function
integration
analytic geometry
calculus
geometry
complex analysis
topology

Hmm, I got stumped on a problem that I thought was going to be super-routine, but I ran into some trouble.

The family is all analytic functions on the open set $ U$ such that $ \int_U |f(x \plus{} iy)| \,dx \,dy < 5.$

I THINK I found a solution in this way:
By switching over to polar coordinates in the integral $ \int_{|z \minus{} a| < R} f(x \plus{} iy) \,dx \,dy$ I can express it in terms of $ f(a)$ and $ R$, and by choosing sets suitably I can show that the family is uniformly bounded in a neighborhood of every point in $ U$. Now, does this seem reasonable at ALL that the integral can be expressed in terms of $ f(a)$ and $ R$? Maybe I made some really dumb over-simplification.

Also, is there a simple solution that I am missing?

Ok here is my solution for reference. I was not aware of the formula I deduce, it is well known to everyone else? Is there a kind of residue theorem for area integrals? Also, can someone come up with a direct less explicit argument?

Let $ a \in U$, $ f$ holomorphic in $ U$ and suppose $ \{|z \minus{} a| \leq R\} \subset U$. Then
$ \int_{|z \minus{} a| \leq R} f(x \plus{} iy) \,dx dy \equal{} \int_{|z| \leq R} f(x \plus{} iy \plus{} a) \,dxdy \equal{}$
$ \int_0^Rr\left(\int_0^{2\pi}f(re^{i\theta} \plus{} a) \,d\theta\right)dr \equal{} \int_0^R r2\pi f(a) \, dr \equal{} \pi R^2f(a)$

So if $ f$ belongs to the family we get $ |f(a)| \leq \frac {5}{\pi R^2}$. It's now a standard matter to conclude the family is normal


Given $F_{0}= 0, F_{1}= 1$, and $F_{n+2}= F_{n+1}+F_{n}$ (the familiar Fibonacci series), prove that

$gcd(F_{n}, F_{m}) = F_{gcd(n, m)}$.

[hide="Hint"] Prove the lemma that if $a | b$ then $F_{a}| F_{b}$. [/hide]

Interesting.

I think I've seen this question asked before, except in more specific  terms when $m$ and $n$ are given and it asks for $gcd(F_{n}, F_{m})$.

I want to see the solution, will some one do it please?  :lol:

hint

[hide]

Let b = ka, then you induct on  to show the . After that it's a proof by contradiction (assume d = max element of naturals that divides both F_a and F_b, then d must be F_gcd(a,b) otherwise blah).

[/hide]


Find $n \in \mathbb{N}$ such that $(x+k)^n-x^n-k^n=0$,$k \in \Re ,k \ne 0,x = ke^{\frac{{2\pi i}}{3}}$

[size=75][color=green]fixed by a mod[/color][/size]

[quote="10000th User"][quote="felipesa"]Find $n \in \aleph$ such that $(x+k)^n-x^n-k^n=0$,$k \in \Re ,k \ne 0,x = ke^{\frac{{2\pi i}}{3}}$[/quote]
$n \in \aleph$ ??? What is $\aleph$? Transfinite number $\aleph_0=|\mathbb{N}|$?[/quote]
$n$ is a natural number

[quote="chess64"]... you mean $n\in \mathbb{N}$?[/quote]
Yes,$n\in \mathbb{N}$

The equation is equivalent to

$(1+e^{i2\pi/3})^n-e^{i2n\pi/3}-1=0$, or

$e^{in\pi/3}-e^{i2n\pi/3}-1=0$

This produces the system

$\begin{cases}\cos{{n\pi}\over 3}-\cos{{2n\pi}\over 3}-1=0\\ \sin{{n\pi}\over 3}-\sin{{2n\pi}\over 3}=0\end{cases}$

The second equation transforms into

$\sin{{n\pi}\over 3}(1-2\cos{{n\pi}\over 3})=0$, which gives ${{n\pi}\over 3}=k\pi\lor {{n\pi}\over 3}=\pm{\pi\over 3}+2k\pi, k\in\mathbb{Z}$. This in turn gives $n=3k\lor n=6k\pm 1$

Substituting these values into the first equation, we see that it's satisfied only by $n=6k\pm 1$.

Hence, the desired set is $n\in\{6k+1,6k+5|k\in\mathbb{Z}_0^{+}\}$


number theory
prime factorization
prime numbers
superior algebra
superior algebra unsolved

How does one describe J(nZ) in terms of prime factorization of n?
J(nZ) = intersection of all maximal ideals of a commutative ring R with unity that contains nZ, where n>1 and Z=integer.

maximal ideals in Z = prime numbers. now everything should be clear ;).

I forgot to mention that I know maximal ideals of Z are prime but does that mean J(nZ) is empty since a prime number will never intersect with other prime numbers? I might have gotten confused with definition of J(nZ). Thank you.


function
calculus
derivative
limit
real analysis
real analysis unsolved

Let $f \in D'(R)$ (see previous threads on distributions, or generalized functions) be such that $f(x+h) = f(x)$ where $h>0$.

Show that f is constant.

--------------

I imagine that there's some way to use the definition of the derivative here, but I'm not sure that it will work with distribtions.

Also note that if $f'=0$ (in the sense of distributions) then f is a constant.

If you know that $f'=0$ implies that $f$ is constant, the rest is straightforward. Let $\varphi$ be any test-function. Let $\varphi_h(x)=\frac 1h(\varphi(x+h)-\varphi(x))$. One can check that $\varphi_h\to\varphi'$ in $\mathcal D$ (this requires some work, of course, but it is a statement about smooth functions, not one about distributions). But then $(f,\varphi')=\lim_{h\to 0}(f,\varphi_h)=\lim_{h\to 0}0=0$, i.e., $f'=0$.


calculus
integration
logarithms
trigonometry
calculus computations

$\int\sqrt{x^{2}+7x+6}dx$. I've tried substitution, but didn't work.

[quote="Stewie"]There's something wrong about the first hint, I think.[/quote]
No,it is true at all!  :wink:

The procedure behind N.T.TUAN's hint: complete the square, then do a trig (or hyperbolic) substitution.


[quote="Peter"]Let $p$ be a prime number of the form $4k+1$. Suppose that $r$ is a quadratic residue of $p$ and that $s$ is a quadratic nonresidue of $p$. Show that $p=a^{2}+b^{2}$, where \[a=\frac{1}{2}\sum^{p-1}_{i=1}\left( \frac{i(i^{2}-r)}{p}\right), b=\frac{1}{2}\sum^{p-1}_{i=1}\left( \frac{i(i^{2}-s)}{p}\right).\] Here, $\left( \frac{k}{p}\right)$ denotes the Legendre Symbol.[/quote]
Quite nice!! Solution is not very hard to be found,but I just wonder how can somebody find the such an identity originally?
[b]Proof[/b]:Firstly,
Denote \[M_{r}=\frac{1}{2}\sum^{p-1}_{i=1}\left( \frac{i(i^{2}-r)}{p}\right)\], then,
if $r,t$ are both quadrati residues, suppose that
$r\equiv u^{2}t (mod p)$
\[\frac{1}{2}\sum^{p-1}_{i=1}\left( \frac{i(i^{2}-r)}{p}\right)= a=\frac{1}{2}\sum_{k=ju,1 \le j\le p-1}\left( \frac{k(k^{2}-r)}{p}\right)=\frac{1}{2}\sum^{p-1}_{j=1}\left( \frac{ju(j^{2}u^{2}-u^{2}t)}{p}\right)=\frac{1}{2}\left(\frac{u}{p}\right ) \sum^{p-1}_{j=1}\left( \frac{j(j^{2}-t)}{p}\right)\]
then we find that $M_{r}^{2}= M_{t}^{2}$
By the same way ,we can prove that $M_{r}^{2}=M_{t}^{2}$,if $r,t$ are both not quadratic residues.
$M_{0}=0$ is obvious.
So \[(a^{2}+b^{2})\frac{p-1}{2}=\sum_{r=0}^{p-1}M_{r}=\sum_{r=0}^{p-1}(\frac{1}{2}\sum^{p-1}_{i=1}\left( \frac{i(i^{2}-r)}{p}\right))^{2}= \frac{1}{4}\left (\sum_{r=0}^{p-1}\sum^{p-1}_{i=1}\left (\left( \frac{i(i^{2}-r)}{p}\right)\right)^{2}+\sum_{i\not=j}\sum_{r=0}^{p-1}{\left( \frac{ij(j^{2}-r)(i^{2}-r)}{p}\right)\right)=\frac{1}{4}\left (\frac{(p-1)(2p-1)}{2}+\sum_{i\not=j}\sum_{r=0}^{p-1}{\left( \frac{ij((r-2^{-1}(i^{2}+j^{2}))^{2}-4^{-1}(i^{2}-j^{2})^{2})}{p}\right)\right)=}}\]
\[\frac{1}{4}\left (\sum_{i\not=j}\sum_{r=0}^{p-1}{\frac{(p-1)(2p-1)}{2}+\sum_{i\not=j}\left(\frac{ij}{p}\right)\sum_{r=0}^{p-1}{\left( \frac{(r^{2}-4^{-1}(i^{2}-j^{2})^{2})}{p}\right)\right)}=}\]
\[\frac{1}{4}\left(\frac{(p-1)(2p-1)}{2}+\sum_{i\not=j,i^{2}-j^{2}\not \equiv 0}-\left(\frac{ij}{p}\right)+\sum_{i=-j}{\left(\frac{ij}{p}\right)}(p-1))\right)\](becuase $sum_{r=0}^{p-1}{\left( \frac{r^{2}-k^{2}}{p}\right)}=-1,(k\not =0),sum_{r=0}^{p-1}{\left( \frac{r^{2}-k^{2}}{p}\right)}=p-1,(k=0),$)
I have no patience to continue at all,but the rest is easy.......

Same approach as the above, but completed and with working display. In what follows, $\displaystyle\sum_i$ represents summing over all residue classes $i$ modulo $p$.

[b]Lemma:[/b] For a fixed $i$, $\displaystyle\sum_j \left( \dfrac{j(j+i)}{p} \right)$ is $-1$ if $i\not\equiv 0$ and $p-1$ otherwise.

[b]Proof:[/b] If $i\equiv 0$ the result is clear so assume otherwise.

First we'll count the number of distinct pairs of QRs $u^2, v^2\neq 0$ with $u^2-v^2\equiv i$. This is equivalent with $i \equiv (u+v)(u-v) \iff u \equiv \dfrac{1}{2} \left( t + \dfrac{i}{t} \right), v \equiv \dfrac{1}{2} \left( t - \dfrac{i}{t} \right)$. Since $u,v\neq 0$ we must have $t\not\equiv \pm \sqrt{\pm i}$. Furthermore, switching $t\iff \frac{i}{t}$ and $t\iff -t$ yields the same pair of QRs $u^2, v^2$, so putting everything together, we have $\frac{p-1}{4}$ solutions if $\left( \frac{i}{p} \right)=-1$ and $\frac{p-5}{4}$ solutions otherwise, which we can express as $\dfrac{p-3}{4} - \dfrac{\left( \frac{i}{p} \right)}{2}$.

Now we count the number of distinct pairs of non-QRs $x,y\not\equiv 0$ with $x-y\equiv i$. Multiply both sides by some non-QR $j$- then since $xj, yj$ are both QRs, this is equivalent now to $a^2-b^2\equiv ij$. By the previous paragraph, the number of distinct pairs $x,y$ is therefore $\dfrac{p-3}{4} - \dfrac{\left( \frac{ij}{p} \right)}{2} = \dfrac{p-3}{4} + \dfrac{\left( \frac{i}{p} \right)}{2}$.

Now note that $\left( \dfrac{j(j+i)}{p} \right)$ is $0$ for two values of $j$ and $1$ if $j+i, j$ are either both nonzero QRs or non-QRs differing by $i$, and by the above there are a total of $\dfrac{p-3}{2}$ such $j$ for which this occurs, so the remaining $\dfrac{p-1}{2}$ values of $j$ yield a value of $-1$. Therefore the entire sum is $\dfrac{p-3}{2}-\dfrac{p-1}{2}=-1$ as desired.

Now let $f(r) =\left( \displaystyle\sum_{i=1}^{p-1} \left( \dfrac{i(i^2-r)}{p} \right)\right)^2$. Clearly $f(0)=0$ and when $x\not\equiv 0$, we have $f(x^2r) =\left( \displaystyle\sum_{i=1}^{p-1} \left( \dfrac{i(i^2-x^2r)}{p} \right)\right)^2 = \left(\displaystyle\sum_{j=1}^{p-1} \left( \dfrac{jx((jx)^2-x^2r)}{p} \right)\right)^2 = \left(\displaystyle\sum_{j=1}^{p-1} \left( \dfrac{j(j^2-r)}{p} \right)\right)^2 \left( \dfrac{x^3}{p} \right)^2 = f(r)$, so $f$ is the same across all QRs and across all non-QRs.

Now we expand $$\displaystyle\sum_j f(j) = \displaystyle\sum_j \displaystyle\sum_{u,v} \left( \dfrac{u(u^2-j)}{p} \right) \left( \dfrac{v(v^2-j)}{p} \right) = \displaystyle\sum_{u,v} \left( \dfrac{uv}{p} \right) \displaystyle\sum_j \left( \dfrac{(u^2-j)(v^2-j)}{p} \right) = \displaystyle\sum_{u^2-v^2=i} \left( \dfrac{uv}{p} \right) \displaystyle\sum_j \left( \dfrac{j(j+i)}{p} \right).$$

Clearly if $uv\equiv 0$ then the corresponding term does not contribute to the sum, so we may WLOG assume $u,v\not\equiv 0$. Then there are $2(p-1)$ pairs of residues $(u,v)$ modulo $p$ for which $u^2\equiv v^2$, in which case by the lemma applied on the inner summation with $i\equiv 0$ we get a total of $2(p-1)(p-1)$ from these terms. In the remaining pairs $(u,v)$, we know by the lemma applied on the inner summation with $i\not\equiv 0$ that the remaining terms sum to $$-\displaystyle\sum_{u\not\equiv 0} \displaystyle\sum_{v\not\equiv u,-u,0} \left( \dfrac{uv}{p} \right) = -\displaystyle\sum_{u\not\equiv 0} \left( -\left( \dfrac{u^2}{p} \right) - \left( \dfrac{-u^2}{p}\right) + \displaystyle\sum_v \left( \dfrac{uv}{p} \right)\right)= -\displaystyle\sum_{u\not\equiv 0} (-2) = 2(p-1),$$ hence the original sum totals to $2p(p-1)$.

Now since we know from earlier that $f(r)$ depends only on $\left( \frac{r}{p} \right)$, we have $a^2+b^2 = \dfrac{1}{4} \left( f(r) + f(s) \right) = \dfrac{1}{4} \dfrac{2p(p-1)}{\frac{p-1}{2}} = p$, so we are done.


geometry
geometric transformation
rotation
exterior angle

[quote="236factorial"]What is the sum of the measures of the exterior angles in a 99 sided figure?[/quote]
[hide]
Isnt the sum of all exterior angles in any convex polygon 360 degrees?
[/hide]

[quote="B4k4ka"]Isnt the sum of all exterior angles in any convex polygon 360 degrees?[/quote]

No... just kidding  :D You are right.

Wait...Why is it 360? Sorry if this is a dumb question... :?  :blush:

That's a good question.. I don't know exactly why. Maybe it has something to do with a complete rotation is 360.

well take a triangle for example.

the sum of all the angles is 180.

if its equalateral, then the exterior angle of one angle is 180-60=120. and if it's not equilateral you can do the same thing.

There are three sides so you do what i did above 3 times. So it's (180x3)-180=X
so you get 540-180=360

so all triangles have exterior angle of 360.

Let's try it with quadrilaterals.

(180x4) - 360 = 720-360=360

It's also probably pertaining to the fact that the sum of the interior angles is (n-2)x 180, where n is the number of sides and angles.

When you calculate the exterior angle of an angle, you do 180-that angle.

so the sum of any polygon's exterior angles is 180n - 180(n-2) which simplifies to be 2x180=360.

argh... i hate proofs, and i just did one without noticing... grr... :mad:

Tnx that makes sense now... If i had to guess I would have guessed 360 before, but now I know why! :)   :idea:

check out this web page.

http://www.intermath-uga.gatech.edu/tweb/gwin1-01/dimsdale/extang/extang.htm

thanks for showing me that website.  I learned something new.


Three distinct points $ A,B,C$ are found, in this order, on a line. Let $ G$ be a circle which passes through $ A$ and $ C$ with its center found outside the line $ AC$ and let $ P$ be the intersection point of the tangents to $ G$, drawn to the points $ A$ and $ C$.Let $ Q$ be the intersection between $ G$ and the segment $ (BP)$.Prove that the intersection between the bisector of angle $ \angle AQC$ and the line $ AC$ doesn't depend on choosing the circle $ G$.

Extend $ PQ$ to meet $ G$ at $ Q'$.  Define $ x: \equal{} \angle{QAC}$, $ y: \equal{} \angle{QCA}$, $ u: \equal{} \angle{Q'AC}$, $ v: \equal{} \angle{Q'CA}$, and $ t: \equal{} \angle{PAC} \equal{} \angle{PCA}$.

Using the law of sine with triangle $ APQ'$, one gets
\[ \frac {AP}{\sin(y)} \equal{} \frac {PQ'}{\sin(t \plus{} u)}\,.
\]
Applying the law of sine with triangle $ CPQ'$, one has
\[ \frac {CP}{\sin(x)} \equal{} \frac {PQ'}{\sin(t \plus{} v)}\,.
\]
Since $ AP \equal{} CP$, one obtains
\[ \frac {\sin(x)}{\sin(y)} \equal{} \frac {\sin(t \plus{} v)}{\sin(t \plus{} u)}\,.
\]
It is easy to show that $ u \plus{} v \plus{} t \equal{} \pi$.  That is, $ \sin(t \plus{} v) \equal{} \sin(u)$ and $ \sin(t \plus{} u) \equal{} \sin(v)$.  This proves that
\[ \frac {\sin(x)}{\sin(y)} \equal{} \frac {\sin(u)}{\sin(v)}\,. ...(*)
\]
Suppose that $ QX$ is the angular bisector of $ \angle{AQC}$ with $ X \in AC$.  Also, $ Q'X'$ bisects $ \angle{AQ'C}$ with $ X' \in AC$.  Therefore, using the law of sine with $ AQX$ and $ CQX$, one gets
\[ \frac {AX}{XC} \equal{} \frac {\sin(x)}{\sin(y)}\,.
\]
Similarly,
\[ \frac {AX'}{X'C} \equal{} \frac {\sin(u)}{\sin(v)}\,.
\]
From (*), we conclude that $ X \equal{} X'$.

Now, let the external angular bisector of $ \angle{AQC}$ meet $ AC$ at $ Y$.  Then, $ XY$ is a diameter of the circle passing through $ X$, $ Q$, $ Y$, and $ Q'$.  Consequently,
\[ YB\cdot{BX} \equal{} QB\cdot{BQ'} \equal{} AB\cdot{BC}\,. ...(1)
\]
Since $ (C,X,A,Y)$ makes a harmonic $ 4$-tuple, then
\[ \frac {CX}{XA} \equal{} \frac {CY}{YA}\,. ...(2)
\]
These conditions force $ X$ to be a fixed point, depending only on $ AB$ and $ AC$.

To prove the above statement, let $ M$ be the midpoint of $ AC$.  Without loss of generality, suppose $ B$ is closer to $ A$ than to $ C$.
\[ AB\cdot{BC} \equal{} YB\cdot{BX} \equal{} (YM\minus{}BM)(BM\minus{}XM)\,.\]
Also,
\[ \frac{CX}{XA} \equal{} \frac {CY}{YA} \equal{} \frac{AM\plus{}YM}{YM\minus{}AM}\,.\]
Thus,
\[ \frac{CX}{XA} \equal{} \frac{AM\plus{}\left(BM\plus{}\frac{AB\cdot{BC}}{BM\minus{}XM}\right)}{\left(BM\plus{}\frac{AB\cdot{BC}}{BM\minus{}XM}\right)\minus{}AM} \equal{} \frac{BC\plus{}\frac{AB\cdot{BC}}{BM\minus{}XM}}{\frac{AB\cdot{BC}}{BM\minus{}XM}\minus{}AB}\,.\]
That is,
\[ \frac{XC}{AX} \equal{} \frac{BC}{AB}\left(\frac{AB\plus{}BM\minus{}XM}{BC\minus{}BM\plus{}XM}\right) \equal{} \frac{BC}{AB}\left(\frac{AX}{XC}\right)\,.\]
Consequently
\[ \frac{AX}{XC}\equal{}\sqrt{\frac{AB}{BC}}\,.\]

PS.  Thanks to "The QuattoMaster 6000" in the next reply.  I have been aware that there might be a simple solution.


In a right triangle, the sum of the lengths of the legs is 17 mm. The area of the right triangle is 30 square millimeters. Find the number of millimeters in the length of the hypotenuse.

Using knowledge of Pythagorean Triples, we get $ 5$,$ 12$,$ 13$. Thus the answer is $ 13$.

[hide="In case you want a non-memorization solution..."]  Let the triangle have legs lengths $ a$ and $ b$, and hypotenuse length $ c$.  Since the two legs of a right triangle are perpendicular, one of them will be the base and the other will be the height.  Thus, the area of the triangle is $ \frac{ab}{2}=30\implies ab=60$.  We are given that $ a+b=17$, so \begin{align*}(a+b)^2
&=a^2+2ab+b^2\\
&=a^2+b^2+2(60)\\
&=a^2+b^2+120\\
&=17^2\\
&=289\end{align*}  Thus, $ a^2+b^2=169$.  Remember that in a right triangle, $ a^2+b^2=c^2$.  Thus, \begin{align*}c
&=\sqrt{c^2}\\
&=\sqrt{a^2+b^2}\\
&=\sqrt{169}\\
&=\boxed{13}\end{align*}[/hide]
However, memorizing Pythagorean Triples makes this process faster.  In a real situation, I most likely would have done it the way \$LaTeX\$ did it.


How do you find the centroid of a triangle please help. I couldn't find any online sources

You mean given the coordinates of the vertices of the triangle, find the coordinates for the centroid?

It's simply the average of the three points: $ \left(\frac{x_1\plus{}x_2\plus{}x_3}{3},\frac{y_1\plus{}y_2\plus{}y_3}{3}\right)$

It's easy to find on google, just search "calculating triangle centroid" or "triangle centroid formula" or something.

Centroid of a triangle Definition:
The point through which all the three medians of a triangle pass is called centroid of the triangle and it divides each median in the ratio 2:1.

To put it in more general terms, the centroid is the intersection of the lines connecting the midpoints of each side and their opposite vertex.

The reasoning for this is that centroid is considered the center of mass

Like if you cut out a triangle and hold the triangle at the point of the centroid, it should balance.

With the same reasoning of midpoint, we get the average of the three coordinates for Centroid.

This is very easy.  This can be done by drawing all the medians from all the three angles.
Medians bisect the angle and the side opposite to it.  So make sure ur medians are right, and the point where all of them meet is the centroid.

Average the X coordinates (divide by 3) and average the Y coordinates, and that is the centroid.

The centroid is the line connecting the midpoints of a triangle and the opposite vertices, it is also the center balance of the triangle


symmetry
geometry
circumcircle
angle bisector
perpendicular bisector

Here's a problem you guys might like:

The Pseudotheorem: All triangles are isosceles.

Pseudoproof. Let ABC be a triangle, and let O be the intersection of the internal angle bisector of A with the perpendicular bisector of BC, as in the figure.
Let D,Q,R be the feet of perpendiculars from O to BC, CA, AB, respectively. By
symmetry across OD, OB= OC, while by symmetry across AO, AQ = AR and OQ = OR.
Now the right triangles <ORB and <OQC have equal legs OR = OQ and equal hypotenuses
OB = OC, so they are congruent, giving RB = QC. Finally, we conclude AB = AR+RB=AQ+QC=AC.

[hide]
The proof fails at AB = AR+RB=AQ+QC=AC, because R and Q are not necessarily between AB and AC, respectively.

This counterpoint depends on the point that O is never inside ABC, and that, if outside ABC, that exactly one of either R or Q cannot possibly be between AB and AC respectively.

First just use the Angle Bisector Theorum to demonstrate that O cannot be inside ABC: without loss of generality if A<B, then the angle bisector will intersect C at a point closer to A, without yet crossing the perpendicular bisector of C.  Thus, O is outside ABC.  Only if A=B does O lie on side BC, and even then is not inside ABC.  In this sense the figure is misleading.

Now take triangle ABC with the same 'instructions' and labelling as before, only with O outside ABC. AOR and AOQ are still congruent, so angles AOR and AOQ are congruent.  Likewise, BOR and COQ, and BOD and COD, are congruent.  Because AO is not perpendicular to BC (otherwise it cannot have intersected OD), there is an obvious angle of AOD.  The only way for BOR and COQ to be congruent is for one of R or Q to be closer to A than B or C, respectively, and for one to be further away.

What ultimately happens is that ORB and OQC, while congruent, are not oriented the same way: one will have its right angle outside ABC.  Thus, rather than AB = AR+RB=AQ+QC=AC, we arrive with, if AB<AC, AB=AR-RB and AC=AQ+QC.
[/hide]

Nice Phillip :). I knew you would get it...

Also, a slightly harder problem regarding that previous one (I'm hiding it because it give the entire original problem away)
[hide]
Show that ABOC is cyclic, i.e. that O is on the circumcircle of ABC.

(this is obviously another way to show that O is outside of ABC)
[/hide]

EDIT: fixed to Philip :)

By the way, my name has only one "L."  The way 'Philip' is spelled differs from person to person: it's very confusing.

[hide]
Let's define some names. Let point E be the intersection of angle bisector AO and side BC.  Let $ \angle{BCA}\equal{}\alpha$, $ \angle{ABC}\equal{}\beta$, $ \angle{ACB}\equal{}\gamma$, $ \angle{AOD}\equal{}x$, and $ \angle{OBC}\equal{}\angle{OCB}\equal{}\theta$. Without loss of generality, lets say side AB<AC, so that B is between A and R.

To prove the quadrilateral is cyclic, we need to show that $ \alpha \equal{} 2\theta$.  Not only will these two directly show it is cyclic because each includes a side and diagonal connecting two opposite points, but it will also suffice to prove that the opposite angles add to 180:
$ \alpha \plus{} 180 \minus{} 2\theta \equal{} 180$
$ 2\theta \plus{} \beta \plus{}\gamma \equal{} 180$

First consider $ x$. $ \triangle{DEO}$ is a right triangle, so $ \angle{DEO}\equal{}90\minus{}x\equal{}\angle{AEB}$. This means that $ \angle{AEC}\equal{}90\plus{}x$ as these angles are supplementary. Considering the triangles $ \triangle{ABE}$ and $ \triangle{AEC}$, we find:
$ \alpha /2 \plus{}\beta \plus{}90\minus{}x\equal{}180$
$ \alpha /2 \plus{}\gamma \plus{}90 \plus{} x\equal{}180$
which after some manipulation and combination, yields:
$ x\equal{}\frac{\beta\minus{}\gamma}{2}$.

Now consider the right $ \triangle{QOC}$. $ \angle{QOC}$ is obviously equal to $ 90\minus{}\gamma\minus{}\theta$.  Because this triangle is congruent to $ \triangle{ROB}$, $ \angle{ROB}\equal{}90\minus{}\gamma\minus{}\theta$.  Now consider $ \triangle{ARO}$:
$ \alpha /2 \plus{} 90 \plus{}\angle{AOR}\equal{}180$
$ \alpha /2 \plus{} 90 \plus{}(\angle{BOD}\minus{}x \plus{}\angle{ROB})\equal{}180$
$ \alpha /2 \plus{} 90 \plus{} (90\minus{}\theta)\minus{}x \plus{}(90\minus{}\gamma\minus{}\theta)\equal{}180$
$ \alpha /2 \minus{}2\theta\minus{}\gamma\minus{}x\plus{}90\equal{}0$
$ \alpha\minus{}2\theta \plus{}(90\minus{}\alpha /2 \minus{}\gamma \minus{}\beta /2 \plus{} \gamma /2)\equal{}0$
$ \alpha\minus{}2\theta\plus{}(90\minus{}\frac{\alpha \plus{}\beta \plus{}\gamma}{2})\equal{}0$

Therefore:
$ \alpha \equal{} 2\theta$
and we're done.
[/hide]

Nice :). One last thing: Where is O on that circle? (this solves all three problems at once)

[hide="My Solution"]
We shall show that O lies on the midpoint of arc AB on the circumcircle of ABC. Let the angle bisector of A hit the circle at D. Since DAB=DAC, BD=DC. Hence D lies on the midpoint of arc ABC. However, the perpendicular bisector of side BC obviously hits arc BC at its midpoint, since BC is a chord. Hence this point O, being the intersection of point of the angle bisector of A and the perpendicular bisector of BC, lies on the midpoint of BC.[/hide]

[quote="Anaxerzia"] Since DAB=DAC, BD=DC.[/quote]

I lost you about here. Can you explain this?

You need to remember your arc measures. A measure of an arc is twice the measure of the angle that intercepts it. So we have $ \frac{1}{2}DC\equal{}\angle DAC\equal{}\angle DAB\equal{}\frac{1}{2}BD$, so $ DC\equal{}BD$.


I'm having trouble making a table. I want the table to have lines separating the rows and columns and I want the day number column to appear.

Here's what I typed in:

\begin{tabular}[t]{|l|cccccccccc|}
\multicolumn{11}{c}{Work Schedule}\\\hline
Day Number&\1&\2&\3&\4&\5&\6&\7&\8&\9&\10\\
Richard&W&W&W&W&O&W&W&W&W&O\\
Naoki&W&W&W&W&W&W&W&O&O&W\\
\end{tabular}

Up until I put this in, I didn't get any errors.
How can I post the log file? I don't know where it is.

[quote="aufha"]How can I post the log file? I don't know where it is.[/quote]It's called IntroNoFinal.log and is in the same directory as IntroNoFinal.tex along with lots of other files created by LaTeX. In TexnicCenter you can get there easily using Tools, Windows Explorer.

[quote="aufha"]Day Number&\1&\2&\3&\4&\5&\6&\7&\8&\9&\10\\[/quote] Get rid of all the \. LaTeX is objecting as it thinks that \1, \2 etc are commands
Add \hline at the end of the table to put in the bottom line.

About the log question. Attached is the what I see after windows explorer. How do I know which is the log file?- which is it?

It's probably the text document in the first column. But it's not obvious because you can't see the file extensions which would make it obvious as it's called IntroNoFinal.log.

In explorer go to Tools, Folder Options, View, untick [i]Hide extensions for known file types[/i] then OK and lo and behold you will see IntroNoFinal.log

Everyone should do this as it is easy to be fooled otherwise by dodgy files called something like clickme.txt which is actually clickme.txt.exe and does something nasty. I wish Microsoft had made that the default.


Let $ x_{1},x_{2},...,x_{241}\in N$ have sum $ 360$. Can we always divide them into three sets, each with equal sum?

By $ N$, I denote the set of positive integers  :)

[quote="amitie"]surely we could just have $ x_{1}=360$ and $ x_{2}=x_{3}=...=0$?

thus we could never have 3  equal sets...[/quote]

$ x_{n}\in N$ so $ x=0$ is not a posibility.

[hide="Solution"]We will show that it's possible by putting all the $ x_{i}$ into the three sets, starting with the biggest.  First note that no $ x_{i}$ can be greater than 120, otherwise the sum of all the elements would be greater than 360.

If we had less than 122 ones, then the sum of all the $ x_{i}$ would be at least $ 121+2\cdot 120$, which is greater than 360.  Therefore we have at least 122 ones, so the sum of the non-one $ x_{i}$ is at most 238.  So we can have at most three $ x_{i}$ greater than 60.  We can easily take care of these, just put them in different sets.

Now let's take care of all the $ x_{i}$ such that $ 2\le x_{i}\le 60$.  Just put the biggest number left into the set with least sum.  Suppose for the sake of contradiction that this algorithm fails, so that some $ x_{i}=k$ puts the sum of a set over 120.  Then before we put it into a set, all three sets have sums of at least $ 121-k$.  Therefore the sum of all the non-one $ x_{i}$ is at least $ k+3(121-k)=363-2k\ge 363-2\cdot 60>238$.  But the sum of the non-one $ x_{i}$ is at most 238, so we have a contradiction.  Therefore we can successfully put all the non-one $ x_{i}$ in the set this way.

So now we have a bunch of ones and three sets with sums of at most 120.  Clearly we can put the ones in the sets so that their sums are all 120, so we're done.[/hide]


inequalities
search
inequalities unsolved

[quote="colorfuldreams"]1) $ a,b,c$ are positive real number
Proof
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{c \plus{} a} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c}$

2) $ a,b,c$ are positive real number and $ abc \equal{} 1$
Proof
$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a}\leq \frac {1}{2 \plus{} a} \plus{} \frac {1}{2 \plus{} b} \plus{} \frac {1}{2 \plus{} c}$

3 $ a_i,b_i$ are positive real numbers
Proof
$ \sum_{cyc}{\frac {a_i^{n \plus{} 1}}{b_i^n}} \geq \frac {({\sum_{cyc}a_i})^{n \plus{} 1}} {({\sum_{cyc}b_i})^{n}}$[/quote]

Third is  an apllication Holder's inequality. Just  bring the denominator on the LHS.

[quote]$ a,b,c$ are positive real number, prove:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq\frac{a\plus{}b}{c\plus{}a}\plus{}\frac{b\plus{}c}{a\plus{}b}\plus{}\frac{c\plus{}a}{b\plus{}c}$[/quote]

[url=http://www.mathlinks.ro/viewtopic.php?p=1624644#1624644]Solution[/url]

Solution of the second:

If we take;

$ x \equal{} a \plus{} b \plus{} c$ , $ y \equal{} ab \plus{} bc \plus{} ca$

Inequality is equivalent to

$ (x \minus{} 1)y^2 \plus{} (3x^2 \plus{} 6x \minus{} 3)y \geq 5x^2 \plus{} 24x \plus{} 27$

Since $ x,y \geq 3$ and $ y^2 \geq 3x$ which are true by well-known inequalities $ (ab \plus{} bc \plus{} ca)^2 \geq 3abc(a \plus{} b \plus{} c) \geq 9(abc)^{\frac {4}{3}}$

$ 3xy \geq 27$ and  $ (x \minus{} 1)3x \plus{} (3x^2 \plus{} 3x \minus{} 3)\sqrt {3x} \leq (x \minus{} 1)y^2 \plus{} (3x^2 \plus{} 3x \minus{} 3)y$

Now if we prove that  $ (x \minus{} 1)3x \plus{} (3x^2 \plus{} 3x \minus{} 3)\sqrt {3x} \geq 5x^2 \plus{} 24x$ then the proof is completed.

The last inequality is equivalent to;

$ 9\lambda^4 \minus{} 2\lambda^3 \minus{} 6\lambda \minus{} 1 \geq 0$ where $ \lambda \equal{} \sqrt {\frac {x}{3}}$

Since $ \lambda \geq 1$

$ 9\lambda^4 \minus{} 2\lambda^3 \minus{} 6\lambda \minus{} 1 \equal{} (2\lambda^4 \minus{} 2\lambda^3) \plus{} (6\lambda^4 \minus{} 6\lambda) \plus{} (\lambda^4 \minus{} 1) \geq 0$ and the solution is done.  :wink:

We may assume that $ c \equal{} \min \left\{ {a,b,c} \right\}$
we have,
\[ \begin{array}{l}
 \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \minus{} 3 \equal{} \frac{a}{b} \plus{} \frac{b}{a} \minus{} 2 \plus{} \frac{b}{c} \plus{} \frac{c}{a} \minus{} \frac{b}{a} \minus{} 1 \\
  \equal{} \frac{{(a \minus{} b)^2 }}{{ab}} \plus{} \frac{{(b \minus{} c)(a \minus{} c)}}{{ac}} \\
 \end{array}\]
And
 \[ \begin{array}{l}
 \frac{{a \plus{} b}}{{c \plus{} a}} \plus{} \frac{{b \plus{} c}}{{a \plus{} b}} \plus{} \frac{{c \plus{} a}}{{b \plus{} c}} \minus{} 3 \equal{} \frac{{c \plus{} a}}{{b \plus{} c}} \plus{} \frac{{b \plus{} c}}{{c \plus{} a}} \minus{} 2 \plus{} \frac{{a \plus{} b}}{{c \plus{} a}} \plus{} \frac{{b \plus{} c}}{{a \plus{} b}} \minus{} \frac{{b \plus{} c}}{{c \plus{} a}} \minus{} 1 \\
  \equal{} \frac{{(a \minus{} b)^2 }}{{(b \plus{} c)(c \plus{} a)}} \plus{} \frac{{(b \minus{} c)(a \minus{} c)}}{{(a \plus{} b)(c \plus{} a)}} \\
 \end{array}\]

Because of $ c \equal{} \min \left\{ {a,b,c} \right\}$
, it’s clear that
\[ \frac{{(a \minus{} b)^2 }}{{ab}} \plus{} \frac{{(b \minus{} c)(a \minus{} c)}}{{ac}} \ge \frac{{(a \minus{} b)^2 }}{{(b \plus{} c)(c \plus{} a)}} \plus{} \frac{{(b \minus{} c)(a \minus{} c)}}{{(a \plus{} b)(c \plus{} a)}}\]

Thus the inequality is proved

[quote="colorfuldreams"]1) $ a,b,c$ are positive real number
Proof
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{c \plus{} a} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c}$
[/quote]

It is equivalent to
\[ \sum_{cyc}\frac{a^2\plus{}bc}{a^2\plus{}ab}\geqslant 3\mbox{,}\]
and after applying AM-GM, it suffices to prove that
\[ \prod_{cyc}(a^2\plus{}bc)\geqslant\prod_{cyc}(a^2\plus{}ab)\mbox{,}\]
which is obvious after multiplying out  :P .

[quote="colorfuldreams"]1) $ a,b,c$ are positive real number
Proof
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{c \plus{} a} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c}$

[/quote]

[hide="Prove."]

$ LHS \minus{} RHS$

$ \equal{} \frac {(\sum{a^4c^2} \minus{} \sum{a^3bc^2}) \plus{} (\sum{a^3b^3} \minus{} 3a^2b^2c^2)}{abc(b \plus{} c)(c \plus{} a)(a \plus{} b)}$.

With AM-GM inequalities, we have

$ \sum{a^3b^3} \minus{} 3a^2b^2c^2 \ge 0$,

$ 6\sum{a^4c^2} \equal{} 4\sum{a^4c^2} \plus{} \sum{b^4a^2} \plus{} \sum{c^4b^2} \ge 6\sum{a^3bc^2}$.

We've done. #

[/hide]

your post's inequalities most are famous...

[quote="Martin N."][quote="colorfuldreams"]1) $ a,b,c$ are positive real number
Proof
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{c \plus{} a} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c}$
[/quote]

It is equivalent to
\[ \sum_{cyc}\frac {a^2 \plus{} bc}{a^2 \plus{} ab}\geqslant 3\mbox{,}\]
and after applying AM-GM, it suffices to prove that
\[ \prod_{cyc}(a^2 \plus{} bc)\geqslant\prod_{cyc}(a^2 \plus{} ab)\mbox{,}\]
which is obvious after multiplying out  :P .[/quote]

I think this method works for the general case with $ n$ positive variables, i.e. for proving $ \sum_{cyc} \frac {a}{b} \geq \sum_{cyc} \frac {a\plus{}c}{b \plus{} c}$

by virtue of $ \frac ab\minus{}\frac{a\plus{}c}{b\plus{}c}\equal{}\frac{b^2\plus{}ac}{b(b\plus{}c)}\minus{}1$.

But is there an easy way to prove \[ \prod_{cyc}(b^2 \plus{} ac)\geqslant\prod_{cyc}(b^2 \plus{} bc)\mbox{  ?}\]
e.g. writing it as \[ \prod_{cyc}(b^2 \plus{} ac)\geqslant\prod_{cyc}(ab\plus{}ac)\] and using some sort of re-arrangement argument :?:  :maybe:

[quote="spanferkel"][quote="Martin N."][quote="colorfuldreams"]1) $ a,b,c$ are positive real number
Proof
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{c \plus{} a} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c}$
[/quote]

It is equivalent to
\[ \sum_{cyc}\frac {a^2 \plus{} bc}{a^2 \plus{} ab}\geqslant 3\mbox{,}\]
and after applying AM-GM, it suffices to prove that
\[ \prod_{cyc}(a^2 \plus{} bc)\geqslant\prod_{cyc}(a^2 \plus{} ab)\mbox{,}\]
which is obvious after multiplying out  :P .[/quote]

I think this method works for the general case with $ n$ positive variables, i.e. for proving $ \sum_{cyc} \frac {a}{b} \geq \sum_{cyc} \frac {a \plus{} c}{b \plus{} c}$

by virtue of $ \frac ab \minus{} \frac {a \plus{} c}{b \plus{} c} \equal{} \frac {b^2 \plus{} ac}{b(b \plus{} c)} \minus{} 1$.

But is there an easy way to prove
\[ \prod_{cyc}(b^2 \plus{} ac)\geqslant\prod_{cyc}(b^2 \plus{} bc)\mbox{ ?}\]
e.g. writing it as
\[ \prod_{cyc}(b^2 \plus{} ac)\geqslant\prod_{cyc}(ab \plus{} ac)\]
and using some sort of re-arrangement argument :?:  :maybe:[/quote]

A bit dirty but see [url=http://www.mathlinks.ro/viewtopic.php?p=1626975#1626975]here[/url] the post #30 by alex2008.

But if you refer to this
[quote="alex2008"]
So it remains to prove:
\[ \prod (a^2 \plus{} bc)\ge abc\prod (a \plus{} b)\]
Now we prove
\[ (a^2 \plus{} bc)(b^2 \plus{} ca)\ge ab(c \plus{} a)(b \plus{} c)\Leftrightarrow a^3 \plus{} b^3\ge ab^2 \plus{} a^2b\Leftrightarrow (a \plus{} b)(a \minus{} b)^2\ge 0\]
Multiplying the similars we are done
[/quote]
it won't help for the general case because we don't have $ (a \plus{} c)$ terms on the RHS if $ n > 3$  :maybe:  :maybe:

[quote="colorfuldreams"]how about question 2? :([/quote]

Search for Bulgaria 1997

[quote="colorfuldreams"]1) $ a,b,c$ are positive real number
Proof
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{c \plus{} a} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c}$[/quote]

[hide="Solution (well, almost)"]
\[ \frac 1 b \plus{} \frac 1 c \ge \frac 4 {b \plus{} c}\]
Adding the similar inequalities,
\[ \frac 1 a \plus{} \frac 1 b \plus{} \frac 1 c \ge \sum{\frac 2 {b \plus{} c}}\]

\[ \implies (a \plus{} b \plus{} c) \left(\frac 1 a \plus{} \frac 1 b \plus{} \frac 1 c\right) \ge (\sum{b \plus{} c}) \left(\sum{\frac 1 {b \plus{} c}}\right)\]

\[ \implies \sum{\frac a b} \plus{} \sum{\frac b a} \ge \sum{\frac {a \plus{} c}{b \plus{} c}} \plus{} \sum{\frac {b \plus{} c}{a \plus{} c}}\]
Which means that at least one of the following inequalities holds
\[ \sum{\frac a b} \ge \sum{\frac {a \plus{} c}{b \plus{} c}}\]

\[ \sum{\frac b a} \ge \sum{\frac {b \plus{} c}{a \plus{} c}}\]
If the second inequality holds, just swap $ a$ and $ c$.[/hide]

I wish, I wish with all my heart that this solution was right. Oh well, at least it looks nice  :)

[hide="EDIT"]I realized this was wrong a minute after posting it, but I didn't want to remove it because it looked nice :P  and also could serve as a reminder of what not to do with cyclic sums.[/hide]

[quote="keyree10"]
I wish, I wish with all my heart that this solution was right. Oh well, at least it looks nice  :)[/quote]
But sorry, I think it doesn't work  :(  Imagine you want to prove it for given $ a,b,c$...

[quote="keyree10"]
Which means that at least one of the following inequalities holds
\[ \sum{\frac a b} \ge \sum{\frac {a \plus{} c}{b \plus{} c}}\]

\[ \sum{\frac b a} \ge \sum{\frac {b \plus{} c}{a \plus{} c}}\]
If the second inequality holds, just swap $ a$ and $ c$.
[/quote]
... so maybe, you can only prove it for $ (c,b,a)$ instead, which is not the same inequality  :maybe:

With a similar method you could "prove" $ \sum \frac a{a\plus{}b}\ge\frac 32$ because $ \sum \frac a{a\plus{}b}\plus{}\sum \frac b{a\plus{}b}\equal{}3$.


Suppose that $x, y,$ and $z$ are positive integers with $xy=z^2 +1$. Prove that there exist integers $a, b, c,$ and $d$ such that $x=a^2 +b^2$, $y=c^2 +d^2$, and $z=ac+bd$.

Corrected copy of my solution on MathLinks, using Gaussian integers:

Write $xy=z^2+1=(z+i)(z-i)$.
Now factor $z+i = \pi_1\cdot \pi_2\cdot ...\cdot \pi_n$ into Gaussian primes and get ${z-i = \overline{z+i} = \overline{\pi_1\cdot \pi_2\cdot ...\cdot \pi_n} = \overline{\pi_1} \cdot \overline{ \pi_2}\cdot ...\cdot \overline{\pi_n}}$ (the primes $\pi_i$ are not necessary distinct). Now by uniqueness of factorisation (and $xy=z$), we w.l.o.g. get that $x=(\pi_1\cdot \pi_2\cdot ...\cdot \pi_k)(\overline{\pi_1}\cdot \overline{\pi_2}\cdot ...\cdot \overline{\pi_k})=(a+bi)(a-bi)=a^2+b^2$ and $y=(\pi_{k+1}\cdot \pi_{k+2}\cdot ...\cdot \pi_n)(\overline{\pi_{k+1}}\cdot \overline{\pi_{k+2}}\cdot ...\cdot \overline{\pi_n})=(c-di)(c+di)=c^2+d^2$.
Now $(a+bi)(c-di)=(z+i)$ by definition, which gives $z=ac+bd$.

[quote="Peter"]Won't there be a nice elementary solution?[/quote]
Actually, there is one elementary solution that uses Fermat's method of finite descent.

Can you post it for us? Additional solutions are always nice. :)

[quote]Suppose that $ x,y$ and $ z$ are positive integers with $ x^2 \plus{} 1 \equal{} yz$.Prove that there exist integers $ a,b,c,d$,such that $ x \equal{} ac \plus{} bd,z \equal{} a^2 \plus{} b^2,y \equal{} c^2 \plus{} d^2$.
[/quote]
   -Suppose to the contrary that there exist some triplets $ (x,y,z)$ that satisfies to equation $ xy \equal{} z^2 \plus{} 1$,but they can not be represented in a needed form.
   -Let's take among of them a triplet with minimum sum,i.e $ x \plus{} y \plus{} z$ reaches minimum value.

    WLOG assume $ y > z$,obviously $ x < y$ and $ x\geq z$.If $ x \equal{} z$ then $ y \equal{} 2,z \equal{} 1,x \equal{} 1$,for this triplet we can take $ a \equal{} 1,b \equal{} 0,c \equal{} 1,d \equal{} 1$.
    -Now we can assume that $ y > x > z$:
    Then consider triplet $ (x \minus{} z,y \plus{} z \minus{} 2x,z)$,these numbers are obviously positive and satisfies to equation $ xy \equal{} z^2 \plus{} 1$,but their sum is equal to $ y \plus{} z \minus{} x < y \plus{} z \plus{} x$,hence there exist positive integers $ a,b,c,d\in\mathbb{N}$,such that
$ x \minus{} z \equal{} ac \plus{} bd$,$ z \equal{} a^2 \plus{} b^2$,$ y \minus{} 2x \plus{} y \equal{} c^2 \plus{} d^2$,but this system implies that
$ x \equal{} a(a \plus{} c) \plus{} b(b \plus{} d)$,$ y \equal{} (a \plus{} c)^2 \plus{} (b \plus{} d)^2$,$ z \equal{} a^2 \plus{} b^2$.Now it is enough to take
$ a_0 \equal{} a,b_0 \equal{} b,c_0 \equal{} a \plus{} c,d_0 \equal{} d \plus{} b$.Thus contradiction obtained.

An alternative [url=http://www.mathlinks.ro/viewtopic.php?p=1758748#1758748]solution[/url] using Minkowski's Theorem :)

Let's take $a,b,c,d$ pairwisely co-prime such that $ac-bd=1$.Bezout's identity confirms that  such $a,b,c,d$ exists.Now using $\text {Brahmagupta-Fibbonacci Identity}$ we get that $x,y$ both are bi-squares i.e. sum of two squares.So let $x=a^2+b^2,y=c^2+d^2$.Then $xy=(ad+bc)^2+(ac-bd)^2$ by the identity.So now let's take $z=ad+bc$.

[quote="ZetaX"]I don't see where you have proven anything related to the problem: You start with special $a,b,c,d$ and define $x,y,z$ from them. That's completely the wrong order.[/quote]
I wrote about $a,b,c,d$ first but it will come after we deduce $x,y$ are sum of two-squares.We need to show the existence of such $a,b,c,d$,and I have found out a triple of $(x,y,z)$ to satisfy this.Hope it's enough.

[quote="Erken"][quote]...Then consider triplet $ (x \minus{} z,y \plus{} z \minus{} 2x,z)$,these numbers are obviously positive and satisfies to equation $ xy \equal{} z^2 \plus{} 1$,...[/quote][/quote]

The triplet should be $(x, x+y-2z, y-z)$.

@vexation : can you explain your way please ?

About Fermat's method, one could make some progress by noticing the identity: ((x+y+2z)/2).((x+y-2z)/2)-((x-y)/2)^2=xy-z^2. With this one, we may prove that all possible counterexamples with z as small as possible have this number being odd. Could one prove this last case?

Hey Erken! Could you post the reference from which you`ve found this elementary proof?

So I have the proof of the case where "z"is even. With this, we easily note that every prime p such that (p-1)/4 is integer, is a sum of two integer Square numbers

@escadainfinita
If I am not mistaken I came up with this solution based on the Fermat's Descent Procedure which served as a key idea to solving one of the famous IMO 30 years back .There is a huge variety of similar problems out there utilising the very same idea.

P.S: It has been almost 8 years since I posted that solution :)

I am posting the statement I have proved. Suppose [i]x, y, z[i] are positive integers such that: xy=z^2+1, z being even with z the smallest value which is able to lead to a factorization, which shows a counterexample to the assertion of the problem. One can find positive integers: a, b, c such that: (1) x= (a+b+2c)/2; (2) y= (a+b-2c)/2; (3)    z= (a-b)/2, since: x+y>2z, because this turns into: (x-z)^2+1>0. Plugging these relations into the first one we get: ab= c^2+1. Now: c< (a-b)/2, because this yields to: a^2-6ba+b^2+4>0, which is true, since the polynomial: f(x) = x^2-6bx+b^2+4 has the discriminant being: 32b^2-16, which is always positive for positive integer b. Now, c is even, since: x-y is divisible by 4 (because both x, y are congruent to 1 (mod 4)). Since z is the smallest value which leads to some factorization which would shows a counterexample, we have: a= s^2+t^2; b= v^2+u^2; c= sv+tu, some integers: s, t, u, v. Plugging these into the formulas (1); (2); (3), we get: a= ((s+v+t+u)/2)^2+((s+v-t-u)/2)^2; b= ((s-v+t-u)/2)^2+((s-v-t+u)/2)^2; c= (s^2+t^2-u^2-v^2)/2= MP+NQ, where: M= (s+v+t+u)/2; N= (s+v-t-u)/2;
P= (s-v+t-u)/2; Q= (s-v-t+u)/2. Since a, b have the same parity, we easily get that {s,t,u,v} can be particioned in two subsets, forming two pairs, each of them having elements of same parity. This imples: M, N, P, Q, are integers. We have reached a contradiction, which shows there is no counterexample. This solves a particular case of the problem, namely: if x, y, z, are positive integers e xy=z^2+1, z being even, there exists integers a,b,c,d with: x= a^2+b^2; y= c^2+d^2; z=ac+bd
Note: notice this yields a very simple proof of the fact that every prime p such that (p-1)/4 is an integer, is the sum of the squares of two integer numbers. Just observe that for such primes p, p is a divisor of: [((p-1)/2)!]^2+1 (in view of the fact that: [(p-1)!+1]/p is an integer, for every prime p) and (p-1)/2 is even, since p>4. Apllying the statement I proved, we have: p=a^2+b^2, for some integers: a,b. :D [b] Does anyone has the proof for the case of z being odd?[/b]

Now, the uniqueness of: a,b, can be explained by other problem from the divisibility section

Observe that if we can show that $x,y$ are sum of two squares then $z$ takes the required form.Choose any prime divisor $p$ of $z^2+1$ then $p \equiv 1 \mod 4$. By Fermat's Christmas theorem, $\exists m,n \in \mathbb{N}$ such that $p=m^2+n^2$. Repeat this for all primes dividing $z^2+1$. Use the identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ repeatedly to express $x,y$ as sum of two squares. Some manipulation therefore leads to the conclusion.


Let $ a, b, c$ positive real numbers such that $ x^2z\plus{}y^2x\plus{}z^2y\equal{}3xyz$. Prove or disprove that

$ (x^3\plus{}y^2z)(y^3\plus{}z^2x)(z^3\plus{}x^2z)\le \frac{1}{2}(x^2z\plus{}y^2x\plus{}z^2y)^3$

_____
 :?:

[quote="Ligouras"]Let $ a, b, c$ positive real numbers such that $ x^2z \plus{} y^2x \plus{} z^2y \equal{} 3xyz$. Prove or disprove that

$ (x^3 \plus{} y^2z)(y^3 \plus{} z^2x)(z^3 \plus{} x^2z)\le \frac {1}{2}(x^2z \plus{} y^2x \plus{} z^2y)^3$

_____
 :?:[/quote]
I think that this is VERY EASY . May be you mean anoter inequality?

[quote="Ligouras"]Let $ a, b, c$ positive real numbers such that $ x^2z \plus{} y^2x \plus{} z^2y \equal{} 3xyz$. Prove or disprove that

$ (x^3 \plus{} y^2z)(y^3 \plus{} z^2x)(z^3 \plus{} x^2z)\le \frac {1}{2}(x^2z \plus{} y^2x \plus{} z^2y)^3$

_____
 :?:[/quote]
$ x^2z \plus{} y^2x \plus{} z^2y \equal{} 3xyz \Rightarrow x\equal{}y\equal{}z$ ;)

[quote="arqady"][quote="Mathias_DK"][quote="Ligouras"]Let $ a, b, c$ positive real numbers such that $ x^2z \plus{} y^2x \plus{} z^2y \equal{} 3xyz$. Prove or disprove that

$ (x^3 \plus{} y^2z)(y^3 \plus{} z^2x)(z^3 \plus{} x^2z)\le \frac {1}{2}(x^2z \plus{} y^2x \plus{} z^2y)^3$

_____
 :?:[/quote]
$ x^2z \plus{} y^2x \plus{} z^2y \equal{} 3xyz \Rightarrow x \equal{} y \equal{} z$ ;)[/quote]
No! It's known that  "$ a, b, c$ positive real numbers" :wink:  :D[/quote]
Haha! It's a trick inequality :P :P

[quote="Mathias_DK"][/quote][quote="arqady"][quote="Mathias_DK"][quote="Ligouras"]Let $ a, b, c$ positive real numbers such that $ x^2z + y^2x + z^2y = 3xyz$. Prove or disprove that

$ (x^3 + y^2z)(y^3 + z^2x)(z^3 + x^2z)\le \frac {1}{2}(x^2z + y^2x + z^2y)^3$

_____
 :?:[/quote]
$ x^2z + y^2x + z^2y = 3xyz \Rightarrow x = y = z$ ;)[/quote]
No! It's known that  "$ a, b, c$ positive real numbers" :wink:  :D[/quote][quote="Mathias_DK"]
Haha! It's a trick inequality :P :P[/quote]

___________
 :D

[quote="Nak"][quote="Ligouras"]Let $ a, b, c$ positive real numbers such that $ x^2z \plus{} y^2x \plus{} z^2y \equal{} 3xyz$. Prove or disprove that

$ (x^3 \plus{} y^2z)(y^3 \plus{} z^2x)(z^3 \plus{} x^2z)\le \frac {1}{2}(x^2z \plus{} y^2x \plus{} z^2y)^3$

_____
 :?:[/quote]
I think that this is VERY EASY . Maybe you mean anoter inequality?[/quote]

Dear NAK, I am satisfied of your answer. This inequality is the resolution of another very important inequality. I waited for feeling that the inequality is correct. Thank you very much


modular arithmetic
greatest common divisor
number theory unsolved
number theory

find all integers m and n such that $ 2^m\plus{}3^n$ is a square integer

All solutions $ (m,n,x)$ are given by $ (3,0,\pm 3),(0,1,\pm 2),(4,2,\pm 5)$. The proof is based on an exhaustive case analysis:

[b]Case 1:[/b] $ n\leq 0\leq m$ implies $ 1\equal{}3^{\minus{}n}(x^2\minus{}2^m)\Longrightarrow n\equal{}0<m\minus{}1$ and $ 2^m\equal{}(x\plus{}1)(x\minus{}1)\Longrightarrow x$ is odd and $ 2^{m\minus{}2}\equal{}\frac{x\plus{}1}{2}\frac{x\minus{}1}{2}$. Since the factors on the RHS have greatest common divisor of $ 1$, we have $ \frac{x\plus{}1}{2}\in\{\pm 1\}$ or $ \frac{x\minus{}1}{2}\in\{\pm 1\}$ and hence, $ x\equal{}\pm 3$ and $ m\equal{}3$

[b]Case 2: [/b]$ m\leq 0\leq n$ implies $ 1\equal{}2^{\minus{}m}(x^2\minus{}3^n)\Longrightarrow m\equal{}0<n$ and $ 3^n\equal{}(x\plus{}1)(x\minus{}1)\Longrightarrow x$ is even. Since the factors on the RHS have greatest common divisor of $ 1$, we have $ x\plus{}1\in\{\pm 1\}$ or $ x\minus{}1\in\{\pm 1\}$ and this gives $ x\equal{}\pm 2,n\equal{}1$

[b]Case 3:[/b] $ m,n<0$ implies $ 2^{\minus{}m}\plus{}3^{\minus{}n}\equal{}x^22^{\minus{}m}3^{\minus{}n}$. Since $ 2^{\minus{}m}$ is not divisible by $ 3^{\minus{}n}$, we get a contradiction.

[b]Case 4:[/b] $ m,n>0$ implies $ 6\not|x$ is odd and therefore $ 1\equiv x^2\equiv 2^m\equiv (\minus{}1)^m\bmod{3}\Longrightarrow m$ is even. Without restriction we can assume $ x>0$. Substitution of $ m\equal{}2k$ with $ k>0$ gives $ 3^n\equal{}(x\plus{}2^k)(x\minus{}2^k)$ and $ 2\not| gcd(x\plus{}2^k,x\minus{}2^k)|gcd(2x,2^{k\plus{}1})\equal{}2$, i.e. the factors on the RHS are coprime. This yields $ x\equal{}1\plus{}2^k$ and $ 3^n\equal{}1\plus{}2^{k\plus{}1}$. From this we conclude $ 1\equiv 3^n\equiv (\minus{}1)^n\bmod{4}\Longrightarrow n$ is even, too. Replacing $ n\equal{}2t$ with $ t>0$ translates into $ 2^{k\minus{}1}\equal{}\frac{3^t\plus{}1}{2}\frac{3^t\minus{}1}{2}$. Since the factors on the RHS are coprime, we must have $ 3^t\minus{}1\equal{}2\Longrightarrow t\equal{}1\Longrightarrow k\equal{}2$ and finally $ n\equal{}2,m\equal{}4,x\equal{}\pm 5$.

Suppose that $ A \equal{} 2^m \plus{} 3^n \equal{} x^2$

If $ m\leq 1$, $ m \equal{} 0$ since $ x^2 \equiv 0,1 \pmod{3} \forall x$. Now $ A \equal{} 1 \plus{} 3^n \equal{} x^2 \Rightarrow (x\minus{}1)(x\plus{}1) \equal{} 3^n$. Since $ \gcd(x\minus{}1,x\plus{}1) | 2$, $ x\minus{}1 \equal{} 1$ and $ x\plus{}1\equal{}3^n$, ...

If $ m \geq 2$, $ 2^m \equiv 0 \pmod{4}$. Hence $ A \equiv (\minus{}1)^{n} \pmod{4}$. This leads us to $ n$ is even. ...

I think that this way saves us some cases.


According to the World Almanac, in 1990, there were 41,223,804 students enrolled in full time day schools and there were 2,397,351 classroom teachers for these students. On average, how many pupils were there per teacher? Round your answer to the nearest tenth.

If you divide the one number by the other (the larger one by the smaller one), you get approximately $ \boxed{17.2}$ pupils per teacher as your answer.

Hi all, I found this problem on FTW when i choose Countdown problems only.
My question is: how would you do this in a Countdown round?
Would you just have to divide it on paper?

That is my question too. Hopefully after this bump someone will answer

I do not think this problem can be answered within 45 seconds without using a calculator.  Does anyone have a solution that does not require the use of a calculator?

You can try rounding, but it may not always work.

Doing $\frac{41200000}{2400000}$, or $\frac{412}{24}$ gives you $17.16$ repeating, or $17.2$ to the nearest tenth.


function algebra polynomial integration induction calculus derivative

Is there a non zero continuous function $ f : \mathbb{R}\to \mathbb{R}$, s.t. for all $ P\in \mathbb{R}[X]$, $ \int_{\mathbb{R}}|f(t)P(t)|dt<\plus{}\infty$ and $ \int_{\mathbb{R}}f(t)P(t)dt\equal{}0$?

it should suffice to have this equality for all $ P \in \mathbb{Q}[X]$, which is countable: $ \{P_1, P_2 , \ldots \}$
Construct a continuous function $ f$ such that $ f(n)\equal{}0$ for all $ n$ by induction:
- first construct it on $ [0;1[$ such that $ \int_{[0;1]} f(x) P_1(x) dx \equal{} 0$
- then on $ [0;2]$ such that $ \int_{[0;2]} f(x) P_1(x) dx\equal{} \int_{[0;2]} f(x) P_2(x) dx \equal{} 0$
- then on $ [0;3]$ etc..
one can moreover assume that $ |f(x)| < \exp(\minus{}x)$, say.

Actually, it's an orthogonality relation- just powers of $ x$ would be sufficient. There is also certainly no solution among the compactly supported functions.

Assuming that $ \int_{\mathbb{R}}|f(x)P(x)|\,dx < \infty$ for all polynomials is half of the restriction that goes into the Schwartz space; it's the right restriction on growth, but we have no guarantee of smoothness. Assume that smoothness anyway, so that $ \widehat{f}$ is also in this space and the inverse transform works well.
We have $ \int_{\mathbb{R}}x^nf(x)\,dx \equal{} 0$ if and only if $ \left.\frac {d^n}{ds^n}f'(s)\right|_{s \equal{} 0}$ Choose $ \widehat{f}$ to be a rapidly decaying nonzero $ C^{\infty}$ function with all of its derivatives zero at zero, and $ f$ its inverse Fourier transform.

One example: $ \widehat{f}(s) \equal{} \begin{cases}\exp\left(\frac {1}{ \minus{} s \plus{} s^2}\right) & 0 < s < 1 \\
\exp\left(\frac {1}{s \plus{} s^2}\right) & \minus{} 1 < s < 0 \\
0 & \text{otherwise}\end{cases}$
(This was chosen real and even so that $ f$ would be as well)
We then get
$ f(x) \equal{} \int_{\mathbb{R}}\widehat{f}(s)e^{2\pi ixs}\,ds \equal{} \int_{\mathbb{R}}\widehat{f}(s)\cos(2\pi xs)\,ds \equal{} 2\int_0^{\infty}\widehat{f}(s)\cos(2\pi xs)\,ds$
$ \equal{} 2\int_0^1\exp\left(\frac {1}{ \minus{} s \plus{} s^2}\right)\cos(2\pi xs)\,ds$

Your choice of conventions for the Fourier transform may change the details of this example, but the essence stays the same.

[quote="alekk"]
one can moreover assume that $ |f(x)| < \exp( \minus{} x)$, say.[/quote]
Really? ;)

[hide="Uh-huh."] If $ |f(x)|<e^{\minus{}a|x|}$, $ \widehat{f}$ is analytic on a strip in $ \mathbb{C}$ around the real axis. If its derivatives are all zero at some point, analyticity makes $ \widehat{f}$ identically zero, and thus $ f$ is zero.
We can (and must) have decay faster than any polynomial, but not as fast as any exponential.
[/hide]

OK, alekk's post is just fundamentally wrong.

[quote="jmerry"]
We have $ \int_{\mathbb{R}}x^nf(x)\,dx \equal{} 0$ if and only if $ \left.\frac {d^n}{ds^n}f'(s)\right|_{s \equal{} 0}$ [/quote]
I don't understand well if and only if  "what" ?

What about
$ I_{n}\equal{}\int_{0}^{\plus{}\infty }t^{n}e^{\minus{}\sqrt[4]{t}}\sin \left( \sqrt[4]{t}\right) dt\equal{}0$
change var
$ x\equal{}t^{\frac{1}{4}}$
give
$ I_{n}\equal{}4\int_{0}^{\plus{}\infty }x^{4n\plus{}3}\sin \left( u\right) e^{\minus{}u}du\equal{}0$

yes you are right  :clap2: , but you have no explanation, how  you  have solved it?
in your case it's hard to explain, since we just must to know such integrals.
jmerry  wrote full explanation  how he construct needed function.