Extremal combinatorics
number theory
Perfect Squares
Subset
IMO Shortlist

Assuming that [b]praneeth[/b]'s example is correct,we only need to establish that having eleven numbers in a such set is impossible.
Consider two cases:
1.One is in the set:The following two numbers can not be in the set simultaneously:
1.$ 2$ and $ 8$
2.$ 3$ and $ 12$
3.$ 4$ and $ 9$
Also if $ 5$ and $ 10$ are both in the set it follows that neither of $ 2$ and $ 8$ can be in the set.The same claim holds for pair $ 7$ and $ 14$.It means that $ 2$ and $ 8$ are not in the set,while numbers:
$ 1,5,6,7,10,11,13,14,15$ are in the set.Hence $ 12$ can not be in the set,but then $ 3\times 15\times 5\equal{}15^2$
2.One is not in the set.I need to think on this case a little bit more

Well we know $ 1$ isn't in the set from the post above. If $ 2$ is in the set then our set cannot contain both elements of any of the following pairs:
$ 3,6$
$ 4,8$
$ 5,10$
$ 6,12$
$ 7,14$
So if we have an 11-element set, $ 2$ cannot be in it.

So $ 1$ and $ 2$ are not in the set, and our set cannot fully contain any of the following triples:
$ 15,10,6$
$ 7,8,14$
$ 3,9,12$

So, that gives us 3 elements not in the set. With 1 and 2 not in the set either, we can have at most 10 elements in the set.

lol:
It's pretty funny that our solution consists from three parts suggested by three different people...

[quote="orl"]$ M$ is a subset of $ \{1, 2, 3, \ldots, 15\}$ such that the product of any three distinct elements of $ M$ is not a square. Determine the maximum number of elements in $ M.$[/quote]
Answer: 3.
Example: $ M \equal{} \{1,4,9\}$.

Assume $ |M| \geq 4$. Assume to the contrary that $ M$ contains at least 4 elements. Consider 4 elements $ a, b, c, d \in M$. Numbers $ abc$, $ abd$, and $ acd$ are perfect squares. Therefore, $ (abc)(abd)/(acd) \equal{} ab^2$ is a square of a rational number. Thus $ a$ is a perfect square. Similarly, $ b$, $ c$, and $ d$ are perfect squares. But there are only 3 distinct perfect squares between 1 and 15: 1, 4, and 9

Allnames, what happens when you add $ 5$ to your set?

Um, Allnames seems to be assuming that the problem involved sets such that the product of any three elements IS a perfect square, when it should be IS NOT a square.

Anyway, I don't see what happens when you add $ 5$ to the set, except that the set will no longer satisfy either condition.

sorry,because of my bad English

I tried it,and i found the result is $ 10$
But i think the problem that I solved is quite interesting

Hi! (excuse me if I make an english mistake  :maybe:  :P , I'm portuguese)

Isn't the right solution 11??
{1, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}

See you on the next post :)

Then we form a subset of co-prime integers excluding squares,i.e.
$$M=\{4,2,3,5,7,11,13\}$$
We now then see the possibility of a square in the set too.
$$M=\{1,2,3,5,7,9,11,13\}$$
Then we search for square free integers.
$$M=\{1,2,3,5,6,7,9,11,13\}$$
Oof other than case bash?

I have decided to dedicate my first ever post on AoPS to this problem because after 12 long years, no one has found the solution
$$M=\{3, 4, 5, 6, 7, 9, 10, 11, 13, 14\}$$
And there is not 1, not 2, not 3, but $\textbf{4}$ missing solutions!
$$M=\{1, 4, 5, 6, 7, 10, 11, 12, 13, 14\}$$
$$M=\{1, 5, 6, 7, 9, 10, 11, 12, 13, 14\}$$
$$M=\{4, 5, 6, 7, 9, 10, 11, 12, 13, 14\}$$
Here is my solution.
Let $S$ denote the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$. We will first simplify $S$ to a multiset $S'$. Note that for each element of $S$, if we multiply or divide it by a perfect square, whether its product with $2$ other elements is a perfect square or not does not change. Thus, for each number $n$ in $S$, if there is a perfect square $k^2>1$ which divides it, we replace it with $\frac{n}{k^2}$. This yields
$$S'=\{1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15\}$$
or
$$S'=\{1, 1, 1, 2, 2, 3, 3, 5, 7, 11, 13, 2\times 3, 2\times 5, 2\times 7, 3\times 5\}$$
(I have prime factorised the numbers for easier reference)
Define $M'$ similarly for $M$. We will prove that the upper bound of $|M'|$ is $10$ by proving we need to eliminate at least $5$ elements from $S'$ to satisfy the no perfect-square condition. Let $(x, y, z)\Rightarrow -1\hspace{0.2cm}(n)$ denote that because of the perfect-square triplet $(x, y, z)$, we have to eliminate at least $1$ element, leaving us with at most $n$ elements remaining.
$(1, 1, 1)\Rightarrow -1\hspace{0.2cm}(14)$
$(2, 7, 2\times 7)\Rightarrow -1\hspace{0.2cm}(13)$
Now we diverge into $2$ cases.
Case $1$: $2$ was removed because of $(2, 7, 2\times 7)$
Since there are $2$ $'2'$s, that means 2 elements were removed leaving us with $12$ elements.
$(2\times 3, 2\times 5, 3\times 5)\Rightarrow -1\hspace{0.2cm}(11)$
$(1, 3, 3)\Rightarrow -1\hspace{0.2cm}(10)$ (We either have to remove both $'1'$s because of the double $3$, or remove one $'3'$.)
Case $2$: $'2'$ was kept
$(2, 5, 2\times 5)\Rightarrow -1\hspace{0.2cm}(12)$
$(2, 3, 2\times 3)\Rightarrow -1\hspace{0.2cm}(11)$
Note that for the triplet $(2, 3, 2\times 3)$, if $'3'$ was removed that means $2$ elements were removed instead of $1$, leaving us with $10$ elements and we are done. Otherwise, the element $2\times 3$ was removed. Then,
$(1, 3, 3)\Rightarrow -1\hspace{0.2cm}(10)$ (Reason is similar to Case $1$)
Thus $|M'|\le 10$ and we are done.
As a bonus we will prove that $M'=\{1, 1, 3, 5, 7, 11, 13, 2\times 3, 2\times 5, 2\times 7\}$ is the only possible solution for $M'$. It then follows that there are $6$ solutions for $M$ - the $\{1, 1\}$ in $M'$ can become $\{1, 4\}$, $\{1, 9\}$, or $\{4, 9\}$ in $M$, and the $\{3\}$ can become $\{3\}$ or $\{12\}$.
In Case $1$, if we have $|M'|=10$ it means we are keeping one $'3'$ and one $'5'$, which means out of the $3$ elements $2\times 3$, $2\times 5$, and $3\times 5$, we have to remove $3\times 5$. The resultant $M'$ is the solution above.
In Case $2$, having $|M'|=10$ means that we must keep both $'2'$s and $2$ $'1'$s. But the triplet $(1, 2, 2)$ shows that this is impossible.
We are done. :)

Dang this problem is bash...

Here goes

The answer is $\boxed{10}$. It is easy to see that this is achievable(and not too hard to find) that a set that works is 1,3,5,6,7,9,10,11,13,14. It is easy to check that this works by eliminating numbers based on common factor(for example, for $5$ to possibly be a perfect square, $10$ would also have to be part of the 3 element product, and then there a only a couple of cases to check).

We now show that $10$ is optimal. Notice that at most $2$ elements in each of the sets {1,4,9}, {3,5,15}, {2,6,12}, and {7,8,14} can be used. This gives us a bound of $11$, still not strong enough. Now hound in on the two sets {2,6,12} and {3,5,15}. If we can somehow show that among these two sets, at most $3$ elements can be used, then we will be done. To show this, suppose we could choose at least $4$ elements among these two sets somehow. Then clearly this would be by choosing two from one set and two from another. The possible sets of four elements are {2,6,3,5} {2,6,3,15} {2,6,5,15} {2,12,3,5} {2,12,3,15} {2,12,5,15} {6,12,3,5} {6,12,3,15} {6,12,5,15}. Notice the first two sets are invalid because they contain {2,3,6}. Also notice that the 4th, 5th, 7th, and 8th cases are bad because they contain {3,12}, which means that none of elements in the set {1,4,9} can be used, already strong enough to prove the bound. Also notice that the 6th and 9th cases are bad because they contain {5,12,15} which product to a perfect square. Therefore, the only possible set that can have more than 10 elements is {2,6,5,15}. However, notice that this set means that in the set {7,8,14}, I must choose the 8, or else 7,14, and 2 would product to a perfect square, but with a 2 and an 8, then {1,4,9} would not work, already strong enough. So now we are done.

$\color{magenta} \boxed{\textbf{SOLUTION N1}}$

Let, $|M|$ denotes the number of elements in Set $M$

$\color{red} \textbf{CLAIM 1 :}$ If $|M|=11$ doesn't work then $|M|=12,13,14,15$ also doesn't work
$\textbf{Proof :}$ Consider any $11$ elements from $M \ge 12$ We must find $3$ elements $a,b,c$ from $M$ such that $abc=x^2$ as we assumed $|M|=11$ doesn't work that is for every $|M|=11$ we will get three elements whose product is a perfect square $\blacksquare$

$\color{red} \textbf{CLAIM 2 :}$ $|M|=11$ doesn't work
$\textbf{Proof :}$
Consider,
$$A_1=(2,1,8)$$
$$A_2=(2,3,6)$$
$$A_3=(2,4,8)$$
$$A_4=(2,5,10)$$
$$A_5=(2,6,12)$$
$$A_6=(2,7,14)$$
$$A_7=(2,8,9)$$

Let, $2 \in M$
Then We have $6,8 \notin M$
Also one of $(5,10)$ and one of $(7,14)$ can't be in $M$
Then Consider $T={1,4,9}$ All of these must be in $M$ but $1\times 4\times 9=36=6^2$ Contradiction!

So, $2 \notin M$

Now,
Consider,
$$B_1=(3,1,12)$$
$$B_2=(3,2,6)$$
$$B_3=(3,4,12)$$
$$B_4=(3,5,15)$$
$$B_5=(3,6,8)$$
$$B_6=(3,9,12)$$

Let, $3\in M$
Then we have $6,12 \notin M$
And one of $(5,15)$ not in $M$
Again consider $T=(1,4,9)$ all of these must be in $M$ Contradiction!

So, $3\notin M$

We want to find two more elements which can't be in $M$

Consider
$$T=(1,4,9)$$
$$N=(5,8,10)$$
$$K=(5,12,15)$$
$$L=(7,8,14)$$

Let, $5\in M \implies$ one of $(8,10)$ and one of $(12,15)$ can't be in $M$
But Then again consider $T=(1,4,9)$ Contradiction!

So, $5 \notin M$

Now, Consider $T,L$
We we have $a\notin M, a\in T$ then $L=(7,8,14) \in M$ Contradiction!
And if we have $b\notin M, b\in L$ then $T=(1,4,9) \in M$ Contradiction!

Therefore We get, $2,3,5 \notin M$ and one of $(7,8,14)$ and one of $(1,4,9)$ not in $M \blacksquare$

From these We can construct $|M|=10$ elements cancelling $1,2,3,5,7$
$$M=(4,6,8,9,10,11,12,13,15) \blacksquare$$


http://mathworld.wolfram.com/PocklingtonsCriterion.html

is there an error, or I don't understand something ?

taking $ a\equal{}1$ (integer...) I'll have $ GCD(a^k\plus{}1,N)\equal{}GCD(2,N)\equal{}1$ for all $ k,p$.

And so $ 2kp\plus{}1$ will be prime for all $ k,p$

See [url=http://www.physicsforums.com/archive/index.php/t-128512.html]this thread[/url].  Apparently the Mathworld entry is missing two conditions.


Για $ x,y,z>0$ να αποδείξετε ότι :

$ \frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(y\minus{}z)^2}\plus{}\frac{1}{(z\minus{}x)^2}\geq \frac{4}{xy\plus{}yz\plus{}zx}$

Βρήκα μια αρκετά ενδιαφέρουσα λύση . Έστω z ο μικροτερος από τους αριθμούς .
Σταθεροποιώ τις διαφορές x-y ,y-z ,z-x και αφαιρώ από τους x,y,z την ποσότητα z δλαδή τελικά θα φτάσουμε (x-z,y-z,0).

Μετά το μετασχηματισμό το αριστερό μέλος δεν μεταβάλλεται ενώ $ xy\plus{}yz\plus{}zx\geq (x\minus{}z)(y\minus{}z)$ άρα το δεξί αυξάνεται .
Επομένως αρκεί να δείξουμε το πρόβλημα για $ x,y\geq z\equal{}0$

Τότε θέλουμε να δείξουμε ότι $ xy(\frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{x^2}\plus{}\frac{1}{y^2})\geq 4$

ή  αν $ k\equal{}\frac{x}{y}$  έχουμε νδο $ \frac{k}{(k\minus{}1)^2}\plus{}k\plus{}\frac{1}{k}\geq 4$

ή ισοδύναμα $ \frac{(k^2\minus{}3k\plus{}1)^2}{(k\minus{}1)^2k}\geq 0$ που ισχύει  :)

Silouan, μήπως θεωρείς κάποια γενικότερη περίπτωση για τη λύση σου (x,y,z > 0 και άρα δεν μπορείς να θέσεις z=0); Απ' ότι καταλαβαίνω, την αποδεικνύεις για όταν δύο από τα $ x,y,z$ είναι θετικοί και το τρίτο μη αρνητικό, του οποίου είναι υποπερίπτωση το να είναι όλα θετικά. Αλλά τότε δεν μπορείς να εφαρμόσεις τις αρχές της συμμετρίας, γιατί οι μεταβλητές δεν ορίζονται με τον ίδιο τρόπο. Τι κάνεις τελικά; :huh: Η λύση κατά τα άλλα είναι αξιοθαύμαστη.  :lol:


limit
calculus
derivative
logarithms
calculus computations

Does the series $ \sum \left[1\minus{}\left(1\minus{}\frac {1}{n^2}\right)^n\right]$ converge?

Hi :)!!

$ [.]$ mean the part integr ????

OR

 you want to say that the series $ \sum_{n \geq 1} \left(1 \minus{} \left( 1 \minus{} \frac {1}{n^2} \right)^n \right)$ converge?

and thinks

[hide]From some numerical investigations, I guessed that the series was limit comparable to $ \sum_{n \equal{} 1}^\infty \frac {1}{n}$. To check this I computed the following limit:

$ \lim_{n \to \infty} \frac {1 \minus{} \left(1 \minus{} \frac {1}{n^2}\right)^n}{\frac {1}{n}} \equal{} \lim_{x \to 0^ \plus{} } \frac {1 \minus{} (1 \minus{} x^2)^{1/x}}{x}$
Using L'Hopital's Rule and logarithmic differentiation, this limit becomes
$ \equal{} \lim_{x \to 0^ \plus{} } \minus{} (1 \minus{} x^2)^{1/x} \left( \minus{} \frac {1}{x^2}\ln (1 \minus{} x^2) \plus{} \frac {1}{x} \cdot \frac { \minus{} 2x}{1 \minus{} x^2} \right)$

Each of these parts of the limit can be checked to exist and have the following values:
$ \lim_{x \to 0^ \plus{} } \minus{} (1 \minus{} x^2)^{1/x} \equal{} \minus{} 1$
$ \lim_{x \to 0^ \plus{} } \minus{} \frac {1}{x^2}\ln (1 \minus{} x^2) \equal{} 1$
$ \lim_{x \to 0^ \plus{} } \frac {1}{x} \cdot \frac { \minus{} 2x}{1 \minus{} x^2} \equal{} \minus{} 2$

Therefore the limit exists (and is not zero) -- so the series in question is limit comparable to the Harmonic Series, and hence diverges.[/hide]

we can prove with taylor series that:

$ 1 \minus{} (1 \minus{} \frac {1}{n^2})^n \sim \frac {1}{n}$


What is the remainder when 13^13 +5  is divided by 6?

Is there any good way to approach this problem mentally? Just working on speed and some new techniques...  :lol:

I cannot see the number...but, for any number when divided by 6:

check the remainder when its divided by 3

check if it is even or odd

combine these two to get what the number is in mod 6

If you want me to explain in more detail feel free to ask

$ 13^{13} \plus{} 5 \equiv 1^{13} \plus{} 5 \equiv 6 \equiv \boxed{0} \pmod 6$.

When you're doing this mentally, try to recognize any patterns between the numbers. Here, $ 13$ is one more than a multiple of $ 6$, and $ 1$ raised to any power is $ 1$.


Yes, I applied. I'm probably not going to get in. Is the application process extremely competitive?

:lol: Someone replied.
Well they told us for 2006 that there were 400 applicants and only 100 made it.
So, in that sense, no it is not very competitive.

thanks! I suppose it depends on what country I'm from.

[quote="AstroPhys"]I went to this program last summer, and it was beyond awesome.  It is held at Waterloo University in Canada in conjunction with Perimeter Institute.  Anyone from around the world can apply.

Is anyone applying to this year's program, or has anyone been there in the past?[/quote]

What is young in this context? High school students I suppose?

[quote]The program is open to students currently enrolled in Grade 11[/quote]

However, the deadline has passed.

It also happens to be where Astrophys got his avatar.
[url]http://www.perimeterinstitute.ca/index.php?lang=en[/url]

color, yes, by young I meant High School Students.

stuck, the deadline did pass and this year's participants are listed on the website.

ISSYP is under Outreach, High School Students.


Consider the number $ x\equal{}\sqrt{8\minus{}2\sqrt{15}}\minus{}\sqrt{8\plus{}2\sqrt{15}}$. Compute $ x^2$.

hello, using the binomial formula we get
$ x^2\equal{}8\minus{}2\sqrt{15}\plus{}8\plus{}2\sqrt{15}\minus{}2\sqrt{64\minus{}4\cdot 15}$
$ x^2\equal{}16\minus{}4\equal{}12$
Sonnhard.

Also:

\[ \sqrt{8\minus{}2\sqrt{15}} \equal{} \sqrt{(\sqrt{5}\minus{}\sqrt{3})^2}\equal{}\sqrt{5}\minus{}\sqrt{3}\]

\[ \sqrt{8\plus{}2\sqrt{15}} \equal{} \sqrt{(\sqrt{5}\plus{}\sqrt{3})^2}\equal{}\sqrt{5}\plus{}\sqrt{3}\]

therefore \[ x\equal{}\sqrt{5}\minus{}\sqrt{3}\minus{}\sqrt{5}\minus{}\sqrt{3} \implies \minus{} 2\sqrt{3} \implies x^2\equal{}12\]

that is exactly how they did it, except they skipped the obvious steps.


geometry
circumcircle
geometric transformation
reflection
trapezoid
trigonometry
abstract algebra

Let $ ABCDE$ be a cyclic pentagon such that $ AC\parallel DE$, and let $ M$ be the midpoint of the segment $ BD$. If $ \angle AMB \equal{}\angle BMC$, then prove that the line $ BE$ bisects the segment $ AC$.

Let $o$ be the circumcircle of $ABCDE$, let $l$ be the perpendicular bisector of $AC$. Points $A,C$ and $E,D$ are pairwise symmetrical about $l$.
Points $M',B'$ are symmetric to $M,B$ about $l$ - respectively. Circle $O$ passes throught points $A,M,M',C$. Let's name point $D$ as the point on circle $O$ where lines $EM'$ and $CM$ intersect (they intersect because $arcDC=arcAD$).
But also point $D$ is the point of intersection of lines tangent to $o$ in points $A$ and $C$. So $AC$ is the polar line, but the polar line must pass through the point of intersection of $EB$ and $DB'$. But because these lines are simmetric about $l$ thus they must pass through the middle of $AC$, so indeed line $EB$ bisects $AC$.

let A' be reflection of C across perpendicular bisector of BD, then A,M,A' collinear.
BM^2=BM.MD=AM.MA=AM.MC, so AM/MB=BM/MC so AMB~BMC.
<MAC=<MBC(similarity)=<DBC=<DAC. subtract <MAC from both sides and get <DAM=<CAB=a.
let <MAB=<CMB=b. Now CA'||BD as A' is the reflection of C across perpendicular bisector of BD.
so <CA'A=<BMA=b.
<CDB=<CBA=a. <ACB=<ADB=<ADC-<CDB=<CAA'-<CDB=b-a. <MCD=<BMC-<MDC=b-a.
so AMD~DMC~ABC(AA)

thus CD/AD=DM/AM=BC/AB.
BC/AB=sin<ABE/sin<CBE      (1)
AB/AX=sin<AXB/sin<ABE      (2)
BC/CX=sin<AXB/sin<CBE      (3)

(1)(2)/(3) yields AX=CX.

-Sam.

oh yeah i should explain how i got (1):

AC||DE and ACDE is cyclic, so ACDE is an isosceles trapezium, so CD=AE and AD=CE.
note also <ACE=<ABE and <CAE=<CBE. i used sine rule on ACE to get the trig expression.

Who is Sam...

Well, I am what not I am yet am who I am.

BOmb

[quote="bomb"][color=darkred]Let $ ABCDE$ be a cyclic pentagon such that $ AC\parallel DE$, and let $ M$ be the midpoint of the segment $ BD$. If $ \angle AMB = \angle BMC$, then prove that the line $ BE$ bisects the segment $ AC$.[/color][/quote]
[color=darkred][b]The your proposed problem is very nicely and no difficultly (yet no easily)![/b][/color]

[color=darkblue][b]I hope that and the my proof will be identically, i.e. nicely, shortly and somehow easily.[/b][/color]

[color=darkblue][b][u]Proof.[/u][/b] Denote the circumcircle $ w = C(O)$ of the given pentagon and the intersections $ \{\begin{array}{c}N\in AC\cap BE \\
\ S\in AC\cap BD \\
\ T\in AC\cap OM\end{array}$. Therefore, the point $ T$ is harmonical conjugate of the point $ S$ w.r.t. the pairs of points $ \{A,C\}$. From $ OM\perp BD$ obtain that the line $ BD$ is the polar of the point $ T$ w.r.t. the circle $ w$. Thus, $ T\in BB\cap DD$ and $ \frac {SA}{SC} = \frac {TA}{TC} = (\frac {BA}{BC})^{2}$, i.e. the point $ S$ is the foot of the $ A$- symmedian in the triangle $ ABC$. But $ AC\parallel ED\Longrightarrow\widehat{ABN}\equiv\widehat{CBS}$. In conclusion, from the [b]Steiner's theorem[/b] obtain that the point $ N$ is the middle of the diagonal $ [AC]$.[/color]

[color=darkblue][b]Remark.[/b] The "kernel" of the above proof is the following cleaner enunciation :[/color]

[color=darkred][b]Let $ w = C(O)$ be the circumcircle of $ \triangle ABC$ and let $ P\in w$ be the point for which the middlepoint $ M$ of the segment $ [AP]$
has the property $ \widehat{PMB}\equiv\widehat{PMC}$. Prove that the line $ BC$ separates the points $ A$, $ P$ and the lines $ AA$, $ OM$, $ BC$ are concurrently.[/b][/color]

[color=indigo][b]Thank you, Virgil Nicula! Your proof is very nice and short! My proof have a little different.[/b][/color]

[quote="bomb"][color=darkred]Let $ ABCDE$ be a cyclic pentagon such that $ AC\parallel DE$, and let $ M$ be the midpoint of the segment $ BD$. If $ \angle AMB \equal{} \angle BMC$, then prove that the line $ BE$ bisects the segment $ AC$.[/color][/quote]
[color=indigo][u][b]Proof.[/b][/u] (I'll use [b]Virgil Nicula's[/b] notations and his result: $ T \equal{} BB\cap DD$)

We have: the point $ T$ is harmonical conjugate of the point $ S$ w.r.t. the pairs of points $ \{A,\,C\}$ and $ TD$ is a tangent of the circle $ (O)$, so:
$ TS\cdot TN \equal{} TC\cdot TA \equal{} TD^{2} \equal{} TM\cdot TO\Longrightarrow\triangle TMS\sim\triangle TNO$ $ \Longrightarrow ON\perp TN$

And hence $ N$ is the midpoint of $ AC$ [/color]

the harmonic range solution screams out at me, but does anybody want to offer a complex number solution?

sexy solution
$ BE$ bisect $ AC$ iff $ [BAE] \equal{} [BCE]$([ ] mean area). Since $ \sin\angle BAE \equal{} \sin\angle BCE$, it is suffice to show $ AB\cdot AE \equal{} BC\cdot CE$ or $ AB\cdot CD \equal{} BC\cdot AD.($since $ AE \equal{} CD,CE \equal{} AD).$ Let $ AM$ meet circle at $ X$. Then $ \angle CMB \equal{} \angle XMD$. Hence $ CX\parallel BD$. Since  $ AX$ bisects $ BD$, $ [ABX] \equal{} [ADX].$ Hence $ AB\cdot BX \equal{} AD\cdot DX$. Hence $ AB\cdot CD \equal{} AD\cdot BC$

Let $ AC\cap BD\equal{}\{Q\}$. Let $ O$ the circumcircle of $ ABCD$ and let $ OM\cap AC\equal{}\{S\}$.
Then because $ \angle AMB\equiv \angle BMC$ and $ BM\perp MS$, we obtain that $ S,Q,C,A$ is a harmonic division.
Let $ T$ and $ T'$ be the tangents from $ S$ to the circumcircle of $ ABCD$. Let $ AC\cap TT'\equal{}Q'$.
Then because $ TT'$ is the polar of $ S$, we obtain $ S, Q', C, A$ to be a harmonic division. But this means $ Q\equal{}Q'$.
Furthermore, $ OS\perp TT'$ and $ OS\perp BD$. This means that $ TT'$ and $ BD$ are parallel. But $ TT'\cap BD\equal{}Q\equal{}Q'$, therefore $ TT'\equiv DB$, thus $ S$ is the pole of $ BD$, which is equivalent to the fact that $ ABCD$ is a harmonic quadrilateral, i.e. $ AB\cdot CD\equal{} AD\cap BC$.


friends but why are thre 2 friend and 2 teachers.

I found it by serching in the net. If I am not mistaken I was looking for the IMO shortlistes  :P .

I found it by looking up problem solving teqneques.

A friend told me.
You should change it to Mathlinks/AoPS and delete the extra friend and teacher tags.

I was told about it a training camp somewhere, can't remember when...

Why should it be changed to mathlinks/aops? It's fairly self-explanatory that it means both, and people ask the same sort of questions saying just AoPS (no mathlinks), like http://www.mathlinks.ro/Forum/viewtopic.php?highlight=aops&t=113304 for example.

GOOGLE

It was a dark and stormy night on September 28, 2006......

The day of the Naturalistic Fall Startup Event.....  :o

Our teacher told my friend and I to enroll in teh Math Counts Series.

i was looking for the junior tests for the JBMO ...

I have seen this some time before..

http://www.mathlinks.ro/Forum/viewtopic.php?t=113304

I know it from AoPS but I don't go on MathLinks

I asked the vice-principle if I could take a harder math course and she recommended me AoPS.


induction
group theory
abstract algebra
superior algebra
superior algebra theorems

$A,B,C$ are three groups. $A\times B\cong A\times C$. Do we have $B\cong C$?

Of course not :).

Just take $A=\mathbb Z_2\times\mathbb Z_2\times\ldots,\ B=\mathbb Z_2,C=\mathbb Z_2\times\mathbb Z_2$.

[quote="Leva1980"]But i the case of finite groups it is true![/quote]
Can you write your proof?

First of all, observe that for every groups $M,N,L$ we have $
Hom(M,N \times L) \cong Hom(M,N) \times Hom(M,L) $, so $ \sharp
Hom(M,N \times L) = \sharp Hom(M,N) \cdot \sharp Hom(M,L) $. So,
because of this, in our case we have $ A \times B \cong A \times C
$ , so for every another group $ D $ , $ \sharp Hom(D,A \times B)
= \sharp Hom(D,A \times C) $, so for every finite group $D$, $
\sharp Hom(D,B) = \sharp Hom(D,C) $. Now, by induction on number
of elements in $ D $, we prove that number of injective morphisms
from $ D $ to $ B $ equals to number of injective morphisms from $
D $ to $ C $. Indeed, denote the number the injective
homomorphisms from $ M $ to $ N $ by $ Hominj(M,N) $. Then for
every finite group $ D $, $ \sharp Hom(D,B) = \sum \sharp
Hominj(D/N,B) $, where summation goes through all normal subgroups
$ N \vartriangleleft D $. Also, $ \sharp Hom(D,C) = \sum \sharp
Hominj(D/N,C) $. Now, in the summation we have the trivial normal
subgroup $ {e} \vartriangleleft D $ and all others. For every
other (nontrivial) subgroup $ N \vartriangleleft D $, by
induction, we have $ \sharp Hominj(D/N,B) = \sharp Hominj(D/N,C)
$. Also we know that $ \sharp Hom(D,B) = \sharp Hom(D,C) $ .
Therefore $ \sharp Hominj(D,B) = \sharp Hominj(D,C) $ . So we have
proved it for every finite group $ D $. Now take $ D = B $. Of
course $ Hominj(D,B) $ is nonempty, because it contains the
identity. So $ Hominj(D,C) $ is nonempty. So $ Hominj(B,C) $ is
nonempty, so we have a monomorphism $ f: B \rightarrow C $. But it
is automatically an isomorphism, because $ B $ and $ C $ have the
same number of elements. So $ B $ and $ C $ are isomorphic.


1) When I was viewing the Resources, some images are marked "External_image". When I view with Quote where the image link is and try to retrieve the image, in which the image is linked to somewhere in ML, the image no longer exists... Is there a way to retrieve back those old images?

2) For example, like this announcement topic
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=28406[/url]
you will see that there is supposed to be the FAQ image on there but it shows a boxed text of "External_image" and I am sure the FAQ image [u]is[/u] an image in ML but considered external? Is this some bug where we can no longer incorporate external images, or even images within Mathlinks???


Find at least one positive integer solution of the equation:
$k^{57}+l^{18}=m^{13}$.


I need help proving the following:
1) Given $ \forall n \in N$,
    a) $ \left( 1\plus{} \frac{1}{n}\right)^{n} < \left( 1\plus{} \frac{1}{n\plus{}1}\right)^{n\plus{}1}$
    b) $ \left( 1\plus{} \frac{1}{n}\right)^{n} < 3\minus{}\frac{3}{n\plus{}2}$
    c) $ \left( \frac{a\plus{}b}{2}\right)^{n} \le \frac{a^n\plus{}b^n}{2}$  for $ a,b>0$
    d) $ \left( \frac{n(n\plus{}1)^3}{8}\right)^{n} \ge (n!)^4$, for $ n \in N$ and $ n\ne 1$
    e) $ \frac{a^n}{b\plus{}c}\plus{}\frac{b^n}{a\plus{}c}\plus{}\frac{c^n}{a\plus{}b}\ge \frac{3}{2}\left(\frac{a^n\plus{}b^n\plus{}c^n}{a\plus{}b\plus{}c}\right)$, for $ a,b,c>0$

Are you looking for an elementary solution?

Note that $ \lim_{n\to \infty} \left(1\plus{}\frac{1}{n}\right)^n \equal{} e$, which leads me to believe this is a calculus question...

Ocha, the inequalities satisfy for all $ n$ natural numbers. So, the interest is not in the limit.

[hide="c"]Taking the $ n$th root of each side, we see this is true by the power mean inequality.[/hide]

[quote="kcstevonne"]I need help proving the following:
1) Given $ \forall n \in N$,
          c) $ \left( \frac {a \plus{} b}{2}\right)^{n} \le \frac {a^n \plus{} b^n}{2}$  for $ a,b > 0$
    [/quote]

[hide="c."]
 c) $ \left( \frac {a \plus{} b}{2}\right)^{n} \le \frac {a^n \plus{} b^n}{2}$  for $ a,b > 0$

WLOG let $ a\ge b, s\equal{}\frac{a\plus{}b}{2}, d\equal{}\frac{a\minus{}b}{2}$

Now proving L.H.S $ \ge$ R.H.S.

$ \frac {a^n \plus{} b^n}{2}\equal{}\frac{(s\plus{}d)^n\plus{}(s\minus{}d)^n}{2}$

$ \equal{}\frac{(s^n\plus{}{}_nC_1s^{n\minus{}1}d\plus{}{}_nC_2s^{n\minus{}2}d^2\plus{}\cdots\plus{}d^n)\plus{}(s^n\minus{}{}_nC_1s^{n\minus{}1}d\plus{}{}_nC_2s^{n\minus{}2}d^2\plus{}\cdots\plus{}(\minus{}1)^nd^n)}{2}$

$ \equal{}s^n\plus{}{}_nC_2s^{n\minus{}2}d^2\plus{}{}_nC_4s^{n\minus{}4}d^4\plus{}\cdots\plus{}\left[d^n\plus{}(\minus{}1)^nd^n\right]\ge s^n\equal{}\left( \frac {a \plus{} b}{2}\right)^{n}$

$ (\because s>0 ,d\ge 0)$

The equality holds when $ a\equal{}b$
[/hide] :lol:

[hide="solution e"]
Assume w.l.o.g $ a\ge b \ge c$, which implies $ a^n\ge b^n \ge c^n$ and $ \frac {1}{b \plus{} c} \ge \frac {1}{a \plus{} c} \ge \frac {1}{a \plus{} b}$

So by Chebyshev
$ \sum \frac {a^n}{b \plus{} c} \ge \frac {1}{3} (a^n \plus{} b^n \plus{} c^n)\left(\frac {1}{b \plus{} c} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{a \plus{} b}\right)$

So it suffices to prove that

$ (a \plus{} b \plus{} c) \left(\frac {1}{b \plus{} c} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{a \plus{} b}\right) \ge \frac {9}{2}$

re-writing as;

$ ((a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a))\left(\frac {1}{b \plus{} c} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{a \plus{} b}\right) \ge 9$

This is true by Cauchy-Schwars  :lol:
[/hide]

[hide="for b"]

\[ {(1 \plus{} \frac {1}{n})}^n < e < 2,8\] for all positive reals so trying 1,2,3,4,5,6..,15 gives result : S i know it s very ugly : S

[/hide]


Nessler's Reagent gives Brown ppt of Iodide of millon's base with ammonia solution and in presence of NaOH.

Does it give any ppt or colour in acidic solution? :huh:  :huh:

this question came in one of our exams.......

@carcul--can you please try!!!!!!!!!!

1) What is Nessler's Reagent?

2) What is Millon's base?

(Explain this by yourselves, do not provide just links.)

Nessler's Reagent is K2HgI4.

Millon's base is NH2-Hg-O-Hg-I (This is in alkaline solution)

So will it give any colour in acidic solution? :help:

the precipitate is basic mercury(II) amido-iodine.
hence it cannot be precicpitated in acidic medium.

also bases are req. to displace nh3(which reacts with nesslers reagent) from ammonium salts. acids cant do that.


projective geometry
geometry unsolved
geometry

Dear All My Friends,
In convex quadrilateral $ ABCD$: $ P\equal{}AC \bigcap BD$. There are four points $ K$, $ L$, $ M$, $ N$ on sides $ AB$, $ BC$, $ CD$, $ DA$ respectively and these four points are concyclic on a circle $ \omega$. Moreover $ K$, $ P$, $ M$ are collinear, $ L$, $ P$, $ N$ are collinear. Suppose the circle $ \omega$ intersects sides $ AB$, $ BC$, $ CD$, $ DA$ again at $ K_{1}$, $ L_{1}$, $ M_{1}$, $ N_{1}$ respectively.
Prove that: $ K_{1}$, $ P$, $ M_{1}$ are collinear, $ L_{1}$, $ P$, $ N_{1}$ are collinear.
Best regards,
Bui Quang Tuan

Dear Mathlnikers,
Please post proofs for this interesting problem.  :D

Dear friends, the following solution was presented by our IMO'08 contestant: [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=23362]Nguyen Pham Dat[/url] in a Vietnamese Mathematics Forum.

Let $ Q$ be the intersection of $ AB$, $ CD$ and $ R$ the intersection of $ AD$, $ BC$. We have $ Q(DAPR)$ is a harmonic bundles, so we conclude that $ QP$ and $ QR$ divide $ KM$ harmonically. Similarly, $ RP$, $ RQ$ divide $ LN$ harmonically, so we conclude that $ QR$ is the polar line of $ P$ w.r.t. the circle $ (KLMN)$, i.e., $ P$ lies on the polar line of $ Q$ relative to $ (KLMN)$.

Now, assume that $ K_1M_1$ intersects $ KM$ in a point $ S$ ($ S\neq P$). It's well-known that $ S$ lies on the polar line of $ Q$ w.r.t. $ (KLMN)$, therefore we conclude that $ PS\equiv KM$ is the polar line of $ Q$ relative to $ (KLMN)$, or we can say that $ QK$, $ QM$ are tangent to $ (KLMN)$, so $ K\equiv K_1$, $ M\equiv M_1$. This completes our solution.


If $ 3 \times 11 \times 13 \times 21 \equal{} 2005 \plus{}b$, what is the value of $ b$?

$ 3 \cdot 11 \cdot 13 \cdot 21$ is equal to $ 9009$.  Subtract $ 2005$ from that, to get the answer of $ \boxed{7004}$.


Starting with benzene, show how you could prepare 3-ethyltoluene.

(Source: adapted from Problem 15.19, Solomons 7th edition, page 700)

Carcul

Sorry to bring u old topic :blush:
Just wanted to answer

1. React with HCOCl in presence of AlCl3 to get Ph-CO-H
2. React with C2H5Cl in presence of AlCl3
3. Wolff Kishner reduction

formylation with HCOCl isnot possible, as it is very unstable. But you can use dry Hcl and CO, along with cu2cl2(gattermann modification).

[quote="Carcul"]No one cares to solve these organic synthesis problems?[/quote]
 :rotfl:  :rotfl:  :rotfl:  :P

[quote="Carcul"]On the other hand, formylation with HCOF is possible.[/quote]

What is the reason that HCOF can be formylated while HCOCl cannot?

[quote="Carcul"]Because HCOF is more stable.[/quote]

But why?

Try to propose some reasons first by yourself.

If $ \ce{\minus{} F}$ is involved it usually has something to do with the high electronegativity of fluorine. So the carbonyl carbon would be highly electron deficient. I wonder if that's part of the reason.

[quote="BanishedTraitor"]If $ \ce{ \minus{} F}$ is involved it usually has something to do with the high electronegativity of fluorine. So the carbonyl carbon would be highly electron deficient. I wonder if that's part of the reason.[/quote]

Even i thought that way. But carbon already has partial + charge due to O. F would only increse the + charge on C. So, infact, HCOF should be more unstable... :maybe:

Hint: which of the halobenzenes is more reactive and why?

[quote="chemrock"]Probably more effective resonance in HCOF than the others because the lone pair on -F can easily jump into the 2p orbitals of C. Whereas in the others they jump from higher energy level to lower 2p level.[/quote]

That seems like a good reason but wouldn't that mean that RCOF is always more stable that RCOX, where X is not F
but then any compound like -C(Cl)=CH- should be very unstable... :maybe:

Why very unstable? Even though the double bond character is low still the sp2-p is very effective in alkenes hell_ever. :huh:


Source: UNCC 1995 HS Math Contest

Find the minimum value of 1 o 2 o 3 o 4 o 5 o 6 o 7 o 8 o 9
where o is either a "+" or a "*".

(A) 36
(B) 40
(C) 44
(D) 45
(E) 84

I don't quite get what the problem is asking since the answer is so obvious, but [hide]I think the answer is C, 44, because 2-9 added becomes 44 and 1 should rather be multiplied than added.[/hide]

Yes, well, that's the answer.  Maybe by making the problem so easy, it seems like it's actually harder than it is.

...
And I didn't post b/c I thought it was too easy...

[quote="Syntax Error"]indeed.[/quote]
I have this feeling that "indeed" is your newest personal phrase...

if you talk to me on aim, i say it like every other word. okay, maybe not. but still. the fact still stands


Prove that $ 3(\sin \theta \minus{} \cos \theta)^4 \plus{} 6(\sin \theta \plus{} \cos \theta)^2 \plus{} 4(\sin^6 \theta \plus{} \cos^6 \theta) \equal{} 13$.

I am only beginning to study trigonometry, so I would appreciate a detailed solution.

Thanks!

EDIT: There's another problem I don't get: Prove that $ (\sin \theta \plus{} \mathrm{cosec} \ \theta)^2 \plus{} (\cos \theta \plus{} \sec \theta)^2 \minus{} (\tan^2 \theta \plus{} \cot^2 \theta) \equal{} 7$.

Ah, I see. I just didn't know that $ a^3\plus{}b^3\equal{}(a\plus{}b)(a^2\minus{}ab\plus{}b^2)$.

Are there any other factorizations like the above that I should know of?

Ummm... can you please post the solution? The hints aren't helping me... :oops:

[hide] Expanding, we have
$ \sin^2\theta\plus{}2\plus{}\csc^2\theta\plus{}\cos^2\theta\plus{}2\plus{}\sec^2\theta\minus{}\tan^2\theta\minus{}\cot^2\theta$  $ (1)$

Now, we know that $ \sin^2\theta\plus{}\cos^2\theta\equal{}1$.

So,
$ \frac{\sin^2\theta}{\sin^2\theta}\plus{}\frac{\cos^2\theta}{\sin^2\theta}\equal{}\frac{1}{\sin^2\theta}$

$ \implies 1\plus{}\cot^2\theta\equal{}\csc^2\theta \implies \csc^2\theta\minus{}\cot^2\theta\equal{}1$.

Similarly dividing by $ \cos\theta$, we have

$ \tan^2\theta\plus{}1\equal{}\sec^2\theta \implies \sec^2\theta\minus{}\tan^2\theta\equal{}1$.

Now,going back to $ (1)$, we can arrange it as

$ \sin^2\theta\plus{}\cos^2\theta\plus{}\sec^2\theta\minus{}\tan^2\theta\plus{}\csc^2\theta\minus{}\cot^2\theta\plus{}4$

$ \implies 1\plus{}1\plus{}1\plus{}4\equal{}\boxed{7}$.[/hide]


integration
calculus
calculus computations

Show that $ \sum_{0}^{\infty}(\minus{}1)^n(n\plus{}1)^{\minus{}1/2}$ is conditionally convergent and that cauchy product of this series with itself diverges.

[hide="hint"]
maximum of $ (x\plus{}1)(j\minus{}x\plus{}1)$ is at $ x\equal{}\frac{1}{2}j$ so $ (n\plus{}1)(j\minus{}n\plus{}1)\le (\frac{1}{2}j\plus{}1)^2$[/hide]

The conditional convergence of the original series is standard.

The Cauchy product is $ \sum_{n = 0}^{\infty}( - 1)^n\sum_{k = 0}^n\frac1{\sqrt {k + 1}\sqrt {n - k + 1}}.$

In order to estimate the inner sum, let us consider only $ k$ such that $ \frac {n}4 < k < \frac {3n}4.$

Then $ k + 1 < \frac {3n}4 + 1$ so $ \frac1{\sqrt {k + 1}} > \frac1{\sqrt {\frac {3n}4 + 1}} = \frac {2}{\sqrt {3n + 4}}.$

Similarly, in the same range we have $ \frac1{\sqrt {n - k + 1}} > \frac {2}{\sqrt {3n + 4}}.$

Since all of these terms appear in the sum, and at least $ \frac {n}2 - 1$ terms lie in that range, we get that

${ \sum_{k = 0}^n\frac1{\sqrt {k + 1}\sqrt {n - k + 1}} > (\frac {n}2 - 1})\cdot \frac {4}{3n + 4}.$

This remains greater than some positive constant as $ n\to\infty.$

Hence, the Cauchy product diverges because its terms do not tend to zero.

As a side note, that restriction doesn't do much. The terms in the middle are the small ones, and thus the full sum is at least $ (n \plus{} 1)\cdot\frac2{n \plus{} 2}$.
The actual values are approximately $ \int_0^1\frac1{\sqrt{x(1\minus{}x)}}\,dx\equal{} \int_{\minus{}1}^1\frac1{\sqrt{1\minus{}t^2}}\,dt\equal{}\pi$

This factor of three (or $ \frac{3\pi}{2}$) makes absolutely no difference to the argument.


I am confused about this question, although its supposed to be the easiest one ever.

I assembling a jigsaw puzzle, a "move" is defined as connecting two pieces of the puzzle together. How many moves would it take to put together a jigsaw puzzle containing 150 pieces?

Does it specify, like, if it is a 15 by 10 puzzle... Or am I just reading it wrong?

149? You pick the first piece, and then it takes one move for each subsequent piece.

it takes 1 move to put 2 pieces together so the answer is 149 because it just is. There is an explanation for it but i learned this and it works for tournament questons too.

That still counts as one move.

Even if you clump pieces together, it will still always take 149 moves for a 150 piece puzzle no matter how you do it, since you will eventually need to add 149 pieces to a single piece.

[quote="AIME15"]Matt, that isn't the best solution :P

Hmm...if I have pieces like so:

| |
|

And I want to put one so it's like

| |
| |

Doesn't that count as two moves, because I'm connecting two pieces, then two pieces?[/quote]

If we use AIME15's definition, then there are 15*9+14*10=135+140=275 "connections", so 275 moves.
If we use the "piece" definition, then there are 150 pieces, and 149 moves.

Well if it was AIME's definition then it would change depending on the dimensions. For example with a 1 by 150 puzzle (strange!) it would be 149 and for a 15 by 10 one it would be larger.

With every move, you have one less piece as you put 2 together.
Beginning: 150 pieces
Ending: 1 piece
So there are 149 moves.

I have another problem.

What is the value of the sum $ \frac{1}{2}$ + $ \frac{1}{6}$ + $ \frac{1}{12}$ + ... + $ \frac{1}{n(n\plus{}1)}$ + ... + $ \frac{1}{9900}$? Express you answer as a common fraction.

I can't figure out the shortcut.  :(

Hint: $ \frac1{n(n\plus{}1)}\equal{}\frac1n\minus{}\frac1{n\plus{}1}$.

Now, add the terms together. Notice anything? :)

Or, you could always add a few terms at a time and hope you find a pattern, which there is in this problem. Then, if you felt like it, you could prove it using induction.

ohhh.

$ \frac {99}{100}$ !

I found another tricky problem...

If $ x^2 \minus{} x \minus{} 1 \equal{} 0$, what is the value of $ x^3 \minus{} 2x \plus{} 1$?

The sum of the coefficients of the second equation is 0, so x=1 is a root. You can perform synthetic division (or long divide) to find $ x^3\minus{}2x\plus{}1\equal{}(x\minus{}1)(x^2\plus{}x\minus{}1)$. Now, solve in terms of $ x^2$ and substitute.

but $ ( x \minus{} 1 ) ( x^2 \minus{} x \minus{} 1 ) \equal{} x^3 \minus{} 2x^2 \plus{} 1$ ...

 :huh:

[quote="xpmath"]The sum of the coefficients of the second equation is 0, so x=1 is a root. You can perform synthetic division (or long divide) to find $ x^3 \minus{} 2x \plus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} x \minus{} 1)$. Now, solve in terms of $ x^2$ and substitute.[/quote]

I still don't get it. Shouldn't it be $ x^2 \minus{} x \minus{} 1$ instead of $ x^2 \plus{} x \minus{} 1$?

$ x^2\minus{}x\minus{}1\equal{}0\implies x^2\equal{}x\plus{}1\implies x^3\equal{}x^2\plus{}x$, so

$ x^3\minus{}2x\plus{}1\equal{}x^2\minus{}x\plus{}1\equal{}\boxed2$.

[quote="math154"]$ x^2 \minus{} x \minus{} 1 \equal{} 0\implies x^2 \equal{} x \plus{} 1\implies x^3 \equal{} x^2 \plus{} x$, so

$ x^3 \minus{} 2x \plus{} 1 \equal{} x^2 \minus{} x \plus{} 1 \equal{} \boxed2$.[/quote]

  :idea: OOOOOHHHHHH

But where did the 2 come from?

 :blush:

Add two to both sides of $ x^2\minus{}x\minus{}1\equal{}0$.

[quote="xpmath"]Add two to both sides of $ x^2 \minus{} x \minus{} 1 \equal{} 0$.[/quote]

 :idea:

Oh! So...

$ x^2 \plus{} x \minus{} 1 \equal{} 2x$

2x(x - 1) = 2(1) = 2 !!

Basically, math154 had $ x^2\equal{}x\plus{}1$. He then multiplied by x to get $ x^3\equal{}x^2\plus{}x$. Since we want to use the information we're given, we can add $ \minus{}2x\plus{}1$ to both sides to get $ x^3\minus{}2x\plus{}1\equal{}x^2\minus{}x\plus{}1\equal{}x^2\minus{}x\minus{}1\plus{}2\equal{}0\plus{}2\equal{}2.$

Thanks everyone! I'm a little slow on this stuff but I get it now!   :lol:

This is like one of those "Think about it" problems. In Intro: Counting and Probability, except it isn't probability or counting.


inequalities
AMC
USA(J)MO
USAMO
inequalities unsolved

[quote="Fibfreak"]Let $n$ be an integer greater than $3$ and let $a_1, a_2, \cdots , a_n$ be real numbers such that

$a_1 + a_2 + \cdots + a_n \ge n$ and ${a_1}^2 + {a_2}^2 + \cdots + {a_n}^2 \ge n^2$.

Prove that $max \{ a_1, a_2, \cdots , a_n \} \ge 2$.[/quote]
To prove this it is enough to have just the latter of the given inequalities.

The generalized mean inequality gives
\[ \max \{ a_1, a_2, \cdots , a_n \} \geq \left(\frac{{a_1}^2 + {a_2}^2 + \cdots + {a_n}^2}{n}\right)^{1/2} \geq n^{1/2} \geq 2.  \]

[quote="Fibfreak"]Let $n$ be an integer greater than $3$ and let $a_1, a_2, \cdots , a_n$ be real numbers such that

$a_1 + a_2 + \cdots + a_n \ge n$ and ${a_1}^2 + {a_2}^2 + \cdots + {a_n}^2 \ge n^2$.

Prove that $max \{ a_1, a_2, \cdots , a_n \} \ge 2$.

i just cant do it... of course for positive real numbers its almost trivial but for all real numbers i find it hard[/quote]
that's no random stuff. I remember seeing this on sum big contest like usamo or russian mo or something like that.it's nice though

[quote="maxal"]To prove this it is enough to have just the latter of the given inequalities.

The generalized mean inequality gives
\[ \max \{ a_1, a_2, \cdots , a_n \} \geq \left(\frac{{a_1}^2 + {a_2}^2 + \cdots + {a_n}^2}{n}\right)^{1/2} \geq n^{1/2} \geq 2.  \][/quote]
this only works for positive numbers, not negative as well

If there are negative numbers allowed, then the mean inequality should be fixed as
\[ \max \{ |a_1|, |a_2|, \cdots , |a_n| \} \geq \left(\frac{{a_1}^2 + {a_2}^2 + \cdots + {a_n}^2}{n}\right)^{1/2} \geq n^{1/2} \geq 2.  \]

Now suppose that $a_1, \dots, a_k$ are non-negative and $a_{k+1},\dots, a_n$ are negative.

Assume that all non-negative numbers are less than $2$. Then $\max \{ |a_1|, |a_2|, \cdots , |a_n| \}$ is achieved at a negative number implying that $k\leq n-1$.

Let $P=a_1+\dots+a_k$ and $N=-a_{k+1}-\dots - a_n$. Then $P - N\geq n$ meaning that $N\leq P-n< 2k-n$.
Note that
$n^2 \leq {a_1}^2 + {a_2}^2 + \cdots + {a_n}^2 < 4k + N^2 < 4k + (2k-n)^2$ implying $k>n-1$, a contradiction.
Therefore, at least one of $a_1, \dots, a_k$ is greater or equal to $2$.


linear algebra
matrix
vector
linear algebra unsolved

Is it true that for every matrix $A$
 $A^TAb=0 \Rightarrow Ab=0$ where $b$ is one dimensional vector
If so what's the proof ?

The condition implies that $b^TA^TAb=0$, so $\langle Ab,Ab\rangle=0$. Since the standard inner product is positive definite, $Ab$ is zero.

Oh, nice trick... or maybe better - idea :)


Prove that there doesn't exits a function $ f: \mathbb R\rightarrow\mathbb R$ for which f(0)>0 and $ f(x\plus{}y)\ge f(x)\plus{}yf(f(x))$

Please check the proof below.
http://www.mathlinks.ro/viewtopic.php?t=238282


inequalities calculus derivative function

Let $ \{a,b,c\}\subset[0,1].$ Prove that
\[ \frac{a}{1\plus{}bc}\plus{}\frac{b}{1\plus{}ac}\plus{}\frac{c}{1\plus{}ab}\plus{}abc\le\frac{5}{2}\]

Changing the problem $ \{a,b,c\} \subset(0,1)$

Let $ F\equal{}b c a\plus{}\frac{a}{b c\plus{}1}\plus{}\frac{c}{a b\plus{}1}\plus{}\frac{b}{a c\plus{}1}$

The derivative test tell us that $ F$ is convex for a (and for the others 2 variables)
$ \frac{\partial ^2 F}{\partial a^2}\equal{}\frac{2 c b^2}{(a b\plus{}1)^3}\plus{}\frac{2 c^2 b}{(a c\plus{}1)^3}>0$

So a intuitive idea behind Convex Functions.

The maximum of a convex function in a given interval will be on the extremes of this interval. (assuming not picewise functions and indefinied points)

So the maximum is on $ a,b,c\equal{}0$ or $ a,b,c\equal{}1$

Using that, the maximum is on $ a,b,c\equal{}1$ and $ F \equal{} \frac{5}{2}$


Not having the time to think about it seriously, let me just kill it:

Since a + b + c = 1, the inequality can be rewritten as

$\frac{a\left( b+c\right) ^{2}}{a\left( a+b+c\right) +bc}+\frac{b\left( c+a\right) ^{2}}{b\left( a+b+c\right) +ca}+\frac{c\left( a+b\right) ^{2}}{c\left( a+b+c\right) +ab}\leq a+b+c$.

But this is "clear", since

$\left( a+b+c\right) -\left( \frac{a\left( b+c\right) ^{2}}{a\left( a+b+c\right) +bc}+\frac{b\left( c+a\right) ^{2}}{b\left( a+b+c\right) +ca}+\frac{c\left( a+b\right) ^{2}}{c\left( a+b+c\right) +ab}\right) $
$=\frac{\left( bc-ca\right) ^{2}+\left( ca-ab\right) ^{2}+\left( ab-bc\right) ^{2}}{\left( b+c\right) \left( c+a\right) \left( a+b\right) }\geq 0$.

  Darij

[quote="ThAzN1"]Suppose that $a,b,c>0$ are positive reals such that $a+b+c=1$.  Prove that

$\frac{a(b+c)^2}{a+bc}+\frac{b(c+a)^2}{b+ca}+\frac{c(a+b)^2}{c+ab} \leq 1$[/quote]

another solution but maybe it is similar to darij's.
Replace a=1-b-c we can rewrite the LHS

$ \frac{a(1-a)^2}{(1-b)(1-c)}+ \frac{b(1-b)^2}{(1-a)(1-c)}+ \frac{c(1-c)^2}{(1-b)(1-a)} $
 We denote x=1-a, y=1-b,z=1-c, the inequality becomes
$ \frac{(1-x)x^2}{yz}+\frac{(1-y)y^2}{xz}+\frac{(1-z)z^2}{yx} \leq\ 1 $
with $  x+y+z=2 $
This inequality is equivalent to $ x^4+y^4+z^4 +xyz(x+y+z) \geq\ x^3(y+z)+y^3(z+x)+z^3(x+y) $
or $ x^2(x-y)(x-z)+y^2(y-x)(y-z)+z^2(z-y)(z-x) \geq\ 0 $ (Schur's inequality)

Inequality is equivalent to
\[
a+b+c-\frac{b (a+c)^2}{b (a+b+c)+a c}-\frac{c (a+b)^2}{c (a+b+c)+a b}-\frac{a (b+c)^2}{a (a+b+c)+b c} \geq 0
\]
and after bringing all terms to common denominator it is equivalent to proving that $f(a,b,c)\geq 0$ where
\[
f(a,b,c)=a^2 b^2-a^2 b c+a^2 c^2-a b^2 c-a b c^2+b^2 c^2
\].
Inequality is symmetric and therefore we may assume that $a \geq b \geq c$. Then there exist non-negative real numbers $x$ and $y$ such that $a=c+x+y$ and $b=c+x$. But then
\[
f(a,b,c)=f(c+x+y,c+x,c)=c^2 x^2+c^2 x y+c^2 y^2+2 c x^3+3 c x^2 y+c x y^2+x^4+2 x^3 y+x^2 y^2 \geq 0
\]


Sorry,    [img]http://www.picturerack.com/ims/guest/13159.gif[/img]

Do we assume were trying to solve the end of that link in intgers? :)

Can a moderator clean this up?  There were two failed attempts at posting this question.


probability
geometry
real analysis
calculus
integration

If $ x$ and $ y$ are positive real numbers less than $ 2$, what is the probability the $ x^{2}\plus{}y^{2}<4$?
[size=75]From [i]High School Math Contests[/i].[/size]

Is there a way to do this without the geometric intepretation?

There's the integral.  Do you think that's any easier?

It boils down to the geometric interpretation anyway.  Because $ x$ and $ y$ are allowed to vary continuously, this is not a combinatorial problem at all, and in order for the probability to even be well-defined, probability has to be defined with respect to area ([url=http://en.wikipedia.org/wiki/Lebesgue_measure]Lebesgue measure[/url]) anyway.

Thanks t0rajir0u .  :)
Could someone post the integration solution?

For a fixed $ x$, the probability that $ x^2 \plus{} y^2 < 4$ is the probability that $ 0 < y < \sqrt{4 \minus{} x^2}$ given that $ 0 < y < 2$, which is $ \frac{ \sqrt{4 \minus{} x^2} }{2}$.  Hence the "average" probability is

$ \frac{1}{2 \minus{} 0} \int_{0}^{2} \frac{ \sqrt{4\minus{}x^2}}{2} \, dx$

which is just the expected circle-area answer.

Or we could do the tricky double integral that ends up being the same thing...

$ \frac{1}{A(R)}\cdot \int\int_R f(x)\, dA$ where $ R$ is the region $ 0<x,y<2$...and $ f(x)\equal{}0$ if $ x^2\plus{}y^2\ge 4$, else it is 1

this is obviously reduces to the same thing as all the other approaches, but it generalizes...


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

At first I was surprised I got an IMO #3 (or did I?), and then I noticed that [url=http://imo-official.org/year_problem_r.aspx?year=1996&column=p4&order=desc]imo-official[/url] suggests it wasn't much of a number 3. Darn.

Put $ m \equal{} n \equal{} 0$ in to get $ f(0) \equal{} 0$. Put $ m \equal{} 0$ in to get $ f(f(n)) \equal{} f(n)$. Now the given equation reduces to $ f(m\plus{}f(n)) \equal{} f(m)\plus{}f(n)$.

We note now that $ f(n) \equal{} 0$ for all nonnegative $ n$ is a solution. Now assume it away. Therefore there exist positive integers $ a,b$ such that $ f(a) \equal{} b$. Since $ f(f(a)) \equal{} f(a)$, $ f(b) \equal{} b$. This implies that if we disregard $ f(n) \equal{} 0$, $ f$ has at least one nontrivial fixed point.

Suppose $ a$ is positive and a fixed point of $ f$, i.e. $ f(a) \equal{} a$. Put $ m \equal{} n \equal{} a$ in to the equation to get $ f(2a) \equal{} 2a$. We see that induction will guarantee that $ f(ka) \equal{} ka$ for all positive integers $ k$.

Let $ a$ be the smallest positive integer such that $ f(a) \equal{} a$. We now prove that if $ f(b) \equal{} b$, $ a\mid b$. First, let $ d$ be such that $ d \equal{} (a,b)$. Then there exist positive integers $ x,y$ such that $ ax\minus{}by \equal{} d$. Now put $ m \equal{} d, n \equal{} by$ in. This makes the LHS $ f(d\plus{}f(by)) \equal{} f(d\plus{}by) \equal{} f(ax) \equal{} ax$. The RHS is $ f(d)\plus{}f(by) \equal{} f(d)\plus{}by$. Setting these equal, the result is $ ax \equal{} f(d)\plus{}by$. Then $ f(d) \equal{} ax\minus{}by \equal{} d$. So $ f(d) \equal{} d$. But $ d\leq a$ and $ a$ is the minimal fixed point. Therefore $ d \equal{} a$, $ (a,b) \equal{} a$, and thus $ a$ divides $ b$.

So if $ a$ is the smallest positive integer such that $ f(a) \equal{} a$, we now have that the image of $ f$ is exactly the multiples of $ a$ including $ 0$. Our (new) given equation, $ f(m\plus{}f(n)) \equal{} f(m)\plus{}f(n)$, can now be rewritten as $ f(m\plus{}ka) \equal{} f(m)\plus{}ka$ for all nonnegative $ m$ and $ k$. Additionally, $ f$ outputs only multiples of $ a$.

Consider $ f(b)$ for any positive $ b$ less than $ a$. (If no such $ b$ exists, this means $ a \equal{} 1$ and we get the solution $ f(n) \equal{} n$) Set $ f(b)$ to any arbitrary multiple of $ a$, and do this for all $ 0 < b < a$. Putting $ m \equal{} b$ gives $ f(b\plus{}ka) \equal{} f(b)\plus{}ka$. This defines $ f$ for all positive integers. We check to see that all of these solutions work, no matter what multiple of $ a$ we set each $ f(b)$ to.

The end answer: Either $ f(n) \equal{} 0$ for all nonnegative integers $ n$ or the following. Let $ a$ be any positive integer, and let $ x_{1}, x_{2},\ldots x_{a\minus{}1}$ be any sequence of nonnegative integers. Then we get a new solution of $ f$: $ f(0) \equal{} 0$, $ f(ka) \equal{} ka$ for all positive integers $ k$, and $ f(b\plus{}ka) \equal{} a(k\plus{}x_{b})$ for all positive integers $ k$ and positive integers $ b < a$.

new solution ,post 15
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=60429&p=2943710#p2943710

i'm enough lazy to avoid reading [b]mellowmelos's solution[/b] ,but at the first glance i saw that the solution's got its first idea_minimality of..._ anyway i hope i used something different  to get the answeres... :)


Give $ A \equal{}\underbrace{444\ldots 444}\limit_{m}, B \equal{}\underbrace{111\ldots 111}\limit_{2m}$
Prove : \[ A\plus{}B\plus{}1\equal{}C^2, C\in \mathbb{Z}\]

$ \boxed{\boxed{\underbrace{999\ldots 999}\limit_{n}\equal{}10^n\minus{}1}}$

$ \Longrightarrow A\equal{}\frac 49(10^m\minus{}1),\ B\equal{}\frac 19(10^{2m}\minus{}1)$.

$ \therefore A\plus{}B\plus{}1\equal{}\left(\frac{10^m\plus{}2}{3}\right)^2$, since $ 10^m\plus{}2\equal{}(9\plus{}1)^m\plus{}2$ is divisible by 3, we have $ C$ is integer.

[quote="kunny"]$ \boxed{\boxed{\underbrace{999\ldots 999}\limit_{n} \equal{} 10^n \minus{} 1}}$

$ \Longrightarrow A \equal{} \frac 49(10^m \minus{} 1),\ B \equal{} \frac 19(10^{2m} \minus{} 1)$.

$ \therefore A \plus{} B \plus{} 1 \equal{} \left(\frac {10^m \plus{} 2}{3}\right)^2$, since $ 10^m \plus{} 2 \equal{} (9 \plus{} 1)^m \plus{} 2$ is divisible by 3, we have $ C$ is integer.[/quote]

Thanks you very much!!!!!!!!!!!!!!!!!!!!


geometry
circumcircle
ratio
perimeter
geometry proposed

Excircle of triangle $ABC$ corresponding vertex $A$, is tangent to $BC$ at $P$. $AP$ intersects circumcircle of $ABC$ at $D$. Prove \[r(PCD)=r(PBD)\] whcih $r(PCD)$ and $r(PBD)$ are inradii of triangles $PCD$ and $PBD$.

What do you mean by R? Surely not the circumradius?

  darij

Possibly he means the inradii $r_{b},\ r_{c}$ of $\triangle PBD,\ \triangle PCD$ are equal. Let $s_{b}, s_{c}$ be the semiperimeters and $\triangle_{b},\ \triangle_{c}$ areas of these 2 triangles. Their D-altitude is common, hence their areas are in the ratio of their bases.

$\frac{r_{b}}{r_{c}}= \frac{\triangle_{b}}{\triangle_{c}}\cdot \frac{s_{c}}{s_{b}}= \frac{BP}{CP}\cdot \frac{s_{c}}{s_{b}}$

ABDC is cyclic, $BP \cdot CP = AP \cdot DP$ and triangles $\triangle ABP \sim \triangle CDP$ are similar, $\frac{AB}{AP}= \frac{CD}{CP},$ and $AB \cdot CP = AP \cdot CD.$ Adding the 2 products,

$(AB+BP) \cdot CP = (CD+DP) \cdot AP$

AP is the perimeter splitter of $\triangle ABC$ from A, $AB+BP = AC+CP.$ Substituting this,

$(AC+CP) \cdot CP = (CD+DP) \cdot AP$

Adding $AP \cdot CP$ to both sides,

$(AC+AP+CP) \cdot CP = (CP+CD+DP) \cdot AP$

$\frac{AC+AP+CP}{AP}= \frac{CP+CD+DP}{CP}$

Triangles $\triangle PBD \sim \triangle PAC$ are similar with similarity coefficient $\frac{BP}{AP}$ and so are their perimeters,

$\frac{BD+BP+DP}{AC+AP+CP}= \frac{BP}{AP}$

Substituting this to the previous equation,

$\frac{BD+BP+DP}{BP}= \frac{CP+CD+DP}{CP}$

$\frac{s_{b}}{BP}= \frac{s_{c}}{CP},\ \ \frac{BP}{CP}\cdot \frac{s_{c}}{s_{b}}= 1$

and it follows that the inradii $r_{b}= r_{c}$ are equal.

[quote="Omid Hatami"][color=darkred][size=117]The $A$- excircle of the triangle $ABC$ is tangent to the side $[BC]$ at the point $P$.
The line $AP$ intersects again the circumcircle of the triangle $ABC$ at the point $D$.
Prove that the inradius of the triangles $BDP$ and $CDP$ are equally.[/size][/quote][/color]
[color=darkblue][size=117][b]Proof[/b] (in the my opinion, although alike with [u]Yetti's[/u], it is a bit more clearly).

For the triangle $XYZ$ denote the semiperimeter $s(XYZ)$ and the area $[XYZ]\ .$

$\{\begin{array}{ccc}AB+BP=AC+CP & \Longrightarrow & s(PAB)=s(PAC)\\\ PCD\sim PAB & \Longrightarrow & \frac{s(PCD)}{s(PAB)}=\frac{PC}{PA}\\\ PAC\sim PBD & \Longrightarrow & \frac{s(PAC)}{s(PBD)}=\frac{PA}{PB}\end{array}$ $\Longrightarrow$ $\frac{[PDB]}{[PDC]}=\frac{PB}{PC}=\frac{r_{1}}{r_{2}}\cdot\frac{s(PBD)}{s(PCD)}$ $\Longrightarrow$

$\frac{r_{1}}{r_{2}}=\frac{PB}{PC}\cdot\frac{s(PCD)}{s(PBD)}=\frac{PB}{PC}\cdot\frac{(PCD)}{(PAB)}\cdot\frac{(PAB)}{(PAC)}\cdot\frac{(PAC)}{(PBD)}=\frac{PB}{PC}\cdot\frac{PC}{PA}\cdot 1\cdot\frac{PA}{PB}=1$ $\Longrightarrow$ $\boxed{\ r_{1}=r_{2}\ }\ .$

[b]Remark.[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=50559 .
[b]Generally,[/b] for any point $P\in (BC)$ there is the relation $\boxed{\ \frac{r_{1}}{r_{2}}=\frac{s(PAB)}{(PAC)}\ }\ !$
[u]Can somebody ascertain $r_{1}$ only in accordance with the elements of the triangle[/u] $ABC\ ?$

[b][u]Yetti[/u][/b], now in the your proof you didn't use the remarkable relation $\frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{r}{r_{1}r_{2}}+\frac{2}{h}\ .$[/size][/color]


integration
inequalities
real analysis
real analysis unsolved

[quote="Cezanne123"]How do we evaluate $\sum_{k = 1}^{50}{\frac{1}{{k^{2}}}}$ to 12 decimal places and show that the answer is correct?

(without calculating all the terms since that method does not guarantee that the answer is correct)[/quote]

I guess we can write,
$\sum_{k=1}^{50}\frac{1}{k^{2}}= 1+\sum_{2}^{50}\frac{1}{k^{2}}\leq 1+\int_{1}^{50}\frac{dt}{t^{2}}$

For 12 decimal places accuracy, you pretty much have to add them all up. Standard floating point arithmetic is good enough for that.

You can compute this sum by estimating the value of
\[\sum_{k=51}^{\infty}\frac{1}{k^{2}}.\]
One can show that
\[\sum_{k=n}^{\infty}\frac{1}{k^{2}}= \frac{1}{n-\frac{1}{2}+\frac{1}{12n}+\frac{1}{24n^{2}}-\frac{1}{720n^{3}}+O(\frac{1}{n^{4}})}.\]
Therefore,
\[\sum_{k=51}^{\infty}\frac{1}{k^{2}}\approx 0.01980133322535.\]
Finally,
\[\sum_{k=1}^{50}\frac{1}{k^{2}}= \sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{k=51}^{\infty}\frac{1}{k^{2}}\approx \frac{\pi^{2}}{6}-0.01980133322535 \approx 1.62513273362.\]
However, if you just want to compute this specific sum, it is easier to directly add up all 50 terms.

Thanks for all the help, but what if all 50 terms are added up and evaluated to 12 decimal places, how do we prove that is the correct answer?

With 50 terms, you should evaluate each one to 14 decimal places, not 12, to get the answer correctly. As to the "proof", just write all the corresponding inequalities (like, say, $0.11111111111111<\frac 1{3^{2}}<0.11111111111112$) and add them up.

In standard floating point arithmetic, a number is represented as $(1+a\cdot 2^{-52})\cdot 2^{b}$ for some integers $a$ and $b$. $2^{52}$ is approximately $2\cdot 10^{-16}$, so we have between 15 and 16 digits of accuracy after the decimal point for numbers between 1 and 2. By fedja's post, this is more than enough.


vector
trigonometry
geometry
parallelogram
search
analytic geometry
complex numbers

First off, I don't have a problem I need help solving; rather an explanation of how the formula for the vector product (or cross product) is derived.
I know the formula is: $ a$ x $ b = |a\parallel{}b|sin \;\theta\;\^n$
where $ a$ and $ b$ are two vectors, $ \theta$ is the angle between them, and $ \^n$ is a unit vector normal to $ a$ and $ b$, and I know how to calculate the vector product using the formula but what I don't understand is why it works and, in particular, where does the sine of the angle come in?

I think that's just the definition of the cross product. The cross product $ \mathbf a\times\mathbf b$ has a magnitude equal the area of the parallelogram formed by $ \mathbf a$ and $ \mathbf b$, which from the triangle area formula $ \frac{ab\sin C}{2}$, is clearly $ |\mathbf a| |\mathbf b|\sin\theta$.

grn_trtle
yeah, I can see that the area of the parallelogram gives us $ |a\parallel{}b|sin\;\theta$, but I can't see how this should equate with the modulus of the vector that is perpendicular to $ a$ and $ b$ that the vector product gives us. Can you shed any more light?

I think that the area is an application of the vector product, being its modulus, and the cross-product (or vector product) is called the vector area. Hope this is helpful.

Again, I think it's just a definition for the sake of convenience (the expression $ ab\sin\theta$ pops up quite a bit; for example, the definition of torque and magnetic field).

I could be wrong though

It's not really a definition of convenience. The cross product came about in a very natural way. The cross product was found when mathematicians were looking for ways to extend "multiplication" to higher dimensional numbers. If you think of complex numbers as two-dimensional numbers, then they have a natural multiplication between them. The product of complex numbers has all of the properties that you would like in a multiplication. It is commutative, associative, distributive, obeys the zero property, has a multiplicative identity, and multiplicative inverses for every complex number except zero.

A mathematician named Hamilton looked for a similar way to multiply three dimensional numbers (3D-vectors), but he was unsuccessful. We now know that no such multiplication exists having all of the properties of regular multiplication. But he was semi-successful in finding a way to multiply four dimensional numbers (though he had to sacrifice the property of commutivity to make it work). He called these numbers and their way multiplying [url=http://en.wikipedia.org/wiki/Quaternion]Quaternions[/url].

Imbedded in this way of multiplying 4D vectors are elements of the dot product and the cross product of 3D vectors.

While the cross product is called a "product", it is missing a lot of the properties that a full-on "product" should have. For example, it is not commutative or associative, and it has no multiplicative identity. (In fact, the only reason we call it a product at all is because it has the distributive property). But it does appear naturally within the product of quaternions, and it also appears naturally in many physical contexts.

So while it may appear as an ad hoc definition at first, it does have a more theoretical (and historical) framework behind it.

If you do a google search for quaternions and cross products you will find a lot of interesting information. Here's one file to get you started:
http://www.math.ilstu.edu/akmanf/newwebsite/HistoryMath/quaternions.pdf

Thanks guys, but how would you go about proving this vector product formula?

[quote="gauss202"]It's not really a definition of convenience. The cross product came about in a very natural way. The cross product was found when... (long post snipped so that people don't have to scroll around so much)[/quote]

Yeah, I agree, though I still consider $ \mathbf{a}\times\mathbf{b}\equal{}|\mathbf{a}| |\mathbf{b}|\sin\theta$ a "definition of convenience." The $ \times$ is just a notation, isn't it?

@jimi: I don't think you [i]can[/i] prove it (or are "supposed to").

[quote="grn_trtle"][quote="gauss202"]It's not really a definition of convenience. The cross product came about in a very natural way. The cross product was found when... (long post snipped so that people don't have to scroll around so much)[/quote]

Yeah, I agree, though I still consider $ \mathbf{a}\times\mathbf{b} \equal{} |\mathbf{a}| |\mathbf{b}|\sin\theta$ a "definition of convenience." The $ \times$ is just a notation, isn't it?[/quote]

I, too, am in agreement.

@jimi. While we cannot prove the definition itself, we can show that all the other properties of the cross product arise from its definition. For example, we have $ <a,b,c> \times <d,e,f> \equal{} <bf\minus{}ce, dc\minus{}af, ae\minus{}db>$ because $ i\times i\equal{}j\times j\equal{}k\times k\equal{}0$ and $ i\times j\equal{}j\times k\equal{}k\times i\equal{}1$ and $ j\times i\equal{}k\times j\equal{}i\times k\equal{}\minus{}1$ (this follows directly from the definition, as $ \sin 0^{\circ}\equal{} 0$ and $ \sin 90^{\circ}\equal{}1$).

Question: Why does the cross product operation produce a vector necessarily perpendicular to the two vectors being crossed (how does this follow from the definition)?

The [i]definition [/i]of the cross product is $ \mathbf{a}\times\mathbf{b} \equal{} |\mathbf{a}| |\mathbf{b}|\sin\theta \mathbf{\hat{n}}$, where $ \mathbf{\hat{n}}$ is the normal vector to the plane spanned by $ \mathbf{a},\mathbf{b}$, whose direction is defined so that vectors $ \{\mathbf{a}, \mathbf{b}, \mathbf{a} \times \mathbf{b} \}$ form a right-handed coordinate system (the right-handedness is by convention).

As for the Cartesian coordinate-wise formula, you can show it is perpendicular just by calculating the dot product with each of the original two vectors to be $ 0$.


This is probably the longest game of hurt and heal.
The rules are the same from the previous one.
And, the participants are the ALL the people from the previous games, who lost.
So, everyone in this game starts with 15 points.
You may hurt twice, and heal once.

ProtestanT-15
D3m0n Shad0w-15
Treething-15
jli-15
#H34N1-15
13375P34K43V312-15
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-15
anirudh-15
randomdragoon-15
bubka-15
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

#H34N1 = anirudh.

If there is reasonable suspicion that someone is using alt accounts, can something be done to restore the "fairness"?

Also, don't forget that posting within one hour of your last post is not permitted.

Good luck to everyone. :)

Hurt anirudh, heal Leet

ProtestanT-15
D3m0n Shad0w-15
Treething-15
jli-15
13375P34K43V312-16
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-15
anirudh-13
randomdragoon-15
bubka-15
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt ccy twice, heal treething

ProtestanT-15
D3m0n Shad0w-15
Treething-16
jli-15
13375P34K43V312-16
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-13
anirudh-13
randomdragoon-15
bubka-15
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Lol, I didn't know that. And, if alts are going to be used, then their post will not count. UNLESS, they post within an hour limit from there original account.

When do you think this will end (in how many days/weeks)?

I forgot to say: Let's try to avoid messing things up this time because it gets really confusing when a bunch of people all start accusing each other of cheating somehow.

Hurt anirudh, heal ccy

ProtestanT-15
D3m0n Shad0w-15
Treething-16
jli-15
13375P34K43V312-16
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-14
anirudh-11
randomdragoon-15
bubka-15
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

ganging up, are we???

Hurt Treething hurt 1337 heal anirudh

ProtestanT-15
D3m0n Shad0w-15
Treething-15
jli-15
13375P34K43V312-15
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-14
anirudh-12
randomdragoon-15
bubka-15
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt bubka,ccy, heal treething

ProtestanT-15
D3m0n Shad0w-15
Treething-16
jli-15
13375P34K43V312-15
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-13
anirudh-12
randomdragoon-15
bubka-14
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt anirudh, hurt Treething, heal ccy

ProtestanT-15
D3m0n Shad0w-15
Treething-15
jli-15
13375P34K43V312-15
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-14
anirudh-11
randomdragoon-15
bubka-14
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt bubka, ccy, heal treething

ProtestanT-15
D3m0n Shad0w-15
Treething-16
jli-15
13375P34K43V312-15
mustafa-15
i_like_pie-15
BanishedTraitor-15
whorthawholebean-15
ccy-13
anirudh-11
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Double hurt i_like_pie, heal 1=2

ProtestanT-15
D3m0n Shad0w-15
Treething-16
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-13
BanishedTraitor-15
whorthawholebean-15
ccy-13
anirudh-11
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt anirudh, heal ccy

ProtestanT-15
D3m0n Shad0w-15
Treething-16
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-13
BanishedTraitor-15
whorthawholebean-15
ccy-14
anirudh-9
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt ccy,aniurdh, heal treething

ProtestanT-15
D3m0n Shad0w-15
Treething-17
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-13
BanishedTraitor-15
whorthawholebean-15
ccy-13
anirudh-8
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Fine. I'll join the chain gang.

Double hurt ccy, heal i_like_pie

ProtestanT-15
D3m0n Shad0w-15
Treething-17
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-14
BanishedTraitor-15
whorthawholebean-15
ccy-10
anirudh-8
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt anirudh, heal ccy

ProtestanT-15
D3m0n Shad0w-15
Treething-17
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-14
BanishedTraitor-15
whorthawholebean-15
ccy-11
anirudh-6
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt Anirudh twice.

ProtestanT-15
D3m0n Shad0w-15
Treething-17
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-14
BanishedTraitor-15
whorthawholebean-15
ccy-11
anirudh-4
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

Hurt anirudh, heal ccy

ProtestanT-15
D3m0n Shad0w-15
Treething-17
jli-15
1=2-16
13375P34K43V312-15
mustafa-15
i_like_pie-14
BanishedTraitor-15
worthawholebean-15
ccy-12
anirudh-2
randomdragoon-15
bubka-13
jackdillion-15
lingomaniac88-15
CheeseisGood-15
Klebian-15
Insanely Bored-15
pianoforte-15

lingo--, Klebian+

jli - 14
Klebian - 40
pianoforte - 3
lingomaniac - 5

lingo--, Klebian+

jli - 14
Klebian - 41
pianoforte - 3
lingomaniac - 3

Who dies first? :P

pianoforte--, jli+

jli - 15
Klebian - 41
pianoforte - 1
lingomaniac - 3

pianoforte-, lingomaniac-, jli+

jli - 16
Klebian - 41
pianoforte - 0
lingomaniac - 2

kill lingomaniac heal jli

jli: 17
Klebian: 41

Well, it's either jli or kleb; let's just kill jli because it's easier.

jli--, Kleb+

jli: 15
Klebian: 42

Noo
Klebian--, Jli+

jli: 16
Klebian: 40

Yees
jli--, Klebian+

jli: 12
Klebian: 42

Should we just call it a day?

jli: 4
Klebian: 46

jli--, Klebian+

jli: 2
Klebian: 47

Nope, we have to officially declare Klebian the winner.


modular arithmetic
algebra
polynomial
quadratics
summer program
PROMYS
floor function

Does the equation $ x^2\plus{}y^2\equal{}p$ have a integer solution, where $ p$ is a prime and $ p\equiv1\pmod4$?

Hint: You can use the Wilson's Theorem $ (p\minus{}1)!\equiv\minus{}1\pmod p.$

[quote="ifai"]Does the equation $ x^2 \plus{} y^2 \equal{} p$ have a integer solution, where $ p$ is a prime and $ p\equiv1\pmod4$?

Hint: You can use the Wilson's Theorem $ (p \minus{} 1)!\equiv \minus{} 1\pmod p.$[/quote]

Yes, consider $ x\equal{}1, y\equal{}2$

Or do you want "all solutions"

there are $ \infty$ solutions to that equation
note that all primes of the form $ 4k\plus{}1$ are odd.
First we will se that all odd numbers of form $ 4k\plus{}1$ are sume of $ 2$ squares.
Supose that $ a^2\plus{}b^2\equal{}m$. if $ m$ is odd, $ m\equal{}2k\plus{}1$.
$ a^2\plus{}b^2\equal{}2m\plus{}1$. $ a$ or $ b$ must be odd, that because $ 2m\plus{}1$ is odd. Let $ a\equal{}2k\plus{}1$ and $ b\equal{}2q$.
$ (2k\plus{}1)^2\plus{}(2q)^2\equal{}2m\plus{}1$
$ 4k^2\plus{}4k\plus{}1\plus{}4q^2\equal{}2m\plus{}1$
$ 4(k^2\plus{}k\plus{}q)\equal{}2m$
so $ 4\mid 2m\rightarrow 2\mid m\rightarrow m\equal{}2t$ $ t\in\mathbb{Z}$.
we get $ 2m\plus{}1\equal{}4t\plus{}1$.
Its obvious that the set of the numbers of the form $ 4t\plus{}1$ contains the set of prime numbers of the form $ 4j\plus{}1$ $ \blacksquare$.

I think you've made a mistake, not all numbers of the form $ 4m\plus{}1$ can be expresses as the sum of two squares. (Although pairs of odd and even squares are).

Consider $ 21$

His proof is a circular reasoning...
[quote="pelao_malo"]First we will [b]see[/b] that all odd numbers of form $ 4k \plus{} 1$ are [b]sum[/b] of $ 2$ squares. [/quote]where the bolded words are grammatical corrections.

And then he concluded
[quote="pelao_malo"]we get 2m+1=4t+1. [/quote]

Also noted by SimonM that not all odd numbers are primes and hence the proof fails.
EDIT:
Read carefully the question:
[quote="ifai"]Does the equation $ x^2 \plus{} y^2 \equal{} p$ have a integer solution, where $ p$ is a prime and $ p\equiv1\pmod4$?

Hint: You can use the Wilson's Theorem $ (p \minus{} 1)!\equiv \minus{} 1\pmod p.$[/quote]
This means for all primes $ p$ of the form $ p\equiv1\pmod4$, not just a few solutions, where they are only for $ p\equal{}5,17,37$

This is a very famous result.

[hide="Discussion"] It is well-known that $ p \equiv 1 \bmod 4 \Leftrightarrow \exists x : p | x^2 \plus{} 1$ (short proof:  the polynomial $ x^{ \frac{p\minus{}1}{2}} \minus{} 1$ contains precisely the quadratic residues $ \bmod p$, and $ \minus{}1$ is clearly a root when $ p \equiv 1 \bmod 4$).  Now, there are one of two ways to go (a sketch):

1)  $ p | (x \plus{} i)(x \minus{} i)$ but $ p$ cannot divide either, hence $ p$ is composite in the Gaussian integers.

1)  Consider the lattice generated by $ (x, 1)$ and use [url=http://mathworld.wolfram.com/MinkowskiConvexBodyTheorem.html]Minkowski's Theorem[/url] (generalized to an arbitrary lattice) on the circle of radius $ \sqrt {p}$.

As I recall, there is a more elementary proof that I've forgotten.  [/hide]

im sorry xD
im chilean i dont know good english
 :maybe:
and i am bad with math
bye  :(

Another theorem about the same thing:

Prove that for each prime there is at most one pair of positive integers $ (a,b)$ satisfying $ a\ge b$ and $ a^2\plus{}b^2\equal{}p$.

Here's what I have...I had to solve it before going to bed.  :P

[u]Solution.[/u] Consider the lattice spanned by the set $ \Lambda([0,p],[1,y])$ over $ \mathbb{Z}$.  Consider the disk $ \mathcal{C} \equal{} \{(x,y)|x^2 \plus{} y^2 < 2p\}$.  By the extended Minkowski Theorem, since $ \text{Vol}\mathcal{C} \equal{} 2p\pi > 2^2|\det\Lambda| \equal{} 4p$, $ \exists (i,j)\in\mathbb{Z}^2$ such that $ i(0,p) \plus{} j(1,y) \equal{} (j,ip \plus{} jy)\in\mathcal{C}$,that is, $ j^2 \plus{} (ip \plus{} jy)^2 < 2p$.  Considering $ j^2 \plus{} (ip \plus{} jy)^2\bmod p$, $ j^2(1 \plus{} y^2)\bmod p$.  But since $ p\equiv 1\pmod 4$, there exists $ y$ such that $ y^2\equiv \minus{} 1\pmod p$, that is, $ j^2 \plus{} (ip \plus{} jy)^2 \equal{} p$.  And the conclusion follows. $ \Box$

[u]Remark.[/u] The proof of Minkowski's Theorem in $ \mathbb{Z}^2$ is simple, and the geenralized case can be proved using calculus.

[quote="bertram"]Another theorem about the same thing:

Prove that for each prime there is at most one pair of positive integers $ (a,b)$ satisfying $ a\ge b$ and $ a^2 \plus{} b^2 \equal{} p$.[/quote]

Does anyone have a proof for this theorem? :)

the original proof (attributed to Fermat, I think) uses the fact that $ -1$ is a QR $ \pmod{p}$ to get $ x,k$ such that $ x^2+1=kp$, so that we have a solution to $ x^2+y^2=kp$. as long as $ k>1$, a solution is then constructed for a smaller $ k$, so the process stops at $ k=1$.

another proof uses this result:

LEMMA. for every real number $ r$ and any positive integer $ n$ there exists a rational number $ \frac{u}{v}$ in lowest terms such that $ |r-\frac{u}{v}|\leq \frac{1}{v(n+1)}$ and $ 1\leq v\leq n$.

first note that there's an integer $ a$ such that $ a^2\equiv -1\pmod{p}$ (since $ p\equiv 1\pmod{4}$). now using $ r=\frac{-a}{p}$ and $ n=\lfloor \sqrt{p} \rfloor$ in the above lemma, we get that there are $ u$ and $ v$ such that

\[ | \frac{-a}{p}-\frac{u}{v}|\leq \frac{1}{v(\lfloor \sqrt{p} \rfloor+1)} < \frac{1}{v\sqrt{p}},
\]

with $ 1\leq v<\sqrt{p}$.

now set $ x=av+pu$ and notice that the above inequality translates to $ |x|<\sqrt{p}$, from which it follows that $ 0<x^2+v^2<2p$. on the other hand, $ x\equiv av\pmod{p}$, so $ x^2+v^2\equiv a^2v^2+v^2=v^2(a^2+1)\equiv 0\pmod{p}$, which means we must have $ x^2+v^2=p$.

The fact that there is exactly one pair is simple.

Once you get that $ p$ is not prime in the Gaussian integers, it must be precisely the product of two primes, since its norm is $ p^2$. (By unique factorization, $ p\equal{}(x\plus{}yi)(x\minus{}yi)$ for one pair (essentially) of $ x,y$ because these are both prime since $ x^2\plus{}y^2\equal{}p$, and thus they have prime norm.)

A more general form of the original poster's theorem is that an odd number $ n$ may be expressed as the sum of two squares if and only if all of its prime divisors are $ 1 \mod 4$.

[quote="Hamster1800"]A more general form of the original poster's theorem is that an odd number $ n$ may be expressed as the sum of two squares if and only if all of its prime divisors are $ 1 \mod 4$.[/quote]

Not quite; the correct statement is that the exponent of every prime congruent to $ 3 \bmod 4$ in the prime factorization of $ n$ is even.

Ah, of course! The simple cases got me :(

[quote="Hamster1800"]The fact that there is exactly one pair is simple.

Once you get that $ p$ is not prime in the Gaussian integers, it must be precisely the product of two primes, since its norm is $ p^2$. (By unique factorization, $ p \equal{} (x \plus{} yi)(x \minus{} yi)$ for one pair (essentially) of $ x,y$ because these are both prime since $ x^2 \plus{} y^2 \equal{} p$, and thus they have prime norm.)

A more general form of the original poster's theorem is that an odd number $ n$ may be expressed as the sum of two squares if and only if all of its prime divisors are $ 1 \mod 4$.[/quote]

[hide="My spammy but elementary solution"]

Suppose that $ a^2\plus{}b^2\equal{}c^2\plus{}d^2\equal{}p$ with $ a\ge b>0$ and $ c\ge d> 0$. Then $ p^2\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2\equal{}(ac\minus{}bd)^2\plus{}(ad\plus{}bc)^2$ and $ (ac\plus{}bd)(ad\plus{}bc) \equal{} a^2cd\plus{}b^2cd\plus{}abc^2\plus{}abd^2 \equal{} p(ab\plus{}cd)$ so either $ ac\plus{}bd\equal{}p\implies ad\minus{}bc\equal{}0$ or $ ad\plus{}bc\equal{}p\implies ac\minus{}bd\equal{}0$ (this case implies $ a\equal{}b$ and $ c\equal{}d$ and the statement is trivial). Then $ ad\minus{}bc\equal{}0\implies \frac{a}{c}\equal{}\frac{b}{d}\implies \frac{a^2}{c^2}\equal{}\frac{b^2}{d^2}\equal{}\frac{a^2\plus{}b^2}{c^2\plus{}d^2}\equal{}1$ so $ a\equal{}c$ and $ b\equal{}d$.

[/hide]

[quote="t0rajir0u"]scorpius (Jason Bland) used Minkowski's in four dimensions to prove the Four Square Theorem at PROMYS, I believe.  :)[/quote]
I've heard (haven't heard the details myself though) that you can prove it using Minkowski's for 2-D vectors and the inequality $ \pi^{2}>8$.

[quote="t0rajir0u"]scorpius (Jason Bland) used Minkowski's in four dimensions to prove the Four Square Theorem at PROMYS, I believe.  :)[/quote]

Certainly, the proof of Four Squares Theorem is a simple problem in the Geometry of Numbers.  Here is what I have deduced with the use of Minkowski's Theorem in $ \mathbb{R}^4$.

[u]Theorem.[/u] Every nonnegative integer can be expressed as the sum of four integer squares.

[u]Proof.[/u] Notice that $ (a_1^2 + b_1^2 + c_1^2 + d_1^2)(a_2^2 + b_2^2 + c_2^2 + d_2^2) = \parallel{}(a_1 + b_1i + c_1j + d_1k)(a_2 + b_2i + c_2j + d_2k)\parallel{}^2$ can be expressed as the sum of four squares, where $ a + bi + cj + dk\in\mathbb{H}$.  Therefore, it is sufficient to prove that every prime number $ p$ can be expressed as the sum of four integer squares.

Consider the sets $ \Lambda = \{(1,0,u, - v)^T,(0,1,u,v)^T,(0,0,p,0)^T,(0,0,0,p)^T\}$ and $ \mathbb{S}^3 - \partial\mathbb{S}^3 = \{(a,b,c,d): a^2 + b^2 + c^2 + d^2 < 2p\}$, where $ \partial B$ is the boundary of a set $ B$.

The volume of $ \mathbb{S}^3 - \partial\mathbb{S}^3$, $ \text{Vol }(\mathbb{S}^3 - \partial\mathbb{S}^3) = \frac {\pi^2(\sqrt {2p})^4}{2} = 2\pi^2p^2$ (can be proved with calculus).  Since $ \text{Vol }(\mathbb{S}^3 - \partial\mathbb{S}^3) = 2\pi^2p^2 > 2^4p^2 = 2^4|\det\Lambda|$, by Minkowski's Theorem, there exists $ (a,b,c,d)\in\mathbb{Z}^4$ such that $ a^2 + b^2 + c^2 + d^2 < 2p$, where
\[ \left(\begin{array}{c}a \\
b \\
c \\
d\end{array}\right) = a\left(\begin{array}{c}1 \\
0 \\
u \\
- v\end{array}\right) + b\left(\begin{array}{c}0 \\
1 \\
v \\
u\end{array}\right) + x\left(\begin{array}{c}0 \\
0 \\
p \\
0\end{array}\right) + y\left(\begin{array}{c}0 \\
0 \\
0 \\
p\end{array}\right).
\]
Consider $ a^2 + b^2 + c^2 + d^2$ modulo $ p$, that is,
\[ \begin{eqnarray*}a^2 + b^2 + c^2 + d^2 & = & a^2 + b^2 + (au + bv + xp)^2 + ( - av + bu + yp)^2 \\
& \equiv & (a^2 + b^2)(1 + u^2 + v^2)\equiv 0\pmod p\end{eqnarray*},
\]
since $ 1 + u^2 + v^2\equiv 0\pmod p$ for some integers $ u,v$.

Ergo, $ a^2 + b^2 + c^2 + d^2 = p$, as required. $ \Box$

Hopefully, I have not made any mistakes.

I think I have to work harder and harder... :wacko:
Thank you all!

The following is a simple proof to this theorem.

[u]Proof.[/u] Notice that if $ p\equal{}4k\plus{}1$, then let $ a\equal{}1\cdot 2\hdots 2k$.  Then, $ a^2\equiv 1\hdots 2k(p\minus{}2k)\hdots (p\minus{}1)\equiv \minus{}1\pmod p$.  Consider the set $ S\equal{}\{ax\minus{}y|x,y\in [0,\lfloor \sqrt{p}\rfloor]\cap \mathbb{Z}\}$. Since $ |[0,\lfloor \sqrt{p}\rfloor]\cap \mathbb{Z}|\equal{}(\lfloor \sqrt{p}\rfloor\plus{}1)^2>p$, there exist $ x_1,y_1,x_2,y_2$ such that $ ax_1\minus{}y_1\equiv ax_2\minus{}y_2\pmod p\iff a(x_1\minus{}x_2)\equiv y_1\minus{}y_2\pmod p$ $ \implies a^2(x_1\minus{}x_2)^2\equiv \minus{}(x_1\minus{}x_2)^2\equiv (y_1\minus{}y_2)^2\pmod p$ $ \iff (x_1\minus{}x_2)^2\plus{}(y_1\plus{}y_2)^2\equiv 0\pmod p$. But $ 0<(x_1\minus{}x_2)^2\plus{}(y_1\plus{}y_2)^2<2p$.  And the conclusion follows. $ \Box$


function
linear algebra
matrix
linear algebra unsolved

Let $ f~: ~\mathcal{M}_{n}(\mathbb{K})\to\mathcal{M}_{n}(\mathbb{K})$, ($ \mathbb{K}$ is a field, but $ \mathbb{K}\not\equal{}\mathbb{F}_{2}$).
Assume that for all $ A,B, C\in\mathcal{M}_{n}(\mathbb{K})$, $ f(ABC)\equal{}f(ACB)$.
Prove the $ f$ is constant over $ \mathcal{M}_{n}(\mathbb{K})\setminus\mbox{GL}_{n}(\mathbb{K})$.
P.S.: $ f$ is not assumed to be linear.

Choose an arbitrary invertible matrix $ A$ and set $ C=A^{-1}$ to observe $ f(B)=f(AA^{-1}B)=f(ABA^{-1})$, i.e. $ f$ is constant on each equivalence class of matrices in $ M_{n}(\mathbb{K})$. Since each $ B\in M_{n}({\mathbb{K})\setminus GL_{n}(\mathbb{K})}$ has the eigenvalue zero we find a representative (in block form) $ B'=\left(\begin{array}{cc}X & 0\\0 & 0\end{array}\right)$ in the equivalence class of $ B$.

Now set $ A=\left(\begin{array}{cc}I & I\\ 0 & 0\end{array}\right),C=\left(\begin{array}{cc}I & 0\\ -I & 0\end{array}\right)$ (applying the same block decomposition as for $ B'$) and calculate: $ f(B)=f(B')=f(AB'C)=f(ACB')=f(0)$. Since $ B$ was arbitrary, $ f$ must be constant on $ M_{n}({\mathbb{K})\setminus GL_{n}(\mathbb{K})}$.

found a small bug in my earlier proof  :blush: , we need to perform a minor case distinction and perform a sub-decomposition of the matrix block form in the final step. For that identify $ X\in M_{r}({\mathbb{K})}$ with $ r<n$.

Case 1:  $ 2r\leq n$

Set $ A=\left(\begin{array}{cc}I_{r}& (I_{r}\;0)\\ 0 & 0\end{array}\right),C=\left(\begin{array}{cc}I_{r}& 0\\ \left(-I_{r}\atop 0\right) & 0\end{array}\right)$ and conclude as in the original proof $ f(B)=\ldots = f(0)$

Case 2:  $ 2r> n$

Set $ A=\left(\begin{array}{cc}I_{r}&\left(0\atop I_{n-r}\right)\\ 0 & 0\end{array}\right),C=\left(\begin{array}{cc}I_{r}& 0\\(0\;-I_{n-r}) & 0\end{array}\right)$ and decompose $ X=\left(\begin{array}{cc}X' & *\\ * & *\end{array}\right)$ with $ X'\in M_{2r-n}(\mathbb{K})$ and calculate $ f(B)=f(B')=f(AB'C)=f(ACB')=f\left(\left(\begin{array}{cc}X' & *\\ 0 & 0\end{array}\right)\right)=f\left(\left(\begin{array}{cc}X' & 0\\ 0 & 0\end{array}\right)\right)$ the last step because of the equivalence of the involved matrices.

Since $ 2r-n<r$ we can repeatedly apply this procedure until we fall into case 1. This completes the proof.

Hi hjbrasch, your proof isn't valid because
a non invertible matrix $ B$ isn't necessarily similar to a matrix of the form $ B'$; example:$ B\equal{}\begin{pmatrix}0&0\\1&0\end{pmatrix}$ or a Jordan nilpotent block.


inequalities
logarithms
inequalities unsolved


calculus
integration
geometry
rectangle
complex analysis

Hi all

I'm trying to prove $ \int_0^{\infty} \frac{x}{sinh(x)} \equal{} \frac{\pi^2}{4}$. I almost have it figured out but can't quite work out one final part. I tried a contour integral, which is like a rectangle with sides bounded by $ x\equal{}\plus{}/\minus{}N$ and $ y\equal{}0$ and $ y\equal{}\pi$, but with a semi circle centered at $ i\pi$ of radius $ \frac{\epsilon}{N}$ which leaves the residue outside of the path of integration (Hopefully you know what I mean). We get that as $ N \rightarrow \infty$, the path integral goes to 4 times $ \int_0^{\infty} \frac{x}{sinh(x)}$ plus the limit of the integrals around the semicircles. Now I know that the residue at $ i\pi$ is $ 2\pi^2$ so I want to show that this the integral around the semicircle is exactly half of the integral around the full circle, but I am not sure exactly how to do this. Any advice would be greatly appreciated. Thanks

Prove it in general. Just assume that $ f$ has a simple (sic!) pole at $ a$, write $ f(z)\equal{}\frac c{z\minus{}a}\plus{}g(z)$ where $ g$ is bounded near $ a$ and integrate over your half-circle.


In the plane, we consider a circle $K$ with center $S$ and radius $1$ and a square $Q$ with center $M$ and side length $2.$ Let $XY$ be the hypotenuse of a right isosceles triangle $XYZ.$ Find the locus of the vertex $Z$ of triangle $XYZ,$ when the point $X$ moves in the whole circle $ K $ (i. e. in the boundary and the interior) and the point $Y$ moves in the whole square $Q$ (i. e. in the boundary and the interior) ?

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


function
search
algebra unsolved
algebra

Prove that there is a function f from the set of all natural numbers
into itself such that $f(f(n)) = n^2$ for all $n$ is positive integers

Yes, and this is called the "search function"  :P :
http://www.mathlinks.ro/Forum/viewtopic.php?t=15306
http://www.mathlinks.ro/Forum/viewtopic.php?t=43674