The convex hexagon have the property that on his three odd sides we can constructe three rectangles with some same vertexes,as it is shown in the  picture.Prove that we can constructe suche three rectangles on the other three sides too.

This figure is quite classical.

Suppose that $MNPQRS$ is the hexagon satisfying this condition. It can be shown that for any point $T$ you will have $TM^{2}+TP^{2}+TR^{2}= TN^{2}+TQ^{2}+TS^{2}$.

Now we just need to work backward by noticing that given 4 points $A, B, C, D$ such that for any point $T$, the equality $TA^{2}+TC^{2}= TB^{2}+TD^{2}$ holds, then $ABCD$ is a rectangle.


Solve for $ x,y$ such that $ 2x > y > x$ ,if
$ 2(2x\minus{}y)^2 \equal{} (y\minus{}x)$

[quote="geniusbliss"]Solve for $ x,y$ such that $ 2x > y > x$ ,if
$ 2(2x \minus{} y)^2 \equal{} (y \minus{} x)$[/quote]

[hide]Let $ z\equal{}y\minus{}x$, so $ 0<z<x$ and $ 2(x\minus{}z)^2\equal{}z$. Solving for $ z$ using the quadratic formula gives:
\[ z\equal{}\frac{4x\plus{}1\pm\sqrt{8x\plus{}1}}4\]
The positive sign gives $ z>x$, so take the negative sign. For $ z$ to be an integer, $ 8x\plus{}1\equal{}(4k\plus{}1)^2$ for some $ k$. Solving for $ x$ gives $ x\equal{}2k^2\plus{}k$ for some $ k$, so $ z\equal{}2k^2$, so $ \boxed{(x,y)\equal{}(2k^2\plus{}k,4k^2\plus{}k)}$ for $ k\in\mathbb{N}$
[/hide]


[quote="toanhocmuonmau"]Let $a,b,c \in [0,2\sqrt{3}-3]$ such that $a+b+c=1$. Prove that
\[12abc \ge \frac{4(a-bc)(b-ca)(c-ab)}{(a+b)(b+c)(c+a)}+(a+b)(b+c)(c+a) \]
:)[/quote]
does someone have a nice solution for it?  :?:


function
functional equation
algebra
IMO Shortlist

Let $ T$ denote the set of all ordered triples $ (p,q,r)$ of nonnegative integers. Find all functions $ f: T \rightarrow \mathbb{R}$ satisfying
\[ f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\
1 + \frac{1}{6}(f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\
+ f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\
 + f(p,q + 1,r - 1) + f(p,q - 1,r + 1)) & \text{otherwise} \end{cases}
\]
for all nonnegative integers $ p$, $ q$, $ r$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Note that $ g(x,y,z): \equal{} \frac {3xyz}{x \plus{} y \plus{} z}$, where $ (x,y,z) \in T$, is a solution.  Let $ h: \equal{} f \minus{} g$.  Then, $ h$ satisfies: for $ (p,q,r) \in T$,
\[ h(p,q,r) \equal{} \begin{cases} 0\,, & \text{if} \; pqr \equal{} 0, \\
\tfrac{1}{6}\Big(h(p \plus{} 1,q \minus{} 1,r) \plus{} h(p \minus{} 1,q \plus{} 1,r) & \\
\plus{} h(p \minus{} 1,q,r \plus{} 1) \plus{} h(p \plus{} 1,q,r \minus{} 1) & \\
\plus{} h(p,q \plus{} 1,r \minus{} 1) \plus{} h(p,q \minus{} 1,r \plus{} 1)\Big)\,, & \text{otherwise.} \end{cases}
\]
Now, on the plane $ x \plus{} y \plus{} z \equal{} k$, if $ (p,q,r) \in T$ does not lie on the boundary (i.e., $ pqr > 0$), then $ h(p,q,r)$ is the average of the values of $ h$ at the neighboring points of $ (p,q,r)$ on this plane.  This implies that the extreme values of $ h$ restricted on this plane are attained at the boundary.  Therefore, both the maximum and the minimum of $ h$ restricted on the plane $ x \plus{} y \plus{} z \equal{} k$ are zero.  This result holds for any nonnegative integer $ k$.  Thus, $ h$ is identically zero, and, consequently,
\[ f(x,y,z) \equal{} \frac {3xyz}{x \plus{} y \plus{} z}
\]
is the only solution to this functional equation.

[quote="peregrinefalcon88"]What about $ g(x,y,z): =\frac{3xyz}{x+y+z} +k(x+y+z)$? i think your proof saying the min and max are zero is wrong, at x+y+z=k the function h is constant so the function above should be the solution?[/quote]

Does your function satisfy $g(x,y,z)=0$ if $xyz=0$?  My only mistake is failing to mention that $f(0,0,0)=0$, as $f(x,y,z)=\frac{3xyz}{x+y+z}$ is not defined when all its arguments are zero.

you are correct, i thought the function would have the same value everywhere for x+y+z=k but i forgot that clearly some of the values are 0 making everything 0 as in your solution.

[hide=Solution]We claim that the only solution is $f(x, y, z) = \frac{3xyz}{x+y+z}$(with $f(0, 0, 0) = 0$. We can easily check that it works, since
$(p+1)(q-1)r+(p-1)(q+1)r=2pqr-2r$ so analogously for the other $2$ sums, we have the sum is $\frac{18pqr-6p-6q-6r}{p+q+r}$, so when we divide by $6$ and add $1$, we indeed get $\frac{3pqr}{p+q+r}$, as desired.

Now, note that it must be the only solution, because the function is equivalent to the expected value of the number of moves to get one coordinate equal to $0$ starting from $(p, q, r)$ and choosing uniformly at random whether to move to $(p+1, q-1, r), (p-1, q+1, r), (p, q-1, r+1), (p, q+1, r-1), (p+1, q, r-1), (p-1, q, r+1)$.[/hide]

What is the motivation behind constructing 3xyz/ (x+y+z) in the first place? How do you notice this?

@above for small values of $x+y+z$ one can actually pin down the value of $f$ at every point. Doing this enough makes the pattern evident and the solution guessable.

Note that $f(x,y,z)=\frac{3xyz}{x+y+z}$ works.

Now, suppose there was some other $g(x,y,z)$ that worked. Letting $h=f-g$, we see that $h(x,y,z)=0$ if $xyz=0$ and otherwise is simply the average of its six neighbors in the plane its in that is normal to $(1,1,1)$. In any such plane $x+y+z=c$ with $x,y,z\ge 0$, there are a finite number of points and the boundary is all $0$. Therefore, if $h(x,y,z)$ is the minimum value, then it must be equal to all its neighbors, so proceeding in this way gives that $h$ is constant and so $h=0$. Thus, $g=f$, as desired.

[b]Remark[/b] This solution is motivated by electrostatics.

It is easy to see that $f(x,y,z) = \frac{3xyz}{x+y+z}$ works. If some other function $g(x,y,z)$ worked, then $f-g$ is the average of its ``neighbors''. Consider the minimum of $f-g$; it cannot be the average unless all its neighbors are also the minimum. Continuing in this way, we get that $f-g$ is a constant. By the boundary conditions, we then know $f-g=0$. So $g=f$. Hence no other solutions exist.

Solution with quite a few others:

First, how do we find a function that works? We try to write down $f(x,y,z)=xyz$, but this fails, as we get by using the second condition
\[f(x,y,z)=xyz-\frac 13(x+y+z)+1\]
It's not clear how to construct another function in $x,y,z$ which is zero when $xyz=0$ when trying to add another factor on. For example, $pqr-k(p+q+r)$ doesn't even work, because we would get $\frac 13=0$. Adding a constant term won't help either. The best thing to try is multiplying or dividing. In addition, we want the $x+y+z$ and $1$ to cancel, so we would ideally divide (as the 1 is not included in the division), and thus too by a multiple of $x+y+z$. This gives us:
\[f(x,y,z)=\frac{xyz}{k(x+y+z)}\implies f(x,y,z)=\frac{xyz-\frac 13(x+y+z)}{k(x+y+z)}+1\]
Another way to see we should choose $x+y+z$ is because we want the denominator to be the same, to avoid complications, and the best thing to try is $x+y+z$. This works perfectly given $k=\frac 13$, so we choose
\[f(x,y,z)=\frac{3xyz}{x+y+z}\]
which works (other than $(x,y,z)=(0,0,0)$, where we already know it is 0).

Now, how do we resolve the rest? The most intuitive thing to do is to do the dumb thing - try $g=f-\frac{3xyz}{x+y+z}$, and this helps because we now have that $g$ is the average of its $6$ neighbors in the same plane ($x+y+z=k$, $k$ is an integer):
[asy]
import cse5;
import olympiad;
size(5cm);
filldraw((3,0)--(0,3)--3*dir(60)--cycle, blue*0.3+white, blue);
D((-5,0)--MP("x",(5,0),N), black, Arrows);
D((0,-5)--MP("z",(0,5),E), black, Arrows);
D(5*dir(240)--MP("y",5*dir(60),dir(150)), black, Arrows);
[/asy]
You can visualize it as such:
[asy]
import olympiad;
import cse5;
pointpen=black;
size(5cm);
D(D(dir(0)) -- D(dir(60)) -- D(dir(120)) -- D(dir(180)) -- D(dir(240)) -- D(dir(300)) -- cycle, blue);
D((0,0),red);
[/asy]
The red dot is the center, and is the average of the six black points around it (this is in the equilateral plane of $x+y+z=k$, for fixed $k$). Now, that means that we can't have a maximum. Indeed, let $\mathcal S=\{(x,y,z)\in\mathcal T\mid x+y+z=k\}$, and assume $(x_0,y_0,z_0)$ was the local maximum. However, we note
\[g(x,y,z)=\frac 16\sum g(\text{neighbor})\leq\frac 16\sum g(x,y,z)=g(x,y,z)\]
so all inequalities are equalities and $g(x,y,z)=g(\text{neighbor})$. Continuing in this way, we see $g$ is constant over $\mathcal S$, but as the value of $g$ at $(k,0,0)$ is $0$, we get $g=0$ and thus we're done.

Note: the reason it's done this way is because it's very intuitive to try to construct a function, which turns out to be very hard, and the core of the problem. The rest of the problem is finished because it suspiciously looks like an average, which it turns out to be.

Note that for each $k$ such that $x+y+z=k$, there is a unique value for $f(x, y, z)$ because we have that all such values make one different equation. This means that we have that there is a unique solution to $f$. We note that if $f(x, y, z) = \frac{3xyz}{x+y+z}$, then $$\sum_{\text{sym}}f(x+1, y, z-1)=\sum_{\text{sym}}\frac{3(x+1)y(z-1)}{x+y+z} = \frac{18xyz}{x+y+z}-6 = 6\left(f(x, y, z)-1\right)$$ which satisfies the given condition, meaning that $f(x, y, z) = \boxed{\frac{3xyz}{x+y+z}}$ is our only solution.

didn't find the actual function (oops)
but can someone check if my proof that only one function exists is right?

basically, define a game with three players, P, Q, R, where P, Q, and R have p, q, r points respectively
then each move, a loser and winner are picked randomly, and the loser gives a point to the winner

then f(p, q, r) is just the expected value till someone hits 0 pts

since this value can't vary for the same value p, q, r, f(p, q, r) only has one possible function

OMG HI CENTS

We claim that $f(p,q,r)=\frac{3pqr}{p+q+r}$ for all positive integer $p,q,r$, and $f(p,q,r)=0$ when at least one of $p,q,r$ is zero is the only function that works. It is easy to check that it works, so we focus on proving that it is the only one.

Given a point $(p,q,r)$, let it's [i]coordsum[/i] $S$ be $p+q+r$. Notice that the given equation for $f(p,q,r)$ only includes points with the same coordsum. Using the equations for every point with positive integer coordinates with coordsum $S$, we can get a system of $n$ variables in $n$ equations where $n$ is the number of points with coordsum $S$ and positive integer coordinates. Consider the lattice points with nonnegative integer coordinates on the plane $$x+y+z=S.$$ Notice that these lattice points form an equilateral lattice, with the points with at least one coordinate zero making the outside equilateral triangle. This is illustrated with the diagram below. Now consider the following question:

Suppose an ant starts on the point $(p,q,r)$ on the graph of $x+y+z=S$. Every move, the ant moves to one of the six lattice points closest to it. What is the expected number of moves that the ant will take to reach a point with at least one coordinate zero?

Notice that this is precisely $f(p,q,r)$, as writing out the states equations needed to solve this, we find that these are exactly the same as the system of $n$ equations in $n$ variables earlier. Therefore the value of $f(p,q,r)$ is fixed for all positive integers $p,q,r$, and therefore at most one solution can exist. As we have already found this solution, we are done.

Note that there is at most one solution because $f$ is uniquely determined as the expected number of moves to reach $xyz=0$ under a defined set of possible moves.

I now claim $f(x,y,z)=\frac{3xyz}{x+y+z}$ where $f(0,0,0)=0$ works. If $pqr=0$, then $f(p,q,r)=0$. Next, the recursive relation gives
\[f(p,q,r) = 1+\frac{3(p+1)(q-1)r+3(p-1)(q+1)r+3(p-1)q(r+1)+3(p+1)q(r-1)+3p(q+1)(r-1)+3p(q-1)(r+1)}{6(p+q+r)}\]
\[=1+\frac{3(6pqr-2p-2q-2r)}{6(p+q+r)}=\frac{3pqr}{p+q+r}\]
Thus, this choice of $f$ works.

The answer is $\boxed{f(x,y,z)=\frac{3xyz}{x+y+z}}$.

im sobad

Note that whatever function it is, it must be unique, since the function is asking the following:

On the plane $x+y+z=p+q+r,$ starting with point $(p,q,r)$ what is the expected value of the number of moves of length $\sqrt{2}$ the point needs to make in order to have $pqr=0.$ There is only one answer to a question like this. Thus, since \[f(p,q,r)=\frac{3pqr}{p+q+r}\] works, we're done.

Note that $\boxed{f(p,q,r)=\begin{cases}\frac{3pqr}{p+q+r}&\text{if }pqr\ne0\\0&\text{if }pqr=0\end{cases}}$ is a solution. We claim it is the only one.

Let $g(p,q,r)=f(p,q,r)-\frac{3pqr}{p+q+r}$ if $pqr\ne0$. The assertion is equivalent to the fact that $g(p,q,r)$ is the average of $g(p+1,q-1,r)$, $g(p-1,q+1,r)$, $g(p-1,q,r+1)$, $g(p+1,q,r-1)$, $g(p,q+1,r-1)$, and $g(p,q-1,r+1)$ if $pqr\ne0$.
Let $(x,y,z)\in T$ be any triple such that $g(x,y,z)$ is maximal. Then:
$$g(x,y,z)=\frac{g(x+1,y-1,z)+g(x-1,y+1,z)+g(x-1,y,z+1)+g(x+1,y,z-1)+g(x,y+1,z-1)+g(x,y-1,z+1)}6\le\frac{6g(x,y,z)}6=g(x,y,z)$$
so equality holds and these values of $g$ are all equal. By induction $g$ must be constant, and since $g(0,0,0)=0$, $g(p,q,r)=0$ proving our claim.

The answer is $f(x,y,z)=\frac{3xyz}{x+y+z}$ only, which can be verified to work with direct substitution. Now, note that $f$ is unique, since $f(x,y,z)$ counts the expected number of moves necessary to make one variable zero from $(x,y,z)$, where each move consists of subtracting $1$ from one element and adding $1$ to a different one. $\blacksquare$

We can find that $\boxed{f(p,q,r)=\frac{3pqr}{p+q+r}},$ is a solution and if $pqr=0,$ then $\boxed{f(p,q,r)=0},$ is another fitting solution.
-----------

[b][u]Claim: No other solution exists. [/b][/u]
[i]Proof.[/i] Assume the contrary that $f_1(p,q,r)$ is another solution. Also let $f(p,q,r)=f_1(p,q,r)+f_2(p,q,r).$ Then $f_2(p,q,r)$ satisfies, $f_2(p,q,r)= \begin{cases} \tfrac{1}{6}\{f_2(p + 1,q - 1,r) + f_2(p - 1,q + 1,r) & \\ + f_2(p - 1,q,r + 1) + f_2(p + 1,q,r - 1) & \\ + f_2(p,q + 1,r - 1) + f_2(p,q - 1,r + 1)\}.\end{cases}$ We have that all values of $f_2(p,q,r)$ is equal to the 6 points around it which implies that for any particular $p+q+r$, $f_2(p,q,r)$ is constant. So for some values of $f_2$ are always $0\implies f_2 \equiv 0.$ And the claim follows. $\blacksquare$

We claim the only solution is $f(x,y,z)=\frac{3xyz}{x+y+z}$ if not all of $x,y,z$ are $0$ and $0$ otherwise. Assume there is another solution $g(x,y,z)$. Then $h=g-f$ is the average of all of its neighbors. Let the minimum of $h$ be achieved at some point. Then all its neighbors are also the minimum and continuing in this fashion we get $h$ is constant and we have $h(0,0,0)=0$, so $f=g$ and we are done.

We claim that the answer is $$f(p,q,r)=\frac{3pqr}{p+q+r}$$ unless $p=q=r=0$, then $f(0,0,0)=0$. This clearly works, and is unique because this is equivalent to the answer to the following problem via states:
Let Alice, Bob, and Charlie have $p,q,r$ chips, respectively. A move involves a random person out of the three giving a chip to another random person, and the game ends when someone has no chips left. Find the expected number of moves before the game ends.
This expected value is clearly unique, so we are done.

Note that $f(p,q,r)$ is the expected number of moves starting from $(p,q,r)$ needed to reach a coordinate axis if in each move, it is moved in the three directions a permutation of $(1,0,-1)$, so the function is obviously unique. Since $$f(p,q,r)=\frac{3pqr}{p+q+r}$$ works, it is the only solution

The following rephrasing of the problem trivializes it:
[quote=Rephrasing, from WOOT]At the beginning of a game, three players start with $a$, $b$, and $c$ chips, respectively. On each turn, two players are picked at random, and the first player gives a chip to the second. This continues until one of the players runs out of chips. Find the expected value of the number of turns until the game stops.[/quote]
Such a function that models the expected value is clearly unique, and furthermore $$f(p, q, r) = \frac{3pqr}{p+q+r}$$ works, so it is the only solution.

Note that the problem is equivalent to $3$ players playing a game with $p,q,r$ chips, and a random player giving a chip to another random player on each move, and $f(p,q,r)$ is the expected number of moves before the game ends. Then there is clearly only one function for this expected value, and $$f(p,q,r)=\frac{3pqr}{p+q+r}$$ works, it is the only solution. $\blacksquare$

The solution is unique, the equation is an expected value equation. It is not hard to check $f(x,y,z) = \frac{3xyz}{x+y+z}$ works.

I claim the solution is unique - indeed, $f(x,y,z)$ can be expressed as the following: [quote=rephrased problem]Suppose that 3 piles have $x,y,$ and $z$ chips respectively. Suppose in a move, one randomly selects a pile and moves a single chip from that pile to any other pile. Then $f(x,y,z)$ counts the expected number of moves until one of the piles becomes empty.[/quote]

Clearly, this works, and since this expected value is unique, $f(x, y, z)$ is unique. One can now easily check that \[f(x,y,z) = \frac{3xyz}{x+y+z}\] is the unique solution.

We will show that the only solution is $f(p,q,r)=\frac{3pqr}{p+q+r}.$

Suppose there were two distinct solutions $f(p,q,r)$ and $f'(p,q,r).$ Let $g(p,q,r)=f(p,q,r)-f'(p,q,r).$ Then, we have $g(p,q,r)=0$ if $pqr=0$ and $g(p,q,r)$ is the average of its six neighbors in the plane $x+y+z=p+q+r.$ We must have $g(p,q,r) \ne 0$ for some $(p,q,r).$ Consider the points $(x,y,z)$ satisfying $x+y+z=p+q+r$ and with $|g(x,y,z)|$ maximized, and of those, consider one with minimal $x.$ Let this point be $(i,j,k).$ Notice that $i \ne 0.$ Then, notice that for any $a,b,c,d,e,f$ we have $|a+b+c+d+e+f| \le |a|+|b|+|c|+|d|+|e|+|f|,$ or the absolute value of their average is less than or equal to the average of their absolute values. However, notice that $|g(i-1,j,k+1)|<|g(i,j,k)|$ by our choice of $(i,j,k),$ and so the average of the absolute values of the outputs of $f$ across the neighbors of $(i,j,k)$ in the plane $x+y+z=p+q+r$ is less than $|g(i,j,k)|.$ However, the absolute value of their average is by definition $|g(i,j,k)|,$ implying that $|g(i,j,k)|<|g(i,j,k)|,$ which is a contradiction. Therefore, $f$ is unique.

We can easily check $f(p,q,r)=\frac{3pqr}{p+q+r}$ indeed works, so we are done.

Many solutions notice that this is the transition equation for the expected number of steps for a game with 3 players with respectively $p,q,r$ chips in front of them, and randomly every turn someone gives another person a chip, with the game ending when some one player has 0 chips left.

I'll write out how you would compute the answer form given this fact, without guesswork.

Let $(X_n,Y_n, Z_n)$ be the number of chips in front of each player after $n$ steps, where we initialize $(X_0, Y_0, Z_0)=(p,q,r)$. We now note that $S_n=X_nY_nZ_n+\frac{1}{3}n(p+q+r)$ is a martingale. Indeed,

$$\mathbb{E}[X_{n+1}Y_{n+1}Z_{n+1}+(n+1)(p+q+r)/3]$$
$$= \frac{1}{6}((X_n+1)(Y_n-1)Z_n+(X_n-1)(Y_n+1)Z_n+X_n(Y_n+1)(Z_n-1)+X_n(Y_n-1)(Z_n+1)+(X_n-1)Y_n(Z_n+1)+(X_n+1)Y_n(Z_n-1))+(n+1)(p+q+r)/3$$
$$=X_{n}Y_{n}Z_{n}+n(p+q+r)/3$$.

Let $T$ be the stopping time. Then, by the optional stopping time theorem, we have $\mathbb{E}[S_T]=\mathbb{E}[S_0]$, and so
$$\mathbb{E}[X_TY_TZ_T]+\mathbb{E}[T](p+q+r)/3=X_0Y_0Z_0+0\iff$$
$$0+\frac{p+q+r}{3}\mathbb{E}[T]=pqr +0\iff$$
$$\mathbb{E}[T] = \frac{3pqr}{p+q+r}$$

as in all the other solutions.

We claim that the answer is $f(x, y, x) = \boxed{\frac{3xyz}{x+y+z}}$

We can see that this satisfies the first criteria and the second is also true.

We claim that there is only $1$ possible value for $f(x, y, z)$. which obviously completes the problem

FTSOC assume that $f(x, y, z)$ can have $n > 1$ different values. Let $f(x-1, y+1, x)$ take on $p > 1$ different values and $f(x-1, y, z+1)$ take on $q > 1$. Notice that these are symmetric thus, WLOG let $n \leq p \leq q$. Notice that we have \[f(x, y, z) = f(x-1,y+1,z) + f(x+1,y-1,z)+ f(x-1,y,z+1) + f(x+1,y,z-1)+ f(x,y-1,z+1) + f(x,y+1,z-1)\]

Thus the only way that we can have $n \leq p \leq q$ is when $n = p = q > 1$. However, we still have at least $p+q-1$ different possible values for $n$. Thus there exist no such triples $n, p, q$ unless $n = p = q = 1$. $\blacksquare$


Well quite interesting facts which i learned recently :

Let us devide the set of natural numbers to $n$ different arthmetic sequances like $P_1,...,P_n$ such that
$P_i \cap P_j=\emptyset,\cup P_i=\mathbb{N}$

Let $P_i: : A_{i+1}=A_i+d_i$ .

1.Prove that

$\sum_{i=1}^n \frac{1}{d_i}=1$

2.prove that 2 of $d_i$s are equal .

This one already appeared several times on ML.
1) follows immediately by density considerations.
2) is equivalent to a famous problem of an old russian olympiad which states that if we divides the set of vertices of a regular $n$-gon into disjoint sets, each of them being formed by the vertices of some regular polygon then, at least two of these regular sub-polygons are isometric. The key step is to consider a minimal sub-polygon.

Pierre.

Both of them follow immediately using formal series.  ;)


quadratics
algebra
binomial theorem
number theory unsolved
number theory

Find all positive integer (x, m, n) when they satisfy

$ (x \plus{} 2)^{2m} \equal{} x^n \plus{} 2$.

[hide]We claim there are no solutions in positive integers.

First, if $ n \equal{} 1$, there are clearly no solutions. Now, we show that $ x$ is odd. If $ x$ were even, taking this equation mod 4, since $ 2m, n > 1$ , we get $ 0 \equiv 2 \mbox{ mod 4}$. Therefore, $ x$ is odd.

In addition, $ n$ is odd because the difference of two quadratic residues mod 4 cannot be 2.

Now, we claim that if $ (2^k v \plus{} 1)^{2m} \equal{} (2^k v \minus{} 1)^n \plus{} 2$ for any positive integer $ k$ and any positive integer $ v$, then $ v$ must be even. Consider this equation mod $ 2^{k \plus{} 1}$. By the binomial theorem, $ (2^kv \plus{} 1)^{2m} \equiv 1 \plus{} (2m)2^k v \equiv 1 \mbox{ mod } 2^{k \plus{} 1}$. Since $ n$ is odd, by the binomial theorem, the RHS is $ 2 \plus{} (2^k v n \minus{} 1) \equiv 1 \plus{} 2^k vn \mbox{ mod } 2^{k \plus{} 1}$. Equating the two, $ vn \equiv 0 \mod 2$. Since $ n$ is odd, $ v$ is even, as desired.

Since $ x$ is odd, we may let $ x \equal{} 2u \minus{} 1$. We observe that setting $ k \equal{} 1$ and $ v \equal{} u$ is exactly our original equation. Thus, by our claim, $ u$ is even. But then, setting $ k \equal{} 4$ and $ v \equal{} u/2$ again gives our original equation, and by our claim, $ u/2$ must also be even. We repeat this process indefinitely; it follows that every power of 2 divides $ u$, so $ u \equal{} 0$, a contradiction. Thus, we are done.[/hide]

Hopefully I didn't mess up. I must say, this is a prety nice problem.


wat do u mean? I don't understand the problem...

[quote="anirudh has successfully translated and"]Find the time on a 12-hour clock in which the degree of the minor angle of that time equals 42.96.[/quote]

[hide]first take off the 42.
you are left with .96 of a degree.
multiply .96 by 60 to get 57.6
then take off the 57 to get .6
multiply .6 by 60 to get 36.
so 42.96 degrees = 42 deg 57 min and 36 sec
[/hide]

btw, anirudh, your "translation" has a problem
because there an infinite amoun of answers, the time
on the clock could be [b]absolutely[/b] anything, as long as the minor angle is 42.96

I haven't touched this is such a while...  :blush:

but yes, minutes and seconds are ways of writing angles, correct? Not only time?

Convert to what? Radians?

That wasn't mentioned so I translated it.

[quote="mathgeniuse^ln(x)"][quote="jenny221"]Convert the angle measures to degrees, minutes, and seconds 42.96'[/quote]

We know that 42.96=42 degress+blah

Now we know that .96 degrees equals.96*60=57.6 minutes

0.6 minutes=36 seconds, so the answer is

$42.96^\circ=42^\circ 57' 36''$[/quote]
Actually, the original text looks like 42.96 [i]minutes[/i] not degrees. But I'm not sure...
[hide].96 minutes = 57.6 seconds
$42'57.6''$[/hide]


Two towns, A and B, are connected by a road. At sunrise, Pat begins biking from A to B along this road, while simultaneously Dana begins biking from B to A. Each person rides at a constant speed, and they cross paths at noon. Pat reaches B at 5 pm, while Dana reaches A at 11:15 pm. When was sunrise?

[hide] well label the point where they met as C, since they started at the same time we know it took the same time for them to get there.  Now after noon it took one 5 hours and one 45/4 hours to get to their respective destinations.  And since the rates were constant, we know the ratio of speeds was the same before and after they met at C, so let x be the amount of time traveled before noon.  5/x = x/(45/4)  so x^2 = 225/4 and x = 15/2 so 7 and a half hours before noon is 4:30 am.[/hide]

[hide]Let us call the time elapsed from $A$ to $C$ as $x$ now, since Pat starts at $A$ and her speed is constant, the ratio of her traveling from $A$ to $C$ and $C$ to $B$ should be constant. The same should occur for Dana from $B$ to $C$ and $C$ to $A$. For this reason we can set up the following proportion: $\frac{x}{5}$ $=$ $\frac{45/4}{x}$. Using cross products we get $x^{2}$ $=$ $\frac{225}{4}$. So, since it has to be a positive value for $x$, we get $x$ $=$ $\frac{15}{2}$ $=$ $7.5$. $12$ $-$ $7.5$ $=$ $4.5$. So sunrise was at $4: 30$ AM[/hide]


logarithms
real analysis
real analysis solved

compute the limit:
log_p (2^(1/p^n)+3^(1/p^n)+...+(p^n+1)^(1/p^n))

thank you very much! :D

I think he wants the limit of this sequence when n->00

I might be wrong. The limit of this thing is obviously 00 because
k^(1/p^n)>1 for all k>1, so the sum after the logarithm is > p^n because it has p^n terms, all greater than 1, so
log_p(that thing)>log_p(p^n)=n -> 00.

Maybe we have to find a different limit.

Maybe the sequence is w(n)=ln_{p^n}(thing)=ln(thing)/ln(p^n)


geometry
circumcircle
incenter
geometry proposed

[color=darkblue]Let $ O$ , $ I$ , $ I_a$ be the circumcenter, the incenter and the $ A$ - exincenter of $ \triangle ABC$ respectively.

The perpendicular line from $ I$ on the line $ AO$ cut the lines $ AB$ , $ AC$ in the points $ M$ , $ N$ respectively.

The perpendicular line from $ I_a$ on the line $ AO$ cut the lines $ AB$ , $ AC$ in the points $ P$ , $ Q$ respectively.

Prove that $ MN \plus{} PQ \equal{} MP \plus{} NQ$ , i.e. the quadrilateral $ MPQN$ is circumscriptibly.[/color]   [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=197053][size=117][color=red][b]See and here ![/b][/color][/size][/url]

AI meet the circumcircle in D. So $ CD\equal{}ID$ cause $ \angle ICD \equal{} \angle CID \equal{} \frac {\alpha}2 \plus{} \frac {\gamma}2$. Cause $ \angle ICI_a\equal{}\angle IBI_a \equal{} \frac{\pi}2$ we have that $ BICI_a$ is cyclic with center the point where the axes of CI meet $ II_a$ and so D. Then the line PQ and the line MN have the same distance from D. Call H the projection of D on MN and K of D on AB. Then the triangles $ \triangle DBK$ and $ \triangle DIH$ are equals cause $ \angle DIH \equal{} \frac{\alpha}2 \minus{} 90 \plus{} \beta \equal{} 90 \minus{} \frac{\alpha}2 \minus{} \gamma \equal{} \angle BDK$. So the line MN and the line MP have the same distance from D. Then MPQN is circumscriptible in a circle with center D.


Well I'm so interested in Math Olympiad but I don't have enough knowledge in it.Can u give me some website which provides Graph theory exercises and solutions please? By the way my English is not very good. :blush:
           Thank you  :lol:


algebra
polynomial
integration
inequalities
trigonometry
algebra proposed

Problem. (i) Show that there exists a constant $c>1$ such that, for any polynomial
\[
f(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{C}[x],
\]
there exist a point $z$ on the unit circle $|z|=1$ such that $|f(z)| \geq c \sqrt{|a_0a_n|}$.

(ii)* Does $c = 2$ work? [It is easy to see that $c \leq 2$.]

I like very much this one, although I do not understand why it took so long to find a solution for the first question. In fact, using Parseval's identity we have ${\sum_{k=0,n}{|a_k|^2}=\frac{1}{2\pi}\cdot\int_{-\pi}^{\pi} {|f(e^{it})|^2}dt}$. We find that $ ||f||^2_{|z|=1}\geq\sum_{k=0,n}{|a_k|^2}$, which shows that $c=\sqrt{2}$ works very well. Unfortunately, I do not have a solution for $c=2$, although I think the inequality is true for this value. What is your solution for $c=2$, vess?

Actually, $c=2$ works as well (but note that this is optimal possible, as the example of $z^n-1$ shows.) This, as it is so well-known, is true in the case of monic polynomials with roots on the unit circle (another way to put this is that, for any $n$-gon $A_1A_2\cdots A_n$ inscribed in the unit circle, there exists a point $M$ on that circle such that $MA_1 \cdot MA_2 \cdots MA_n \geq 2$, the extremal cases being only the regular $n$-gons.) Therefore, if we consider the transformed polynomial
\[
\widehat{f}(z) := \prod_{f(\alpha)=0} (z - \alpha/|\alpha|),
\]
it is sufficient to show that:

Claim. For any $z$ on the unit circle,
\[
|f(z)| \geq \sqrt{|a_0a_n|} |\widehat{f}(z)|.
\]

Proof.   Let the roots of $f(z)$ be $r_je^{i x_j}$. We have:
\[
\frac{|f(z)|^2}{|\widehat{f}(z)|^2} = |a_n|^2 \prod_j \frac{(1+r_j^2) - 2r_j \cos{x_j}}{2(1 - \cos{x_j})} \geq |a_n|^2 \prod_j \frac{2r_j(1-\cos{x_j})}{2(1-\cos{x_j})} = |a_n|^2 \prod_j r_j = |a_n| \cdot |a_0|,
\]
as claimed.

Because, as remarked earlier, $\sup_{|z|=1} |\widehat{f}(z)| \geq 2$, the assertion holds with $c=2$.

--Vesselin


does anyone know the results of bosnia? :blush:

Bosnian results :

 Nermin Hodzic  12
 Amna Hadziabdic  12
 Edin Boskovic  4
 Milenko Milicevic  7
 Momcilo Kovac  4
 Aleksandar Mladjenovic  1

 :maybe:


To promote her school's annual Kite Olympics, Genevieve
makes a small kite and a large kite for a bulletin board display.
The kites look like the one in the diagram.

[asy]for ( int x = 0; x <= 6; ++x )
{
    for ( int y = 0; y <= 7; ++y )
    {
        dot((x,y));
    }
}
draw((0,5)--(3,7)--(6,5)--(3,0)--cycle);[/asy]

For her small kite, Genevieve draws the kite on a grid with the points one inch apart.  For the large kite
she triples both the height and width of the entire grid.
\\ \\
What is the number of square inches in the area
of the small kite?

[quote="GameBot"]What is the number of square inches in the area
of the [u][b]small [/b][/u]kite?[/quote]

Simple: the small kite is 1/9 the size of the big kite

If you want the actual numerical answer, then just copy from FantasyLover, except don't multiply by 9 so 21. It's kind of like stealing food from a honey badger "Thanks, stupid! Thanks for doing all the work! I'll just take the scraps!"

Sidenote
I am sorry if I offended [i]FantasyLover[/i] by saying [quote="ME"]"Thanks, stupid! ..."[/quote] I did not mean to say that in a pejorative manor, I was just attempting to compare two situations because I thought that it would be comical, but in reality, the situations are rather tangential. I hope that I did not offend you and apologize if I did.

Actually Fantasy Lover is right. It is 189

[quote=FantasyLover]the area of her kite is 6*7/2=21.

and she triples both side and height, so it becomes

21*3*3=189

answer : 189[/quote]

[hide=Solution]You would actually divide that by 9 which is 21 so
[b][u]Answer: 21[/u][/b][/hide]


[quote="harazi"]Salutare. Ia uitati o problema tare draguta propuse de (cine altul) Laurentiu Panaitopol intr-o gazeta matematica foarte veche:
    Sa se afle functiile reale f pentru care $f(x)/x$ tinde la 1 pentru x foarte mare si astfel incat $f(4x)=f(2x)+f(x)+x$[/quote]

Perhaps I misunderstand the problem, but it is trivial if you write
If $f(x) = g(x) + x$   (with  $g(x) = o(|x|)$)  and consider $g(c2^n)$ for constant $c$.
$g(n)$ is a polynomially bounded solution of the Fibonacci recursion for positive [i] and negative[/i] $n$, so $g = 0$.

[quote="harazi"]I think that easy problems can be very nice also. If it were a difficult problem, I would have posted it in the unsolved forum.[/quote]

It is not meant as an insult.  Rather, I was not certain that I understood the problem.

No, I did not interpret it as an insult. And I think you understood the problem. I can translate it in english, if you want to be sure, but I think it's ok.

[quote="fleeting_guest"]
$g(n)$ is a polynomially bounded solution of the Fibonacci recursion for positive [i] and negative[/i] $n$, so $g = 0$.[/quote]
Could you explain this more clear please? All I get is $g(2^nc) = a (1-sqrt(5)/2)^n + b(1+sqrt(5)/2)^n$, where $a$ and $b$ are constants.


trigonometry
geometry
circumcircle
function
cyclic quadrilateral

$ \boxed{1}$ Given a cyclic quadrilateral $ ABCD$ such that $ AB \equal{} AD \equal{} 1,\ \angle{BAD} \equal{} 90^\circ,\ \angle{BAC} \equal{} \theta$. Let $ E$ be the point of the intersection $ AC$ and $ BD.$

(1) Express the lengths of the line segments $ AE,\ BE,\ CE,\ DE$ in terms of $ \theta$.

(2) Express $ \sin \angle{AED}$ in terms of $ \theta$.

(3) Let $ P,\ Q,\ R,\ S$ be circumcenter of triangles $ ABE,\ BCE,\ CDE,\ DAE$ respectively. Find the area of the quadrilateral $ PQRS.$

$ \boxed{2}$ For real constants $ a,\ b$, consider a function $ f(x) \equal{} x^3 \plus{} 2x^2 \minus{} (a^2 \minus{} 3a \minus{} b)x \plus{} a^2 \plus{} ab.$

(1) Factor $ f(x).$

(2) Find the condition of $ a,\ b$ for which an equation $ f(x) \equal{} 0$ has double root.

(3) Find the condition of $ a,\ b$ for which $ f(x)\geq 0$ holds for all $ x\geq 0$, sketch the condition in the $ a \minus{} b$ plane.

[quote="kunny"]
$ \boxed{2}$ For real constants $ a,\ b$, consider a function $ f(x) \equal{} x^3 \plus{} 2x^2 \minus{} (a^2 \minus{} 3a \minus{} b)x \plus{} a^2 \plus{} ab.$

(1) Factor $ f(x).$

(2) Find the condition of $ a,\ b$ for which an equation $ f(x) \equal{} 0$ has double root.

(3) Find the condition of $ a,\ b$ for which $ f(x)\geq 0$ holds for all $ x\geq 0$, sketch the condotion in $ a \minus{} b$ plane.[/quote]

$ \boxed{2}$
[hide="(1)"][b](1)[/b] $ f(\minus{}a)\equal{}0$ , so
$ f(x)\equal{}[x\plus{}a][x^2\plus{}(2\minus{}a)x\plus{}(a\plus{}b)]$ [/hide]

[hide="(2)"] [b](2)[/b][b]Case 1:[/b] $ g(x)\equal{}x^2\plus{}(2\minus{}a)x\plus{}(a\plus{}b)\equal{}0$ has distinct real roots with one of which $ \equal{}\minus{}a$

, i.e. Discriminant$ \equal{}a^2\minus{}8a\plus{}4\minus{}4b>0...........(i)$ and $ g(\minus{}a)\equal{}2a^2\minus{}a\plus{}b\equal{}0...........(ii)$

Substitute $ (ii)$ into $ (i)$ , $ (i)$ becomes $ (3a\minus{}2)^2>0$

So $ (a,b)\equal{}(a,a\minus{}2a^2)$ with $ a \neq \frac{2}{3}$ is the solution for this case.

[b]Case 2:[/b] $ g(x)\equal{}0$ has double root $ \neq \minus{}a$, i.e.

Discriminant$ \equal{}a^2\minus{}8a\plus{}4\minus{}4b\equal{}0...........(iii)$ and $ g(\minus{}a)\equal{}2a^2\minus{}a\plus{}b \neq 0...........(iv)$

Put $ (iv)$ into $ (iii)$ we get $ (3a\minus{}2)^2 \neq 0$

So $ (a,b)\equal{}(a,\frac{a^2}{4}\minus{}2a\plus{}1)$ with $ a \neq \frac{2}{3}$ is the solution for this case. [/hide]

(2) In Case 1, $ D>0$ is needless. In any case of 1, 2, $ a\neq \frac{2}{3}$ is needless.

Isn't that too much for humanities? I wonder how is the real mathematics major  :huh:

I mean Humanities by the total name of the faculty of Law, Economics, Literature, Psychology etc.


How many four-figure numbers ($ \overline{abcd}$) are there satisfying $ a\plus{}d\equal{}b\plus{}c$?

well, normally we use combinatorics or transformations in these, but they dont work here.
i've a totally different approach, so please comment on this.
[i]it would be hard to understand now, but i'll attach a table tomorrow. [/i]
[u]solution[/u]$ \bullet$take the numbers in blocks of 10 i.e.1001-1010,1011-1020...9991-9999
now exclude the multiples of 10, so we r left with blocks of 9.[u]case I[/u]here, first digit is 1 and so, max(a+d)=10. and, that in every group of 9 at most one can satisfy the question's condition. in range(1001-1099) all groups satisfy the condition but this is not valid for all. e.g.if c+d=17, its not possible. so we get the following table:
if 'a=1' , the 'b' are
                          0  1  2  3  4  5  6  7  8  9
no.of 'c' satisfying in the same order: 10 10 9  8  7  6  5  4  3  2

for 'a=2 ,                   10 10 10 9  8  7  6  5  4  3

                              and so on....
for 'a=9'                           all are 10

summing it all up 64+72+79+...100=780(maybe calc. mistake)
$ \bullet$now, consider, the multiples of . last digit is 0. so, it all depends on a. here also series of numbers satisfied in different 'a'
2,3,4...10. adding it=54.adding both=834

now, i dont think a table is requred.but if any step is unclear, let me know.

[hide="solution"]the solution is 98[/hide]

please give a logical explanation with your answers
and i found some problem with my solution


Okay I'm taking Jan. 27 SAT in a week. I'm pretty nervous and OK at the same time because I've been through these big tests before. Anyway, my school is offering the next SAT at our school (for Jan. SAT, I'm driving an hour to test center) but the deadline is Feb. 2, a week after my SAT.

I think that it takes about 3-4 weeks for your scores to show up in the online. Well, I'm not sure if I should register for SAT anyway even I get good score. Like if I get good score, I can cancel the test right? I'm not sure what's the consequence for that though. Any idea?

Also, if I do take it, do my SAT scores get averaged? Or they pick the best? Or they view all? I use "they" here to refer colleges. Thanks!

Pretty sure you can cancel a test in advance and nobody really cares.

As to your second question, it depends on the college, and you'll need to look this up for each college you're looking at.  Some colleges only look at the latest, others will take the best you've done in each section... it depends.

[quote]Also, if I do take it, do my SAT scores get averaged? Or they pick the best?[/quote]

Colleges never average SAT I scores. Some colleges (most of the most selective colleges) give you the benefit of your highest section scores, section by section. Other colleges (a lot of state universities) give you the benefit of your best single-sitting scores. NO college makes it disadvantageous to take the SAT I more than once.

Good luck with the test. You might as well take it twice, if you have already paid for the first test date.

[quote="tokenadult"]NO college makes it disadvantageous to take the SAT I more than once. [/quote]

I would like to believe that there is really no disadvantage. But I also find it hard to believe that admissions personnel (at least at selective universities) can reliably ignore negative information that is put before them.   Would they really treat a given one-shot high scorer the same as an applicant who started with dismal marks and then took 5 SAT sessions to climb to the same end result?  Particularly if grades were discrepant with the final SAT scores?

Also, is there a similar no-penalty policy for SAT II's, APs, ACT and other  exams, or just for SAT-1?

It's really true that there is no disadvantage. It is also true that test scores are less influential in college admission in the United States than in almost any other country. Most AoPSers probably do more to increase their chances of getting into college by having some significant community involvement than by trying for a higher test score than the first one they obtained.

[quote="tokenadult"]That would be especially unlikely to happen, it seems to me, if the lower score was the earlier score, which was the concern of the OP in this thread.[/quote]

It would seem so.  Reporting a higher score after a lower score shows signs of improvement, which is certainly indicative of a trend that an admissions officer would want to have continue in college :)

I thought it went by really fast. Besides, it's only 3 hours and 45 minutes.

[quote="star99"]ONLY?  as compared to what?[/quote]

the USAMO of course.

Perhaps I didn't find it long because I was looking forward to it.

The SAT I is more shorter, and much less crucial for college admission, than the college admission tests in most other countries. One reason a lot of AoPSers get very nervous about the SAT I (which is really not worth getting nervous about) is that they have foreign-born parents who remember how college admission works in countries where entrance tests are

1) longer

2) harder

and

3) much more important for deciding who gets into college. By comparison, the SAT I is

1) short

2) easy

and

3) only moderately important for deciding who gets into college.

yeah...in  india, there are several tests which play a deciding factor.  there is a test of 9th and 10th grade material, and there is a test of 11th and 12th grade material.  and then there are tests depending what college and what field you want to go to.

re: SAT carrying less weight than exams in many countries outside the US,  tokenadult is absolutely correct

At the beginning of this thread, the original poster ("OP") asked a polite, informational question that is a fairly frequently asked question on all online forums about college admission, a question whose answer I've looked up many times on different college Web sites.

[quote="Silverfalcon"]Also, if I do take it, do my SAT scores get averaged? Or they pick the best? Or they view all? I use "they" here to refer colleges. Thanks![/quote]

I replied, based on my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]research on this issue[/url].

[quote="tokenadult"]Colleges never average SAT I scores. Some colleges (most of the most selective colleges) give you the benefit of your highest section scores, section by section. Other colleges (a lot of state universities) give you the benefit of your best single-sitting scores. NO college makes it disadvantageous to take the SAT I more than once.[/quote]

The last sentence in my reply was disagreed with by one forum participant. I asked for an explicit counterexample.

[quote="tokenadult"]I'd like anyone to show even one example in which a student who reports both a low score (from the point of view of a particular college's applicant pool) and a high score (ditto) fails to get in SOLELY BECAUSE OF THE ADDITIONALLY REPORTED LOW SCORE. I have no reason to believe that that ever happens, because no one has ever pointed me to an example. That would be especially unlikely to happen, it seems to me, if the lower score was the earlier score, which was the concern of the OP in this thread.[/quote]

That request did not turn up any counterexamples to my first statement. Indeed, I will now offer a confirming set of multiple examples (which can be verified by another AoPS participant of long standing) that even the hardest case of getting a lower score subsequent to a high first score doesn't matter. A math coach who has posted here on AoPS since at least 2004, and whom I have had the pleasure of meeting in person, closely follows the admission results of all the top-scoring math students in his state. Most MATHCOUNTS state team members he knows (and he knows quite a few) have had peak scores, scores of 800, on the SAT I math section by age thirteen. Several of those students go on to study very advanced university-level math by high school age. Those students retake the SAT I at high school age, and actually do WORSE on the math section after taking calculus and courses beyond calculus, because they are no longer fresh and in practice on SAT-level math. But those students are admitted to all the most highly selective universities in the United States, and none of those students are rejected because their SAT I math scores actually have a downward trend over time.

This thread then went off onto some side issues and some metadiscussion about moderation, which I don't treat as on-topic on this forum. The side discussion included the correct statement that it is inexpedient for a high school student to have high test scores but low high school grades. It is my best knowledge and belief that the OP in this thread does not have that problem, but works diligently to get good grades in school.

Before this thread gets any farther into being hijacked, I will use my moderator powers to delete posts and lock it, so that the simple answer to the OP's legitimate simple question is not obscured. Hundreds of people submit more than one SAT score to the most highly selective colleges in the United States, and still get in, so this is entirely not an issue to worry about. I appreciate the participation of everyone who stayed on topic here, and I wish all of the college applicants (this year, or in future years) all the best in gaining your test scores, high school grades, and other admission credentials.


1.whati s perfluoro benzene.
2.a hydrocarbon $ C_{5}H_{10}$ does not react with $ Cl_{2}$ in dark but reacts with it in light to give a single mono-chlorinated compound.find the hydrocarbon

[hide="Answer"]1. "Perfluoro" means that all hydrogens of a given compound were substituted by fluorine atoms.

2. The molecular formula tells us that the compound is an alkene or a cyclic alkane. The fact that it doesn't react with chlorine in the dark means that it cannot have a double bond, and so the unknown is a cyclic alkane. As we get only one monochlorinated compound, that means that all hydrogens of the unknown are equivalent (otherwise, we would obtain more than one monochlorinated product). The only structure that match all these requirements is then cyclopentane.[/hide]


linear algebra
matrix
vector
LaTeX
topology

Given an example of a linear transformation R2 -> R2 such that R(T) = N(T).
I don't know where to start, is there a general method for a question like this?

thank you

I assume that by $ R(T)$ and $ N(T)$ you mean the range (column space) and null space of $ T,$ respectively? Note that this would imply that $ T^2\equal{}0.$ In other words, $ T$ must be a nilpotent matrix.

What do nilpotent $ 2\times 2$ matrices look like? Let's start with another question: What can the range of a $ 2\times 2$ matrix be? It can be two-dimensional, but then $ T$ would be invertible and this couldn't possibly be nilpotent. Or it can be zero-dimensional. But that would mean $ T$ is the zero matrix, with range $ \{0\}$ and null space everything, and that's not what is being asked for.

The only possibility left is to find a rank-1 nilpotent matrix. And every nonzero nilpotent $ 2\times 2$ matrix must have rank 1.

So we've narrowed it down to this: do you know any examples of nonzero nilpotent matrices?

[color=green][Redundant quote deleted][/color]

thank you for the reply. yes I do mean the range and null space. I didn't know the name "nilpotent matrix" so I had to look it up.
good call with the T^2 = 0, I think I could have the answer now.. I think upper triangular matrices with its diagonals being all 0's are "nilpotent", so we have in R2
|01|
|00|

but I'm not sure how to express this as a linear transformation?

I think it means T: R2 -> R2 where T(a,b) = (a,0)? the projection onto the x axis?

thanks a lot for helping me

[quote="Nukular"]Actually it would be $ T(a,b) \equal{} (b,0)$.[/quote]

Ok, I see it. Because the matrix is always ordered like (a,b) from bottom down so
|01|
|00|

means that you get (a = 0, b = 0) and the next column should be (0,1) in the standard basis but it is 1 0 which means it gets swapped..

I have another question though.. when we talk about linear transformations, are we always just considering the basis elements?

thank you guys so much

Linear transformations are determined by the images of basis elements. If $ e_1,\dots,e_n$ form a basis for a space $ V$, then $ T : V \to W$ is given once we know $ T(e_1),\dots,T(e_n)$; we have that $ T(\sum_i a_i e_i) \equal{} \sum_i a_i T(e_i)$ by linearity. This is why we can represent a linear transformation by a matrix; we give all the data of $ T$ in terms what $ T$ does to a basis for $ V$, writing each $ T(e_i)$ as a sum of basis elements for $ W$.

In the simple case where $ V \equal{} W \equal{} \mathbb{R}^2$, the matrix

$ \left( \begin{array}{rr} a & b \\ c & d \end{array}\right)$

corresponds to the transformation $ T$ where $ T(x,y) \equal{} (a,b)x \plus{} (c,d)y \equal{} (ax\plus{}cy,bx\plus{}dy)$.

When we do computations it can be simpler to use the matrices; on the other hand, sometimes we may prefer to use linear transformations and abstract algebra -- it depends on context.

[color=green][Redundant quote deleted][/color]

 ah, I see thanks for clearing that up.

so in R2 we had x = a(1,0)  + b(0,1)

but in the transformation, we have T(1,0) and T(0,1) = (0,0) and (1,0) so T(x) = a(0,0) + b(1,0)
so, when we say "linear transformation in terms of a matrix", its just the basis elements because if we know the basis elements, we can derive all other vectors

sorry if I just basically repeated everything you said, I just want to make sure I truly understand

thank you I think I understand now. The notations and terminologies can throw me off..

[quote="Nukular"]... sometimes we may prefer to use linear transformations and abstract algebra -- it depends on context.[/quote]
I'd turn that "sometimes" into "seldom," but that's just my bias. I rather strongly prefer to think in matrices, and I see non-matrix language as just gumming up the machinery of understanding.

mche: Yes, $ \begin{bmatrix}0&1\\0&0\end{bmatrix}$ is just about the easiest example I can think of. (Note: hover your mouse over that or click on it to see the $ \text{\LaTeX}$ code for production a matrix - something well worth knowing if you want to type things here.)

Nothing stops you from using $ \begin{bmatrix}0&-36\\0&0\end{bmatrix}.$ That's also nontrivial and nilpotent. And there are others that aren't triangular, for instance $ \begin{bmatrix}1&1\\-1&-1\end{bmatrix}.$ There are many examples.

[color=green][Moderator says: you're not getting it. Do not quote the previous message. Stop using the "quote" button and use the "reply" button instead, unless you have a very good reason, and even so, only quote selectively.][/color]

 okay, thanks for telling me about the math font thing

I just noticed something that might be confusing to me.. the target space is a vector space, but T(basis elements) is not the basis for the target space is it?  it's strange because you can express T(x) as a linear combination of T(0,1) and T(1,0) (basis elements for R2).. but (0,0) and (0,1) is not a basis for R2. but I guess that that just means (0,0) and (0,1) are linear combinations of the target space's basis?

like
 $ \begin{bmatrix}0 & \minus{} 36 \\
0 & 0\end{bmatrix}.$
is not a basis but you will be able to express all T(x) as linear combinations of these.. never mind, it isn't confusing - I just forgot for a second that R(T) is not always all of W when T: V -> W

thank you so much, I appreciate the help

[quote="Kent Merryfield"]I'd turn that "sometimes" into "seldom," but that's just my bias. I rather strongly prefer to think in matrices, and I see non-matrix language as just gumming up the machinery of understanding.[/quote]

I would find it really awkward to write down boundary maps of chain complexes in algebraic topology using matrices, especially when you don't necessarily have the dimensions of the spaces you are dealing with at hand (especially since they are often infinite-dimensional.)


If $ a,b$ and $ c$ are natural numbers,prove that $ \left(\frac{3abc}{ab \plus{}bc \plus{}ca} \right)^{a^2 \plus{}b^2 \plus{}c^2} \geq \left(\sqrt[a]a  \sqrt[b]b \sqrt[c]c \right)^{abc}$

$ \left(\frac {3abc}{ab \plus{} bc \plus{} ca}\right) \ge 1$

$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$

$ \left(\frac {3abc}{ab \plus{} bc \plus{} ca}\right)^{a^{2} \plus{} b^{2} \plus{} c^{2}}\geq\left(\frac {3abc}{ab \plus{} bc \plus{} ca}\right)^{ab \plus{} bc \plus{} ca} \equal{} \left(\frac {a(bc) \plus{} b(ac) \plus{} c(ab)}{ab \plus{} bc \plus{} ca}\right)^{ab \plus{} bc \plus{} ca}\geq(a^{bc}b^{ca}c^{ab})$

I can not see how inequality is proved ?


calculus
integration
logarithms
calculus computations

According to this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215725]post[/url] your integral equals $ e\minus{}\int_0^\infty \frac{dx}{\Gamma(1\plus{}x)}$ .


geometry
geometric transformation
reflection
rectangle
combinatorics proposed
combinatorics

Two players, Red and Blue, play in alternating turns on a 10x10 board. Blue goes first. In his turn, a player picks a row or column (not chosen by any player yet) and color all its squares with his own color. If any of these squares was already colored, the new color substitutes the old one.

The game ends after 20 turns, when all rows and column were chosen. Red wins if the number of red squares in the board exceeds at least by 10 the number of blue squares; otherwise Blue wins.

Determine which player has a winning strategy and describe this strategy.

Second player wins by reflecting first player's each move in the main diagonal. After each second player move, the bottom-left half will be antisymmetric (red reflects to blue, vice versa) to the top-right half, these both excluding the main diagonal. So without the main diagonal, the players have equal score. But the second player owns the whole main diagonal, which is 10 points.

Basically the same with other words.(May be considered as an overkill, by haven't noticed what is mention above about the diagonal)

[hide]
Winner: Red player.
Strategy: paint perpendicular to the inmediately before painting.

Let $ k$ be number of rows the Blue painted.

We reorder the board to have all, blue's at the bottom an the red's at the top, each player own a rectangle with sides $ k$ and $ 10\minus{}k$, there are two squares of sides $ k$ and $ 10\minus{}k$ being shared, but as Red goes second he owns , respectively, $ 1,2,...,k$ at each turn, squares, the same in the square with side$ 10\minus{}k$.

And so the punctuation of Blue is $ k(10\minus{}k)\plus{} \frac{k^2\minus{}k}{2}\plus{}\frac{(10\minus{}k)^2\minus{}(10\minus{}k)}{2}$ and the punctation of Red is $ k(10\minus{}k)\plus{}\frac{k^2\plus{}k}{2}\plus{}\frac{(10\minus{}k)^2\plus{}(10\minus{}k)}{2}$, thus the difference in points in this way is always $ 10$
[/hide]

 :)

My way of solution involves simetry as well, but I didn't use the diagonal but the line between de $4th$ and $5th$ row and the line between the $4th$ and $5th$ column, so if [color=#FF0000]red[/color] plays a row , then [color=#0000FF]blue[/color] w'd have to play it's respective simetric row, and with the columns [color=#0000FF]blue[/color] proceeds the same... since there are 20 moves, the last move will be for [color=#0000FF]blue[/color], and this way he/she will ensure that at least the game will be tied, or [color=#0000FF]blue[/color] will win by one tile

The problem with that argument is that Blue starts first, not Red. Also, you didn't prove the conclusion (Blue can ensure that the game is tied or win for Blue), and also you missed the fact that Blue cannot win by one tile (the difference of tiles must be even).

Ok I did a complete different problem, my bad


trigonometry algebra polynomial quadratics logarithms function domain

[quote="Thorn"]Find an algebraic solution to:

$1+2\sin x = \sin(2x)$

This is either much harder than your typical textbook trig equation or there is some clever trick I didn't see.  I ended up solving a 4th degree polynomial, but perhaps there is a nice solution.[/quote]

I presume you are supposed to find values for $x$ which make the statement true?

There is a deeper trick that you most likely haven't been exposed to yet.
However it involves a technique the use of which I have been greatly maligned for in this forum. So in the interest of heeding the Master's advice not to cast my pearls before swine ... I shall only venture part way toward a complete solution.

What you will obtain as a result is a pair of identical quadratic equations - one in the complex variable $z,$ and the other in its complex conjugate $z^{\ast}.$ Since these variables are complex conjugates, you need only solve one of the quadratic equations to obtain the solutions to the other as well.

Here's the part that you'll recognize immediately:

\begin{eqnarray*}1\text{quadratic equation }&\mapsto& 2\text{ solutions}\\ 2\text{quadratic equations }&\mapsto& 4\text{ solutions}\end{eqnarray*}

Of the four solutions, one pair of solutions differs from the other pair only in terms of sign.

The trick is to recognize that $\sin x =\frac{e^{ix}-e^{-ix}}{2i}$

One can then let $e^{ix}= z$ and $e^{-ix}= z^{\ast}.$

Making the appropriate substitutions we find

$1-\frac{1}{2i}[(z^{2}-2z)-(z^{\ast}^{2}-2z^{\ast})] = 0.$

If one recognizes that $z^{2}-2z$ and $z^{\ast}^{2}-2z^{\ast}$ are complex conjugates of one another then one may let $z^{2}-2z = y$ and recast the above equation as

$1-\frac{1}{2i}[y-y^{\ast}] = 0.$

$\frac{1}{2i}[y-y^{\ast}] = \sin(-i\log y)$

$1-\sin(-i\log y) = 0$

$\sin(-i\log y) = 1$

$-i\log y = \frac{\pi}{2}$

$\log y = i\frac{\pi}{2}$

$y = e^{i\frac{\pi}{2}}= i$

Now substitute y back into the equation from whence it came to obtain:

$z^{2}-2z = i$

$z^{2}-2z-i = 0$

Solve the above quadratic equation for $z.$

Recall that $z = e^{ix}$

$\boxed{\therefore x =-i\log z}$

The person rides again.

From $1-{1\over 2i}[y-y^{*}]=0$ we get only $y=a+i$, where $a$ is any real, and NOT $y=i$

[quote="Dr. No"][quote="t0rajir0u"][quote="Dr. No"]$\frac{1}{2i}[y-y^{\ast}] = \sin(-i\log y)$

$1-\sin(-i\log y) = 0$

$\sin(-i\log y) = 1$[/quote]

Can you show us how you obtained $\Im (y) = i$?[/quote]
I'd be glad to as a private sidebar but I've been warned not to discuss complex variables in the HSB forum. (The Pythagoreans are trying to keep the $\pi$ all to themselves again I suspect. :) )[/quote]
ingenious reply   :)
(especially the HSB forum part :wink: )

[quote="ritchjp"]I found a solution, but I had to do it graphically. The curve that is generated from 4A^2+4A+1=0 intersects the x-axis at two points. It intersects approximately at (-.25,0) and (-.9,0).
If you go to coolmath.com and go to their graphing calculator and key in the above equation, it will generate the curve for you.

http://www.coolmath.com/graphit/index.html

Then you should be able to find out what x is.[/quote]
Not to kick the piano bench out from under you when you're getting ready to play, son, but its a fourth order equation, not a quadratic, you need to find a couple more roots there.

[quote="Farenhajt"]From $1-{1\over 2i}[y-y^{*}]=0$ we get only $y=a+i$, where $a$ is any real, and NOT $y=i$[/quote]
Good try, but not even close. When you see a difference of complex conjugates, especially in this problem, - think $\sin$ function.

The expression ${1\over 2i}[y-y^{*}]$ is the definition of $\sin$ for some angle as yet undetermined. That's why I made the substitution $y = z^{2}-2z$ in the first place.

[quote="Dr. No"]

There is a deeper trick that you most likely haven't been exposed to yet.
However it involves a technique the use of which I have been greatly maligned for in this forum. [b]So in the interest of heeding the Master's advice not to cast my pearls before swine [/b]... I shall only venture part way toward a complete solution.

...

Making the appropriate substitutions we find

$1-\frac{1}{2i}\left[\left(z^{2}-z\right)-\left(z^{\ast}^{2}-z^{\ast}\right)\right] = 0.$

[/quote]

the first quoted step is incorrect... furthermore, your solution does not produce values that satisfy the equation, (i also believe you mean $\ln$ not $\log$)

also i would like to comment that there is only one pair of complex conjugate roots, the other two roots are real,

now to the bolded comment...i find that needlessly condescending, your solution is full of flaws, one could hardly qualify this as "pearls"

[quote="Dr. No"]
Good try, but not even close. When you see a difference of complex conjugates, especially in this problem, - [size=150]think[/size] $\sin$ function.
[/quote]

Hey, lunatic, you DARE to teach me anything?  :lol:  :lol:

Missed your antipsychotics for today, huh? Now be a good mollusc and return to your shell... That's right...  :roll:

Altheman,

I dropped a factor of two in the collection of like terms. That's a mistake, but full of flaws? Hardly.

As for the "pearls before swine" remark that was annoyance, not condescension. In another thread in the HSB forum, I had pointed out an ambiguity in a particular problem statement, I then attempted to resolve it by extending the domain to the complex numbers. That effort unleashed an unending torrent of criticisms, some of them valid, but most of them complete rubbish.

I like solving problems. I come here to share what I've learned, to learn new things myself, and hang out with very bright people. But if most of them are going to behave like jackasses, it's just not worth it.

[quote="jhcreinhardt"]it's better to post things that most of us can understand (and is right) than try to overwhelm us with things that you hope no one will challenge.[/quote]
Sound advice. Though I don't mind having my work questioned, I often make petty arithmetic mistakes, I think its criminally rude to hurl insults in the process.

its just a suggestion..that whole swine thing wasnt an insult?

Dr No – Don’t want to seem  to be unkind to you but it is not “petty arithmetic mistakes” that are being criticized , but  what many think are serious flaws, disguised in impressive sounding lingo (what Prof Feynman used to call “faking rigorousness ”)

So, what, I believe,  “unleashed an unending torrent of criticisms”, was not because you “ had pointed out an ambiguity in a particular problem statement, [and]  then attempted to resolve it by extending the domain to the complex numbers” but “pearls” like:
[hide][quote]it is impossible for negative real number, whatever power you raise it to, to have a real logarithm.[/quote]
Or [quote]
The fact that $\lbrace a,b,c\rbrace \in \mathbb{Z_{+}}\implies a = 3,\,\, b = 2,\,\, \text{and }c = 1 \text{ or }a = 3,\,\, b = 1, \text{ and }c=2.$[/quote][/hide]
(No need to rehash the old arguments, These are just a few examples, one can find many)

Sorry to go of topic, and apologies if comments seem too harsh (these are my opinions only), but I believe, these kind of posts are not helpful and derails the discussion. I do believe mods should intervene to these kind of posts too.

Getting back to the topic:

To solve $x^{4}+x+(1/4)=0$ [hide] (or similar equation), one can use numerical methods (or use calculator like TI-86/89 etc)
You can also  (Method is standard and can be found in good math book )

You  rewrite it as $(x^{2}+px+q)(x^{2}-px+r)$
Giving: ($r+q = p^{2}$
$r-q = 1/p$
$qr = 1/4$
This gives $(p^{6}-p^{2}-1 = 0$  A cubic equation in $p^{2}$
(which I assume you know how to solve)

The solution is $p^{2}= \frac{(108+12 \sqrt{69})^{1/3}+(108-12 \sqrt{69})^{1/3}}{6}\approx 1.3247$
Now you know, p,q,r and rest is “easy” .
You have $(x^{2}-px+\frac{p^{3}+1}{2p})(x^{2}+px+\frac{p^{3}-1}{2p}) = 0$
Two quadratic equations, value of $p$ ($\approx \sqrt{ 1.3247}$ ) given above)
are now ready to be solved.

 [/hide]
Hope this helps.

[quote="Dr. No"]

$\frac{1}{2i}[y-y^{\ast}] = \sin(-i\log y)$

[/quote]

i don't believe you can make this step, $y$ and $y*$ do not have  the same absolute value (i.e. you cannot solve the equation $e^{ix}=y$ and $e^{-ix}=y*$ unless $|y|=1$, which is not the case

ex. $z=a+bi$ and $z*=\frac{a^{2}+b^{2}}{a+bi}$, the sine substitution is only valid if $a^{2}+b^{2}=1$

Pardon the interuption of flow here, I'm going to respond to things in somewhat of a reverse order... some things are more important than others.

Altheman,

First, I don't make the claim that $y = e^{ix}$ ... $z = e^{ix}$ so $x$ as a variable is already spent.

Second, if while in college you manage to bash yourself in the skull with a good complex analysis text enough times, the notions that $\left|\cos \phi\right| \le 1$ and $\left|\sin \phi\right| \le 1$ will soon leave your mind. (These have to be the case only if $\phi$ is a real number. If $\phi$ is complex all bets are off.)

Consider, $e^{i\phi}$ where $\phi = a \pm ib,$ then

$e^{i\phi}= e^{i\left(a \pm ib\right)}= e^{\mp b}\cdot e^{ia}.$

With their magnitudes no longer restricted sine and cosine merely become (complex) functions like any others: They each have a magnitude and a phase.

That said, consider:

\begin{eqnarray}y &=& \cos\q+i\sin q \\ y^{\ast}&=& \cos q-i\sin q\end{eqnarray}

Adding (1) to (2) results in:

$\left[ y+y^{\ast}\right] = 2\cos q$

While subtracting (2) from (1) results in:

$\left[y-y^{\ast}\right] = 2i\sin q$

That's how I did it.

Edit:

In any case, this discussion is obviously going nowhere, since both sides are being childishly unwilling to discuss the merits of Dr. No's approach straightforwardly.  I don't see a reason for this thread to continue.

@Dr. No:  i understand that the extension of sin so there is no absolute value restriction, but regardless, the substitution does not work

$e^{i\theta}=4+i$ and $e^{-i\theta}=4-i$

by your logic, there is suitable $\theta$ such that both of those are true,

hence, $i\theta=\ln (4+i)+2k\pi$  ($k\in\mathbb{Z}$) and $i\theta=-\ln(4-i)$

$\ln (4+i)+2k\pi=-\ln 4-i$
$\ln (4+i)(4-i)=-2k\pi$

evidently, this does not work the reason that it usually works is because $z\cdot z*=1$, so $\ln (1)=0\implies k=0$, however, when the absolute value is not $1$, then the substitution does not work

@junggi: why spam like that?

@t0rajir0u: i feel that i am having a productive experience with this topic, but the fact that his answer (even edited) does not produce correct solutions, means that we should discuss what is wrong with the solution

To all,

I'm going to have to continue this later. I have to go out of town for the next couple of days.

Regards,
Dr. No

[quote="ritchjp"]I made an error in my previuos post. It should read 4A^4 + 4A + 1. But when I use the graphing calculator I still come out with the curve I did before and it still intersects the x-axis at the coordinates I specified. By the way, those were the only two points where it intersected the x-axis.[/quote]

Yes, as has been mention before,  there are  only two real roots (about -0.8968 and -0.2542)  and two complex ($0.5755.. \pm 0.8750..i$)

(For example, you can see my earlier post, for these values)

Dr NO:
[hide]Speaking of “bash[ing] yourself in the skull with a good complex analysis text enough times” and noticing “.... is complex all bets are off”  you “consider”:

[quote]\begin{eqnarray}y &=& \cos\q+i\sin q \\ y^{\ast}&=& \cos q-i\sin q\end{eqnarray}

Adding (1) to (2) results in:

$\left[ y+y^{\ast}\right] = 2\cos q$ … etc ..[/quote]

if  $q$ , as you said, is not necessary real,  the second part  simply  makes no sense. Complex conjugate of $a+ib$ , is $a-ib$  when  $a$ and $b$ ARE REAL.  For complex values, -let me use your own words -  “all bets are off” even if one “bashes oneself with a textbook” ]

Or am I missing something?

Also: what exactly you mean by::

[quote]What you will obtain as a result is a pair of identical quadratic equations - one in the complex variable  and the other in its complex conjugate  Since these variables are complex conjugates, you need only solve one of the quadratic equations to obtain the solutions to the other as well. ...
Of the four solutions, one pair of solutions differs from the other pair only in terms of sign... [/quote]

Hmm… “identical quadratic equation” .. (shouldn’t they produce identical roots? ).. and what exactly is a quadratic equation in “complex conjugate?” .. and how a complex number "pair" differs "only in terms of  *sign*"

There are several places  in your posts where meaning, to put it lightly, is unclear.
It may be worth your while to edit/rewrite/   those posts so that they are useful to others.
(Just my thoughts)
[/hide]

[quote="junggi"]Dr. No, help me with this solution approach, similar to yours. I don't see anything wrong with it, but I seem to be getting different answers.  :maybe:
We are solving the equation $4x^{4}+4x+1=0$, correct?(see 1337's post)
Ok so,
$4x^{4}+4x+1=0$
$\iff (4x^{4}+4x+1)\cdot0=0\cdot0$
$\iff 0=0$
If we treat $0$ as a real number here, clearly we cannot do anything at all with it. However, if we treat $0$ as a complex number, we can add $x$ to the LHS and $1$ to the RHS.
So we get the solution
$x=1.$
Plugging in to the original equation,
$4\cdot1^{4}+4\cdot1+1=0$
$\iff 9=0$
So $x=1$ indeed is a solution.
Similarly, we get $x=\sqrt[\pi]{\pi!^{e}},x=\frac{\phi(i\pi)}{0},x=\sigma(\Gamma(\pi^{ei})+\varphi)$ where $\phi,\sigma,\Gamma$ are those functions, and $\varphi$ is the constant.
Plugging them in, we get they all are solutions.
What did I do wrong  :maybe: ?[/quote]

:rotfl: :rotfl: :rotfl: :rotfl: :rotfl:

[hide="almost a solution"]We can put it in terms of sine and cosine:

$1+2\sin{x}=2\sin{x}\cos{x}$

$\sin^{2}{x}-2\sin{x}\cos{x}+\cos^{2}{x}=-2sin{x}$

$\left(\sin{x}-\cos{x}\right)^{2}=-2sin{x}$

That indicates that the sine is negative. Therefore,

$pi<x<2pi$

and

$0>\sin{x}\geq-1$

If $\sin{x}=-1$, then x=270. When we plug that in, we get

$-1=0$, which is not true. Therefore,

$0>\sin{x}>-1$[/hide]

stuck at what may be the last step...   :fool:

[quote="Gyan"]

Dr NO:
Speaking of “bash[ing] yourself in the skull with a good complex analysis text enough times” and noticing “.... is complex all bets are off”  you “consider”:

[quote]\begin{eqnarray}y &=& \cos\q+i\sin q \\ y^{\ast}&=& \cos q-i\sin q\end{eqnarray}

Adding (1) to (2) results in:

$[ y+y^{\ast}] = 2\cos q$ … etc ..[/quote]

if  $q$ , as you said, is not necessary real,  the second part  simply  makes no sense. Complex conjugate of $a+ib$ , is $a-ib$  when  $a$ and $b$ ARE REAL.  For complex values, -let me use your own words -  “all bets are off” even if one “bashes oneself with a textbook” ]

Or am I missing something?

Also: what exactly you mean by::

[quote]What you will obtain as a result is a pair of identical quadratic equations - one in the complex variable  and the other in its complex conjugate  Since these variables are complex conjugates, you need only solve one of the quadratic equations to obtain the solutions to the other as well. ...
Of the four solutions, one pair of solutions differs from the other pair only in terms of sign... [/quote]

Hmm… “identical quadratic equation” .. (shouldn’t they produce identical roots? ).. and what exactly is a quadratic equation in “complex conjugate?” .. and how a complex number "pair" differs "only in terms of  *sign*"

There are several places  in your posts where meaning, to put it lightly, is unclear.
It may be worth your while to edit/rewrite...[/quote]
Oh, good grief ...

\begin{eqnarray}e^{iq}&\equiv& \cos q+i\sin q\\ e^{-iq}&\equiv& \cos q-i\sin q\end{eqnarray}

$\Rightarrow \sin q \equiv \frac{1}{2i}[e^{iq}-e^{-iq}]$, regardless of whether $q$ is real.

Making no assumption as to whether $x$ is real or not, the original problem statement

$1+2\sin x = \sin 2x$ becomes

$1-\frac{1}{2i}[(e^{i2x}-2e^{ix})-(e^{-i2x}-2e^{-ix})] = 0$

letting $z = e^{ix}$ and $u = e^{-ix}$

We have $1-\frac{1}{2i}[(z^{2}-2z)-(u^{2}-2u)] = 0$

Now letting $z^{2}-2z = e^{iq}$ and $u^{2}-2u = e^{-iq}$

$1-\frac{1}{2i}[e^{iq}-e^{-iq}] \equiv 1-\sin q = 0$

$\Rightarrow q = \sin^{-1}(1) = \frac{\pi}{2}$

Substituting back

\begin{eqnarray*}z^{2}-2z = e^{i\frac{\pi}{2}}&=& i\\u^{2}-2u = e^{-i\frac{\pi}{2}}&=&-i\end{eqnarray*}

When I wrote that these equations were identical I meant that they had identical forms when cast in terms of complex conjugates. We no longer assuming that they are complex conjugates, resulting in twice as much work to arrive at the same destination, but oh well...

By completing the square for both equations:

\begin{eqnarray*}(z-1)^{2}&=& 1+i\\(u-1)^{2}&=& 1-i\end{eqnarray*}

Solving the above system:

\begin{eqnarray*}z &=& 1 \pm \sqrt{1+i}\\ u &=& 1 \pm \sqrt{1-i}\end{eqnarray*}

Substituting for $z$ and $u$:

\begin{eqnarray*}e^{ix}&=& 1 \pm \sqrt{1+i}\\ e^{-ix}&=& 1 \pm \sqrt{1-i}\end{eqnarray*}

Solving the above for $x$:

\begin{eqnarray*}x &=&-i\log(1 \pm \sqrt{1+i})\text{ or }\\ x &=& i\log(1 \pm \sqrt{1-i})\end{eqnarray*}

[quote="Dr. No"]

letting $z = e^{ix}$ and $u = e^{-ix}$

...

Now letting $z^{2}-2z = e^{iq}$ and $u^{2}-2u = e^{-iq}$

[/quote]

this does not work...evidently $z=\frac{1}{u}$, hence,

multiply the second pair of equations, and you get:

$(z^{2}-2z)(z^{-2}-2z^{-1})=1$.....(1)

however, you start with $2i=z^{2}-2z-(z^{-2}-2z^{-1})$ (i.e. after the trig substution...)

i cannot see how you can justify (1), you cannot make the second substitution

[quote="Dr. No"]
Oh, good grief ...[/quote]
Good grief indeed!

So now  $y$  and $y^{\ast}$  “becomes” $e^{iq}$ and $e^{-iq}$  !!

Good grief indeed!

(which BTW are NOT complex conjugate of each other if $q$ is not real – but that small factor need not matter when pages and pages of  Dr. No's nonsensical ramblings, (at least in my opinion)  goes on and on and on  – Good Grief !!!)

And just when I  am happy when  I can  find  a few lines, in the above post  without a  flaw’ or error eg:

[quote]$\sin q \equiv \frac{1}{2i}\left[e^{iq}-e^{-iq}\right]$, regardless of whether $q$ is real.[/quote]

That is true, of course, nothing earth-shaking, just how some define sin function, (of course $e^{iq}$ and $e^{-iq}$  now  need not be complex conjugate of each other)  but just a few lines later  Dr NO comes out again  with what he calls   a “pearl before swine”

[quote]Now letting $z^{2}-2z = e^{iq}$ and $u^{2}-2u = e^{-iq}$[/quote]

How do you get that?  ( Hint: No need to show  pages and pages of  complex analysis  type mumbo-jumbo, just multiply  LHS of these two and see if you get $1$)

[b]Good Grief Indeed! [/b]

Also, Dr NO -  BTW have you substituted the value of x you got into the original equation, just to see if the value indeed makes sense.  (Hint:  One value, as I said before, and can be easily verified,  is about $(-0.257 (\approx-14.7^{o}) )$  which one of your answer has that numerical value? -  Just curious)

This is, I hope, my last post .. Hope  Dr NO does not come out with  pages and pages ... just answer/clarify the two points I raised:

Dr. No - Per my last suggestion, have you  tried  to substitute the value of x in the original equation? BTW  near the end ,  just before you write the value of $x$  ..

[quote="Dr. No"]
Substituting for $z$ and $u$:

\begin{eqnarray*}e^{ix}&=& 1 \pm \sqrt{1+i}\\ e^{-ix}&=& 1 \pm \sqrt{1-i}\end{eqnarray*}

Solving the above for $x$ ....
[/quote]

Hmm ... multiply these (Right had sides) of two. Do you get $1$?  (product of $e^{ix}$ and $e^{-ix}$  ought to be  1 ? right? )  :o  :roll:

Sorry, it seems me that I do not get why such long discussion goes.
The equation isn't too hard.

Observation of curves $y=1+2\sin x$ and $y=\sin 2x$  gives that it has exactly 2 roots for $t=\tan \frac{x}{2}$

Substitution $\sin x = \frac{2t}{1+t^{2}}$ and $\cos x = \frac{1-t^{2}}{1+t^{2}}$  gives an equation

$t^{4}+8t^{3}+2t^{2}+1=0$
(maple gives 2 roots $t_{1}=-7.739425778, t_{2}=-0.6217106179$)

which is separatable in 2 ones:

(*) $(t^{2}+4t-7+2\lambda) \pm 2\sqrt{\lambda}(t+2-\frac{7}{\lambda}=0$

where $\lambda \approx 4.75$ is unique real root of the cube equation

(**) $(2-\frac{7}{\lambda})^{2}=\lambda-7+\frac{12}{\lambda}$

In total we have $x=2\arctan t_{1,2}+2k\pi$, where $t_{1,2}$ are roots of (*) with the sign plus as in contrary it has no real solution.
(now maple gives 2 roots $t_{1}=-7.739425778, t_{2}=-0.6217106181$)

P.S. The maple checking is added.

Altheman,

You're right, the second substitution, $e^{iq}$ and $e^{-iq}$, over specifies the system to be solved.

Nonetheless, I've found an approach that works

[hide="Solution"]

Starting with the original equation $1+2\sin x = \sin 2x$

Rewriting it in terms of complex exponentials

$\Rightarrow 1+2\left(\frac{e^{ix}-e^{-ix}}{2i}\right)-\left(\frac{e^{i2x}-e^{-i2x}}{2i}\right) = 0$

Multiplying both sides of the resulting equation by $2i e^{i2x}$...

$2i e^{i2x}+2 e^{i3x}-2 e^{ix}-e^{i4x}+1 = 0$

Rearranging terms (and changing signs)...

$e^{i4x}-2 e^{i3x}-2i e^{i2x}+2 e^{ix}-1 = 0$

Making the substitution: $z=e^{ix}$

$z^{4}-2z^{3}-2i z^{2}+2 z-1 = 0$

From this point I cheated (not really cheated, as I believe the problem said to find an algebraic solution; I did exactly that.)

Plugging the above polynomial into MATLAB (TM) gives the following results

>> z = roots(p)

z =

   2.1373 + 0.9744i
  -0.9672 - 0.2542i
   0.4425 - 0.8968i
   0.3874 + 0.1766i

>> x = -i*log(z)

x =

   0.4277 - 0.8540i
  -2.8846 - 0.0000i
  -1.1125 - 0.0000i
   0.4277 + 0.8540i

Back substituting these values for $x$ into the original equation shows the LHS and RHS are equal to four significant figures (five significant figures in the cases of the real zeros.)
[/hide]

[quote="Gyan"][quote="Dr. No"]
Oh, good grief ...[/quote]
Good grief indeed!

So now  $y$  and $y^{\ast}$  “becomes” $e^{iq}$ and $e^{-iq}$  !!

Good grief indeed!

[/quote]
"Good grief" was an expression of exasperation with myself as much as with anyone else.

You are correct, $e^{iq}$ and $e^{-iq}$ aren't necessarily complex conjugates of one another. (They are complex conjugates only if $q$ itself is a real number.) Nevertheless, $e^{iq}$ and $e^{-iq}$ define the trig functions $\cos q$ and $\sin q$ regardless of whether $q$ is a real number... hence, "good grief." (The series expression of $e^{iq}$ equals the sum of the series expressions for $\cos q$ and $i\sin q.$)

Furthermore, $e^{iq}\times e^{-iq}\equiv 1$ again regardless of whether $q$ is a real number.

Pardon my insight, but when I first took on this problem I recognized the expressions $e^{i2x}-2e^{ix}$ and $e^{-i2x}-2e^{-ix}$ as complex conjugates of each other; meaning that their difference could be characterized as a sine function in its own right. (It is not true that they are complex conjugates, but only because $x$ is not a real number.)

The point of mathematics is identifying and exploiting patterns, as far as I'm concerned, not satisfying picayune personal obsessions with the obvious!

LATEX NOT WORKING :(

1+ 2sinx = 2sinx sqrt(1-sin^2 x)
square both sides
1+4sinx = -4sin^4x

so you know there is general solution for 4th degree polynomial (ugly)
and take arcsin done

Locked. Sorry I didn't catch the arguing earlier.


Hey guys,
I come across this problem while trying to solve another, it's kind of like: how many ways are there to that four integer numbers {a, b, c, d} will always sum up to a particular number N. For instance, say N = 4, then you've got one way is {4, 0, 0, 0}, another is {3, 1, 0, 0}, then {2, 1, 0, 1}...Plz help! ;-)

The problem is equivalent to finding the number of ways to divide a stick with length N by 3 cuts. So the answer is $\binom{n+1}{3}$

Oh, cool...Thanx...I think should be N+3, how come N+1?

I think this problem is related to Stirling's numbers of the second order. :idea:


Circles A, B, C, and D are mutually tangent. The radius of D is 2 and the radius of A is 1. If D contains the other three circles,  which are externally tangent to each other, and $B\cong C$, then what is the radius of circle B?

[hide]Draw the correct diagram(as it stated). and connect the center of circle A to the intersection of circles B & C, the line is a tangent line to circles B & C. Therefore, you get right triangles. Use pythagorean theorem, you get the radius of B is $\boxed{\frac{8}9}$[/hide]


Prove that $ \tan^2{\left(\frac{\pi}{7}\right)} \plus{} \tan^2{\left(\frac{2\pi}{7}\right)} \plus{} \tan^2{\left(\frac{3\pi}{7}\right)} \equal{} 21$.

\begin{eqnarray*}
\tan 7\theta
& = & \frac{\mathrm{Im} (\cos \theta + i \sin \theta)^7}{\mathrm{Re} (\cos \theta + i \sin \theta)^7} \\
& = & \frac{\cos ^7 \theta - 21 \cos ^5 \theta \sin ^2 \theta + 35 \cos ^3 \theta \sin ^4 \theta - 7 \cos \theta \sin ^6 \theta}{7 \cos ^6 \theta \sin \theta  - 35 \cos ^4 \theta \sin ^3 \theta + 21 \cos ^2 \theta \sin ^5 \theta - \sin ^7 \theta} \\
& = & \frac{\tan ^7 \theta - 21 \tan ^5 + 35 \sin ^3 \theta - 7 \tan \theta}{7 \sin ^6 \theta - 35 \sin ^4 \theta + 21 \sin ^2 \theta - 1} \\
\end{eqnarray*}
So,
\[ \tan ^7 \theta - 21 \tan ^5 + 35 \sin ^3 \theta - 7 \tan \theta = 0 \iff \theta = 0, \frac{\pi}{7}, \frac{2\pi}{7}, \frac{3\pi}{7}, \frac{4\pi}{7}, \frac{5\pi}{7}, \frac{6\pi}{7}\]
Therefore,
\[ \sum_{k=0}^6 \tan^2 \frac{k \pi}{7} = \left( \sum_{k=0}^6 \tan \frac{k \pi}{7} \right)^2 - 2\sum_{0 \le i < j \le 6} \tan \frac{i \pi}{7} \tan \frac{j \pi}{7} = 0^2 - 2 \cdot (-21) = 42\]
Thus,
\[ \sum_{k=0}^6 \tan^2 \frac{k \pi}{7} = \tan 0 + 2\sum_{k=1}^3 \tan^2 \frac{k \pi}{7} = 42 \iff \color{red}\sum_{k=1}^3 \tan^2 \frac{k \pi}{7} = 21\]

Here is my solution, without the use of complex numbers (though it still might be a bit overkill) :

[hide]We use the equivalent expression

$ \tan^2\frac{\pi}{7}\plus{}\tan^2\frac{2\pi}{7}\plus{}\tan^2\frac{4\pi}{7}$

Notice that $ \frac{\pi}{7}\plus{}\frac{2\pi}{7}\plus{}\frac{4\pi}{7}\equal{}\pi$ so that we have

$ \tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\equal{}\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}$

So, our expression is equivalent to

$ \left(\tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\right)^2\minus{}2\left(\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\plus{}\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\plus{}\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\right)$

$ \equal{}\left(\tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\right)^2\minus{}2\left(\tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\right)\left(\cot\frac{\pi}{7}\plus{}\cot\frac{2\pi}{7}\plus{}\cot\frac{4\pi}{7}\right)$

$ \equal{}\left(\tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\right)\left(\tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\minus{}2\cot\frac{\pi}{7}\minus{}2\cot\frac{2\pi}{7}\minus{}2\cot\frac{4\pi}{7}\right)$

We have that $ \tan\theta\minus{}\cot\theta\equal{}\minus{}2\cot2\theta$ so that the expression is equivalent to

$ \left(\tan\frac{\pi}{7}\plus{}\tan\frac{2\pi}{7}\plus{}\tan\frac{4\pi}{7}\right)\left(\minus{}2\cot\frac{2\pi}{7}\minus{}2\cot\frac{4\pi}{7}\minus{}2\cot\frac{8\pi}{7}\minus{}\cot\frac{\pi}{7}\minus{}\cot\frac{2\pi}{7}\minus{}\cot\frac{4\pi}{7}\right)$

$ \equal{}\minus{}3\left(\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\right)\left(\cot\frac{\pi}{7}\plus{}\cot\frac{2\pi}{7}\plus{}\cot\frac{4\pi}{7}\right)$

$ \equal{}\minus{}3\left(\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\plus{}\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\plus{}\tan\frac{4\pi}{7}\tan\frac{\pi}{7}\right)$

$ \equal{}\minus{}3\frac{\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\cos\frac{4\pi}{7}\plus{}\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}\cos\frac{\pi}{7}\plus{}\sin\frac{4\pi}{7}\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}{\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}$

$ \equal{}\minus{}3\frac{\sin\frac{\pi}{7}\left(\sin\frac{2\pi}{7}\cos\frac{4\pi}{7}\plus{}\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}\right)\plus{}\left(\minus{}\cos\frac{6\pi}{7}\plus{}cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\right)\cos\frac{\pi}{7}}{\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}$

$ \equal{}\minus{}3\frac{\sin^2\frac{\pi}{7}\plus{}cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\plus{}\cos^2\frac{\pi}{7}}{\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}$

$ \equal{}\minus{}3\minus{}3\frac{1}{\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}$

$ \equal{}\minus{}3\minus{}3\frac{2\sin\frac{\pi}{7}}{\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}$

$ \equal{}\minus{}3\minus{}3\frac{4\sin\frac{\pi}{7}}{\sin\frac{4\pi}{7}\cos\frac{4\pi}{7}}$

$ \equal{}\minus{}3\minus{}3\frac{8\sin\frac{\pi}{7}}{\sin\frac{8\pi}{7}}$

$ \equal{}\minus{}3\minus{}3(\minus{}8)\equal{}21$

[/hide]


Este adevarat ca $ \lim(1\plus{}\frac1{2^2}\plus{}\frac1{3^2}\plus{}...\plus{}\frac1{n^2})\equal{}\frac{\pi^2}6$ ?
E corect sau confund cu altceva?

see Evaluating zeta(2)

here

http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/rjc.html


trigonometry algebra function domain quadratics AMC

Solve for all x in domain[0,2:pi:)

cos2x/(sin3x-sinx)=1.

some basic formulas: cos 2x=1-2sin:^2:x and sin 3x = 3 sin x - 4 sin:^3:x.
Multiplying the denominator to the RHS, subtracting it over to the left hand side, and making the substitutioni sin x =a, I get 4a:^3:-2a:^2:-2a+1. I tried synthetic division, and lucky me: the first root I tried, 1/2, worked. It left behind the quadratic 4a:^2:-2=0, which is easy to solve as a=:pm:1/:rt2:. So, we now have the equations to solve:
sin x = 1/2, sin x =1/:rt2: and sin x=-1/:rt:2. The last two give :pi:/4, 3:pi:/4, 5:pi:/4 and 7:pi:/4, while the first one gives x=:pi:/6 and 5:pi:/6.

I don't know why no one seems to bother memorizing sum-to-product formulas. This problem is trivial.

i know
it is extremely trivial

just use sum-to product on the denominator and ur basically done


this algebra problem is a toughie.
solve for $ y$ in terms of $ x$ and $ n$:

$ q\minus{}x\equal{}ny^2$
$ 4q\minus{}n\equal{}xy$

i have a solution i will post soon...but i am not sure if it is right.  please explain answers.

Standard elimination???

$ q - x = ny^2\implies4q - 4x = 4ny^2$ (1)
$ 4q - n = xy$ (2)
(1)-(2) $ n - 4x = 4ny^2 - xy\implies(4n)y^2 + ( - x)y + (4x - n) = 0$.
By the quadratic formula, we have $ y = \frac {x\pm\sqrt {x^2 - 64xn + 16n^2}}{8n}$ unless I made a stupid mistake...

edit: It was actually a typo...fixed.

your stupid mistake is...
you forgot to distribute the negative
in the unsimplified form of the quadratic equation:
$ y\equal{}\frac{x\pm\sqrt{x^2\minus{}4(4n(4x\minus{}n))}}{8n}$
you have remember to distribute $ \minus{}4$, not $ 4$.
so:
$ y\equal{}\frac{x\pm\sqrt{x^2\minus{}64nx\plus{}16n^2}}{8n}$
but otherwise, its right. :winner_first:


geometry
geometric transformation
rotation
group theory
abstract algebra
reflection
vector

prove that $SO_3(\mathbb{R})$ (group of the positive isometries  of $\mathbb{R}^3$) is simple.

They are precisely the rotations around axes. By similarity with matrices in $SO_3$ we can turn any rotation of angle $\alpha$ into any other one, so if a normal subgroup contains one rotation of angle $\alpha$, it contains all of them. Now, it also contains those of angle $2\pi-\alpha$, so it contains those of angle $\pi$ as well (by continuity), i.e. the reflections in axes. It's easy to get any rotation of any angle we want by multiplying reflections in various axes, so we have all the transformations in $SO_3$.

I might have missed something, but something like that definitely works.

Excuse me for being dumb, Grobber, but I really don't get how you can change any rotation into any other by similarity with matrices from $SO_3(R)$.

I think Grobber meant that we can obtain any rotation of angle $\alpha$. ( if $R$ is a rotation of angle $\alpha$ and axis $u$, consider a positive isometry $f$ that maps $u$ to $v$, where $v$ is an arbitrary vector : then $foRof^{-1}$ is the rotation of angle $\alpha$ and axis $v$). Thus we also get the rotations of angle $2 \pi - \alpha$. But I don't understand how we can get the rotations of angle $\pi$ with those previous rotations ?

I'm not sure exactly what I meant (I was in a hurry, as I am now, and it's been a hell of a day for me :)), but I suppose I wanted to do something along the following lines (forget about that $2\pi-\alpha$ thing):

We can get all the rotations of angle $\alpha$ (indeed, alekk, this is what I meant: not all rotations, but all rotations of angle $\alpha$). Now compose a rotation of angle $\alpha$ and fixed axis with some rotation of angle $\alpha$ whose axis varies continuously. By taking the same axis, but oriented backwards, we get the rotations of angle $0$, so we have $0$ and $2\alpha$. The range of angles we get is thus an arc on the circle representing the interval $[0,2\pi)$ (I mean the circle we get by gluing the ends of that interval) which contains $0$, because of the continuous movement of the axis of the second rotation, so we must get some $\frac\pi n$ for some large $n$ (or some $-\frac\pi n$), meaning that we get $\pi$ (the $n$'th power of a rotation of angle $\frac\pi n$).

I believe this is what I wanted to say, but then sort of wandered off with that $2\pi-\alpha$ thing :).

Am I on the right track now?


inequalities
algebra
binomial theorem
inequalities solved

It's indeed true by some advanced math, though I'd like to see some elegant proof.

The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216]national contest problem[/url] was a special case when $ n\equal{}2$.

We consider the case n>=2.
WLOG, assume that x = max{x, y, z}.
We have  y^n.z <= x^(n-1).y.z, z^n.x <= z.x^n and z^n.x <= z^2.x^(n-1). In other hand, (n-1)/n >= 1/2 and we have (n-1)z/n >= z/2.
So  P = x^n.y + y^n.z + z^n.x <= x^n.y + x^(n-1).y.z + z^n.x/2 + z^2.x^(n-1)/2 = x^(n-1)(x+z)(y+z/2) <= x^(n-1)(x+z)(y+(n-1)z/n)
= n^n(x/n)...(x/n)((x+z)/n)(y+(n-1)z/n) (*)
The last expression has n-1 times (x/n) and by AM-GM (*) <= n^n((x+y+z)/(n+1))^(n+1) = n^n/(n+1)^(n+1).

The equality holds when x = n/(n+1), y = 1/(n+1) and z = 0.

This generalized inequality was published in Vietnamese journal Mathematics and Youth for 1996 year.

The sln is elemetary, but complicated.

Namdung

One can also prove by similar method that for $m,n>1$ we have

$a^mb^n+b^mc^n+c^ma^n \leq \frac{m^mn^n}{(m+n)^{m+n}}$.

[Moderator edit: I've just "created" a topic for this inequality: http://www.mathlinks.ro/Forum/viewtopic.php?t=30337 ]

Another similar approach:

 Take $ x \equal{} \text{max}\left\{x,y,z\right\}$ and we prove that $ x^ny \plus{} y^nz \plus{} z^nx\leq (x \plus{} \frac {z}{2})^n(y \plus{} \frac {z}{2})$. Write this as $ y[ (x \plus{} \frac {z}{2})^n \minus{} x^n] \plus{} \frac {z(x \plus{} \frac {z}{2})^n}{2}\geq y^nz \plus{} z^nx$. We split this into $ (x \plus{} \frac {z}{2})^n \minus{} x^n\geq y^{n \minus{} 1}z$ and $ (x \plus{} \frac {z}{2})^n\geq 2xz^{n \minus{} 1}$. The first one follows from the fact that $ (x \plus{} \frac {z}{2})^n \minus{} x^n\geq nx^{n \minus{} 1}\cdot\frac {z}{2}\geq y^{n \minus{} 1}{z}$, while the second one can be written as $ (\frac {1}{2} \plus{} \frac {x}{z})^n\geq 2\cdot\frac {x}{z}$ and it�s obvious from the binomial theorem and the fact that $ x\geq z$. Now, using AM-GM we conclude instantaneously that $ (x \plus{} \frac {z}{2})^n(y \plus{} \frac {z}{2})\leq \frac {n^n}{(n \plus{} 1)^{n \plus{} 1}}$. A beautiful problem.

Set $a=x+z/2$, $b=y+z/2$. Prove that $a^nb\geq x^n y + y^n z + z^n x$.
$ an+bn=a+a+....a+nb\geq\sqrt[n+1]{a^nbn}$.
 Then $a^nb\leq \frac{n^n}{(n+1)^n*(n+1)}$.

[quote="StRyKeR"]Let $ x$, $ y$, $ z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that

$ x^n y \plus{} y^n z \plus{} z^n x \leq \frac {n^n}{\left(n \plus{} 1\right)^{n \plus{} 1}}$

for every integer $ n\geq 2$. In this inequality, equality can hold only if one of the variables is equal to $ 0$.[/quote]

   Assume $ x \equal{}$ max$ (x,y,z)$
We have $ y\leq x \Rightarrow y^{n}z\leq x^{n \minus{} 1}yz$
              $ z\leq x \Rightarrow z^{n}x\leq z^{2}x^{n \minus{} 1}\leq zx^{n}$
              $ n > 1 \Rightarrow \frac {n \minus{} 1}{n}\geq \frac {1}{2}\Rightarrow \frac {n \minus{} 1}{n}z\geq \frac {z}{2}$
 $ \Rightarrow x^{n}y \plus{} y^{n}z \plus{} z^{n}x\leq x^{n}y \plus{} x^{n \minus{} 1}yz \plus{} \frac {1}{2}z^{n}x \plus{} \frac {1}{2}z^{n}x$
  $ \leq x^{n}y \plus{} x^{n \minus{} 1}yz \plus{} \frac {zx^{n}}{2} \plus{} \frac {z^{2}x^{n \minus{} 1}}{2}$
  $ \equal{} x^{n \minus{} 1}(x \plus{} z)(y \plus{} \frac {z}{2})$
  $ \leq x^{n \minus{} 1}(x \plus{} z)(y \plus{} \frac {n \minus{} 1}{2}z)$
Appiling AM-GM inequality:
  $ x^{n \minus{} 1}(x \plus{} z)(y \plus{} \frac {n \minus{} 1}{2}z)$
  $ \leq \frac {(x \plus{} y \plus{} z)^{n \plus{} 1}}{(n \plus{} 1)^{n \plus{} 1}}n^{n} \equal{} \frac {n^{n}}{(n \plus{} 1)^{n \plus{} 1}}$
  $ \Rightarrow$ As desire  :)  :)


calculus
integration
trigonometry
real analysis
real analysis unsolved

set the intervalle of definition then take $ x \equal{}  \arctan(u^{2})$ :huh:

[hide]Make $ \tan x \equal{} u^2$ to get $ P \equal{} \int\frac{1}{\sqrt{\tan x}}\,dx \equal{} 2 \int \frac{1}{1\plus{}u^4}\,du$ and this last primitive was been posted already several times.[/hide]


Ten football clubs compete annually for a cup which is to become the absolute property of the club which wins it for three years in succession. Assuming that all the clubs are of equal skill, find the chance that last year’s winners, not having won the previous year, will ultimately win the cup.

Let $ p$ be the probability that the last year's winners win the cup (after they've won the cup).

If they win again (probability $ \frac1{10}$), then they have a $ \frac1{10}$ chance of winning the cup.

If they win again (probability $ \frac1{10}$), they have a $ \frac9{10}$ chance of not winning the next game, after which the team that won that game has probability $ p$ to win the cup, and all other teams have a  $ \frac{1\minus{}p}9$ probability to win the cup.

If they don't win (probability $ \frac9{10}$), then, since at least one team must win, the team that won that game has probability $ p$ to win the cup, and all other teams have a $ \frac{1\minus{}p}9$ chance to win the cup.

Therefore, $ p\equal{}\frac1{10}\cdot\frac1{10}\plus{}\frac1{10}\cdot\frac9{10}\cdot\frac{1\minus{}p}9\plus{}\frac9{10}\cdot\frac{1\minus{}p}9\equal{}\frac1{100}\plus{}\frac{1\minus{}p}{100}\plus{}\frac{1\minus{}p}{10}$

$ \implies100p\equal{}1\plus{}1\minus{}p\plus{}10\minus{}10p\implies111p\equal{}12\implies p\equal{}\boxed{\frac4{37}}$.

Thanks I was doing all correct but not dividing by 9 :|

Thanks again !!


Hi, everybody.... I have question, it may be simple but I want to learn the solition... Problem is the following;

There are 6 people and they all have their own hats. Nobody will take his own hat. How many different situations can we get?

Let count the number of situations that somebody take his hat. Let it number $ C$
Let $ C_n$ the number of situations that exactly $ n$ people fixed take their hat.
If there are $ k$ people then $ C_n\equal{}(k\minus{}n)!$
Using the principle of Inclusion-Exclusion:
$ C\equal{}\binom{n}{1}C_1\minus{}\binom{n}{2}C_2\plus{}\binom{n}{3}C_3\minus{}...^\plus{}_\minus{}\binom{n}{n}C_n$

The number that we want is $ n!\minus{}C$.
For $ n\equal{}6$
$ 6!\minus{}6!\plus{}\frac{6!}{2!}\minus{}\frac{6!}{3!}\plus{}\frac{6!}{4!}\minus{}\frac{6!}{5!}\plus{}\frac{6!}{6!}\equal{}245$