hello, do you mean $ 64Rr\le(a\plus{}b\plus{}c)^2\plus{}20r^2$?
Sonnhard.

[quote="thanhnam2902"]Let $ ABC$ is a triangle, let $ R, r$ are radius of circum circle and incribed circle of $ ABC$ triangle respectively, let $ AB \equal{} c, BC \equal{} a, CA \equal{} b$.

Prove that: $ \boxed {64r.R \le (a \plus{} b \plus{} c)^2 \plus{} 20r^2}$[/quote]
This is known as "Blundons inequality". If $ s \equal{} \frac {a \plus{} b \plus{} c}{3}$ then $ 9IG^2 \equal{} s^2 \plus{} 5r^2 \minus{} 16Rr \ge 0$ which is clearly equivalent to your inequality.


inequalities
probability
inequalities unsolved

1. Let $ a_{1},a_{2},......a_{n}$ satisying $ \sum_{i \equal{} 1}^{n}a_{i} \equal{} 0,\sum_{i \equal{} 1}^{n}a_{i}^{2} \equal{} n$.Prove that $ \frac {1}{n}\sum_{i \equal{} 1}^{n}a_{i}^{4}\geq 1 \plus{} \frac {1}{n^{2}}\sum_{i \equal{} 1}^{n}a_{i}^{3}$
2. Let a,b,c,d be four positive real numbers. Prove that $ (a \minus{} b)(b \minus{} c)(c \minus{} d)(d \minus{} a) \plus{} (a \minus{} c)^{2}(b \minus{} d)^{2}\geq 0$
3. Let a,b,c be nonnegative real numbers satisfying $ a \plus{} b \plus{} c\leq \frac {3\sqrt {3}}{2}$.Prove that $ \sqrt {a \plus{} b^{3}} \plus{} \sqrt {b \plus{} c^{3}} \plus{} \sqrt {c \plus{} a^{3}}\leq \frac {3\sqrt {7\sqrt {3}}}{2\sqrt {2}}$

for the first one $ \sum_{i\equal{}1}^{n}a_{i}^{2}\equal{}0 \Leftrightarrow a_1\equal{}a_2\equal{}...\equal{}a_n\equal{}0$ then $ \frac{1}{n}\sum_{i\equal{}1}^{n}a_{i}^{4}\leq 1\plus{}\frac{1}{n^{2}}\sum_{i\equal{}1}^{n}a_{i}^{3}$ :!:  :?:

[quote="duythuc_lqd"]1. Let $ a_{1},a_{2},......a_{n}$ satisying $ \sum_{i \equal{} 1}^{n}a_{i} \equal{} 0,\sum_{i \equal{} 1}^{n}a_{i}^{2} \equal{} 0$.Prove that $ \frac {1}{n}\sum_{i \equal{} 1}^{n}a_{i}^{4}\geq 1 \plus{} \frac {1}{n^{2}}\sum_{i \equal{} 1}^{n}a_{i}^{3}$
[/quote]
In this question ; I am assuming that $ a_i$'s are all/at least one of them is complex; because if
$ \sum a_i^2 \equal{} 0$ when $ a_i\in\mathbb{R}$ we always have $ a_1 \equal{} \cdots a_n \equal{} 0$.
 :huh:
EDIT: I did not see Peine's post ; before I posted it.  Extremely sorry.

[quote="kunny"]Don't mind it, everyone sometime makes a mistake. :)

If possible, this phrase may be mistake? :lol:[/quote]
Nope, probability of this statement being a mistake is $ \equal{} \frac {1}{infinity }$. :rotfl:

Hey, boy, you should do pay attention how to talk to [b]adults[/b]. :mad:

The second one:
Let $ x\equal{}a\minus{}b,y\equal{}b\minus{}c,z\equal{}c\minus{}d,u\equal{}d\minus{}a$,then $ x\plus{}y\plus{}z\plus{}u\equal{}0$
And we have
$ (a\minus{}c)^2(b\minus{}d)^2\equal{}(x\plus{}y)^2(y\plus{}z)^2\equal{}(x\plus{}y)(y\plus{}z)(z\plus{}u)(u\plus{}x)$
$ \equal{}(xyz\plus{}yzu\plus{}zux\plus{}uxy)(x\plus{}y\plus{}z\plus{}u)\plus{}(xz\minus{}uy)^2$
$ \equal{}(xz\minus{}yu)^2$
it remains to prove$ \equal{}(xz\minus{}yu)^2\plus{}xyzu\geq0$
But $ \equal{}(xz\minus{}yu)^2\plus{}xyzu\equal{}x^2z^2\plus{}y^2u^2\minus{}xyzu\geq2|xyzu|\minus{}xyzu\geq|xyzu|\geq0$
Q.E.D

[quote="duythuc_lqd"]
2. Let a,b,c,d be four positive real numbers. Prove that $ (a \minus{} b)(b \minus{} c)(c \minus{} d)(d \minus{} a) \plus{} (a \minus{} c)^{2}(b \minus{} d)^{2}\geq 0$
[/quote]

A stronger one holds in fact:

$ 4(a \minus{} b)(b \minus{} c)(c \minus{} d)(d \minus{} a) \plus{} (a \minus{} c)^{2}(b \minus{} d)^{2} \equal{} [(a \minus{} b)(c \minus{} d) \minus{} (a \minus{} d)(b \minus{} c)]^2\geq 0$ for all reals $ a,b,c,d$.  :lol:

Equivalently with $ x,y,z$ as above (the way to solve it without even a pencil :) )
\[ (x \plus{} y)^{2}(y \plus{} z)^{2} \minus{} 4xyz(x \plus{} y \plus{} z) \equal{} [y(x \plus{} y \plus{} z) \plus{} xz]^2 \minus{} 4[y(x \plus{} y \plus{} z)][xz] \equal{} [y(x \plus{} y \plus{} z) \minus{} xz]^2\ge0.
\]
:lol:


[quote="i_like_pie"]5 standard dice are rolled simultaneously. You get two 5's, two 4's and a 1. You want to get a "full house" by getting another 4 or 5 in at most 2 rolls of the last die. What is the probability of you succeeding?

Note: Based on Yahtzee, so those of you who play might already know.[/quote]
[hide]Complementary Counting
There's a $ \frac{2}{3}$ chance of misrolling. So, after two turns there's a $ \frac{4}{9}$ chance of not rolling a 2 or a 3, so our probability is $ \frac{5}{9}$[/hide]


function
calculus
integration
real analysis
real analysis unsolved

Could someone help me with this problem?

I need to construct a continuous function f: R into R such that its improper integral is zero, but which is unbounded as x approaches negative infinity and x approaches infinity.

Let $ f_n$ such that $ \forall_{x\in [n \minus{} \frac 1{2^n},n \plus{} \frac 1{2^n}]}f_n(x) \equal{} n cos(2^{n\minus{}1}\pi x), \ \forall_{x\notin [n \minus{} \frac 1{2^n},n \plus{} \frac 1{2^n}]}f_n(x) \equal{} 0$ . Manifestly $ F(x) \equal{} \sum_{n \equal{} 1}^\infty (f_n(x) \minus{} f_n( \minus{} x))$ satisfies .. :cool:


Καλωσορίζοντας τον νέο φίλο μας [b][size=100]george2[/size][/b], βάζω μόνο το σχήμα, αντί κάποια άλλη υπόδειξη.

Κώστας Βήττας.

[quote="george2"]Δίνεται ορθογώνιο τρίγωνο ΑΒΓ (Η γωνία Α είναι ορθή) και ΑΔ το ύψος του. Από σημείο Ε της ΑΒ φέρνουμε παράλληλη προς την ΒΓ που τέμνει το ύψος ΑΔ στο Ζ.Στη συνέχεια φέρνουμε την κάθετη στην ΖΓ στο σημείο Ζ που τέμνει την ΑΒ στο Η.Να αποδείξετε ότι ΑΗ=ΒΕ.[/quote]

Σύμφωνα με την υποδειξη του Κώστα ,

  τα τρίγωνα $ CAH, CDZ$  είναι όμοια διότι $ \angle CZH \equal{} \angle ZAB \equal{} \angle ACB$ , δηλαδή $ \angle ACH \equal{} \angle DCZ$( ή διότι από το εγράψιμο πάλι είναι $ \angle CHA \equal{} \angle CZD$ , όπως εύστοχα σημειώνει ο Κώστας. .

Eπομένως   :

 $ \frac {DZ}{DC} \equal{} \frac {AH}{AC}$

 Αλλά

 $ \frac {AC}{DC} \equal{} \frac {AB}{AD}$

Άρα

   $ HA \equal{} DZ \cdot \frac {AC}{DC} \equal{} DZ \cdot \frac {AB}{AD} \equal{} \frac {DZ}{AD} \cdot AB \equal{} \frac {EB}{AB} \cdot AB \equal{} EB$

Κώστα , σε ευχαριστούμε πολύ και για το σχήμα :) !

Μπάμπης

Fernoume apo to Z parallili pros tin AB pou temnei tin BC sto P. Tote gwnia(ZPC)=gwnia(ZHC) kai ara oi per. kikloi twn trigwnon ZPC kai ZHC einai isoi. Alla  gwnia(ACH)=gwnia(ZCP) opote AH=ZP=BE.

(Euxaristw kai egw ton kurio Vitta gia to sxima tou).

Ευχαριστώ τον κύριο vittasko για το καλωσόρισμα αλλά και για το σχήμα του.
Η λύση μου:
Έφερα παράλληλη στην ΑΒ που τέμνει τις ΑΓ και ΓΒ στα Μ,Κ αντίστοιχα.
Το ΖΚΒΕ είναι παραλληλόγραμμο.Έφερα την ΑΚ.
Στο τρίγωνο ΑΖΓ τα ύψη ΖΜ και ΓΔ τέμνονται στο Κ.(άρα Κ ορθόκεντρο του τριγώνου ΑΖΓ)Έτσι ΑΚ κάθετη στη ΓΖ άρα ΖΗ παράλληλη της ΑΚ.Άρα το ΖΗΑΚ είναι παραλληλόγραμμο. Έτσι προκύπτει ότι ΑΗ=ΕΒ.

[quote="staretak"]Fernoume apo to $ Z$  parallili pros tin $ AB$ pou temnei tin $ BC$ sto $ P.$ Tote $ \angle ZPC \equal{} \angle ZHC$ kai ara oi per. kikloi twn trigwnon $ \bigtriangleup ZPC$ kai $ \bigtriangleup ZHC$ einai isoi. Alla  $ \angle ACH \equal{} \angle ZCP$ opote $ AH \equal{} ZP \equal{}   BE.$[/quote]
Ωραία λύση [b][size=100]staretak[/size][/b]. :)

Κώστας Βήττας

[quote="george2"]...Έφερα από το $ Z$ παράλληλη στην $ AB$ που τέμνει τις $ A\Gamma$  και $ \Gamma B$  στα $ M,$ $ K,$ αντίστοιχα.

Το $ ZKBE$ είναι παραλληλόγραμμο. Έφερα την $ AK.$

Στο τρίγωνο $ \bigtriangleup AZ\Gamma$ τα ύψη $ ZM$ και $ AD,$ τέμνονται στο $ K$ $ ($ και άρα το $ K,$ είναι το ορθόκεντρο του τριγώνου $ \bigtriangleup AZ\Gamma$ $ ).$

Έτσι η $ AK$ είναι κάθετη στη $ \Gamma Z$ και άρα $ ZH\parallel AK.$ Άρα το $ ZHAK,$ είναι παραλληλόγραμμο και έτσι προκύπτει ότι $ AH \equal{} EB.$[/quote]
Γιώργο, νομίζω ότι αυτή είναι η πιο όμορφη λύση.  :)

Κώστας Βήττας.


algebra
polynomial
linear algebra
matrix
vector
linear algebra unsolved

Suppose that $A,B$ are two $nxn$ matrices in an arbitrary field that verify $rank(A-B)=1$ and have caracteristic polynomials $X^n+1, X^n-1$. Prove that the group generated by these two matrices has at most $(2n)!$ elements.

May be I am wrong, but if I didn't make stupid mistakes,
this group must always have $2^{n} n$ elements.

I think you are wright Leval1980, it is $n2^n$

Do you know the year they use this problem at CCP ?
thank you

This problem was the last question of the CCP second math examination in 2001.

There were four parts in this problem (which was about companion matrices), and the last one studied those matrices $ A, B$ satisfying $ \text{rank}(A \minus{} B) \equal{} 1.$

Especially, the aim of this part was to prove the following result $ (A, B \in \mathrm{GL}_n(\mathbb{K})) :$
$ (\text{rank}(A \minus{} B) \equal{} 1$ and $ \chi_A, \chi_B$ coprime$ )$ $ \iff$ $ (\exists P \in \mathrm{GL}_n(\mathbb{K}) \ \ / \ \ A \equal{} P^{ \minus{} 1} C_A P$ and $ B \equal{} P^{ \minus{} 1}C_BP).$
(where $ \chi_M$ denotes the characteristic polynomial of the matrix $ M$ and $ C_M$ denotes the companion matrix of $ ( \minus{} 1)^n \chi_M.$)

Then, the 23rd question was :
[color=blue]Let $ a$ and $ b$ be two automorphisms of a $ \mathbb{K} \minus{}$vector space $ V$ with dimension $ n$ satisfying $ \text{rank}(a \minus{} b) \equal{} 1, \chi_a(X) \equal{} ( \minus{} 1)^n(X^n \plus{} 1)$ and $ \chi_b(X) \equal{} ( \minus{} 1)^n(X^n \minus{} 1).$
Using a group action, show that the group $ G$ generated by $ u$ and $ v$ is finite with cardinal less than or equal to $ (2n)!$[/color]

As $ \chi_a$ and $ \chi_b$ are easily seen to be coprime, we may apply the result I've stated above and get that there is some basis $ \mathcal{B}$ of $ V$ such that
$ \text{Mat}_{\mathcal{B}}(a) \equal{} C_A \equal{} \begin{pmatrix} 0 & \ldots & & \ldots & 0 & \minus{} 1 \\
1 & \ddots & & & & 0 \\
0 & \ddots & & & & \vdots \\
\vdots & \ddots & & & \\
\vdots & & & \ddots & \ddots & \vdots \\
0 & \ldots & \ldots & 0 & 1 & 0 \end{pmatrix}$ and $ \text{Mat}_{\mathcal{B}}(b) \equal{} C_B \equal{} \begin{pmatrix} 0 & \ldots & & \ldots & 0 & 1 \\
1 & \ddots & & & & 0 \\
0 & \ddots & & & & \vdots \\
\vdots & \ddots & & & \\
\vdots & & & \ddots & \ddots & \vdots \\
0 & \ldots & \ldots & 0 & 1 & 0 \end{pmatrix}$
But those matrices are clearly in $ \mathcal{O}_n(\mathbb{Z}),$ which is a group with cardinal $ 2^n n!,$ so that $ |G| \leq 2^n n!,$ we're done. $ ((2n)! \geq 2^n n!))$

Moreover, it is true that, as noticed by [b]Leva1980[/b], $ |G| \equal{} 2^n n,$ but I'll leave this to the (interested) reader. :)


Let $ a,b,c$ are non-negative real number. Find the mind value and max value (if these exist) of

$ P\equal{}\sqrt{\frac a{4a\plus{}b\plus{}c}}\plus{}\sqrt{\frac b{4b\plus{}c\plus{}a}}\plus{}\sqrt{\frac c{4c\plus{}a\plus{}b}}$

[quote="dduclam"]Let $ a,b,c$ are non-negative real number. Find the mind value and max value (if these exist) of

$ P \equal{} \sqrt {\frac a{4a \plus{} b \plus{} c}} \plus{} \sqrt {\frac b{4b \plus{} c \plus{} a}} \plus{} \sqrt {\frac c{4c \plus{} a \plus{} b}}$[/quote]
We will prove that $ P \leq \sqrt {\frac {3}{2}}$
Note that $ P \leq \sqrt {3.\sum_{cyc} \frac {a}{4a \plus{} b \plus{} c}}$
We need prove that $ \sum_{cyc} \frac {a}{4a \plus{} b \plus{} c} \leq \frac {1}{2}$
We can assume that $ a \plus{} b \plus{} c \equal{} 3$ then this inequality becomes $ \sum_{cyc} \frac {a}{a \plus{} 1} \leq \frac {3}{2}$
or $ \sum_{cyc} \frac {a \minus{} 1}{a \plus{} 1} \leq 0$
We can suppose that $ a \geq b \geq c$ then $ a \minus{} 1 \geq b \minus{} 1 \geq c \minus{} 1$ and $ \frac {1}{a \plus{} 1} \leq \frac {1}{b \plus{} 1} \leq \frac {c}{c \plus{} 1}$ and by Chebyshep we have done!
Equality occurs when $ a \equal{} b \equal{} c$

The infimum is $ \frac{1}{2}$, reached by (1,0,0). WLOG assume $ a\plus{}b\plus{}c \equal{} 1$, and note that
$ \sqrt{\frac{a}{3a\plus{}1}} \plus{} \sqrt{\frac{b}{3b\plus{}1}} \ge \sqrt{\frac{a\plus{}b}{3(a\plus{}b)\plus{}1}}$
by squaring it.

[quote="chien than"] We will prove that $ P \leq \sqrt {\frac {3}{2}}$
Note that $ P \leq \sqrt {3.\sum_{cyc} \frac {a}{4a \plus{} b \plus{} c}}$
We need prove that $ \sum_{cyc} \frac {a}{4a \plus{} b \plus{} c} \leq \frac {1}{2}$
We can assume that $ a \plus{} b \plus{} c \equal{} 3$ then this inequality becomes $ \sum_{cyc} \frac {a}{a \plus{} 1} \leq \frac {3}{2}$
or $ \sum_{cyc} \frac {a \minus{} 1}{a \plus{} 1} \leq 0$
We can suppose that $ a \geq b \geq c$ then $ a \minus{} 1 \geq b \minus{} 1 \geq c \minus{} 1$ and $ \frac {1}{a \plus{} 1} \leq \frac {1}{b \plus{} 1} \leq \frac {c}{c \plus{} 1}$ and by Chebyshep we have done!
Equality occurs when $ a \equal{} b \equal{} c$[/quote]

nice solution,thank you chien than :D

Another nice and similar inequality of mine is

\[ \frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b} \le \frac{1}{3}.\]

Equality holds for $ a=b=c$ or $ (a,b,c)\~(2,1,0).$

This implies

\[ \sqrt{\frac{a}{4a+4b+c}}+\sqrt{\frac{b}{4b+4c+a}}+\sqrt{\frac{c}{4c+4a+b}} \le 1.\]

[quote="hungkhtn"]Another nice and similar inequality of mine is
\[ \frac {a}{4a + 4b + c} + \frac {b}{4b + 4c + a} + \frac {c}{4c + 4a + b} \le \frac {1}{3}.
\]
Equality holds for $ a = b = c$ or $ (a,b,c)\~(2,1,0).$

This implies
\[ \sqrt {\frac {a}{4a + 4b + c}} + \sqrt {\frac {b}{4b + 4c + a}} + \sqrt {\frac {c}{4c + 4a + b}} \le 1.
\]
[/quote]
We have:
$ \sum_{cyc} \sqrt{\frac{a}{4a+4b+c}} \leq \sqrt{3.\sum_{cyc} \frac{a}{4a+4b+c}}$
Suppose that $ a+b+c=3$ then we need prove:
$ \sum_{cyc} \frac{a}{a+b+1} \leq 1$
or $ a^2b+b^2c+c^2a+abc \leq 4$
 :P


Let`s somebody suggest us some interesting pH problems ( or any analytical problems ) from chemistry olympiads around the world.

[b]The pH of Buffer Solutions Problem[/b]

1. An amount of $ \text{NH}_3$ gas $ (27^{\text{o}}\text{C}, 3 \text{atm})$ are bubbled into a $ 1$ L solution of $ \text{H}_2\text{SO}_4$, so the pH changes from $ 1$ to $ 9$. Calculate the volume of $ \text{NH}_3$ gas that has been bubbled into that solution.
$ K_b \text{ NH}_3 \equal{} 10^{ \minus{} 5}$
$ R \equal{} 0.082 \text{ L atm}\text{ mol}^{ \minus{} 1}\text{ K}^{ \minus{} 1}$

2. A $ 7.15$ gram sample of hydrated sodium carbonate, is dissolved into water to give $ 250$ mL of solution. Several $ 150$ mL of this solution was added into $ 200$ mL of $ 0.4$ M $ \text{HF}$ solution, and the pH of this mixture was found to be $ 3$. Find the correct formula of the hydrate.
$ K_a \text{ HF} \equal{} 6 \times 10^{ \minus{} 4}$

3. A buffer solution was made by dissolving $ 2.45$ grams of $ \text{H}_2\text{SO}_4$ into $ 500$ mL of a $ \text{NH}_3$ solution. A $ 1.12$ liter of $ \text{HBr}$ gas (STP) was bubbled into the buffer solution. After this process, the pH of that solution was found to be $ 9$. Determine the pH of the original $ \text{NH}_3$ solution.
$ K_b \text{ NH}_3 \equal{} 10^{ \minus{} 5}$

4. A $ 2.75$ gram of gaseous mixture containing acetylene $ (\text{C}_2\text{H}_2)$ and propene $ (\text{C}_3\text{H}_6)$ was combusted with an exact amount of oxygen. After the cooling process, the formed gas was bubbled into $ 300$ mL of $ 0.3$ M $ \text{NaHCO}_3$ solution, and the pH changed to $ 6$. Determine the composition of the mixture.
$ K_a \text{ H}_2\text{CO}_3 \equal{} 4.5 \times 10^{ \minus{} 7}$


superior algebra
superior algebra unsolved

[quote="wiosna"]Show that Weyl algebra K[x][y.d] is a simple ring iff characteristic K =0[/quote]

Suppose first that $ \text{char} K \equal{} p > 0.$ Then $ x^p$ will be a central element because $ yx^p \equal{} x^py \plus{} d(x^p) \equal{} x^py \plus{} px^{p \minus{} 1} \equal{} x^py.$ Thus, in this case, the (two sided) ideal

generated by $ x^p$ is a non-zero proper ideal of $ A \equal{} K[x][y,d],$ and hence $ A$ is not simple.

So let $ \text{char} K \equal{} 0.$ Suppose $ (0) \neq I \neq A$ is a two sided ideal of $ A.$ If $ I \cap K[x] \neq \{0 \},$ then choose $ z \equal{} a_mx^m \plus{} \cdots \plus{} a_0 \in I \cap K[x], \ a_m \neq 0,$ of minimal degree.

If $ m \equal{} 0,$ then $ z \equal{} a_0 \in K,$ and hence $ 1 \in I$ resulting in the contradiction $ I \equal{} A.$ If $ m \geq 1,$ then look at $ yz \minus{} zy \equal{} ma_mx^{m \minus{} 1} \plus{} (m \minus{} 1)a_{m \minus{} 1}x^{m \minus{} 2} \plus{} \cdots \plus{} a_1 \in I,$

which is impossible by minimality of $ m$ and the fact that $ ma_m \neq 0.$ So we may assume that $ I \cap K[x] \equal{} \{0 \}.$ Let $ 0 \neq z \equal{} r_ny^n \plus{} r_{n \minus{} 1}y^{n \minus{} 1} \plus{} \cdots \plus{} r_0 \in I, \ r_j \in K[x],$

and $ r_n \neq 0,$ be an element of minimal degree. Clearly $ n \geq 1.$ Now since $ y^j x \equal{} xy^j \plus{} jy^{j \minus{} 1},$ we have: $ zx \minus{} xz \equal{} nr_ny^{n \minus{} 1} \plus{} \text{terms with lower degree} \in I.$ So $ nr_n \equal{} 0,$

by minimality of $ n,$ which is impossible because $ n \geq 1$ and $ r_n \neq 0.$


Given that $\sum a_n$ diverges, show that $\sum\frac{a_n}{1+a_n}$ diverges as well.  $a_n>0$

Any ideas?

If there are infinitely many $a_n>1$, then there are infinitely many $\frac{a_n}{1+a_n}>\frac 12$, and you're done. If, on the other hand, for large enough $n$ we have $a_n\le 1$, then for large enough $n$ we have $\frac{a_n}{1+a_n}\ge\frac{a_n}2$, and, again, we're done.


Lebesgue had suggested the following Diophantine equation

x^2 - y^3 = 7

Let's try!

Namdung

Posted already, see

http://www.mathlinks.ro/viewtopic.php?t=494

em... it's not quite the same problem, is it?? :?


advanced fields
advanced fields unsolved

[quote="countdownmath"]Use a suitable inversion to show that, if $l$ is Poincare line represented by a semicircle with diameter $PQ$ on the $x$-axis, then the locus of points of given distance $d$ from $l$ and on a given side of $l$ is represented by a Euclidean circular arc connecting $P$ and $Q$.[/quote]

Let $l$ be a bowed geodesic in the Poincare Plane. Use the inversion $I_{P,PQ}$ then this will map $l$ to the straight geodesic through $P$. Now the locus of points which are equidistant from $I(l)$ are a pair of Euclidean rays $R_{1},R_{2}$ emanating from $P$ from either side and equal angles. Then if you take the inversion back, that is, $I(R_{1}),I(R_{2})$ you get two semi-circles passing through $P,Q$.

[quote="countdownmath"]thanks for the reply sylow..
but is it possible to be more specific about what the inversion actually is? (i.e. using formula, or construction)[/quote]

I did, the inversion through point P, having the radius equal to PQ.

o, sorry, i wasn't aware of the notation. Thanks.

[quote]Now the locus of points which are equidistant from $I(l)$ are a pair of Euclidean rays $R_{1},R_{2}$ emanating from $P$ from either side and equal angles. Then if you take the inversion back, that is, $I(R_{1}),I(R_{2})$ you get two semi-circles passing through $P,Q$.[/quote]

I don't quite follow this part of your argument... How would you actually visualize the pair of Euclidean rays equidistant from $I(l)$ from $P$?
Also, why is it that we get the two semicircles passing through $P,Q$ from $I({R_{1}}), I({R_{2}})$? I mean, isn't that exactly the same as $l$?


My question was that the solution said that the probability that the five WinnerBalls match is $ \frac {1}{\binom {44}{5}}$. Shouldn't it be $ \frac {1}{\binom {49}{5}}$ since the problem says among 44 OTHER numbered balls? The problem should be edited to remove other or clarified.

The other key word is [b]among[/b], meaning the five WinnerBalls are picked from those 44 balls.  The "other" is meant to distinguish between the 44 from which the WinnerBalls are drawn and the 27 from which the MegaBall is drawn.

English is so weird. Thanks for the clarification.

[quote="GameBot"]The lottery in our state consists of two drawings.  First, a MegaBall is picked from among 27 numbered balls.  Second, five WinnerBalls are picked from among 44 other numbered balls. To win the lottery, you must pick the MegaBall number correctly and also pick the numbers on the five WinnerBalls (but you don't need to get the order right for the WinnerBalls).  What is the probability that the ticket I hold has the winning numbers?[/quote]
I feel dumb. I didn't really understand how to approach this problem. Maybe I misunderstood it...I don't know.  :blush:

[hide="So would the answer be"] $ \frac{1}{\binom{44}{5}}+\frac{1}{\binom{27}{1}}$[/hide], [hide="or would it be"] $ \frac{1}{\binom{44}{5}}\times{\frac{1}{\binom{27}{1}}}$?[/hide]

[hide="Answer"]
It should be $ \frac{1}{\binom{44}{5}}\times{\frac{1}{\binom{27}{1}}}$ which equals ${\frac{1}{29,322,216}}$.[/hide]

[hide="Complicated things made simple:"]
The probability that the MegaBall matches is $ \frac{1}{27} $, The probability that the 5 WinnerBalls match is $ \frac{1}{\binom{44}{5}} $. So my chances of winning are $ \left(\frac{1}{27}\right)\times\left(\frac{1}{\binom{44}{5}}\right) =\boxed{\frac{1}{29,\!322,\!216}} $.[/hide]

[quote=stevenydc]is there a lot heavy duty calculations like that on AMC12? cuz this one is pretty easy but i made stupid mistakes on calculations...[/quote]

lol I just used a calculator :P this would never be on an AMC ... or at least I hope


limit
topology
Functional Analysis
real analysis
real analysis unsolved

Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be continious bijection s.t. $\lim_{||x|| \to \infty} ||f(x)|| = \infty$ then $f$ is homemorphism.

We prove that if $f: \mathbb{R}^n \to \mathbb{R}^m$ is a continuous mapping with the property that $|f(x)| \to \infty$ when $|x| \to \infty$ then $f$ is
a closed mapping. This proves the stated theorem. Let $F \subset \mathbb{R}^n$ be closed and $y \in \overline{fF}$. There exists $M > 0$ so that
$|x| > M \Longrightarrow |f(x)| > |y| + 1$. Let $F^* = \{x \in F: |x| \le M\}$. This set is closed in $F$ which is closed in $\mathbb{R}^n$
which means that $F^*$ is closed in $\mathbb{R}^n$. It is obviously bounded and so, by Heine-Borel, compact.  Now note that
$f(F \setminus F^*) \cap B(y, 1) = \emptyset$. We also note that $\overline{fF} = \overline{f([F \setminus F^*] \cup F^*)} = \overline{f(F \setminus F^*)} \cup \overline{fF^*}$ and $y \not \in \overline{f(F \setminus F^*)}$.
This means that $y \in \overline{fF^*}$. Because $f$ is continuous, $fF^*$ is compact, especially closed. Now we have that
$y \in fF^* \subset fF$. We have proved that $\overline{fF} = fF$ which means that $f$ is a closed mapping.

Here's a theorem due to Banach and Mazur, I believe, which is a good deal stronger :):

Let $E,F$ be Banach spaces, and let $f: E\to F$ be a local homeomorphism. Then the following assertions are equivalent:

(a) $f$ is a proper map (i.e. the preimage of every compact set is compact);

(b) $f$ is a closed map (the image of every closed set is closed);

(c) $f$ is a homeomorphism.

A problem very similar to this one has been discussed on the forum before (the topic was started by harazi). The difference was that in that problem $f$ was a local diffeomorphism (and the Banach spaces $E,F$ were actually $\mathbb R^n$ for some $n$). Anyway, the proof in that thread (if you can find it :)) basically works for this result as well.

Clearly, a continuous map from $\mathbb R^n$ to itself which goes to infinity at $\infty$ (such as the one in the problem posted by eugene) is proper, and any continuous bijection $f: \mathbb R^n\to\mathbb R^n$ is a local homeomorphism, so this is a particular case of the result cited above.

Thank you very much.
Grobber, could you please tell me where i can find a proof of this nice theorem by Banach and Mazur: in Functional Analysis textbooks i have i can't find it.

I've been looking through some of he problems in the "Real Analysis" section today, and I only saw your reply now, eugene :). So, sorry for the delay I guess.

You can modify the proof in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=space+covering&t=74822]this[/url] thread. Obviously, (c) implies both (a) and (b), so we're interested in (a) iff (b) and (a) or (b) implies (c).

(a) $\Rightarrow$ (b) is proven in the last post in that topic, second paragraph.

(b) $\Rightarrow$ (a):

Let $K\subset F$ be a compact set, and ssume $f^{-1}(K)$ is not compact. We can then extract a sequence $(x_{n})_{n}$ from $f^{-1}(K)$ such that $\|x_{n}-x_{m}\|\ge c>0,\ \forall m\ne n$ for a constant $c$. Let $z\in K$ be a limit point of $\{f(x_{n})\}$ (such a limit point exists because $K$ is compact). Around each $x_{n}$, there is a ball $B_{n}$ of radius $<\frac{c}{3}$ which is mapped homeomorphically onto its image by $f$. Choose $y_{n}\in B_{n}$ such that $f(y_{n})\ne z$, but $f(y_{n})\to z$. The closed set $\{y_{n}\}$ is mapped onto a set which is not closed (its limit point $z$ does not belong to it), and this contradicts (b).

For (a), (b) $\Rightarrow$ (c) again, slightly modify the proof in that other thread. The same idea (proving that $f$ is a covering map from $E$ onto $F$) works practically word for word.


algebra
polynomial
geometry
algebra proposed

Find all the polynomials P(x,y) such that for every a,b,c,d we have that
P(a,b)P(c,d)=P(ac+bd,ad+bc). :rotfl:  ;)

Third day no solution I knoe it is difficult. Tomorrow I will post my solution

What actually do you want, man?  I see it is in the proposed section area, thus you're not hurried to see a solution? And if you really want to know, it's not that difficult, but long to write. So I prefer to direct you to kalva site and read there. It is given in some old Balkan mathematical olympiad. These repeated posts with "I'll post my solution" and so on are not very welcomed here.

I only said something. This is no reason to speak to me like this! :mad: Why you always (from the first day I was logged in) tease and scold me? I make you something?

I think that you understand what I said :mad:  :(

:rotfl:  :rotfl:  :rotfl:  :rotfl: this is funny....sorry to interrupt, no offence intended, but this IS pretty funny, just read it.... :rotfl:  :rotfl:  :rotfl:  :rotfl:

silouan ,you are better to calm down .
Harazi is moderator of the Forum, and his is one of the best guys here.
Actually he never want to insult someone or teasing or ... ,you know ,lots of times ,different guys said they will post there solution tommorow an nothing happend that why he was saying [quote]repeated posts with i will send my solution[/quote]
Be sure there is no fight here.
And i apologize for intrupting

PS.I will try to find a solution for this ,at least for doing something usefull

Here you are :
[url]http://www.kalva.demon.co.uk/balkan/bsoln/bsol882.html[/url]
Harazi said where the solution was ,Didnt he ?
But also can we see your solution ,silouan?
I am interested.

After all these I want to say SORRY to all guys for my behaviour.First of all a big SORRY to Harazi. :oops:

My solution
Lemma 1
Let the polynomial P(x,y) then we have
a) If P(x,x)=0 for any$ x\in R,C$ then $x-y| P(x,y)$
b)If P(x,-x)=0 for any$ x\in R,C$ then $x+y| P(x,y)$

Lemma 2
We suppose all the polynomials P(x,y), Q(x,y) with coefficients at R,C then we have
a)If $P(x,y)Q(x,y)\equiv 0 $and Q(x,y) is not a 0 ponynomial than $P(x,y)\equiv 0$
b)If $P(x,y)Q(x,y)\equiv 0 $then $P(x,y)\equiv 0$ or $Q(x,y)\equiv 0 $

now to the solution
we put a=b=c=d=0 at the equation we want to solve and we have
P(0,0)P(0,0)=P(0,0) <=> P(0,0)=1 ar P(0,0)=0
If P(0,0)=1 we have P(x,y)=1
So the constant polynomial P(x,y)=1 is a solution to our problem.
If P(0,0)=0 then $P(x,y)\equiv 0 $so this is an anothar solution to our problem.
At the last case if we put at the equation which want to solve a=b=c and c=-d=w we will have
P(x,x)P(w,-w)=0 and with Lemma 2
P(x,x)=0 or P(w,-w)=0 and from the Lemma 1 we have that $x-y| P(x,y) $or $x+y| P(x,y)$
so we could write $P(x,y)=(x-y)^m(x+y)^nQ(x,y)$   (2) where m,n the major that the (2) holds.
We use the P(x,y) as it is at (2) is the equation we want to solve.
And after calculations we have that $Q(a,b)(Q(c,d)=(a-b)^m(c-d)^n(a+b)^n(c+d)^n$
so Q(a,b)(Q(c,d)=Q(ac+bd,ad+bc)
So Q(x,y)=1 because in the second case we have contradiction
So $P(x,y)=(x-y)^m(x+y)^n$

What are your opinions Harazi and Lumos for my solution?

Sorry, I don't have the time to read it carefully, but at first glance seems the same with the solution I knew.

Well ,I think it is logical way of thinking for this problem.
Flat and smooth.
Nice solution.


[quote="nguoivn"]Given $ a_{1},...,a_{n}$ are in $ [0,1]$. Prove that:
\[ \dfrac{a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \plus{} n.a_{1}.a_{2}....a_{n}}{(a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n})^2} \ge \dfrac{1}{n \minus{} 1}\]
:)[/quote]
Let $ a_{1},...,a_{n}$ such that LHS is minimal. We can assume wlog $ a_1\le\cdots\le a_n$.

Suppose $ a_2 < 1$. Now $ a_1a_2 \minus{} (a_1 \plus{} a_2 \minus{} 1)\cdot 1 \equal{} (1 \minus{} a_1)(1 \minus{} a_2) > 0$, so changing $ (a_1,a_2)\to (a_1 \plus{} a_2 \minus{} 1,1)$ [hide][i]Note that $ a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}\ge n \minus{} 1$ (otherwise the ineq holds obviously) and thus $ a_{1} \plus{} a_{2}\ge1$ [/i][/hide] makes the product and thus the LHS smaller, as the sum remains constant. Contradiction. So we must have $ a_{2} \equal{} ... \equal{} a_{n} \equal{} 1$ and for this case the inequality is trivial. Equality holds for $ a_1 \equal{} 0$. :)


I have the question & the answer, but I don't understand how to get from one to the other (i.e. the process).  :blush:  :oops:

Question: Simplify [img]http://i118.photobucket.com/albums/o116/arachne8/question.jpg[/img]

Answer: [img]http://i118.photobucket.com/albums/o116/arachne8/answer.jpg[/img]

Can someone help me out with the process?

Kind regards,
Elizabeth

$ (p\minus{}q)^2\equal{}(q\minus{}p)^2$, so apply it here.

$ \sqrt{(q\minus{}p)^2\plus{}(p\minus{}q)^2}$=$ \sqrt{(p\minus{}q)^2\plus{}(p\minus{}q)^2}$=$ \sqrt{2(p\minus{}q)^2}$ because $ (p\minus{}q)^2\equal{}(q\minus{}p)^2$


LaTeX
modular arithmetic
geometric series

A poor woman goes to the rich emperor of China and gives a marvelous dancing performance. The emperor congratulates her and is so impressed that he will grant her anything she wishes. The woman asks for one penny. The emperor begins to laugh, but the woman goes on. She says that every day for the next 30 days, the emperor must give her triple of what she received the previous day. How many dollars and cents will the woman have after 30 days?

[quote="gauss1181"][quote="hm29168"]So it's basically [hide]$ 3^0 + 3^1 + ... + 3{}^3{}^0$ cents[/hide][/quote]
Well, in simplest form, that'd be $ 3^0 + 3^1 + ... + 3{}^3{}^0$ = $ 1 - 3{}^3{}^0 / ( - 2)$ = $ 1.03 * 10{}^1{}^4$ cents
[/quote]
For future reference, use[code]\cdots[/code]for $ \cdots$ instead of $ ...$, [code]\frac{x}{y}[/code]for $ \frac{x}{y}$, and [code]\times[/code]for $ \times$ instead of $ *$.

[hide]So it's $ 3^0\plus{}3^1\plus{}\cdots\plus{}3^{28}\plus{}3^{29}\plus{}3^{30}$. Now, we could sum this up using a geometric series approach, or use the fact that $ 1\plus{}x\plus{}x^2\plus{}\cdots{x^n}\equal{}\frac{x^{n\plus{}1}\minus{}1}{x\minus{}1}$, which is also derived from using a geometric series.
So, here it's $ \frac{3^{31}\minus{}1}{2}$ cents.
Hmm, dollars and cents is going to be somewhat tricky without a calculator. Let's take it (mod 100) to find the cents. Since $ 3^{20}$ is 1 (mod 100), we can write this as $ 3^{11}\pmod{100}$. Now, this is $ 81*81*27$, which is also $ 361*27$ (mod 100) which is 47 (mod 100). So the cents is 23.

Uh, I'll just skip the dollars part. [/hide]

@xpmath - Good idea:  She gets $ \$3088366981419.73$  Also, remember that division doesn't always give you just one answer in mods

Awwwww.
I must've been pretty bored to do all that stuff.

Isn't this like one of those get rich quick schemes were you ask for one cent on the first day, 2 on the second, 4 on the third, ect? Any way, a lot of middle schoolers would somply whip a calculator and do 1 times 3, 3 times 3, ect 30 times.

a big number.
probably more than the emperor has.


This isn't really combinatorics, but it's cute and I can't think of a better place to put it.

For reference, the finite version:  $ N$ prisoners are lined up in a row, all facing the same direction, and given hats that are either black or white.  They are told that, starting from the back of the row (the prisoner who can see everyone else), they are to guess their hat color, and if they are correct, they are set free.  Otherwise, they stay in prison.  What is the strategy that maximizes the number of prisoners guaranteed to be set free, and what is that strategy?

The infinite version:  now there are a countable number of prisoners.  Every prisoner can see all but finitely many of the other prisoners (the ones in front of him).  Does there exist a strategy that guarantees that all but finitely many of the prisoners can be guaranteed to be set free?

[url=http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle]This page[/url] seems to give a pretty good description of different variants of the problem, including yours.


hi please help, i'm trying to solve this Initial Value Problem:

dy/dx  = x^2(y^2 + y - 2)
y(1) = 0

thank you

-Trevor

Whoops.  But it's so easy to misread text.

There's a very obvious method here: separation of variables. Integrate both sides of Moubinool's expression and solve for $y$ (if possible).


function
limit
calculus
calculus computations

Is there a function $ f: R\rightarrow R$ such that for any other function $ g: R\rightarrow R$ ,the limit

$ \lim_{x\rightarrow 1}[f(x)\plus{}g(x)]$ does not exist?

no there is not. suppose is, then choose $ g(x)\equal{}\minus{}f(x)$ and limit will equal to $ 0$ and exist.


superior algebra
superior algebra theorems

Find the number of elements in the set {$a \in S_{4}| a(3) = 3$}

That is a factorial.

The stabilizer of a point in $S_{n}$ is isomorphic to $S_{n-1}$; there's a natural correspondence between those permutations and permutations of the other $n-1$ elements.


vector
calculus
category theory
linear algebra

We have three orthonormal vectors       $ \vec i_1 , \vec i_2, \vec i_3$ , and we know  wich are the components of an arbitrary  vector $ \vec A$  in this base, explicitly:

$ \vec A \equal{} (\vec A \bullet \vec i_1) \vec i_1 \plus{} (\vec A \bullet \vec i_2) \vec i_2 \plus{} (\vec A \bullet \vec i_3) \vec i_3$

If now we want to generalize to a base that is orthogonal , but is NOT normalized, we can divide by the modulus, and we came back to the first case, so we have:

$ \vec i_1 \equal{} \frac {\vec e_1}{e_1} , \vec i_2 \equal{} \frac {\vec e_2}{e_2}, \vec i_3 \equal{} \frac {\vec e_3}{e_3}$

So now the expression of $ \vec A$ is:

$ \vec A \equal{} \frac {\vec A \bullet \vec e_1}{e_1^2} \vec e_1 \plus{} \frac {\vec A \bullet \vec e_2}{e_2^2} \vec e_2 \plus{} \frac {\vec A \bullet \vec e_3}{e_3^2} \vec e_3$  Equation 1

In the next case the base will be noncoplanar, not orthogonal and the vectors won't be normalized, I am following the book from Borisenko and Taraponov about tensor calculus, they introduce the reciprocal bases to solve this problem to arrive at this expression:

$ \vec A \equal{} (\vec A \bullet \vec e^1) \vec e_1 \plus{} (\vec A \bullet \vec e^2) \vec e_2 \plus{} (\vec A \bullet \vec e^3) \vec e_3$

Where the reciprocal base is defined as:

$ \vec e^1 \equal{} \frac {\vec e_2 \times \vec e_3}{\vec e_1 \bullet (\vec e_2 \times \vec e_3 )}$

e^2, and e^3 have a similar definition in terms of e_1, e_2 and e_3.

Well the question is ¿why do we need to introduce the reciprocal base to solve this problem? ¿what is the benefit of using it?.

I don't know why you don't calculate the components of A using equation 1 . You can project A to the vector e_1, e_2 and e_3  using the dot product and you find the components of A ,in a similar way as you have done before with the other basis ¿is there something that impede you to make the calculation in this way?.

And you have another  option, simply using this equality:

$ \vec A \equal{} \vec A$

First A expressed in the orthonormal base , and then expressed in the more general noncoplanar, nonorthogonal and not normalized :

$ (\vec A \bullet \vec i_1) \vec i_1 \plus{} (\vec A \bullet \vec i_2) \vec i_2 \plus{} (\vec A \bullet \vec i_3) \vec i_3 \equal{} A^1 \vec e_1 \plus{} A^2 \vec e_2 \plus{} A^3 \vec e_3$

And you have three equations wich can be easily solved , so you wold not need introduce other base ¿is anything wrong in this last equation?

The book then introduce the concept of covariant and contravariant components of a vector, and I would like to understand why do you need the reciprocal base, and what it is.

¿do you have any suggestions? :maybe:

Please tell me something, though it was a stupid question that a kid of 5 years can understand :surrender:


Show that if every point in the plane is colored either red or blue, there exist some two points that are one unit apart.

Two points of the same color or two points of opposite color?

If your problem is what I think it is, then just consider an equilateral triangle.

Otherwise, there are trivially two points that are one unit apart....

i think the proof goes sumtin lyk tis:

u have n boxes and n + 1 pigeons. put n of the pigeons in each of the boxes. the remaining pigeon must go into one of the already fitted boxes.

And there's the proof. Its only the base case though.

:rotfl:

Life would be much easier if I could solve every pigeonhole problem that awy.  Alas, my friend, most of the time you must specify what the boxes and the pigeons are.

No one solved yet?

[hide]
If the plane is either all red or all blue, answer is trivial.
Pick an arbitrary point, wlog a red one.  Draw the circle of points 1 unit away from the point.  If any point on this circle is red, we're done.  So assume they are all blue.  In this case, consider any arbitrary point on the circle, draw a new circle of radius one from it.  This circle will intersect the first circle twice, so we know two points on the second circle are blue, and are 1 unit away from each other.  QED
[/hide]

[quote="t0rajir0u"]Hmm.  Then both possible versions of the problem are true![/quote]

Rather silly of me to say this, since the plane could be monochrome.   :oops:

[hide="simpler solution that no one bothered to write up yet"]
Consider a unit equilateral triangle, with three vertices that are colored either red or blue. By pigeonhole some two must be the same color, and everything is one unit apart so we're done.[/hide]

how about this...prove that in that same plane, there exists a monochrome equilateral triangle side=1


let $ X$ be the one-point  union of the torus $ T\equal{}S^{1}\times S^{1}$  and $ S^{1}$

the problem is to compute the homology groups of  $ X$, I know that the fundamental group of X is $ \pi_{1}(X)\equal{}(\mathbb{Z}\times \mathbb{Z})*\mathbb{Z}$, and then the first homology group $ H_{1}(X)\equal{}\mathbb{Z}^{3}$,  but I would like to see how to compute $ H_{1}(X)$ by using May-Vietorics sequence . Here is what I have gotten so far :

$ \ldots H_{1}(T\cap S^{1})\to H_{1}(T)\times H_{1}(S^{1})\to H_{1}(X)\to H_{0}(T\cap S^{1})\to\ldots$ which is  :

$ \ldots 0\to (\mathbb{Z}\times \mathbb{Z})\times \mathbb{Z}\stackrel{\partial}{\rightarrow}  H_{1}(X)\stackrel{\partial}{\rightarrow}  \mathbb{Z}\to0\ldots$

I still don't know how the boundary maps  :

$ \partial : (\mathbb{Z}\times \mathbb{Z})\times \mathbb{Z}\to H_{1}(X)$
$ \partial : H_{1}(X)\to \mathbb{Z}$

It would be great if some one can show me how these boundary maps behave , and it would be more than awesome if you could describe these maps . Thanks so much

Take a CW decomposition and calculate the cellular homology. There is a very explicit formula for the boundary maps in terms of the attaching maps.

Gockel, thanks so much for your reply

Could you please show me what you said? I am sorry that I am just beginner and I am not familiar with CW decomposition  :( !

The first map you ask about isn't technically a "boundary" map. In general, that is the map $ \phi: H_k(U)\oplus H_k(V)\to H_k(X)$. Let the inclusions be denoted $ i: U\hookrightarrow X$ and $ j: V\hookrightarrow X$ then we have the explicit mapping $ \phi (u,v) \equal{} i_*(u)\minus{}j_*(v)$.


LaTeX
arithmetic sequence
number theory unsolved
number theory

I think that this problem is one theorem of erdos, and i don't know if have one elementary proof..

Prove that the product of four or more consecutive terme of a co-prime arithmetic progression (a+br) can't be a power.

If you can prove for one arithmetic progression with r = 1, please, post also!

http://mcraeclan.com/MathHelp/PuzzleSequenceOfSquares4.htm

a tip

[quote="yagaron"]We can proove this using 2 things.

${f(n)=a+b n}$

Is a Coprime Arithmetic Progression

${f(1) f(2) f(3) f(4)=b^{4}.k^{2}}$
${(a+b) (a+2 b) (a+3 b) (a+4 b)=b^{4}.k^{2}}$

Solving For ${a=\frac{1}{2}b \left(-5\pm \sqrt{5\pm 4 \sqrt{k^{2}+1}}\right)}$

Using the Triple Pitagoric, we can proove that $a,b,k$ cannot be integers simultaneously.[/quote]

i don't understand nothing that you say, my friend!
and i have one intuition that this aren't more useful then your last post.  :wink:

[quote="sylow_theory"]I see an elegant elementary solution, but apparently the margin of the LaTeX is too small to contain it.[/quote]
nice one! :rotfl:

[quote]i don't understand nothing that you say, my friend!
and i have one intuition that this aren't more useful then your last post.[/quote]

I make the product of 4 consecutives terms of arithmetic progression with a,b coprimes

since there is not $1+k^{2}=j^{2}$ such tahtat k,j are integers. the proof is: If $k$ is integer, $a$ is not integer.

Since $a$ need to be integer the proof is donne

but the best solution was from our friend Fermat

[quote="yagaron"]
but the best solution was from our friend Fermat[/quote]

I know this is a little of topic, but I strongly believe that Fermat did have the answers to his theorems.

I do believe Too.

Let's make a topic

"How believe that Fermat has the proof of his Theorem?"

I think the ficticious proof he have was more geometric then algebraic

he


H Hilbert, E subset. Prove:
$(E^\perp)^\perp$ is the smallest closed subspace containing E.

Turns out this one is easy, just need to use the fact every closed subspace has a supplement.


USAMTS conics ellipse AMC AIME USA(J)MO USAMO

Yay! Something to keep me busy while waiting for PUMaC. It shouldn't be hard for me to improve from last year since I got a 102 (I am hoping to improve by 30 points maybe). Probably we should not talk about answers until at least Friday (the day by which sponsors must mail answer sheets in). Also, slightly off topic, I WANT MY USAMTS SCORE RIGHT NOW! (sry caps)
Edit: does anyone else feel that their post count is significantly lower than what it should be?

Yeah, the post numbers all went down. They are no longer counting the posts in forums such as this. Just "major" forums. (My words.)

Good luck tomorrow, I'm going to show Claus your UGA post so that maybe he'll quit saying that he "failed" a test. He is only in 8th grade after all! :-)

Fulton

[quote="Twiz"][quote="Arvind_sn"]dood i got 16 rite and 3 wrong and 6 blank...is that bad? :!:[/quote]
Hmm... so I suppose its so much for waiting until friday to discuss scores. It should be okay, I'm sure everyone's taken it by now. 108 shouldn't be a bad score. I got 102 last year and ended up going to the awards banquet - and I felt it was maybe slightly harder this year than last year. How have you already checked your answers by the way? I *believe/hope* I got 22 right, 2 wrong (careless mistakes), and 1 blank for a total of 134 (if so I achieved my goal of improving by 30 points). One of my errors was $ \minus{} 55 \minus{} \minus{} 5 \equal{} \minus{} 60$. The other was writing something in big letters on my paper so I wouldn't forget about it and then forgetting about it :lol: The other 22 check out with the amazing power of my ti voyage 200 and with spending the entire ride home checking them.[/quote]

Two things:

1. We were supposed to wait until Friday??? Who said?

2. We could use calculators??? My proctor said we couldn't use them!!! If we could I would have blasted the test apart with my 89 which I brought to the test anyway but couldn't use...

No, I mean checking with my calculator afterwards. And I just figured we should wait until friday because that's when schools send in the answer sheets.

What's second round like, how much are problems worth, and how many should I get to get at least an Honorable Mention?

This should help:
http://math.kennesaw.edu/~ckoppelm/
The composite score is calculated by adding your round one score and three times your round two score. The composite last year was 153.

I don't remember if Charlie posted an "official" start discussion time, but waiting until Friday before discussing the problems seems to be a good idea. So let's hold the discussion to scores, relative difficulty, etc. until Friday.

Thanks,
Tom

Don't worry, Arvind, that score should qualify you for the next round (although you will have to compensate--I have good reason to believe that I just barely missed the top 30 last year).

BTW: I am somewhat dissapointed with my score (108 pts., 16 right, 2 wrong--I had a cold), but I guess we can't be perfect (looks like two of us will have to compensate).

[quote="gtf"]let's hold the discussion to scores, relative difficulty, etc. until Friday.[/quote]
Yay, you finally agreed with something I said :lol:

[quote="Criticalline1859"]you will have to compensate[/quote]
Hmm... I got a 102 last year, and I don't believe I compensated much (I only got a 29 on the proof round), but I still made the top 30 by a wide margin...

:huh: Umm... Anyway, the point is to try to do as well as you can (which I shouldn't bother saying too often since we already know that).

P. S. (((((((((((((((((((((((I really like parenthesis. (maybe I like factoring???... or something...))))))))))))))))))))))))

P. P. S. ...............I really like ellipses too (maybe I like the conic section--or something)...............

P. P. P. S. --------I really like to use dashes too (...maybe...)

P. P. P. P. S. I etc... (ack--this is getting out of hand)

P. P.... ...P. P. S. I somewhat like recursive functions.

P. P.... ...P. P. P. S. Yes, yes, I know... corny math joke. :D

Correction (before I get corrected): I got a 110.

so given it's 5 hours to Friday:

what was the answer to 24 with the square-arithmetic-sequence thing?

Results are out. Check with your coach/teacher.

Fulton

Wow, that was quick! I got [censored] place, which is pretty much what I expected. I think I need to work on my proof writing skills. :)

I would say not to post your place so that we don't spoil Mr. Koppelman's fun in seeing our suprise at who wins. Everyone who is invited should try particularly hard to attend; last year they had a _very_ nice luncheon that you would greatly enjoy. I know I greatly enjoyed it. The more we show our thanks by attending the ceremonies and letting Mr. Koppelman have fun, the higher the probability that KSU will continue to sponsor this delightful contest.

[quote="solafidefarms"]I would say not to post your place so that we don't spoil Mr. Koppelman's fun in seeing our suprise at who wins.[/quote]I think that sentence was directed mostly at me. :lol:
Okay, my earlier post is edited.

What the heck I got !xobile (i.e. censored) place! Oh well, but that itself was not expected  :lol:

btw... when and in which building is the awards ceremony going to be held tomorrow
(I lost my invitation somewhere)

It is in Clendenin Building, from 12:00 to 3:30.

Yay!  Hurray for brute-forcing and getting partial credit! :rotfl:

Hi everyone,

I don't remember the question number but the geometry question was basically just a direct theorem called Miquel Point theorem.

Have a look here.

[url]http://mathworld.wolfram.com/MiquelPoint.html[/url]

Hope you are enjoying the marathon.

 [b]Have a good winter break[/b].

I also found that problem in the 2000 BAMO (Bay Area Mathematical Olympiad) before the awards ceremony last year (I told Billy, but I'm not sure he remembers). At least, It was the special case with the midpoints of the sides. However, they did give the general solution.

Anyway, Happy Holidays to all.


function
algebra
polynomial
limit
number theory
least common multiple
induction

Let n be a positive integer, and let $a_1,a_2,\ldots,a_n$ be rational numbers. Suppose that $a_1^m+a_2^m+\cdots+a_n^m$ is an integer for all positive integers m. Prove that $a_1,a_2,\ldots,a_n$ are themselves integers.

I'm pretty sure this isn't the traditional proof and I'm not sure if it works either...

[hide]
Let $a_k = d_k+\frac{r_k}{c_k}$ where $r_k < c_k$ and $r_k, d_k$ are integral.

For any term, we have

$a_k^m = \left(d_k+\frac{r_k}{c_k}\right)^m$.

Taking any term after the first in the binomial expansion, we have

$d_k^x\frac{r_k^{m-x}}{c_k^{m-x}}\binom{m}{x}$

But take $m \to \infty$ and we have the exponential part $\frac{r_k^{m-x}}{c_k^{m-x}}$ decreases faster than $\binom{m}{x}$ increases (which can be bounded above by $m^x$, a power function), and we have all the fractional parts tending toward zero, which means at some point they cannot add up to an integer. Not completely sure if this argument holds, but as far as I can tell it does.

Hence we must have $r_k = 0$ which means $a_k$ are all integers.[/hide]

From first impressions i imagine that we can use a polynomial with the $a_i$ as roots, and use gauss's theorem that all rational roots are integers if all the leading co-efficient is 1 and the others and themselves integers...

[quote="paladin8"]I'm pretty sure this isn't the traditional proof and I'm not sure if it works either...

$d_k^x\frac{r_k^{m-x}}{c_k^{m-x}}\binom{m}{x}$

But take $m \to \infty$ and we have the exponential part $\frac{r_k^{m-x}}{c_k^{m-x}}$ decreases faster than $\binom{m}{x}$ increases (which can be bounded above by $m^x$, a power function), and we have all the fractional parts tending toward zero, which means at some point they cannot add up to an integer. Not completely sure if this argument holds, but as far as I can tell it does.

Hence we must have $r_k = 0$ which means $a_k$ are all integers.[/quote]

hmm aren't we trying to show that

$\lim_{m\to \infty} \frac{r_k^{m-x}}{x_k^{m-x}} \binom{m}{x} = 1$ ? therefore leaving you with just $d_k^x$?

cuz if the limit of that as it approaches infinity is $0$, then wont that make the whole expression:

$\lim_{m \to \infty} d_k^x\frac{r_k^{m-x}}{c_k^{m-x}}\binom{m}{x} = 0$? unless the $d_k^x$ increases faster than $c_k^{m-x}$ decreases or something.....

hmm i ono if ne of that made ne sense

could anyone also explain gauss's theorem a lil more? since mathworld doesnt really have much on it...

yea... there has to be a clever non-calc involved solution ne ways

i thought about making like.... $a_k = \frac{p_k}{q_k}$ and showing that all the $q_i = 1$ or something but i ono how to continue

[quote="probability1.01"]Well his point is that the fractional parts exist but tend towards 0. But intuitively, it doesn't make sense...[/quote]

I would say intuitively it does make sense... basically I'm saying that as $m \to \infty$ we have $\sum a_k^m \to \sum d_k^m$ so like at some point $m$ will be large enough such that $\sum d_k^m < \sum a_k^m < \left(\sum d_k^m\right)+1$. The part I'm not completely sure about is whether or not the fractional parts actually tend toward zero...

hmm wait actually u do want to show that:

$\lim_{m \to \infty} d_k^x\frac{r_k^{m-x}}{c_k^{m-x}}\binom{m}{x} = 0$

this way the expansion of

$a_k^m = \Big(d_k + \frac{r_k}{c_k}\Big)^m = d_k^m$

 since all the fractional parts are just 0 as $m \to \infty$

but can u prove that $\frac{1}{c_k^{m-x}}$ decreases faster that $d_k^{x}r_k^{m-x}\binom{m}{x}$?

[quote="Altheman"][hide="some ideas"]$a_i=\frac{p_i}{q_i}$ where $p_i$ and $q_i$ are coprime positive integers

clearing the fractions, we have

$p_1^m*q_2^m*...q_n^m+p_2^m*q_1^m*q_3^m*...q_n^m+...+p_n^m*q_1^m*...*q_{n-1}^m=k*q_1...q_n$

if we look in $\mod q_i$ for $i=1,2,...n$ then we see that $(q_2*q_3*...q_n) \equiv 0\mod q_1$, etc

i don't know where to go from here...there might be some kind of descent think using the common factors, and show that if q_i divides it, then q_i^n divides it for all n, and the only way for that to work is if $q_i=1$[/hide][/quote]

I'd say that [hide="thats too complicated..."]
You don't really need to assign distinct denominators to each fraction - just take the $lcm$.

Then we write $a_i = \frac{b_i}{d}$ where $b_i, d \in \mathbb{N}$ and $\sum_{i=1}^{n} b_i^m \equiv 0 \bmod d^m$.

Consider the sequence $c_i \equiv b_i \bmod d$ such that $\sum_{i=1}^{n} c_i \equiv 0 \bmod d$ (this follows from the $m=1$ case) and all $c_i < d$.  We can write $b_i = c_i + dm_i$ for some integers $m_i$ which are absolutely irrelevant.  Expanding the original $b_i$ summation with binominal theorem, we can show that $\sum_{i=1}^{n} b_i^m \equiv 0 \bmod d^m \implies \sum_{i=1}^{n} c_i^m \equiv 0 \bmod d^m$ by strong induction.

Notice that the LHS grows more slowly than the RHS - that is, $\lim_{m \to \infty} \sum_{k=1}^{n} \frac{c_i^m}{d^m} = 0$ and hence there can be no solution unless $d = 1$.

(This seems fairly similar to Paladin's proof, but a little more straightforward, I think? :) )
[/hide]

[quote="Altheman"]i just realized, this is obvious if you just use induction on the number of terms[/quote]

Err, how is it obvious... just because we know it is true for some $n$ doesn't mean we know anything about the case $n+1$... $a_1^m+\cdots+a_{n+1}^m$ is an integer doesn't tell us anything about $a_1^m+\cdots+a_n^m$ so I don't see a good way to make an induction step...

sorry, i was tired, it made sense at first, to say you have $\sum_1^n a_i^m=k$, then add $a_{n+1}^m$, and that has to be an integer, but it is incorrect

Well, now that we've solved the original problem, I propose a generalization and a follow-up:

1.  $\sum_{i=1}^{n} a_i^m$ is an integer for an infinite number of values of $m \in \mathbb{N}$ (not necessarily all integers).  Is it true that $a_i$ must still be integers?

2.  If the answer to the previous question is 'yes,' find the set of values $a_1, a_2, ... a_n$ that maximizes the number of values of $m$ for which $\sum_{i=1}^{n} a_i^m$ is an integer, but such that none of the $a_i$ are integral, and give the number of values of $m$ for which this occurs.

:D

[quote="t0rajir0u"]1. ... for an infinite number of values of $m \in \mathbb{N}$ (not necessarily all integers)...[/quote]

Please, clarify this: Are there values in $\mathbb{N}$ that are not integers?

He means that there is an infinite amount of integers $m$ such that the term is an integer, whereas it's not necessary to be integral for other values.

[quote="t0rajir0u"]1.  $\sum_{i=1}^{n} a_i^m$ is an integer for an infinite number of values of $m \in \mathbb{N}$ (not necessarily all integers).  Is it true that $a_i$ must still be integers?[/quote]

Easy counterexample.  Take $n = 1$ and $a_1 = \sqrt{2}$.

The numbers must be rational (cf. the initial post).

A more rigorous solution, courtesy of Nukular (from a previous session of the Olympiad problem solving class):

[hide]
Lemma: If $b_1, b_2, \dots, b_n$ are nonzero integers and $(c,b_i) = 1$  $\forall i$, and
$c^m | b_1^m + b_2^m + \dots + b_n^m$, then $c = 1$.

Proof:
Let p be a prime that divides c and k any positive integer.  Then for m sufficiently large,

\[ p^k | b_1^m + b_2^m + \dots + b_n^m  \].

So, let us pick an m such that $\phi(p^k) = p^{k-1}(p-1) | m$.

Then

$0 \equiv b_1^m + b_2^m + \dots + b_n^m \equiv n \pmod{p^k}$.

So we conclude that $p^k | n \quad \forall k \qquad \Rightarrow \Leftarrow \text{QED}$

Now for the main body of the proof.

We proceed by induction.

If n = 1, this is trivial since then we have $a_1$ is itself an integer.

Otherwise, assume $a_i$ is nonzero for each i, and for each $i$ write $a_i = \frac{b_i}{c_i} \quad \text{with}\quad (b_i,c_i) = 1 \quad \text{and} \; c_i > 0$.
Assume WLOG $c_1 \ge c_2 \ge \dots \ge c_n$.

Now $\forall m \; \exists k_m \in \mathbb{Z} \quad\text{such that}$

\[ a_1^m + a_2^m + \dots + a_n^m = k_m  \].

then

\[ (c_na_1)^m + (c_na_2)^m + \dots + (c_na_{n-1})^m = k_mc_n^m - b_n^m \in \mathbb{Z}  \]

So we can conclude from inductive hypothesis that

\[ (c_na_1), (c_na_2), \dots ,(c_na_{n-1}) \in \mathbb{Z}  \]

So $c_i | c_n \; \forall i$.

But since $c_n \le c_i \; \forall i$, we can conclude $c_n = c_i \; \forall i$.

Then the condition that the sum of the mth powers of the $a_i$ is integral for each m is equivalent to $c_n^m | b_1^m + b_2^m + \dots + b_n^m$, for all m. We conclude by the lemma that $c_n = 1$ and therefore $a_i \in \mathbb{Z} \; \forall i \quad QED$.
[/hide]

Impressive.  I was hoping that a more rigorous solution would take care of the generalization that I gave, but your solution relies on picking $m$ such that $\phi(p) | m$ which clearly doesn't have to be doable in the generalized problem... I really would like someone to tackle the new problem!  :)

Somewhat similar problem:

Let $x_1, \dots, x_k$ be distinct rational numbers greater than or equal 1, $c_1, \dots, c_k$ be non-zero real numbers, and $\{m_n\}$ be a sequence of integers such that
\[ \lim_{n\to\infty} \left( m_n - \sum_{j=1}^k c_j x_j^n\right) = 0.  \]
Prove that all $x_1, \dots, x_k$ are integers.


若干年前, 我校一位校友考上了清華大學(物理系). 到了今天, 我班的一位同學亦成功地保送到北京大學(物理學院), 成為我校第二位能進中國大陸的頂尖大學的學生.

我有感而發, 便想問大家個問題: 像內地的北京大學, 清華大學; 或者香港的香港大學, 中文大學等等, 都被公認為好的大學. 不知在大家心中, 怎樣的大學才稱得上"好"? 它們要具備甚麼條件? 若不考慮"錢"這個問題, 你最希望能進哪所大學? 為甚麼?

最後, 祝願各位準中學畢業生(或者尚未到畢業班的同學)都能考上理想的大學!

[quote="JahrJsuiJcokie"][quote="ifai"]JahrJsuiJcokie, after reading this essay, what is your [b]own[/b] opinion or conclusion?[/quote]
你的[b]理解力[/b]，让在下感到无比的惊讶！难道研习数学，就会让你的国文水平，如此[b]低下[/b]。我想，我之所以沿用繁体字库，或是盛行古文字语，乃是我认为汉语言有其无比之概括与精练；我们面对专业术语的[color=cyan]泛英语化[/color]，而低下我们卑微的头颅吗？！
我们要为一门国际化的语言，而舍弃我们民族的标识吗？
你对于[color=cyan]大学之道[/color]，到底是在于“[color=red]存在的质疑[/color]”，还是意于“[color=green]无知的认同[/color]”呢？！
你既不独立，也难于附和，却是在徘徊。
我无以指引，只能以文相告。  :?
我会令设主题，关于一类多项式“旬二周元解”的公式表达详末。[/quote]

你是在說你的問題，還是在說大學？一所好的大學，不一定要研究你的問題吧！

[quote="jin"]:stretcher:  :huh: What do you mean by such a long lecture?Is it related to the university?[/quote]
jin,yoou al be right.And  i have lost in LOVEIN RVIVER.

[quote="Soarer"][quote="dingdongdog"]请问一下：在香港大学、香港科技大学和北京大学中，哪个大学的数学系最好？
或者说，学术氛围最好？[/quote]

個人意見：
科大數學系教授最強，出國最易。
中大(在香港)學生最強。
北大... 不用說吧，學生太強；但個人認為學習壓力太大，要狂K題目，到以後做研究不一定好。[/quote]

本科阶段狂K题目绝对不是坏事
和以后最研究好坏也没必然联系
如果有的话，也是以正相关居多
当然超级牛的可以不狂K

要说出国
北大最有氛围
香港学校的UG出国氛围和大陆名校感觉没法比
至于容易与否，就不太好说了

Soarer, 你现在在哪个大学?

香港科大。

出國方面嘛，內地的情況我不知道。香港來說應該科大往外國當交流生的機會最多。因為有一些本地學生實力很爛，所以內地的來到一般機會比較大。當交流生回來，出了一兩份paper，學習成績又不錯，那phd去名校機會就大了。今年數學系的就有一個收到berkeley, yale, caltech, ucla, courant institute 等八個offer的牛人。

另外，在香港想狂K題目不容易啊.... 因為誘惑太多，一般都很難做得到....

另外, 我認為在本科唸數學的, 狂K題目不如多學些不同領域的數學知識. 做研究時想多學點就難了. 另外, 不斷做題只會令自己變成人肉計算機, 影響思維的進一步發展, 導致日後做不出好的研究. 就像在北大數學系這狂K題目的風氣裡, 何曾造就出一個對數學的發展有一定影響的數學家?

[quote="ifai"]香港的學生實力爛? 據我所知, 香港學生自入學便要接受一個又一個的公開試. 能考上(香港的)大學的, 大概也有一定程度吧? 我們澳門的學生, 可算是全亞洲最輕鬆的了. 學習沒有甚麼壓力, 不太差的學生都不愁沒大學收. 另外, 香港大部分學生都選擇留港升學, 內地生多數只能靠高考進大學, 一般澳門學生則可選擇大陸, 台灣或者留澳升學, 擇選比你們都多. 不過, 最大的缺點就是所得太少, 結果離開澳門後, 競爭力會大降.[/quote]

不是香港學生實力不行.. 各個系收生標準不同嘛..
至於研究，我覺得yace 說的也有道理，除非你超強，不然現在有狂K題目的必要... 其實，「何曾造就出一個對數學的發展有一定影響的數學家?」和K題目沒有太大關係。K題目只是代表打好基礎，再讀上去就看各自造化。

中學時代的確有必要狂K題目......

話說我考澳門大學入學試數學科時, 今年所出的題目不太難, 但就是做得不熟, 結果考得不太好. 可是, 這已成過去...

剛過去的星期六,日考台灣, 我第一志願報台大數學. 嘗試做過台灣的數學試題, 我覺得數學是用來收復其他科的失地的, 因為只要小心點不犯小毛病, 100分是不難得到的. 但今年實戰, 恐怕連100/2都沒有!!! 聽聞今年起台灣試題都會根據香港高考的方向出題, 我們這些澳門考生就成了犧牲品... 同樣地, 這亦已成定局....

Therefore, 為了將內地聯考考好, 可真要下些苦功去K題!

Soarer, 我想請問香港每年大概有多少人考台灣的(國立)大學?

相信台灣的生源主要是澳門, 今年台灣這樣搞都不知為甚麼. 各科試題都是怪怪的(中文除外), 英文作文題更是看不懂, 完全不知有何可寫(A Rolling Stone Gathers No Moss, 當時我連字面意思也不知!)

唉, 今次台灣的分數, 大概是 中文~物理>化學>數學>英文. 可真無奈+痛苦...

不清楚。

我沒有聽說過有中學同學去了台灣呢。

[quote="Soarer"]不清楚。

我沒有聽說過有中學同學去了台灣呢。[/quote]
就是嘛! 可這裡大部分的學生, 不是留澳, 就是到內地或台灣升學.
今年澳門考台人數大概1400, 相信香港考生總沒有澳門的1/5.
真不知台灣聯招會那邊怎麼想. 唉~ 亦只怪自己學藝不精!


modular arithmetic
number theory proposed
number theory

Find all triples $ (x,y,z)$ of positive integers such that:
$ (x+y)(1+xy)=2^{z}$

The only solutions are ($ 1$,$ 2^{n}-1$,$ 2n$) and ($ 2^{n}-1$,$ 2^{n}+1$,$ 3n+1$) for $ n\ge 1$.

Suppose we had a different solution. We know that $ x+y$ and $ xy+1$ are both powers of two and $ xy+1\ge x+y$, since this is equivalent to $ (x-1)(y-1)\ge 0$. Suppose that $ x+y=2^{n}$. Then since $ xy+1\ge x+y$ is a power of two, we know that $ xy\equiv-1 \pmod{2^{n}}$. Since $ y\equiv-x \pmod{2^{n}}$, we see $ x^{2}\equiv 1 \pmod{2^{n}}$. But then $ x^{2}-1\equiv (x-1)(x+1)\equiv 0 \pmod{2^{n}}$. Now $ x$ is obviously odd, so exactly one of $ x-1$ and $ x+1$ is divisible by 2 but not 4. For the product to be divisible by $ 2^{n}$, the other must be divisible by $ 2^{n-1}$. But the only possibilities for $ x$ are then $ 1$, $ 2^{n-1}\pm 1$, and $ 2^{n}-1$, which yield the solutions given above.


modular arithmetic
number theory
Diophantine equation

When Nehru after consulting with his advisers, proposed a bill to reframe the currency system of the country, many of the other ministers voted against it.

What Nehru wanted was that only 5 and 3 rupee coins to exist. He claimed that with 3 and 5 rupee coins, there was always a way to get [b]any[/b] natural number of coins.

Is Nehru right?
If yes, prove it. If not, which is the least number which can not be given with 3 and 5 rupee coins.

By the Chicken McNugget Theorem, the least value that cannot be made is 7.

Google the theorem if you don't already know what it is.

First of all, the Chicken McNugget theorem looks for the greatest number which cannot be expressed as a linear combination of coprime $ m$ and $ n$ with nonnegative integers as coefficients. As ThinkFlow stated, any number greater than $ 7$ can be expressed as such for $ m \equal{} 3, \; n \equal{} 5$. But Nehru is right, we can also obtain other values as well because trading system allows change. Such as:

$ 1 \equal{} 2\cdot 5 \minus{} 3 \cdot 3$
$ 2 \equal{} 1\cdot 5 \minus{} 1\cdot 3$
$ 3 \equal{} 1\cdot 3$
$ 4 \equal{} 2\cdot 5 \minus{} 2 \cdot 3$
$ 5 \equal{} 1\cdot 5$
$ 6 \equal{} 2 \cdot 3$
$ 7 \equal{} 2 \cdot 5 \minus{} 1 \cdot 3$

For example: you need to pay 4 rupees, you would give 2 coins of 5 rupees and get 2 coins of 3 rupees change.

Also, we can explain this by considering diophantine equation
\[ 3x \plus{} 5y \equal{} c
\]
We don't need to find solutions, just prove that for any integer $ c$ exists infinitely many integer solutions.

Theorem:[i] For linear diophantine equation $ ax \plus{} by \equal{} c$ there is an integer solution if and only if $ gcd(a, b)|c$. If there is one integer solution, there are infinitely many.[/i]

Since $ gcd(3, 5) \equal{} 1$ the equation $ 3x \plus{} 5y \equal{} c$ has infinitely many integer solutions for any integer $ c$.

Sorry, I overlooked the fact that the Chicken McNugget looks at nonnegative coefficients, not all coefficients. :blush:

[quote="mathwizarddude"][quote="Flame"]Also, we can explain this by considering diophantine equation
\[ 3x \plus{} 5y \equal{} c
\]
We don't need to find solutions, just prove that for any integer $ c$ exists infinitely many integer solutions.

Theorem:[i] For linear diophantine equation $ ax \plus{} by \equal{} c$ there is an integer solution if and only if $ gcd(a, b)|c$. If there is one integer solution, there are infinitely many.[/i]

Since $ gcd(3, 5) \equal{} 1$ the equation $ 3x \plus{} 5y \equal{} c$ has infinitely many integer solutions for any integer $ c$.[/quote]
Can you prove this theorem?[/quote]

Proof:

First let $ gcd(a,b)\equal{}d$. If the given equation has solutions, obviously $ d|c$. Now observe congruence
\[ ax \equiv c \pmod b\]
This congruence has solutions (if you want the proof of that, I can write it too). If one solution is $ x_1$, then all solutions of the congruence are given with $ x \equiv \frac{b}{d} \cdot k \pmod b$, where $ k\equal{}0, 1, 2 \cdots, d\minus{}1$. Thus all the solutions of the linear diohantine equation $ ax\plus{}by\equal{}c$ are given with $ x\equal{}x_1\plus{}\frac{b}{d}\cdot t, t \in Z.$ Put it in diophantine equation to get $ by\equal{}c\minus{}ax_1\minus{}\frac{ab}{d}\cdot t\equal{}by_1\minus{}\frac{ab}{d}\cdot t \implies y\equal{}y_1\minus{}\frac{a}{d}\cdot t$. Now it's clear there are infinitely many solutions.

Note that this is proof from college textbook and is probably not suitable for intermediate high school.


geometry
circumcircle
trapezoid
rectangle
cyclic quadrilateral
geometry solved

Let ABDC be cyclic. There exists point E in plane s.t ED is perpendicular to BD and EA is 90 degrees to AC. Ther exist F in plane s.t. FC is perpend. to AC and FB is perpend. to BD. Let M be point of interection of AD and BC. Dtermine the necssary+sufficient condition for <FMC=<EMD to occur.

Bomb

The point $E$ is the intersection of normals to the diagonals $BD, AC$ erected at the points $D, A$. Similarly, the point $F$ is the intersection of normals to the diagonals $BD, AC$ erected at the points $B, C$. WLOG, assume that $AB > CD$ ($AB = CD$ is impossible, because the sides $DA, BC$ would be parallel and they would not intersect at the point $M$). The triangles $\triangle AED \sim \triangle BFC$ are similar, because

$\angle ADE = 90^o - \angle ADB = 90^o - \angle ACB = \angle BCF$

$\angle DAE = 90^o - \angle CAD = 90^o - \angle CBD = \angle CBF$

$\angle AED = \angle BFC$

The first 2 pairs of angles are equal, because the angles $\angle ADB = \angle ACB$, $\angle CAD = \angle CBD$ span the same arcs $AB, CD$ of the quadrilateral circumcircle $(O)$, the last pair because the line pairs $AE \parallel CF$ and $DE \parallel BF$ ($AE, CF \perp AC$, $BF, DE \perp BD$). Hence, their external angles $\angle EDM = \angle FCM$ are equal. Thus the necessary and sufficient condition for the angles $\angle EMD, \angle FMC$ to be equal is similarity of the triangles $\triangle EMD, \triangle FMC$. Iff this is the case, then the triangles $\triangle EAM, \triangle FBM$ are also similar and we get

$\frac{MA}{MD} = \frac{MB}{MC}$

Since $MA \cdot MD = MB \cdot MC$ is power of the point $M$ to the circle $(O)$, multiplying and dividing these 2 equations yields

$MA^2 = MB^2,\ \ MA = MB$

$MD^2 = MC^2,\ \ MD = MC$

Together with the power of the point $M$ to the circle $(O)$, this is equivalent to

$DA = MA - MD = MB - MC = BC$

i.e., the cyclic quadrilateral is an isosceles trapezoid with $DA = BC, AB \parallel CD$ (but not a rectangle, because the point $M$ would not exist).

Thanks, but I think a slight misunderstanding may have occurred.
Sorry for the dodgy notation, the actual question is :
ABCD is cyclic. M is the intersection of the diagonals AC,BD, while E is defined as the point the point of interesction with EA perpendicular to AB and EC perpendicular to CD and F is defined as FB perpendicular to AB and FD  is perpendicular to CD. Then RTP is, determine when <EMC=<FMB.
But based on Yettis proof, I thgink its esseniaslly the same. [color=red][Moderator edit: Indeed, this is just what Yetti proved, but applied to the cyclic quadrilateral ADBC instead of ABCD.][/color]

BOMB


When doing AMC's, I'm sure we often employ a few cheap techniques and effective test-taking strategies.

There are two that I know that I and others commonly use:

1. Multiple choice analysis. Say it's a problem in which we are asked to find, say, 1 less than a particular value. The test creators, knowing that we may forget to subtract 1, will put 1 more than the actual answer on the test. Thus, if we see 8 as an answer and 9 as an answer, and there are no more consecutive numbers as answer choices, we may conclude the answer is 8.

2. Filling in our own information. A great example is #23 on this year's AMC 10A. We had a convex quadrilateral ABCD, with AE the intersection of AC and BD. We were given the side lengths of AB, CD, and AC, and that the areas of triangle AED and BEC were the same. We were asked to find AE. I, being cheap, assumed that we were working with an isosceles trapezoid, and the answer came trivially from there.

I also often make careless mistakes. After I finish each problem, next to the problem, I put a percent telling how certain I am of its correctness. I also check my algebra as I go along.

What do you do?

I solve the problems. If I can't solve it, I leave it blank. Unless it's the only problem I didn't solve, in which case I guess.


number theory
least common multiple
superior algebra
superior algebra unsolved

find all k s.t. there is element of order k in Sn

All elements can be written as a product of disjoint cycles, where a cycle is written as taking one element in the cycle to the next, etc. For example:
In S4, the element that takes 1 to 3, 2 to 4, 3 to 1, and 4 to 2 can be written {1 3}*{2 4}, where {1 3} is the element that takes 1 to 3 and 3 to 1, and {2 4} is the element that takes 2 to 4 and 4 to 2.
These cycles are commutative with each other so long as they are disjoint. Therefore, the order of the element is equal to the product of the orders of the disjoint cycles. Therefore, all possible orders are those that can be written as a1*a2*...*ak where the sum of the a's is n.
For example, in S4, there are elements of order:
1 1 1 1 = 1
2 1 1 = 2
3 1 = 3
4 = 4
2 2 = 4
And that is all.
In S5, there are elements of order:
1 1 1 1 1 = 1
2 1 1 1 = 2
3 1 1 = 3
4 1 = 4
5 = 5
2 2 1 = 4
3 2 = 6
=Bmsiv-

you wrote-Therefore, the order of the element is equal to the product of the orders of the disjoint cycles.
 sholdnt it be lcm?
 and are these all possible k`s?


The $ 600$ students at King Middle School are divided into three groups of equal size for lunch.  Each group has lunch at a different time.  A computer randomly assigns each student to one of the three lunch groups.  The probability that the three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately:

\[ \textbf{(A)}\ \frac{1}{27} \qquad
\textbf{(B)}\ \frac{1}{9} \qquad
\textbf{(C)}\ \frac{1}{8} \qquad
\textbf{(D)}\ \frac{1}{6} \qquad
\textbf{(E)}\ \frac{1}{3}
\]

[hide="Click for solution"]
Al must be assigned to one of the lunch groups.  The probability that Bob is assigned to the same lunch group is approximately $ \frac{1}{3}$, and the probability that Carol is assigned to that same group is also approximately $ \frac{1}{3}$.  Thus the probability that all three are assigned to the same group is approximately $ \frac{1}{3} \times \frac{1}{3}\equal{}\frac{1}{9}$, or $ \boxed{\textbf{(B)}}$.
[/hide]

[hide]There is a $\frac{1}{3}$possibility that a student would be assigned to a certain lunch group, so we multiply $\frac{1}{3}$ times $\frac{1}{3}$ times $\frac{1}{3}$ to get $\frac{1}{27}$. However, this applies to three circumstances each for one lunch group. So add the three $\frac{1}{27}$'s to get $\frac{1}{9}$, or $\boxed{B}$[/hide]

I think you should wait a bit before giving the solution
edit: wait this was posted in 2009

1986 AJHSME/24 :P

Interesting how the concept of "computer" is used...

[quote=yackspit]what would the answer be if you had to compute the exact probability?[/quote]

[hide=Solution]The location of the Al doesn't Matter. Bob has a ((600/3)-1)/(600-1) chance of being sorted into the same group. Carol has a ((600/3)-2)/(600-2)) chance of being sorted into the same group. Answer = 199/599*198/598 = 19701/179101[/hide]


linear algebra
matrix
linear algebra unsolved

If $A$ is a square matrix and its largest singular value is smaller than $1$, is  $A+I$ singular?

No. $(I+A)^{-1}=I-A+A^{2}-A^{3}+\cdots$, and the series converges by a norm calculation.


I am in the BC just now, and i was wondering if this made sense:

$\sum_{n=1}^\infty \frac{1}{\sqrt{2n^2+3}}$ define convergence/divergence

requires us to use the limit comparison. However, if we denote such $k=2n^2+3$, the question becomes clearer with p-series having $p=\frac{1}{2}$ which indicates that the series diverges. Limit Comparison also yields divergence. However, my teacher says that since the $+3$ is there denoting $k$ is invalid and thus my proof is wrong. I am not sure if he is right or not.

You're headed down a wrong path, and the "$+3$" doesn't have much to do with anything.

Let me take your argument to a ridiculous extreme:

Consider $\sum_{n=1}^{\infty}\frac1{\sqrt{n^4}}.$

Now let $k=n^4.$  So we compare this to $\sum_k\frac1{\sqrt{k}},$ which diverges.

That is an absurd conclusion, since the original series is simply the convergent series $\sum_n\frac1{n^2}.$

The problem with your "let $k=g(n)$" for some function of $n$ was that that function $g$ (in your case, $g(n)=2n^2+3$) did not map $\mathbb{N}$ onto $\mathbb{N}.$ It does you no good to contemplate some thin subset.

Limit comparison is the great "de-clutterer." It's a method for thinking rapidly. Look at that problem again:

$\sum_n\frac1{\sqrt{2n^2+3}}$

For large $n$, the "$+3$" isn't going to matter, so we have some constant multiple of $n^2.$ Take the square root of that, and what we're looking at might as well be some constant multiple of $\sum_n\frac1n.$ That's the harmonic series, which diverges.

The whole thought process takes just seconds - far, far less time than it took me to type that. You could follow with some written formalities, citing the limit comparison test specifically and showing that comparing to the harmonic series is appropriate under that theorem - but the problem is already decided, this is just the mop-up phase.


calculus
derivative
calculus computations

Is there any formular to calculate $ \frac{\mathrm{d}^n}{\mathrm{d}x^n} \frac{u(x)}{v(x)} $ in term of $u^{(i)}(x) and
$v^(i)(x)$ ? :)

there's Leibniz formula for (u*v)<sup>(n)</sup>,
write u/v as u * (1/v), then apply Leibniz formula?

but i want it in term of v(i). can you express (1\v)(i) by v(i)? :?


hi!
prouve that [url=http://www.servimg.com/image_preview.php?i=2&u=11334236][img]http://i22.servimg.com/u/f22/11/33/42/36/bb165f10.gif[/img][/url]
for n>1.

But $ \dbinom{n}{2n} \equal{} \frac {n!}{(2n)!( \minus{} n)!}$. Isn't there any problem?

[hide="Suggestion"]Actually, is should be $ \dbinom{2n}{n}$. Using Stirling's approximation,

$ n! \geq \left(\frac {n}{e}\right)^n \sqrt {2 \pi n}$

you can prove your inequality.[/hide]


Determine, with proof, all functions f :R→R
such that f(1) = 1,
and
f(xy + f(x)) = xf(y) + f(x)
for all x,y ∈ R.

Heh, I remember this one.

[hide]
$x=1, y=0$ shows us that $f(0)=0$.
$x=1, y=-1$ gives us $f(-1)=-1$.

$x=1, y=n$ gives us $f(n+1) = f(n)+1$. With this, we can easily show by induction that $f(k)=k$ where $k$ is a natural.
Similarly with $x=-1, y=n$, we get $f(-n-1)=-f(n)-1$, and we can extend the above result to all integers.

From here:
Let $y={{n}\over{x}}$, giving us:
$n+x=xf({{n}\over{x}})+f(x)$.
$n=xf({{n}\over{x}})$

$f({{n}\over{x}}) ={{n}\over{x}}$

Hence, $f(x)=x$ for all the reals.

Unfortunately, under the pressure of the actual competition, my head stopped working, and the last part completely evaded me (and I just solved it there in 5 minutes). :(
[/hide]

I suppose he wanted to write this:
$n+f(x) = xf(\frac{n}{x})+f(x)$
n must be a natural, but x can be a real.

Maybe the easiest way is to put $x=\frac{1}{y}$:
$f(1+f(\frac{1}{y})) = \frac{1}{y}f(y)+f(\frac{1}{y})$
$1+f(\frac{1}{y}) = \frac{1}{y}f(y)+f(\frac{1}{y})$
$1 = \frac{1}{y}f(y)$
$f(y)=y$

[quote="paladin8"][quote="rjt"]From here:
Let $y={{n}\over{x}}$, giving us:
$n+x=xf({{n}\over{x}})+f(x)$.
$n=xf({{n}\over{x}})$

$f({{n}\over{x}}) ={{n}\over{x}}$
[/quote]

But you have assumed $x$ is an integer here, or you wouldn't know that $f(x) = x$. So you have only proved it for the rationals, right?[/quote]

:blush: Oops.

If we let $y=0$, we get $f(f(y))=f(y)$.

Let $x=0, y=f(y)$ to get $f(f(y)+1)=f(f(y))+1=f(y)+1$. Then proceed as edriv said.


trigonometry
geometry
trig identities
Law of Sines

Let A be a point on a circle, AB, AC, AD be three chords in order, then: \[AC \cdot \sin{\angle{BAD}}=AB \cdot \sin{\angle{CAD}}+AD \cdot \sin{\angle{CAB}}\]

[hide]By Ptolemy's theorem on $ABCD$, $AC\cdot BD = AB\cdot CD+AD\cdot BC$. By the Law of Sines on $BCD$, $\frac{BD}{\sin\angle BAD}= \frac{BC}{\sin\angle CAB}= \frac{CD}{\sin\angle CAD}$, and the result follows.[/hide]


function algebra polynomial integration trigonometry calculus probability

For each positive integer $ n$, evaluate the sum
\[ \sum_{k\equal{}0}^{2n}(\minus{}1)^{k}\frac{\binom{4n}{2k}}{\binom{2n}{k}}\]

hello, we have $ \sum_{k\equal{}0}^{2n}(\minus{}1)^{k}\frac{\binom{4n}{2k}}{\binom{2n}{k}} \equal{}\frac{1\plus{}e^{2in\pi}}{2\minus{}4n}$.
Sonnhard.

Now can someone prove the Wolfram Alpha answer? ;)

Why does that matter?  Anyway, $ e^{2in\pi} \equal{} 1$.

[quote="blahblahblah"]How is knowing the answer ever 'worthless'?[/quote]
It isn't an answer to be important.
As for us, "[b]Why[/b] does it become the solution?" is important.

Therefore only the answer does not become the essential solution.
So, only an answer is [b]worthless[/b].

Do you think that only an answer by the calculation with the CAS and the quotation from the site is more important?
It feels me worthless.

I'll argue by analogy.

Let's say I knew a reasonable amount of mathematics, but had no clue as to how to evaluate the series

$ \sum^\infty_{j\equal{}1} \frac{1}{j^2}$

Say I didn't even know how to begin to approach this problem.  If someone then told me that the answer is $ \pi^2/6$, don't you think that this would suggest some methods that one might use?  Methods related to trigonometric functions, perhaps?

I'm not saying that an answer alone suffices, but in many cases knowing it alone greatly aids the solution process.

[quote="blahblahblah"]I'm not saying that an answer alone suffices, but in many cases knowing it alone greatly aids the solution process.[/quote]

I see.
But, does not the reply only for answers think in inappropriateness?
I think that a calculation with the PC or the information called the quotation of the site is necessary at least.
[b]If different[/b], proof should be able to reply.

[quote="Kouichi Nakagawa"][quote="Dr Sonnhard Graubner"]hello, this is not from Wolframalpha.
Sonnhard.[/quote]
You did not prove it and did calculate with a CAS?
(A CAS (computer algebra system) is a software program that facilitates symbolic mathematics. For example, Mathematica, Maple etc.)
If so, your post is worthless.
Please prove it.[/quote]

Based on your definition of "worthlessness", you have made worthless posts previously too.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1522134#1522134

This is a Math forum, there is no need to be hostile. If you think you can do better, demonstrate the point by solving the problem and not by whining.

IndoChina, that thread is completely different.  Since that problem reduces to a polynomial, one reasonable approach is using a computer solution; once you know enough answers, it is relatively easy to verify that you have found them all.

On the other hand, it's basically impossible for me to convert Dr. Graubner's post into a proof.  Had he posted a WZ-pair that we could use to verify his answer, it would be a little different. (though still somewhat unsatisfying)

The problem with simply posting an answer like this is not that the knowledge is worthless--far from it--but rather that it clutters the forum.  When a difficult problem remains at zero posts, it is much more likely to get answered *someday*, and the more posts that are added, the more someone will have to read later to even *know* that the problem is not solved. (this is to some extent a shortcoming of the forum itself--I would like to see a "flag as solved" button)

In any case, neither Kouichi or myself intended to be hostile, but probably there is some frustration over seeing a 2% solution posted--I know that I have spent a significant amount of time on this problem, and I could easily post a page of intermediate work.  But it is very hard for me to know if my approaches are anything but dead-ends, just as it is hard for me to know if simply knowing the answer will shed any light on the manipulations necessary to get there.

The fact that he did not simplify his answer also increases the perception that he is not presenting us with real thought about the problem. (though it does shed light on the case where $ n$ is a half-integer, which is nice, though trivial)

@LydianRain: About your "flag as solved" idea. It's good I agree, however I just wanted to know why the suggestion when there is a Solved Problems subforum. I guess it takes a significant amount of the moderator's time to look into each topic and see whether or not it has been solved, and then move it? Mhmm at any rate, as one voice as already spoken I just want to start a chain of supporters. Hopefully we get such a button soon =)

Edit: Also, [url=http://mathforum.org/library/drmath/view/56163.html]Answer[/url] which I accidently found looking up the Fundamental Theorem of Combinatorial Enumeration ^_^

[quote="Fermat's Little Turtle"]@LydianRain: About your "flag as solved" idea. It's good I agree, however I just wanted to know why the suggestion when there is a Solved Problems subforum. I guess it takes a significant amount of the moderator's time to look into each topic and see whether or not it has been solved, and then move it? Mhmm at any rate, as one voice as already spoken I just want to start a chain of supporters. Hopefully we get such a button soon =)[/quote]
Well, there's also the issue of "open problems" vs. "unsolved problems".  Many times problems are posted in "unsolved" by people who know the solution, which is fine by me but technically violates the description of the forum.  So the notion of a "solution" is pretty nuanced.  For any question that can be solved in a straightforward manner, I consider the thread closed when someone has provided a good hint.  For many other threads, I want to see a complete, written-up solution.

I just had a thought, though.  What would complement a "flag as solved" button nicely would be a voting system that would allow you to express your desire to see a particular problem solved in full.  Sometimes I go back through old zero-response threads when I want a hard problem to solve, it would be neat if I could see the ones people were really curious about.

I'd vote for this thread; I still really want to know how to do this sum.

I don't know if you saw my edit than. I got back to it a few minutes later and you were probably typing yours up. Anyways the solution to the sum is in my post above.
And also the voting thing could be abused. Many times we get very straight forward questions and hints almost equal to solutions, but the OP cusses anyways and starts asking for a full solution without trying. Those would easily get at least a vote, unnecessarily. You know those very same people you dislike....the ones asking for homework question solutions instead of help :P

Mhmm I think a new thread should be started for this discussion. I am not sure where it should go though. Also if you want a hard question (to me at least), [url=http://www.mathlinks.ro/viewtopic.php?t=284398]enjoy =)[/url]. It's on counting the edges in a convex n-gon. I know there is a solution using a recurrence relation, but I am going for a more direct solution of just finding each individual term (of Euler's Formula).

Interesting, it's quite like a WZ-pair solution...

[b]Lemma:[/b]  $ \int_{0}^{1} t^a (1 \minus{} t)^b \, dt \equal{} \frac{1}{(a\plus{}b\plus{}1) {a\plus{}b \choose b}}$.

It follows that $ (\minus{}1)^k \frac{ {4n \choose 2k} }{ {2n \choose k} } \equal{} (\minus{}1)^k (2n\plus{}1) {4n \choose 2k} \int_{0}^{1} t^k (1 \minus{} t)^{2n\minus{}k} \, dt$.  Let $ t \equal{} x^2$ and define

$ F_{4n}(x) \equal{} \sum_{k\equal{}0}^{2n} {4n \choose 2k} (\minus{}x^2)^k (1 \minus{} x^2)^{2n\minus{}k}$,

which is equal to

$ \frac{ (\sqrt{1\minus{}x^2} \plus{} ix)^{4n} \plus{} (\sqrt{1\minus{}x^2} \minus{} ix)^{4n}}{2}.$

Letting $ x \equal{} \sin \theta$ it follows that $ F_{4n}(x) \equal{} \cos 4n \theta$.  The original sum becomes the integral $ \minus{}(2n\plus{}1) \int_{0}^{\frac{\pi}{2}} \cos 4n \theta \, \sin 2 \theta \, d \theta$ and the rest is computation, for example by the product-to-sum formula.

I ran into that lemma while thinking about this, tried to use it but failed to get that far; bravo.

Do you know how to prove that lemma, by the way?

[quote="LydianRain"]Do you know how to prove that lemma, by the way?[/quote]

One proof is [url=http://en.wikipedia.org/wiki/Beta_function]here[/url] (substitute in the (known) values of the beta function at the integers to get t0r's expression)

My favorite proof is probabilistic.  Suppose we pick $ a \plus{} b \plus{} 1$ points $ t, t_1, ... t_{a\plus{}b} \in [0, 1]$ uniformly.  What's the probability that $ a$ of the points appear to the left of $ t$ and $ b$ appear to the right of $ t$?

Clearly this is $ {a\plus{}b \choose a} \int_{0}^{1} t^a (1 \minus{} t)^b \, dt$.  On the other hand, this is also the probability that the first point we choose is also the $ (a\plus{}1)^{st}$ point in left-to-right order, which is $ \frac{1}{a\plus{}b\plus{}1}$ by uniformity.

The generating function analogue of this proof is straightforward: compute $ \int_{0}^{1} (xt \plus{} (1 \minus{} t))^{a\plus{}b} \, dt$.

Hmm funny. I can't figure out the clearly part, t0r. What does the integral represent in that form?
Unless its some sort of variation of the binomial distribution... :|

$ t^a (1 \minus{} t)^b$ represents the probability that, for [b]fixed[/b] $ t$, $ a$ chosen points are to the right of $ t$ and $ b$ chosen points are to the left of $ t$.  We integrate over $ [0, 1]$ because $ t$ is chosen uniformly and multiply by $ {a\plus{}b \choose a}$ because we need to choose which of $ t_1, ... t_{a\plus{}b}$ need to be on the left and which need to be on the right.

It is troublesome to make a note in English.
So I write an answer in Japanese.

$ (x^2\plus{}2ix\minus{}1)^n$ と $ (x^2\plus{}1)^n$ に着目して, 多項定理を適用すると, 求める和は,
\[ \sum _{p\equal{}0}^{2n} \frac{\sum _{q\equal{}0}^p \frac{(2n)!}{(p\minus{}q)!(2q)!(2n\minus{}p\minus{}q)!}1^{p\minus{}q}(2i)^{2q}(\minus{}1)^{6\minus{}p\minus{}q}}{\frac{(2n)!}{(p)!(2n\minus{}p)!}1^p1^{2n\minus{}p}}\]
これを整理すると,
\[ \sum _{p\equal{}0}^{2 n} \sum _{q\equal{}0}^p \frac{(\minus{}1)^p 4^q (2 n\minus{}p)! p!}{(2 n\minus{}p\minus{}q)! (p\minus{}q)! (2 q)!}\]
また,
\[ \sum _{q\equal{}0}^p \frac{4^q }{(2 n\minus{}p\minus{}q)! (p\minus{}q)! (2 q)!} \equal{} \frac{(4n)!}{(2n)!(4 n\minus{}2 p)!(2 p)!}\]
これは, $ n$ を固定して $ p$ を動かして推測することにより, $ \frac{1}{(4 n\minus{}2 p)!(2 p)!}$ が帰納的に求まる.
これを代入して $ p$ を固定することにより, $ \frac{(4n)!}{(2n)!}$ が帰納的に求まる.
よって,
\[ \sum _{p\equal{}0}^{2 n} \sum _{q\equal{}0}^p \frac{(\minus{}1)^p 4^q (2 n\minus{}p)! p!}{(2 n\minus{}p\minus{}q)! (p\minus{}q)! (2 q)!} \equal{} \sum _{p\equal{}0}^{2 n} \frac{(4n)! (\minus{}1)^p (2 n\minus{}p)! p!}{(2n)!(4 n\minus{}2 p)!(2 p)!}\]
ここで,
\[ \sum _{p\equal{}0}^{2n} \frac{(\minus{}1)^p (2 n\minus{}p)! p!}{(4 n\minus{}2 p)!(2 p)!} \equal{} \frac{1}{(4 n)! \left(\frac{1}{(2 n)!}\minus{}\frac{1}{(2 n\minus{}1)!}\right)} \equal{} \frac{(2 n)!}{(1\minus{}2 n) (4 n)!}\]
これも帰納的に求まる.
また, [url=http://www.research.att.com/~njas/sequences/A119837]A119837[/url] - OEIS を見るからに, これらの関連を探れば組み合わせ的に意味を照らし合わせて証明ができるかもしれない.
故に,
\[ \sum _{p\equal{}0}^{2 n} \frac{(4n)! (\minus{}1)^p (2 n\minus{}p)! p!}{(2n)!(4 n\minus{}2 p)!(2 p)!} \equal{} \frac{(4 n)! (2 n)!}{(2 n)! (1\minus{}2 n) (4 n)!} \equal{} \boxed{\frac{1}{1\minus{}2n}}\]

PS $ \sum _{q\equal{}0}^p \frac{4^q }{(2 n\minus{}p\minus{}q)! (p\minus{}q)! (2 q)!} \equal{} \frac{(4n)!}{(2n)!(4 n\minus{}2 p)!(2 p)!}$ も組み合わせ的に意味を照らし合わせて証明ができるかもしれないが, 帰納的に証明したので, 興味がある方は組み合わせ的な意味付けをしてみてください.

Upon request from Darij Grinberg, I shall attempt making a translation.

[quote="Kouichi Nakagawa"]
Looking at $ (x^2 \plus{} 2ix \minus{} 1)^n$ and $ (x^2 \plus{} 1)^n$ and applying the multinomial theorem, the sum we are looking for is
\[ \sum _{p \equal{} 0}^{2n} \frac {\sum _{q \equal{} 0}^p \frac {(2n)!}{(p \minus{} q)!(2q)!(2n \minus{} p \minus{} q)!}1^{p \minus{} q}(2i)^{2q}( \minus{} 1)^{6 \minus{} p \minus{} q}}{\frac {(2n)!}{(p)!(2n \minus{} p)!}1^p1^{2n \minus{} p}}
\]
Rearranging this,
\[ \sum _{p \equal{} 0}^{2 n} \sum _{q \equal{} 0}^p \frac {( \minus{} 1)^p 4^q (2 n \minus{} p)! p!}{(2 n \minus{} p \minus{} q)! (p \minus{} q)! (2 q)!}
\]
and
\[ \sum _{q \equal{} 0}^p \frac {4^q }{(2 n \minus{} p \minus{} q)! (p \minus{} q)! (2 q)!} \equal{} \frac {(4n)!}{(2n)!(4 n \minus{} 2 p)!(2 p)!}
\]
For this, fixing n and making conjectures from changing p around, $ \frac {1}{(4 n \minus{} 2 p)!(2 p)!}$ can be figured out inductively.
Substituting this and fixing $ p$, $ \frac {(4n)!}{(2n)!}$ can be figured out inductively.
Accordingly,
\[ \sum _{p \equal{} 0}^{2 n} \sum _{q \equal{} 0}^p \frac {( \minus{} 1)^p 4^q (2 n \minus{} p)! p!}{(2 n \minus{} p \minus{} q)! (p \minus{} q)! (2 q)!} \equal{} \sum _{p \equal{} 0}^{2 n} \frac {(4n)! ( \minus{} 1)^p (2 n \minus{} p)! p!}{(2n)!(4 n \minus{} 2 p)!(2 p)!}
\]
Here,
\[ \sum _{p \equal{} 0}^{2n} \frac {( \minus{} 1)^p (2 n \minus{} p)! p!}{(4 n \minus{} 2 p)!(2 p)!} \equal{} \frac {1}{(4 n)! \left(\frac {1}{(2 n)!} \minus{} \frac {1}{(2 n \minus{} 1)!}\right)} \equal{} \frac {(2 n)!}{(1 \minus{} 2 n) (4 n)!}
\]
This, too, can be figured out inductively.
Furthermore, from looking at [url=http://www.research.att.com/~njas/sequences/A119837]A119837[/url] - OEIS, I think it might be possible to construct a proof by considering the combinatorial meaning of the sum if you search for similarities to this.
Consequently,
\[ \sum _{p \equal{} 0}^{2 n} \frac {(4n)! ( \minus{} 1)^p (2 n \minus{} p)! p!}{(2n)!(4 n \minus{} 2 p)!(2 p)!} \equal{} \frac {(4 n)! (2 n)!}{(2 n)! (1 \minus{} 2 n) (4 n)!} \equal{} \boxed{\frac {1}{1 \minus{} 2n}}
\]
PS: Since I have proven $ \sum _{q \equal{} 0}^p \frac {4^q }{(2 n \minus{} p \minus{} q)! (p \minus{} q)! (2 q)!} \equal{} \frac {(4n)!}{(2n)!(4 n \minus{} 2 p)!(2 p)!}$ inductively, although it also might be provable by considering its combinatorial meaning, I would welcome interested persons to try attaching a combinatorial meaning to it.[/quote]
I apologize for any mistakes that might have crept in; my experience in reading mathematical literature in Japanese is, for all means and purposes, zero and I didn't even take my time to read the problem nor to try actually understanding the proposal.  :maybe:

Hi negativezero :)

Thank you for translating my post into English.
I am a Japanese and very weak in English.

I made [url=http://www.mathlinks.ro/viewtopic.php?p=1545929#1545929]t0rajir0u's proof (Evaluate the sum #17)[/url] with notes.
If you need, I'll post it.
Do anybody want it?


Camy made a list of every possible distinct five-digit positive integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list?

The average of 1, 3, 4, 5 and 9 is 4.4. So the average number in the list is $ 44000\plus{}4400\plus{}440\plus{}44\plus{}4.4\equal{}48888.4$. Since there are a total of $ 5!\equal{}120$ numbers on the list, their sum is $ 48888.4\times{120}\equal{}\boxed{5866608}$

I put all the digits in each place, & added, so I added (for the sake of simplicity) 99999+55555+44444+33333+11111, getting 244442. Then, since there are 4! ways to choose which digit is in each place, I multiplied the sum by 24, getting 5,866,608.

[quote=Greenlantern09]it is 1199976[/quote]

Hide, and explain your answer.

[hide]
Note that there are $4! = 24$ numbers ending in 1, since we have 4 choices for the 10s digit, 3 choices for the 100s digit, 2 choices for the 1000s digit, and 1 choice for the remaining digit. Thus there are also 24 numbers ending in each of 3, 4, 5, 9, and the total contribution of ones digits to the sum is $24 (1 + 3 + 4 + 5 + 9) = 528$. But we can make a similar argument about the contribution of the digits in the other places (10s, 100s, etc.), so our total sum is $528 + 5280 + \ldots + 5280000 = 528 (1 + 10 + \ldots + 10000) = 528\cdot 11,111 = \boxed{5,\!866,\!608}$. [/hide]

[hide]Solution:
Note that there are $4! = 24$ numbers ending in 1, since we have 4 choices for the 10s digit, 3 choices for the 100s digit, 2 choices for the 1000s digit, and 1 choice for the remaining digit. Thus there are also 24 numbers ending in each of 3, 4, 5, 9, and the total contribution of ones digits to the sum is $24 (1 + 3 + 4 + 5 + 9) = 528$. But we can make a similar argument about the contribution of the digits in the other places (10s, 100s, etc.), so our total sum is $528 + 5280 + \ldots + 5280000 = 528 (1 + 10 + \ldots + 10000) = 528\cdot 11,111 = \boxed{5,\!866,\!608}$[/hide]

Why isn't the average of the numbers just 13459+95431/2? Smallest+largest numbers/2?

That is not how you calculate the mean of the numbers in a set (unless there are only 2 numbers).

[quote=mewto][quote=PertBanking]I'm not so much concerned with the solution to this problem (i.e. I got it right), I'm just concerned with the apparent lack of entertainment outlets that Camy has at her disposal.  What kind of person makes a 120-item-long list of numbers in her spare time?  Poor girl.[/quote]

I second this. Someone needs to tell her about the wonders of minecraft.[/quote]

Why the bump?

There are $5! = 120$ ways to make a 5-digit number/order 5 numbers. There are $120/5 = 24$ places (not necessarily distinct) where each digit can be in certain orderings. The sum of the possible digits is 22. So $22 \cdot 24 = 528$ is the sum of the digits. We need to find the sum of the numbers, so we multiply that by $10,000 + 1,000 + 100 + 10 + 1 = 11,111$ to get our answer of $\boxed{5866608}$.

[quote=Greenlantern09]it is 1199976[/quote]

Wait, wasn't this like the answer to a late MATHCOUNTS Chapter problem this year??

There is another way to solve the problem, but it is a bit harder. You can imagine it adding up into the ones digit, the tens digit, the thousands digit, and the hundred thousands digit, as the 528 adds up. So then we put 8 in the ones place, and so on. It ends up into 5,866,608.

[hide]We look at the digit $1$ only: It appears $4!$ times in the ten thousands digit, thousands digit and so on. Similarly for $3,4,5,9$. Hence the answer is $4! \times 11111 (\text{ to represent the number in their respective digits }) \times (1+3+4+5+9) = 5866608$.[/hide]

[quote=titanium7]That was hard.[/quote]

It was indeed


In the prime factorization of $ 40!$, how many times does the factor 5 occur?

[quote="vallon22"]one way is listing out all the factors, but that would take too long
however, we can list out only the factors that have 5's

5, 10, 15, 20, 25, 30, 35, 40

in each of those, the number 5 occurs once, except in 25, in which they occur twice

so we have 7 times + 2 times = 9 times

OR

we can divide 40 by 5, and we get the answer 8
then we divide 40 by 25, and we get the answer 1
the next power of 5 is 125, and 40 divided by 125 is not a whole number

therefore, the number 5 occurs 8+1 or 9 times in the prime factorization of 40!

our answer is still 9[/quote]

Hello, I'm here just to add my opinion about the easiest method, but I think this method here "therefore, the number 5 occurs 8+1 or 9 times in the prime factorization of 40!" is the best method. $\dfrac{40}{5}=8$ but remember $25=5^2$ so $8+1=\boxed{9}$. :)


Let $ (i, j, k)$ be an ordered triple of positive integers such that $ i \plus{} j \plus{} k \equal{} 23$.
For example $ (1, 1, 21)$, $ (21, 1, 1)$ and $ (6, 9, 8)$. For each triple we can compute the
product $ P$ of $ i, j$ and $ k$. In our three examples this gives $ P \equal{} 21, 21$ and $ 432$
respectively. Let $ S$ be the sum of the $ P$ for all allowed triples $ (i, j, k)$
$ S \equal{} 21 \plus{} 21 \plus{} 432 \plus{}...$ .
Show that $ 23$ divides $ S$.

This problem is almost exactly the same as an HMMT individual problem from two years ago.

[hide=" Hint unrelated to t0rajir0us"]You can actually find the sum cleanly.  The method for finding the number of solutions in positive integers to $ i \plus{} j \plus{} k \equal{} 23$ is well-known.  Try to modify the method to get what you want.[/hide]

[quote="t0rajir0u"][hide="Hint"] The relevant generating function is $ (x \plus{} 2x^2 \plus{} 3x^3 \plus{} 4x^4 \plus{} ...)^3$ (can you see why?). [/hide]
For an interesting question that is in some sense the opposite of this, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196088&search_id=1977562434]this thread[/url].[/quote]

Okay so how does the generating function help?

[hide="Generating Functions Solution"]$ (x \plus{} 2x^{2} \plus{} 3x^{3} \plus{} \cdots )^{3} \equal{} x^{3}(1 \plus{} x \plus{} x^{2} \plus{} \cdots )^{6}$

We want the coefficient of $ x^{23}$ in this, which is the same as the coefficient of $ x^{20}$ in $ (1 \plus{} x \plus{} x^{2} \plus{} \cdots )^{6}$.  Since the coefficient of $ x^{k}$ in $ (1 \plus{} x \plus{} x^{2} \plus{} \cdots )^{n}$ is $ \binom{n \plus{} k \minus{} 1}{n \minus{} 1}$, the coefficient we want is $ \binom{25}{5}$.[/hide]

Or as in the HMMT problem, you can also view selecting 5 objects from 25 in an unconventional way: for any selection of 5 objects, the second and the fourth can be set as "barriers". Then we have effectively divided the 25 objects into $ i$ objects, 1 barrier, $ j$ objects, 1 barrier, and $ k$ objects. So there are $ ijk$ ways to select 5 objects with the 2nd and 4th objects as barriers.

The sum of all possible products $ ijk$ can be seen to equal to the number of ways to select 5 objects out of 25, so the answer is $ \binom {25}{5}$ and it easily shown to be divisible by 23.