just to make a point, I'll do the next hour as well.  After 80 minutes, those 64 split into 4 for a total of 256.  Then those split into 1024, and then 4096 after 2 hours.  Hence if $ x$ is hours,
$ b(x)\equal{}64^{x}$

[quote="JRav"]1) Say you start off with 1 bacterium.  Then after 20 minutes there will be 4.  After 20 more minutes, those 4 will divide into 4 more for a total of 16 bacteria.  After another 20 minutes, those 16 split into 4, for a total of 64 bacteria.  Just to make a point, I'll do the next hour as well.  After 80 minutes, those 64 split into 4 for a total of 256.  Then those split into 1024, and then 4096 after 2 hours.  Hence if $ x$ is hours,
$ b(x) \equal{} 64^{x}$[/quote]

Oh true. What about 4^3x? would that work as well?

And about my other answers are they all correct?

That's the same thing so either is fine.  For the second one, neither 27 nor 0 are right.  Just plug them back in and see.

$ \left(\frac{2}{3}\right)^{2x\plus{}3}\equal{}\left(\frac{27}{8}\right)^{3x\minus{}2}\equal{}\left(\frac{2}{3}\right)^{\minus{}3(3x\minus{}2)}\equal{}\left(\frac{2}{3}\right)^{\minus{}9x\plus{}6}$
$ \Rightarrow 2x\plus{}3\equal{}\minus{}9x\plus{}6\Rightarrow 11x\equal{}3\Rightarrow x\equal{}\frac{3}{11}$

[quote="JRav"]That's the same thing so either is fine.  For the second one, neither 27 nor 0 are right.  Just plug them back in and see.

$ \left(\frac {2}{3}\right)^{2x \plus{} 3} \equal{} \left(\frac {27}{8}\right)^{3x \minus{} 2} \equal{} \left(\frac {2}{3}\right)^{ \minus{} 3(3x \minus{} 2)} \equal{} \left(\frac {2}{3}\right)^{ \minus{} 9x \plus{} 6}$
$ \Rightarrow 2x \plus{} 3 \equal{} \minus{} 9x \plus{} 6\Rightarrow 11x \equal{} 3\Rightarrow x \equal{} \frac {3}{11}$[/quote]

Oh no I wrote 0.27, the decimal.

Then that's correct.

For the rest...

[hide="2B"]
$ 3^{x \plus{} 3} \equal{} 13574\implies\boxed{x \equal{} \minus{} 3 \plus{} \log_3{13574}\approx5.661554709\ldots}$.
[/hide]
[hide="3"]
$ f(x) \equal{} \log_5(3 \minus{} x)\implies f(f^{ \minus{} 1}(x)) \equal{} x \equal{} \log_5(3 \minus{} f^{ \minus{} 1}(x))\implies5^x \equal{} 3 \minus{} f^{ \minus{} 1}(x)$

$ \implies\boxed{f^{ \minus{} 1}(x) \equal{} 3 \minus{} 5^x}$.
[/hide]
[hide="4"]
A: True, because it decreases as $ x$ increases.
B: False; I assume the ordinate is the "$ y$-intercept," and $ f(0) \equal{} \boxed{11}\ne8$.
C: True. Since $ \left(\frac12\right)^{4x \minus{} 1}$ never equals $ 0$ but tends to it, $ f(x)$ has asymptote $ y \equal{} 5$.
D: True. $ \left(\frac12\right)^{4x \minus{} 1}$ is never negative for real $ x$.
E: I don't exactly understand it, but I'll assume it's true because the origin is not part of the graph.
[/hide]
[hide="5"]
Yes, that's one possible answer. Another is $ \boxed{y \equal{} \log_5(8x \plus{} 45)}$.
[/hide]
[hide="6"]
It is compounded twice every year, so the function is $ \boxed{f(x)\equal{}\$1500(1.024)^{2x}\equal{}\$1500(1.048576)^x\equal{}\$1500\left(\frac{16384}{15625}\right)^x}$, so $ f(10)\equal{}\$2410.407\sim \equal{}\boxed{\$2410.41}$ (just a computational error on your part).
[/hide]
[hide="7"]
Yeah, just plug them in; your numbers are correct.
[/hide]
[hide="8"]
$ 8(\log_4{32} \plus{} \log_28) \minus{} (\log_2{16})^2 \equal{} 8(\log_{2^2}{2^5} \plus{} 3) \minus{} (4)^2 \equal{} 8\left(\frac52 \plus{} 3\right) \minus{} 16$

$ \equal{} 8\cdot\frac {11}2 \minus{} 16 \equal{} 44 \minus{} 16 \equal{} \boxed{28}$.
[/hide]

Edit: Solution to $ 6$ added. Thanks, AIME15.

Math154-quote the post to find what number 6 wants.

6- Martha brought a bond for 1500.00 at an annual rate of 4.8%, compounded semi annually. Find the rule that describes the value as a function of the number of years. Using the rule, find the value in 10 years. I found the rule was y = 1500. 1.024^2x, so in 10 years y = 1500 . 1.024^20 = 1901,47

[quote="helpmeplz"]Hey guys I just hope people here could help me. I've done all the problems I need to do, but this really counts for a lot and I need every correct answer I can get. If you find any faults, please tell me. I really appreciate it guys.

I wrote out all the questions, and at the bottom of each I wrote my answers. Please guide me guys. I need to ace this. [/quote]
Although it seems you got a majority of these correct, you should still be more careful in what you're doing.  Several of your mistakes are careless and could have been easily avoided.


function
Euler
inequalities
algebra unsolved
algebra

Find all funtion $ R \to R$ such that $ f^2 (x) \equal{} f(x^2 )$.

Although I think it very easy but I can't find solution! Help me!  :(  :(

Thank you verry much!

$ f$ - continuous?
If not, we cannot solve it, because a lot of "bad" functions is sutisfies this equations, and they are not continuous.

[quote="Kirill"]$ f$ - continuous?[/quote]

Yes, that funtion continuous in R. :)  :)

Constants $ f(x) \equal{} 0$ and $ f(x) \equal{} 1$ are not very interesting.

First of all, we want to find $ f(x)$ for $ x \geq 1$.
Let $ f$ - non constant on $ [1,\infty)$ and $ f(k) \neq 0;1$ for some $ k > 1$. It's easy to prove, that $ f(k) > 0$, so $ f(k) \equal{} k^r$ for some real r.
Let $ f(x) \equal{} g(x) \plus{} x^r$. Your equality became as:
$ g(x)(g(x) \plus{} 2x^r) \equal{} g(x^2)$ and $ g(k) \equal{} 0$.

So, we have that $ g(\sqrt {k}) \equal{} g(k) \equal{} 0$ (it's impossible that $ g(\sqrt {k}) \equal{} \minus{} 2k^{r/2}$ - because $ f(\sqrt {k}) > 0$).
So, $ g(^{2^p}\sqrt {k}) \equal{} g(k) \equal{} 0$ - for all $ p$ - nonegative integer. From continuosly we have that $ g(1) \equal{} 0$.

Suppose exist $ l > 1$ such that $ g(l) \neq 0$.
Then we have that $ g(^{2^p}\sqrt {l}) \neq 0$, but $ g(1) \equal{} 0$. Contradiction.

So, $ g(x) \equal{} 0$ for all $ x \geq 1$ and $ f(x) \equal{} x^r$ for $ x \geq 1$.

Use this idea to prove that $ f(x) \equal{} x^m$ for $ 0 \leq x \leq 1$ and construct the answer. Remember, that $ r \neq m$.

O, maybe my solutions falls somewhere...

can you please explain your solution a bit more clearly?

Sorry, but my solution fails in this lines:
[quote="Kirill"]
Suppose exist $ l > 1$ such that $ g(l) \neq 0$.
Then we have that $ g(^{2^p}\sqrt {l}) \neq 0$, but $ g(1) \equal{} 0$. Contradiction.
[/quote]
There is no contradiction here.

Generally: Find $ f: \set{R} \rightarrow \set{R}$ such that $ f$ is continous and $ f(x).f(y)=f(x.y)  \forall  x,y \in \set{R}$.

[quote="duy_nguyen"]Generally: Find $ f: \set{R} \rightarrow \set{R}$ such that $ f$ is continous and $ f(x).f(y) = f(x.y) \forall x,y \in \set{R}$.[/quote]
Yes it is old,the result is $ f(x)=x^a,a\in R$
and more if exist $ c=const$ satisfying $ |F(x)|\le c$ (not condition $ f$ continous the result is similar

[quote="duy_nguyen"]Generally: Find $ f: \set{R} \rightarrow \set{R}$ such that $ f$ is continous and $ f(x).f(y) = f(x.y) \forall x,y \in \set{R}$.[/quote]
I don't think it is generally,these problems are complete different and your problem is very old and esay,but Euler 's problem is harder! :wink:

[quote="L_Euler"]Find all funtion $ R \to R$ such that $ f^2 (x) \equal{} f(x^2)$[/quote]
$ f$ has independent solutions on $ [0,1]$ and $ [1,\infty)$ if we ignore the continuous patch at $ x\equal{}1$. Here is a solution for the interval [0,1]. The techniques used in the proof should be analogous for $ [1,\infty)$. But I haven't checked that in detail. I believe that an appropriate choice of the values $ f(\minus{}0.5)$ and $ f(\minus{}2)$, e.g., should determine $ f$ on $ (\minus{}\infty,0)$.

If $ f$ is not constant on [0,1], then there is some $ \alpha>0$ such that $ f(x)\equal{}x^\alpha$ for all $ x\in[0,1]$.
[hide="Proof"]Set $ I\equal{}[0,1]$. Certainly $ f|I\ge 0$. It holds $ f(I)\subset I$: Otherwise there is some $ x\in I$ with $ f(x)>1$. Then $ x$ is neither 0 nor 1 and $ x^{2^n}\to 0$ and $ f(x)^{2^n}\to\infty$ for $ n\to\infty$, but the sequence $ f(x^{2^n})\equal{}f(x)^{2^n}$ is convergent, by continuity of $ f$. Contradiction.

Since $ f(I)$ is a non-trivial sub-interval of $ I$ (f is non-constant and continuous), there is some $ x_0\in (0,1)$ with $ 0<f(x_0)<1$.

If $ f(x_0)<x_0^\alpha$ for $ \alpha>0$, then $ f(x)<x^\alpha$ for all $ x\in (0,1)$. The assertion with the inequality sign reversed is true as well and can be shown in a similar way. [hide="Proof"]By continuity there is some $ \varepsilon>0$ such that the inequality extends to $ U\equal{}(x_0\plus{}\varepsilon,x_0\plus{}\varepsilon)\subset I$, i.e. $ f(x)<x^\alpha$ for all $ x\in U$. Then there is $ n_0\in\mathbb N$ such that $ U^{2^n}$ and $ U^{2^{n\plus{}1}}$ intersect ($ n\ge n_0$). This gives $ x_1\in (0,1)$ such that the interval $ (0,x_1)$ is contained in the union of the $ U^n$. The inequality clearly extends to $ (0,x_1)$. Now playing the same trick by taking the $ 2^n$-th roots extends the inequality from $ (0,x_1)$ to $ (0,1)$.[/hide]
Now fix $ \alpha>0$ with $ f(x_0)\equal{}x_0^\alpha$. For every $ \varepsilon>0$ we have $ x_0^{\alpha\plus{}\varepsilon}<f(x_0)<x_0^{\alpha\minus{}\varepsilon}$ which implies $ x^{\alpha\plus{}\varepsilon}\le f(x)\le x^{\alpha\minus{}\varepsilon}$ for all $ x\in I$. Dropping $ \varepsilon\downarrow 0$ we get the result.[/hide]


topology
advanced fields
advanced fields unsolved

show that if a space $X$ deformation retract to a point $x \in X$ . then for each neighborhood $U$  of $x$ in $X$ there exists a neighborhood $V \subseteq U$  of $x$ such that the inclusion map $V \hookrightarrow U$ ia nullhomotopic. :blush:

Let $r: X\times I\to X$ be a deformation retract, so that for $y\in X$, $r(y,0)=y$, $r(y,1)=x$, and for $t\in I$, $r(x,t)=x$.

Then $r^{-1}(U)$ is an open neighborhood of $\{x\}\times I$.  Use compactness of $I$ to show that $r^{-1}(U) \supset V\times I$ for some neighboorhood $V\subset U$ containing $x$.  Now show that the restriction $r: V\times I\to U$ gives a homotopy from $i: V\hookrightarrow U$ to the constant map at $x$.


group theory
abstract algebra
superior algebra
superior algebra solved

What are the conditions on $G$ such that there exists $n$ such that $G$ is isomorphic  to a subgroup of $A_n$ (the alternate group of $n$ elements) ?

The only condition is that $G$ is finite.
Every group with $n$ elements can be embedded in $S_n$ and $S_n$ can be embedded into $A_{n+2}$.

same as for being subgroup of Sn
that the group is finite

you can see this by the fact that Sn is isomorphic to a subgroup of A(n+2)
just map every even permutation in Sn to itself in A(n+2)
and map every odd permutation in Sn to itself*the transposition (n+1   n+2)

nice result !

How to prove $G$ finite is isomorphic to a subgroup of $S_n$ ?
Group action ?

woops, it seems i almost simultaneously answered with someone else

anyway : consider a group G   with n elements
then with every element you can associate a permutation on the elements of the group G itself, induced by left multiplication with every element
so if that element is a, you get a multiplication sending e to a, a^-1   to a, a to a^2,......

you can check that this map of elements to G , to permutations on G, is a morphism, that is injective
so, our group G is indead isomorphic to the subgroup of Sn

(this is however a very theoretical result, you really don't wanna handle the group S4 as a subgroup of S24, because that is way too big, you wanna find 'nicer' representations than that)


Here's a (well-known) quickie that many of you will solve in less than 15 minutes ...  :)

Let $f(n)$ denote the length of a maximal list $L$ of $n$-term sequences of zeros and ones such that any two different sequences in $L$ differ in at least three positions. Show that $f(n) \geq \frac{2^{n-1}}{n}$.

--Vesselin

Indeed.
For any sequence $S=(a_1,\cdots,a_n)$ consider $|S|=\sum_{i=1}^n i a_i$ $(mod$ $2n)$.
It's easy to see that if two sequences $A,B$ have at most two distinct entries, then $|A|\neq |B|$.
Hence the class $C_i$ of sequences $S$ with $|S|=i$ is a good list.
But one of this lists will have at leat $\frac {2^n}{2n}=\frac {2^{n-1}}n$ elements.

Am I wrong or $f(n) \leq \frac {2^n} n$ ?
I didn't think about it, but is there a known equivalent of $f(n)$? (I suppose that the closed form of $f(n)$ is not known, no?)

Pierre.

I think $f(n)\leq \frac{2^n}{n+1}$.
For a binary sequence $b$ consider $S_b$ the set of all binary sequences $b'$ with $|b-b'|\leq 1$. Then $|S_b|=n+1$.
Also if $L$ is a good list then for any $X,Y \in L$ we get $S_X$, $S_Y$ are disjoint.
Hence $2^n\geq |\bigcup_{X \in L} S_X|=|L|(n+1)$ and so $|L| \leq \frac {2^n}{n+1}$

That was exactly my idea, but I forgot the initial string  ;)  to get only $n$ instead of $n+1$.

Pierre.


have a look at this rareeeeee picture :D

i believe this was taken when Valentin was changing the format  :D
N even i was searching fer it n the slow server even more bugged me  :rotfl:   :D

Dei did you put that in MS Paint and edit it????Knowing mythili ranganathan she might have even did it. :P:D

now surely not :D nor did mythili do it :rotfl:
rohit was right :lol:


geometry
geometric transformation
homothety
ratio
circumcircle
trapezoid
projective geometry

(a) let $ D: AO_1 \cap C_1$ and $ C: BO_2 \cap C_2$. Cause T is the internal center of similitude of $ C_1$ and $ C_2$ we have $ BC \parallel DA$, so AB pass trought the external center of similitude of $ C_1$ and $ C_2$ that we call Z.
(b) call M the midpoint of AB. With an homothety with center Z and ratio $ \frac{r_1\plus{}r_2}{2r_1}$, A go in M, and so the locus of point M when A moves on $ C_1$ is a circle with diameter $ O_1O_2$.

My opinion:
a) Let
$C_1 \left( {K,R} \right)\;and\;C_2 \left( {L,r} \right)$
the circles. Let TM the public tangent of the circles,
${\rm M} \in {\rm A}{\rm B} \Rightarrow KA\parallel LB\;\left( {\mathop {KTA}\limits^ \wedge   = \mathop {MTB}\limits^ \wedge   = \mathop {TEB}\limits^ \wedge  ,and\,\vartriangle KTA,\vartriangle LBE\;isosceles\;triangles} \right).$
The point
$S \equiv KL \cap AB$
is constant because is the sender of omothesia of the circles
$C_1 \left( {K,R} \right)\;and\;C_2 \left( {L,r} \right).$
$H \equiv AB \cap C_2 ,TE = 2r.$
b) Te KABL is trapezium. Let Q the mint point of the KL, R the mint point of the KL, when $QR = \frac{{R + r}}{2}...$

S.E.Louridas

nice and easy Problem.
[hide="My solution"]
(a). Let $O_1$ and $O_2$ be centers of $C_1$ and $C_2$. It's obvious that $O_1$ , $O_2$ and $T$ are collinear, an easy angle chasing gives that $O_1A \parallel O_2B$, let $AB$ meet $O_1O_2$ at $S$, then from Thales theorem, $\frac{SO_1}{SO_2}=\frac{O_1A}{O_2B}=\frac{TO_1}{TO_2}$ hence $S$ is the harmonic conjugate of $T$ with respect to $O_1,O_2$.
(b). Obvious $MA=MB=MT$, then $O_1A^2+MT^2=O_1T^2+AM^2$ then $MO_1 \perp AT$ similary $MO_2 \perp BT$, thus $\angle{O_1MO_2}=90$ therefor, the locus of $M$ is the circle with diametre $O_1O_2$.
[/hide]

Another solution:

Let the common external tangent of the two circles be $CD, CD$ cuts $O_{1}O_{2}$ at $K; O_{1}O_{2}$ cuts the two circles at another points $G$ and $E$. $AB$ cuts the two circles at another points $N$ and $M$.
Since $\angle CO_{1}G = DO_{2}T$ and both $\Delta CO_{1}G$ and $DO_{2}T$ are isoscales, $\angle KCG = 1/2.\angle CO_{1}G = 1/2.\angle DO_{2}T = \angle DEG$, hence $CGED$ is cyclic.
As $AT\perp TB, AG\parallel TB$, thus easy to see that $AO_{1}\parallel BO_{2}$ and $AGEM$ is cyclic.
$\angle KCA = 1/2.\angle CO_{1}A = 1/2.\angle DO_{2}B = \angle DMA$, thus $CAMD$ is cyclic.
As $CGED$ is cyclic, $KC. KD = KG. KE$, hence $K$ is on the radical axis of the circumcircle of $CAMD$ and the circumcircle of $AGEM$, so $K$ must be on their common chord $AM$.
Answer to the first question: All such lines $AB$ pass through $K$.
Let $P$ be the midpoint of $AB$. Since $AGTB$ is a trapezium, $PO_{1}\parallel AG$, similarly, $PO_{2}\parallel AT$, hence $\angle O_{1}PO_{2} = 90^{\circ}$.
Answer to the second question: The midpoint $P$ of $AB$ lies on the circle with diameter $O_{1}O_{2}$.

Another way is with the all-migthy coordinates.

Part a.

Let $O_1$ and $O_2$  be the centers of the circles, then we can say that w.l.o.g. $T=(0,0)$, and $O_1=(-y,0)$ and $O_2=(x,0)$, where $y$ is a positive real number.
Then we have that:
$$c_1:(a+y)^2+b^2=y^2$$
$$c_2:(a-x)^2+b^2=x^2$$

Now let $A=(m,n) \in c_1$ and let $l_1,l_2$ be the lines $AT$ and $BT$, respectively.
We easily have that:
$$l_1:y=\frac{n}{m}x$$
$$l_2:y=-\frac{m}{n}x$$
this we have since $A \in l_1$ and $l_1 \perp l_2$.

Now let $B=(p,q) \in c_2$, since we have that $B \in l_2$ we must have that $q=-\frac{m}{n}p$, and we have that:
$$(p-x)^2+q^2=x^2$$
this easily implies that:
$$\left(1+\frac{m^2}{n^2}\right)p^2-2px=0$$
since $p \neq 0 $, we get that:
$$p=\frac{2xn^2}{m^2+n^2}$$

Now we need to get the line $AB$.
We easily get by calculation that:
$$\overline{AB}=l_3:y=-\frac{2xmn+m^2n+n^3}{2xn^2-m^3-mn^2}x+\frac{2xn(n^2+m^2)}{2xn^2-m^3-mn^2}$$
now let's look at the intersection with the $x$ axis.
We have that the point of intersection has the coordinates:
$$x'=\frac{2x(n^2+m^2)}{2xm+m^2+n^2}=-2\frac{xy}{x-y}$$
we used that $m^2+n^2=-2my$.
Thus we see that the point of intersection with the $x$ axis has now dependency on the points $A$ and $B$, only on th radii of the circles. Thus we have that the lines of type $AB$ intersection all in one point.

Part b.

We have that $M$ is the circumcenter of $ATB$. and we easily have that $O_1M$ is the perpendicular bisector of $AT$ and $O_2M$ is the perpendicular bisector of $BT$.
Thus we have that $\angle O_1MO_2 = 90$, thus we have that $M$ moves along a circle with diameter of $O_1O_2$.

[quote=brianchung11]Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$.

(a) Show that all such lines $ AB$ are concurrent.
(b) Find the locus of midpoints of all such segments $ AB$.[/quote]

Let $O_1 $ and $O_2$ be the centers of $\mathcal{C}_1$ and $\mathcal{C}_2$, respectively. Denote by $A'$ and $B'$ the antipodes of $A$ and $B$ in $\mathcal{C}_1$ and $\mathcal{C}_2$, respectively. Let $M$ be the midpoint of segment $AB$, and let $N$ be the midpoint of segment $A'B'$. Also, let $X=AB\cap A'B'$.
Due to the angle condition, it holds $A-T-B'$ and $B-T-A'$ collinear.
By simple angle chasing using the common tangent, it holds $\measuredangle BB'T=\measuredangle A'AT$, so $AA'\parallel BB'$.
Also, $O_1-T-O_2-X$ are concurrent, because
$$
-1=(A,A';BB'\cap AA',XT\cap AA')=(A,A';\infty_{AA'},XT\cap AA') \Longrightarrow O_1-T-X \thickspace \text{collinear.}
$$
a) All lines $AB$ concur in $AB\cap A'B'$, which is a fixed point, because
$$
-1=(A,A';\infty_{AA'},O_1)\stackrel{B}=(X,T;O_2,O_1)
$$
b) The locus is the circle with diameter $\overline{O_1O_2}$, except for $O_1$ and $O_2$, because
$$
\measuredangle ATB=90^o \Longrightarrow AT\perp BT \Longrightarrow AB'\perp BA' \Longrightarrow MO_2\perp MO_1.
$$

[quote=brianchung11]Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$.

(a) Show that all such lines $ AB$ are concurrent.
(b) Find the locus of midpoints of all such segments $ AB$.[/quote]

Took less than 7 mins but its so fun when u realice what is the problem asking for.
[b](a)[/b] Let $O_1,O_2$ the centers of $C_1,C_2$ respectivily, then clearly $O_1,T,O_2$ are colinear and by angle chase u have $\angle TO_1A+\angle TO_2B=180$ which means that $O_1A \parallel O_2B$, now let $O_1T \cap C_1=C$ and $O_2T \cap C_2=D$ then $\angle CAT=\angle ATB=\angle TBD=90$ so $CA \parallel TB$ and $BD \parallel TA$ meaning that $\triangle CAT \sim \triangle DBT$ and since $C,D,T$ are fixed we have that all lines $AB$ pass through the center of those similarities, which is the exsimilicenter of $C_1,C_2$.
[b](b)[/b] Since $M$ is the center of $(TAB)$ we have that $O_1M \perp AT$ and $O_2M \perp BT$ so $\angle O_1MO_2=90$ becuase of the rectangle formed by the lines $O_1M,O_2M,AT,BT$ and this tells that the locus of all the points $M$ is the circle with diameter $O_1O_2$ thus we are done :D


Two different circles and a triangle lie in the same plane.  What is the greatest number of points that can belong to at least two of the three figures?

[hide="ANSWER"]$ 14$[/hide]

*Is there a better/more reliable way than just drawing random figures?

**Each time I do this type of question, I get a different answer (usually wrong).  Any suggestions?  Thanks.

[quote="unimpossible"]well the triangle can intersect both circles, so 12 intersection poitns there.

the 2 circles can intersect eachother twice, so 12+2=14[/quote]

I understand the second part about the two circles intersecting.  However, for the first part, why is it $ 12$?

if you think about it, it is possible for each side of the triangle to intersect the circle at two points.

If all 3 sides of the triangle intersect one circle, there are 6 points.

If all 3 sides of the triangle intersect both circles, there are 12 points.

Here is a not-so-pretty but nevertheless rough sketch of those 6 intersection points
[img]11744[/img]

And two circles can intersect at most 2 points.

EDIT#1: eh image disabled?
EDIT#2: i have to put the 'url' tags to show the image...
[url][/url]


Let $ x,y\in R$ and $ x \minus{} 3\sqrt {x \plus{} 1} \equal{} 3\sqrt {y \plus{} 2} \minus{} y$.

Find maximum of $ P \equal{} x \plus{} y$

[quote="thanhnam2902"]Let $ x,y\in R$ and $ x \minus{} 3\sqrt {x \plus{} 1} \equal{} 3\sqrt {y \plus{} 2} \minus{} y$.

Find maximm of $ P \equal{} x \plus{} y$[/quote]
$ P \equal{} x \plus{} y \equal{} 3\left(\sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 2}\right)\leq3\sqrt {2(P \plus{} 3)},$ which gives $ P\leq9 \plus{} 3\sqrt {15}.$
The value $ 9 \plus{} 3\sqrt {15}$ holds when $ x \equal{} y \plus{} 1 \equal{} \frac {10 \plus{} 3\sqrt {15}}{2}.$

Let $ x \minus{} 2\sqrt {x \plus{} 1} \equal{} \sqrt {y \plus{} 9} \minus{} y.$
Find the maximum value of $ x\plus{}y.$

[quote="arqady"][quote="thanhnam2902"]Let $ x,y\in R$ and $ x \minus{} 3\sqrt {x \plus{} 1} \equal{} 3\sqrt {y \plus{} 2} \minus{} y$.

Find maximm of $ P \equal{} x \plus{} y$[/quote]
$ P \equal{} x \plus{} y \equal{} 3\left(\sqrt {x \plus{} 1} \plus{} \sqrt {y \plus{} 2}\right)\leq3\sqrt {2(P \plus{} 3)},$ which gives $ P\leq9 \plus{} 3\sqrt {15}.$
The value $ 9 \plus{} 3\sqrt {15}$ holds when $ x \equal{} y \plus{} 1 \equal{} \frac {10 \plus{} 3\sqrt {15}}{2}.$

Let $ x \minus{} 2\sqrt {x \plus{} 1} \equal{} \sqrt {y \plus{} 9} \minus{} y.$
Find the maximum value of $ x \plus{} y.$[/quote]
Hi arqady. Thank you for your solution. I have just change "maximm" ----> "maximum". Thank you!

We use the same idea as mastering Bunhiacovski inequality
$ x \minus{} 2\sqrt {x \plus{} 1} \equal{} \sqrt {y \plus{} 9} \minus{} y.$
which is equivalent to
$ P\equal{}x\plus{}y \equal{}2\sqrt {x \plus{} 1}\plus{} \sqrt {y \plus{} 9} \le \sqrt{5(P\plus{}10)}$


Let a,b,c,d is real number prove that
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}ab\plus{}cd \geq\sqrt{3}(ac\minus{}bd)$

[quote="blackmouse"]Let a,b,c,d is real number prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} ab \plus{} cd \geq\sqrt {3}(ac \minus{} bd)$[/quote]

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} ab \plus{} cd \geq\sqrt {3}(ac \minus{} bd)$  <->   $ (a\plus{}\frac{b\minus{}\sqrt{3}c}{2})^{2}\plus{}(d\plus{}\frac{c\plus{}\sqrt{3}b}{2})^{2}\geq0$

[quote="zaya_yc"][quote="blackmouse"]Let a,b,c,d is real number prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} ab \plus{} cd \geq\sqrt {3}(ac \minus{} bd)$[/quote]

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} ab \plus{} cd \geq\sqrt {3}(ac \minus{} bd)$  <->   $ (a \plus{} \frac {b \minus{} \sqrt {3}c}{2})^{2} \plus{} (d \plus{} \frac {c \plus{} \sqrt {3}b}{2})^{2}\geq0$[/quote]
Thanks good idea :P


http://matharchive.org/mml/index.html
does not work anymore. it says not found  :huh:

MML has a new statistician and a new website (http://www.mmleague.net/)  If the new statistician is interested, I will be providing him with 5-10 years of MML meets to put online there.  I'm not going to be storing the meets on matharchive.org if there's an MML website that can easily hold them.  (The [i]old[/i] MML website was on Geocities, which has download limits, so it wasn't practical to put the meets online there.)


Prove, that lines $ GJ, LI, HK$ are concurrent


integration
calculus
calculus computations


Find integer solution of    x.y.z=x^3+y^3+z^3

This problem is an old problem solved by Werner Mnich in 1957. In fact, he proved that, for all non-zero integer $d$,  the more general equation $x^3+y^3+z^3 = dxyz$ has no integer solution $(x,y,z)$ such that $xyz \neq 0.$

I can give a solution for the case $d=1$ but it is quite long and uses tools like $\mathbb {Z} [j]$ and infinite descent. So, if somebody has another way to do it, he is welcome... ;)

Pierre.


The first four digits of a certain positive integer $n$ are $1137$. Prove that the digits of $n$ can be shuffled in such a way that the new number is divisible by 7.

Lets suppose that the rearanged unumber is $m=10^k.Q+R$, $R<10^k$, $999<Q<10^4$. The congruence
$m\equiv 0 (mod 7)$ is equivalent to
$10^k.Q\equiv -R (mod 7)$, when multiplying by $5^k$ we get
$Q\equiv -R.5^k (mod 7)$. So it's enough to show that Q can be congruent to any residue mod n.
That's true because
$3171\equiv 0 (mod 7)$,
$1317\equiv 1 (mod 7)$,
$1731\equiv 2 (mod 7)$,
$1137\equiv 3 (mod 7)$,
$1173\equiv 4 (mod 7)$,
$1713\equiv 5 (mod 7)$,
$1371\equiv 6 (mod 7)$.
This solution is too trivial to be a right solution of problem 3, but can't find any mistake in it.


The heat of combustion of fructose, C6H12O6, is -2812 kJ/mol. If a fresh golden delicious apple weighing 120 g contains 16.0 g of fructose, what caloric content does the fructose contribute to the apple?

1) Find the molar mass of fructose.

2) If x is the value obtained in 1), then the heat of combustion of fructose is $ \frac { \minus{} 2812 (kJ/mol)}{x\,(g/mol)} \equal{} \frac { \minus{} 2812}{x}\,(kJ/g)$

3) Now for 16.0 g of fructose the heat of combustion is...


inequalities
trigonometry
inequalities proposed

Consider $ n \in \mathbb{N}^*.$ Prove that: $ \minus{}2 < \sin 1\plus{}\sin 2\plus{}...\plus{}\sin n<2.$

[quote="moldovan"]Consider $ n \in \mathbb{N}^*.$ Prove that: $ \minus{} 2 < \sin 1 \plus{} \sin 2 \plus{} ... \plus{} \sin n < 2.$[/quote]
Because $ \minus{}\frac{1}{2}\tan{\frac{1}{4}} < \sin 1 \plus{} \sin 2 \plus{} ... \plus{} \sin n < \frac{1}{2}\cot{\frac{1}{4}}.$


The binary operation $ \#$ is defined as $ a \# b \equal{} 2a\minus{}3b$. What is the value of
$ (4 \# 5) \# 6$?

We first have to use the $ (4\#5)$ part because of the parentheses, and you plug them in: $ (2(4) \minus{} 3(5) \equal{} \minus{} 7$. Then we have to use $ ( \minus{} 7\#6)$. We again plug it in and we get $ (2( \minus{} 7) \minus{} 3(6) \equal{} \minus{} 32$.
The answer is $ \boxed{ \minus{} 32}$


What is the coefficient of $x^{3}y^{8}$ in the expansion of $(x-2y^{2})^{7}$?


trigonometry
function
integration
real analysis
real analysis unsolved

find sum  of sequence :
$ \sum_{0}^{n} \frac {\cos nx}{n!}$
  and  $ T \equal{} \sum_{1}^{n} \sin kx$
  final :
 $ cos \frac {\pi}{101} \plus{} cos \frac {3\pi}{101} \plus{} ... \plus{} cos\frac {99\pi}{101}$
  pro 4 /
find sum;
$ 3^{\sum_{k \equal{} 1}^{n} \frac {k}{3^k}}$
найти сумму ряда :
  $ \sum_{n \equal{} 0}^{\infty} \frac {( \minus{} 1)^n(n \plus{} 1)}{(n \plus{} 2)n!2^{n \minus{} 1}}$
pro 5/
 find sum of sequence :
$ 1 \plus{} \frac {x^3}{3!} \plus{} \frac {x^6}{6!} \plus{} \frac {x^9}{9!} \plus{} ...$
pro 6 :let : $ A \equal{} 1 \plus{} 1/\sqrt {2} \plus{} 1/\sqrt {3} \plus{} ... \plus{} 1/\sqrt {36}.$  and $ B \equal{} 1  \plus{} 1/\sqrt [3]{2} \plus{} ... \plus{} 1/\sqrt [3]{27}.$.
 prove that :
$ A > B$

First of all, your notation is strange.. When defining a sum, you should write the first and the last number of the interval you sum on because from your notation it isn't quite obvious what needs to be calculated. Still, I'll presume you sum up to $ \infty$.

1)

After writing the $ \cos$ in exponential form, use the fact that:

$ \sum^{\infty}_{n\equal{}0}\frac{x^n}{n!}\equal{}e^x$

You should get as a result: $ e^{\cos\,x}\cos \sin\,x$

2)

Again, rewrite the $ \sin$ function in exponential form and use the fact that:

$ \sum^{n}_{i\equal{}1}q^n\equal{}\frac{q^{n\plus{}1}\minus{}q}{q\minus{}1}$

You should get the result: $ \sin^2 \frac{n\,x}{2}\cos\frac{x}{2}\plus{}\frac{1}{2}\sin{n\,x}$

3)

The sum can be represented as:

$ \sum^{50}_{n\equal{}1}\cos \left(\frac{\pi}{101}(2\,n\minus{}1)\right)$

Replace the $ 50$ value with a variable and use the same method like in the previous question to derive a closed form, plug back the $ 50$ value in, do some trig transformations and you should derive to the result: $ \frac{1}{2}$

4)

Define:

$ f(x)\equal{}\sum^{n}_{k\equal{}1}k x^k$

Then, you could write:

$ f(x)\equal{}x \frac{d}{d\,x} \sum^{n}_{k\equal{}1} \int k x^{k\minus{}1} \,dx$

$ \equal{}x \frac{d}{d\,x}  \sum^{n}_{k\equal{}1} x^k$

$ \equal{}x \frac{d}{d\,x} \frac{x^{n\plus{}1}\minus{}x}{x\minus{}1}$

$ \equal{}x \frac{1\minus{}x^n}{(1\minus{}x)^2}\plus{}n\frac{x^{n\plus{}1}}{x\minus{}1}$

Your question can be written as:

$ 3^f(\frac{1}{3})$

that is

$ 3^{\frac{3}{4}\minus{}\frac{1}{4}3^{\minus{}n}(3\plus{}2n)}$

5)

Defining:

$ f(x)\equal{}\sum^{\infty}_{n\equal{}0}\frac{x^n}{n!} \frac{n\plus{}1}{n\plus{}2}$

and using:

$ \int (n\plus{}1) x^n \,dx\equal{}x{n\plus{}1}$

$ \frac{d}{dx}\frac{x^{n\plus{}2}}{n\plus{}2}\equal{}x^{n\plus{}1}$

$ \sum^{\infty}_{n\equal{}0}\frac{x^n}{n!}\equal{}e^x$

$ \sum f(n) x^n\equal{}x^a\sum f(n) x^{n\minus{}a}$

you can write:

$ f(x)\equal{}\frac{2}{x^2} e^x (x^2\minus{}x\plus{}1)\minus{}\frac{2}{x^2}$

and your problem can be defined as $ f(\minus{}\frac{1}{2})\equal{}\frac{14}{\sqrt{e}}\minus{}8$

6)

I'm not sure how should one approach this, but Mathematica gives the following result:

$ \frac{1}{3}e^x\plus{}\frac{2}{3}e^{\minus{}\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2}x\right)$

7)

It's not... $ A\approx 10.6228$ and $ B \approx 12.693$

One way to approach the next to the last problem (the one the original poster called #5 and milin called #6): this series is the solution to the following differential equation initial value problem:

$ y'''\minus{}y\equal{}0;\ y(0)\equal{}1,y'(0)\equal{}0,y''(0)\equal{}0.$


topology
superior algebra
superior algebra solved

This is as easy as hello.

Let $a\in(0,1)$ and $(x,y),(x',y')\in X\times Y$. Then $a(x,y)+(1-a)(x',y')=(ax+(1-a)x',ay+(1-a)y')\in X\times Y$ by the hypothesis that $X$ and $Y$ are convex.


Find all $f: Z\rightarrow Z$ such that
$f(m+8)\leq f(m)+8$ and $f(m+11)\geq f(m)+11$ for all $m\in Z$


It seems like a quite a few people are reading this but can anybody reply?

I am not quite sure of the answer to this problelm...
These are the choices for the question below,
a. 30 protons and 39 neutrons
b. 31 protons and 31 neutrons
c. 31 protons and 39 neutrons
d. 31 protons and 47 neutrons
e. 32 protons adn 48 neutrons

1. Gallium-69 and Gallium-70 are stable isotopes.  Of the five nuclei listed which two are the most stable?  Explain your answer.
2.  Of the five nuclei listed, which two are most likely to emit beta radiation?  Expalin your answer

I think the answers to the first question are A and C since they have the closest amount of difference in mass... but not quite sure if this is the correct reasoning.
To the second one,   I think it is D and E since atoms that have too many neutrons emit Beta raditaion. Again,,, not sure.

These are from Prentice Hall Chemistry book.. I am independently studying with no answers and noone to ask questions so I am just posting here. Does anyone know a good website that answers questions. like these kind...
for example I know there is mathnerds for asking math question
Thank you soooo mUch!

To get you started, re-read the 1st question and see if there is something you didn't use so far, something "left over?"

Yes, it is the first part of the question which contains two statements, which for the purposes of this question we take as totally true.

Is this "totally true" information applicable to any of the possible answers?  Could it only be by chance that Gallium-69 and Gallium-70 have the same atomic weight as answers a. and c.?

To start with the 2nd question, are the most stable nuclei (answer to question 1) the best canidates for the most likely to emit beta radiation?  This may lead you to close the circle a bit around the answer to the second question.

I hope this little discussion starts the solutions rolling.


Let $ a, b, c > 0$ and $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that: $ \boxed{\frac{a}{2a^2\plus{}bc}\plus{}\frac{b}{2b^2\plus{}ca}\plus{}\frac{c}{2c^2\plus{}ab}\ge abc}$

Here is one of the methods:
    Divide both sides by abc and Let x=ab, y=bc, z=ca , so x+y+z=3
And the equation becomes ∑1/(2xz+y^2)≥1
    mutiplying L.H.S. by (x+y+z)^2 and R.H.S. by 9 and let 2xz+y^2=j , 2xy+z^2=k, 2yz+x^2=l
Then the equation becomes ∑(j+k+l)/j ≥9
    which is clearly true by expanding two sides and rearrangement inequality.

No need for a rearrangement inequality.
∑(j+k+l)/j ≥9  <=> k/j+l/j+j/k+l/k+l/j+l/k≥6 which is a AM-GM inequality. :oops:


Dear Mathlinkers,
let ABC be a triangle, I the incenter, 2 the circle passing through A, I, C,
3 the circle passing through A, I, B,
U a point on 3, D the symmetric of AB wrt AU, W the second point of intersection of D with 2, V the midpoint of the arc WC which doesn't contain A.
Prove that IU is orthogonal to IV.Sincerely
Jean-Louis

In my figure $ U,I,V$ lie on the same line $ \Leftrightarrow \angle UIV=180^{\circle}$

It is easy to prove that $ P$ and $ N$ are the circumcentres of $ AIB, AIC$, respectively, where $ P=CI \cap C(O)$ and $ N=BI \cap C(O)$

Let $ \angle UAB=x \Rightarrow$ $ \angle UIA=\frac {\widehat {UA}} {2} =\frac {360^{\circ}-2x-A-B} {2}$ $ =180^{\circ}-x-\frac {A} {2}-\frac {B} {2}$

$ \angle AIC=180^{\circ}-\frac {A} {2}-\frac {C} {2}$

$ \angle {CIV}=\frac {\widehat{VC}} {2}=\frac {\widehat{WC}} {4}=\frac {\angle WAC} {2}=\frac {A-(180^{\circ}-2x)} {2}=\frac {A} {2}-90^{\circ}+x$

Finally, $ \angle UIV=\angle UIA+\angle AIC+\angle CIV=180^{\circ}$

Dear Mathlinkers,
sorry for my typo...
If U is on the arc BI, then IU is orthogonal to IV.
Sincerely
Jean-Louis

It’s quite easy to see that angle VAU= angle A/2 (1), angle AUI= angle B/2, angle AVI=angle C/2, but the angle VIU is exactly the sum of previous angles, being the sum of 2 external angles of the triangles AUI and AVI.

Best regards,
sunken rock


geometry
pigeonhole principle
geometry proposed

The $K$ convex $n$-gon is inside a square which has sides of 1 unit. Show that we can choose 3 of $K$-s vortices so that the area of the triangle made by these vortices is smaller than $\frac{60}{n^{3}}$ units.


Where do I get Geometry by IF Sharygin in India, preferably TN?
Akashnil or someone having it, can you tell me ?

Ask the olympiad guys from chennai da. There are lots of seniors. Contact them through your professor.

You are Rakesh from IIT chennai rite?
Anand sir told me that you were correcting the chem papers.

Ok. Seniors in the sense?
Like G Vignesh?
I tried sending pms to him both in AOPS and JOi in summer hols. He never sent me even a single word reply and I am determined not to ask him.
I asked Rohit kannan and he claims he wants Arthur Engel to put 'vetti' scene.
Who else can I ask?

And professor refers to whom? Anand sir or Rajesh sir?

yes I am the same rakesh.
Send vignesh the mail thro normal mail
I think both kk sir and rajesh sir can guide you in the right direction. Apart from that you also have other seniors in the 12th and 11th who qualified to IMO-TC. Contact them.

[quote="Aravind Srinivas L"]Like G Vignesh?
I tried sending pms to him both in AOPS and JOi in summer hols. He never sent me even a single word reply and I am determined not to ask him.
[/quote]
better be determined not to ask me too :P

shut up aravind!!!!!
U r just too much....... U dont appreciate good humor.... You are a nerd.. ok fine but dont expect everyone to be like you....
At least respect people for their age...

sry aravind if i hurt u
i was just angry with ur casual remarks abt madness
i revised my entire post
and guys slap me if i ever again come into such controversies :D  :rotfl:
LOL!

lets close this topic unless anyone finds a way to get Sharygin.

This forum is a sh1t. Kiddos spamming at every oppurtunity.Grow up.

[hide]This might look like another spam but its needed.[/hide]

another spam but it was needed! :P
[hide]
[quote="befuddlers"]
and guys slap me if i ever again come into such controversies :D[/quote]
how can we slap u :rotfl:
[/hide][/hide]

Now what did he say about me that made you all go mad?  :roll:   :rotfl:  I know this is spam. Par, can you please move this post to the general discussion forum?  :|

shud we repeat it again??? :rotfl:

i'll say  its gud u didnt see it :P  :rotfl:


The US Open at Flushing Meadows, NY begins August 31!

Please vote, sorry its only for men's singles.

lol roger federer is in the finals guys.

federer:99.99999999999...% win.

Wierd...
You would think that Nadal has a chance. Itd be nice if Roddick won, but I doubt that.

federer can beat roddick 7-6, 7-5, 4-6, 5-7, and 19-21 xD

Like, other players besides the ones listed.

I haven't seen Nadal against Federer in a while.... But, I'm still going for Federer....

[quote="cgyao15"]i'm leaning toward Djokovic but Federer and Rodrick are still options.[/quote]

Federer. DUUUHHY

Actually now Djokovic, Federer, and Del Potro are in.
And Nadal and gonzalez still playing.

Gee WHAT HAPPENED TO RODDICK THIS TIME??
And why are there 2 us open threads?

Don't do that too often.
Del Potro only moved up 1 ranking from winning USOPen against FEDERER OMG.

But that's fine I'm mostly a Fed/Murray/Roddick fan

And how did Murray lose to Cilic in 3 sets??
Does cilic have a great serve or something?

I'm not sure.

But all the great players this US Open sucked for some reason.

And, the rankings are decided by points.

Winning the US Open gives you maybe 50000 points?

 and Federer has like a 1000000 point difference from Del Potro so...

Wow. I didn't know that. Well, looks like Roddick had significantly more than Del Potro?

not really.

maybe by only like 45000 points?

My friend said Federer believed that he could play to his 40s.

Well Jimmy Connors made it to USOpen Semis when he was 39!!!


function geometry trigonometry ratio probability

Although these USC tests already have solution documents, there are alternate (and probably better) solutions to some of the [url=http://www.math.sc.edu/contest/problems.html]problems[/url]

This thread is for any problems you see that can be solved in an alternate way from the way USC solved it.

[hide="USC 1996 #23"]Since we already have a minimal of $2, 2, 3, 4$, we want to find the number of solutions to $a+b+c+d=20-(2+2+3+4)=9$. The number of solutions is $\binom{12}{3}=220$.[/hide]

[hide="USC 1996 #25"]The solution had to do with Fibonacci numbers or something which I think is a lot less efficient than this way:

We want to find the number of distinct partitions of $6$ in which each number it is partitioned into is either $1$ or $2$.

Thus, we have:
$(2,2,1,1)$ and there are $\frac{4!}{2!2!}=6$ ways to do this.

$(2,2,2)$ and $1$ way to do this.

$(1,1,1,1,1,1)$ and $1$ way to do this.

$(2,1,1,1,1)$ and $\frac{5!}{4!}=5$ ways to do this.

Thus our answer is $6+1+1+5=13$ ways to walk up the stairs.[/hide]

[hide="USC 1996 #28"]Although the solution uses the same technique that I did, I think that there simplification of the congruences was a little tedious.  Here's a simpler way. As before, we want to find $10^{45}\mod{1996}$.

$(10^{5})^{9}\equiv (200)^{9}\equiv (200^{3})^{3}\equiv 32^{3}\equiv 832 \mod{1996}$. The answer follows from there. Note: This problem is very similar to #4 [url=http://www.math.sc.edu/~filaseta/contests/cont99/cont99.pdf]here[/url] [/hide]

[hide="USC 2001 #28"]Here's a more efficient way to do this problem. Because $[\triangle{ABC}]$ is $1$ and constant. The answer we are looking for is simply $\frac{[\triangle{DEF}]}{[\triangle{ABC}]}$. Assume $\triangle{ABC}$ is an equilateral triangle with sidelength $5$ and area $\frac{25\sqrt{3}}{4}$. It follows that $\triangle{DEF}$ is also equilateral with sidelength $\sqrt{2^{2}+3^{2}-2(2)(3)(\cos{60^\circ})=\sqrt{7}}$. The area ratio is thus $\frac{\frac{7\sqrt{3}}{4}}{\frac{25\sqrt{3}}{4}}=\frac{7}{25}$.[/hide]

[hide="USC 2001 #30"]Here's a less complicated way to do this problem. Notice that if we keep inscribing consecutive circles in the triangle, the diameters of the circles (and thus the height of the triangle) will be:

$4+2+1+1/2+... \infty=\frac{4}{1/2}=8$.

Let the triangle be $ABC$ with the altitude of $8$ being $AX$. Let the center of circle with radius $2$ be $O$, and let it's tangent point to the triangle ($\not=X$) be $P$. We have that $\triangle{ACX}\sim \triangle{AOP}$. $AO=6$ so we have that $AP=\sqrt{32}$. Thus, we have that $\frac{8}{CX}=\frac{\sqrt{32}}{2}\Longrightarrow CX=2\sqrt{2}$ and $BC=4\sqrt{2}$. Thus the area is $16\sqrt{2}$.[/hide]

eh, i don't like the way USC solves NT problems...

[hide="USC 2003 #24"]

$N \equiv 2 \mod{5}$
$N \equiv 3 \mod{7}$
$N \equiv 4 \mod{9}$

Start with the first two equations:

We have:

$N-2 \equiv 0 \mod{5}$
$N-2 \equiv 1 \mod{7}$

So we have $N=15+35k+2$ which must be $\equiv 4 \mod{9}$.

Looking in $\mod{9}$, we have:

$N=6+8k+2 \equiv-1-k \equiv 4 \mod{9}\Longrightarrow k \equiv-5 \equiv 4 \mod{9}$.

Thus, the smallest $k=4$.

So the smallest value of $N=15+2+35(4)=157$ and the sum of the digits is $13$.[/hide]

[hide="USC 2003 #25"]
We long divide $\frac{n^{2}+4}{n+3}$ continuously and find that the answer is the number of integers which satisfy $\frac{n+3}{13}$ being an integer. There are $7$ values in the interval.[/hide]

[hide="USC 2003 #29 "]$\phi 31=30$. Therefore, $n=30$ is the smallest $n$ that satisfies the condition. And indeed, $5^{30}+30 \equiv 1-1 \equiv 0 \mod{31}$.[/hide]

WHOOOOOOO...I QUADRUPLE POSTED!!!!

[hide="USC 2004 #20"]Let $n$ be the number of boys or girls.

We have:

$\binom{n}{2}\binom{12-n}{1}+\binom{n}{1}\binom{12-n}{2}=\frac{n(n-1)(12-n)}{2}+\frac{n(12-n)(11-n)}{2}=n(12-n)(1)=320 \Longrightarrow n(12-n)=32 \Longrightarrow n=4,8$. The answer is $4$.[/hide]

[hide="USC 2004 #25"]The numbers are in the forms $1+7k$ or $2+11p$.

If a number $N$ satisfies both these sequences, then:

$N \equiv 1 \mod{7}$
$N \equiv 2 \mod{11}$

Thus:

$N-1 \equiv 0 \mod{7}$
$N-1 \equiv 1 \mod{11}$

And thus, $N=57+77q$. It follows that there are $25$ values of $N$ in the given interval.

[/hide]
[hide="USC 2004 #26"]
We want to find how many ways we can arrange three numbers in eight spaces without being adjacent. Consider the string:

_ _ _ _ _ _ _ _

There are five numbers that are not $3$.

They can be arranged as such:

_ X _ X _ X _ X _ X _

We must choose three of the _ s, so we have $\binom{6}{3}$ and since we can't use $3$s for the remaining five spots, we have $\binom{6}{3}5^{5}$.

The total number of ways to have three $3$s is $\binom{8}{3}5^{5}$.

Our probability is thus $\frac{5}{14}$.[/hide]

[hide="USC 2005 #20"]Let's start by condidering the cases of how many cars contain a certain "starting" color.

Case 1: We choose one car for one color. Let the remaining two colors be $x$ and $y$. We can have one $x$ and four $y$s, two $x$s and three $y$s, or three $x$s and one $y$.

This can be done in $\binom{5}{1}(\binom{4}{1}+\binom{4}{2}+\binom{4}{3}=70$ ways.

Case 2: We choose two cars for one color.

This can be done in $\binom{5}{2}(\binom{3}{1}+\binom{3}{2})=60$ ways.

Case 3: We choose three cars for one color.

This can be done in $\binom{5}{3}(\binom{2}{1})=20$ ways.

The total ways is thus $70+60+20=150$.[/hide]
[hide="USC 2005 #27"]
Let's let West have tulips. Then we consider the following cases:

Case 1: North, South, and East are all different flowers and none are tulips.

This can be done in $3! \binom{4}{3}=24$ ways.

Case 2: East contains tulips as well, but North and South both contain different flowers, none of which are tulips.

This can be done in $2! \binom{4}{2}=12$ ways.

Case 3: North and South have the same flowers (none of which are tulips), and East has different flowers flowers from tulips and those of North and East.

This can be done in $\binom{4}{1}\binom{3}{1}=12$ ways.

Case 4: North and South have the same flowers (none of which are tulips), and East and West both have tulips.

This can be done in $\binom{4}{1}=4$ ways.

The number of ways in which we start of with tulips is thus $24+12+12+4=52$.

But, there are five starting flowers (including tulips), thus the total number of ways is $52 \cdot 5=260$.[/hide]

[quote="mad_skillz_aops"]Oh wow, you hexagruple posted, or whatever 6 is.[/quote]

indeed :wink:, but now I have to start over :( and there's only one contest left...

[hide="USC 2006 #29"]Note that:

$(\cos{\theta}+i\sin{\theta})^{3}=\cos^{3}\theta+3\cos^{2}\theta i\sin{\theta}-3\cos{\theta}\sin^{2}\theta-i\sin^{3}\theta$.

Extracting the real parts, we see that:

$\cos{(3\theta)}=\cos^{3}\theta-3\cos{\theta}\sin^{2}\theta=4\cos^{3}\theta-3\cos{\theta}$.

Thus, we want to find the minimum of $4\cos^{2}\theta+\cos{\theta}-3$ and using the formula $x=-b/2a$, we find the minimum is $-\frac{49}{16}$.[/hide]


Stie cineva unde pot gasi si eu rezultatele de la Titeica pe internet?Multumesc anticipat!

Nu rezultatele nu s-au afistat!Dar eu le stiu pe cele de la clasa a VII-a! Daca vrei ti le zic!

Perfect!Multumesc.Btw, care e numele tau adevarat?Ai fost si tu la concurs?Ce loc ai luat?Si in ce clasa esti?


How do I insert equations into forum messages?[/hide]

u can write them out or use LATEX, go to the LATEX thread in the index to find out more.


geometry
incenter
circumcircle
angle bisector
perpendicular bisector
geometry solved

ABC is an acute-angled scalene triangle with incenter I and orthocenter H.
BB' and CC' are medians.
The line IB' meets the line AB at B'', and the line IC' meets the line CA at C''.
The line B''C'' meets the line BC at K.
The circumcenter of BHC is O.
Show that A, I and O are collinear if and only if area BKB'' = area CKC''.

Help !

I even don't manage to prove this :(

You know, I always freak in geometry when there are too much data given ...
as is the case here :?

I will show that A, I and O are collinear if and only if  < A = 60°.

If < A = 60° then < CIB = 120°. In fact < CIB = < A + < B/2 + < C/2 = < A/2 + < A/2 + < B/2 + < C/2 = 30° + 90°. Then I is on c(CHB). Let K be the (second) point where AI intersect c(CHB). It holds that < CIK = < A/2 + < C/2 and that < CKI = < CBI = < B/2 therefore < ICK = 90° and we are done.

If A, I, O are collinear let's call F the (second) point where AC intersect c(CHB). It result for simmetry tha < FOB = 2 < AOB. But < FOB = 2 < C then < C = < AOB. Then ABOC is ciclic from which we have that < A + < BOC = 180° and, given that, < BOC = 2< A we can conclude that < A = 180°/3.

[quote="grobber"]I think I've got it, but I'll just post this hint: try to prove that both conditions occur iff A=60 deg.[/quote]It is time to complete solution.

The problem is not too difficult since the greatest part of the solution is already posted on the forum. Here is an extension of a problem which makes it easier to prove:

[color=blue][b]Extended problem.[/b] Let ABC be an acute-angled scalene triangle with incenter I and orthocenter H.
Let B' and C' be the midpoints of the sides CA and AB, respectively. Let B" be the point of intersection of the lines IB' and AB, and let C" be the point of intersection of the lines IC' and CA. Let the line B"C" meet the line BC at K.
Denote by O the circumcenter of triangle BHC.

Prove that the following four assertions are all equivalent:

[b](1)[/b] The points A, I and O are collinear.
[b](2)[/b] We have A = 60°.
[b](3)[/b] We have [AC"B"] = [ABC].
[b](4)[/b] We have [BKB"] = [CKC"].

Hereby, the abbreviation [$P_1P_2...P_n$] means the area of an arbitrary polygon $P_1P_2...P_n$.[/color]

[i]Solution.[/i] In order to prove that the assertions [b](1)[/b], [b](2)[/b], [b](3)[/b], [b](4)[/b] are all equivalent, it is enough to prove the equivalences [b](1) $\Longleftrightarrow$ (2)[/b], [b](2) $\Longleftrightarrow$ (3)[/b] and [b](3) $\Longleftrightarrow$ (4)[/b].

The equivalence [b](2) $\Longleftrightarrow$ (3)[/b] was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=3689 post #4 [b](a)[/b], so we needn't prove it here again.

The equivalence [b](3) $\Longleftrightarrow$ (4)[/b] is very easy to see: The point B" can lie either inside the segment AB, or on the extension of this segment beyound B; the point C" can lie either inside the segment CA, or on the extension of this segment beyound C. Hence, we can distinguish between four cases:

[b]Case 1.[/b] The point B" lies inside the segment AB, and the point C" lies inside the segment CA.
[b]Case 2.[/b] The point B" lies inside the segment AB, and the point C" lies on the extension of the segment CA beyound C.
[b]Case 3.[/b] The point B" lies on the extension of the segment AB beyound B, and the point C" lies inside the segment CA.
[b]Case 4.[/b] The point B" lies on the extension of the segment AB beyound B, and the point C" lies on the extension of the segment CA beyound C.

In each of the Cases 1 and 4, the line B"C" does not intersect the segment BC; in other words, the point of intersection K of the lines B"C" and BC lies outside the segment BC, and none of the two equations [AC"B"] = [ABC] and [BKB"] = [CKC"] can be fulfilled, since one of the two triangles AC"B" and ABC is completely contained inside the other one, so that either [AC"B"] > [ABC] or [AC"B"] < [ABC], but never [AC"B"] = [ABC], and one of the two triangles BKB" and CKC" is completely contained inside the other one, so that either [BKB"] > [CKC"] or [BKB"] < [CKC"], but never [BKB"] = [CKC"]. Hence, in Cases 1 and 4, none of the two conditions [b](3)[/b] and [b](4)[/b] can hold, and thus, it is enough to consider the Cases 2 and 3. Since these two cases are completely analogous to each other, we will only consider Case 2.

In this case, [AB"KC] = [ABC] - [BKB"] and [AB"KC] = [AC"B"] - [CKC"], so that [ABC] - [BKB"] = [AC"B"] - [CKC"]. Thus, [AC"B"] = [ABC] if and only if [BKB"] = [CKC"]. This proves the equivalence [b](3) $\Longleftrightarrow$ (4)[/b].

Remains to show the equivalence [b](1) $\Longleftrightarrow$ (2)[/b]. This equivalence was very nicely proven by Sprmnt21 in post #4; here is a slightly (but only slightly) different proof:

Since I is the incenter of triangle ABC, the line AI is the angle bisector of the angle CAB. Let this line AI intersect the circumcircle of triangle ABC at a point Q (apart from the point A); then, we have < QAB = < QAC because of the line AI being the angle bisector of the angle CAB. But if two chords of a circle have equal chordal angles, then they must have the same length; hence, the two chords QB and QC of the circumcircle of triangle ABC, having equal chordal angles < QAB = < QAC, must have the same length. So we have QB = QC; in other words, the point Q lies on the perpendicular bisector of the segment BC. Thus, the point Q is a common point of the line AI with the perpendicular bisector of the segment BC. Since the triangle ABC is scalene, the perpendicular bisector of the segment BC doesn't pass through the point A, so that the line AI cannot coincide with the perpendicular bisector of the segment BC; hence, it follows that the point Q is the [i]only common point[/i] of the line AI with the perpendicular bisector of the segment BC.

Now, the point O, being the circumcenter of triangle BHC, must also lie on the perpendicular bisector of the segment BC; hence, this point O will lie on the line AI if and only if O = Q. But O = Q holds if and only if < QBC = < OBC (in fact, the points O and Q already lie on a common given line, namely on the perpendicular bisector of the segment BC, and the condition < QBC = < OBC puts them on another common line, namely on a line through the point B, so that they must be equal, being common points of two distinct lines). Thus, the point O lies on the line AI if and only if < QBC = < OBC.

Now, since the point O is the circumcenter of triangle BHC, the central angle theorem yields < BOC = 2 (180° - < BHC) (since 180° - < BHC is the acute chordal angle of the chord BC in the circumcircle of triangle BHC). On the other hand, by the sum of angles in triangle BHC, we have 180° - < BHC = < HBC + < HCB. If Y is the foot of the altitude of triangle ABC from the vertex B, then, in the right-angled triangle BYC, we have < YBC = 90° - < YCB; in other words, < HBC = 90° - C. Similarly, < HCB = 90° - B. Thus, 180° - < BHC = < HBC + < HCB = (90° - C) + (90° - B) = 180° - B - C = A (since A + B + C = 180° by the sum of angles in triangle ABC). Hence, < BOC = 2 (180° - < BHC) = 2A. Since the triangle BOC is isosceles (in fact, OB = OC, since the point O is the circumcenter of triangle BHC), we thus have $\measuredangle OBC=\frac{180^{\circ}-\measuredangle BOC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$.

On the other hand, since the point Q lies on the circumcircle of triangle ABC, we have < QBC = < QAC, and since the point Q lies on the line AI, which is the angle bisector of the angle CAB, we have $\measuredangle QAC=\frac{A}{2}$. Thus, $\measuredangle QBC=\frac{A}{2}$.

So, altogether, we have < OBC = 90° - A and  $\measuredangle QBC=\frac{A}{2}$. Thus, < QBC = < OBC holds if and only if $\frac{A}{2}=90^{\circ}-A$, what is equivalent to $\frac32A=90^{\circ}$, or, in other words, to A = 60°.

Since we know that the point O lies on the line AI if and only if < QBC = < OBC, we thus can conclude: The point O lies on the line AI if and only if A = 60°. In other words, the points A, I and O are collinear if and only if A = 60°. This proves the equivalence [b](1) $\Longleftrightarrow$ (2)[/b].

Thus, the extended problem is completely solved.

  Darij


geometry
vector
rotation
geometric transformation
geometry unsolved

Problem 1
ABC be a triangle. For any line L wich is not parallel to any of its sides,
consider the centroid G_L of the triangle determined by the intersection
of L with AB,BC,and CA.
Find the locus of G_L as L varies.
---------------------------------------------------------------------------------
Problem 3
For three coplanar congruent circles wich pass through a common
point, let S be the set of points wich are interior to at least two of
the circles. How should the three circles be placed so that the area
of S is minimized ?
----------------------------------------------------------------------------------
Problem 4
Let A1,A2,A3,B1,B2,B3 be coplanar points such that AiBj = i+j
for any i,j in {1;2;3} What can one say about these six points ?
---------------------------------------------------------------------------

Hey, I can't understand sth abt prb 1: G_L is the centroid of WHICH triangle? Is it the one formed by the 3 pts of intersection between L and the 3 sides of ABC? I was confused by the fact that you called it 'triangle', because those 3 pts are actually collinear (they are on L).

Yes the three points are colinear, the triangle is degerate.
The cenroid G_L of a degenerate triangle is the vector average
of the three points.

[quote="Moubinool"]Problem 1
ABC be a triangle. For any line L wich is not parallel to any of its sides,
consider the centroid G_L of the triangle determined by the intersection
of L with AB,BC,and CA.
Find the locus of G_L as L varies.[/quote]
It is clear $G_L$ cannot coincide with vertices $A,B,C$.
a) Take an arbitrary point $G$ not lying on lines $AB$, $BC$ and $CA$. Choose some direction and rotate it constructing corresponding lines. You will see that you can choose a proper initial direction and right rotations in such a way that from continuity arguments you obtain $G_L$ passes $G$ for some line $L$.
b) For $G$ lying on $AB$, $BC$ or $CA$ you can explicity construct the requred line $L$.

Myth, I believe this problem was posted one more time by Pierre. It had some additional questions which Moubinool did not post. For instance, it asked the same question for equilateral triangles. I can't remember every detail right now, but I have the distinct feeling this was not the answer, although nobody gave me any feedback when I posted what I believe to be a solution.

I'll try to find it.

Edit:

Here it is: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=inellipse&t=14871[/url]

It seems, I have mistaken with interior of $ABC$. :blush:


geometry
angle bisector
geometry proposed

Let $ C$ be a circle in plane. Let $ C_1$ and $ C_2$ be nonintersecting circles touching $ C$ internally at points $ A$ and $ B$ respectively. Let $ t$ be a common tangent of $ C_1$ and $ C_2$ touching them at points $ D$ and $ E$ respectively, such that both $ C_1$ and $ C_2$ are on the same side of $ t$. Let $ F$ be the point of intersection of $ AD$ and $ BE$. Show that $ F$ lies on $ C$.

Let $ F$ be the intersection of $ AD$ and $ \mathcal C$. Then $ \frac {C_1D}{CF} \equal{} \frac {AC_1}{AC} \equal{} \frac {R_1}{R}$ gives that $ CF \parallel C_1D$.

Since $ C_1D \parallel C_2E$, so $ CF \parallel C_2E$. Since $ \frac {C_2E}{CF} \equal{} \frac {BC_2}{BC}$, so $ B$, $ E$ and $ F$ are collinear.

Therefore the intersection of $ AD$ and $ BE$ lies on the circle $ \mathcal C$

Let DE intersect the circle C at M and N; it is then well known that AD is the internal angle bisector of $ \angle MAN$. Similarly BE is the angle bisector of $ \angle MBN$ and F is the middle of the arc MN.

Best regards,
sunken rock


algebra
polynomial
function
invariant
topology
complex analysis
complex analysis unsolved

Consider a family of polynomials
$p_t=a_0(t)X^n+a_1(t)X^{n-1}+...+a_{n-1}(t)X+a_n(t),$ with $a_i$ continuous functions, $t\in [0,1].$ If $p_t$ have no zeros on the unit circle for all $t\in [0,1]$ then $p_0$ and $p_1$ have the same nimber of zeros in the interior of the unit circle :lol:

The number of zeros of a polynomial $P$ withou roots of the unit circle inside the circle is equal to the degree of the map $f_P=\frac{P}{|P|}$ defined on the unit circle. Since the degree is invariant to homotopy, and we have a homotopy from $p_0$ to $p_1$ defined by $t\mapsto p_t$, which induces a homotopy from $f_{p_0}$ to $f_{p_1}$, it means that the degrees of $f_{p_0},f_{p_1}$ are equal, and hence the numbers of zeros of $p_0,p_1$ inside the circle are equal.


1/x + 1/[3(x^3)] + 1/[5(x^5)] + 1/[7(x^7)] + .... = ?

Is this a taylor series for something? I've looked at a bunch of lists and can't find anything that matches it

thanks

Taylor series are polynomials.  All of your exponents there are negative.

As it stands, this is equal to $\text{arctanh}\left( \frac{1}{x}\right)$, but it's not a Taylor series.

Wouldn't it have been easier to write that $\LaTeX$ as a forum post?

Anyway, yes, that is the way you would go about expressing $\text{arctanh}$ in terms of more elementary functions.


m* means outer measure here.

catatheodory criterion checks how the outer measure
splits over all sets. e.g:
E is measurable iff for any set A, we have
m*(A) = m*(A/\E) + m*(A-E)

there is an improved criterion, which we only need to check
how the outer measure splits over one set: that is,
If B is measurable and E < B, then E is measurable iff
m(B) = m*(E) + m*(B-E).

Please give a proof, thanks.
( <-- is difficult: if m*(E)<+inf, then easy, but I don't know how to prove it if m*(E)=+inf.  )

Look up Cezar Lupu's posts ... there is a topic on that matter, to which I replied.

I searched forum, can't find another problem about caratheodory condition.
Is this improved condition correct? please give some detail. Thanks.

if m*(E)=inf, then m(B)=inf, so the given equation is trivia. So this problem only works for finite m*(E).


AMC
AMC 12
MATHCOUNTS
AIME
search
vector
geometry

[quote="lsgl764"]I predict that either Bobby Shen of Texas or Justin Ahmann of Indiana will win writing round in 2008.

Indiana will won team since they had three 7th graders this year.[/quote]
Umm... you're predicting a YEAR before nationals? They MIGHT not even make it. They could have a bad day at states. You never know, and thats why its not a good idea to start predicting a YEAR(about) early.

thats if that kid from indiana even makes nationals.

youll see.  you already mentioned the other 2 7th graders returning, and there are aobut 10 other people that are just as good.

I would have to agree that you can't make any accurate predictions until State or later. Plus, the tests this year were easy, and if they're more difficult (or even easier) next year, then entirely different people might win the written round. So don't make any predictions based on this year's results.

[quote="lsgl764"]Indiana will won team since they had three 7th graders this year.[/quote]

06-MA had 3 7th, 1 8th-6th place
07-MA had 1 7th, 3 8th-6th place

All 3 7th graders might not even make it back. I learned that the hard way...

wow. 2008 already?
anyways, texas for team round, wildcard for written winner

... there are 10 people as good as us in Indiana?? Justin Ahmann didn't even break 100 on the AMC12, and he already learned AP statistics and Calc... fanatic is gonna get pissed at me for this... but seriously? Wtf?? he didn't even pass AMC 12?!?! -_-...

[quote="CircleSquared"]Nah, your coach told us that you had a 5th grader who is going to pwn next year  :maybe:

I'm scared.[/quote]

Well, he might go to nationals, but I've seen him in other (non-Mathcounts) competitions... he isn't [u]that[/u] great. He would have done really well on this year's easy test, though... If it's hard, he won't stand much of a chance. Much as I hate to say things like this about a potential Washington team member, his aptitude seems slightly overrated to me. He's also very annoying, from what I've heard about him, and probably wouldn't work well on a team... and one person alone can't make a team.[/quote]

[quote="perfect628"][quote="vector2101"]Well, I'm sure [u]you[/u] would say that... (observe location) I personally think that Texas will place high, but I still have no idea--there were, after all, some 8th graders on the team this year. One thing I'm sure of: the Washington team won't be as good (even after their relatively poor showing this year)--three 8th graders this year (including myself, if you haven't guessed by now).[/quote]

omg, why can't there be any good 7th, 6th, or even 5th graders, who are better than the 8th graders who are leaving? It's no indication as to how a state will do next year.[/quote]

Technically, you're right. However, 8th graders, in general, have had more experience than younger students in math competitions. So I would say that it's not a good indication, but it's actually (sadly) one of the best indications that we have, without going into a detailed analysis of state scores, work ethics, etc.

Apologies for any offense... none was intended.

[quote="lsgl764"]I predict that either Bobby Shen of Texas or Justin Ahmann of Indiana will win writing round in 2008.

Indiana will won team since they had three 7th graders this year.[/quote]

??? Who's Bobby Shen?  :huh:

A really good 6th grader who managed to win Texas state through CD. And he was one of the top this year. Some kind of prodigy who kind of creeps/freaks me out. Seriously

First off, AMC/AIME success DOES NOT translate into mathcounts. I would use my own sucky AIME but good mathcounts performance as an example, but i didn't break into the top 12.

Lol @ "fanatic is going to be pissed at me for this"... haha. i'm pissed at you for much more important things  :wink:  :P .

Anyhow, justin did a great job and I [i]expect[/i] him to do just as well next year and will cheer him on, but I almost feel like it was a fluke. Either that or I was a flop. I think it was more "I was a flop", because justin and I had tied on every single target we had ever done, and then suddenly he has 8 and I have 5. Justin and I were extremely close in skill leve, either of us winning on any given day, and suddenly kaboom he wins when it counts =/. Altho we did tie sprint.

Honestly, I'd be afraid of lyndon next year. Justin has pure skill, but lyndon has killer instinct... but i could be wrong. Lyndon's also much better at higher level contest math; it was cause of him we got #10 right. Although it was cause of Justin we got #9 right (on the team round). Did I do anything??? :(.

Maybe I might have time to get used to it since I'm going to mathpath.

But it's in Colorado Springs, not Denver. (close enough)

whaddaya mean "actually might"? Have I ever lied before?

[quote="indianamath"][quote="kstan013"]uhh...what's the point of predicting the winner of a math contest in 11 months when
a) you don't even know who's competing, exactly
b) one stupid mistake could kill you drastically. Had the test been taken again, I'll bet $\$$20 that Justin Ahmann would not win, although all he does is math, apparently. lol
c) you haven't considered that the thin Denver air will cause everyone to hyperventilate and pass out during the contest, leaving only the Colorado and Wisconsin kids standing[/quote]
 actually, the actual third statement actually might actually make an actual difference.

actually, not much though.[/quote]
IT WILL. TRUST ME. EVERYONE SHOULD BRING A FRESH SUPPLY OF OXYGEN!!!1!

What? We're climbing Mt. Everest?

But, as kstan013 said, this is probably too early to predict the winner for next year's nationals.

[quote="binonunquineist"]Bobby will win!!!

This year he just kinda messed up on Target because he's not very familiar with algebratic-geometry and stuff like Target #5 & 6. But he'll be awesome at countdown. He could probably have even beaten me in countdown this year had me made it :( and i was actually worried that he would. He has the awesome advantage of still being little and not being able to care less about competitions so he doesn't get any nervousy factors. That's probably how he beat me out at states too.[/quote]

lol, algebratic-geometry? I hope you did not mean algebraic geometry, because algebraic geometry is like the second beastliest subject to ever exist (second only to complex analysis). String Theory comes in a close third.

Anyway, I was thinking that, when I saw you win, I was thinking about how Bobby would have fared in countdown. Likewise, Allen and Wenyu both lost in their state countdowns, which were both official.

Basically 3 out of the top 4 were first written, second countdown at state.

And to the comment of Bobby being little, I would feel so intimidated by him in countdown because I would never want to have the shame of losing to a kid two feet shorter than me at countdown.  :P

[quote="13375P34K43V312"]
And to the comment of Bobby being little, I would feel so intimidated by him in countdown because I would never want to have the shame of losing to a kid two feet shorter than me at countdown.  :P[/quote]

How tall are you ? Are you the real mathmatician in pink ?

Losing to Bobby is no shame ! Even the greatest like Kevin Chen has been losing to Bobby once or twice before.

Why don't we put david lu into the mix?  He was 30th at nats this year and beat me in state countdown 3-0.  I think he'll make countdown  :D

maybe a underdog unknown may win it next year.  Who knows?

I believe that I will win next year, come to think of it.

Nathan Salo could win. He is really fast, but I think he got freaked out this year.

Lyndon Ji could definitely win countdown, I'd like to think I was pretty fast and lyndon could take me punch for punch (except I got super adrenaline rush on stage and usually did better)

Sorry, but I'm going to complain, as kstan did, about a suspendously time-wasting topic. Guys, spend your time _studying_ instead of jabbering about who is going to win next year. This topic's a great way to ensure you won't do as well as you could.


Let $f(x) = a^x$ where a is a positive number and g(x) is inverse function of f.
How many roots the equation $f(x) = g(x)$ may have?

$a>0$ and $a\neq 1$

If $0<a<1$ they have only one intersection...
If $a>1$, no intersection.

[quote="liyi"]If a>1, no intersection.[/quote]
You might want to recheck that. There is a number $b>1$ such that:
If $1<a<b$, two intersections.
If $a=b$, one intersection.
If $a>b$, no intersection.


$ a$ is a real number. $ x_1$ and $ x_2$ are the distinct roots of $ x^2 \plus{} ax \plus{} 2 \equal{} x$. $ x_3$ and $ x_4$ are the distinct roots of $ (x \minus{} a)^2 \plus{} a(x \minus{} a) \plus{} 2 \equal{} x$. If $ x_3 \minus{} x_1 \equal{} 3(x_4 \minus{} x_2)$, then $ x_4 \minus{} x_2$ will be ?

$\textbf{(A)}\  \frac {a}{2} \qquad\textbf{(B)}\  \frac {a}{3} \qquad\textbf{(C)}\  \frac {2a}{3} \qquad\textbf{(D)}\  \frac {3a}{2} \qquad\textbf{(E)}\ \text{None}$

[hide]The equations simplify to $ x^2\plus{}(a\minus{}1)x\plus{}2\equal{}0$ and $ x^2\minus{}(a\plus{}1)x\plus{}2\equal{}0$

Hence $ x_1\plus{}x_2\equal{}\minus{}a\plus{}1,x_3\plus{}x_4\equal{}a\plus{}1$.

So, $ x_1\equal{}\minus{}a\plus{}1\minus{}x_2,x_3\equal{}a\plus{}1\minus{}x_4$. Plugging that into the given condition, we get

$ 2a\minus{}x_4\plus{}x_2\equal{}3(x_4\minus{}x_2)\iff 4(x_4\minus{}x_2)\equal{}2a\iff x_4\minus{}x_2\equal{}{a\over 2}$, so the answer is $ A$.[/hide]


function calculus derivative trigonometry

Use a graphing utility to graph the function f(x)=-2sinx+cosx .. It doesn't say the inteval so I dont know what to do

I think we're supposed to find the intercepts, first derivative and second derivative, critical numbers and point of inflection .. and etc to graph this function...

please help .. thank you

That equation can be written $\sqrt{5} \cos \left( x+\arccos \frac{\sqrt{5}}{5} \right)$. That should make it easier to graph.

Graph one period.  The function has the same period as $\cos x$ and $\sin x$, so the period is $2\pi$, which means you should probably graph $0 \leq x \leq 2\pi$ and $-3 \leq y \leq 3$ with a graphing utility.  (The maximum won't be 3 - as demonstrated, and as can be proven, the maximum is $\sqrt 5$.

can you help me find the y intercept  x intercept ..critical numbers and stuff please thankx

[quote="I"]That equation can be written $\sqrt{5} \cos \left( x+\arccos \frac{\sqrt{5}}{5} \right)$.[/quote]
[hide]$y$-intercept: Plug in $x=0$, get 1
$x$-intecepts: $\sqrt{5} \cos \left( x+\arccos \frac{\sqrt{5}}{5} \right)=0\Rightarrow x+\arccos \frac{\sqrt{5}}{5}=\left(n+\frac{1}{2}\right) \pi$. Then the $x$-intercepts are given by $\left(n+\frac{1}{2}\right) \pi-\arccos \frac{\sqrt{5}}{5}$ for any integer $n$.
period: $2\pi$
maxes/mins: maximum of $\sqrt{5}$ at $x=2n\pi -\arccos \frac{\sqrt{5}}{5}$, minimum of $-\sqrt{5}$ at $x=(2n+1)\pi -\arccos \frac{\sqrt{5}}{5}$. Again, $n$ is any integer.
points of inflection: the $x$-intercepts[/hide]

[quote="scorpius119"][quote="I"]That equation can be written $\sqrt{5} \cos \left( x+\arccos \frac{\sqrt{5}}{5} \right)$.[/quote][/quote]
how can you know the equation can be written like above

Expanding it out gives $\sqrt{5}\left( \cos x \cos \arccos \frac{\sqrt{5}}{5} -\sin x \sin \arccos \frac{\sqrt{5}}{5}\right)=\sqrt{5}\left( \frac{\sqrt{5}}{5} \cos x-\frac{2\sqrt{5}}{5} \sin x \right)$
$=\cos x-2\sin x$.

[quote="Kle"][quote="I"]That equation can be written $\sqrt{5} \cos \left( x+\arccos \frac{\sqrt{5}}{5} \right)$.[/quote]
how can you know the equation can be written like above[/quote]

Here's how you can actually go from the given equation to what he had (not the other way around...):

[hide]First of all, the motivation for what we are doing is that we have the difference of two trig functions but would like to express them as one single trig function.

Let $y=\cos x - 2\sin x$.  This looks a little bit like the formula for the cosine of the sum of angles: $\cos(\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta.$  If we multiply both sides by $A$ (as you will see later, this accounts for the amplitude of the function) we get \[ Ay=A\cos x-(2A)\sin x. \]

In order for this to match up with our expression for cosine angle sum, we must have $A=\cos \theta$ and $2A=\sin \theta$ for some angle $\theta$.  Using the Pythagorean identity we have \[ \sin^2\theta + \cos^2\theta =5A^2=1\Rightarrow A=\pm\frac 1{\sqrt{5}}. \]

So $y=\pm\frac 1{\sqrt{5}}\cos(\theta+x).$  then you can just solve for $\theta$ and plug it in and you get what scorpius had.[/hide]


Arizona increased the number of characters on the license plate from 6 to 7. So I was thinking, how many different combinations can be made with the 6 chracters in form AAA000 where A's are alphabets and 0's are numbers. I was originally thinking 26^3*10^3, but I think this only works when the order doesn't matter. However, the from has to has the three letters then the three numbers. I help to convince my dad that the number is not 36^10 -_-;(I have no idea how he got that)

No, you're right, it's $ 260^{3}$. You can make $ 26^{3}$ different license plates that have the same array of numbers, and $ 10^{3}$ different license plates that have the same array of letters; therefore, you can make $ 26^{3} \cdot 10^{3}$ different license plates in total. If you still have doubts, look at the case where you count the different license plates you can make of the form $ ANNN$, where $ A$ is any letter in the English alphabet, and $ N$ is any digit. Note that you can make $ 26,000$ different license plates.

Likewise, there are $ 26^{4} \cdot 10^{3}$ and $ 26 \cdot 10^{4}$ possible different license plates of the form $ AAAANNN$ and $ AAANNNN$, respectively.

You are definitely correct. Choose each of the characters in order. There are 26 possibilities for the first, 26 for the second, 26 for the third, 10 for the fourth, 10 for the fifth, and 10 for the last slot, multiplying all possibilities together gives the answer.

Your dad's method would be correct (and try explaining this to him) if there were 26 + 10 (both letters AND numbers) options for each slot, and there were ten slots.

This is not the case in your problem, as we only have 6 slots, and there are different options for some slots.

[quote="jhangil"]Really? Thanks for your help, but I still have one more question. So if the order of the numbers and letters didn't matter would it cause the answer to be 36^6? Also the newspaper said that by increaing one more number thus making it AAA0000 it provides 106,480,000 more combinations. Maybe they have it wrong but 26^3*10^4-26^3*10^3 is 158,184,000.
P.S. I didn't know that there were over 17 million cars in Arizona ^^;[/quote]
There are probably some combinations that are either reserved for police/firefighter units, or maybe even censored if it's something inappropriate, which would account for the missing extra combinations.

@Jhangil's question number 2

[hide]maybe break it into cases - for the letters there are $ 26C3$ ways to pick the combinations with all distinct, plus the $ 26 * 25$ ways to pick the ones with 2 letters the same, plus the 26 combinations with all three equal $ \implies 3276$  ways to pick the letters. Repeating this with the numbers we get 3276 x 220 = [b]720720[/b] ways[/hide]


function
modular arithmetic
search
number theory
relatively prime
algebra proposed
algebra

Let a,b,c be pairwise coprime positive integers. Find the number of positive integers that could not be expressed under the form

k.ab +l.bc+m.ca where k,l,m are greater or equal to zero

http://www.mathlinks.ro/Forum/viewtopic.php?p=801444#801444

outline of solution already exists
see the resources section for other problems from Vietnam TST.

ok,I'll write a ful solution here:

[b]let $ a,b,c$ be pairwise coprime positive integers.an integer $ n\geq 1$ is said to be [i]stubborn[/i] if it cannot be written in the form $ n \equal{} bcx \plus{} cay \plus{} abz$ for any positive integers $ x,y,z$.determine,as a function of $ a,b$ and $ c$,the number of stubborn integers.[/b]

[b]solution.[/b] we claim that any integer $ n$ can be written in the form $ bcx \plus{} cay \plus{} abz$ where $ x,y,z$ are integers with $ 0 < y\leq b$ and $ 0 < z\leq c$,where $ x$ is possibly negative.because $ a$ and $ bc$ are coprime,we can write $ n \equal{} an' \plus{} bcx_0$ for some integers $ n',x_0$.because $ b,c$ are coprime, $ n' \equal{} cy_0 \plus{} bz_0$ for some integers $ y_0,z_0$.hence $ n \equal{} bcx_0 \plus{} cay_0 \plus{} abz_0$.choosing integers $ \beta,\gamma$ such that $ 0 < y_0 \plus{} \beta b\leq b$ and $ 0 < z_0 \plus{} \gamma c\leq c$,we find that

$ n \equal{} bc(x_0 \minus{} \beta a \minus{} \gamma a) \plus{} ca(y_0 \plus{} \beta b) \plus{} ab(z_0 \plus{} \gamma c)$

of the desired form.
observe that any positive integer less than $ bc \plus{} ca \plus{} ab$ is clearly stubborn.on the other hand,we claim that every integer $ n > 2abc$ is not stubborn.
given such an integer,write $ n \equal{} bcx \plus{} cay \plus{} abz$ with $ 0 < y\leq b$ and $ 0 < z\leq c$ then

$ 2abc < bcx \plus{} cay \plus{} abz\leq bcx \plus{} cab \plus{} abc \equal{} bcx \plus{} 2abc$

implying that $ x > 0$ as needed.
next,we prove that exactly half the positive integers in $ S \equal{} [bc \plus{} ca \plus{} ab,2abc]$ are stubborn.to do so,it suffices to prove that $ n\in S$ is stubborn if and only if $ f(n) \equal{} (2abc \plus{} bc \plus{} ca \plus{} ab) \minus{} n$ is not stubborn.for the "only if" direction,suppose that $ n$ is stubborn and write $ f(n) \equal{} bcx \plus{} cay \plus{} abz$ with $ 0 < y\leq b$ and $ 0 < z\leq c$.if $ x$ were not positive,then we could write $ n \equal{} bc(1 \minus{} x_0) \plus{} ca(b \plus{} 1 \minus{} y_0) \plus{} ab(c \plus{} 1 \minus{} z_0)$,with $ 1 \minus{} x_0,b \plus{} 1 \minus{} y_0$ and $ c \plus{} 1 \minus{} z_0$ positive_but this is impossible because $ n$ is stubborn.therefore $ x > 0$ and $ f(n)$ is not stubborn.
to prove the "if" direction,suppose for the sake of contradiction that $ f(n)$ is not stubborn and that $ n$ is not stubborn as well.write $ f(n) \equal{} bcx_0 \plus{} cay_0 \plus{} abz_0$ and $ n \equal{} bcx_1 \plus{} cay_1 \plus{} abz_1$ for positive integers $ x_i,y_i,z_i$ then

$ 2abc \equal{} bc(x_0 \plus{} x_1 \minus{} 1) \plus{} ca(y_0 \plus{} y_1 \minus{} 1) \plus{} ab(z_0 \plus{} z_1 \minus{} 1)$

write $ x \equal{} x_0 \plus{} x_1 \minus{} 1$ and define $ y$ and $ z$ similarly.taking the above equation modulo $ a$ shows that $ 0\equiv bcx\pmod a$.because $ bc$ is relatively prime to $ a$,$ x$ must be divisible by $ a$,implying that $ x\geq a$.similarly $ y\geq b$ and $ z\geq c$ thus $ 2abc \equal{} bcx \plus{} cay \plus{} abz\geq 3abc$,a contradiction.
to summarize: the $ bc \plus{} ca \plus{} ab \minus{} 1$ positive integers less than $ bc \plus{} ca \plus{} ab$ are stubborn,every integer greater than $ 2abc$ is not stubborn, and half of the $ 2abc \minus{} (bc \plus{} ca \plus{} ab) \plus{} 1$ remaining positive integers are stubborn.this yields a total of

$ bc \plus{} ca \plus{} ab \minus{} 1 \plus{} \frac {2abc \minus{} (bc \plus{} ca \plus{} ab) \plus{} 1}2 \equal{} \frac {2abc \plus{} bc \plus{} ca \plus{} ab \minus{} 1}2$

stubborn positive integers.

That is the official solution. When i took part in this TST, i heard that there was another solution basing on geometry. Anyone knows?

some time ago I posted a generalization of this
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=787949636&t=147909
you could try.

It 's not at all a generalization!!!!

The crucial point of this TST question is to discover that n is stubborn if and only if  2.abc +ab+bc+ca -n is not stubborn (*)

This question worth 7 points, if u show that all the integer greater or equals 2.abc are not stubborn, u get only 1 point. Anyone discover
(*) will have complete solution ( or with a few minor errors) which worths 7 points

what are you talking about? in the other thread I asked for the number of "generalized stubborn" numbers. So there is a generalization.


conics
parabola
geometry
perpendicular bisector
geometry unsolved

Let $ P$ be an arbitrary point of the plane such that $ d(P,d) \plus{} PA \equal{} MN \ \ (1)$

We will first find the locus of $ P$ and then we will intersect this locus with the line $ r$

From $ (1)$, it is clear that $ \boxed{d(P,d) \leq MN\text{ and }PA \leq MN}\ \ (2)$

On the half-plane that it is defined by $ d$ and $ A$, we construct line $ q\parallel d$ such that $ d(q,d)\equal{}MN$
Then it is $ d(P,q) \plus{} d(P,d) \equal{} MN$

Now, from $ (1)$ we get $ d(P,q) \equal{} PA$

So the locus of $ P$ is the parabola with directrix $ q$ and focus $ A$

Now we must find the intersection points between the parabola and the line $ r$

[color=brown][size=125]Problem[/size]
Given the directrix $ q$ and the focus $ A$ of a parabola, and a line $ r$, construct the intersection points by a ruler and a compass[/color]

If $ r\perp q$ then it's easy. Let $ B\in r \cap q$, then we bring the perpendicular bisector of $ AB$, we intersect it with $ r$, and this is the point we're looking for

So suppose that $ r,q$ are not perpendicular
Through $ A$ we draw a perpendicular line to $ r$. This line intersects $ r$ at $ H$ and the directrix at $ C$
We draw a circle $ (C)$ with radius $ CH^2\minus{}HA^2$. This circle intersects $ q$ at the points $ E',F'$
Finally, we construct the points $ E,F$ on $ r$ such that $ E',F'$ are the projections of $ E,F$ onto $ q$

$ E'E^2 \equal{} EC^2 \minus{} E'C^2 \equal{} (EH^2\plus{}CH^2) \minus{} (CH^2\minus{}HA^2) \equal{} EH^2 \plus{}HA^2 \equal{} EA^2$,
so $ E$ belongs to parabola, similarly for $ F$

[color=blue]Notes:
1) If $ CH\equal{}HA$ we have one common point, if $ CH<HA$ there's no solution
2) If some of the points $ E,F$ does not satisfy the conditions $ (2)$ then it is rejected[/color]


How do I manually change all margins to 1in in LaTeX?

Hi,

you could use the [url=http://ctan.org/pkg/geometry]geometry-package[/url]:
[code]\usepackage[margin=1in,nohead]{geometry}[/code]
See its [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]documentation[/url].

Stefan


inequalities
function
inequalities proposed

The correct variant is:

$\frac{(a-b)(a-c)}{a^3+abc}+\frac{(b-c)(b-a)}{b^3+abc}+\frac{(c-a)(c-b)}{c^3+abc}\geq0$

[quote="silouan"]can anyone continue this?[/quote]
Valentin-Schur again ;)

Which form of Vornicu-Shur? (It has many types)

The most basic form.
[quote="Arne"]Vornicu-Schur is the "correct" name :D[/quote]
 :blush:  :P  :wallbash_red:

We have:
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{b+c}{a^{2}+bc}+\frac{c+a}{b^{2}+ca}+\frac{a+b}{c^{2}+ab}\Leftrightarrow\]
\[\frac{1}{a}-\frac{b+c}{a^{2}+bc}+\frac{1}{b}-\frac{c+a}{b^{2}+ca}+\frac{1}{c}-\frac{a+b}{c^{2}+ab}\geq0\Leftrightarrow\]
\[\frac{(a-b)(a-c)}{a^3+abc}+\frac{(b-c)(b-a)}{b^3+abc}+\frac{(c-a)(c-b)}{c^3+abc}\geq0.\]
The last one is true, because is Vornicu-Shur inequality for convexe function $f(x)=\frac{1}{x^3+abc}.$
[b]Vornicu-Shur[/b] inequality is:
\[f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)\geq0,\]
for all $a,b,c\in R$ and $f: R\rightarrow R_+$, increase function on R, or convexe function on R.

Noting that when $a \ge b \ge c$ then $a^3+abc \ge b^3+abc \ge c^3+abc$ is also okay.


ABC is a triangle right angled at A. E and D are points on BC such that CE^2 + BD^2 = DE^2.
Prove that [u]/[/u] EAD = 45 degree.  :?: [/u]

But the other problems were only for isoceles right triangles, and this is for any right triangle. Is the result still true? I don't think it is.

No, it is not true!

best regards
sunken rock