algebra
polynomial
function
number theory
number theory open

Given a finite field $ \mathbb{F}_q$ and an enumeration $ x_1, ... x_q$ of the elements of $ \mathbb{F}_q$ along with $ y_1, y_2, ... y_q \in \mathbb{F}_q$, does there always exist a polynomial $ P(x) \in \mathbb{F}_q[x]$ such that $ P(x_i) \equal{} y_i \forall i$?  What is it?  What about over an infinite field of positive characteristic?

[quote="t0rajir0u"]Given a finite field $ \mathbb{F}_q$ and an enumeration $ x_1, ... x_q$ of the elements of $ \mathbb{F}_q$ along with $ y_1, y_2, ... y_q \in \mathbb{F}_q$, does there always exist a polynomial $ P(x) \in \mathbb{F}_q[x]$ such that $ P(x_i) \equal{} y_i \forall i$?[/quote]

Of course, there does. Lagrange interpolation on distinct points works over any field (and even over a commutative ring with $ 1$ if the distances between the points are all invertible).

[quote="t0rajir0u"]What is it?[/quote]

The one of smallest degree has the form $ P\left(X\right)\equal{}\minus{}\sum_{i\equal{}1}^q y_i\prod_{j\neq i}\left(X\minus{}x_j\right)$, as you can see using Wilson's theorem (more precisely, its generalization to $ \mathbb{F}_q$).

  darij

My mistake.  I was under the impression that Lagrange interpolation might require dividing by the characteristic for some reason.

So I guess the question I [i]really[/i] wanted to ask is about $ \mathbb{Z}/(p^n)$ with $ n > 1$.

[quote="t0rajir0u"]So I guess the question I [i]really[/i] wanted to ask is about $ \mathbb{Z}/(p^n)$ with $ n > 1$.[/quote]

Interesting question, but a hard one, I fear. Of course, a necessary condition for a function $ f: \mathbb{Z}\slash\left(p^n\right)\to\mathbb{Z}\slash\left(p^n\right)$ to be representable as a polynomial is that for any $ \alpha\in\left\{0,1,...,n \minus{} 1\right\}$, we have $ f\left(x_i\right) \minus{} f\left(x_j\right)\in p^{\alpha}\mathbb{Z}\slash\left(p^n\right)$ for any $ i$ and $ j$ such that $ x_i \minus{} x_j\in p^{\alpha}\mathbb{Z}\slash\left(p^n\right)$. For $ n \equal{} 1$ and for $ \left(p,n\right) \equal{} \left(2,2\right)$, this is also a sufficient condition, but this doesn't trivially generalize, even not in the case of $ f$ being a permutation: For $ p > 2$ and $ n > 1$, the $ p^n \minus{} 1$-th finite difference $ \sum_{k \equal{} 0}^{p^n \minus{} 1}\left( \minus{} 1\right)^k\binom{p^n \minus{} 1}{k}P\left(k\right)$ of any polynomial $ P$ is zero modulo $ p^n$ (in fact, the $ p^n \minus{} 1$-th finite difference is a polynomial all of whose coefficients are divisible by $ \left(p^n \minus{} 1\right)!$, but $ \left(p^n \minus{} 1\right)!$ is divisible by $ p^n$), what in most cases does not hold for every permutation of $ \mathbb{Z}\slash\left(p^n\right)$ (at least, it doesn't hold for $ p > 2$). For $ p \equal{} 2$ and $ n > 2$, we can argue similarly using the $ p^n \minus{} 2$-th finite difference (which, applied to the polynomial, will be $ 0$ modulo $ p^n$ because it will be a polynomial with its coefficients being divisible by $ \left(p^n \minus{} 2\right)!$, which also happens to be divisible by $ p^n$). It is not really clear what conditions we should add to ensure the representability of our permutation as a polynomial of degree $ p^n \minus{} 1$.

  darij

Back to the first question and answer:
When we want to do an interpolation in a finite field GF(p), I know that we must use the lagrange interpolation formula
and perform the multiplication and division and addition " modulo p ".
Is it enough ? or other calculations must be performed too ?

For example (for 2 points) : y = y1 (x-x2) / (x1-x2) + y2 (x-x1) / (x2-x1)
Is it enough to calculate the inverse of (x1-x2( and also (x2-x1) and perform the modulo multiplications......etc. ?

Thanks.


complex analysis
function
algebra
polynomial
integration
calculus
absolute value

[color=blue] [b] Problem. [/b]
 Let $f: \mathbb{C}\to\mathbb{C}$ an holomorphic function such that exists $a, b>0$ and $R>0$ such that
 \[ |f(z)|\geq a|z|^{b},  \] for any $z\in\mathbb{C}$, $|z|\geq R.$
 Prove that $f$ is a polynomial function of degree $\geq b$.

 [/color]

Since $f(z)\to\infty$ as $z\to\infty$, $f$ has a pole at $\infty$, and is meromorphic on the whole extended plane. All such functions are rational functions. Since $f$ has no finite poles, $f$ is a polynomial. Now, the size comparison immediately shows that $f$ has degree at least $b$.

Jmerry can you  explain a little more? I have just started to learn complex analisys and I don't think that I have understood completely your solution. :)

Consider $g(z)=\frac1{f(1/z)}.$ Then on some punctured neigborhood of zero, $|g(z)|\le c|z|^b$ for some $c.$ Thus $g$ has a removable singularity at $0,$ and furthermore, removing that singularity would give us $g(0)=0.$ Then there is an integer $m\ge b$ such that $g(z)=z^mh(z)$ where $h$ is holomorphic on a neighborhood of $0$ and $h(0)\ne 0.$

We now have that for sufficiently large $|z|,\,f(z)=z^m[h(z^{-1})]^{-1}.$ Since $h$ tends to a nonzero limit, we have $|f(z)|=O(|z^m|)$ as $|z|\to\infty.$

Now we can use one of Cauchy's estimates. For any given $z,$ let $C_r$ be the circle centered at $z$ of radius $r.$

$f^{[m+1]}(z)=\frac{(m+1)!}{2\pi i}\int_{C_r}\frac{f(\zeta)}{(\zeta-z)^{m+2}}\,d\zeta.$

Take the limit as $r\to\infty.$ From above, we have that $|f(\zeta)|=O(r^m).$ The demominator in that integral has absolute value $r^{m+2}$ and the arclength of the circle is $2\pi r.$ Thus the whole expression is $O(r^{-1})$ and goes to zero as $r\to\infty.$

We conclude that $f^{[m+1]}(z)=0$ for all $z.$ This means that $f$ is a polynomial of degree no greater than $m.$ Going back to the beginning of the problem, we realize that $f$ must actually be a polynomial of degree exactly $m,$ for the $m\ge b$ that we determined above.

This is standard theory; it's basically just what jmerry said.


It takes Bryan one hour to dig a hole that is $ 1.4$ meters wide,
$ 1.5$ meters long and $ 0.5$ meters deep. At the same rate, how
many hours will it take Bryan to dig a hole that is $ 2.8$ meters
wide, $ 3.0$ meters long and $ 1.0$ meters deep?

Since all the dimensions are all doubled, the time will take $ 2\cdot 2\cdot 2\equal{}8$ times as long compared to before, or 8 hours

Another way is to use proportions.

The volume of the first hole is $ 1.4 \times 1.5 \times 0.5 \equal{} 1.05$. The volume of the second hole is $ 2.8 \times 3 \times 1 \equal{} 8.4$.

$ \frac{1.05}{1} \equal{} \frac{8.4}{x}$

$ 1.05x \equal{} 8.4$

$ x \equal{} \boxed{8}$


algebra
polynomial
calculus
integration
algebra proposed

Do there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers and a non-constant polynomial $P(x)$ such that $a_{m}+a_{n}=P(mn)$ for every positive integral $m$ and $n?$

[i]Proposed by A. Golovanov[/i]

[quote="mathmanman"]Do there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers and a non-constant polynomial $P(x)$ such that $a_{m}+a_{n}=P(mn)$ for every positive integral $m$ and $n?$[/quote]

It does not :

$m=1$ $\Rightarrow $ $a_{n}=P(n)-a_{1}$
So we have $P(m)+P(n)-2a_{1}=P(mn)$ and for example $2P(n)-2a_{1}=P(n^{2})$ which can't be true $\forall n$ (take $n\rightarrow+\infty$) if $P(x)$ is non constant.

--
Patrick

[quote="mathmanman"]Do there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers and a non-constant polynomial $P(x)$ such that $a_{m}+a_{n}=P(mn)$ for every positive integral $m$ and $n?$[/quote]
We get $a_{n}=P(n)-a_{1}, \ a_{1}=P(1)/2.$ If $deg(P)=k$, then for big n $|a_{k}|n^{k}/2=c_{1}n^{k}<|P(n)|<2a_{k}n^{k}=c_{2}n^{k}$.
Therefore $|P(n^{2})|>c_{1}n^{2k}>2c_{2}n^{k}>|P(n)|+|P(n)|$ - contradition.

It is just equivalent to a polynomial which satisifes.$P(m^2)+P(n^2)=2P(mn)$,but taking $m=1$,due to unequal degrees,$P$ doesn't exist.


You should try putting those questions on some other forum.  :D They certainly aren't MATHCOUNTS level.

[quote="unknown_pattern18"]why is it not a mathcount? i dont get it plz xplain[/quote]
It's too hard. ;)  :D

MATHCOUNTS is a competition meant for middle schoolers.  It usually has mostly word problems, and geometry.  In this forum, we post questions on the difficult middle school level, but thanks for joining AoPS.  :D

imagine (base-1) is (b-1) because i couldnt put maths symbol for base i wrote the base.  think of it as a new symbol that is not used in the equation. like phi,theta,delta....... but its different to the b that is used in the equation

[quote="unknown_pattern18"]i dont know how to write the problem, its not a small problem  i have worked on it for 2 years.but anyways i wil just write two of them, any help is appreciated.

here is the first problems.
 bn^b=t                   make b the subject of the formula

i am new to this forum i dont know how to write it in mathematical form but let me know if u didnt undrestand any part.

the other problems is this

d=((a-1)n^(base-1))+n(b-1)+n(c-1)+n(d-1)+............+n(y-1)+z
make n the subject of  formula

 :idea:  :D[/quote]
I think this is what the problem means:
Solve for b in terms of n and t:
$bn^b=t$

true
Solve for b in terms of n and t: or in other word make b the subject
bn^b=t

There is no elementary expression for the solution.  In certain special cases you might be able to determine the values of $b$ which work and prove there are no others, but there is no "nice" function or formula which will give you the exact answer in general.  (One could, and I'm sure someone has, defined such a function, but giving it a name won't help you evaluate it.)

Your second question seems to be presented in an odd fashion.  As written, why don't you collect all of the later terms into a single linear and constant term?  Also, why all the minus-ones?  What makes your question different from the following:
Solve for $n$ if $d = a n^e + bn + c$.  In this case, the same comment applies as above.  In particular, we can easily choose the parameters to make the equation $n^5 - n = -1$, an equation which is known to have no solutions expressible in radicals.


Let $f,g \in \mathbb{Z} [X]$ s.t. degree $f,g \geq 1$ and $\gcd (f,g) = 1$. Prove that $f(a) | g(a)$ happens only for finitely many integer values of $a$.

Nice problem

From Bezout we can find $u, v \in \mathbb{Z}[X]$ such as $u \cdot f + b \cdot g = 1$
So if $f(a)|g(a)$ then $f(a) | 1$ which implies $f(a) = \pm 1$ which has only finite number roots.

[quote="tµtµ"]From Bezout we can find $u, v \in \mathbb{Z}[X]$ such as $u \cdot f + b \cdot g = 1$[/quote]

Are you sure this is true? I thought that $u,v \in \mathbb{Q} [X]$. (maybe i'm talking gibberish)

Anyway, even in this case your solution remains true: we multiply the equality by convenient $k$ to get $f(a) | k$, which again has only finitely many solutions.


A very classic problem.

Simplify: $ \sqrt{33 \minus{} 8 \sqrt{17}}$

Let's guess that the answer is $ \sqrt{a} \minus{} \sqrt{b}$.

Then, we have

$ (a\plus{}b) \minus{} 2\sqrt{ab} \equal{} 33 \minus{} 2\sqrt{16\times17}$

$ a\plus{}b\equal{} 33$

$ ab \equal{} 16\times 17$

Therefore,$ a \equal{} 17$, $ b \equal{} 16$.

So the answer is $ \sqrt{17} \minus{} \sqrt{16}$

Which can be simplified to:

$ \sqrt{17}\minus{}4$  :lol:

When I solve this type of problem, I always set the RHS to be $ a\plus{}b$. I mean, it's fine to use $ \sqrt{a} \minus{} \sqrt{b}$ but yeah, that's just me. You still got the right answer though.

Really?

I usually set the RHS to be $ a\plus{}b\sqrt{c}$, where c is the number in the radical of the original problem, which in this case is 17.

Let this be equal to $ a\plus{}b\sqrt{c}$. We can safely assume that $ c\equal{}17$.

Therefore, $ a\plus{}b\sqrt{17}\equal{}\sqrt{33\minus{}8\sqrt{17}}$.

Squaring,

$ a^2\plus{}17b^2\plus{}2ab\sqrt{17}\equal{}33\minus{}8\sqrt{17}$.

Matching terms,

$ a^2\plus{}17b^2\equal{}33$

$ 2ab\equal{}\minus{}8 \implies ab\equal{}\minus{}4$.

Solving, we get $ a\equal{}4$ and $ b\equal{}\minus{}1$.

So our answer is $ 4\minus{}\sqrt{17}$

Checking, $ 4\minus{}\sqrt{17}\equal{}\sqrt{33\minus{}8\sqrt{17}} \implies 16\plus{}17\minus{}8\sqrt{17}\equal{}33\minus{}8\sqrt{17}$,

Which is correct.

Just make sure C is prime.

Try $ \sqrt{5\plus{}2\sqrt{6}}$

Whoops, error.

Squaring gives $ 2$ solutions, not just $ 1$.

We also could have $ \minus{}(4\minus{}\sqrt{17}\equal{}\sqrt{17}\minus{}4$.

This solution is the ONLY correct one, because the square root of a number

in its usual context can't be negative.

@pac-man the answer is $ 5 \plus{} 2\sqrt{6} \equal{} (\sqrt{2} \plus{} \sqrt{3})^2$


[quote="Interval"]Going into the final exam, which will count as 2/3 of the final grade, Terri has to test scores of 86, 80, 84 and 90.

a) What score does Terri need on the final in order to earn a B, which requires an average score of 80?

b) What does she need to earn an A, which requires an average of 90?[/quote]

Let's see...

[hide] The average of her scores so far is (86+80+84+90)/4, or 85. This counts as 1/3 of the final grade. She wants her final grade to be 80 for (a), so the average of her test grades before (85) and twice her final grade now must be 240. So she needs (240-85)/2, or 77.5 on her final test. For (b) the sum must be 270, so 92.5.[/hide]

Not 100% sure, though.


How many positive integers less than 1991 are of the form $x^n$, where $x \ge 1$ and $n \ge 2$ are integers?

[quote="tarquin"][hide]Counting
$2^2 \rightarrow 2^{10}$
$3^2 \rightarrow 3^6$
$5^2 \rightarrow 5^4$
$7^2 \rightarrow 7^3$
$11^2 \rightarrow 11^3$
Then all the others are just prime squares: $13, 17, 19, 23, 29, 31, 37, 41, 43$.
Add them up: $9 + 5 + 3 + 2 + 2 +9 = 30$.[/hide][/quote]

why did you not count the 6's, or the 10's, or 14, 15, 18, or (1)etc???

[hide]i got:
1
$2^2 \rightarrow 2^{10}$
$3^2 \rightarrow 3^6$
$5^2 \rightarrow 5^4$
$6^2 \rightarrow 6^4$
$7^2 \rightarrow 7^3$
$10^2 \rightarrow 10^3$
$11^2 \rightarrow 11^3$
$12^2 \rightarrow 12^3$
and all numbers 13-44 that aren't perfect squares or cubes.
so thats 1 + 9 + 5 + 3 + 3 + 2 + 2 + 2 + 2 + (32-4) = 57??? [/hide]

Hmmm...here's what I did.  :?

[hide="Solution?"]For each value of n, there are the following possible number of values for x:

n=2;     x=44
n=3;     x=12
n=4;     x=6
n=5;     x=4
n=6;     x=3
n=7;     x=2
n=8;     x=2
n=9;     x=2
n=10;     x=2

However, the numbers where n=4,6,8,10 have already been counted under n=2. Likewise, the numbers where n=6,9 have already been counted under n=3.

Also, for n=3, x=1,4,9 have already been counted as perfect squares. And for n=5, x=1,4 have already been counted. And for n=7, x=1 has already been counted.

This means that the total number of distinct integers is $(44+12+4+2)-(3+2+1)=56$?[/hide]


calculus
trigonometry
geometry
geometry unsolved

Consider all the triangles $ABC$ which have a fixed base $AB$ and whose altitude from $C$ is a constant $h$.  For which of these triangles is the product of its altitudes a maximum?

This problem can be tackled using calculus but the following approach is more direct. Let $h_a$ and $h_b$ be the altitudes from $A$ and $B$, respectively.

Then $AB\cdot h\cdot AC\cdot h_b\cdot BC\cdot h_a=8[ABC]^2=(AB\cdot h)^3$

which is a constant. So the product $h_a\cdot h_b\cdot h$ attains its maximum when the product $AC\cdot BC $ attains minimum.

Since $AC\cdot BC\cdot \sinh = BC\cdot h_a=2[ABC]$, which is constant, $AC\cdot BC$ attains its minimum when $\sin C$ reaches its maximum. There are two cases:

1) $h\le\frac{AB}{2}$. Then there exists a triangle $ABC$ which has a right angle at $C$, and for precisely such a triangle $\sin C$ attains its maximum namely $1$.

2) $h>\frac{AB}{2}$. In this case the angle at $C$ is acute and assume its maximum when $AC=BC$.

[hide="Another approach"]
Let the altitude from C meet AB at D, and let AD=a, BD=b, so that $a+b$ is constant. Then,

$AC=\sqrt{a^2+h^2}, BC=\sqrt{b^2+h^2}$

and so the product of the altitudes (noting that the area is constant) is

$h(\frac{(a+b)h}{\sqrt{a^2+h^2}})(\frac{(a+b)h}{\sqrt{b^2+h^2}})=\frac{(a+b)^2h^3}{\sqrt{(a^2+h^2)(b^2+h^2)}}$.

As $(a+b)^2h^3$ is constant, maximizing this is equivalent to minimizing

$(a^2+h^2)(b^2+h^2)=a^2b^2+h^2(a^2+b^2)+h^4=a^2b^2-2h^2ab+h^2(a+b)^2+h^4$.

As $h^2(a+b)^2+h^4$ is fixed, this is equivalent to minimizing

$a^2b^2-2h^2ab=(ab-h^2)^2-h^4$, which is equivalent to minimizing $(ab-h^2)^2$.

Clearly $(ab-h^2)^2$ attains its minimum at 0, at which point $ab=h^2$. Note that this is possible iff $\frac{a+b}{2} \geq \sqrt{ab} \implies a+b \geq 2h$.

Furthermore, this implies that D lies in a circle on diameter AB going through C by Power of a Point. In particular, this means that ABC is right.

If $a+b<2h$, then we must maximize $ab$ given $a+b$, or $a=b$ implying that ABC is isosceles. We can write this more concisely as

$h \leq \frac{AB}{2} \implies \triangle ABC~\text{right}, h>\frac{AB}{2} \implies \triangle ABC~\text{isosceles with AC=BC}$.[/hide]

Product of altitudes $= c^2\sin A\sin B$, thus we need to maximise $\sin A\sin B$. We have $\frac{abc}{4R} = \frac{1}{2}\cdot c\cdot h = $ constant say $k$. Thus $\sin A\sin B = \frac{k}{c\cdot R}$ $\implies$ we need to maximise $\frac{1}{R}$ or minimise $R$.

[asy]
import olympiad;
import geometry;
unitsize(3cm);

pair C = dir(130), B = dir(210), A = -1/B;
triangle tri = triangle((point) A, (point) B, (point) C); label("A", A, A);
label("B", B, B*dir(10));
label("$C'$",C,C);
draw(tri);

pair M = midpoint(A--B); dot(M);

pair D = A+C-B; pair E = foot(C,A,B); pair F = A+C-E; pair G = 2F-D; pair C_1 = (G+C)/2; dot(C_1); pair D_1 = 2C-C_1; draw(C_1--M,red);draw(D--D_1,blue);
label("$C$",C_1,C_1); label("M",M,M); draw(A--C_1);draw(B--C_1); pair O = circumcenter(C_1,A,B); dot(O);label("O",O,O*dir(90)); pair O_1 = circumcenter(C,A,B); dot(O);label("$O'$",O_1,O_1*dir(-90)); dot(O_1);

[/asy]

Let $C$ be such that $AC = BC$ and $C'$ be a point on line parallel to $AB$ through $C$ and $C'\not\equiv C$. Let $O$ be the circumcenter of $ABC$ and $O'$ be the circumcenter of $ABC'$.

If $CO<CO'$ then $CO<C'O'$ and $AO>AO'$. Thus $CO<CO'=AO'<AO=CO$ which is not possible. Thus we have that $AO<AO'$ for all $C'$ which gives that smallest circumradius is when $AC = BC$.


algebra
polynomial
calculus
derivative
real analysis
real analysis solved

Let $P$  be a polynomial of degree $n$ having the roots $x_1,x_2,...,x_n,$ where   $-\infty < x_1 < x_2 <...<x_j<...<x_n < \infty .$

If $y=y(x_1,x_2,...,x_n)$  is a root of the derivative $P'$ , show that
 \[ 0 < y'(x_j) < 1 \; \; ,\; \; \mbox{for}\; \; \; j\in \{1,2,...,n\}. \]
[i]Note: [/i]   $\displaystyle y'(x_j): = \displaystyle \frac{\partial y(x_1,...,x_j,...,x_n)}{\partial x_j} \; .$

The basic ideea is $\displaystyle P'(x) = P(x)*\sum_{i=1}^{n}\frac{1}{x-x_i}$ . All root $Y$ of$ P'$ is not in $\{x_1,x_2,...,x_n\}$, because $x_1,x_2,...,x_n$ are distinct, so$ P(Y) \neq 0$.

 $ 0=P'(Y)=P(Y)*\sum_{i=1}^{n}\frac{1}{Y-x_i} \Longrightarrow \sum_{i=1}^{n}\frac{1}{Y-x_i} = 0 $ $\Longrightarrow \frac{1}{(Y-x_j)^2}  - Y'_{x_j}*\sum_{i=1}^{n}\frac{1}{(Y-x_i)^2}= 0 $ $\Longrightarrow Y'_{x_j} = \frac{\frac{1}{(Y-x_j)^2}}{\sum_{i=1}^{n}\frac{1}{(Y-x_i)^2}} \Longrightarrow  0 < Y'_{x_j}  < 1 $

 :cool:


Adrian teaches a class of six pairs of twins. He wishes to set up teams
for a quiz, but wants to avoid putting any pair of twins into the same
team. Subject to this condition:
i) In how many ways can he split them into two teams of six?
ii) In how many ways can he split them into three teams of four?

I think I understand what's happening, i.e. we want to create restricted subsets, but when I do the computation it always seems strange. I think I'm missing something here, any solutions?

[quote="bobbym"]Hi;

Isn't there only one way for the first one?[/quote]

Don't think so, consider you had only two pairs of twins, $ A_1$, $ A_2$, $ B_1$, $ B_2$, from which you could construct two sets of two teams:
1. $ \{A_1,B_1\}, \{A_2,B_2\}$
2. $ \{A_1,B_2\}, \{A_2,B_1\}$

and so if you had six pairs, I'm sure you could create more than just the one set o' teams ;P

Hi;

Not with two teams of 6 people from 6 twins. If you can't have any twins together: He doesn't say that we should consider the twins as different.

aa bb cc dd ee ff

Team 1) a,b,c,d,e,f

Team 2) a,b,c,d,e,f

What else?

Heh, I actually don't know if the twins are different or not :)\

guess i'll have to find out.


I've heard rumors on this site that they send this book to all USAMO qualifiers. Is this true?

Also, what kind of information does this book cover, and when do they send it?

Sorry, double post. Mods, could you delete this one?

You can delete it yourself, you know.  Just click the x button at the bottom right-hand corner.  Yes, I'm wondering the same question about the bok--SOMEONE ANSWER PLEASE

[quote="probability1.01"]If I'm not stupid (40% chance) and not forgetful (20% chance) then they do.[/quote]

Don't worry. You're neither stupid nor forgetful (48% chance).

What if there is a correlation between stupid and forgetful?

Hm... I interpreted that as a 40% chance of not being stupid and a 20% chance of not being forgetful (at least that's how the numbers would work for me).

Yes, that is what I meant.  Apparently the 8% chance that both are true occured!


Prove that for every integer $ n\geq 1$ there exist a polynomial $ P(x)$ with integer coefficients such that $ P(1),P(2),...,P(n)$ are distinct powers of two.

Posted before, for example $ P(x)\equal{}\sum_{k\equal{}0}^n \binom{x}{k}.$

Rust, you'll notice that your choice of $ P(x)$ does not have integer coefficients.  ;)

Yes.
But I give hear solution for $ P(i)\equal{}a^{k_i}$ for any integer $ a>1$. I don't want repeat it.

Please give a correct solution, Rust  :wink:


i think this is right?

[hide]

First off, from sum of an infinite geo series:

$\displaystyle\frac{b_1}{1-r} = \sqrt2 + 1$

$b_1 = (1-r)(\sqrt2 +1)$

Then we know:

$b_1 = a_1^2$ [i](1)[/i]
$b_1r = (a_1+d)^2 = a_1^2 + 2a_1d + d^2$[i] (2)[/i]
$b_1r^2 = (a_1+2d)^2 = a_1^2 + 4a_1d + 4d^2$ [i](3)[/i]

Subtracting (1) from (2), you have $b_1(r-1) = 2a_1d + d^2$

Subtracting (2) from (3), you have $b_1r(r-1) = 2a_1d + 4d^2$

Subtracting those two, you have: $(r-1)^2(b_1) = 3d^2$ [i](4)[/i]

Subtracting (1) from (3), you have $b_1(r-1)(r+1) = 4a_1 + 4d^2 = 4d(a_1 + d) = 4d \sqrt{b_1r}$ [i](5)
[/i]
Dividing (5) into (4), you have:

\[ \displaystyle\frac{r-1}{r+1} = \frac{3d}{4\sqrt{b_1r}}\]

Simplifying you have:

$16(b_1r)(r-1)^2 = (r+1)^2(9d^2) \Rightarrow 16(1-r)(\sqrt2 + 1)(r)(r-1)^2 = (r+1)^2(9d^2)$ [i](6)[/i]

Finally, Dividing (4) into (6), you get:

$16r = 3(r+1)^2$
$0 = 3r^2 - 10r+3$
$r = 3$ or $\displaystyle\frac{1}{3}$

Plugging the values of $r$ into (4), we see that $3$ is erroneous (taking us into the imaginary numbers), so:

$(r-1)^2(1-r)(\sqrt2+1) = 3d^2$

$\displaystyle\frac{8(\sqrt2+1)}{81} = d^2$

$\sqrt{\displaystyle\frac{8(\sqrt2+1)}{81}} = \frac{2\sqrt{2\sqrt2+2}}{9}d$

(We can toss out the negative possibility for $d$ because it specifies in the problem that $a_2 > a_1$, so $d > 0$)

Then, to find $a_1$, we have:

$a_1 = \sqrt{b_1} = \sqrt{(1-r)(\sqrt2+1)} = \sqrt{\displaystyle\frac{2}{3}(\sqrt2+1)}$

[/hide]


For each positive integer $k$, prove that there is a perfect square of the form $n2^{k}-7$, where $n$ is a positive integer.

http://www.mathlinks.ro/Forum/viewtopic.php?t=90294
Solutions are identical...


inequalities
geometry
area of a triangle
Heron\u0027s formula

If a,b,c are the lengths of the sides of a triangle, prove that

$ abc \geq (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$

hello,substituting $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ in your inequality, we get
$ (x\plus{}y)(x\plus{}z)(y\plus{}z)\geq8xyz$ wich is true by AM-GM.
Sonnhard.

Thanks.  Is there any way to use the fact that a,b,c form the sides of a triangle?  It's just the right hand side looks a lot like Heron's formula, and the left hand side looks a bit like the formula for the area of a triangle, S = abc/4R.

hello, yes it is. From $ R\geq2r$ we have with $ R\equal{}\frac{abc}{4A}$,$ r\equal{}\frac{A}{s}$ and $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$
$ sabc\geq8A^2$ and from here you will get your inequality.
Sonnhard.

For whatever reason, the right hand side looks familiar. I think its part of the proof for herons formula.

Possibly, but it is more likely the alternative form, $ A\equal{}\frac14\sqrt{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}$.

A lot of these substitutions are standard. You just have to know when to use them...

[quote="hsiljak"][quote="Wickedestjr"][quote="Dr Sonnhard Graubner"]hello,substituting $ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$ [/quote]

Good solution. I just have a small question. How did you know that you had to do this?[/quote]

It's the most common technique used with inequalities dealing with sides of triangle. It transforms the geometrical inequality in pure algebraic form.

Technical note no. 1: $ x,y,z$ are segments on triangle sides between incircle tangential points and vertices of the triangle.
Technical note no. 2: note that such a substitution is based on triangle inequality...[/quote]

Can it be used to solve any inequality?

Are you always changing it into x + y, x + z, and y + z, when there are three variables?

@Sonnhard - What is AM-GM? Sorry I have never heard of that before.

[quote="Wickedestjr"][/quote][quote="hsiljak"][quote="Wickedestjr"][quote="Dr Sonnhard Graubner"]hello,substituting $ a = y + z,b = x + z,c = x + y$ [/quote]

Good solution. I just have a small question. How did you know that you had to do this?[/quote]

It's the most common technique used with inequalities dealing with sides of triangle. It transforms the geometrical inequality in pure algebraic form.

Technical note no. 1: $ x,y,z$ are segments on triangle sides between incircle tangential points and vertices of the triangle.
Technical note no. 2: note that such a substitution is based on triangle inequality...[/quote][quote="Wickedestjr"]

Can it be used to solve any inequality?

Are you always changing it into x + y, x + z, and y + z, when there are three variables?

@Sonnhard - What is AM-GM? Sorry I have never heard of that before.[/quote]
That substitution only is used when a,b,c are the sides of a triangle, since it's not necessarily true that any three numbers can be broken up that way and still positive (which is crucial for [url=http://en.wikipedia.org/wiki/AM-GM]AM-GM[/url]).

[quote="Machodog"]If a,b,c are the lengths of the sides of a triangle, prove that

$ abc \geq (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$[/quote]
This inequality is true for a,b,c>0. We can easily prove it by using AM-GM and assume that a=max(a,b,c).One more note about this inequality is that it's an alternative form of Schur's inequality, which is popular on Olympiad inequality problems.
About the substitution:It's a famous substitution and you can see how incredibly useful it is if you do the last IMO problem in 1980s(sorry I don't remember specifically what year, but it's around 1980-1986). You can also read the history of that problem(about a 14 German guy).

There is another nice solution. Squaring,
$ a^2b^2c^2 \ge (a \plus{} b \minus{} c)^2(b \plus{} c \minus{} a)^2(c \plus{} a \minus{} b)^2$
But $ (a \plus{} b \minus{} c)(c \plus{} a \minus{} b) \equal{} (a \plus{} (b \minus{} c))(a \minus{} (b \minus{} c)) \equal{} a^2 \minus{} (b \minus{} c)^2 \le a^2$
Multiplying cyclically gives the desired result.

[quote="Bugi"][quote="nickolas"]Its just shur's inequality[/quote]

This is a thread old for months. And you wrote nothing new, see ghjk's last post. So, don't post if you have nothing new to say.[/quote]
Is this your site?

I think this a well-known problem

is there any other method to prove this problem? :D

if you mean the first inequality, there are many solution, it's a result of Schur,
[b]solution 1,[/b] result ok Schur inequality Schur.
[b]solution 2,[/b] case1, suppose that $ a,b,c$ are not lengthes of side of triangle, and suppose that $ a \equal{} max\{a,b,c\}$
then $ a \plus{} b \minus{} c\geq0$ and $ a \minus{} b \plus{} c\geq0$ and$ b \plus{} c \minus{} a\leq0$ then $ (a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)\leq 0 \leq abc$
case2, if $ a,b,c$ are lengthes of side of triangle proved by Dr Sonnhard Graubner.
[b]solution 3,[/b] see the second post of Dr Sonnhard Graubner
[b]solution 4,[/b] using this, $ a\geq \sqrt {a^2 \minus{} (b \minus{} c)^2} \equal{} \sqrt {a \plus{} b \minus{} c}\sqrt {a \minus{} b \plus{} c}$
then $ b\geq \sqrt { \minus{} a \plus{} b \plus{} c}\sqrt {a \plus{} b \minus{} c}$ and $ c\geq \sqrt {a \minus{} b \plus{} c}\sqrt { \minus{} a \plus{} b \plus{} c}$ with product you find,
$ (a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c) \leq abc$
for the moment,I don't find any other solution.  :D

Let a,b,c be the sides of a triangle. Can you prove that there exist always 3 numbers x,y,z which satisfy the equalities:

$ a\equal{}x\plus{}y$
$ b\equal{}x\plus{}z$
$ c\equal{}y\plus{}z$

[quote="hsiljak"]Technical note no. 1: $ x,y,z$ are segments on triangle sides between incircle tangential points and vertices of the triangle.
Technical note no. 2: note that such a substitution is based on triangle inequality...[/quote]


Since some people here think I am trying to get you guys to do my homework (which is completely incorrect), can someone please explain to me how to solve the following function OR a similar problem with different values:

A function $ f$ satisfies the following conditions: $ f(5)\equal{}20$, $ f'(5)\equal{}2$, and $ f''(x)<0$, for $ x\geq5$. Which of the following are possible values for $ f(7)$?

$ 26, 20, 24, 28, 22$

Have you noticed that the other calculus problems you've posted have been moved to the calculus forum?  Or the very first thread at the top of this forum that is titled "PLEASE do not post calculus problems in this forum!"?  (Don't go reposting this problem -- a moderator should be along to move it shortly -- but in the future, do try to figure out and follow the posting guidelines when you start posting.)  (This may also have something to do with your atrocious rating  :wink:  That, and that you post homework exercises without providing any of your own ideas or workings.)

[hide]The function increases past $ x\equal{}5$ however the graph is concave down. If you follow the tangent line to $ x\equal{}7$ you reach a height of 24 and therefore $ f(7)$ must be smaller than $ 24$ making it $ 22$ and $ 20$.[/hide]

[quote="JoeBlow"]A nearly identical question with the numbers changed:

You begin walking at 4 miles per hour, and as you tire you gradually slow down over the next two hours.  What's the furthest you could have gone in those two hours?  In particular, could you have possibly gone a distance of 10 miles?  8 miles?  How about 6 miles?[/quote]

Assuming you only walked for two hours, wouldn't the only possible answer be 6 miles, otherwise you wouldn't have slowed down over the two hour period.


Prove that

$ \frac{r_br_c}{a^2}\plus{}\frac{r_cr_a}{b^2}\plus{}\frac{r_ar_b}{c^2}\geq\frac{(5R\minus{}r)(2R\minus{}r)}{3R^2}$

[quote="Ligouras"]Prove that

$ \frac {r_br_c}{a^2} \plus{} \frac {r_cr_a}{b^2} \plus{} \frac {r_ar_b}{c^2}\geq\frac {(5R \minus{} r)(2R \minus{} r)}{3R^2}$[/quote]

The following inequality is equivalent to

$ 18ac^3b^2 \plus{} 18b^3a^2c \plus{} 18a^3cb^2 \plus{} 18ab^3c^2 \minus{} 87a^2b^2c^2 \plus{} 18ba^2c^3 \plus{} 18a^3bc^2$

$ \minus{} 20a^4bc \minus{} 20acb^420abc^4 \plus{} 2b^5c \plus{} 4a^2b^4 \plus{} 2bc^5 \plus{} 4a^2c^4 \plus{} 2ca^5 \plus{} 2c^5a \plus{} 4b^2c^4 \plus{} 2ba^5$

$ \plus{} 2b^5a \minus{} a^6 \minus{} b^6 \minus{} c^6 \plus{} 2c^3b^3 \plus{} 4a^4b^2 \plus{} 4c^2a^4 \plus{} 2a^3c^3 \plus{} 4b^4c^2 \plus{} 2b^3a^3\geq 0$

$ \Longleftrightarrow 12\sum (x^4(x \minus{} y)(x \minus{} z)) \plus{} 48\sum(x^3(y \plus{} z)(x \minus{} y)(x \minus{} z))$

$ \plus{} 44(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\plus{} 32\sum(y^2z^2(x \minus{} y)(x \minus{} z))$

 $ \plus{} 188xyz\sum(x(x \minus{} y)(x \minus{} z)) \plus{} 76xyz\sum((y \plus{} z)(x \minus{} y)(x \minus{} z))\geq 0$


modular arithmetic
Euler
function
number theory
totient function
prime factorization

[quote="xhro23"]009 po ba?[/quote]

Can you post a solution? I got a different answer.

try to use euler theorem, extension to fermat theorem.. but euler used it for any n.
you can see it here http://en.wikipedia.org/wiki/Euler's_theorem

lets proceed :D
a phi(o) = 1  mod 1000.

then we count how many number are coprimes of 1000 from 1 to n.  we will end up 1000(1-1/2 -1/5 + 10) = 400,
means we subtract all those divisbile by 2, 5 and add up those numbers divisible by 2and 5 because we counted twice.

so a^400= 1 mod 1000 where a must be integer coprime to 1000 which 9 or 3^2 incidentally..
so

3^400 = 1 mod 1000..
which means 9^200 must end in 001 to make the last 3 digit 000 upon subtracting 1..

so i agree with jvenezuela.. :D

regards

Yes, Euler's generalization works also. Nice solution.

By the way, there is a fast way to evaluate Euler's totient function (the phi function). Consider the prime factorization of $ n \equal{} p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}\dots$, where each $ p_k$ is a distinct prime in the prime factorization of $ n$ and each $ e_k$ is non-zero. Then we have:

$ \phi (n) \equal{} n\left(\frac {p_1 \minus{} 1}{p_1}\right)\left(\frac {p_2 \minus{} 1}{p_2}\right)\dots \left(\frac {p_k \minus{} 1}{p_k}\right) \dots$

In our case, $ 1000 \equal{} 2^3\cdot 5^3$, so:

$ \phi (1000) \equal{} 1000\left(\frac {1}{2}\right)\left(\frac {4}{5}\right) \equal{} 400$


function
number theory
prime factorization

For all irational numbers, there is a very close rational number. Ex/ $ \sqrt{7}$ on a graph could just as well be $ \frac{10583}{4000}$. For this reason, your problem is ambiguous unless you meant $ f(x) \equal{} 1\;\;\forall x\in\mathbb{Z}$ (all integers)

[quote="Alicea"]Who can draw the graphic of the following function: f(x) =1 if x is rational                                                                           and                                                                      f(x) =0 if x is irrational?[/quote]

This question is ridiculous as it would literally FOREVER to graph anything that is remotely credible of this equation.

If you mean

${ f(x)=\{\begin{array}{c}1,\forall{x}\in\mathbb{Z}\\ \\ 0,\forall{x}\not\in\mathbb{Z}\end{array}}$

Then you would just graph the lattice points on $ y=1$ and graph $ y=0$ with empty circles in the lattice points, to signify that they are not being graphed.

[quote="MellowMelon"]I'm guessing that the function the poster gave is indeed what was meant, though I have no idea why he/she would ask someone to draw it. It does have other interesting properties.

Prime factorization, there's no ambiguity in the function. It's just not continuous at any point and pretty much impossible for the most powerful supercomputer imaginable to graph. But every real number is definitely rational or irrational, so it is well-defined.[/quote]

Supercomputer? Here's a good approximation:

[asy]size(100); draw((-5,0)--(5,0)); draw((-5,1)--(5,1));[/asy]

[quote="MellowMelon"][quote="Hamster1800"]Supercomputer? Here's a good approximation:

[asy]size(100); draw((-5,0)--(5,0)); draw((-5,1)--(5,1));[/asy][/quote]

Okay, but that is simplifying some stuff out, such as the fact that the function is nowhere continuous. I still stand by my statement. :P[/quote]

Graphically, though, this is perfectly accurate (up to the resolution of the screen) regardless of the scale, since the rationals are dense in the reals :P


calculus
derivative
algebra unsolved
algebra

Let $ P(x) \equal{}(x\minus{}x_1)(x\minus{}x_2)...(x\minus{}x_n)$and $ Q(x)\equal{}P(x)(\frac{1}{x\minus{}x_1}\plus{}\frac{1}{x\minus{}x_2}\plus{}...\plus{}\frac{1}{x\minus{}x_n})$ and let $ y_1,y_2,...,y_{n\minus{}1}$ are roots of $ Q(x)$.Prove that $ min|y_i\minus{}y_j|<min|x_i\minus{}x_j|$.

If the $ \text{min}$ refers to the minimum absolute distance between any pair of roots, the problem is false. If $ P(x) \equal{} x^3 \minus{} x$, then $ Q(x) \equal{} 3x^2 \minus{} 1$ (note $ Q$ is always the derivative of $ P$), and $ \text{min} | y_i \minus{} y_j| \equal{} \frac{2}{\sqrt{3}} > 1 \equal{} \text{min} | x_i \minus{} x_j |$.


geometry
rectangle
rhombus
analytic geometry
ratio

In rectangle $ ABCD$, $ G$ and $ H$ are trisection points of $ AD$ with $ G$ between $ A$ and $ H$, and $ E$ and $ F$ are trisection points of $ BC$ with $ E$ between $ B$ and $ F$.  $ BG$ intersects $ AE$ at $ P$ and $ AF$ at $ S$, and $ BH$ intersects $ AE$ at $ Q$ and $ AF$ at $ R$.  IF $ AB\equal{}360$ and $ BC\equal{}450$, compute the area of $ PQRS$.

(I have typed out some relations which the original problem represented on a diagram).

The region in question is a rhombus, so the area is equal to $ \frac {1}{2}d_1d_2$
The first diagonal is 75 units (the vertical diagonal) by the trisection given.
The second one can be found with coordinates to be 120 units.
Therefore the area is $ 4500$

$ R$ is the center of $ AHFB$, so $ R$ is the midpoint of $ EG$. $ P$ is the center of $ ABEG$, so $ PR$ is perpendicular to $ EG$. Extend $ PR$ and let $ J$ be its intersection with $ AB$. Note that $ JE$ is diagonal of the rectangle $ BERJ$. We have that $ Q$ is the centroid of $ JER$. The medians divide the triangle into six smaller triangles with equal area, so the area of $ PQR$ is $ \frac{1}{6}.(\frac{180.150}{2})\equal{}2250$. This is also the area of $ RSP$, and the desired area is the sum of that areas, that is $ 4500$.

[hide]
[asy]pair A,B,C,D,EE,F,G,H,P,Q,R,SS,T;
A=(0,360); B=(0,0); C=(450,0); D=(450,360);
EE=(150,0); F=(300,0); G=(150,360); H=(300,360);
P=(intersectionpoints(A--EE,B--G))[0];
Q=(intersectionpoints(A--EE,B--H))[0];
SS=(intersectionpoints(A--F,B--G))[0];
R=H/2;
T=A/2;
draw(A--B--C--D--A);
draw(A--EE);
draw(A--F);
draw(B--G);
draw(B--H);
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",EE,S);
label("$F$",F,S);
label("$G$",G,N);
label("$H$",H,N);
label("$P$",P,NW);
label("$Q$",Q,S);
label("$R$",R,E);
label("$S$",SS,N);
label("$T$",T,W);
draw(R--T);
draw(P--Q--R--SS--P,linewidth(1));[/asy]
Since the diagonals of $ PQRS$ are perpendicular to each other, its area is $ \frac{1}{2}d_1d_2$.  Let $ T$ be the foot of the altitude of $ \triangle ABR$ from $ R$.
$ ABFH$ is a rectangle whose diagonals intersect at $ R$, so $ TR\equal{}150$.  $ ABEG$ is a rectangle whose diagonals intersect at $ P$, so $ TP\equal{}75$.  Thus $ PR\equal{}75\equal{}d_1$.
Notice that $ \overline{BG}$ and $ \overline{AR}$ are medians of $ \triangle ABH$.  Thus $ S$ is the centroid of $ \triangle ABH$.  A centroid cuts a median into two pieces whose lengths are of the ratio $ 2: 1$ (this can easily be proven using [url=http://mathcircle.berkeley.edu/mpgeo/mpgeo.html]mass points[/url]), so $ \frac{AS}{SR}\equal{}2\Rightarrow \frac{AR}{SR}\equal{}3$.  Since $ \triangle ABR\sim\triangle SQR$, $ \frac{AB}{SQ}\equal{}\frac{AR}{SR}\equal{}3\Rightarrow SQ\equal{}120\equal{}d_2$.
Thus the area of $ PQRS$ is $ \frac{1}{2}(75)(120)\equal{}\boxed{4500}$.
[/hide]


Find all pair of primes $ p,q$ such that

$ \frac{p^{2n\plus{}1}\minus{}1}{p\minus{}1}\equal{}\frac{q^3\minus{}1}{q\minus{}1}$ for some integer $ n>1$

[quote="indybar"]Find all pair of primes $ p,q$ such that

$ \frac {p^{2n \plus{} 1} \minus{} 1}{p \minus{} 1} \equal{} \frac {q^3 \minus{} 1}{q \minus{} 1}$ for some integer $ n > 1$[/quote]
[hide="Solution"]
Expand out, so we have that $ p^{2n}\plus{}p^{2n\minus{}1}\plus{}\cdots \plus{} 1 \equal{} q^2 \plus{} q\plus{}1$.  Thus, \[ p^{2n}\plus{}p^{2n\minus{}1}\plus{}\cdots \plus{} p \equal{} p\cdot (\frac{p^{2n}\minus{}1}{p\minus{}1}) \equal{} q(q\plus{}1)\] Clearly, $ q>p^n$ since $ p^{2n}\plus{}p^{2n\minus{}1}\plus{} \cdots \plus{} p > p^{2n} \plus{} p^n$ unless $ n\equal{} 2n\minus{}1$, which is only possible if $ n\equal{}1$, which is a contradiction by the problem conditions.  Thus, $ q\ge p^n\plus{}1$.  If $ q \equal{} p^n \plus{} 1$, then $ p^{2n} \plus{} 3p^n \plus{} 2 \equal{} p^{2n} \plus{}p^{2n\minus{}1} \plus{} \cdots p$, meaning that $ 3p^n \plus{} 2 \equal{} p^{2n\minus{}1} \plus{} p^{2n\minus{}2} \plus{} \cdots \plus{} p$.  The left side is congruent to $ 2\bmod p$ and the right is congruent to $ 0\bmod p$, so $ p\equal{}2$.  This means that $ 3\cdot 2^n \plus{} 2 \equal{} 2^{2n\minus{}1} \plus{} 2^{2n\minus{}2} \cdots \plus{} 2$.  Thus, $ 2^{n\plus{}1} \plus{} 2^n \equal{} 2^{2n\minus{}1} \plus{}2^{2n\minus{}2} \cdots \plus{} 2^2$.  If $ n> 2$, then $ 2^{n\plus{}1} < 2^{2n\minus{}1}$ and $ 2^{2n\minus{}2} > 2^n$, so the Right Hand Side is greater the Left Hand Side, which is impossible.  This means that $ n\equal{}2$ and $ p\equal{}2$.  It follows that $ \frac{p^{2n\plus{}1} \minus{} 1}{p\minus{}1} \equal{} 2^5\minus{}1\equal{}31$, which yields that $ q\equal{}5$.

If $ p^n \plus{} 1 < q$, then notice that $ q\ne p$ (or else $ 2n\plus{}1 \equal{} 3\implies n\equal{}1$, which is impossible) so we have that $ p^{2n} \minus{} 1 \equiv 0\bmod q$.  Thus, $ p^n \plus{} 1\equiv 0\bmod q$ or $ p^n \minus{} 1\equiv 0\bmod q$.  Both, however, are false since $ p^n \plus{} 1 < q$.  We conclude that the only solution is $ (p, q, n) \equal{} (2, 5, 2)$. [/hide]

Alternative solution:

We observe that $ q>p^n$
Then we observe that the Left hand side simplifies to $ 1\plus{}p\plus{}p^2..................\plus{}p^{2n}$
And the right hand side is $ 1\plus{}q\plus{}q^2$
Simplifying we get  $ p\plus{}p^2..................\plus{}p^{2n}$ =  $ q\plus{}q^2$
So we get $ p(p\plus{}1)$ divides $ q\plus{}q^2$
Which implies p divides q+1 (since q is a prime p cannot divide q)
So let $ q\plus{}1\equal{}p^n\plus{}pl$ where l is a natural number
Then  $ q\equal{}p^n\plus{}pl\minus{}1$
Multiplying we get $ q\plus{}q^2 \equal{} p^{2n} \plus{}2lp^{n\plus{}1}\minus{}p^n\plus{}(pl)^2\minus{}pl$
Which implies $ p^{2n} \plus{}2lp^{n\plus{}1}\minus{}p^n\plus{}(pl)^2\minus{}pl \equal{} p\plus{}p^2..................\plus{}p^{2n}$
We see that l<p otherwise the left hand side does not contain any term of p
So we see that the next highest power of the left hand side is n+1
But this should equal the second greatest power of the right hand side which is 2n-1
Which implies $ n\equal{}2$
But since the co-efficient of p is 1 on the right hand side but -1 on the left hand side we observe that
$ p^2\minus{}p\equal{}p$
which implies $ p\equal{}2$
Plugging in we can also verify that  $ p^{2n} \plus{}2lp^{n\plus{}1}\minus{}p^n\plus{}(pl)^2\minus{}pl \equal{} p\plus{}p^2..................\plus{}p^{2n}$
holds when n is 2 and p is 2.
So we get $ 2\plus{}2^2\plus{}2^3\plus{}2^4$=$ q\plus{}q^2$
Solving we get q is 5.
Thus the only solution is $ p\equal{}2$,$ n\equal{}2$,$ q\equal{}5$


trigonometry
geometry
trig identities
Law of Sines
geometry open

Hi...........

Given a triangle ABC with AB = AC =a+b and BC = a. Traced to a segment from B determining a point in AC, point D, where the DA = a and DC = b. Knowing that ABD = 10 °, find the internal angles of the triangle.

thank

We know $ \angle C \equal{} (90 \minus{} A/2)$ and $ \angle D \equal{} A \plus{} 10$
By Law of Sines on ABD, $ \frac {sinA}{BD} \equal{} \frac {sin10}{a}$
By Law of Sines on CDB, $ \frac {sin(90 \minus{} A/2)}{BD} \equal{} \frac {sin(A \plus{} 10)}{a}$
Dividing, $ \frac {sinA}{cos(A/2)} \equal{} \frac {sin10}{sin(A \plus{} 10)}$
Guessing, we can find that A=20.

but there is another answer?

why you came to 20 degrees so easy?

I had to guess (I first tried multiples of 10), but it is easily proved.
$ \frac {sin20}{cos 10} \equal{} \frac {2sin10cos10}{cos10} \equal{} 2sin10 \equal{} \frac {sin10}{1/2} \equal{} \frac {sin10}{sin30}$.
Can anyone solve my equation without guessing? My calculator shows 20 is the only answer.


If I write $ B=I+A+\cdots+A^{k}$, inductively I get:

$ B \cdot \left[ \begin{array}{c}1 \\%Error. "noalign" is a bad command.
{\medskip}1 \\%Error. "noalign" is a bad command.
{\medskip}1 \end{array}\right] = \left[ \begin{array}{c}k+1 \\%Error. "noalign" is a bad command.
{\medskip}k+1 \\%Error. "noalign" is a bad command.
{\medskip}k+1 \end{array}\right]=[v3]$

$ B \cdot \left[ \begin{array}{c}1 \\%Error. "noalign" is a bad command.
{\medskip}1 \\%Error. "noalign" is a bad command.
{\medskip}0 \end{array}\right] = \left[ \begin{array}{c}2007 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}+1 \\%Error. "noalign" is a bad command.
{\medskip}2007 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}+1 \\%Error. "noalign" is a bad command.
{\medskip}2007 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}-k \end{array}\right]=[v2]$

$ B \cdot \left[ \begin{array}{c}1 \\%Error. "noalign" is a bad command.
{\medskip}0 \\%Error. "noalign" is a bad command.
{\medskip}0 \end{array}\right] = \left[ \begin{array}{c}2006 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}+1 \\%Error. "noalign" is a bad command.
{\medskip}2006 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}-k \\%Error. "noalign" is a bad command.
{\medskip}2006 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}-k \end{array}\right]=[v1]$

Therefore, $ B=[v1 \hspace{2 mm}v2 \hspace{2 mm}v3] \cdot \left[ \begin{array}{ccc}1&-1&0\\%Error. "noalign" is a bad command.
{ \medskip}0&1&-1\\%Error. "nolaign" is a bad command.
{ \medskip}0&0&1\end{array}\right]$

This gives:

$ b_{11}=2006 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}+1$
$ b_{22}=\frac{k(k+1)}{2}+k+1$
$ b_{33}=-\left(2007 \cdot \frac{k(k+1)}{2}-\frac{k(k-1)}{2}-k \right)+k+1$

Now just substitute in $ k=2006$!

I will now wait for the "easy" solution...

I would find the eigenvalues and eigenvectors for $ A$ then use that

$ D=Q^{-1}AQ$ where $ D$ is a diagonal matrix with the eigenvalues for its entries and $ Q$ is the matrix formed by making each column a different eigenvector.
so $ A^{n}=QD^{n}Q^{-1}$ (this is easy to prove). so from here you have a nice formula for each entry in $ A^{n}$ so now you will just have a sum to evaluate. (it shouldn't be that difficult since it looks like it will be some sort of geometric series or a variation of that.)
This should not be too difficult since $ A$ is only a 3x3, but it will still take some effort to find the eigenvalues just because of how $ A$ looks, but I could be wrong.

Since $ A$ is the sum of the identity and a rank 1 matrix, it's not going to diagonalize.
Since $ A-I$ is a rank 1 matrix with trace zero, it's nilpotent. We can triangularize it: $ A=\begin{bmatrix}1&0&1\\-2005&1&1\\0&0&1\end{bmatrix}^{-1}\cdot \begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\cdot \begin{bmatrix}1&0&1\\-2005&1&1\\0&0&1\end{bmatrix}^{-1}$

The details aren't too hard from here; it's essentially the same as edoo's calculations.


[quote="4everwise"]If a box contains two red balls and one yellow, find the probability of drawing a red and a yellow if two balls are drawn.[/quote]
[hide="answer"]
1-P(getting 2 reds)=1-2/3=1/3[/hide]

[quote="math92"][hide]
the probablity of getting a red is 2/3 and the probabilty of getting a yellow is 1/3 so 2/3*1/3=2/9...that looks wrong.[/hide][/quote]

Remember that the balls were not replaced.

[quote="math92"]oh yeah, sorry.
i still can't get the answer :([/quote]

After the first ball is drawn, there is only two balls left. Go from there.

[quote="math92"][hide]
The chance of drawing a red first is 2/3
the chance of then drawing a yellow is 1/2
so 2/3*1/2=[b]1/3[/b]
is that right?
[/hide][/quote]

Yup :yup: you've got the replacement concept.


$ \frac{x^2\minus{}4x\plus{}3}{x^2\plus{}1}<1$.
What is the range of values of $ {x}$ in this inequality.

[hide="Solution"]$ x^2 \plus{} 1 > 1 > 0$, so the given inequality is equivalent to $ x^2 \minus{} 4x \plus{} 3 < x^2 \plus{} 1$, which is equivalent to $ \minus{} 4x \plus{} 2 < 0$, so $ \boxed{x > \frac {1}{2}}$, or $ \boxed{\left(\frac{1}{2}, \infty\right)}$.[/hide]

hello, your inequality is equivalent to
$ \frac{x^2\minus{}4x\plus{}3}{x^2\plus{}1}\minus{}1<0 \Leftrightarrow \frac{\minus{}2x\plus{}1}{x^2\plus{}1}<0$ so we get
$ x>1/2$ because of $ x^2\plus{}1>1>0$
Sonnhard.

Thats what i got implying the book made a mistake.
One more:For what range of values of $ {x}$ is $ \frac{x^2\minus{}5x\plus{}6}{x\minus{}1}>0$.
I want to see an analytical way.


Express $ 0.234543234543234543\cdot\cdot\cdot$ as a fraction in lowest terms.  (The "rising and falling" pattern continues, indicated by the ellipsis.)

Note:  This given number is in base ten, by the way.

$ \frac{234543}{999999}$

the way you solve it is like this:

$ x\equal{}.234543234543...$
$ 1000000x\equal{}234543.234543234543...$
subtract (since the decimal tends to infinity) and divide to get the answer above. unless i put too many or too little 9s

Hm...this seems to simplify to $ \frac {2113}{9009}$ (I mean base $ 55$, by the way).

Very nice and creative problem, Arrange your tan!

By the way, try to use \ldots or \cdots next time instead of \cdot\cdot\cdot.

Extension (From my mock MathCounts test): Express $ 4.134134134\dots_5$ as a common fraction in base 10.

[quote="math154"]

By the way, try to use \ldots or \cdots next time instead of \cdot\cdot\cdot.[/quote]
Yes, this is important.

My ellipsis is wrong as it's asymmetrical, for one thing.  And it might
also be wrong in other aspects.

I want to remember to use "\ldots" instead of what I typed.


geometry
incenter
search
trigonometry
geometry proposed

$I$ is the incenter of the triangle $ABC$ and $A'$ is the center of the excircle opposite $A$. The bisector of angle $BIC$ meets the side $BC$ at $A''$. The points $B', C', B'', C''$ are defined similarly. Show that the lines $A'A'', B'B'', C'C''$ are concurrent.

[quote="vittasko"]The lines $AA'',$ $BB'',$ $CC'',$ are concurrent, as well known and easy to prove.

The lines $AA',$ $BB',$ $CC',$ are concurrent at incenter $I,$ of $\bigtriangleup ABC.$ That is the given triangle $\bigtriangleup ABC,$ is the cevian triangle of $\bigtriangleup A'B'C',$ with respect to the point $I.$

So, because of the points $A'',$ $B'',$ $C'',$ lie on $BC,$ $AC,$ $AB$ respectively, such that the lines $AA'',$ $BB'',$ $CC'',$ to be concurrent, we conclude that the triangles $\bigtriangleup A'B'C'$ and $\bigtriangleup A''B''C'',$ are perspective $($ as well known at least to me and not difficult to prove $).$

Hence, the lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent at one point and the proof is completed.

Kostas Vittas.[/quote]

Do u use transitivity?Or otherwise how exactly do u prove that A'B'C' and A"B"C" are perspective?

[quote="Nick Rapanos"]
Do u use transitivity?Or otherwise how exactly do u prove that A'B'C' and A"B"C" are perspective?[/quote]

Dear Nick, this is a well known result  coming from the past and probably somebody may give us a link in the forum, about it’s proof.

[b][size=117]PROPOSITION.[/size][/b] – A triangle $\bigtriangleup ABC$ is given and let $\bigtriangleup A'B'C'$ be, the cevian triangle of $\bigtriangleup ABC,$ with respect to an arbitrary point $P.$ If $\bigtriangleup A''B''C'',$ is the cevian triangle of $\bigtriangleup A'B'C'$ with respect to an arbitrary point $P',$ then the triangles $\bigtriangleup ABC$ and $\bigtriangleup A''B''C'',$ are perspective.

I will search my archive, because I don’t remember how $($ by Menelaus and Ceva theorems I think $),$ to prove it and if we will have not any reference until tomorrow, I will post here a proof.

Kostas Vittas.

Nice solution...
But ,what do you be believe : Could we use this fact as a well known in a contest or not ? :maybe:

@Nick : Η προβολική που λέγαμε χτυπάει ξανα .... :)

[quote="silouan"]
But ,what do you be believe : Could we use this fact as a well known in a contest or not ? :maybe: [/quote]

In this case of N.T.TUAN’s problem, the solution I presented is a direct application of another proposition, not including in the main chapters of geometry theory. So, I think that when it is used as a main Lemma in a contest, must be proved.

In the forum, I believe also that it is necessary to be given the proofs of Lemmas, in order to all of us preserve them in the future.

Kostas Vittas.

@Silouan : Ποια Προβολική χτυπάει ξανά; Αν επιτρέπεται, θέλω να ξέρω και εγώ.

[quote="vittasko"].....
[b][size=117]PROPOSITION.[/size][/b] – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup A'B'C'$ be, the cevian triangle of $ \bigtriangleup ABC,$ with respect to an arbitrary point $ P.$ If $ \bigtriangleup A''B''C'',$ is the cevian triangle of $ \bigtriangleup A'B'C'$ with respect to an arbitrary point $ P',$ then the triangles $ \bigtriangleup ABC$ and $ \bigtriangleup A''B''C'',$ are perspective.
.....
[/quote]

Please see the [b][size=100]Cevian Nests theorem[/size][/b], at http://www.mathlinks.ro/Forum/viewtopic.php?t=6579

Kostas Vittas.

[quote="N.T.TUAN"][color=darkred]$ I$ is the incenter and $ I_{a}$ is $ A$- exincenter of the triangle $ ABC$. The bisector of $ \angle BIC$ meets the side $ BC$ at $ A'$. The points $ B'$,  $ C'$ are defined similarly. Show that the lines $ A'I_{a}\cap B'I_{b}\cap C'I_{c}\not\in \emptyset$.[/color][/quote]
[color=darkblue][b]Proof.[/b] $ A_{1}\in I_{a}A'\cap I_{b}I_{c}$ $ \implies\over{\mathrm{easily}}$ $ \frac{I_{a}B}{I_{a}C}=\frac{\cos\frac{C}{2}}{\cos\frac{B}{2}}$ and $ \boxed{\frac{A_{1}I_{b}}{A_{1}I_{c}}=\frac{I_{a}I_{b}}{I_{a}I_{c}}\cdot\frac{I_{a}B}{I_{a}C}\cdot\frac{IC}{IB}}$ a.s.o.
[hide="Proof of the previous relation."]Indeed, $ \left\{\begin{array}{ccc}\triangle I_{b}I_{a}I_{c}& : & \frac{A_{1}I_{b}}{A_{1}I_{c}}=\frac{I_{a}I_{b}}{I_{a}I_{c}}\cdot\frac{\sin\widehat{A_{1}I_{a}I_{b}}}{\sin\widehat{A_{1}I_{a}I_{c}}}\\\\ \triangle BI_{a}C & : & \frac{A'B}{A'C}=\frac{I_{a}B}{I_{a}C}\cdot\frac{\sin\widehat{A_{1}I_{a}I_{c}}}{\sin\widehat{A_{1}I_{a}I_{b}}}\\\\ \triangle BIC & : & \frac{A'C}{A'B}=\frac{IC}{IB}\end{array}\right\|$ $ \implies$ $ \frac{A_{1}I_{b}}{A_{1}I_{c}}=\frac{I_{a}I_{b}}{I_{a}I_{c}}\cdot\frac{I_{a}B}{I_{a}C}\cdot\frac{IC}{IB}$.[/color][/hide]


Where can I get good e-books on subjects like Cosmology,Concept of time,Relativity,etc.Please post the link.
Thanks in advance.

I'm rather sure that downloading these eBooks would be illegal. Buy them, or visit the library. Besides, eBooks aren't as good to read as real books. Hard on the eyes.

Some how all ebook downloading site is illegal, but use these legals
1.archives.org
2.Project Gutenberg


quadratics
geometry
function
algebra
quadratic formula

Hello, this is a neater continuation of alton66's marathon. (See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1188760#1188760.)

The rules are as follows:

1. Give complete solutions.

2. After solving, post a new problem. However, if you are unsure if your answer is correct, verify it first.

3. Make sure they're MOEMS level!

4. Have fun!

Problem 1:

$ a \zeta b \equal{} (a \minus{} b)(2a \plus{} 2b)$ and $ a \star b \equal{} (b \minus{} a)(2a \plus{} 2b)$. Find $ z$ if $ 4 \zeta (5 \star z) \equal{} \minus{} 936$.

I would say that's not MOEMS level because I would be surprised if a problem of that type appeared on, say, the 2009 MOEMS test. However, feel free to solve it if you want. Meanwhile...

OK, here goes:  (I've replaced z with x for writing convenience)

$ 4 \zeta ((x \minus{} 5)(10 \plus{} 2x)) \equal{} \minus{} 936$
$ 4 \zeta (10x \plus{} 2x^2 \minus{} 50 \minus{} 10x) \equal{} \minus{} 936$
$ 4 \zeta (2x^2 \minus{} 50) \equal{} \minus{} 936$
$ (4 \minus{} (2x^2 \minus{} 50)(2(4) \plus{} 4x^2 \minus{} 100) \equal{} \minus{} 936$
$ (4 \minus{} 2x^2 \plus{} 50)(8 \plus{} 4x^2 \minus{} 100) \equal{} \minus{} 936$
$ ( \minus{} 2x^2 \plus{} 54)(4x^2 \minus{} 92) \equal{} \minus{} 936$

let $ a \equal{} x^2$

$ \minus{} 8a^2 \plus{} 184a \plus{} 216a \minus{} 4968 \equal{} \minus{} 936$
$ \minus{} 8a^2 \plus{} 400a \minus{} 4032 \equal{} 0$
$ a^2 \minus{} 50a \plus{} 504 \equal{} 0$

finally a quadratic equation :) applying the formula

$ \frac {50 \pm \sqrt {2500 \minus{} 2016}}2$
$ \frac {50 \pm 484}2$

The two solutions for a are $ 267$ and $ \minus{} 217$

Since $ a \equal{} x^2$
When $ a \equal{} 267 \rightarrow x \equal{} \sqrt {267}$
and $ a \equal{} \minus{} 217 \rightarrow x \equal{} \sqrt { \minus{} 217}$

Could of made a typo somewhere, but I think thats right.

-Scoff- Well, I suppose this will be neater...

Problem: A man was in jail and trying to escape when he cape up with a plan. He stole some pieces of meat to get by the gaurd dogs. For the first dog he gave half his toal pieces and 1 more. For the second he gave half his remaining and 1 more. For the third dog he gave half his remaining  pieces and 1 more. How many pieces did the prisoner start with?

The answer to my problem was actually $ \boxed{6}$, but it's okay.

[u]Alton's Problem
[/u]
$ \frac{1}{2}x \plus{} 1 \plus{} \frac{1}{4}x \plus{} 1 \plus{} \frac{1}{8}x \plus{} 1 \equal{} x$

$ \frac{7}{8}x \plus{} 3 \equal{} x$

$ 3 \equal{} \frac{1}{8}x$

$ 24 \equal{} x$

I think I'm wrong, so I won't post another problem.

It's wrong, there is a simpler not so algebraic way of doing it.

Do we need to know how many pieces he was left with after the third guard dog? Because otherwise it could be anything!

Assuming the prisoner had $ x$ pieces of meat left, we can build the sequence by alternating add 1 and multiply by 2.

$ x$
$ 2(x + 1)$
$ 2(2(x + 1) + 1)$
$ 2(2(2(x + 1) + 1) + 1)$

So the amount he started with, in terms of $ x$, is
\begin{align*}2(2(2(x + 1) + 1) + 1) & = 2(2(2x + 2 + 1) + 1) \\
& = 2(2(2x + 3) + 1) \\
& = 2(4x + 6 + 1) \\
& = 2(4x + 7) \\
& = 8x + 14\end{align*}

So if he used up all his pieces, then the answer would be $ \boxed {14}$. If he had only one piece left, the answer would be $ \boxed {22}$.

[b]New problem[/b]
Suppose we had the figure below.

[asy]filldraw(scale(2) * unitcircle, grey);
filldraw(shift(W) * unitcircle, white);
filldraw(shift(E) * unitcircle, white);[/asy]

Which consists of a large circle of radius 4 and two smaller circles of equal size which are tangent to each other and the large circle. Find the area of the shaded region.

Edit: by the way, the error to the answer to isabella2296's problem was that in the quadratic formula step:

$ \frac {50 \pm \sqrt{2500 - 2016}}2$

He forgot to square root the 484! :o

So the possible values for $ a$ were actually:

$ \frac {50 \pm 22}2 = 25 \pm 11$
$ a = 36\ \text{or}\ 14$

So $ x$ could've been $ 6$ or $ \sqrt{14}$. :)

If the larger circle has a radius of 4, then the two smaller circles have a radius of 2. The total area of the two white circles is $ 4\pi \plus{} 4\pi$ or $ 8\pi$. The large circle's total area is $ 16\pi$, so $ 16\pi \minus{} 8\pi \equal{} \boxed{8\pi}$.

New Problem:

Jane took 4 tests. Her scores were: 86, 97, 78, and 94. She wants to have an average of at least 90. What is the least score she needs to get on her next test?

[quote="Yongyi781"]
Edit: by the way, the error to the answer to isabella2296's problem was that in the quadratic formula step:

$ \frac {50 \pm \sqrt {2500 \minus{} 2016}}2$

He forgot to square root the 484! :o

So the possible values for $ a$ were actually:

$ \frac {50 \pm 22}2 \equal{} 25 \pm 11$
$ a \equal{} 36\ \text{or}\ 14$

So $ x$ could've been $ 6$ or $ \sqrt {14}$. :)[/quote]

Ohh yeah...that's right. Thanks :)

::   answer to isabella's problem:

let x be what she gets on her final test.

Then:

$ \frac {86 \plus{} 97 \plus{} 78 \plus{} 94}5 \plus{} \frac {x}5 \equal{} 90$
$ \frac {355}5 \plus{} \frac {x}5 \equal{} 90$
$ 71 \plus{} \frac {x}5 \equal{} 90$
$ \frac {x}5 \equal{} 19$
$ \boxed{x \equal{} 95}$

Jane needs to get atleast a 95 to get a 90 avg.

 New Problem:

Find the sum of $ 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} ... \plus{} 97 \plus{} 99$

[hide]She needs to get a score of at least 95.

Her current total s 355, and the toal she needs to get is 5X90=450. 450-355=95.[/hide]

Oh and how do you chose what the link says when you use the hide button?

Blackswordsman beat you to it.

For the hide thing, put (hide=Blah Blah Blah) blah blah blah (/hide). Normally you put "[" and "]" but I used parentheses so it wouldn't show up as a hide.

For Black's problem: basically, we're finding the sum of the first $ 49$ integers, so we do $ 49^2$ which is $ \boxed{2401}$.

Sally has $ 20$ apples. $ 5$ of them are rotten. How many more clean apples does she need to make her rotten percentage $ 20 \%$?

Thanks for the hide thing. I was in the middle of posting when black posted  :o :(  :mad: . Also, why does it seem my solutions are so much less complex compared to others? I type a lot less.

It's okay. Some people give a lot of detail, that's all. Oh, so can you or someone else please answer my problem?

Using a proportion:   ? = 100% and 5 = 20% gives

$ \frac{5 \cdot 100}{20} \equal{} \frac{500}{20} \equal{} 25$

She needs 5 more apples.

New Problem:

$ 100^2 \minus{} 99^2 \plus{} 98^2 \minus{} 97^2 \plus{}...\plus{} 2^2 \minus{} 1^2 \equal{} ?$

Note the difference of squares. Rewrite it as $ (199)(1)\plus{}(195)(1)\plus{}(191)(1)\plus{}...\plus{}(3)(1)$. So, it is basically $ 199\plus{}(199\minus{}4)\plus{}(199\minus{}8)\plus{}...\plus{}(199\minus{}196)$. There are 50 199s and 49 "subtraction parts." The subtraction parts can be rewritten as 4(49)(50)/2=4900. The 50 199s is 9950. So, our answer is 9950-4900=5050. Hmm. I think I made my answer too complicated.

Next Problem:

Suppose it is Monday. What day of the week will it be in 4352 days?

$ 4352 \div 7 \equal{} 621  R  5$

So our answer is 5 days after Monday, or $ \boxed{\text{Saturday}}$.

New Problem:

$ f(4x\plus{}6) \equal{} x^2 \plus{} 14x \minus{} 7$

Find $ f(3)$.

oh...then i dont know how to solve it =[

[quote="isabella2296"]
New Problem:

$ f(4x \plus{} 6) \equal{} x^2 \plus{} 14x \minus{} 7$

Find $ f(3)$.[/quote]

Erm...i think this is right.
Because we want $ f(3)$, we have to find a value of $ x$ such that $ 3\equal{}4x\plus{}6$.  Solving, $ 4x\equal{}\minus{}3 \Longrightarrow x\equal{}\frac{\minus{}3}{4}$.  Now plug into the function: $ \frac{9}{16}\minus{}{42}{4}\minus{}7 \Longrightarrow \frac{\minus{}271}{16}$.  I think thats the answer, but i'm pretty sure i screwed up somewhere.

New problem

If 12 a + 6 b = 96 , what is the value of a/2 + b/4 ?

Note that $ \frac {12a \plus{} 6b}{24} \equal{} \frac {a}{2} \plus{} \frac {b}{4}$. So, we divide 96 by 24 to get 4.

New Problem:

If the angle measure of a pentagon is $ 79^\circ$, $ 105^\circ$, $ 102^\circ$, $ 2x$, and $ x$, find the mean of the exterior angles.

EDIT: I edited the numbers to make it plausible.

Do you want the answer in terms of $ x$?

If not:

The sum of the angles in a regular pentagon is 540, so each angle is 108 degrees.  THe exterior angles mean the supplement to each of the angles, so they average at 82 degrees ( i think)

Oh woops. Sorry. I had to edit the numbers since I realized it wasn't plausible. (I think one of the angles was over 360). Also, 82 would be wrong


induction
function
strong induction
number theory open
number theory

Prove that every positive integer can be written as the sum of distinct integer powers of 3, 4, and 5.
(e.g. 29 = 3^3 + 3^0 + 4^0, 145 = 5^3 + 4^2 + 3^1 + 5^0. For the purposes of this problem, the zero powers of 3, 4, and 5 are distinct.)

Let $ f(n)$ denote the $ n$th smallest number that may be expressed as a power of one of $ \{3,4,5\}$, so $ f(1) = f(2) = f(3) = 1$. Let $ P_n(k) = 1$ if $ k$ may be written as the sum of the elements of a subset of $ \{f(1), \ldots, f(n - 1)\}$ [eg, satisfies conditions of problem statement], and $ 0$ otherwise; then we claim by strong induction that $ P_n(k) = 1$ for all $ k$ satisfying $ 0 < k \le 2f(n)$, and $ n > 1$. Obviously, if $ P_m(k') = 1$, and $ k - k'$ may be written as a sum of the elements of a subset of $ \{f(m + 1), \ldots, f(n)\}$, then $ P_n(k) = 1$.

Let $ g_{k,n}(s) = k - \left[\sum_{i = 1}^s f(n - i)\right]$; if $ 0 < g_{k,n}(s) \le 2f\big(n - (s + 1)\big)$, then by inductive hypothesis, $ P_{n - (s + 1)}\left(g_{k,n}(s)\right) = 1$, so $ P_{n - 1}(k) = 1$. Note that\begin{align*}f(n) \le 3f(n - 1), 4f(n - 2), 5f(n - 3), 9f(n - 4),\end{align*} and writing out our result for $ s = 0,1,2,3$ gives:\begin{align*}0 \le g_{k,n}(0) \le 2f(n - 1) \ & \Longrightarrow\ P_{n - 1}(k) = 1,\ \forall k: 0 < k \le 2f(n - 1) \\
0 \le g_{k,n}(1) \le 2f(n - 2) \ & \Longrightarrow\ P_{n - 1}(k) = 1,\ \forall k: f(n - 1) < k \le f(n - 1) + 2f(n - 2) \\
0 \le g_{k,n}(2) \le 2f(n - 3) \ & \Longrightarrow\ P_{n - 1}(k) = 1,\ \forall k: f(n - 1) + f(n - 2) < k \le f(n - 1) + f(n - 2) + 2f(n - 3) \\
0 \le g_{k,n}(3) \le 2f(n - 4) \ & \Longrightarrow\ P_{n - 1}(k) = 1, \\
\forall k: f(n - 1) + f(n - 2) + f(n - 3) < k & \le \frac {f(n)}{3} + \frac {f(n)}{4} + \frac {f(n)}{5} + \frac {2f(n)}{9} = \frac {181f(n)}{180} \\
& \le f(n - 1) + f(n - 2) + f(n - 3) + 2f(n - 4) \end{align*}whence $ P_{n - 1}(k) = 1$ for all $ 0 < k \le f(n)$. Furthermore, for all $ k$ satisfying $ f(n) < k \le 2f(n)$, we have $ P_{n - 1}(k - f(n)) = 1$ and thus $ P_n(k) = 1$, and our problem statement is true.

Okay, since I don't completely understand your solution (I'm only a junior, you're already in university... :D), I'll try and analyze it:

[quote="azjps"]Let $ f(n)$ denote the $ n$th smallest number that may be expressed as a power of one of $ \{3,4,5\}$, so $ f(1) = f(2) = f(3) = 1$. Let $ P_n(k) = 1$ if $ k$ may be written as the sum of the elements of a subset of $ \{f(1), \ldots, f(n - 1)\}$ [eg, satisfies conditions of problem statement], and $ 0$ otherwise; then we claim by strong induction that $ P_n(k) = 1$ for all $ k$ satisfying $ 0 < k \le 2f(n)$, and $ n > 1$. Obviously, if $ P_m(k') = 1$, and $ k - k'$ may be written as a sum of the elements of a subset of $ \{f(m + 1), \ldots, f(n)\}$, then $ P_n(k) = 1$. [/quote]

This part I understand. You're defining the functions, etc., and claiming that for some $ f(n)$, any integer up to twice that number should always be summable as the $ f$ of numbers less than it. I also get the $ k'$ part.

[quote]Let $ g_{k,n}(s) = k - \left[\sum_{i = 1}^s f(n - i)\right]$;[/quote]

So for some integer $ k$, $ g_{k,n}(s)$ is $ k - f(n - 1) - f(n - 2)$, etc. down to $ f(n - s)$.

[quote]if $ 0 < g_{k,n}(s) \le 2f\big(n - (s + 1)\big)$, then by inductive hypothesis, $ P_{n - (s + 1)}\left(g_{k,n}(s)\right) = 1$, so $ P_{n - 1}(k) = 1$.[/quote]

Ah, so $ g_{k,n}(s)$ is the $ k'$ we referred to earlier? Okay, continuing...

[quote]Note that
\begin{align*}f(n) \le 3f(n - 1), 4f(n - 2), 5f(n - 3), 9f(n - 4),\end{align*}
[/quote]

Okay, I get that.

[quote]and writing out our result for $ s = 0,1,2,3$ gives:
\begin{align*}0 \le g_{k,n}(0) \le 2f(n - 1) \ & \Longrightarrow\ P_{n - 1}(k) = 1,\ \forall k: 0 < k \le 2f(n - 1) \\
0 \le g_{k,n}(1) \le 2f(n - 2) \ & \Longrightarrow\ P_{n - 1}(k) = 1,\ \forall k: f(n - 1) < k \le f(n - 1) + 2f(n - 2) \\
0 \le g_{k,n}(2) \le 2f(n - 3) \ & \Longrightarrow\ P_{n - 1}(k) = 1,\ \forall k: f(n - 1) + f(n - 2) < k \le f(n - 1) + f(n - 2) + 2f(n - 3) \\
0 \le g_{k,n}(3) \le 2f(n - 4) \ & \Longrightarrow\ P_{n - 1}(k) = 1, \\
\forall k: f(n - 1) + f(n - 2) + f(n - 3) < k & \le \frac {f(n)}{3} + \frac {f(n)}{4} + \frac {f(n)}{5} + \frac {2f(n)}{9} = \frac {181f(n)}{180} \\
& \le f(n - 1) + f(n - 2) + f(n - 3) + 2f(n - 4) \end{align*}
[/quote]

Okay, this stuff is just calculations...

I don't understand where the restrictions come from. It's probably something simple that I can't see, but I can't see it...

*edit* wait, no, I see it now. Because $ g_{k_n}(s)$ is the subtraction of some terms, you need to add those terms back in order to provide the correct inequality.

[quote]whence $ P_{n - 1}(k) = 1$ for all $ 0 < k \le f(n)$. Furthermore, for all $ k$ satisfying $ f(n) < k \le 2f(n)$, we have $ P_{n - 1}(k - f(n)) = 1$ and thus $ P_n(k) = 1$, and our problem statement is true.[/quote]

This part is just proof of the induction.

Alright, so there are two parts that I don't get... if somebody could please explain them, I'd be much obliged.

Sorry for the rather late response, also I'm only a freshmen  :)

I used a bit of condensed notation to avoid having to re-write the statements, esp $ P$ just being that the proposition is true, so my apologies for that (though reading "and therefore the statement is true for this value of $ k$" fifteen times might be tiresome also  :) ).

[quote]Ah, so $ g_{k,n}(s)$ is the $ k'$ we referred to earlier? Okay, continuing... [/quote]

Yes, because we have $ k \equal{} f(n\minus{}1) \plus{} f(n\minus{}2) \plus{} \cdots \plus{} f(n\minus{}s) \plus{}$ some term for which we know can be written as a smaller sum of powers (by induction, some subset of $ \{f(1), \ldots, f(n\minus{}s\minus{}1)\}$). We want to use some larger values of $ \{f(n\minus{}1), \ldots, f(n\minus{}s)$ to "shoot" $ k$ into a region where we already know numbers can be written as a sum of powers, and we have some strong enough bounds on $ f(n\minus{}1), \ldots, f(n\minus{}4)$ to be able to do this. The four computations show that any $ k$ from $ 0 < k \le f(n)$ will work, and then the rest is easy.

Is there any part that needs clarification? (I presume you answered your own question)


secondary mathematics teachers teaching grade schoolers - a boon or a bane?

How about "uncommon"? Where I'm from, people with multiple-subject teaching credentials are in oversupply, while those with single-subject credentials in mathematics are in undersupply. Hence, I don't hear about those with single-subject credentials in elementary schools. I'm sure it happens somewhere, because nearly everything has been tried somewhere, but the population pressures are against it.


hehe studying at the last minute for AIME, does anyone have the session transcript for last year's 2004 AIME A&B discussion? thanks! :] if you do, can you email them as attachments or something to aznfunyun@gmail.com? thank you!

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=27

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=26

There they are. Here's the general page for Math Jam transcripts:

http://artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php


inequalities
calculus
three variable inequality

[quote="can_hang2007"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$, then

$ \frac {7 \minus{} 6a}{2 \plus{} a^2} \plus{} \frac {7 \minus{} 6b}{2 \plus{} b^2} \plus{} \frac {7 \minus{} 6c}{2 \plus{} c^2} \ge 1$.[/quote]
What is your solution, VasC?[/quote]

My solution is the following.

[quote="can_hang2007"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$, then

$ \frac {7 \minus{} 6a}{2 \plus{} a^2} \plus{} \frac {7 \minus{} 6b}{2 \plus{} b^2} \plus{} \frac {7 \minus{} 6c}{2 \plus{} c^2} \ge 1$.[/quote]

In my opinion, it is a very nice inequality but very hard. See my ugly proof at: http://canhang2007.wordpress.com/2008/12/18/hard-one-vasile-cirtoaje/[/quote]
We can prove this inequality using the following remark : if $ a,b>0$ and $ 1\le ab\le 10$ then $ f(a)\plus{}f(b)\ge 2f(\sqrt{ab})$ where $ f(x)\equal{}\frac{7\minus{}6x}{2\plus{}x^2}$.
Indeed, if $ \frac{1}{10}\le a\le 1$ or $ \frac{1}{10}\le b\le 1$ or $ \frac{1}{10}\le c\le 1$, then according to the remark, it suffices to prove the inequality in the case $ b\equal{}c$ or $ c\equal{}a$ or $ a\equal{}b$.
Otherwise, we assume $ a\ge b\ge c$, then $ 0<c<\frac{1}{10}$, hence $ \frac{7\minus{}6c}{2\plus{}c^2}>3$. Adding this inequality to two trivial inequalities $ \frac{7\minus{}6a}{2\plus{}a^2}\ge \minus{}1$ and $ \frac{7\minus{}6b}{2\plus{}b^2}\ge \minus{}1$ we get the result.

Nice solution, ehku. :lol:
 But I think it is not very simple to show that $ 1\le ab\le 10$ yields $ f(a)\plus{}f(b) \ge 2f(\sqrt{ab})$.

[quote="Vasc"]Nice solution, ehku. :lol:
 But I think it is not very simple to show that $ 1\le ab\le 10$ yields $ f(a) \plus{} f(b) \ge 2f(\sqrt {ab})$.[/quote]
I used a technique called "Mixing Variables by Derivative", which is very useful for classical mixing variables, UMV (AC/GC) theorem, SMV theorem,...

Put $ s\equal{}a\plus{}b,p\equal{}ab$ then $ s\ge 2\sqrt{p}$ and \[ f(a)\plus{}f(b)\equal{}\frac{7\minus{}6a}{2\plus{}a^2}\plus{}\frac{7\minus{}6b}{2\plus{}b^2}\equal{}\frac{7s^2\minus{}6(2\plus{}p)s\plus{}14(2\minus{}p)}{2s^2\plus{}(2\minus{}p)^2}\equal{}g(s).\]
We have
\[ g'(s)\equal{}\frac{2(2\plus{}p)(6s^2\minus{}7(2\minus{}p)s\minus{}3(2\minus{}p)^2)}{[2s^2\plus{}(p\minus{}2)^2]^2}\]
\[ \equal{}\frac{2(2\plus{}p)(2s\plus{}3p\minus{}6)(3s\minus{}p\plus{}2)}{[2s^2\plus{}(p\minus{}2)^2]^2} \ge 0\]
because $ 2s\plus{}3p\minus{}6\ge 4\sqrt{p}\plus{}3p\minus{}6 >0$ and $ 3s\minus{}p\plus{}2\ge 6\sqrt{p}\minus{}p\plus{}2 >0$.
Therefore, $ g(s)\ge g(2\sqrt{p})\equal{}2f(\sqrt{ab})$, the remark was proved.
We see that the proof works also for $ \frac{26\minus{}4\sqrt{22}}{9}\le ab\le 20\plus{}6\sqrt{11}$.

Nice, ehku.

But we can prove that $ f(a) \plus{} f(b) \ge 2f(\sqrt {ab})$ without using derivative, since it is equivalent to

$ (s \minus{} p)[(7p^2 \plus{} 24p \minus{} 14)s \plus{} (p^2 \minus{} 2)( \minus{} 3p^2 \plus{} 7p \plus{} 6)] \ge 0$,

where $ s \equal{} \frac {a \plus{} b}{2}$ and $ p \equal{} \sqrt {ab}$, clearly $ s\ge p$ and $ 1\le p \le 3$. We have

 $ (7p^2 \plus{} 24p \minus{} 14)s \plus{} (p^2 \minus{} 2)( \minus{} 3p^2 \plus{} 7p \plus{} 6)\ge (7p^2 \plus{} 24p \minus{} 14)p \plus{} (p^2 \minus{} 2)( \minus{} 3p^2 \plus{} 7p \plus{} 6) \equal{} (2 \plus{} 6p \minus{} p^2)(3p^2 \plus{} 4p \minus{} 6) \ge 0.$

[quote="ehku"][quote="Vasc"]Nice solution, ehku. :lol:
 But I think it is not very simple to show that $ 1\le ab\le 10$ yields $ f(a) \plus{} f(b) \ge 2f(\sqrt {ab})$.[/quote]
I used a technique called "Mixing Variables by Derivative", which is very useful for classical mixing variables, UMV (AC/GC) theorem, SMV theorem,...

Put $ s \equal{} a \plus{} b,p \equal{} ab$ then $ s\ge 2\sqrt {p}$ and
\[ f(a) \plus{} f(b) \equal{} \frac {7 \minus{} 6a}{2 \plus{} a^2} \plus{} \frac {7 \minus{} 6b}{2 \plus{} b^2} \equal{} \frac {7s^2 \minus{} 6(2 \plus{} p)s \plus{} 14(2 \minus{} p)}{2s^2 \plus{} (2 \minus{} p)^2} \equal{} g(s).
\]
We have
\[ g'(s) \equal{} \frac {2(2 \plus{} p)(6s^2 \minus{} 7(2 \minus{} p)s \minus{} 3(2 \minus{} p)^2)}{[2s^2 \plus{} (p \minus{} 2)^2]^2}
\]

\[ \equal{} \frac {2(2 \plus{} p)(2s \plus{} 3p \minus{} 6)(3s \minus{} p \plus{} 2)}{[2s^2 \plus{} (p \minus{} 2)^2]^2} \ge 0
\]
because $ 2s \plus{} 3p \minus{} 6\ge 4\sqrt {p} \plus{} 3p \minus{} 6 > 0$ and $ 3s \minus{} p \plus{} 2\ge 6\sqrt {p} \minus{} p \plus{} 2 > 0$.
Therefore, $ g(s)\ge g(2\sqrt {p}) \equal{} 2f(\sqrt {ab})$, the remark was proved.
We see that the proof works also for $ \frac {26 \minus{} 4\sqrt {22}}{9}\le ab\le 20 \plus{} 6\sqrt {11}$.[/quote]

Great solution, I am really interested in your proof. Very nice idea :) But sometime by the way we couldn't get the nice form of $ f'(s)$. It means that it is not easy to find out the value of $ p,s$ such that $ f'(s)\ge 0$. For example:
\[ \sum \frac {a}{a^3 \plus{} 1} \le \frac {3}{2},abc \equal{} 1
\]
In fact:
\[ \frac {a}{a^3 \plus{} 1} \plus{} \frac {b}{b^3 \plus{} 1} \equal{} \frac {s^2p \plus{} s \minus{} 2p^2}{s^3 \minus{} 3ps \plus{} p^3 \plus{} 1} \equal{} f(s)
\]
Before some caculate simply I got:
\[ f'(s) \equal{} \frac {(2p^4s \minus{} 5p^3 \minus{} 2s^3 \plus{} 3p^2s^2 \minus{} ps^4 \plus{} 2ps \plus{} 1)}{(p^3 \plus{} s^3 \minus{} 3ps \plus{} 1)^2}
\]
which couldn't be factored like above one so I couldn't define the sign of $ f'(x)$  :roll:
Did you find out the better for this technique, ehku?!  :)

[quote=Vasc]If $ a,b,c$ are positive real numbers such that $ abc\equal{}1$, then

$ \frac {7\minus{}6a}{2\plus{}a^2}\plus{}\frac {7\minus{}6b}{2\plus{}b^2}\plus{}\frac {7\minus{}6c}{2\plus{}c^2} \ge 1$.[/quote]
See here:

https://artofproblemsolving.com/community/c6h591208p7837853

A nice generalization.

If $a,b,c,d$ are positive real numbers such that
$$a\ge b\ge 1\ge c\ge d, \ \ \ \ \ abcd=1,$$
then
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}+\frac{7-6d}{2+d^2}\ge \frac 4{3}.$$

For $b=1$ or $c=1$, we get the original inequality.

[quote=Vasc]If $ a,b,c$ are positive real numbers such that $ abc\equal{}1$, then

$ \frac {7\minus{}6a}{2\plus{}a^2}\plus{}\frac {7\minus{}6b}{2\plus{}b^2}\plus{}\frac {7\minus{}6c}{2\plus{}c^2} \ge 1$.[/quote]

The following is also true:
If $ a,b,c$ are non-negative real numbers such that $ a+b+c\equal{}1$, then

$ \frac {7\minus{}6a}{2\plus{}a^2}\plus{}\frac {7\minus{}6b}{2\plus{}b^2}\plus{}\frac {7\minus{}6c}{2\plus{}c^2} \ge \frac{127}{18}$

[quote=arqady]djdkdc, are you sure?[/quote]

Dear arqady if i have forgotten something or i'm wrong tell me.

[quote=djdkdc][quote=Vasc]If $ a,b,c$ are positive real numbers such that $ abc\equal{}1$, then

$ \frac {7\minus{}6a}{2\plus{}a^2}\plus{}\frac {7\minus{}6b}{2\plus{}b^2}\plus{}\frac {7\minus{}6c}{2\plus{}c^2} \ge 1$.[/quote]

Denote $x=lna_{cyc}$and$f(x)=\frac{7-6e^x}{2+e^{2x}}$
so this ineq becomes:
if $x+y+z=1$ then $\sum_{cyc} \frac{7-6e^x}{2+e^{2x}}\ge 1$.because $ \frac{d^2}{dx^2} f(x)=0$ has  one root(one inflection point) so by $SIP$(karamata and jensen theorem), ineq reaches at min iff $$a=b=1/c^2$$.
so it's enough to prove that:
$$\frac{2(7-6a)}{a^2+2}+\frac{(7a^2-6)a^2}{2a^4+1} \ge 1$$$\Longleftrightarrow$  $5a^6-24a^5+32a^4-13a^2-12a+12\ge0(a\neq0)$ $\Longleftrightarrow a> 0$done.:whistling:[/quote]

oh my god...my sol is completely wrong because a silly mistake(it doesn't have one inflection point...).i'm sorry.

[quote=zdyzhj][quote=arqady]djdkdc, this inequality is not so easy.[/quote]

easy to prove, no difficult at all.[/quote]

I agree with you lagrange multi kills it easily... :thumbup:


Consider a word to be [i]pronouncable[/i] if it does not have a string of 3 or more consonants. For example, "athm" is not a pronouncable word, but "mtah" is. How many 4 letter pronouncable words are there?(assume y is a vowel)

[hide="answer"]i got 200976[/hide]

blah, didnt need to put it in calc, but im too lazy to do all that

$ 26*26*26*26\minus{}2*20*20*20*6\minus{}20*20*20*20$


Hey everyone, I thought I would poke my head in here again and share some of my work:

[img]http://www.hypersanity.com/galleries/graphic_design/images/a_carlos_santana_greek_2004.jpg[/img]

[img]http://www.hypersanity.com/galleries/the_beautiful/images/d_Beautiful_Woman_colored_pencil.jpg[/img]

[img]http://www.hypersanity.com/galleries/graphic_design/images/g_SC-Coverconcept1.jpg[/img]

I have a bunch more on my site if anyone wants to see it:

http://www.hypersanity.com

A couple of other pieces are on this site too under "Angelo Vildasol".

http://alphabetofmanliness.com/?u=artists

The second pic is absolutely amazing -- keep up the good work! :)

just perfect...........keep creating  :)


Let $f$ be an increasing mapping from the family of subsets of a given ﬁnite set $H$ into itself, i.e. such that for every $X \subseteq Y\subseteq H$  we have $f (X )\subseteq f (Y )\subseteq H .$  Prove that there exists a subset $H_{0}$ of $H$ such that $f (H_{0}) = H_{0}.$

If $f(H) = H$, we're done.  Otherwise, let $f(H) = H_{1}\subset H$.  Now, if $f(H_{1}) = H_{1}$, we're done.  Otherwise, $f(H_{1}) = H_{2}\subset H_{1}= f(H)$.  And so on.  So for each $i$, either $f(H_{i}) = H_{i}$ or $H_{i+1}= f(H_{i}) \subset H_{i}$.  If there is no instance of the first, we have an infinite sequence of decreasing subsets, which is impossible.

This is a particular case of the following result due to Tarski and Knaster:

Let $\mathcal L$ be a complete lattice, and $f: \mathcal L\to\mathcal L$ an increasing map. Then, the set of fixed points of $f$ is non-empty (and in fact forms a complete lattice).


Μελετώντας τη συμπεριφορά κάποιων συναρτήσεων κατέληξα στην παρακάτω ανισότητα. Η μόνη λύση που έχω είναι με τη βοήθεια της ανάλυσης. Επειδή μου φαίνεται όμορφη ανισότητα σας την παρουσιάζω! ώστε να δωθούν και άλλες λύσεις. ( Δεν την έχω προσπαθήσεις ακόμα να τη λύσω διαφορετικά ).

Υ.Σ.  Επειδή βιάστηκα να ανεβάσω το post ελπίζω να είναι ενδιαφέρουσα...

Έστω x,y,z $ \in [a,a\plus{}1]$  όπου  $ a \in R$

Αποδείξτε ότι                      :   $ 4(x^2\plus{}y^2\plus{}z^2)\plus{}3(2a\plus{}1)^2 \le 4(2a\plus{}1)(x\plus{}y\plus{}z)\plus{}3$

[b][color=darkred]Η ανισότητα γενικεύεται για  n  αριθμούς που ανήκουν στο σύνολο [ α, α+1 ][/color].[/b]

[quote="vangelismak"]Μελετώντας τη συμπεριφορά κάποιων συναρτήσεων κατέληξα στην παρακάτω ανισότητα. Η μόνη λύση που έχω είναι με τη βοήθεια της ανάλυσης. Επειδή μου φαίνεται όμορφη ανισότητα σας την παρουσιάζω! ώστε να δωθούν και άλλες λύσεις. ( Δεν την έχω προσπαθήσεις ακόμα να τη λύσω διαφορετικά ).

Υ.Σ.  Επειδή βιάστηκα να ανεβάσω το post ελπίζω να είναι ενδιαφέρουσα...

Έστω x,y,z $ \in [a,a + 1]$  όπου  $ a \in R$

Αποδείξτε ότι                      :   $ 4(x^2 + y^2 + z^2) + 3(2a + 1)^2 \le 4(2a + 1)(x + y + z) + 3$

[b][color=darkred]Η ανισότητα γενικεύεται για  n  αριθμούς που ανήκουν στο σύνολο [ α, α+1 ][/color].[/b][/quote]
Ισχυουν οι ισοδυναμιες:

$ 4(x^2  + y^2  + z^2 ) + 3(2a + 1)^2  \le 4(2a + 1)(x + y + z) + 3 \Leftrightarrow  \\
 4(x^2  + y^2  + z^2  - (2a + 1)(x + y + z)) + 3((2a + 1)^2  - 1^2 ) \le 0 \Leftrightarrow  \\
4(x^2  + y^2  + z^2  - (2a + 1)(x + y + z)) + 3(2a)(2a + 2) \le 0 \Leftrightarrow  \\
 x^2  + y^2  + z^2  - (2a + 1)(x + y + z) + 3a(a + 1) \le 0 \Leftrightarrow  \\
 x^2  + y^2  + z^2  - (2a + 1)(x + y + z) + 3a(2a + 1) - 3a^2  \le 0 \Leftrightarrow  \\
 (x - a)(x + a) + (y - a)(y + a) + (z - a)(z + a) - (2a + 1)(x + y + z - 3a) \le 0 \Leftrightarrow  \\
 (x - a)(x + a) + (y - a)(y + a) + (z - a)(z + a) - (2a + 1)(x - a) - (2a + 1)(y - a) - (2a + 1)(z - a) \le 0 \Leftrightarrow  \\
 (x - a)(x + a - 2a - 1) + (y - a)(y + a - 2a - 1) + (z - a)(z + a - 2a - 1) \le 0 \Leftrightarrow  \\
 (x - a)(x - a - 1) + (y - a)(y - a - 1) + (z - a)(z - a - 1) \le 0 \\$

Ισχυει ομως:
$ a \le x \le a + 1 \Rightarrow 0 \le x - a \le 1$  [b][color=blue](1)[/color][/b] Ακομη:
$ (1)\Rightarrow - 1 \le x - a - 1 \le 0$ [b][color=red](2)[/color][/b]

Και (1) λογω (2) συνεπαγεται:
$ 0 \ge (x - a)(x - a - 1)$ [b][color=green](3)[/color][/b]

Ομοια και για y,z:
$ 0 \ge (y - a)(z - a - 1)$ [b][color=indigo](4)[/color][/b]
$ 0 \ge (z - a)(z - a - 1)$ [b][color=orange](5)[/color][/b]

Απο (3)(4)(5) συνεπαγεται η ζητουμενη λογω των παραπανω ισοδυναμιων.

Μελετωντας ποιες συμπεριφορες, ποιων συναρτησεων κατεληξες σε αυτην?

Thanks για τη λύση σου ...
Βασικά όσο και αν φαίνεται  περίεργο αυτή προέκυψε από τη $ x^2$ . Η όλη ιδέα ήταν ότι άν πάρουμε ένα διάστημα της μορφής [ α, α+1 ] και βρούμε την εξίσωση της εφαπτομένης της Cf στο σημείο της  (2α+1)/2 , λαμβάνοντας υπόψη και την κυρτότητα της f, προκύπτει ότι  μέγιστη κατακόρυφη απόσταση των διαγραμμάτων της Cf και της εφαπτομένης (ε) είναι 1/4 για οποιαδήποτε επιλογή του α $ \in$ R. Αυτή η ανισότητα έχει προκύψει από το άθροισμα τριών τέτοιων κατακόρυφων αποστάσεων !!! όπου οι τετμημένες των σημείων είναι τα x,y,z που ανήκουν στο σύνολο [ α, α+1 ]. Η ισότητα ισχύει όταν καθένα από τα x,y,z γίνουν α  ή  α+1.

Αρκεί να αποδείξουμε ότι $ 4x^2 \plus{} (2a \plus{} 1)^2 <\equal{} 4(2a \plus{} 1)x \plus{} 1 < \equal{} > 4x^2 \minus{} 4(2a \plus{} 1)x \minus{} 1 \plus{} (2a \plus{} 1)^2 <\equal{} 0$
Θεωρώντας την παράσταση αυτή τριώνυμο ως προς $ x$, παίρνουμε $ D \equal{} 16(2a \plus{} 1)^2 \minus{} 16[ (2a \plus{} 1)^2 \minus{} 1] \equal{} 16 \equal{} 4^2 > 0$
άρα για την παράσταση έχουμε τις ρίζες $ x_1 \equal{} \frac {4(2a \plus{} 1) \minus{} 4}{8} \equal{} a$ και $ x_2 \equal{} \frac {4(2a \plus{} 1) \plus{} 4}{8} \equal{} a \plus{} 1$
η παράσταση θα είναι $ <\equal{} 0$ όταν το $ x$ βρίσκεται μεταξύ των ριζών, δηλαδή όταν $ x_1 <\equal{} x <\equal{} x_2 < \equal{} > a <\equal{} x <\equal{} a \plus{} 1$, που ισχύει από την εκφώνηση.

Άρα $ 4x^2 \plus{} (2a \plus{} 1)^2 <\equal{} 4(2a \plus{} 1)x \plus{} 1$
Ομοίως $ 4y^2 \plus{} (2a \plus{} 1)^2 <\equal{} 4(2a \plus{} 1)y \plus{} 1$
και $ 4z^2 \plus{} (2a \plus{} 1)^2 <\equal{} 4(2a \plus{} 1)z \plus{} 1$
Προσθέτοντας τώρα αυτές τις τρεις σχέσεις κατά μέλη, βγαίνει το ζητούμενο.

Μου άρεσαν οι λύσεις σας πάρα πολύ, αλλά αν τις καλοκοιτάξετε κάνετε και οι δύο απόδειξη της θεωρίας τριωνύμου (πάντως οι σκέψεις σας παραμένουν εντυπωσιακές).

maniopa  η χρήση της θεωρίας του τριωνύμου λύνει αυτό το συγκεκριμένο θέμα ... που προέκυψε από την  $ x^2$. Το θέμα αυτό το έκανα post επειδεί δεν είναι βασανιστικό στις πράξεις ( λύνεται με χρήση θεωρίας τριωνύμου ) και είναι κάπως κομψό. Η γενική ιδέα όμως δεν νομίζω ότι μπορεί να αντιμετωπιστεί με χρήση θεωρίας τριωνύμου. Θέλω να πώ ότι το γενικό δεν είναι η θεωρία του τριωνύμου που στηρίχτηκα πάνω σ΄αυτή για να κατασκευάσω την άσκηση αλλά η συμπεριφορά γενικότερα της $ x^n$ με την εφαπτομένη της σε κάποιο σημείο.


calculus
integration
calculus computations

hello, setting $ t=x+\sqrt{1+x^2}$, then we have $ x=\frac{t^2-1}{2t}$ and $ dx=\frac{t^2+1}{2t^2}dt$ and we get ${ \int\frac{t^{15}}{\frac{t^2+1}{2t}}}\frac{t^2+1}{2t^2}\,dt=\int t^{14}\,dt$.
Sonnhard.


conics
parabola
geometry solved
geometry

I was just looking at my math textbook (gr 10) where a parabola is drawn. It almost looks like a section of a circle. Now my question is what is the maximum portion of any circle that can lie on any parabola. I think there is a fixed value for this question.

Think of the parabola as a cup. Crate a circle of any size and try to fit it inside the parabola. if it dosent fit increase the [b]a[/b] value of the parabola ( y=a(x-h)+k ) until the very bottom of the circle matches the vertex of the parabola.

Is it possible to figure this out?

So wait, are you asking for the largest circle that can lie in the cup of a parabola? Because you can't get a circular curve and a parabolic curve to be exactly the same unless they are infinitely small (i.e. points).

Ohh alright you answered my question there. So why cant a parabola be shaped like a circle what about 3 points the vertex and the 2 points next to it ... is it possible that they can match that part of a circle?

Yes, but only because any 3 points have a circle through them.  Thus the fact that both curves go through these points does not guarantee that the curves are (even locally) the same.


We put $U=U_{0}\sin(\omega t)$ alternating voltage on a plane capacitor whose plates are at distance $l$ from each other. We shoot an electron between the plates into it with $v_{0}$ initial velocity. The phase of the voltage when the electron gets shot in is $\phi$. Find the trajectory( path) of the electron between the plates.

Is $v_{0}$ horizontal?

The force acting on the electron is $F =-eE =-eu/l =-(eU_{0}/l)\sin \omega t$. So now it's a trivial differential equation. :huh: Is this ok?

yes $v_{0}$ is horizontal.
I dont see why you got the $-$.
and the phase is not 0 when the electron entered.

The $-$ is because the electron is negatively charged ($e > 0$).

Well ok, if we say $t = 0$ at the moment the electron enters than
\[F =-(eU_{0}/l)\sin(\omega t+\phi).\]

Since you have all the constants, it's still simple. Is there a catch? :o I think that gravity should be neglected, but even if it isn't it doesn't affect the horizontal motion, so...?


algebra
polynomial
induction
algebra unsolved

Consider the polynomial $W(x) = (x - a)^kQ(x)$, where $a \neq 0$, $Q$ is a nonzero polynomial, and $k$ a natural number. Prove that $W$ has at least $k + 1$ nonzero coefficients.

[quote="grobber"]WLOG, we may assume that $W(0)\ne 0$ (we simply divide $W$ by $x^n$ for the largest possible exponent $n$; the number of non-zero coefficients does not change).

Now we can prove the assertion by induction on the degree of the polynomial. If $n$ is the degree of $W$, the degree of $W'$ is $n-1$, and $(x-a)^{k-1}|W'(x)$. This means, according to the induction hypothesis, that $W'$ has at least $k$ non-zero coefficients, and these correspond to $k$ non-zero coefficients of $W$. Furthermore, the free term of $W$ is a non-zero coefficient which does not appear in $W'$, so we're done.[/quote]

Nice, but to make it formal you need to prove it if $k=1$

let $ \ W(x)=a_{l}x^{l}+..a_{0}$
then see we can express $\ Q(x) =x^{m_{0}}Q_{1}(x) $ where $\ m \geq 0 $
where $\ Q_{1}(0) \neq 0$
so $\ W(x)=x^{m_{0}}(x-a)^{k}Q_{1}(x)=x^{m_{0}}W_{1}(x)$
see $\ W_{1}(x)=(x-a)^{k}Q_{1}(x) $
so $\ W_{1}(0) \neq 0 $
so $ \ a_{m_{0}} \neq 0$
now see $\ W_{1}'(x)=(x-a)^{k-1}Q_{2}(x) $ where Q_{2} can't be identically equal to 0.
again by same logic we can get  $\ Q_{2}(x) =x^{m_{1}}Q_{3}(x) $ where $\ m \geq 0 $
where $\ Q_{3}(0) \neq 0$
then by same logic the coefficient of x^{m_{1}} in the polynomial $ \ W'_{1}(x) \neq 0$
so the coefficient of x^{m_{1}+1} in the polynomial $ \ W_{1}(x) \neq 0$
so $ \ a_{m_{0}+m_{1}+1} \neq 0$
sillly  go on differentiating k-times we get
$\ a_{m_{0}},a_{m_{0}+m_{1}+1},a_{m_{0}+m_{1}+m_{2}+2},....,a_{m_{0}+....m_{k}+k} \neq 0 $
so (k+1) many coefficients are non zero.


inequalities
real analysis
real analysis unsolved

Suppose $a_n$ converges to 0 (real numbers) and the sequence $\frac{a_{n+1}}{a_n}$ is always between $-1$ (possibly equal) and $ 0<r<1$. Does it follow that $\sum_{n\geq 1}{a_n}$ converges?

Yes, I believe it is convergent, but I'm not sure the following is correct.

Take $N$ large enough that $a_n\le t,\ \forall n\ge N$ and some $p\in\mathbb N$. We want to show that if $N$ is large enough, then $|a_n+\ldots+a_{n+p}|$ is small. Break the sum $a_n+a_{n+1}+\ldots+a_{n+p}$ into pieces as follows: if $a_k,a_{k+1},\ldots,a_{k+m}$ form a maximal (wrt inclusion) consecutive subsequence of $a_n,\ldots,a_{n+p}$ s.t. $\frac{|a_{i+1}|}{|a_i|}\in(r,1]$ for all $i\in\overline{k,k+m-1}$, then we take $|a_k+a_{k+1}+\ldots+a_{k+m}|$ out of the sum (I'm trying to write an inequality of the form $|a_n+\ldots+a_{n+p}|\le\sum|S_i|$, where $S_i$ are sums of subsequences of $a_n,\ldots,a_{n+p}$), and if $a_l$ is not part of such a maximal subsequence, then take it out all by itself, as $|a_l|$.

It's easy to see that there is some universal constant $C$ s.t. whenever $a_k,a_{k+1},\ldots,a_{k+m}$ is as described above, we have $|a_k+\ldots+a_{k+m}|\le C|a_k|$, so in the end we will get an inequality of the type $|a_n+\ldots+a_{n+p}|\le |a_n|(1+r+\ldots)+C|a_n|(1+r+\ldots)$, and the conclusion follows, since if $N$ is large enough, $a_n$ with $n\ge N$ will be close to $0$.

I hope it's at least partially understandable, and maybe even correct :).


The scale drawing shown represents a field in the shape of an isosceles trapezoid. What is the area of the actual field? Express your answer in simplest radical form.

[asy]draw((0,-15)--(0,15)--(100,45)--(100,-45)--cycle);
label("2cm",(0,0),W);
label("6cm",(100,0),E);
label("12cm",(50,-33),S);
label("12cm",(50,33),N);
label("Scale: 1cm=10m",(50,-47),S);[/asy]

Height is easily found: $ \sqrt{140}\equal{}2\sqrt{35}$. Area is $ \frac{2\plus{}6}{2}\cdot 2\sqrt{35}\equal{}8\sqrt{35}$, or in meters:
$ \boxed{800\sqrt{35}}$(because of scale).

How to find the height:

$ \frac {6 \minus{} 2}{2} \equal{} 2$

$ 12^2 \minus{} 2^2 \equal{} \sqrt {140} \equal{} 2\sqrt {35}$

The area of a trapezoid with base lengths $ a$ and $ b$ is $ \frac {a \plus{} b}{2} \times h$.

EDIT: Oops, forgot that. Fixed. :)

The area is actually $ \frac{a\plus{}b}{2}\cdot h$, where $ h$ is height.


Θα κάνω μια ερώτηση λίγο άσχετη. Σε ποια σχολή θέλετε να περάσετε όσοι είστε λύκειο ή περάσατε?
(δεν νομίζω να θέλουν όλοι μαθηματικό  :whistling: )
 Εγώ προσωπικά σκέφτομαι για Μηχανολόγων Μηχανικών ΕΜΠ και ελπίζω να κανουν αρκετά μαθηματικά  :roll:

Μαθηματικό 4-ever  :heli:
Να τελειώσω φέτος με το καλό (2 μήνες έμειναν-πού θα παει ) :stretcher:

Και μην ξανακούσω "ελπίζω στη σχολή να βρώ το ένα και το άλλο", αν σε ενδιαφέρει μπες στο site της σχολής και δες το πρόγραμμα σπουδών σας (http://www.mech.ntua.gr/gr you know).
Εδώ μέχρι και το site του μαθηματικού του ΑΠΘ εδεήθησαν να το τελειώσουν (επιτέλους -καιρός ήτανε)
:pilot:

Ναι, είδα το πρόγραμμά τους και τι μαθηματικά έχουν και μπορώ να πω ότι με ικανοποιεί αρκετά αυτό που είδα  :P  (εξάλλου ο λόγος που θέλω την σχολή δεν είναι (τόσο) τα μαθηματικά)

ε, mailo εγω ειμαι στο τμημα των μηχανολογων, ρωτα με οτι θες  :)

Ωραία Κώστα! Θα σου στείλω ΠΜ  :)

Μονο εγω κι ο alexandros_plus παμε τρίτη λυκείου? Οι υπόλοιποι είστε φοιτητές?

Η πλειοψηφια ειναι φοιτητες οπως κι εγω mailo. :)


trigonometry
geometry
absolute value
complex numbers

I'm having trouble simpllifying the following trig expression.

sin1*sin3*sin5...sin179

I simplifyed it to:

(sin1sin3...sin89)^2

All my other attempts at simplification (half angle formulas, sum to product
formulas, etc.) simplifyed into expressions that were impossible to simplify.
Please help!
Thanks.

To evaluate sin1*sin3*sin5*...*sin179, we evaluate sin1*sin2*sin3*...*sin179 and sin2*sin4*sin6*...*sin178 and then divide them at the end. So we shall show that the former is 180/2^179, and the latter is 90/2^89.

We write x^n-1=(x-1)(x-a)(x-a^2)(x-a^3)...(x-a^n), where a is the primitive nth root of unity e^(2i*pi/n)=cos(360/n)+i*sin(360/n). We divide both sides of the equation by x-1 to give 1+x+x^2+x^3+...+x^(n-1)=(x-a)(x-a^2)(x-a^3)...(x-a^n). Now, let x=1, and so we get n=(1-a)(1-a^2)(1-a^3)...(1-a^n). Now we take the absolute value of each side and use the identity |1-e^(i*theta)|=2*sin(theta/2). This gives for the first one 180=2sin1*2sin2*2sin3*...*2sin179, or sin1*sin2*sin3*...*sin179=180/2^179. Similarly, 90=2sin2*2sin4*2sin6*...*2sin178, or sin2*sin4*sin6*...*sin178=90/2^89. Therefore sin1*sin3*sin5*...*sin179=(180/2^179)/(90/2^89)=1/2^89.

Hope this helps.

Very impressive. Can you tell me what book you got the theory of that from? It seems very interesting.

I read it in a solution to a problem from the Bay Area Math Meet Test of Ingenuity. The website is at http://bamm.usfca.edu. It appears that most of the solutions are now gone unfortunately. I'm not sure where else you can find this method, but I think it is very useful for evaluating the product of trig functions.

I appreciate your help. However, I'm an eighth grade math student
and I have little knowledge of the math you used. It was in my math
book in a half-angle formula section. Is there any way you could use
simpler techniques?
If you can't I understand.
Thanks for helping!

Does your trig book deal with complex numbers & such (or roots of unity)?  Zeta's solution does use a half-angle formula (albeit in complex form).

You can get there with the trig you know.

All you need is:

sin x = cos (90 - x) = - cos(90 + x) = etc.

sin 2x = 2 sin x cos x

Here's the next step as a hint:

(sin1sin3...sin89)^2  = (sin1sin3...sin89)(cos89cos87...cos1)

[I thought to do that because of the aforementioned double angle formula for sine; I knew I'd be able to take another step from here.]

Sorry it took me so long to respond.

I've tried the line of reasoning stated earlier, but it always simplifies into something that can't be simplified and I can never get rid of the sines.

I realize this might be a pretty easy problem for most of you out there but obviously I'm missing something here, and I was hoping someone could help me figure it out. :?:
Thanks :D

Not an easy problem, I think.  I'm kind of surprised you found it in a regular trig book.

Next step in hint:

We had:

(sin1sin3...sin89)^2 = (sin1sin3...sin89)(cos89cos87...cos1)

Combine sines & cosines:

(sin1sin3...sin89)(cos89cos87...cos1)  = (sin1cos1)(sin3cos3)....
=(sin2)/2 * (sin6/2) * ... (sin 178)/2

Now, remember that

sin 2x = 2 sin x cos x

is also

sin x = (sin 2x)/(2 cos x)

See what you can do from here...

(Also, if anyone out there knows a faster way to get where I'm going without the slick jump to complex numbers, let us know.)

With your last hint I believe I can finally figure it out.

Tell me if there are any flaws in my proof.

The next step is:

(1/2^45)(sin^2(2)sin^2(6)...sin^2(86))
(By the identity sin x= sin (180-x) for angles less than 180)

Then I use your version of the double angle formula

(1/2^89)((sin^2(4)/cos^2(2))(sin^2(12)/cos^2(6))...(sin^2(172)/cos^2(86))

Then the expression simplifies to:

(1/2^89)((sin^2(4)/cos^2(2))(sin^2(8)/(cos^2(6))...(sin^2(88)/cos^2(86))

A simple application of the identity sin x = cos (90-x) leaves us with:

(1/2^89)

Please tell me if this works.

Thanks for your help

Is it not enough to only say that sin(1) = sin(179), sin(3) = sin(177),...,sin(89)=sin(91) because of the unit circle?


Let X, Y be topological spaces. Let f be a map from X to Y. Show that f is continuous if and only if f of closure of E is a subset of the closure of f(E) for every subset E of X, if and only if the closure of inverse map of f of E is a subset of the inverse map of f of the closure of E, for every subset E of Y.

Show first one!

Assume $ f$ is continuous, and $ U$ is neighborhood of $ f(x)$ with $ (x \in \bar{E})$

cuz $ f$ is continuous, $ f ^{ \minus{} 1} (U)$ is open containing $ x$

cuz $ x \in \bar{E}$ , $ f^{ \minus{} 1}(U)$ intersect $ E$

thus $ U$ intersects $ f(E)$, means $ f(x) \in \bar{f(E)}$ thus $ f(\bar{E}) \subset \bar{f(E)}$

This shows first if-only statement.


calculus
function
integration
derivative
calculus computations

the school year had just started for me, and i am in calculus bc this year.  our teacher mentioned something about having an advantage in having an ti-89 for the ap exam.  I already have a ti-84+, is it necessary for me to acquire a ti-89.  also do they allow two calculators in the ap exam?
thanks a bunch. :lol:

It is certainly not [i]necessary[/i] to have a TI-89, and the test is supposed to be designed so that it has no significant advantage.  Whether or not I'd recommend it is the question and I say it depends on you.  It is certainly a nice tool to have around if you want to check an antiderivative or something, you can just type it in.  With that, though, I saw some people take advantage of that function so frequently that they never became good at integrating by hand.

So its up to you, it could be a worthwhile investment, but if you like math I'm sure you'll do just fine with any calculator.

As for having two, I am pretty sure you can bring a backup calculator but you can't use both at the same time.

And just so you know I took it with my favorite, the TI-86  :)  I got an 89 also, but chose not to use it because it is pretty confusing and I hadn't taken the time to learn how to do all the things I already knew with the 86.  So maybe that's another bad thing.

the only advantage of an 89 is that if can display indefinite integrals while an 84 cannot. however, the ap exam will not have you solve ones that are of such difficulty that an 89 would be "necessary".
really i say just learn to integrate (you would be surprised at the number of calculus students that have trouble with this) and you will do fine.

as for taking derivatives those a pretty straight foward and an 89 is most certainly not needed.

for taylor polnomials an 89 is nice but not necessary since you should have enough time to do it by hand on the exam, plus showing work helps. (something the 89 cannot do for you  :P )

no, AP calc test is relatively easy compared to other AP tests (and to anything in aops).

If you know your stuff you should do alright with ti83 or better.