logarithms
limit
calculus
algebra unsolved
algebra

solve the following equation:

$ \log_{2}(x)\equal{}1\minus{}(x\minus{}2)^{2}$

thanx.

Let $ g(x)\equal{}log_2(x)\plus{}(x\minus{}2)^2\minus{}1$. Then $ g'(x)\equal{}\frac{2(x\minus{}1)^2\plus{}1\minus{}2ln(2)}{xln(2)}, \ g''(x)\equal{}\frac{2\minus{}x^2ln(2)}{x^2ln(2)}$.
Obviosly $ x\equal{}1$ and $ x\equal{}2$ are roots. Therefore they are all roots $ x\ge 1$.
But we had one solution with $ x<1$.

$ \log_2x=1-(x-2)^2\Leftrightarrow \log_2x=-(x-1)(x-3).$
1). $ \log_22 = 1 = 1 - (2 - 2)^2\Rightarrow x = 2\mbox{ - solution!}$
2). $ \log_21 = 0 = 1 - (1 - 2)^2\Rightarrow x = 1\mbox{ - solution!}$
3). $ x > 2\Rightarrow\{\begin{array}{c}\log_2x > \log_22 = 1 \\
1 - (x - 2)^2 < 1\end{array} \Rightarrow \log_2x > 1 - (x - 2)^2$
$ \Rightarrow x\in(2;\infty)\mbox{ not solution.}$

[quote="reason"]solve the following equation:

$ \log_{2}(x) \equal{} 1 \minus{} (x \minus{} 2)^{2}$

thanx.[/quote]

Let $ f(x) \equal{} \log_2(x) \plus{} (x \minus{} 2)^2 \minus{} 1$ and so $ f'(x) \equal{} \frac 1x(2x^2 \minus{} 4x \plus{} \frac 1{\ln 2})$ which has $ 2$ real roots. So $ f(x)$ has at most three roots.

$ \lim_{x\to 0^ \plus{} } \equal{} \minus{} \infty < 0$, $ f(\frac 12) \equal{} \frac 14 > 0$ and $ f(1) \equal{} f(2) \equal{} 0$

So $ f(x)$ has exactly 3 roots :
$ x_1 \equal{} 2$
$ x_2 \equal{} 1$
and $ x_3\in(0,\frac 12)$. Numeric calculus gives $ x_3 \equal{} 0.22548259432668...$


Find in an elementary way all triples of integers $ a,b,c$ such that $ a\cdot 2^n+b| c^n+1$ for all natural number n.

Ok Harazi, here we go:

put $n=2^t$ for big enough $t$, we have that all prime factors of $c^{2^t}+1$ are of the form $2^{t+1}k+1$ right? so we also have that $a2^{2^t}+b=1$ (mod$2^{t+1}$) but it implies for $t$ big enough that $b=1$

So we have that $2^na+1|c^n+1$  for every $n$, and i am a little bit stuck....

(BTW i will try to finish it soon)


Find the negative root(s) of the equation $ (x\plus{}1)^2 \equal{} 2(x\plus{}1)$.

$ x\equal{}\minus{}1$ is an obvious solution. If this is not the case, $ x\plus{}1\equal{}2$, and $ x\equal{}1$. This is also evidenced by letting $ x\plus{}1\equal{}y$.

$ y^2\equal{}2y$
$ y^2\minus{}2y\equal{}0$
$ y^2\minus{}2y\plus{}1\equal{}1$
$ (y\minus{}1)^2\equal{}1$
$ x^2\equal{}1$.

So -1.


Can anyone give a good estimate of where the cut off on Tuesday's AMC-10 is? Is 116 enough?

Do not discuss the tests until Thursday, noon ET.  Not even scores or speculation about cut-offs.


Problem 3
Find the greatest possible positive integer $ x$ for which the number $ A\equal{}2^{182}\plus{}4^{x}\plus{}8^{700}$ is a perfect square of a positive integer.

Not really sure about this but...

[hide]
$ A \equal{} 2^{182} \plus{} 4^x \plus{} 8^{700} \equal{} 2^{182} \plus{} 2^{2x} \plus{} 2^{2100}$

Notice $ (2^a \plus{} 2^b)^2 \equal{} 2^{2a} \plus{} 2^{2b} \plus{} 2^{a \plus{} b \plus{} 1}$
Matching the exponents, we get:

$ 2a \equal{} 182 \Rightarrow a \equal{} 91$
$ 2b \equal{} 2x \Rightarrow b \equal{} x$
$ a \plus{} b \plus{} 1 \equal{} 2100 \Rightarrow 91 \plus{} x \plus{} 1 \equal{} 2100 \Rightarrow x \equal{} \boxed{2008}.$[/hide]

Well as RisingFlame said, but wanna add this to be completely:
Let a>b => 2a>a+b+1>2b(as there are number such as 182, 2100 there wont be a problem with that +1 added at a+b) .

In this case 2a should be 2x as the problem requests the greatest value of it .
As 2100>182(think its self-explanatory  :P ) a+b+1 = 2100, 2b=182. It results 2a=4016 => a=2008=x.

Hope u understand .

cheers,
ZrZ


Find the smallest number with 24 factors that are perfect integral squares.

um... just to let you know, i don't think many people on the middle school forum would know about intergrals since it's a calculus term.  :)

integral here is an adjective meaning which is an integer.

one thing, this is from a pretty high level competition, and another, it has nothing to do with circle area.

first, find the smallest number with 24 factors, then square it.

$24 = 2^3 \cdot 3$

this can either be:

$2^7 \cdot 3^2$
$2^2 \cdot 3^1 \cdot 5^1 \cdot 7^1$
or some other variation

I'm betting on the second one.

actually it's a lot smaller...

$2^3 \cdot 3^2 \cdot 5 = 360$

squaring that gives $\boxed{129600}$

Is that right?


calculus
integration
calculus computations

[quote="nemesis"]What is the value of the integral  :
$\int_0^{\pi/2} sin^{n}x dx$[/quote]

use reduction formula $I_n=\int^{\pi/2}_0 sin^nx dx$

then $\frac{n}{n-1}I_n=I_{n-2}$

[quote="nemesis"]What is the value of the integral  :
$\int_0^{\pi/2} sin^{n}x dx$[/quote]
it's integral of wallis

We can not evaluate this integral, all what we can do is to give some relations satisfied by this integral:

[b]1.[/b] $(n+2)I_{n+2}=(n+1)I_{n}$.

[b]2.[/b] $\forall n \in\mathbb{N}, I_n>0$

[b]3.[/b] $I_n \sim \sqrt{\frac{\pi}{2 n}}$ The most important one!

You can find here.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80515]www.mathlinks.ro/Forum/viewtopic.php?t=80515[/url]


For any positive real numbers $ a,b,c$ such that $ a\plus{}b\plus{}c\equal{}1$, prove that
\[ \frac{1\plus{}a}{1\minus{}a}\plus{}\frac{1\plus{}b}{1\minus{}b}\plus{}\frac{1\plus{}c}{1\minus{}c} \ge 2 \left(\frac{b}{a}\plus{} \frac{c}{b} \plus{} \frac{a}{c}\right)
\]

It's German TST 2005
See here : http://www.mathlinks.ro/viewtopic.php?p=161786#p161786


invariant
superior algebra
superior algebra unsolved

[quote="Robin Hartshorne, [i]Algebraic Geometry[/i], exercise 1.7.5"][b](a)[/b] Show that an irreducible curve $ Y$ of degree $ d > 1$ in $ \mathbb{P}^2$ cannot have a point of multiplicity $ \geq d$.
[b](b)[/b] If $ Y$ is an irreducible curve of degree $ d > 1$ having a point of multiplicity $ d \minus{} 1$, then $ Y$ is a rational curve (Ex. 6.1).[/quote]

The questions are:
[b](1)[/b] Are the curves supposed to be projective or quasi-projective? Is there a really obvious way to see that this is the same?
[b](2)[/b] Is the curve $ Y$ in [b](b)[/b] supposed to lie in $ \mathbb{P}^2$, or not? There is a rather easy idea that works in $ \mathbb{P}^2$ (modulo some "obvious" assertions which I have no idea how to prove), but I don't see how it would hold in $ \mathbb{P}^n$. Is the problem [b](b)[/b] valid in $ \mathbb{P}^n$ at all?

  darij

Again, I answer my own questions.

[b](1)[/b] is not really important, since multiplicities of points, rationality, irreducibility and being a curve (i. e. an algebraic set of dimension $ 1$) are all invariant when passing from an algebraic set to its closure and back.

[b](2)[/b] is not important either, since we can always WLOG assume that the curve $ Y$ lies in $ \mathbb{P}^2$. In fact, let our curve $ Y$ lie in some space $ \mathbb{P}^n$, where $ n\geq 3$, and let $ a$ be a point of multiplicity $ d \minus{} 1$ on the curve $ Y$. Let $ b$ and $ c$ be two points on the curve $ Y$ distinct from each other and also distinct from $ a$, and let $ u$ be a hyperplane passing through the points $ a$, $ b$ and $ c$ (such a hyperplane exists since $ n\geq 3$). Assume that the curve $ Y$ does not completely lie on the hyperplane $ u$. Then, according to the generalized Bezout theorem (Theorem 7.7 in Hartshorne Chapter 1), the number of points where $ Y$ intersects the hyperplane $ u$, counted with multiplicities, equals $ \deg Y\cdot\deg u \equal{} d\cdot 1 \equal{} d$. But this number must be at least $ \left(d \minus{} 1\right) \plus{} 1 \plus{} 1 \equal{} d \plus{} 1$, because the curve $ Y$ intersects the hyperplane $ u$ at the point $ a$ with multiplicity $ \geq d \minus{} 1$, at the point $ b$ with multiplicity $ \geq 1$ and at the point $ c$ with multiplicity $ \geq 1$ (and maybe at some other points). Hence, we get a contradiction, and thus our assumption that the curve $ Y$ does not completely lie on the hyperplane $ u$ was wrong. It follows that the curve $ Y$ completely lies on the hyperplane $ u$, so that we can consider it as a curve in $ \mathbb{P}^{n \minus{} 1}$ rather than $ \mathbb{P}^n$. Inductively proceeding this way, we can finally imbed the curve $ Y$ in $ \mathbb{P}^2$.

I'm leaving the exercise for curves in $ \mathbb{P}^2$ to the reader (the title of the topic being a hint).

  Darij


Let N be an integer which is not a perfect square. Prove or disprove that if N is the nth such positive integer, then N=n+sqrt(n) rounded off to the nearest integer.

For instance, 5 is a third non perfect square starting from one, so 3+sqrt3 is approximately 4.7, rounded off it becomes 5.

My friend asked me this question but i couldn't solve it. Any tips?

A way to practically alway tackle this kind of problem:
Look at what happens when $ n$ grows by $ 1$. There are cases where the value jumps by $ 1$ and sometimes it also jumpes by $ 2$. So it reaches exactly every positive integer once, except those which are left out by a jump of $ 2$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

K be a normal subgroup of a finite group G. Given that, G/K has n elemnts of order n. Then prove that, G has an element of order n.

Thank you... :)

You don't need $ n$ elements of order $ n$, it suffices that there is one. Proving this is rather easy, you may want to show that if $ f : G \to H$ is a surjective homomorphism, then the existence of an element $ x$ of order $ n$ in $ H$ implies the same for $ G$. In fact we can take a power of $ f^{\minus{}1}(x)$.

That is false. $ x$ can have order greater than $ n$ even if $ f(x)$ has order $ n$. It only guarantees that $ x^n$ is in the kernel of $ f$. We do, however, have that the order of $ f(x)$ divides the order of $ x$, so that if $ f(x)$ has order $ n$, $ x$ has order $ mn$, and we can consider $ x^m$ as the element we want.

And if you would have read my post carefully, you would have seen that this is what I said: "we can take a power of $ f^{ \minus{} 1}(x)$."


A rubber band of natural length $l_{o}$ is fixed at end $A$ and the other end B is pulled horizontally with a constant velocity $u_{0}$ at $t = 0$ .An insect $C$ initially located at $B$ now starts mmoving at at a velocity $v_{c}$ relative to the rubber band and towards $A$.Assuming elastic limits , determine the time taken by the insect to reach point $A$

Let $l(t)$ be the length of the band, and let $x(t)$ be the distance $|AC|$. The band stretches with speed $v_{0}$, so $\text dl/\text dt = v_{0}$. The increase of $x$ due to stretching is proportional to $x$, assuming that the band stretches uniformly; thus, we have $\Delta x = x\text dl/l$. The distance covered by the insect in time $\text dt$ is $v\text dt$, so we have
\[\text dx = \Delta x-v\text dt = \frac{x}{l}\text dl-v\text dt = \frac{x}{l_{0}+v_{0}t}v_{0}\text dt-v\text dt \implies\]
\[\implies \text dx = \left( \frac{xv_{0}}{l_{0}+v_{0}t}-v\right)\text dt.\]

Substituting $y\equiv xv_{0}/(l_{0}+v_{0}t)-v$, we find
\[x = \frac{(v+y)(l_{0}+v_{0}t)}{v_{0}},\] and
\[\text dx = \left(\frac{l_{0}}{v_{0}}+t\right)\text dy+(v+y)\text dt,\]
which after separation of variables yields
\[dy =-\frac{vv_{0}}{l_{0}+v_{0}t}\text dt.\]
Integrating this and using $y(0) = v_{0}-v$ we get
\[y = v_{0}-v+v\ln\frac{l_{0}}{l_{0}+v_{0}t}.\]
When the insect reaches $A$ we will have $x = 0$ and $y =-v$, so the needed time can be obtained from the above equation and it is
\[t = \frac{l_{0}}{v_{0}}\left(e^{v_{0}/v}-1\right).\]


Prove that $ \forall n \geq 2$  and $ x \in (0,2) $  we have:
$ (1+ \frac{1}{ \sqrt[n]{x}})(1+ \frac{1}{ \sqrt[n]{2-x}}) \geq 4 $.

cheers! :D :D :cool:


On this Wednesday's PSAT, was there a section where we had to fill out which form we were taking or the date? Because I'm afraid because I didn't do that and therefore my answer sheet would be graded differently since the computer might mistaken it for a different test!! :o  :o  :o Or does each answer sheet have a specific code already on it?

I'm pretty sure neither of those matter; I think in the directions there was a part about comparing the W on the answer sheet to the W on the test.  So it's on the answer sheet already.

Thank you because I was scared about that, but I don't think I compared the two, oh well, I doubt the school made a mistake with the answer sheets though


Cylinder $ B$'s height is equal to the radius of cylinder $ A$ and cylinder $ B$'s radius is equal to the height $ h$ of cylinder $ A$. If the volume of cylinder $ A$ is twice the volume of cylinder $ B$, the volume of cylinder $ A$ can be written as $ n \pi h^3$ cubic units. What is the value of $ N$?

[asy]unitsize(1inch);
path a=(0,-1.2)..(-10,0)--(10,0)..cycle;
path b=(17,-0.9)..(13,0)--(21,0)..cycle;
path c=(0,1.2)..(-10,0)--(10,0)..cycle;
path d=(17,0.9)..(13,0)--(21,0)..cycle;
draw(c);
draw(d);
draw(shift((0,7))*a);
draw(shift((0,7))*c);
draw(shift((0,15))*b);
draw(shift((0,15))*d);
draw((-10,0)--(-10,7));
draw((10,0)--(10,7));
draw((13,0)--(13,15));
draw((21,0)--(21,15));
draw(a);
draw(b);
label("$A$",(0,9),N);
label("$B$",(17,18),N);[/asy]

$ H_b = R_a$
$ R_b = H_a = h$
$ \pi R_a^2h = 2\pi R_b^2H_b$
$ R_a = h$
Our area of $ A$ is $ \pi R_a^2h = \pi h^3\implies N = \boxed1$.

Call cylinder A's volume $2{\pi}{h^2}{r}$. Call cylinder B's volume ${\pi}{r^2}{h}$.

$2{\pi}{h^2}{r} = {\pi}{r^2}{h}$

$r = 2h$

The volume of cylinder A is therefore $2{\pi}{h^2}\times{2h} = 4{\pi}{h^3}$.

Answer: $\fbox{4}$


[quote="thanhnam2902"]$ cos8x \plus{} 4cos4x \plus{} 128cos^8x \equal{} 134$[/quote]
$ \cos 8x \plus{} 4 \cos 4x \plus{} 128 \cos^8 x \leq 1\plus{}4\plus{}128\equal{}133<134$

[quote="Brut3Forc3"][quote="thanhnam2902"]$ cos8x \plus{} 4cos4x \plus{} 128cos^8x \equal{} 134$[/quote]
$ \cos 8x \plus{} 4 \cos 4x \plus{} 128 \cos^8 x \leq 1 \plus{} 4 \plus{} 128 \equal{} 133 < 134$[/quote]

Oh! yes. It's easy by your way.  :lol:


A haeminic protein contains 0,426 per cent iron(Fe).Calculate its minimum molecular mass.

even i took biochemistry classes this semester i still dont know how to start solving this problesms...i remember i once saw problems like this on the romanian national olimpiad short list...so do you have amy idea where i should start?


group theory
abstract algebra
superior algebra
superior algebra solved

Let $F$ be the finite field. show that for every element $a\in F$ there exist two element of $F$ for example $x,y$ such that $x^2+y^2=a$.

Saman Gharib

count the cardinal of $\{x^2 | x \in F\}$ and $\{a-y^2 | x \in F\}$.

the cardinality proof is easier perhaps but here is another one
if the finite field has even order every element is square and it is trivial
let the order q now be odd
if a is square then it is again obvious
if the equation has a solution for a nonsquare a, it is proven for all nonsquares, as every other nonsquare = a multiplied with a square

so suppose now that it has no solution, for every nonsquare a
then x^2+y^2 is always square

the squares thus form an additive subgroup
contradiction, as lagrange implies then that (q+1)/2 divides q

[quote="fredbel6"]the squares thus form an additive subgroup
[/quote]

How about $-1$? Is it always a square?

when your field has an even number of elements it is a square and it is obvious as $-1=1$

when your field has $q=p^h$ elements $-1=x^2$ is solvable iff $q=1$ mod 4
the proof is simple, if you know that for all elements nonzero in the field $x^{q-1}-1=0$ holds

so if $x^2=-1$
then $1=x^{q-1}=(-1)^{\frac{q-1}{2}}$
implying that $q=1$ mod 4

now for the converse, we know $x^{q-1}-1$ can completely be factored in our field
however $x^4-1$ divides it if $q=1$ mod 4 so that also completely factors
but $x^2+1$ divides that so conclusion : it has a root

[quote="fredbel6"]
so if $x^4=-1$
then $1=x^{q-1}=(-1)^{\frac{q-1}{4}}$
implying that $q=1$ mod 4
[/quote]

You obviously mean $2$ not $4$ in $x^4=-1$. and should look like this:
so if $x^2=-1$
then $1=x^{q-1}=(-1)^{\frac{q-1}{2}}$
implying that $(q-1)/2$ is even and  $q\equiv1(mod 4)$

The proof of the converse is quite nice.

Did you really think I was asking for a proof of this? In your proof (for the original problem) you reached a step when you said that if $x^2+y^2$ is never a non-square, then the set of squares is a subgroup. It is clear why it is a monoid/semigroup. I was asking why this monoid/semigroup is closed under taking inverses i.e. why $-1$ is a square?

Doesn't that follow directly from the fact that $p \cdot x^2 =0$ ($p$ is the characteristic here)¿ (so often enough adding squares gives $0$)

Amfulger, you are so correct, I have no idea what went wrong, I will change it

Indeed, that is kind of a hole
ZetaX has a good correction, however : when you work with a finite group and you want to check if a subset is a subgroup, you only need to check closure under addition, and then everything else follows (the finiteness is crucial in the proof)

I hope this corrects it


Does anyone know how to show the following:

[color=green]Suppose that R is a ring with identity, and suppose that 1R = 0R. Show that
R = { 0R }. [/color]

Thx guys.

All of your threads were moved because they were in the wrong section.  I solved your problem here

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=437553&highlight=#437553


integration
function
real analysis
real analysis unsolved

Suppose $ f: \mathbb{R}\rightarrow \mathbb{C}$, $ 2\pi$-periodic and $ f\in C^{\infty}$.

$ f^{\wedge}(n) \equal{} \frac {1}{2\pi}\int_{ \minus{} \pi}^{\pi}f(x)e^{ \minus{} inx}\ dx$ ,$ \ n \in \mathbb{Z}$ and $ s_n(x) \equal{} \sum_{p \equal{} \minus{} n}^{n}f^{\wedge}(p)e^{ipx}$.
Fix a $ k \in \mathbb{N}$.

Prove that for all $ x \in \mathbb{R}$  , $ n^k|f(x) \minus{} s_n(x)| \rightarrow 0$ when $ n \rightarrow \infty$.

I know that $ lim_{|n| \rightarrow \infty}n^kf^{\wedge}(n) \equal{} 0$. I think we need this for the proof, which I cannot prove :p.
I think I need a huge hint.:)

Note: in trying to prove that $ n^k|S_n(x)\minus{}f(x)|\to 0,$ try using that $ n^{k\plus{}2}|\widehat{f}(n)|$ is bounded.

I have not searched very well due to lack of time. Can you please show me the first steps ?
Thanks

Lemma: If $ f\in C^k$ and we let $ M_k\equal{}\max_{x}|f^{[k]}(x)|,$ then
\[ |\widehat{f}(n)|\le \frac{M_k}{|n|^k}\]
Can you prove that?

[quote="Kent Merryfield"]Lemma: If $ f\in C^k$ and we let $ M_k \equal{} \max_{x}|f^{[k]}(x)|,$ then
\[ |\widehat{f}(n)|\le \frac {M_k}{|n|^k}
\]
Can you prove that?[/quote]

Yes, $ f^{[k]}$ is bounded because it is also $ 2\pi$-periodic, so that supremum exists. Thus, we can find $ M_k$ and using $ |\widehat{f}(n)| \equal{} |\frac {\widehat{f}^{[k]}(n)}{n^{k}}|$, the lemma is proven. (we can easily prove that $ |\widehat{f}^{[k]}(n)| \leq M_k$ )

From which, of course, $ |\widehat{f}(n)|\le\frac{M_{k\plus{}2}}{|n|^{k\plus{}2}}.$ After all, we were assuming $ f\in C^{\infty}.$

Which means that $ |S_nf(x)\minus{}f(x)|\equal{}\left|\sum_{|p|>n}\widehat{f}(p)e^{ipx}\right| \le \sum_{|p|>n}\frac{M_{k\plus{}2}}{|p|^{k\plus{}2}}.$

Can you finish from there?

[quote="Kent Merryfield"]From which, of course, $ |\widehat{f}(n)|\le\frac {M_{k \plus{} 2}}{|n|^{k \plus{} 2}}.$ After all, we were assuming $ f\in C^{\infty}.$

Which means that $ |S_nf(x) \minus{} f(x)| \equal{} \left|\sum_{|p| > n}\widehat{f}(p)e^{ipx}\right| \le \sum_{|p| > n}\frac {M_{k \plus{} 2}}{|p|^{k \plus{} 2}}.$

Can you finish from there?[/quote]

Yes, but I assume that $ |S_nf(x) \minus{} f(x)|$ should be replaced by $ |S_n(x) \minus{} f(x)|$ ?

Ah, notation. I'm just used to writing it as $ S_nf,$ thinking of $ S_n$ as an operator - the partial sum operator - acting on the function $ f.$ It all means the same thing.


Team $A$ and Team $B$ play a best of $7$ series. Team $A$ has a $3/4$ probability of beating Team $B$ in each game. What is the probability that the series goes to a seventh game?
The choices are:

$(A)$ $405/2048$       $(B)$ $135/512$     $(C)$ $135/1024$    $(D)$ $1/7$     $(E)$ $135/2048$

[hide]There are 6 choose 3 = 20 ways to arrange the three wins of each team (since that's necessary to have a seventh game). Each of those 20 ways has a $(\frac{3}{4})^3(\frac{1}{4})^3=\frac{27}{4096}$ chance of happening. The total probabililty is $\frac{135}{1024}$. [/hide]

[quote="JavaMan"][quote="236factorial"][hide]There are 6 choose 3 = 21 ways to arrange the three wins of each team (since that's necessary to have a seventh game). Each of those 21 ways has a $(\frac{3}{4})^3(\frac{1}{4})^3=\frac{27}{4096}$ chance of happening. The total probabililty is $\frac{135}{1024}$. [/hide][/quote]

Isn't 6 choose 3 = 20?[/quote]

Yeah, it does, except when I realized that while doing the problem, I forgot to fix the solution  :blush:

Thanks everyone, and thank you for the explanation 236factorial.

dimak11214 and tz, you should post the method with which you achieved the solution. The solution itself is meaningless without an explanation.


geometry
trigonometry
email
geometry proposed

Let $ABC$ an acute triangle such that $ A=60^. $ and $BH=OI$ where $H$ is the orthocenter and $I$ is the incenter.
Determine the measures of the other two angles of the triangle $ABC$. :D

O is the center of circumscribed circle of triangle $ABC$ ?!?!?

:first: You are welcome, Cezar ! (See the problem 8.104 from my book about geometry).

$A=60\Longleftrightarrow OI=2R|sin\frac{B-C}{4}|\ and\ HB=2R|cosB|.\ Thus,$
$HB=OI\Longleftrightarrow |cosB|=|sin\frac{B-C}{4}|\Longleftrightarrow 1+cos(2B)=1-cos\frac{B-C}{2}$ $\ a.s.o.\ (Answer:\ B=80).$
$In\ generally,(\forall)\lambda\in (0,1),A=2\arcsin \lambda \Longleftrightarrow OI^2=(2\lambda -1)^2 R^2+8\lambda R^2\sin^2\frac{B-C}{4}.$

Levi what book ?!? Where can i find it  :D

[b]Site[[/b] ([i]Editura GIL[/i]): http://www.gil.ro
[b]Email[/b] ([i]Mircea Lascu-editor in chief[/i]): gil1993@zalau.astral.ro


function
real analysis
real analysis unsolved

Prove that there is a function $f: \mathbb R\to\mathbb R$ whose graph $G_f=\{(x,f(x))\ |\ x\in\mathbb R\}$ is a dense connected subset of the plane.

I can only assume you're referring to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hahn&t=34104]this[/url], but that example has a completely different purpose. The graphs constructed there were supposed to be [b]dis[/b]connected. Were you referring to some other topic? :?

By considering cuts with a horizontal segment and two vertical rays at the endpoints, we can show that $f$ must take on every real value in every interval.
Just a start...


geometry
geometric transformation
reflection
group theory
abstract algebra
rotation
superior algebra

$ Dih(28)$

Need to find elements of order $ 2, 4, 7$.

$ 28\equal{} 2^2*7$

14 reflections (of order 2) and 14 rotations.

First look at $ n_7$.

$ n_{7}$ $ \equiv$ 1 (mod 7)

A unique Sylow 7-subgroup of order 7. No reflections in this subgroup (as they are of order 2).

There are 7 rotations (including identity).

So, if <x> are rotations and <y> are reflections, then in the Sylow 7-subgroup of order 7 there are only elements generated by x.

$ {1, x^7}$ are of order 2. $ x^2$ is of order 7? No elements of order 4 in  in the Sylow 7-subgroup.

Looking at $ n_2$.

$ n_{2}$ $ \equiv$ 1 (mod 2)

The Sylow 2-subgroup is of order 4.

as we have $ 2^2$, does this mean that there are no elements of order 2 in the Sylow-2 subgroup, but only elements of order 4.

I need to find:

(1) elements of order $ 2, 4, 7$ in Dih(28)
(2) list the Sylow 2-subgroups and the Sylow 7-subgroups.

Not sure if I am going in the right direction with this...

Any help would be appreciated!

Let's think about this problem geometrically. Interpreting Dih(28) as rotations/ reflections in the plane, all reflections have order 2. (You can think of Dih(28) as the group of symmetries of a 14-gon. The reflections have lines spaced 180n/14 degrees apart.) This gives 14 elements of order 2 here. Now rotations can be by 360n/14 degrees, so there are 6 elements with order 14(n=1,3,5,9,11,13), 1 additional element of order 2 (n=7), and 6 elements of order 7 (n even, not 0). There is no element of order 4.

Say x,y generate the group with x a rotation (order 14) and y a reflection.
As you noticed, there is exactly one 7-SSG, $ \langle x^2 \rangle$. A 2-SSG must be a 4-group. (i.e $ C_2\times C_2$) All elements except the identity have order 2. Now since the composition of two distinct reflections is a nontrivial rotation, a 2-SSG must have a rotation. It must be $ x^7$. If it contains a reflection about a line l, it contains a reflection around the line perpendicular to it (play around with this). We have accounted for all elements in 2-SSG. Since there are 7 pairs of perpendicular reflection lines, there are 7 2-SSG's. (7 is odd and divides 28, check)

All of the above can be put purely in algebraic terms. Just write all the elements in terms of x and y, and use the relations in the group to take the place of the geometric arguments.


function
LaTeX
logarithms
ratio
limit
number theory
prime numbers

-Sorry about the lack of LaTeX in this post, if any viewer is willing to format this into LaTeX, I would deeply appreciate it =)-

π(x) is the prime counting function counting the number of prime numbers less than or equal to some real number x. Even though it employs the Greek letter π (pi), it doesn't necessarily relate to pi... Or does it?

My conjecture:
The positive square root of the value of π(x) for x=10^n for any integer n>1, divided by the positive square root of the value of π(x) for x=10^(n-1), approaches pi as n approaches infinity.

Here is a chart of the values I have tested so far, to ten significant digits: (Calculator rounding errors may be present for n=23)
n | sqrt(π(10^n))/sqrt(π(10^(n-1))) | π(x)
02 | 2.500000000 | 25
03 | 2.592296279 | 168
04 | 2.704713698 | 1229
05 | 2.793692768 | 9592
06 | 2.860715775 | 78498
07 | 2.909671841 | 664579
08 | 2.944372685 | 5761455
09 | 2.970768844 | 50847534
10 | 3.002193486 | 455052511
11 | 3.008259549 | 4118054813
12 | 3.021993665 | 37607912018
13 | 3.033469026 | 346065536839
14 | 3.043202797 | 3204941750802
15 | 3.051564793 | 29844570422669

We skip to n=23 since the apparent convergence is very slow at this point.

23 | 3.091357533 | 1925320391606803968923

If anyone can supply more values as to this table or values of pi(x) for x=10^n, I would be very grateful.

By the Prime Number Theorem, $ \pi(n) \sim \frac{n}{\ln n}$, where $ \sim$ means that the ratio of the two approaches 1 as $ n$ goes to infinity.  Thus, $ \lim_{n \to \infty}\frac{\pi(10^{n})}{\pi(10^{n-1})}= \lim_{n \to \infty}\frac{10^{n}/\ln 10^{n}}{10^{n-1}/ \ln{10^{n-1}}}= \lim_{n \to \infty}\frac{10(n-1)}{n}= 10$.  Thus the value you're interested in is not $ \pi$ but $ \sqrt{10}$.  The moral of this story is, not every number a little bit larger than 3 is equal to $ \pi$.

I see how this works. Thank you for pointing this out to me.

How do you close threads on this forum?

You're welcome.  The Prime Number Theorem is pretty cool (in addition to being quite deep and difficult to prove).

There is no formal mechanism for closing threads on this forum -- posts on which no one has commented recently just slowly fall to the bottom.


function
integration
logarithms
calculus
real analysis
real analysis unsolved

suposse if f is a complex measurable function  on X and u is positive measure on X and $\phi (p) =\int_X |f|^p d\mu$ with $0<p<\infty$  Let $E= \{ p: \phi (p)<\infty\}$

Is E necessarily open? Closed? Can E consist of a single point?  Can E be any connected subset of $(0,\infty)$ ? :oops:

Let the underlying measure space be $X=(0,\infty).$

Let $f(x)=\frac1{\sqrt{x}|\ln x|}$ for $x>2$ and for $0<x<\frac12.$ (We just let $f(x)=0$ on the interval $(1/2,2);$ we're not interested in that interval.)

Then $f\in L^p(X)$ only for $p=2.$

Hence: yes, the set $E$ could consist of a single point, and no, $E$ does not have to be open.

The example of $g(x)=\frac1{\sqrt{x+1}},$ which is in $L^p(X)$ for $p>2$ shows that $E$ does not have to be closed.

But $E$ does have to be an interval (connected). We can show that if $f\in L^{p_0}\cap L^{p_1}$ and $p_0<p<p_1,$ then $f\in L^p.$ I'll let someone else supply this proof.

May i know how to show that $ \int_{2}^{\infty} \left( \frac{1}{ \sqrt{x} \ln x } \right)^{p} dx \equal{} \infty$ for $ p \neq 2$?

This expression doesn't seem "integration-friendly" for me  :blush:

Using the integral test, you have to prove that the series $ \sum_{n\geq 2} \left( \frac{1}{ \sqrt{n} \ln n } \right)^{p}\equal{}\infty$ for $ p<2$. But this is a consequence of the condensation test.

[quote="Kent Merryfield"] Let $ f(x) \equal{} \frac1{\sqrt {x}|\ln x|}$ for $ x > 2$ and for $ 0 < x < \frac12.$ (We just let $ f(x) \equal{} 0$ on the interval $ (1/2,2);$ we're not interested in that interval.)[/quote]

Use the other interval: $ \int_0^{\frac{1}{2}} \left(\frac1{\sqrt {x}|\ln x|}\right)^pdx\equal{}\infty$ for $ p>2$.


logarithms
function
integration
calculus
calculus computations

Hi I was wondering if anyone can find the inverse Laplace Transform (without means of complex variables).  Thanks.

Question: Find $ \displaystyle\mathcal{L}^{\minus{}1}\left\{\ln\frac{s\minus{}3}{s\plus{}1}\right\}$.

One approach is to use the power series for the logarithm. For sufficiently large $ s,$

$ \ln\left(\frac {s \minus{} 3}{s \plus{} 1}\right) \equal{} \ln\left(1 \minus{} \frac4{s \plus{} 1}\right) \equal{} \minus{} \sum_{k \equal{} 1}^{\infty}\frac1k\cdot\frac {4^k}{(s \plus{} 1)^k}$

[hide="Incorrect; corrected two posts below."]Take the inverse Laplace transform of that term by term to get

$ f(t) \equal{} \minus{} \sum_{k \equal{} 1}^{\infty}\frac1k\cdot\frac {4^k}{k!}t^{k \minus{} 1}e^{ \minus{} t} \equal{} \minus{} e^{ \minus{} t}\sum_{k \equal{} 1}^{\infty}\frac {4^kt^{k \minus{} 1}}{k\cdot k!}$

This isn't an elementary function, but it can at least be defined as an integral. I get that it is
\[ f(t) \equal{} \minus{} \frac {e^{ \minus{} t}}{t}\int_0^t\frac {e^{4u \minus{} 1}}{u}\,du
\][/hide]

Hi Professor, I just worked out a solution.  I am not sure if it concurs with your answer.
First, let $ \displaystyle F(s)\equal{}\ln\left(\frac{s\minus{}3}{s\plus{}1}\right) \equal{}\ln\left(1\minus{}\frac{4}{s\plus{}1}\right)\equal{}\mathcal{L}\{f(t)\}$.
Then $ \displaystyle F'(s)\equal{}\frac{1}{1\minus{}\frac{4}{s\plus{}1}}\cdot\frac{4}{(s\plus{}1)^2}\equal{}\frac{4}{(s\minus{}3)(s\plus{}1)}\equal{}\frac{1}{s\minus{}3}\minus{}\frac{1}{s\plus{}1}$
Hence, $ \displaystyle\mathcal{L}^{\minus{}1}\{F'(s)\}\equal{}e^{3t}\minus{}e^{\minus{}t}\equal{}\minus{}t\cdot f(t)$
Therefore, $ \displaystyle f(t)\equal{}\frac{e^{\minus{}t}\minus{}e^{3t}}{t}$.

Ah - You're right and I made a mistake.

The inverse Laplace transform of $ \minus{}\sum_{k\equal{}1}^{\infty}\frac1k\cdot\frac{4^k}{(s\plus{}1)^k}$ is

$ \minus{}\sum_{k\equal{}1}^{\infty}\frac1k\cdot\frac{4^k}{k!}\cdot t^{k\minus{}1}e^{\minus{}t}\equal{}\minus{}e^{\minus{}t}\sum_{k\equal{}1}^{\infty}\frac{4^kt^{k\minus{}1}}{k!}$

$ \equal{}\minus{}\frac{e^{\minus{}t}}{t}\sum_{k\equal{}1}^{\infty}\frac{(4t)^k}{k!}\equal{}\minus{}\frac{e^{\minus{}t}}{k}\left(e^{4t}\minus{}1\right)\equal{}\minus{}\frac{e^{3t}\minus{}e^{\minus{}t}}{t}.$

In general, let $ \mathcal{L}[f(t)] \equal{} F(s)$. Then,

$ \mathcal{L}[tf(t)] \equal{} \minus{} \frac {d}{ds} F(s) < \equal{} > tf(t) \equal{} \minus{} \mathcal{L}^{ \minus{} 1} \left[\frac {d}{ds} F(s)\right]$

$ < \equal{} > f(t) \equal{} \minus{} \frac {1}{t} \mathcal{L}^{ \minus{} 1} \left[\frac {d}{ds} F(s)\right]$,

and of course $ f(t) \equal{} \mathcal{L}^{ \minus{} 1} [F(s)]$. If $ F(s) \equal{} \ln \frac {s \plus{} a}{s \plus{} b}$ then

$ \frac {d}{ds} F(s) \equal{} \frac {d}{ds} [\ln(s \plus{} a) \minus{} \ln(s \plus{} b)] \equal{} \frac {1}{s \plus{} a} \minus{} \frac {1}{s \plus{} b}$, and so

$ f(t) \equal{} \minus{} \frac {1}{t} \mathcal{L}^{ \minus{} 1} \left(\frac {1}{s \plus{} a} \minus{} \frac {1}{s \plus{} b}\right) \equal{} \minus{} \frac {1}{t} \left(e^{ \minus{} at} \minus{} e^{ \minus{} bt}\right)$

$ < \equal{} > f(t) \equal{} \frac {e^{ \minus{} bt} \minus{} e^{ \minus{} at}}{t}$.

From this, it also follows that

$ \int_0^{\plus{} \infty} \frac {e^{ \minus{} bt} \minus{} e^{ \minus{} at}}{t} \,dt \equal{} \ln \frac{a}{b}$.


group theory
abstract algebra
topology
advanced fields
advanced fields solved

Let $\mathcal{F}_{2}$ denote the free group on 2 generators. Show that the derived (or commutator) subgroup $[\mathcal{F}_{2},\mathcal{F}_{2}]$ of $\mathcal{F}_{2}$ is not finitely generated.

It’s a corollary of the following problem (which I’ve posted on this forum before, but since it’s been unanswered for such a long time, I’ll prove it here):

Given a finitely generated free group $F$ and a non-trivial normal subgroup $N$ of infinite index, $N$ is not finitely generated.

We know that $N$ is free (all subgroups of a free group are free), and we want to show that it’s not free on finitely many generators. We can use the theory of covering spaces for this:

Let $F$ be free on $n$ generators. Now consider a bouquet of $n$ circles (i.e. $n$ circles which share a common point, which will be the basepoint of the space). This is a graph $X$ whose fundamental group is $F$. Corresponding to $N$, there is a regular covering space $\tilde X$, also a connected graph, whose fundamental group is $N$, which we assume to be finitely generated by $m$ elements. The assumptions we made mean that $\tilde X$ looks like this: an infinite (because the covering space has infinitely many sheets, since $N$ has infinite index) spanning tree, together with $m$ other edges, $e_1,e_2,\ldots,e_m$. Every closed path in $\tilde X$ which maps onto a non-trivial closed path in $X$ must pass through one of the edges $e_i$.

Let $\tilde \alpha$ be a closed path with basepoint $\tilde x\in\tilde X$, which projects (via the covering map $p: \tilde X\to X$) onto a non-trivial closed path $\alpha$ in $X$. Since the covering space is regular (I’m not sure, but I think it’s sometimes called Galois as well; interesting analogy, huh? :)), given any $\tilde x'\in\tilde X$, there is an automorphism $\sigma$ of the covering space which takes $\tilde x$ onto $\tilde x'$. If we take $\tilde x'$ to be very far away from all the edges $e_i$, $\sigma(\tilde\alpha)$ would have to pass through more vertices than $\tilde\alpha$, contradicting the fact that $\sigma$ is an automorphism for our covering space.

That’s it, basically. I might have made some terminology mistakes (or even mistakes in the proof) since I’m rather new to the theory, but I think this is the right idea.

P.S.

It’s clear how to derive the problem in the topic from this: We take $F=\mathcal F_2$, and $N=[\mathcal F_2,\mathcal F_2]$. $N$ Is normal in $F$, and $F/N$ is the free abelian group on two generators, which is infinite, so we can apply what we’ve shown above.


Adding the sides of the diagram given we arrive at a perimeter of $ \boxed{15}$ units.
sorry that it looks unprofessional, is there anyway this can be drawn on asymptote?


Euler
topology
logarithms
analytic geometry
search

I was unsure where to put this, but I have to find out what Mathematician matches each "creation/invention"

If anyone could respond on any matches between the mathematicians and the topics that they created, that would be great. I do know that for every mathematician matches a topic.

MATHEMATICIANS to choose from:

[color=red]Diophantus of Alexandria
Muhammad ibn Musa al-Khwarizmi
John Napier
Rene Descartes
Blaise Pascal
Leonhard Euler
Maria Agnesi
Pierre-Simon Laplace
William Rowan Hamilton
George Boole
Emmy Noether
Srinivasa Ramanujan[/color]

TOPICS to choose from:

[color=red]Family of Curves
Differential Equations
Algebra and Topology
"New" Concept of Zero
Logarithms
Infinite Series
Quaternion Algebra
Indeterminate Equations
Famous Triangle
Logic
[i]i[/i] (I don't know how that looks, but its that "i" that looks italicized with a tail!
Cartesian Coordinate System[/color]

and thanks so much to anyone who helps!

The forum that you posted in is where very experienced math students discuss algebra problems, so I have moved this topic.  Also, if you look around the forums you will find that there is very little discussion of homework.  Members of this board are generally happy to help any other member understand concepts and problem solving techniques.  They even discuss individual problems very often.  You will not however find a lot of people willing to help cut down on your homework time -- particularly for something that you could simply Google and research on your own.

This looks sort of like an Math History quiz :? Nice ...   :)

Pretty easy ...  :)

Diophantus ------- Indeterminate equations
al-Khwarizmi ----- 'New' concept of zero
Napier ------------- Logarithms
Descartes --------- Cartesian coordinate system
Pascal ------------- Famous triangle
Euler -------------- $i$
Agnesi ------------ Family of curves
Laplace ----------- Differential equations
Hamilton ---------- Quaternion algebras
Boole -------------- Logic
Noether ----------- Algebra & topology
Ramanujan ------- Infinite series

I agree with Mr. Crawford here, if it's simply a matter of doing a bit of research, I think you should do that on your own. If your math teacher did not explain a concept well and you'd like some more help on it, by all means let us know.

As a matter of fact, I did decide to do a bit of research. I Googled a few of the topics and was able to come up with the information and proof for probably 3/4 of the names listed. I would like to know, however, how does one find such information?

Again thank you for your help and understanding of my stupidity :)!

One could always do a search on Google for "Mathematicians biographies," which turns up a wonderful website that has hundreds (thousands?) of biographies of mathematicians that you can look through.

That's probably even more work than is really involved. I'm think MathWorld would have all this.


inequalities
geometry
3D geometry
inequalities unsolved

Consider
$\sum_{cyc}a^{5}b^{3}\ge \sum_{cyc}a^{4}bc^{3}$

According to Sung-yoon Kim's conditions, this inequality cannot be proved by a simple application of AM-GM, since $\frac{1}{3}+\frac{3}{5}= \frac{14}{15}< 1$. See
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=585871#585871
for details.

Sung-yoon Kim and spanferkel (and myself, I guess, but they've done most of the work so far) conjecture that such inequalities (that do not satisfy the conditions) do not hold for all $a, b, c > 0$. But obviously this one holds "most of the time." So here are two questions:

1) Can we find a counterexample $(a, b, c)$ with all three nonnegative so that the inequality fails?

2) How well can we characterize the region on which the inequality holds? For example, if we take $(a, 1, 1)$ the inequality reduces to $(a+1)(a-1)^{2}(a^{2}+1) \ge 0$ which holds for $a \ge-1$. If we take $(a, a, 1)$, then it reduces to $a^{3}(a-1)^{2}(a+1)(a^{2}+1)$ and this holds so long as $a \not \in (-1, 0)$! I have no idea what the region might be.

Any other thoughts on this? (If it's not too related, post in the other thread, where we're discussing the general result.)

PS hopefully I didn't make any mistakes in my algebra that led to the above. Please correct me if I did! :)

(a,b,c)=(1,4,30) :D

as to the regions: I have strictly no idea :wallbash:

It was your example that made me find the general solution once I had realized that $(1,4,32)=(1,2^{2},2^{5})$ works as well as a counterexample. But Sung-yoon Kim was more rapid.  :P

As to the regions, note that with $(a,b,c)$, they contain $(\lambda a,\lambda b,\lambda c)$ too. So I think, as far as the "octant" with $a,b,c\ge0$ is concerned, the region where the ineq holds should be the union of three very slim sorts of cones, placed cyclically around the main diagonal $x=y=z$.
Now all this doesn't help us much for tackling the problem for more variables  :(


The classroom is not opening for me. I am behind a firewall and clicked the firewall thing already, still doesn't work. This is the first time it hasn't worked. What's wrong?

What happens when you click the firewall thing?


A merchant offers a large group of items at $ 30\%$ off.  Later,
the merchant takes $ 20\%$ off these sale prices and claims that the
final price of these items is $ 50\%$ off the original price.  What
is the total percentage discount?

The end price is actually $ 0.7 \cdot 0.8 \equal{} 0.56 \equal{} 56\%$ of the original price.

This leaves $ 100\% \minus{} 56\% \equal{} 44\%$ as the total percent discount.

The question asks for the difference so it would be 50-44=6


number theory
inequalities unsolved
inequalities

it's a conjecture or guess that has been proved or something.. at least that's what I heard. And it is one of the names given to people with a certain amount of posts.

[quote="Shouman Das"]can anyone tell me please about the poincare conjecture?
i m very much interested about it,
help me. :rotfl:[/quote]
Post the question to number theory  forum.
You will probably get more replies there then here .


number theory
least common multiple
number theory proposed

Prove that for any two positive integers $ a$ and $ b$ the equation $ LCM(a,a\plus{}5)\equal{}LCM(b,b\plus{}5)$ implies $ a\equal{}b$.

$ gcd(a,a\plus{}5)\equal{}gcd(a,a\plus{}5\minus{}a)\equal{}gcd(a,5)\equal{}1 , 5$
when $ gcd(a,a\plus{}5)\equal{}1$,then $ lcm(a,a\plus{}5)\equal{}a(a\plus{}5)$
and $ lcm(a\plus{}5,a)\equal{}lcm(b\plus{}5,b)\equal{}a(a\plus{}5)$
if $ 5|b$ ,then $ 5|lcm(b\plus{}5,b)$  contradiction
if b is not divided by 5 , then lcm (b,b+5)=b(b+5),
so $ b(b\plus{}5)\equal{}a(a\plus{}5)$,$ a\equal{}b$
if $ gcd(a,a\plus{}5)\equal{}5$,then let $ a\equal{}5a_1$, $ lcm(a,a\plus{}5)\equal{}5a_1(a_1\plus{}1)$
the proof is similar


algorithm
logarithms
summer program
MathPath
algebra
binomial theorem

there is a long and tedious method of calculating them by hand. we spent a day on it in class this school year. biggest waste of 45 minutes in my life........the point is, there is a very long way, but no one cares if you know it. its kind of like doing long division to figure out 567218465464364278978779/64164932647632157643289415.

[quote="236factorial"][i]From middle school topic on square roots: [/i]

Do you know how to find the square root of a number with out a calculator?

Like, what is $\sqrt{501}$?

Divide and average? Algorithms? Differentials (not appropriate)?

I'm spliting this and moving this to the high school forum.[/quote]

I just estimate and most of the times i get it right if its a perfect square, for example, take 1936, and since it ends with a six, you know its square root has to end in a 4 or 6, and you might know that $45^2$ is 2025, so it should be obvious that that the square root of 1936 is 44.

[quote="236factorial"]what if it isn't a perfect square? I'm pretty good at guessing ($\sqrt2=1.4$, etc.) by knowing my perfect squares though.[/quote]

Why not use a calculator when it isnt a perfect square?

What if you are calculator deprived?  :D

[quote]What if you are calculator deprived? [/quote]

but, you would never have to do that for a math problem.

Hello, 236factorial,

Allow me to introduce two ways to solve this problem. There is an elementary way and there's an advanced way (not too much but it's advanced compare to other way). Both ways are assumed to find the square root of an integer to one decimal place.

Let's look at the elementary way.

$\sqrt{35}$

Now, $25 < 35 < 36$ which means $5 < \sqrt{35} < 6$ so the first non-decimal digit of $\sqrt{35}$ is 5. As for the one decimal place, we pick up arbitary digit that seemed to work out (it doesn't have to be the RIGHT one).

I like 8 so I assumed the $\sqrt{35} = 5.8$. Now, we divide the original number (35) by our assumption 5.8. This gives us 6.0 (we only need one decimal place). To find the exact digit, we take the average of our assumption and the quotient from our division. In this case:

$\frac{5.8+6.0}{2} = 5.9$ and indeed, $\sqrt{35} = 5.9160797$ (I won't go to infinite digits) and clearly, our answer is correct.

As for advanced way, I'll in hidden text because this requires understanding of binomial theorem.

[hide="Advanced"]

I think this was something called First-Order Approximation. Write $\sqrt{35}$ as $(30+5)^{\frac{1}{2}}$. We'll only take first two terms of the expansion. Since binomial theorem states that

$(x+y)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} y^1 ...$ (I'll stop here since there's no need to go further).

So, in our case

$(30+5)^{\frac{1}{2}} = \binom{\frac{1}{2}}{0} 30^{\binom{1}{2}} + \binom{\frac{1}{2}}{1} 30^{-\frac{1}{2}} 5^1$

And after finding square root of 30 from our first method, I get (after rounding to about 3 digits to ensure I get the right answer), we get 5.93345.[/hide]

Note that our answer from method 2 came out larger than original because I rounded and only looked at the first two terms.

Hope this helps!

[quote="h78reg"]http://www.jimloy.com/arith/sqrt.htm

imagine that you need to know the square root of 6453.678. would you really want to use this?[/quote]

This is the right method. Wait till my grandpa hears about this. Well, this is the method my grandpa uses, but I find it a little confusing, but if you ask your dad, you'll probably know it.

yes, all our parents and grandparents probably know a method which is similar to this because there were no calculators back then.

Hmm... what about this?

$\log_n501=2 \Rightarrow n^2=501$
$\log_n501=\frac{\log_{10}501}{\log_{10}n}$
$\log_{10}501=\log_{10}3\cdot \log_{10}167$
$\log 167\approx \log 168=3\log 2+\log 3+\log 7\approx 2.2296$
$\log 501\approx .4771+2.2296=2.7067$

Therefore, $501\approx 10^{2.7067}$, so $\sqrt{501}\approx 10^{1.35335}$, whatever that is. :roll:

[b]EDIT:[/b] Upon finding some logs that add up to close to that, $10^{1.35335}$ is a little over 22, and my calculator says that $\sqrt{501}=22.383$. So, not far off. :) Although, you can get more accurate by figuring that $22^2=484$ and $23^2=529$, it's closer to $22^2$ but not much, and therefore must be between $22.3$ and $22.5$. :P

my method is guess and check.
say you want to find the square root of 225.
try 10.  10^2=100 so 10 is too low.  Try 20. 20^2=400, that's too high. Try 15.  15^2=225.

[quote="Silverfalcon"]Only problem you have is that you need calculator to find the log.  ;)[/quote]

Well, when my dad did it that way, he had to memorize logs  :D

I'll attempt to explain the method that h78reg believes is a big waste of time :D. Then one of you can explain why it works. I'm going to attempt to formulate an explanation using better notation (and thus hopefully easier to understand) than what I've found online.

Suppose you have a number in base $10$ notation written as $N=d_1d_2\cdots{d}_{2n}$, where of course the $d_i$ are digits. If the number has an odd number of digits and no decimal point, you may obviously rewrite it with $d_1=0$ at the front. If there is a decimal point in the number and an odd number of total digits, then decide whether to attach the digit $0$ before or after the number in such a way that there will be an even number of digits after (and also before) the decimal point. (All these technical details became painfully necessary when I wrote a program for this in my VB class.)

Now, let $a_n$ be a sequence of two-digit numbers (the first digit may be zero) formed by taking $a_1=d_1d_2,a_2=d_3d_4,\ldots,a_k=d_{2k-1}d_{2k},\ldots$ (EDIT: I left something out: for $i>2n$, let $d_i=0$. First put a decimal place at the end, of course, if there isn't one already.) Let $r_1$ be the greatest integer whose square does not exceed $a_1$. Write $r_1$ as the first digit of $\sqrt{N}$, and let $b_1=r_1$.

To extract the square root of $N$ to as many places as you wish, repeat the following procedure, say for $i=1\to\infty$. $r$ will be the number whose digits are $r_1r_2\cdots{r}_i$, NOT including $r_{i+1}$:

Let $x_i$ be the number $100(a_i-r\cdot{b}_i)+a_{i+1}$, and find the greatest integer $r_{i+1}$ such that $r_{i+1}\cdot{b}_{i+1}\le{x}_i$, where $b_{i+1}$ is defined by $10(2r)+r_{i+1}$. This number, $r_{i+1}$, is the next digit of $\sqrt{N}$.

As for the decimal, if in the number $N$ it was between $a_{j-1}$ and $a_j$, then in the result $r$ it will likewise be between $r_{j-1}$ and $r_j$.

That seemed complicated at least to me. I hope it's correct and understandable, or at least one of those. Now, show that it works.

Another note: the explanations I've found online all show how to write it out, but my explanation lacks this completely. Sorry about that.

You don't need calculators to find logs. I figured out the first 25 on my own. Starting with only the fact that $\log 2\approx .3010$ and $\log 3\approx .4771$, you can do the following:

$\log 4=2\log 2\approx .6020$
$\log 5=\log 10-\log 2\approx .6990$
$\log 6=\log 2+\log 3\approx .7781$
$\log 7^2\approx \log 50=\log 10+\log 5\approx 1.6990 \Rightarrow \log 7\approx .8495$
$\log 8=3\log 2\approx .9030$
$\log 9=2\log 3\approx .9542$
$\log 10=1$
$\log 11^2\approx \log 120=3\log 2+\log 3+\log 5\approx 2.0791 \Rightarrow \log 11\approx 1.0396$
$\log 12=2\log 2+\log 3\approx .6020+.4771=1.0791$
$\log 13^2\approx \log 168=3\log 2+\log 3+\log 7\approx 2.2296 \Rightarrow \log 13\approx 1.1148$

And, my personal favorite (for the method):

$\log 23^3\approx \log 12168=3\log 2+2\log 3+2\log 13\approx 4.0868 \Rightarrow \log 23\approx 1.3623$

One note: when using a method such as squaring or cubing the original log and then finding a number close-by, you'll only really be accurate to two digits, although you can slightly compensate for it, since you know which direction you changed it in. For example, $\log 7\approx .8451$, which is actually a full $.0044$ off from what my method gives. However, I knew that my answer would be high, because I basically added one to the squared log. I could have compensated for this; with some experience, I think I would have known about the right amount to take off to make it closer. For accuracy to two digits after the decimal point, though, this can be used reliably pretty far before you start getting inaccurate.

When I was a freshman in college, the hand calculator had just been invented, but was very expensive. Most of my science/engineering classmates carried slide rules. I was a little different, preferring a table of logarithms. A 4-place log table fits on one sheet of paper - I may actually have been carrying a Xerox copy.

Lets do $\sqrt{501}.$ Now I happen to have memorized that $\log2=.3010,$ which means that $\log5=.6990$, so we'll be near that. But we look it up anyway. Since 501 is three digits, we don't have to interpolate. We find the entry "6998." What that means is $\log5.01=.6998,$ to that many digits of accuracy. (Did I mention that these are all base 10 logarithms? That should be obvious.)

We handle powers of 10 separately. $501=5.01\times10^2$ so $\log501=2.6998.$

Divide by 2 - easy enough to do by hand. In fact you should be able to do that in your head. We get $1.3499.$ Now we look for "3499" in the body of the table. We find "3483" connected to "223" and "3502" connected to "224." Interpolate between them: 3483 to 3502 is a distance of 19; 3499 is 3 down from the top, which is 16 up from the bottom. 3/19 rounds to .1, which is the same as 16/19 rounding to .9. So we have "2239". meaning that $10^{.3499}=2.239,$ or $10^{1.3499}=2.239\times10^1=22.39.$

It turns out that, rounded to four significant figures, $\sqrt{501}=22.38$ (from $22.383$) so an acculation of errors in the table approximations and in the interpolation made for a small error in the fourth significant figure. But at least I could get a fourth significant figure, albeit a little squishy. You're not going to get more than three significant figures out of a slide rule that you could carry.

Of course, I don't actually have a 4-place log table any more, and I was simultating what it would say by using a spreadsheet. Of course, a calculator is better as a practical hand-held tool. But there were some pedagogical advantages.

1. Logarithms are the bane of anyone teaching at the Algebra II/Trig/Precalculus level (and even beyond that). I know a lot of false laws of arithmetic (seen on student's papers over the years), but there may be more false laws of logarithms than everything else put together. If you're using a table of logarithms day in and day out to multiply, divide, find roots, and raise to powers, then you [i]will[/i] know the correct laws of logarithms.

2.  With either the table of logarithms or the slide rule, you must handle the magnitude (the power of 10) separately from the "mantissa" (the part of the logarithm to the right of the decimal point). You naturally write numbers in scientific notation, and you become very aware of orders of magnitude. I think it leaves you with a deeper understanding of the size of numbers. This is a big part of the "common sense" of scientific quantitation.

One more example: $\sqrt{6453.678}$

From the table: $645 \rightarrow 8096$, $646\rightarrow8102$
The gap is 6. we want .367 of that, a little more than a third: we add 2 to get $8098.$
There's a $10^3$ in there, so our logarithm is $3.8098.$
Divide by 2: $1.9049$

From the table: $803\rightarrow9047,$ $804\rightarrow9052.$ 2/5 of the way between them is a 4, so we say $8034.$

We estimate $\sqrt{6453.678}\approx80.34.$

Note that this isn't as good as the answer I gave earlier in this topic (and the fourth significant digit is off by 1 again.) We lost accuracy from the 7 significant figure input to the 4 significant figure answer (well, almost 4), becuase we just can't do better than 4 significant figures with a 4-place table. A more accurate table wouldn't fit on one page.

Please check this site out, it is in MATHPATH!!!
[url=http://www.mathpath.org/Algor/algor.square.root.htm]http://www.mathpath.org/Algor/algor.square.root.htm[/url]
Thank you very much Mr Merryfield. I only see your logarithm method rarely, it was not in my algebra book. I have only seen it in those old algebra books in my library. I think more authors should put it, it makes you understand the use of logarithms better than punching keys in :blush:

In fact, logarithms were originally invented to aid and speed computations. Look up the names "Napier" and "Briggs" in the St. Andrews math history website.

Better trigonometric tables were developed in the 15th and 16th centuries, developed by astronomers, for astronomers, as was always the case with trigonometry. It becames known within the astronomical community that there was a way to use a trigonometric table to use addition  to perform a multiplication. One multiplication would require something like 4 additions or subtractions and three table lookups.

Napier (an eccentric and isolated Scottish nobleman who dabbled in astronomy) set out to produce a new table for the specific specialized purpose of being able to use addition or subtraction to perform multiplication or division. It was he that coined the word "logarithm."

Napier's logarithsm decreased rather than increased, and he did not have the notion of a base for his system, not understanding the connection with exponentiation. If you want to claim that his system had a base, it was .9999999. In practice, that makes his logarithms something akin to natural logarithms. Briggs (an Oxford professor who made the arduous trip to Scotland just to talk to Napier) did understand the connection to exponentiation and the notion of base, and it was Briggs who produced the first base 10 logarithm table. But Briggs's purpose was the same as Napier's: to make computation easier.


Find all the values $0\le \theta < 360^\circ$ for which the complex number $z=\cos \theta+i\sin \theta$ satisfies $z^{28}-z^8-1=0$.

Well two complex numbers with magnitude $1$ can differ by $1$ iff one is $\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$ and the other is $-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$... so work out those cases? It's sort of tedious that way...

[hide="Partial solution?"]Using Demoivre's theorem we get $(\cos 28\theta - \cos 8\theta )+i(\sin 28\theta - \sin 8\theta )=1$, therefore:

$\sin 28\theta - \sin 8\theta =0$
$\cos 28\theta -\cos 8\theta =1$

From the first equation we get $\theta =0^{\circ},40^{\circ},140^{\circ},18^{\circ},36^{\circ},...,342^{\circ}$.

(I probably messed up somewhere up there, though)

I don't know what to do now except test each of these into the 2nd equation...[/hide]
Edit: actually I did screw up doing that, one of them is $\theta =5+10n$, but I think there may be others...can someone continue for me?

[quote="chess64"]\[ \sin 28 \theta + \sin -8\theta = 0 \Longleftrightarrow 2\sin 10\theta\cos 18\theta = 0,  \] \[ \theta \in \{ 0,18,36,\ldots,342 \} \cup \{ 5, 15, 25, \ldots, 355\}.  \] \[ \cos 28\theta - \cos 8\theta = 1 \Longleftrightarrow \sin 18\theta \sin 10\theta = -\frac{1}{2}.  \] This one is harder to analyze... :([/quote]

Not much harder: From the second you get $\sin 10\theta\neq 0$, hence the set of solutions for the first is defined only by $\cos 18\theta=0$, which in turn implies $\sin 18\theta=\pm 1\implies \sin 10\theta=\mp {1\over 2}$, so you have a couple of diofantine equations.


Please note:  As it says on the back cover of the AIME I contest pamphlet:
"Your Exam Manager will receive a copy of the 2006 AIME Solution Pamphlet with the scores."

The scores will be going out in about 2 weeks.

In the meantime, the numerical 3-digit answers are on our website.

Steve Dunbar
Director, MAA American Mathematics Competitions

When will stats be available for the AMCs?

[quote="me@home"]do you mean AIME's?[/quote]

No I think he means the national stats and state stats for the AMC, as in the score distribution, qualifiers, etc.


mathblitz, I don't know why anyone would want to shame an administrator.

[quote="mathblitz"]we all know that rrusczyk made this website and a good portion of us know his first name, BUT I MET HIM YESTERDAY! OMG! hehehe about that i dont know whether he knows me as that little kid in front of him or that obnoxious 11 year old that kept on asking him questions. ANYWAY this is a place where we honor administrators or shame them. POST ALOT PLZ we want to know what you think of the  admins!.[color=red] *cough* maybe the admins will see how easy*cough* it is to find *cough* spammers on this *cough* topic *cough* and ban them. *cough*[/color][/quote]

....u only see dallas ppl and new yorkers yelling out that they met the admins....

go new jersey

go red sox

go fort worth...

I'm from upstate NY and OMG I'VE MET THREE ADMINS! RUSCZYK, PATRICK, AND SATO!


how many natural solutions does the equation:
$\sqrt{x}+\sqrt{y}=\sqrt{1376}$
does have?$(x\le y)$

your answer is incomplete... :D
$(86,744)$ is not the only solution for $(x,y)$ :lol:

Wasn't this like old AHSME problem? Not exactly alike but I remember solving as what he did.

[hide="Just for fun to try"]

$\displaystyle \sqrt{1376} = 4\sqrt{86}$

With $x \leq y$ so if we say $x = a \sqrt {86}$ and $y = (4-a) \sqrt {86}$, we can either have case of:

$(a,b) = (1,3)$ or $(2,2)$.[/hide]

 :D

silverfalcon and xxreddevilzxx, you guys are both correct. yeah there are only two solutions...
i'm not sure if it was an old AHSME problem, cause i got this from IRMO... ;)


I see that you can add icon to topic. What does these icons means (is there any "dictionary" to them)? Is there any icon that means: "there already is a solution" or "this problem was solved"? It will be much easier to look for older posts that aren't solved ...

[quote="RC-7th"]It varies from forum to forum.

The question mark generally means "question".
The star generally means "challange".
The square generally means "note".
The arrow generally means "important",
No icon generally means "None of the icons fits this topic".[/quote]Currently there is no [b]official[/b] definition applied to these icons, regardless of what RC-7th is saying, [b]except[/b] in the new Middle School Category.

We might officially introduce icon using later, when all the software upgrades of the forum that intersect with that part are done. Partially we don't want to introduce them now because it's not really clear that a solved problem should be attacked by a user that sees it for the first time. The fact that it has been solved before should not stop you from solving it and if you want posting your solution also (even if other solutions, some of which are similar to yours have been posted).


Decades ago, Delta Airlines tried to find the most fuel-efficient way for planes to be able to get to all three airports on an equilateral triangle. Though the "obvious" solution would be taking the 2 shorter sides out of the 3. But if they put a hub in the center, they found they would reduce the path by about 13.4 percent. Prove that, in any triangle, taking a point in the middle and making paths to the sides will not reduce the fuel needed using the shortest two sides by over 13.4 percent

[quote="Scrambled"]Decades ago, Delta Airlines tried to find the most fuel-efficient way for planes to be able to get to all three airports on an equilateral triangle. Though the "obvious" solution would be taking the 2 shorter sides out of the 3. But if they put a hub in the middle, they found they would reduce the path by about 13.4 percent. Prove that, in any triangle, taking a point in the middle and making paths to the sides will not reduce the fuel needed using the shortest two sides by over 13.4 percent[/quote]

WHAT IS THE MIDDLE, THE CIRCUMCENTRE?

The middle is any point in the interior of the triangle

is this the right level for the problem?

on second thought, can a mod delete this thread?


integration
trigonometry
real analysis
real analysis unsolved

Prove $ \int_{\minus{}\pi}^{\pi}e^{2\cos{t}}dt\equal{}2\pi \sum_{k\equal{}0}^{\plus{}\infty}(k!)^{\minus{}2}$.

We have that

$ I\equal{}2\int\limits_{0}^{\pi/2}(e^{2\cos x}\plus{}e^{\minus{}2cos(x)})dx\equal{}4\sum\limits_{m\equal{}0}^{\infty}\frac{2^{2m}}{(2m)!}\int\limits_{0}^{\pi/2}(\cos(x))^{2m}dx$

$ \equal{}2\sum\limits_{m\equal{}0}^{\infty}\frac{2^{2m}}{(2m)!}\beta(m\plus{}1/2,1/2)\equal{}2\sum\limits_{m\equal{}0}^{\infty}\frac{2^{2m}}{(2m)!}\frac{\Gamma(m\plus{}1/2)\sqrt{\pi}}{m!}$

$ \equal{}2\pi\sum\limits_{m\equal{}0}^{\infty}\frac{1}{(m!)^{2}}$.

I used the fact that $ \Gamma(m\plus{}1/2)\equal{}\frac{1\cdot 3\cdots (2m\minus{}1)}{2^{m}}\sqrt{\pi}$.

I hope that there are no errors.


i have  a  inequality  not hard! I wanna see  a solution with Cauchy's ineq.

Let   $ a,b,c$ be lengths of sides  triangle $ ABC$.
Prove that:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} <2$

If you denote $ x\equal{}a\plus{}b\minus{}c,y\equal{}b\plus{}c\minus{}a,z\equal{}c\plus{}a\minus{}b$ your inequality can be rewritten as
\[ \frac{x}{2x\plus{}y\plus{}z}\plus{}\frac{y}{2y\plus{}z\plus{}x}\plus{}\frac{z}{2z\plus{}x\plus{}y}>\frac12\]
and I hope it is clear how to use Cauchy to use it.
Nevertheless there is more efficient shortcuts to this  :lol:

here is another proof

$ LHS\equal{} \sum_{cyc}\frac{2a}{b\plus{}c\plus{}b\plus{}c}<\sum_{cyc}\frac{2a}{a\plus{}b\plus{}c}\equal{}2$

 :)  :D

[quote="Hong Quy"]i have  a  inequality  not hard! I wanna see  a solution with Cauchy's ineq.

Let   $ a,b,c$ be lengths of sides  triangle $ ABC$.
Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} < 2$[/quote]
$ \iff \sum_{cyc} \frac {b \plus{} c \minus{} a}{b \plus{} c} \ge 1$
By Cauchy-Schwartz:
$ \sum_{cyc} \frac {b \plus{} c \minus{} a}{b \plus{} c} \ge \frac {(a \plus{} b \plus{} c)^2}{2(a^2 \plus{} b^2 \plus{} c^2)} \ge 1 \iff$
$ 2(ab \plus{} bc \plus{} ca) \ge a^2 \plus{} b^2 \plus{} c^2 \iff$
$ a(b\plus{}c\minus{}a) \plus{} b(a\plus{}c\minus{}b) \plus{} c(a\plus{}b\minus{}c) \ge 0$.

The inequality seems to be famous. [url=http://www.animath.fr/old/cours/inegalites.pdf]See here Page 3.[/url]
But I have never imagined Cauchy-Schwarz can solve it.

Intuitively, $ a\longrightarrow 0,\ b\longrightarrow c$ yields $ L.H.S.<2$.

[quote="Hong Quy"]i have  a  inequality  not hard! I wanna see  a solution with Cauchy's ineq.

Let   $ a,b,c$ be lengths of sides  triangle $ ABC$.
Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} < 2$[/quote]
I have:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} < \frac{a\plus{}b}{a\plus{}b\plus{}c}\plus{}\frac{b\plus{}c}{a\plus{}b\plus{}c}\plus{}\frac{c\plus{}a}{a\plus{}b\plus{}c}\equal{}2$


beautiful

mine is by first proving $\frac{a(3-a^2)}{2-a^2} \geq 2a^2$ which is equivalent to $(a-1)^2(2a+3) \geq 0$


If you are given $ \log 8 \approx .9031$ and $ \log 9 \approx .9542$, then the only logarithm that cannot be found without the use of tables is:

$ \textbf{(A)}\ \log 17 \qquad\textbf{(B)}\ \log \frac {5}{4} \qquad\textbf{(C)}\ \log 15 \qquad\textbf{(D)}\ \log 600 \qquad\textbf{(E)}\ \log .4$

[hide]
Since $ \log_{10}8 \equal{} 3\log_{10} 2,    \log_{10}9 \equal{} 2\log_{10}3$
and  $ \log_{10} 2 \plus{} \log_{10} 5 \equal{} 1$

we can find everyone excepte $ \log_{10}17 \longrightarrow \boxed{A}$
This makes sense because $ 17$ is prime and therefore cannot be made by multiplying/dividing with $ 2,3,10$[/hide]

for finding B, C and D, we need to know log5.

log8 = 0.9031

log(80) = 9.031

log 5 + 4*log(2) = 9.031.

and so, you can find log5.

but cannot find log17 using any combination of 3 and 2, since it is a prime.

so, D is the answer.

[quote="Attila the Hun"]so, D is the answer.[/quote]You mean $ \bf (A)$ is the answer.


here is a problem from Bulgarian MO-2007:

Find the least m for which any five equilateral triangles of the    total area m can cover an equilateral triangle of area 1.

1. already out there: [url]http://www.mathlinks.ro/viewtopic.php?t=167593[/url]

2. why opening such a topic in DL page? (orl, please move it).


A matrix $ A$ is left-invertible if there exists $ B$ such that $ BA \equal{} I$ where $ I$ is the identity matrix.  A is right-invertible if there exists $ C$ such that $ AC \equal{} I$.

(a) Prove that if a matrix is left-invertible, then it corresponds to an injective linear map.
(b) Prove that if a matrix is right-invertible, then it corresponds to a surjective linear map.
[Hint: think about the interpretation of a vector as its own expansion in the canonical
basis of $ R^n$

Sorry, another question:

I saw this on Wikipedia under inverse element:

A matrix A that is $ m \times n$ where $ m > n$ has a left inverse and a right inverse if $ m < n$.  Why is this so?  Why does a matrix where $ m > n$ have a left inverse instead of a right?

Thanks for the help,

An $ m\times n$ matrix with $ m>n$ [b]may[/b] have a left inverse but cannot have a right inverse. You could certainly write examples that don't have either.

The concept of [b]rank[/b] can be quite valuable here, and the theorem that says $ \operatorname{rank}AB\le \min(\operatorname{rank}A,\operatorname{rank}B).$

Suppose $ A$ is $ m\times n$ with $ m>n$. We know that $ \operatorname{rank}A\le n$ - the rank must always be no larger than the "short side" of the matrix.

A left inverse would be an $ n\times m$ matrix $ B$ such that $ BA\equal{}I_n.$ The rank of $ I_n$ is $ n,$ and it is possible to have $ \operatorname{rank}A\equal{}\operatorname{rank}B\equal{}n,$ so this is possible.

A right inverse would be an $ n\times m$ matrix $ B$ such that $ AB\equal{}I_m.$ Now the rank of $ I_m$ is $ m,$ but we must have $ \operatorname{rank}AB\le n,$ so this is not possible.


"All primes are in the forms of 6n+-1"

Is this a fact that we can use to start a proof?

Thanks!

Yep, its a very well known theorem, so its OK to state it. But as Gyan said, never forget to check the 2 exceptions.
(Its easy enough to prove anyway: you can factorise all of 6n, 6n+2, 6n+3, 6n+4).

thanks, that makes life easier proving "that problem"^___^

another question, does that prove it? 3n+1 is not factorable either

Ah, but when you divide a number by 6, you must get a remainder of 0, 1, 2, 3, 4, or 5. In other words, every number is 6n or 6n+1 or 6n+2 or 6n+3 or 6n+4 or 6n+5. (the 6n+5 there being one less than a multiple of 6). We don't need to worry about other things like 3n+1.

0 = 6*0 + 0
1 = 6*0 + 1
2 = 6*0 + 2
3 = 6*0 + 3
4 = 6*0 + 4 (this equals 3*1+1 but that doesn't matter!)
5 = 6*0 + 5
6 = 6*1 + 0
etc.

primes are in the form of 6n+-1, (except for 2,3) but 6n+-1 is not nessesary prime. hm...

i dont htink it is that easy to prove it

Factorable immediately implies not-prime.  That is, 6n + 2 can never be prime for n > 0 because we can factor it as 2(3n + 1), and a prime number has no such factorization.  However, just because you can't factor such an expression in general does not mean that it won't be factorable for specific values of n -- for example, 6n + 1 cannot be factored for a general n, but it factors very nicely in the case n = 4.  The result of this is that we can say every prime greater than 3 must be either 6n + 1 or 6n - 1 for some n (or otherwise we could factor it) but every number of the form (6n + 1) or (6n - 1) need not be prime.


abstract algebra
linear algebra
matrix
advanced fields
advanced fields unsolved

Hello,

does anyone know of an example of an exact sequence of abelian groups

\[0 \to \mathbb{Z}\to G \to \mathbb{Q}\to 0 \]

that does not split?

Thank you

It is known that there are non-trivial extensions (see [url=http://en.wikipedia.org/wiki/Whitehead_problem]Whitehead problem[/url]), and elementary homological algebra shows that an equivalent question is to give a homomorphism $\mathbb Q\to\mathbb Q/\mathbb Z$ which does not have the obvious form (in particular, its kernel is not finitely generated). owk

P.S. It should not be hard to write one down and work out the corresponding extension explicitly. owk

I am not sure what you are saying, as I do not know any homological algebra. Can this be done in an elementary way, i.e., just picking a group $G$ and constructing the maps?

Let $\pi\colon\mathbb Q\to\mathbb Q/\mathbb Z$ be the canonical projection and let $f\colon\mathbb Q\to\mathbb Q/\mathbb Z$ be a "non-obvious" homomorphism (i.e., not of the form $f(x)=\pi(ax)$ for some $a\in\mathbb Q$). Then $G=\{(x,y)\in\mathbb Q^{2}\mid f(x)=\pi(y)\}$ with $G\to\mathbb Q,\ (x,y)\mapsto x$ works. Examples of non-obvious homomorphisms are given by: [hide="nice puzzle"]the $p$-adic principal part for a prime $p$, i.e. $f(a/p^{k}+b/c)=a/p^{k}$ whenever $p\nmid c$[/hide]
owk

Thank you!

I totaly understand now. Where can I find more information about problems like this, and the solution you gave?

Charles Weibel's An Introduction to Homological Algebra is a good reference for these things (sections 3.3-3.5; actually, a special case of exercise 3.5.3 says that all such $G$ are classified by $\hat{\mathbb Z}/\mathbb Z$). owk

Ok...

I thought I saw how to show the sequence was not split exact, but now I am not able to. Following what you said above I am trying to use the fact that $\text{ker}(f)$ is not finitely generated, which implies that $\text{ker}(a)$ is not finitely generated. Then I assume that I have a map $c : \mathbb{Q}\to G$ such that $bc = 1_\mathbb{Q}$, but I am not sure how to complete the argument. For clarification I have a commutative diagram bellow with the maps I am working with.

\[\begin{matrix}0 & \to & \mathbb{Z}& \to & G & \stackrel{b}{\to}& \mathbb{Q}& \to & 0\\ &&\downarrow{\scriptstyle 1_\mathbb{Z}}&&\downarrow{\scriptstyle a}&&\downarrow{\scriptstyle f}\\ 0 & \to & \mathbb{Z}& \to & \mathbb{Q}& \stackrel{\pi}{\to}& \mathbb{Q}/\mathbb{Z}& \to & 0 \end{matrix}\]

Well we have that $\pi a = f b$ which implies that $f = fbc = \pi ac$, so that $f(r) = ac(r)+\mathbb{Z}$. If $ac = 1_\mathbb{Q}$, then I would be done as this implies $\text{ker}(f) = \mathbf{Z}$, but I do not think this is the case. So, I looked at the map $ac$... We have that $c(r) = (r, x_{r})$, as $ac(r) = r$. Now, suppose that $r$ is in the kernel of $ac$. Then, $ac(r) = a(r, x_{r}) = x_{r}$, which implies that $x_{r}= 0$. So, we have that $(r, 0) \in G$ which implies that $r=u/v$ where $p \nmid v$. Also, all such pairs are in the kernel. So, we have found the kernel of $ac$, but it is not finitely generated. How does this lead to a contradiction?

$ac$ is a homomorphism $\mathbb Q\to\mathbb Q$, therefore just multiplication by some rational number. Combined with $f=\pi ac$, this contradicts our assumption on $f$ not having this form. owk

Right! Because it is determined by where it sends 1. Thank you for all your help!


function
trigonometry
calculus
calculus computations

Find restriction at x=0 of function :
$\frac{\sqrt{\frac{cos2x+\sqrt[3]{1+3^{x}}}{2}}-\sqrt[3]{\frac{cos3x+3cosx-ln(x+1)^{4}}{4}}}{x}$

Sorry I don't understand your question  . But my problem is n't wrong

What do you mean by "Find restriction at x=0"?
I first thought that you were asking the limit when x tends to zero, but we don't have an indetermination (the limit would be $+\infty$).

Indeed, the numerator can be evaluated at $x=0$:
$\sqrt{\frac{\cos 2x+\sqrt[3]{1+3^{x}}}{2}}-\sqrt[3]{\frac{\cos 3x+3\cosh-\ln(x+1)^{4}}{4}}=\sqrt{\frac{1+\sqrt[3]{2}}{2}}-1>0$. Could be a typo in there somewhere.  :maybe:
Don't forget to put \ before sin,cos,ln and such.


For what ordered pair $ (a,b)$ are there infinite solutions $ (x,y)$ to the system
\begin{align*}
2ax+2y&=b\\
5x+y&=-3
\end{align*}

We need to make the equations the same. So, make the constant term and the x term both double in the first equation.

$ 2a = 5 \cdot 2$

$ a = 5$

$ b = -3 \cdot 2$

$ b = -6$

$ (5, -6)$


Hello, 1234567890!

[quote]Find the equation for the graph ABC with points at A(0,4); B(6,0); C(-4,0);  D(-6,0).[/quote]

Sorry, I don't understand the problem . . . at all!

The term "the graph ABC" makes no sense.
Are we concerned with that triangle? . . . Then why give four points?

Are we looking for a graph that passes through those four points?
There are only a few [i]zillion[/i] of them . . .

not zillion. More like an infinite.
Maybe the ABC graph is different than the A,B,&C coordinates. Or maybe the D is just a point given to confuse us. Yes, more information would be nice...
Or maybe I'm just not reading the problem right...

Here's a polynomial with those 4 points (although there are an infinite number of polynomials or even functions in general):
$f(x)=(1/16)(x-4)(x+4)(x-6)$

You can get this by setting up f(x)=A(x-4)(x+4)(x-6) [since those are the zeros] then solving for A when x=0 and f(x)=6.

note for the zeros, (x-a) is if a zero is at a.  so a zero at 6 then (x-6)
but is a zero is at -6 then (x+6) because its (x-xint)


Are there any general-case rules for 101 base p?

divisible by 2 if $p$ is odd?
and also $1$ or $2\mod{3}$
(that last one i just made up from the square residue)

He means for dividing a number $k_p$ by $101_p$

Such as, say $871,192,838,932,743,563_p (\mod{101_p})$  :P  I didn't get this one either  :(

Um, I guess I go with the REALLY long long division solution... :(

I did find a divisibility rule for 101.  Take the number 4 digits at a time and add them.  Continue this process with the sum until a 4 digit number results.  If this number is divisible by 101, then so is the initial number.  If you know how to prove divisibility by 11, its not that hard to prove, noting that 10000 = 1(mod 9999) and 101|9999.  But once you get this far, I couldn't think of a systematic way of finding it, so I guess and checked starting with 11 and I guess I was pretty luck since its 12  :P   Oh well, I haven't looked at the solution packet yet, so maybe thats got a good way.

Another way is to divide them in groups of two from the right and alternately add and subtract.


Q: If the first term of a G.P. a1,a2,a3,...... is unity such that  4a2+5a3 is least, then the common ratio of G.P. is

a) -2/5                  b)-3/5                  c)2/5               d)none of these

(note: 1,2,3 in a1,a2,a3 are in subscript)

[hide="Solution"]
$ a_{2}\equal{}ra_{1}$
$ a_{3}\equal{}r^{2}a_{1}$
$ y\equal{}4a_{2}\plus{}5a_{3}\equal{}4ra_{1}\plus{}5r^{2}a_{1}$

To minimize $ y$, set its derivative equal to $ 0$:

$ 0\equal{}4a_{1}\plus{}10ra_{1}$

$ \boxed{r\equal{}\minus{}\frac{2}{5}}$
[/hide]

Is this an iit question????
Anyways thanks for the reply...


Do not discuss the 2008 MATHCOUNTS Chapter Competition until we say it is OK to do so.

Do not discuss problems from the contest.

Do not discuss your score.

Do not mention whether or not you (or anyone else) found it easy or hard.

You may simply state "I came in <whatever> place, and my team came in <whatever> place."  Nothing more than that.

can we say our rank at the school round yet?

[quote="rrusczyk"]You may simply state "I came in <whatever> place, and my team came in <whatever> place."  Nothing more than that.[/quote]

Okay, I came in 1st place, and my team came in 1st place. :D

We are serious about this ban.  One user has already had his/her account disabled for violating it.  You MUST NOT disclose any information about the tests.

[quote="abacadaea"]are we allowed to pm people about the contest who have already taken it?[/quote]

No.  Do not discuss it at all on this site until after all Chapter Contests are finished. We will let you know when it is OK to discuss.

Thanks.

Is my post considered discussing about the contest???

If you did "badly" and you're sad, would that be considered cheating to show people how you felt about the contest?

[quote="tonypr"]What harm would saying our team score do?[/quote]

Let's say that a team got a 66 and that leaked out. Then everyone would not prepare thinking that it is easy, and in reality, it's easy/hard, so they would do good/bad, respectively.

We don't want a whole bunch of 66's, or a whole bunch of 10's.

I think it means that other teams might not prepare as well/badly as they would be preparing if they knew that a team got a 66

i think that just to be on the safe side abbide by the rules and just wait untill specified in the announcement.

[quote="Ihatepie"]well according to the MATHCOUNTS site, the Chapter competitions are from February 1-24, so can we talk about it by Monday?[/quote]
You definitely want to wait until MATHCOUNTS gives AoPS the okay to discuss Chapter and they will post that it is ok.  There could have been a snow storm somewhere and a chapter got cancelled and pushed back to next weekend(mid week).

I'm pretty sure we can discuss the problems, but we shouldn't yet, just in case.
I got first place, and my team was second.

[quote="ajain"][quote="CircleSquared"]I'm fairly certain we can discuss now, as Chapter problems (I think) are appearing in FTW.[/quote]

I have two comments (1) don't give away where the problems are.  (2) The problems aren't available to the public so how will it be on the game?[/quote]

I obviously have not done (1).
I'm also a bad person to ask about (2).

[quote="CircleSquared"]I'm fairly certain we can discuss now, as Chapter problems (I think) are appearing in FTW.[/quote]

There are no Chapter 2008 problems in FTW.  Please continue to not discuss problems and answers until we have been told by MATHCOUNTS that it is OK to do so.  We will post when it is OK.

um.. with the FTW issue, one kid on my team definitely saw a problem.
and he remembered the answer and got it right.
(but it wasn't during countdown)..

Guys, it's the other way around... there have been FTW problems in countdown rounds around the U.S.  :wink:  Unfair if you ask me.

They must have stolen the problem from us!

In all seriousness, we do not have copies of the 2008 chapter competition and therefore have not put any problems in.

Couple things people.

the first is that I have seen the exact same problems apppear twice in different Mathcounts competitions.  Once on a sprint, later on a countdown.  So it is possible MC recycled an old problem that was on FTWalready from a previous year.

the other possibility is that there really are only so many different countdown style questions MC asks at the chapter level.  So two different organizations could have written the same exact question, or two that are awfuly close.

when I sat and watched the countdown there were probably a dozen questions that I had seen before, or thatwere at leastclose enough to ones I'd seen before for me notto be able to tell the difference.  The same I am sure is true about some of the kids whouse this site a lot and are preparing a lot.  If they have an advantage it is one earned from hard work, not because AoPS is doing anything wrong.  On the contrary, they are doing so much right that you see a problem and know exactly what to do.  Maybe it is then time to move on past the simpler problems from countdown style problems......

Now that all Chapters are over, i'm pretty sure that I saw one problem in Chapters while using AoPS (not saying problem or location of problem), and the problem was very specific......

Are there any chapters that have not completed their competition?  As a coach, I like my students to go over the chapter competition before the state.  I do understand if chapter competitions are not over.

We haven't heard yet if all chapters were wrapped up this past weekend or not.  Please continue to not discuss.  You can, of course, discuss it with your own students.


Prove that for any n natural and x in [2-sqrt(2), 2+sqrt(2)] we have 0<(x^2-4x+4)^n+(-x^2+4x-2)^n<=2^n.

thank you very much!
bye bye!

this is very simple! :D :D

Lets denote y=x^2-4x+4 and we have that 0<y^n+(2-y)^n<=2^n.
For x in [2-sqrt(2), 2+sqrt(2)] we have that 0<=y<=2 and (2-y)>=0 so y^n+(2-y)^n<=2^(n-1)y+2^(n-1)(2-y)=2^n.

all done! :D cheers! :D :D


linear algebra
matrix
group theory
abstract algebra
algebra
polynomial
geometry

Let $||..||$ a norm on $C^n$

$N(A)=\sup_{||x||=1} ||Ax||$ norm on $M_n(C)$

$G$ a sub-group of $GL_n(C)$

Suppose for any $A\in G$,  $N(A-I_n)<\sqrt{3}$

Prove that $G={I_n}$

Strange problem. All I found is that every matrix from this subgroup is of the form $I+N$ with N nilpotent, but after that?

Here is what I think: first, if a is an eigenvalue of a fixed matrix A from G, then we clearly have $ abs(a^k-1)<\sqrt{3}$ for all integer k. First,, thuis clearly implies $abs(a)=1$ and then if we write $ a=cost+isint t$ we have $ cos(kt)>1/2$ for all integer k. A simple, but boring study shows that t must be an even multiple of pi, thus in fact a=1 and the claim from the previous post follows.
     Now, aalways with A fixed, we can wrrite $A=I+N$ where N is nilpotent. Since the norm of $ A^k-I$ is at most 2 for all k, it follows that if we note $h_j$ the norm of $ N^j$ we have $ C(k,n-1)h_{n-1}$ is at most $ 2+C(k,1)h_1+...+C(k,n-2)h_{n-2}$; Comparing this as rates of growts of polynomials in k, we find that $ h_{n-1}=0$ and then a finite descent shows that in fact N=0.

Do not write

[quote="harazi"]...it is me who is idiot..[/quote]

in your future post

thx

Ok, done. I was nervous since for about 10 minutes I was looking at the problem without knowing how to continue, after establishing the essential fact that the eigenvalues are all 1. Moubi, I have some questions about this problem: what is with the "special " value $\sqrt{3}$. Are there other values, greater, for which the conclusion also holds? Can we replace this number by any number? Do you know such details?

Yes you got the idea

A possible approach:

1/ See G is bounded, and every element of G is diagonalisable, with eigenvalue of modulus 1

2/ Let $A\in G$, $\khi_{A}(x)=(x-a_1)...(x-a_n)$ prove that there exist $Y_j \in C^n$
$||Y_j||=1$ and $AY_j = a_jY_j$ and conclude $||A^k-I_n||\geq max|a_j^k-1|$

3/ Deduce $|a_j^k-1|<\sqrt{3}$ for every $k\in Z$, $1 \leq j \leq n$

4/ $a_j=e^{iq_j}$ with $- \pi \leq q_{j} \leq \pi$ use 3/ to get $q_j=0$

Finally
$A=I_n$

[quote="harazi"]what is with the "special " value $\sqrt{3}$[/quote]

Here $\sqrt{3}$ is convenient for trigonometrical reason

$|2\sin((kq_j)/2)|<\sqrt{3}$

$\sqrt{3}/2$ gives angle like $2\pi/3$

I've seen that, but the main question is: does the conclusion of the problem change if we change this value into a greater one? Can we find some strange subgroups that are not trivial and that verify the conditions of the problem probably with another,larger, constant?

If we suppose $||A-I_n||<a$ for any $0<a<\sqrt{3}$ then $G={I_n}$

$\sqrt{3}$ is the optimal one

Wow! Could you please explicit and give such a subgroup for any larger value? Thanks.

No, Moubi, in fact I want to suppose that the norm is smaller than a with $ a^2>3$. Reformulated question:
   Is it true that for all a such that $ a^2>3$ there is a non-trivial subgroup G of $GL_n(C)$ such that for any A in G we have that the norm of $A-I$ is smaller than a? This would mean that your value is optimal.

Let the subgroup be $I,\omega I,\omega^2I$ where $\omega$ is a nontrivial cube root of unity.

Ok, thanks a lot. I wonder whether we can find infinite subgroups with this property.  :?

A student told me, it was an oral of Polytechnique 1998
Revue de Mathématiques Spéciales
RMS 9-10, Année 1999, page 1266

Unfortunately, it will be quite difficult to find these magazines, since the oldest they have at CDI are from 2000.

I think there is no infinite subgroup such that:$|A-Id| \leq \sqrt{3}$
Suppose that $G$ is a subgroup such that $|A-Id \leq \sqrt{3}$ for every
$A$. Then it is easy to show that every eigenvalue $a$ of $A$ is in
$\{ 1, w, w^2 \}$ where $w^3=1$. So we can show that $A^3=I+N$ where
$N$ is a nilpotent matrix. As Harazi proved it, this implies that $N=0$.
Hence $A^3=I$ and $A$ is diagonalizable. And if I remember, all subgroups
s.t. there exists $p$, $A^p=I$ for all $A$ are finite.

so I think we can prove that:
let $0<C<2$ a real and $G$ a subgroup of $GL(n,C)$ such that:
for all $M \in G$, $|M-I_n|<=C$. Then $G$ is finite.


integration
function
calculus
derivative
complex analysis
algebra
domain

The question is as follows

For $z \in \mathbb{C}$_ = $\mathbb{C}$\${x \in \mathds{R}: x \leqslant 0$ , define $F(z)$ = $\int_{[}_{_}_{1}_{,}_{z}_{]}$ $\frac{1}{w}$$dw$, where [1,z] is the straight line path from 1 to z .

Prove that $f(z)$=$\frac{1}{z}$, for $z \in \mathds{C}$_   (we were asked to prove F'=f)

 I attempted this question substituting values in a proof of the Antiderivative theorem.

Tried to prove

$lim_{h}_\to_{0}$ $\mid \frac{F(z+h)-F(z)}{h}-f(z) \mid$ $=0$

$F(z+h)-F(z) = \int_{[}_{_}1_{,}_{z}_{+}_{h}_{]}f(w)dw$  -$\int_{[}_{_}1_{,}_{z}_{]}f(w)dw$

= $\int_{[}_{_}1_{,}_{z}_{+}_{h}_{]}f(w)dw$ $+$ $\int_{[}_{_}z_{,}_{1}_{]}f(w)dw$

= $\int_\gamma_{_}n$ $f(w)dw$( = 0 by Goursats Lemma) + $\int_{[}_{_}z_{,}_{z}_{+}_{h}_{]}f(w)dw$

 Then

$\mid \frac{F(z+h)-F(z)}{h}-f(z) \mid$ = $\frac{1}{\mid h \mid}$ $\mid \int_{[}_{_}z_{,}_{z}_{+}_{h}_{]}f(w)dw-f(z)h) \mid$

= $\mid \int_{[}_{_}z_{,}_{z}_{+}_{h}_{]}(f(w)-f(z))dw \mid$

$\leqslant \frac{1}{\mid h \mid}sup_{_}w_\in_{[}_{z}_{,}_{z}_{+}_{h}_{]}\mid f(w)-f(z) \mid h \mid$

$\to 0$ as $h \to 0$.

Therefore the limit is 0 and$F'(z)= f(z) = \frac{1}{z}$

 This looks right and I think that the fact that z has to be positive according to the question doesnt really affect the answer , but I still have doubts .. Can someone please tell me if this is right or if I have done something wrong thanks..[/quote]

You need to use the fact that the function is not defined on the negative real axis, since the anti-derivative theorem (as you call it) has a continuity assumption--and there's a singularity at zero, which you have to employ in the last step (it approaches zero, because $f(w)$ is continuous, and thus can be decomposed into $f(z)+g(w)$ where $g(w)$ approaches 0 as w approaches z, so the supremum goes to zero.  It's critical because the two h-es in your last step cancel out (the $|h|$ should be outside of the supremum, so you have to show that the supremum goes to zero, which holds for a _continuous_ function f.)

Actually, on second thought: The argument isn't so simple...  You do need to note that 1/z is holomorphic away from 0 (thus continuous) and use that in the last step of your proof.  Also, though, the domain given is, critically, simply connected (meaning that any two paths with identical start/end points are homotopic) implies that any two paths with identical start/end points have identical integrals.  Choose an arbitrary path from 1 to z (in the domain--a straight line works, for instance), and then the same path going from 1 to z and then to z+h (a straight line here, since it's in the domain of holomorphicity since h is arbitrarily close)--Goursat's theorem doesn't apply, because in the traditional application of the anti-derivative theorem you define two straight line paths (horizontal, then vertical, which will go over the singularity of 1/z for some paths, so the function is meromorphic--in truth, if you define your paths in the opposite manner, it is possible to use Goursat's because you'll never pass through the singularity... Thus,  invariance under paths allows you to show $F(z+h)-F(z)=$ the integral over the line from z to z+h. From there, you continue, incorporating continuity in your last step.

I don't know if you would call this a conclusion, but I guess what I am getting at in the preceding post is that the correct solution really depends on where you are in class (whether you have proved the homotopy version of the Cauchy integral theorem.)  Otherwise, for the particular case you employ, you should be able to use Goursat, provided you specify that you move vertically first (although even that is particularly ugly, because you can't center the disk at zero, without loss of generality, since that's where you lack continuity.  If you do choose to employ Goursat, if the point given has imaginary part greater than zero, move vertically epsilon to avoid then discontinuity, horizontally to zero, and then employ Goursat, showing that you don't pass over the discontinuity trivially--same idea for imaginary less than zero.  I expect that your teacher wouldn't have given you the problem, though, if you were not taught about the homotopy version of Cauchy, since that's rather ugly.


inequalities
trigonometry
inequalities unsolved

Let $ a,b,c,d > 0$ such that $ (a^2\plus{}b^2)^3\equal{}c^2\plus{}d^2$. Prove that:
\[ \frac{a^3}c\plus{}\frac{b^3}d\ge1\]

nice. thanks =)
on a side note, my first instinct upon seeing the condition was to use trig, with $ a\equal{}k\sin x$, $ b\equal{}k\cos x$, $ c\equal{}k^3\sin y$, $ d\equal{}k^3\cos y$. is there a trig solution?

This problem has been posted before
http://www.mathlinks.ro/viewtopic.php?t=119243

[quote="raveen.ds"]Let $ a,b,c,d > 0$ such that $ (a^2 \plus{} b^2)^3 \equal{} c^2 \plus{} d^2$. Prove that:
\[ \frac {a^3}c \plus{} \frac {b^3}d\ge1\]
[/quote]

It's easy by cauchy-swarchz
We have:
$ LHS \ge \frac {(a^2 \plus{} b^2)^2}{ac \plus{} bd}$
And:$ (ac \plus{} bd)^2 \le (a^2 \plus{} b^2)(c^2 \plus{} d^2)$


ratio
geometry
angle bisector
geometry unsolved

Let ABCD a convex quadrilateral.  P and Q on AD and BC respectively such that: $BP/PC=AQ/QD=AB/CD$
If BA and CD intersect in a point R, prove that the bisectriz $l$ of the angle BRC and PQ are paralells.

The posted problem contains typos, the correct problem is as follows:

Let $ABCD$ be a convex quadrilateral. Let $P$ and $Q$ be points on $BC$ and $AD,$ respectively, such that $\frac{BP}{PC}= \frac{AQ}{QD}= \frac{AB}{CD}.$ If $BA$ and $CD$ intersect at a point $R,$ prove that the internal bisector $l$ of the angle $\angle BRC$ and $PQ$ are parallels.
_________________________________________________________

Assuming $R \equiv BA \cap CD$ is on the opposite side of BC than A, D and $S \equiv BC \cap AD$ on the opposite side of CD than A, B. Let PQ meet AB, CD at K, L. Using Menelaus theorem for $\triangle RBC,\ \triangle RAD$ cut by the common transversal $PQ \equiv KL,$

$\frac{KR}{KB}\cdot \frac{PB}{PC}\cdot \frac{LC}{LR}= 1 = \frac{KR}{KA}\cdot \frac{QA}{QD}\cdot \frac{LD}{LR}$

Reducing this by the common factor $\frac{KR}{LR}$ and by the equal ratios $\frac{PB}{PC}= \frac{QA}{QD},$ we get

$\frac{LC}{KB}= \frac{LD}{KA},\ \ \frac{KA}{KB}= \frac{LD}{LC}$

$\frac{KB-AB}{KB}= \frac{LC-CD}{LC},\ \ 1-\frac{AB}{KB}= 1-\frac{CD}{LC},$

$\frac{AB}{KB}= \frac{CD}{LC},\ \ \frac{LC}{KB}= \frac{CD}{AB}$

Using Menelaus theorem for $\triangle RBC$ again, this time substituing $\frac{PB}{PC}= \frac{AB}{CD}$ and $\frac{LC}{KB}= \frac{CD}{AB},$

$1 = \frac{KR}{KB}\cdot \frac{PB}{PC}\cdot \frac{LC}{LR}= \frac{CD}{AB}\cdot \frac{AB}{CD}\cdot \frac{KR}{LR}= \frac{KR}{LR}$

Thus the triangle $\triangle KRL$ is isosceles with KR = LR, which means that its base $KL \equiv PQ$ is parallel to the external bisector of its angle $\angle KRL,$ identical with the internal bisector $l$ of the angle $\angle BRC.$

Since $K \in AB,\ L \in CD$ and $R \equiv AB \cap CD,$ it would be $K \equiv L \equiv R.$ For $P \in BC,\ Q \in AD,$ $\frac{BP}{PC}= \frac{AQ}{QD}$ is then possible only if $BC \parallel AD$ (easy to show by a contradiction) and $\frac{BP}{PC}= \frac{AQ}{QD}= \frac{AB}{CD}$ only if $PQ$ itself is the internal bisector $l$ of $\angle BRC.$ That is, if P, Q are both inside the segments BC, AD (which is what I assumed). If P, Q are both outside the segments BC, AD, the other possibility is that PQ is the external bisector of $\angle BRC,$ in which case PQ is perpendicular to the internal bisector $l$ of $\angle BRC.$

I see it now that in the general case, when K, L are not identical, and when P, Q are both outside the segments BC, AD, the triangle $\triangle KRL$ comes out isosceles in the same way as in the above proof, except its base $KL \equiv PQ$ is now perpendicular to the internal bisector $l$ of $\angle BRC.$

Take $ X,Y$ on the diagonals $ BD$ and $ CA,$ such that $ \frac {_{\overline{BP}}}{^{\overline{PC}}} \equal{} \frac {_{\overline{BX}}}{^{\overline{XD}}} \equal{} \frac {_{\overline{AY}}}{^{\overline{YC}}} \equal{} \frac {_{\overline{AQ}}}{^{\overline{QD}}}.$ By Thales theorem $ PX \parallel QY \parallel RD$ and $PY \parallel QX \parallel RA.$ Furthermore, we have

$ \frac {\overline{PX}}{\overline{CD}} \equal{} \frac {\overline{PB}}{\overline{BC}} \ , \ \frac {\overline{PY}}{\overline{AB}} \equal{} \frac {\overline{PC}}{\overline{BC}} \Longrightarrow \frac {\overline{PX}}{\overline{PY}} \equal{} \frac {\overline{CD}}{\overline{AB}} \cdot \frac {\overline{PB}}{\overline{PC}} \equal{} \frac {\overline{AB}}{\overline{CD}} \cdot \frac {\overline{CD}}{\overline{AB}}$

$ \Longrightarrow \overline{PX} \equal{} \overline{PY} \Longrightarrow PXQY$ is a rhombus.

As a result, diagonal $ PQ$ of $ PXQY$ bisects $ \angle XPY.$ Since the lines $ PX,PY$ are parallel to  $ RC,RB,$ respectively, it follows that their bisectors $ PQ$ and  $ l$ are parallel as well.

Let be the points $ E,\ F,$ such that $ AQEB,\ CDQF$ to be parallelograms.

So, from $ BE\parallel \equal{} AQ$ and $ CF\parallel \equal{} QD$ we have $ \frac {BE}{CF} \equal{} \frac {AQ}{QD} \equal{} \frac {BP}{PC} \equal{} \frac {AB}{CD}$ $ ,(1)$

From $ (1)$, based on the [b]Thales theorem[/b], we conclude that the points $ E,\ P,\ F,$ are collinear and $ \frac {EP}{PF} \equal{} \frac {BP}{PC}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {EP}{PF} \equal{} \frac {AB}{CD}$ and so, $ QP$ is the angle bisector of the angle $ \angle EQF.$

Because of now, $ \angle EQF \equal{} \angle BRC,$ and also because of the rays of these angles are parallel each other to the same direction, we conclude that their angle bisectors ( as their homologous rays ) are parallel each other and the proof is completed.

Kostas Vittas.


Let $x,y,z,t$ reals greater than $2$. Prove that

$yzt\sqrt{x-2}+ztx\sqrt{y-2}+txy\sqrt{z-2}+xyz\sqrt{t-2} \leq xyzt\sqrt{2}$

Manlio, how many times do I have to say that these trivial inequalities do not appear in GMA? I hope you are not going to post all inequalities that appeared in Gazeta Matematica since they are hundreds.

Hey harazi... why can't he post them all? More ineqs is more fun :D

But indeed, the source should be correct if entered ;)

Cool... another joke inequality (but I needed 10 minutes to understand the joke!).

In fact, dividing the inequality by xyzt, it becomes

$\frac{\sqrt{x-2}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-2}}{z}+\frac{\sqrt{t-2}}{t}\leq \sqrt{2}$.

And this follows from $\frac{\sqrt{x-2}}{x}\leq \frac{\sqrt{2}}{4}$, $\frac{\sqrt{y-2}}{y}\leq \frac{\sqrt{2}}{4}$, $\frac{\sqrt{z-2}}{z}\leq \frac{\sqrt{2}}{4}$ and $\frac{\sqrt{t-2}}{t}\leq \frac{\sqrt{2}}{4}$.

The inequality $\frac{\sqrt{x-2}}{x}\leq \frac{\sqrt{2}}{4}$ can be proven as follows: After squaring, it takes the form $\frac{x-2}{x^{2}}\leq \frac{2}{4^{2}}$. Now multiply with $x^2\cdot 4^2$ and get $\left( x-2\right) \cdot 4^{2}\leq 2\cdot x^{2}$. But this is clear, since $2\cdot x^{2}-\left( x-2\right) \cdot 4^{2}=2\left( x-4\right) ^{2} \geq 0$. Proof complete.

  Darij


Let $ a,b,c\ge 0$ such that $ (a \plus{} b \plus{} c)^2\left( \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} \frac {729}{8}$. Prove that:
\[ 135a^2b^2c^2 \plus{} 16(a^2 \minus{} bc)(b^2 \minus{} ca)(c^2 \minus{} ab)\ge 0
\]
When I solved one of very very nice can_hang2007's inequalities I have created it  :)
[b]can_hang2007's problem[/b]
Let $ a,b,c$ be positive real numbers such that $ (a \plus{} b \plus{} c)^2\left( \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} \frac {729}{8}$
Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \ge \frac {297}{4}
\]
With equality if and only if $ a \equal{} b \equal{} 4c$ and its cyclic permutations.

[quote="zaizai-hoang"]Let $ a,b,c\ge 0$. Prove that:
\[ 135a^2b^2c^2 \plus{} 16(a^2 \minus{} ab)(b^2 \minus{} ca)(c^2 \minus{} ab)\ge 0
\]
 [/quote]

Wrong! Try $ a\equal{}2,b\equal{}1,c\equal{}0$  :wink:

[quote="zaizai-hoang"]Let $ a,b,c\ge 0$. Prove that:
\[ 135a^2b^2c^2 \plus{} 16(a^2 \minus{} ab)(b^2 \minus{} ca)(c^2 \minus{} ab)\ge 0
\]
When I solved one of very very nice can_hang2007's inequalities I have created it  :)
[b]can_hang2007's problem[/b]
Let $ a,b,c$ be positive real numbers such that $ (a \plus{} b \plus{} c)^2\left( \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} \frac {729}{8}$
Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \ge \frac {297}{4}
\]
With equality if and only if $ a \equal{} b \equal{} 4c$ and its cyclic permutations.[/quote]
Sorry but $ a,b,c$ such that $ (a \plus{} b \plus{} c)^2\left( \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} \frac {729}{8}$  :wink: