[quote="236factorial"]If I posted 1000 posts and 2.3% of that is spam, how many posts by me are not spam?[/quote]

[hide]That means that 23 of your posts are spam ($1000*0.023$)

So the posts that aren't spam is $1000-23=977$[/hide]

[hide]
since $1000 \cdot .023=23$ that means 23 of your posts are spam so $\boxed{977}$ are not.[/hide]


In $ \text{\LaTeX}$, how do you box things?

$ \boxed{\text{like this}}$:
[code]$\boxed{\text{like this}}$[/code]

Can you box things in Latex without using text?

Sure. \boxed works in math mode as well: $ 1\boxed{23}45^{6\boxed 78}$

Note: in using TexNic, you must include the amsmath packlage to use the \boxed command.


algorithm algebra polynomial trigonometry function domain calculus

Maybe someone can halp me.I need ebook or something similarly with algorithms like for partial intergration(Su*v=u*v-Svdu)-combination of polynomial and exp or trig function
"u=polynomial ---> dv=exp or trig function
combination of polynomial and log or cyclo function
"u=log or cyclo ---> dv=........so on
I need it for some program where I need lot of math
I'm interest in high school math and college so if you know few titles or links for tutorials for this specifc domain I'll be appreciate to replay to my topic

For example Maple can do symbolical and numerical integration. I'm not sure what kind of integral you have to compute. For example Simpson's rule is explained in many numerical analysis books. For symbolical integration, Risch's algorithm is a general method to integrate functions which are a finite combination of elementary functions.


A line passes through the orthocenter $ H$ of acute triangle $ ABC$ and meets the sides $ AC,BC$ at $ D,E$ respectively. The line through $ H$ perpendicular to $ DE$ intersects the line $ AB$ at $ F$. Prove that $ \frac{DH}{HE}\equal{}\frac{AF}{FB}$.

I think that the point $ F,$ must be the intersection point of $ BC$ instead of $ AB$ and then, to prove $ \frac {DH}{HE} \equal{} \frac {CF}{FB}.$

If yes, it is a well known easy problem, coming from the past.

Kostas Vittas.

Oh, his problem was solved by you about 10 days ago, dear [b]Kostas Vittas[/b]!!! http://www.mathlinks.ro/viewtopic.php?t=203470

@Carlos Tevez: what is the tactic you and your team will use in the coming Champion League final match (vs. Chelsea)?  :D

Thank you a lot dear [b][size=100]April[/size][/b] and sorry dear all my friends, for such that my blindness.  :(

I made the schema and I thought $ D,\ E,$ as the intersection points of $ AC,\ AB...$ respectively. I can’t understand that I forgot the problem I solved a few days ago (old age...   :mad:  )

By the way, let me to present the similar problem I mentioned, with only the schema, because its solution is very easy.

[b][size=100][color=DarkBlue] A line passes through the orthocenter $ H,$ of an acute triangle $ \bigtriangleup ABC$ and meets the sides $ AC,\ AB$ at points $ D,\ E,$ respectively. The line through $ H$ and perpendicular to $ DE,$ intersects the sideline $ BC,$ at point $ F.$ Prove that $ \frac {DH}{HE} = \frac {CF}{FB}$ [/color][/size][/b].

Sorry again, Kostas Vittas.

[quote="April"]

@Carlos Tevez: what is the tactic you and your team will use in the coming Champion League final match (vs. Chelsea)?  :D[/quote]

Come on, we won't need any tactic. It'll be a cakewalk  :D


Let $ a, b, c$ be three sides of a triangle such that $ a\plus{}b\plus{}c\equal{}2$. Prove that  $ \frac{52}{27}\le a^2\plus{}b^2\plus{}c^2\plus{}2abc < 2$.

Thank.

after the same substitution that I used in my first post, we have $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\equal{}2\minus{}2xyz$
we have by AM-GM $ xyz\leq \left(\frac{x\plus{}y\plus{}z}{3}\right)^3\equal{}\frac{1}{27}$ therefor $ 2\minus{}2xyz \geq 2\minus{}\frac{2}{27}\equal{}\frac{52}{27}$
then $ \frac{52}{27}\leq a^2\plus{}b^2\plus{}c^2\plus{}2abc < 2$


a cyclic quadrilateral $ ABCD$ has a circumradius $ R$
show that $ AB\plus{}BC\plus{}CD\plus{}DA < 2R\plus{}AC\plus{}BD$

You've said it's easy Why don't you post your solution ???

Can you show me ?? เอาเลยยยยยยยยยยยยยยยยยย มาเด้ๆๆๆๆๆ ย่าาาาาาาาา!!!!


geometry
3D geometry
tetrahedron
sphere
geometric transformation
reflection
geometry proposed

A question i wonder that : Given a tetrahedron ABCD. Prove or disprove that three spheres (B,BA), (C,CA),(D,DA) cover ( fill up) the tetrahedron ?
May be this is a mediocre question, i don't know.  :(

Yes, they do cover the tetrahedron. Consider $A_b,A_c,A_d$, the reflections of $A$ in $B,C,D$ respectively. These points lie on the spheres $\mathcal S_b=(B,BA),\mathcal S_c=(C,CA),\mathcal S_d=(D,DA)$ respectively. More over, $\mathcal S_x$ has $AA_x$ as its diameter, for all $x\in\{b,c,d\}$. It's easy to show that for a tetrahedron $XYZT$, the spheres of diameters $XY,XZ,XT$ cover te tetrahedron $XYZT$ $(*)$, meaning that, in our case, our spheres cover the tetrahedron $AA_bA_cA_d$, which contains $ABCD$.

In order to prove $(*)$, just notice that for any point $P$ inside or on the border of $XYZT$, at least one of $\angle XPY,\angle XPZ,\angle XPT$ is $\ge\frac\pi 2$.

When i solved this lemma ( for my exercise) , i tried to prove that three spheres above cover the triangle BCD, after that we can have the claim with the convexity of sphere.
Btw, thanks grobber.


Determine all positive integers $n$ such that for every relatively prime $a, b$ that divide $n$ is \[ a+b-1|n \]

SOLUTION: $n$ is a prime power or $n=12$

Suppose the primes dividing $n$ were $p_1<p_2<\cdots<p_k$ with $k\geq 2$, then $p_1+\prod_{i>1}p_i-1$ it's a power of $p_1$, implying that $p_1\mid\prod_{i>1}p_i-1$ but $p_1^2\nmid\prod_{i>1}p_i-1$, and as a consequence we have $p_1^2+\prod_{i>1}p_i-1=p_1\prod_{i>1}p_i^{e_i}$ where at least one of the exponent, lets say $e_j$ is positive. Now we have $p_j\mid p_1^2-1$ and as a consequence $p_j\mid p_1+1$, and so $p_1<p_j\leq p_1+1$, implying that $p_1=2$, $j=2$ and $p_2=3$.
By similar considerations we arrive at the equations $1+\prod_{i>1}p_i=2^a$ and $3+\prod_{i>1}p_i=2\cdot 3^b$, but it is easy to prove that the only solutions to the equation $2\cdot 3^b-2^a=2$ are $2\cdot 3-2^2=2$ and $2\cdot 3^2-2^4=2$, from where there remain just two posibilities: $k=2$, $p_1=2$ and $p_2=3$; and $k=3$, $p_1=2$, $p_2=3$, $p_3=5$.

The second posibility is impossible because then $3+5-1=7\mid n$.

It only remains the first posibility. Now $8+3-1=2\cdot 5\nmid n$ implying that $8\nmid n$, and $2+3-1=4\mid n$. Also, $2+9-1=2\cdot 5\nmid n$ implying that $9\nmid n$. Putting all together results $n=4\cdot 3=12$.


When two consecutive whole numbers are randomly selected, what is the probability that one of them is a multiple of 4? Express your answer as a common fraction.

[hide=Solution]The above is incorrect.
We know that by logic one of the two numbers must be a mutiple of 2 and only one can be a multiple of 2. Since a number can either be:
$2 \pmod{4}$
or $0 \pmod{4}$,
it is a 50-50 chance to pick any one. So our answer is $\frac{1}{2}$[/hide]

[quote=AbhiramJuvvadi]I agree with glory_of_math, the answer is $1/2$.[/quote]

Hide answers please

[quote=sophie8][quote=AbhiramJuvvadi]I agree with glory_of_math, the answer is $1/2$.[/quote]

Why must you revive a 7-year-old post with just a simple "I agree"?[/quote]

I think you mean 7-month, not 7 year.

[quote=sophie8][quote=AIME15]This is $ 2 \left (\frac{3}{4} \times \frac{1}{4} \right )\equal{}\frac{6}{16}\equal{}\boxed{\frac{3}{8}}$.[/quote]

It says 2009.[/quote]

Yes but the last post was on March 2016, I usually count from the last post.

[quote=sophie8]Um, if you're saying that glory_of_math should hide, don't even think about it, but AbhiramJuvvadi, it wouldn't really matter because glory_of_math wouldn't hide his/her answer so no matter what, the answer would still be there.[/quote]

Why wouldn't they hide? The last time that they visited were all in December or January.

Read bluecarneal's Alcumus announcement.

They have visited recently, so it wouldn't be counted really old.

All whole numbers can be written as $4k$, $4k+1$, $4k+2$, or $4k+3$ for some integer $k$. Suppose we look at the smaller of the two consecutive numbers. If it is of the form $4k$, it is a multiple of 4. If it is of the form $4k+3$, the other is $4k+4$, a multiple of 4. If on the other hand, it can be written as $4k+1$ or $4k+2$, neither number is a multiple of 4. So in two of the four equally likely cases, one of the numbers is a multiple of 4, for a probability of $2/4=\boxed{1/2}$.

I used symmetry, either one of them could be a multiple of 4, or neither. So, the probability is [hide=Answer]1/2[/hide]


Dude please check the first question.

I think there is something wrong in the question.  :maybe:  :?:


Εκεί που έλεγα να κάνω καμιά δουλειά , νάσου και μου στέλνει ένας Ρουμάνος καμιά δεκαριά βιβλία και ..πάει ο χρόνος !

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

Αν  $ x,y,z >0$ και $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\equal{}3$ , να αποδείξετε ότι :

   $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \leq (x\plus{}y)(y\plus{}z)(z\plus{}x)$

Μπάμπης

[quote="stergiu"][quote="Γιώργος"][quote="stergiu"]Εκεί που έλεγα να κάνω καμιά δουλειά , νάσου και μου στέλνει ένας Ρουμάνος καμιά δεκαριά βιβλία και ..πάει ο χρόνος !

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

Αν  $ x,y,z > 0$ και $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 3$ , να αποδείξετε ότι :

   $ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \leq (x \plus{} y)(y \plus{} z)(z \plus{} x)$

Μπάμπης[/quote]

Αρχικά έχουμε ότι $ (x \plus{} y)(y \plus{} z)(z \plus{} x)\geq8xyz$. Άρα αρκεί τώρα $ 8xyz\geq(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)$. Έχουμε $ 8xyz\geq(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)\Leftrightarrow8\geq\frac {x \plus{} 1}{x}\frac {y \plus{} 1}{y}\frac {z \plus{} 1}{z} \equal{} (1 \plus{} \frac {1}{x})(1 \plus{} \frac {1}{y})(1 \plus{} \frac {1}{z}) \equal{} 1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} \frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx} \plus{} \frac {1}{xyz} \equal{} 1 \plus{} 3 \plus{} \frac {x \plus{} y \plus{} z \plus{} 1}{xyz}\Leftrightarrow4\geq\frac {x \plus{} y \plus{} z \plus{} 1}{xyz} \Leftrightarrow 4xyz\geq x \plus{} y \plus{} z \plus{} 1$(1)

Όμως από την ανισότητα Andreescu έχουμε ότι $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\geq\frac {3^{2}}{x \plus{} y \plus{} z}\Leftrightarrow3\geq\frac {3^{2}}{x \plus{} y \plus{} z}$ Άρα $ \Leftrightarrow x \plus{} y \plus{} z\geq3$    (2)

Αλλά από την  ΑΜ-GM έχουμε ότι $ x \plus{} y \plus{} z\geq3\sqrt [3]{xyz}\Leftrightarrow3\geq3\sqrt [3]{xyz}$. Οπότε $ xyz\leq1$               (3)

Οπότε αρκεί να αποδείξουμε την (1) η οποία ισχύει από τις (2) και (3).[/quote]

Γιώργο ,
 πολύ καλές σκέψεις. Είναι όμως αργά και ίσως κάνω λάθος : Ρίξε  ξανά μια ματιά στις τελευταίες σχέσεις . Φοβάμαι μήπως σου ξεφεύγει κάτι.Μήπως δηλαδή οι (2) και (3) δεν δίνουν την (1).(μην τη παίρνεις ως δεδομένη την (1) , αλλά ως ζητούμενη).

 Μπάμπης[/quote]

Σωστή η παρατήρησή σας, αλλά ακόμη δεν καταλαβαίνω που βρίσκεται το μοιραίο λάθος. Η φορά της ανισότητας βγαίνει αντίθετη της ζητούμενης. Που έχω κάνει λάθος? :huh:

$ x\plus{}y\plus{}z\geq3$
$ 4\geq4xyz$

Τις προσθέτουμε κατά μέλη και έχουμε $ x\plus{}y\plus{}z\plus{}4\geq4xyz\plus{}3\Leftrightarrow4xyz\leq x\plus{}y\plus{}z\plus{}1$???????

[quote="Γιώργος"]
Όμως από την ανισότητα Andreescu έχουμε ότι $ \frac {1}{x} + \frac {1}{y} + \frac {1}{z}\geq\frac {3^{2}}{x + y + z}\Leftrightarrow3\geq\frac {3^{2}}{x + y + z}$ Άρα $ \Leftrightarrow x + y + z\geq3$    (2)

Αλλά από την  ΑΜ-GM έχουμε ότι $ \boxed{x + y + z\geq3\sqrt [3]{xyz}\Leftrightarrow3\geq3\sqrt [3]{xyz}}$. Οπότε $ xyz\leq1$               (3)

[/quote]

Εδώ είναι το λάθος (αυτό που είναι σε περίγραμμα) :wink:  Το ότι $ x + y + z\geq3$ ΔΕΝ ΣΥΝΕΠΑΓΕΤΑΙ $ xyz\leq1$.

Θα έπρεπε για να ισχύει η σχέση που γράφεις να ήταν $ x + y + z\leq3$

(Εσύ λες ότι αφού $ x + y + z\geq3$ και $ x + y + z\geq3\sqrt [3]{xyz}$ συνεπάγεται ${ 3\geq3\sqrt [3]{xyz}}$ αλλά αυτό είναι λάθος!)

[quote="Αρχιμήδης 6"]Φιλε Γιωργο εσυ αποτι καταλαβα προσπαθεις να αποδειξεις οτι $ (x \plus{} y)(y \plus{} z)(z \plus{} x) > \equal{} 8xyz > \equal{} (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)$ ομως το πρωτο μερος δεν ισχυει γιαυτο σου βγαινει λαθος... :([/quote]   Για να καταλάβω εαν θέλουμε να αποδείξουμε ότι $ a\geq b$ και γνωρίζουμε ότι $ b\geq c$ τότε αρκεί να αποδείξουμε ότι $ a\geq c$, ενώ αν γνωρίζουμε ότι $ a\geq d$ δεν μπορούμε να πάρουμε ως ζητούμενη τη $ d\geq b$? σωστά?

:([/quote]   Για να καταλάβω εαν θέλουμε να αποδείξουμε ότι $ a\geq b$ και γνωρίζουμε ότι $ b\geq c$ τότε αρκεί να αποδείξουμε ότι $ a\geq c$, [/quote]

Aυτό είναι λάθος Γιώργο. Το σωστό είναι οτι αν θές να αποδείξεις ότι
$ a\ge b$ και γνωρίζεις ότι $ b\le c$ τότε αρκεί να.δ.ο. $ a\ge c$.

[quote="Αρχιμήδης 6"]Γιωργο αυτο που εγραψα ηταν τελειος λαθος γιαυτο και το διεγραψα.Τα πραγματα ειναι απλα:αν a>b και b>c τοτε a>c και μονο αυτο... :) Ωραια παντος η ασκηση![/quote]

Η άσκση ωραία είναι αλλά δυστηχώς δεν βρήκα ακόμη λυση. Μου φαίνεται πως πάλι δε θα κοιμηθώ σήμερα  :mad:  :mad:  :mad:  :mad: . Σευχαριστώ πάντως για τη διευκρίνησή σου...

Proove that if $ x,y,z>0$ and $ 1/x\plus{}1/y\plus{}1/z\equal{}3$ then $ (x\plus{}y)(y\plus{}z)(z\plus{}x)>\equal{}(x\plus{}1)(y\plus{}1)(z\plus{}1)$

[color=blue]Efharisto[/color], Αρχιμήδης 6 :)

For $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 3\ (x > 0,\ y > 0,\ z > 0)\Longleftrightarrow xy \plus{} yz \plus{} zx \equal{} 3xyz.$ Let $ x \plus{} y \plus{} z \equal{} a,\ xyz \equal{} b,$ we have $ xy \plus{} yz \plus{} zx \equal{} 3b\ (a > 0,\ b > 0).$

Since $ A.M.\geq G.M.\geq H.M.$ gives $ b\geq 1,$ so

$ (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} (a \minus{} z)(a \minus{} x)(a \minus{} y)$

$ \equal{} a^3 \minus{} (x \plus{} y \plus{} z)a^2 \plus{} (xy \plus{} yz \plus{} zx)a \minus{} xyz$

$ \equal{} 3ab \minus{} b,\ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \equal{} 1 \plus{} a \plus{} 4b.$

Since $ \frac {x \plus{} y \plus{} z}{3}\geq \frac {3}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}}\Longleftrightarrow a\geq 3,$

thus we have $ (x \plus{} y)(y \plus{} z)(z \plus{} x) \minus{} (1 \plus{} x)(1 \plus{} y)(1 \plus{} z) \equal{} (3b \minus{} 1)a \minus{} 5b \minus{} 1\geq (3b \minus{} 1)\cdot 3 \minus{} 5b \minus{} 1$

$ \equal{} 4(b \minus{} 1)\geq 0.$ Q.E.D

This solution is not mine;  it is my father's.
But I thought that it would be a waste not to post it, as it seems to be quite a faster way to reach the same result as kunny. So, there it is:  :lol:

$ \frac {(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {a \plus{} 1}{a \plus{} b}\frac {1 \plus{} b}{b \plus{} c}\frac {1 \plus{} c}{c \plus{} a} \equal{}$

$ \frac {1 \plus{} \frac {1}{a}}{1 \plus{} \frac {b}{a}}\frac {1 \plus{} \frac {1}{b}}{1 \plus{} \frac {c}{b}}\frac {1 \plus{} \frac {1}{c}}{1 \plus{} \frac {a}{c}} \le$ (using that $ x \plus{} 1 \ge \sqrt {x}$) $ \frac {(1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{c})(1 \plus{} \frac {1}{c})}{2\sqrt {\frac {b}{a}}2\sqrt {\frac {c}{b}}2\sqrt {\frac {a}{c}}} \equal{}$

$ \frac {(1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{c})(1 \plus{} \frac {1}{c})}{8}$ $ \le$ (using AM - GM)$ \frac {(\frac {1 \plus{} \frac {1}{a} \plus{} 1 \plus{} \frac {1}{b} \plus{} 1 \plus{} \frac {1}{c}}{3})^3}{8} \equal{} \frac {(\frac {6}{3})^3}{8} \equal{} \frac {2^3}{2^3} \equal{} 1 \Leftrightarrow$

$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \le (a \plus{} b)(b \plus{} c)(c \plus{} a)$ q.e.d.  :D

I only find that, as $ xyz \le 1$, we can reach $ \frac {(1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{c})(1 \plus{} \frac {1}{c})}{8}$ by just using the Huygens inequality.


geometry
analytic geometry
pigeonhole principle

Show that a convex pentagon with lattice points as vertices has area no less than 5/2.

[hide]hmm... by picks theorem, this is equivalent to showing that no convex pentagon with lattice point coordinates can have no lattice points other than the five vertices.

Assume for the sake of contradiction such a pentagon exists...

Consider the lattice points mod 2, there are four points in Z[2]X Z[2], and we are choosing 5 points, so by pigeonhole two must be equivalent mod 2.  And ZOMG their midpoint is a lattice point! and by convexity this must be in the polygon, so by our assumption that lattice point must be a vertex.  But three colinear vertices in a convex polygon? Contradiction.[/hide]

I don't know if the following is rigorous or not, but it seems to work:

[hide] So we want to show that there exists a lattice point in the interior. Draw all vertical lines that contain at least one of the lattice points that are vertices of the pentagon. Note that there must be at least three, or else there will be 3 on the same line, which is bad (either we get concavity or collinearity). Similarly with horizontal lines. Then, of our horizontal and vertical lines, take away the four lines that are at the ends. The point(s) of intersection of the remaining lines must be inside the pentagon?[/hide]

[hide="weird things that my brain comes up with"]
Put one vertex at 0,0 and convert it into an algebraic proof with shoelace?

[/hide]

Pick's Theorem is a much nicer solution than Shoelace.  You'd still have to do some of the same number-theoretic reasoning, anyway.


AMC AIME function Harvard college USA(J)MO USAMO

How do you solve the following two problems from this year's AIME?

1) There are four numbers. The first and the forth have a difference of 30. The first three form a arithmetic sequence and the last three form a geometric sequence. Find the sum of the four terms.

2)How many four digit numbers have the property that the sum of the first two digits equals the sum of the lasst two digits?

These were problems 8 and 9, respectively from this year's AIME .

I'll just give a few hints, and then try to see if you can solve the rest of it.

1) Let the terms be a_1, a_2, a_3, and a_4. Since a_4-a_1=30, the sequences are increasing. Can you prove that a_4-a_3>a_3-a_2? Also, we know that a_3-a_2=a_2-a_1, so what does this tell you about the possible values of a_2-a_1? In the problem they are all integers, so use that freely.

2) Consider the four digit number abcd. We want a+b=c+d. What are the possible values of a+b? Remember that a cannot be equal to zero? Be prepared to do lots of casework. It's not that bad.

1.) Let the four number be a, b, c, and d. abs(a-d)=30 (since the difference between the first and the fourth number is 30), abs(a-b)=abs(b-c) (since the first, second, and the third number form an arithmatic sequence), xb:^2:=c, xb:^3:=d (since the second, the third, and the fourth number form a geometric sequence)
...I got this far...
2.) The combination that a (the sum of first two digits) and b (the sum of last to digits) equals n (a positive integer) is...
If n is even: 2n+2
If n is odd: 2n-1
So the answer is 2n+2:Sigma:n=1 (1:le:n:le:9, n=even natural number)+2n-1:Sigma:n=1 (1:le:n:le:9, n=odd natural number)
[b]=103[/b]

[quote="ComplexZeta"]Since a_4-a_1=30, the sequences are increasing. .[/quote]
It just says there is a difference of 30...

He didn't quote it quite right. I believe on the actual AIME it said that it was an increasing sequence. If I'm wrong, please correct me.

Tare, I believe your solution to #2 is incorrect. I didin't actaully check your work, but a simple casecheck would reveal that 103 is way too small.

I just finished 1... took me about a half hour of variable screwing around with, then i looked at it for a bit, and saw an answer, but then realized that i was thinking that d - x = 4 (DOH... i still cant believe i wasted 30 minutes like this)... so i went back and then the last twenty minutes, ive been screwing around with d - x = 30 and i got the answer. If you just want the answer, its right here : [hide]36[/hide]  I hope that didnt come out as text..... if it did, i am immensely sorry.

Solution (of which i am going to piece together from writing that is all over the place):

[hide]Start off with the sequence going x,y,c,d, where d-x = 30 and x,y,c is arithemtic and y,c,d is geometric.

y + r = yg, therefore r = yg - y and y = r/(g-1). Because x + r = y and yg=c and c-y = r, we can get x + yg - y = r/(g-1). Multiplying by the g-1, we arrive at gx + yg^2 - 2yg -x + y = r. Adding an x and subtracting a y leaves us with x + r -y on the right side, which equals 0. So we have

gx + yg^2 -2yg = 0. We can divide by g because a solution for g as 0 is meaningless, leaving us with yg -2y + x = 0. Hence gy = 2y -x.... x/(2-g)=y.

Now going back to the original equation of d-x = 30, we can get

yg^2 - (2-g)y = 30. So yg^2 -2y +gy = 30, hence y(g+2)(g-1) = 30. Looking at 30's factors, we see that 5(3)(2) works here, hence both y and g = 3. And from that, c =9 d = 27 and x = -3, which makes d-x = 30 work.[/hide]

This one seemed quite a bit easier than i thoguht it would be. I hope its correct..... answer : [hide]360[/hide]

Solution:

[hide]Looking at a few possibilities here, if a+b = 1, then there is 1 possibility for a (0) and b (1), and two for c and d (when c is 1 or when c is 0). Going further, we see that whenever c+d add up to a number, there is n + 1 possibilities for them. This is because the combinations are (0,n  1,n-1

2,n-2  3,n-3... ) until u reach the half way point. But because we can use zero, the half waypoint is the ceiling function of the number halved (5/2 becomes 3). And then because c and d can be interswitched, the number is multiplied by 2 and we get n+1 (if u dont follow this, try out a few cases and it will be clear). Then for a + b, the amount of possibilities is just n because we take away 1 possbility when a is equal to zero. These additions can go up to 18 (99).... so for c+d, the total number of possibilities is

2 + 3 +..... 19, equal to 19(20)/2  -1. Where as a +b is equal to

1+2+3.... 18, equal to 18(19)/2... Adding them up we get 9(19) +10(19) -1, equal to 19^2 -1... equal to 360.[/hide]

The answers are different from yours. I beleive they are 129 for problem 8 and 615 for problem 9.

hmm, jelyman, i believe you original reasoning is correct, though until the last step i don't follow it. Perhaps you missed some cases?

Unless you have already learned counting, the second problem is extremely difficult. I tried it by more primitive methods (drawing a picture, standing on head, etc.) but was unable to solve it. Good luck!

Can someone please show me what i did wrong then, especially for the first one, because those numbers definitely work.

For the second one: this may help you see where you went wrong, hopefully

[hide]

With the equal sums being anything from 1 through 9 we get, 1x2 ways plus 2x3 ways + ... + 9x10 ways. When the equal sums are 10 through 18 we get 9x9 ways, 8x8 ways, .. 1x1 ways, since they must be digits. Overall that makes 615.

[/hide]

I had a friend (harvard guy who went to my hs and did usamo junior/senior year...) explain both to me. First, number 1 has to be pos numbers. It would be helpful if you could provide the full problem in the future. Number 2, i screwed up and he showed me why. (and i also considered the two dependent ugh.....)


Some cities of the country are connected by routes.  For any three cities, there is a route connecting two of them (directly). There are 2000 cities. Show that the cities can be divided between two sons of the king such that each son get 1000 cities and any two cities of one son is connected by a route through only that son's cities.

To me, it seems a bit unclear. I can think of two interpretations, but one leads to a false problem, while the other one leads to a pretty easy one :).

At first, I thought [i]both[/i] sons had to have connected countries, but this is not true. Think of a country having $1999$ cities, each two connected to each other, and one isolated city. This was the first interpretation.

Then, I thought only [i]one[/i] son was required to have a connected country. In this case it's true, and not very difficult. We could proceed as follows:

First notice that the initial graph $G$ (the large country, before it's split) can have at most $2$ connected components. This is because three cities, each one from a different connected component, would contradict the hypothesis. The $\le 2$ connected components contain a total of $2000$ vertices, so there must be one (call it $H$; we could, of course, have $H=G$ if $G$ is connected) with $\ge 1000$ vertices. Now we begin building the first son's country. First we add a city from $H$ to it. Then we add another one, also from $H$, which is connected to the first one. Then another one, connected to at least one of the first two. And so on, until we've given the first son $1000$ cities in $H$. The other $1000$ go to the other son.

Well, it seems I misunderstood the problem. The wording was a bit unclear  :blush:

should be like this:
Some cities of the country are connected by roads. For any three cities, pick any two of them and there will be a route (direct or indirect) connecting the two cities and not passing through the third one. There are 2000 cities. Show that the cities can be divided between two sons of the king such that each son get 1000 cities and any two cities of one son is connected by a route through only that son's cities (the last condition holds for both sons)


geometry
parallelogram
trigonometry
AMC
USA(J)MO
USAMO
angle bisector

Let $ABCD$ be a parallelogram. The interior angle bisector of $\angle ADC$ intersects the line $BC$ in $E$, and the perpendicular bisector of the side $AD$ intersects the line $DE$ in $M$. Let $F= AM \cap BC$. Prove that:

a) $DE=AF$;
b) $AD\cdot AB = DE\cdot DM$.

[i]Daniela and Marius Lobaza, Timisoara[/i]

Can you sketch a graph of this problem??
I must have made some mistakes....[/hide]

Hints:
[hide]
a) obvious
b) law of sines
c) this is 8th grade?!?  :?  It's a wonder the US team can even compete with the Romanian team in the IMO
[/hide]

[quote="probability1.01"]
c) this is 8th grade?!?  :?  It's a wonder the US team can even compete with the Romanian team in the IMO
[/quote]Actually this is 7th grade. The answer is obviously yes, and usually the US Team does much better. The MOSP dramatically increases the knowledge of any US student attending it, thus any previous difference encountered is overcome. In plus the chances of getting 6 very good people out of 300 millions is much greater than to get 6 great students out of 20 millions :)

Yeah, I tend to think the US does well at the IMO due to sheer numbers :).  (sorry for the semi-pun :D)

Ah, part a) has a very nice solution :D  I'll just give a hint:

[hide="Click for hint"]
Look for iscoceles triangles.  In this case, there are some [i]similar[/i] iscocelese triangles :D
[/hide]

Part b quickly follows by the [hide]law of cosines[/hide] :P

EDIT: There's also a super-easy way to do part b) using only [hide]similar triangles[/hide]

[quote="probability1.01"]
c) this is 8th grade?!?  :?  It's a wonder the US team can even compete with the Romanian team in the IMO[/quote]

Hmm, not really.  It seems that we have at least one 7th grader make it to the USAMO every year.  Also, extremely motivated students who do lots of studying can probably get through these problems.  I had little trouble getting through this problem, though I didn't post it because it relied heavily on a diagram which I can't draw on the computer.

If you really care, the basic idea of my solution was:

[hide]
a). We can just find angles and show that FEM and MAD are both iscocelese.  Thus MF = ME and AM =DM, so AM+MF=DM+ME and AF=DE.
b).  We can easily show by AAA that DCE ~ DMA.  So AD/DM = DE/DC.  DC=AB, so AD/DM = DE/AB.  AD * AB = DE * DM.
[/hide]

Yyes this was also my solution. very easy problem.

a)$AM=MD$,$ME=MF$(because $\angle ADM=\angle MEF,\angle MAD=MFE$,hence $ME=MF$),so $AF=DE$.
b)easily get $AMD$ and $CED$ are similar,but $CD=AB$,hence the result.

a) Since $AM=DM$ and $AD\parallel EF$ we have that $AF=DE$

b)Let $K=DE \cap AB$.It is easy to get $AK=AD$  $\angle ADE=\angle CDE=\angle MEC=\angle MFE=180^{\circ}-\angle MKB$. So $MKBF$ is cyclic quadrilateral $\to$ $AM\cdot AF=AK\cdot AB$ $\to$ $AD\cdot AB = DE\cdot DM.$


how is it that inversetan(1/2k^2)=inversetan(2k+1)-inverse(2k-1)?

Simply take the tangent of both sides and use the tangent addition formula.


Can anyone recommand a program for drawing circuit diagrams?

Or how about just physics diagrams in general?

there are a couple out there that are costs $$$. are you interested in them or just free ones>?


modular arithmetic
rotation
combinatorics unsolved
combinatorics

1. The $ n$ holes of a socket are arranged along a circle at equal distances and numbered $ 1,...,n.$ For what $ n$ can the prongs of a plug fitting the socket be numbered such that at least one prong in each plug-in goes into a hole of the same number?

2. Assume a convex $ 2m$-gon $ A_1,...,A_{2m}$. In its interior we choose a point $ P$, which does not lie on any diagonal. Show that $ P$ lies inside an even number of triangles with vertices among $ A_1,...,A_{2m}$.

I'm a little confused by the wording of #1 - how many prongs is a plug supposed to have?

I think the plug has to have $ n$ prongs.

The first can be reformulated as: for which $ n$ there exists a permutation $ \sigma$ s.t. $ \sigma(n)\minus{}n\pmod n$ are different for different $ n$. It is easy to see that a necessary condition is that $ n$ divides $ 1\plus{}\dots\plus{}n$, which means that $ n$ has to be odd. But it is also sufficient, because we can enumerate the prongs in the opposite order, which works.

Where in the problem does it state "$ \sigma(n)\minus{}n\mod n$ are different for different $ n$"?
Also if I'm understanding the problem correctly can't we also have $ \sigma$ in the same order as $ 1,...,n$ (in another word $ \sigma(n)\equal{}n$)? If $ \sigma(n)\equal{}n$ of the plug is fitted to some $ k$ on the socket, then just rotate the plug $ n\minus{}k\mod n$ so that all numbers will fit. :?:

In fact, the phrase should read "$ \sigma(i) \minus{} i\pmod n$ are different for different $ i$". Sorry for confusion.

A basic fact: each prong can be put in the hole with the same number for a unique plug-in of the plug. So necessary and sufficient condition for a numbering $ \sigma$ of the prongs to satisfy the statement is that for no plug-in there are two different prongs $ i,j$ which are plugged into their holes. Obviously the converse of the latter property is $ \sigma(i) \minus{} \sigma(j) \equal{} i \minus{} j\pmod n$, or $ \sigma(i) \minus{} i \equal{} \sigma(j) \minus{} j$.


irrational number
number theory proposed
number theory

[quote="BlueVelvet"][u]Theorem[/u]: there are infinitely many natural numbers $ n\in\mathbb{N}$ such that $ \sqrt {n}$ is irrational.

Now, there are various ways to prove this, including the use of primes and their properties. But what about this one?

[u]Proof[/u]: Assume, to the contrary, that there are only finitely many such numbers, say $ \{n_1,...,n_m\}$ with $ \sqrt {n_i}$ irrational. If $ m: \equal{} \max \{n_1,...,n_m\}$, then $ \sqrt {m}$ is the largest irrational number of the form $ \sqrt {n}$ with natural $ n$. Since the product of a rational number and an irrational number is again irrational, it follows that $ \sqrt {4m} \equal{} 2\sqrt {m}$ is irrational. But $ \sqrt {4m} > \sqrt {m}$, contradicting the assumption that $ \sqrt {m}$ was the largest such number.

I don't see any error in the argument. Is it correct?[/quote]

Sure, but it's a very complicated demo. Why not just say that $ \sqrt{2n^2}$ is irrational for any $ n$ ?

Heh, that has to be the simplest possible proof. Thanks  :D


i figured this out with alot of work, and then mildorf showed me a much simpler approach using something called "dynamic programming"

the long way:
don't think of it as flipping, but having 10 coins, and placing them such that no two heads are besides each other.
you can have at max 5 heads in your sequence.
how many different ways can you arrange the 5 heads and 5 tails? 6C5
(because there are 6 slots where you can place the 5 heads, after you put down the tails)
how many for 4 heads? (7C4)
how many for 3?
and so on til you get to the 0 case
you add those numbers together and you get 144
divide that by 2^10, simplify, and there's your fraction.

so now the dp way...

gimme a sec... lemme work it out again.

ah! i have it!

so we're still looking at the problem as if you have 10 coins and your finding the number of ways to place them so that no two heads are next to each other

alright, so say you have N coins.
if you place the first of the N coins (n) heads up (H), there is only one choice for the next coin to be placed, tails up.
if you place (n) tails up (T),  there are two ways the next coin could be placed.

so placing this nth coin heads up predefines the next term (the n-1 term, since ther are only n-1 coins to place now), because there is only one choice for which side can be up
but placing this nth coin tails up does not predefine the next term, because the (n-1) coin can still be placed either tails up or heads up

what it's like:
H T _
T _

 the _ is the first term that is not predefined

then we allow S(n)=the number of ways to place n coins so that they meets our requirements. [in S(n), pretend the n is the subscript, like the nth term of a sequence]
so S(n)=S(n-2)+S(n-1)
                  H           T

because the H predefines the next (n-1) term, there is only one possibility for (n-1), T.  it is only after we get to the next term after that (n-2) that we get freedom of choice for which side of the coin can be up.
for T, it doesn't predefine anything, so the n-1 term has freedom of choice.

at this point you should be thinking "hey! i've seen sequences like that before!  it's fibonacci!"

and you're right

so now all you need to do is calculate the first two terms and you're set
numbers of ways to have no two heads next to each other if you have 1 coins: 2
if you have two coins: 3

so now you have the first and second terms of youre fibonacci sequence. so fill in the rest

2 3 5 8 13 21 34 55 89 144

oh look! 144 is the 10th term!


geometry
projective geometry
perpendicular bisector
geometry solved

Let H be an arbitrary point on the height CP of the acute triangle ABC. The lines AH and BH intersect BC and AC at M and N, respectively.
[b](a)[/b] Prove that < NPC = < MPC.
[b](b)[/b] Let O be the point of intersection of the lines MN and CP. A line through the point O intersects the sides of quadrilateral CNHM at D and E. Prove that < EPC = < DPC.

The first point can be solved very quickly with projective geometry ($PN\mapsto PM$ is clearly a homography, and it's determined by three particular positions of $H$; since the conclusion is true for $H=P,H=A$, and $H=$the orthocenter of $ABC$, which are easy to check, it must always be true). The second problem is solved in a very similar fashion.

I'm sure there are synthetic solutions as well (there always are :)), but this is what I first thought of.

I've solved using analytic geometry. Denote by [tex](0,0)[/tex], [tex](n,0)[/tex], [tex](m,0)[/tex], [tex](0,1)[/tex] and [tex](0,k)[/tex] the points [tex]P[/tex], [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] and [tex]H[/tex], respectively. Then we have the following:
[tex] BC -> f(x)=-\frac{x}{m} + 1[/tex]
[tex] AH -> f(x)=-\frac{k}{n} x + k[/tex]
[tex] M=BC \cap AH = (\frac{kmn-mn}{km-n}, \frac{km-kn}{km-n})[/tex]
It follows that the angular coefficient of line [tex]PM[/tex] is [tex]\frac{km-kn}{kmn-mn}[/tex]. Similarly, the angular coefficient of line [tex]PN[/tex] is [tex]\frac{kn-km}{kmn-mn}[/tex]. It follows that [tex]NPA=MPB \Rightarrow NPC=MPC[/tex].
Now, notice that [tex]O=MN \cap CP = (0,\frac{2k}{1+k})[/tex]. Assume, WLOG, that the line r([tex]f(x)=ax+\frac{2k}{1+k}[/tex]) intersects [tex]CM[/tex] and [tex]NH[/tex]. Let [tex]E = r \cap CM[/tex] and [tex]D = r \cap NH[/tex]. Then we obtain that the angular coefficient of line [tex]PE[/tex] is [tex] \frac{am+amk+2k}{m-km}[/tex], while the angular coefficient of line [tex]PD[/tex] is [tex]\frac{am+amk+2k}{km-m}[/tex], and the result follows.

For the first one,
Consider a line l that passes through C and it's parallel to AB. Denote by X and Y it's intersection with NP and MP. Then $\frac{CN}{NA}=\frac{CX}{AP}$ and $\frac{BM}{CM}=\frac{BP}{CY}$  and  by Ceva we get That $CX=CY$ and since <PCX=90 we get that <NPC=<MPC.

Maybe somebody is still interested in a synthetic solution of the problem. As there was some confusion with the formulation of the problem, I rewrite it as follows:

[color=blue][b]Problem.[/b] Let P be the foot of the altitude of a triangle ABC from the vertex C, and let H be an arbitrary point on this altitude CP. The lines AH and BH intersect the lines BC and CA at the points M and N, respectively.

[b](a)[/b] Prove that < NPC = - < MPC, where we use directed angles modulo 180°.
[b](b)[/b] Let O be the point of intersection of the lines MN and CP. An arbitrary line through the point O intersects the lines CN and MH at the points D and E, respectively. Prove that < EPC = - < DPC.[/color]

[i]Solution.[/i] Part [b](a)[/b] is the so-called Blanchet theorem and has a number of different proofs. Maybe the easiest one is as follows: Let the lines PM and PN meet the parallel to the line AB through the point C at the points X and Y, respectively. Then, by Thales, $\frac{CX}{BP}=\frac{CM}{BM}$ and $\frac{YC}{PA}=\frac{CN}{AN}$, where we use directed segments. Hence, $CX=BP\cdot\frac{CM}{BM}$ and $YC=PA\cdot\frac{CN}{AN}$, so that

$\frac{CX}{YC}=\frac{BP\cdot\frac{CM}{BM}}{PA\cdot\frac{CN}{AN}}=\frac{BP}{PA}\cdot\frac{AN}{CN}\cdot\frac{CM}{BM}=\frac{BP}{PA}\cdot\frac{AN}{-NC}\cdot\frac{CM}{-MB}=\frac{BP}{PA}\cdot\frac{AN}{NC}\cdot\frac{CM}{MB}$.

But since the lines AM, BN and CP concur (namely, at the point H), the Ceva theorem yields $\frac{BP}{PA}\cdot\frac{AN}{NC}\cdot\frac{CM}{MB}=1$, so that we conclude $\frac{CX}{YC}=1$, and thus CX = YC. Hence, the point C is the midpoint of the segment XY. On the other hand, since the line CP is an altitude of triangle ABC, we have $CP\perp AB$, what, combined with XY || AB, yields $CP\perp XY$. Thus, the line CP passes through the midpoint C of the segment XY and is perpendicular to this segment XY. Hence, the line CP is the perpendicular bisector of the segment XY; in other words, the points X and Y are symmetric to each other with respect to the line CP. Hence, < YPC = - < XPC; in other words, < NPC = - < MPC, and problem [b](a)[/b] is solved.

[b](b)[/b] Since the points M, N, O on the sidelines HA, AC, CH of triangle AHC are collinear, the Menelaos theorem yields

$\frac{HM}{MA}\cdot\frac{AN}{NC}\cdot\frac{CO}{OH}=-1$.

Since the points E, D, O on the sidelines HA, AC, CH of triangle AHC are collinear, the Menelaos theorem yields

$\frac{HE}{EA}\cdot\frac{AD}{DC}\cdot\frac{CO}{OH}=-1$.

Consequently, $\frac{HM}{MA}\cdot\frac{AN}{NC}=\frac{HE}{EA}\cdot\frac{AD}{DC}$.

Since the lines CM, HN and AP concur (namely, at the point B), the Ceva theorem, applied to the triangle AHC, yields

$\frac{HM}{MA}\cdot\frac{AN}{NC}\cdot\frac{CP}{PH}=1$.

Combined with $\frac{HM}{MA}\cdot\frac{AN}{NC}=\frac{HE}{EA}\cdot\frac{AD}{DC}$, this results in

$\frac{HE}{EA}\cdot\frac{AD}{DC}\cdot\frac{CP}{PH}=1$.

This, according to the Ceva theorem, applied to the triangle AHC, yields that the lines CE, HD and AP concur. Let the point of concurrence be called Q. Since the line CH is the altitude of triangle ABC from the vertex C, we have $CH\perp AB$; thus, the line AB is the altitude of triangle AHC from the vertex A. The point P, being the point where this altitude meets the line CH, is simply the foot of this altitude. The point Q, lying on this line AP, can be considered just as some point on this altitude. The lines HQ and CQ intersect the lines CA and AH at the points D and E, respectively. Thus, after the problem [b](a)[/b], we have < EPA = - < DPA. But since $CH\perp AB$, we have < APC = < CPA = 90°, so that < EPC = < EPA + < APC = - < DPA + 90° = - < DPA + < CPA = < CPD = - < DPC, and problem [b](b)[/b] is solved.

  Darij


I know  this is  a silly question  for  you , but i would like to know  how  can i  write  something  (like  (crux) )  at  the  end  of a  line . How  can  i do this ?

Thanks  for your time

The \hfill command pushes everything after it to the end of the line.

But I'm not sure if that's what you're asking about...

Yes  that was  what  i wanted....it works  fine!  Thanks


geometry
parallelogram
number theory proposed
number theory

Let $a,b,c,d$ be integers such that $ad-bc=k>0,(a,b)=(c,d)=1$.Prove that there are exactly k ordered pair of real numbers $(x_{1},x_{2})$ satisfying $0 \leq x_{1},x_{2} <1$ and for which both $ax_{1}+bx_{2}$ and $cx_{1}+dx_{2}$ are integers.

Nice indeed :).

$k$ is the area of the parallelogram $\mathcal P$ having vertices $O=(0,0),U=(a,b),V=(c,d),T=(a+c,b+d)$. $\mathcal P$ has, say, $t$ lattice points in its interior and exactly $4$ (the vertices) on its boundary because of the condition $(a,b)=(c,d)=1$. This means that its area, $k$, is equal to $\frac 42+t-1=t+1$. Now, the number of pairs $x,y\in[0,1)$ s.t. $ax+by,cx+dy$ are integers is equal to the number of lattice points covered by the semi-open parallelogram $[OU)\times[OV)$, and these are the $t$ points inside $\mathcal P$ and the point $O$. In other words, there are $t+1=k$ of them, and this is what we wanted to prove.


suppose that [b]p[/b] is prime and [b]m[/b] is natural number.prove that :
[b](mp)!-m!(p!) m[/b] divisible by [b]p m+3[/b].

It is equivalent to $p^3|\frac{(mp)!}{(p!)^m}-m!=\binom {mp}p\cdot\binom{(m-1)p}p\cdot\ldots\cdot\binom {2p}p\cdot\binom pp-m!$. This follows immediately from $p^3|\binom {kp}p-k,\ \forall k\in\mathbb N$, which has already been discussed on the forum, and is easy enough to prove anyway :).


modular arithmetic
number theory unsolved
number theory

Find all the couples (m,n) such as:

m^((n^5 - n)/60)

is a integer number.

enjoy!

If I got it right, this is really easy.
You have to analyze only the divisibility of $ n$ by $ 4$, and you get:
if $ n\equiv 2\pmod 4$ then $ m$ has to be a perfect square
else $ m$ can be any number

let's see if i've understood...

i consider n^5-n= n(n+1)(n-1)(n^2+1)= a
a is always even

if n=0 mod 4, then a is divisible by 3,4,5, hence is a multiple of 60

if n= 2 mod 4, then a is divisible by 3,2,5, hence m must be a perfect square

if n= 1 mod 4 we repeat the same demonstration considering n+1
if n= 3 mod 4 we repeat the same demonstration considering n-1

right?
tx a lot


invariant
function
analytic geometry
calculus
derivative

The velocity at point $ x$ and at time $ t$ of a fluid in a 1-dimensional flow is $ u(x,t)$. It satisfies the equation
\[ \frac{\partial u}{\partial t} \plus{} u \frac{\partial u}{\partial x} \equal{} f(x,t).\]Show that this equation is invariant under the Galilean transformation
\[ x' \equal{} x \plus{} vt, t' \equal{} t\]where $ v$ is constant, provided $ f(x,t)$ is invariant.

What is the meaning of $ f(x,t)$ being invariant in this problem? Does it mean $ f(x,t) \equiv f(x', t')$?

Since $ t \equal{} t'$ I think we have $ \frac{\partial u}{\partial t} \equiv \frac{\partial u}{\partial t'}$. How do I prove that $ \frac{\partial u}{\partial x} \equiv \frac{\partial u}{\partial x'}$?

No.  $ f(x, t)$ is a function which assigns a certain value to every point in the tx-plane.  If you change coordinates in the plane, $ f(x, t)$ still represents the same function; in particular, it is [b]not[/b] equal to $ f(x', t')$, but to a different function $ g(a, b)$ (in the [b]new[/b] coordinates) with the property that $ g(x', t') \equal{} f(x, t)$.

Thanks, hmm.. But then $ t \equal{} t'$ so don't we have $ f(x,t) \equal{} f(x,t')$? How do I make use of the condition that $ f(x,t)$ is invariant to prove the invariance of the whole equation under the Galilean transformation?

Also, since $ x' \equal{} x \plus{} vt$ would $ \partial x \equal{} \partial x'$ where $ t$ is held constant?

Change coordinates on both sides and show that they're still equal.  Are you familiar with the chain rule for partial derivatives?

Yes I think i know the chain rule, but probably not how to apply it correctly (still confused about all the functions and variables here): Let $ v'$ be the new velocity under the Galilean transformation. Then $ v' \equal{} \frac{dx}{dt'} \equal{} \frac{\partial u}{\partial x} \frac{dx}{dt} \plus{} \frac{\partial u}{\partial t} \equal{} \frac{\partial u}{\partial x} u \plus{} \frac{\partial u}{\partial t}$. Is this correct?

How do you change coordinates when we have no information about what the function $ u(x,t)$ is?

You want to show that the equation, after a change of coordinates, still reads $ \frac{\partial u}{\partial t'} \plus{} u \frac{\partial u}{\partial x'} \equal{} f$.  To do that you need to write out those two partial derivatives in terms of the partial derivatives in the original equation using the chain rule.

Ok, the formulas give $ \frac{\partial u}{\partial t'} \equal{} \frac{\partial u}{\partial t}\frac{\partial t}{\partial t'} \plus{} \frac{\partial u}{\partial x}\frac{\partial x}{\partial t'}$ and $ \frac{\partial u}{\partial x'} \equal{} \frac{\partial u}{\partial t}\frac{\partial t}{\partial x'} \plus{} \frac{\partial u}{\partial x}\frac{\partial x}{\partial x'}$.

But $ \frac{\partial t}{\partial t'} \equal{} 1$, $ \frac{\partial x}{\partial t'} \equal{} v$, $ \frac{\partial t}{\partial x'} \equal{} \frac{1}{v}$, $ \frac{\partial x}{\partial x'} \equal{} 1$.

So $ \frac{\partial u}{\partial t'} \equal{} \frac{\partial u}{\partial t} \plus{} \frac{\partial u}{\partial x}v$ and $ \frac{\partial u}{\partial x'} \equal{} \frac{\partial u}{\partial t}\frac{1}{v} \plus{} \frac{\partial u}{\partial x}$. Substitution doesn't lead to something nice..

[quote="aidan"]$ \frac {\partial x}{\partial t'} \equal{} v$[/quote]
Are you sure?

But $ x' = x + vt$. So, holding $ x$ constant, $  \frac {\partial x'}{\partial t} |_x = v$. Is there a mistake?

$ \frac{\partial x'}{\partial t}$ is not $ \frac{\partial x}{\partial t'}$.

Oops, I mean $  \frac{\partial x}{\partial t'}|_{x'} = v$ because $ x = x' - vt = x' - vt'$.

Read what you just wrote more carefully.

But I don't know what the mistake is: $ x \equal{} x' \minus{} vt'$ so $ x$ is a function of $ x', t'$. Holding $ x'$ constant and differentiating $ x$ wrt $ t'$ should give the result.. could you kindly point out the mistake :)

Edit: Ok, I see missed out a minus sign - $ \frac{\partial x}{\partial t'} \equal{} \minus{}v$. Stupid mistake  :blush:

New approach:

Let $ U(x', t') \equal{} u(x(x',t'),t(x',t'))$.

Then $ \frac{\partial x}{\partial x'} \equal{} 1, \frac{\partial x}{\partial t'} \equal{} \minus{}v, \frac{\partial t}{\partial x'} \equal{} 0, \frac{\partial t}{\partial t'} \equal{} 1$ (hope these are correct)

Making substitutions using the chain rule, I get $ \frac{\partial U}{\partial x'} \equal{} \frac{\partial u}{\partial x} \, , \, \frac{\partial U}{\partial t'} \equal{} \minus{}v \frac{\partial u}{\partial x} \plus{} \frac{\partial u}{\partial t}$.

Substitutions then give $ \frac{\partial u}{\partial t} \plus{} u \frac{\partial u}{\partial x} \equal{} \frac{\partial U}{\partial t'} \plus{} v \frac{\partial u}{\partial x} \plus{} U \frac{\partial U}{\partial x'} \equal{} \frac{\partial U}{\partial t'} \plus{} v \frac{\partial U}{\partial x'} \plus{} U \frac{\partial U}{\partial x'}$. Does this show the invariance?

You haven't put an $ f$ in your equations yet.

Earlier you said $ f(x,t) \equal{} g(x', t')$ for some function $ g$.

So $ \frac{\partial U}{\partial t'}\plus{}v\frac{\partial U}{\partial x'}\plus{}U\frac{\partial U}{\partial x'} \equal{} g(x',t')$? This doesn't seem to show invariance though  :maybe:


advanced fields
advanced fields unsolved

I dont know if this problem's place is here but.. here it goes.
if it has to be asked in a different place please tell me where and sry :blush:
Here it is:
Given a finite alphabet $\Sigma$
$L=\(\{xu^{2}\)y\textbar x,u,y\(\epsilon\Sigma^{* }\) u\(\neq\varepsilon\) \}$
1) If $|\Sigma|=2$ prove that $L$ is regular
2) if $|\Sigma|>2$ prove that $L$ is not regular

10x

This is a problem that belongs to theoretical computer science, so it belongs in the computer science forum.  In the $|\Sigma| = 2$ case, you should be able to construct the generating regular expression directly.

Also, \in looks nicer than \epsilon to denote set membership.


Only one number between 33000 and 34000 with units digit 6 is a perfect square. What is the sum of the digits of this number?

[hide]33856

answer is 25
I used educated guess and check like realizing the number  being squared must have a units digit of 4 or 6 .etc.[/hide]

[hide]184 x 184 = 33856

sum of it's digits = 25[/hide]

i used guess and chek to get [hide=" my answer"]$184^2=33856$ $3+3+8+5+6=25$[/hide]

A way to do it without guess-and-test:

[hide="Answer"]$\sqrt{33000}=\sqrt{3\cdot 11\cdot 10^3}=10\sqrt{330}\approx 180$
$\sqrt{34000}=\sqrt{2\cdot 17\cdot 10^3}=10\sqrt{340} \Rightarrow \sqrt{324}<\sqrt{340}<\sqrt{361}$
$\sqrt{340}$ is about halfway in-between (since it's an exponential function), so $\sqrt{34000}\approx 185$. For the square to end in a $6$, the square root must end in a $4$ or a $6$, so it must be $\boxed{184}$.[/hide]

This might be more complicated, but it's more of an algebraic way to do it.

[hide]

$\sqrt{340}\approx \sqrt{324}=18$ .So you know the first two digit . Now the last digit can be either $4$ or $6$ .  You can just BF out both $184^2$ and $186^2$ or another way to eliminate $186^2$ is to notice that $185^2$ can be compute in such way

$185^2=100(18\times 19) +25=100\times 342 +25$ which impies $186^2$ is out . :)

[/hide]


floor function
geometry
perimeter
inequalities
vector
calculus
integration

Draw a $2004 \times 2004$ array of points. What is the largest integer $n$ for which it is possible to draw a convex $n$-gon whose vertices are chosen from the points in the array?

2003*4

A line segment of length k passes through at most $\lfloor k \rfloor$ lattice points. (*)   Consider the square around the outside of the array.  It has 2003*4 points.

If one has a convex figure with more lattice points, its convex hull is contained within or on the square, and its perimeter is smaller  (you can think of the smaller figure as coming from the square through a series of cuts, each cut by the triangle inequality decreasing the perimeter.)

Then, by (*), this smaller figure with smaller perimeter passes through at most a number of points less than 2003*4.

Thus 2003*4 is maximal.

I think you misread the problem. The question asks for the largest number of vertices a polygon could have, not the maximum number of lattice points on the boundary. This problem is quite difficult. Only one person solved it, I think.

cauchyguy;if you know the solution;please post it

The answer is 561.

First, note that, if we trace the perimeter of the polygon clockwise, we only move to the right on the upper section of the polygon, and only to the left on the lower section.  So the total movement horizontally is at most 2003 * 2 = 4006.  Similarly, the total movement vertically is at most 4006, and the total movement period is at most 8012.

Let n be the number of vertices;  consider the n vectors that move each vertex to the next vertex clockwise.  Clearly no two vectors can be pointed in the same direction.  Let us list all possible vectors (i,j) with integral coordinates in order of "length" |i| + |j|, except that for each collection of vectors V_i,j = {(ci, cj), c in N} with GCD (i, j) = 1, we only choose the vector (i, j).  So we are only choosing the vectors (i, j) with i and j relatively prime.

The first four vectors are (1, 0), (-1, 0), (0, 1), and (0, -1).  The number of vectors of length m is four times the number of such vectors with both coordinates positive;  the latter number is phi(m).  So the are 4 phi(m) vectors of each length m >= 2.

After some calculation, we find that

4 + 4 Sum [m = 2 to 21] m phi (m) = 7988.
4 + 4 Sum [m = 2 to 21] phi (m) = 560.

In other words, the minimal sum of the lengths of the a collection of 560 integral vertices with no scalar multiples is 7988. Note that by symmetry, the sum of these 560 vectors is zero;  therefore we can construct a convex polygon by ordering them in terms of the angle they make with the positive x-axis.  Further, by symmetry, the total movements in the horizontal and vertical directions are the same, and equal 7988/2 = 3884.  So the polygon has height and width 1997, and so fits within a 2004 by 2004 lattice.

As 8012 - 7988 = 24, there cannot be a convex polygon of 562 vertices or more, since that would require a minimum total movement of 7988 + 22 * 2 = 8032.

But we can make a convex polygon with 561 vertices:  for example, remove the vectors (5, 16) and (10, -11), and add the vectors (-1, 21), (-5, -17), and (21, 1).  Both sets of vectors add up to (15, 5), and the sum of the lengths is 12 greater for the second set in both the horizontal and vertical directions.  So, we can form a convex polygon from our new set using the same procedure as before, and the height and width increase to 2003, so it fits perfectly in a 2004 by 2004 lattice.

The hardest part of the problem seems to be the construction for $n=561$, so I've included some (quite sketchy) motivation for that below. I'm not really a big fan of these kinds of detail-oriented/random construction problems, though.

[hide]Let $f(\langle{x,y}\rangle) = |x|+|y|$ and $S_k = \phi(1)+\cdots+\phi(k)$, $T_k = 1\phi(1)+\cdots+k\phi(k)$ for $k\ge1$.

Fix $n\ge4$ and suppose there exists an $n$-gon $P = A_1A_2\ldots A_n$ (ordered clockwise), WLOG (for convenience) with minimal area. Then letting $\vec{v_i} = \overrightarrow{A_iA_{i+1}}$ (indexing mod $n$), we see that (1) $v_i$ must have coprime coordinates for all $i$; (2) for any coprime nonnegative integers $a,b$, the vectors $\langle\pm a,\pm b\rangle$ each appear at most once; (3) when centered at the origin, $v_1,v_2,\ldots,v_n$ run clockwise in that order.

For $j>0$, there exist at most $4\phi(j)$ indices $i$ with $f(v_i) = j$ (this is a ``coincidence'' for $j=1$). If $4S_m \le n < 4S_{m+1}$ ($m\ge1$), then
\begin{align*} 4(2003) \ge 2X+2Y &= \sum_{i=1}^{n}f(v_i) \ge (m+1)(n-4S_m) + \sum_{k=1}^{m}k(4\phi(k)) \\ &= (m+1)(n-4S_m) + 4T_m, \end{align*}where $X,Y$ are the horizontal and vertical ``diameters'' of $P$, respectively. ($X$ is the sum of the $x$-coordinates of all $v_i$ with nonnegative $x$-coordinates.)

The first 22 values of the totient function are \[ 1,1,2,2,4,2,6,4,6,4,10,4,12,6,8,8,16,6,18,8,12,10, \] which allow us to compute $S_7=18$, $S_{11} = 42$, $S_{15}=72$, $S_{21}=140$, $T_7=91$, $T_{11}=236+91=327$, $T_{15}=408+327=735$, $T_{17}=400+735=1135$, $T_{20}=610+1135=1745$, $T_{21}=1997$, $T_{22}=2217>2003$. In particular, $m\le 21$. Furthermore, if $n\ge 4S_{21}$, then $4(2003-1997) \ge 22(n-4S_{21})$ implies $n-4S_{21} \le 1$, so $n\le 4\cdot140 + 1$.

If $n=560$, we have an obvious construction: take the $560 = 4S_{21}$ coprime vectors $w_i$ with $1\le f(w_i) \le 21$, and label them $v_1,\ldots,v_n$ in clockwise order. Then by symmetry, $\sum v_i = 0$ and the corresponding polygon $P$ has $X=Y=T_{21}=1997$.

If $n=561$, we cannot use all of the $v$ with $f(v)\le21$, or else the $v_i$ cannot sum to zero. Thus we may (a) replace two $f(v)=21$ with $f(v)=22$ (so we have three $22$); (b) replace one $f(v)=21$ with a $f(v)=22$ or $f(v)=23$, or (c) replace one $f(v)=20$ with a $f(v)=22$. (b), (c) seem a bit restrictive (as we only get two vectors to play with, although in some sense we do get ``close'' with things like $(16,-6),(-6,17)$, remove $(10,11)$), so we try (a) instead. The key here is to think in terms of the difference $2003-1997=6$: if we remove a vector $\langle{r,s}\rangle$ with $r>0$, this gives us an additional $r$ ``room'' to work with in the positive $x$-direction (so $r+6$ total). Furthermore, in (a) we have to end up with $X=Y=2003$, so we need to use exactly $r+6$ total.

One natural choice is to remove $\langle{a,21-a}\rangle$ and $\langle{-(21-b),-b}\rangle$ with $a,b>0$ and take $\langle{a+6,\pm(16-a)}\rangle$, $\langle{\pm(16-b),-(b+6)}\rangle$. Then it's easy to see we must have $-(16-a)$ and $+(16-b)$, so the final vector must be $\langle{-11,11}\rangle$.

Similarly, if we try to use $\langle{a+6,-(16-a)}\rangle$ and $\langle{-(a-5),27-a}\rangle$, we're left with $\langle{a+b-32,10-a-b}\rangle$. But and these three vectors (which must have both coordinates odd, since $22$ is even) add up to $\langle{a+b-21,21-a-b}\rangle$, we must have $a+b-21,21-a-b$ odd. This in turn forces $a+b-32,10-a-b$ both even.

Thus we'll want our two removed vectors to share a sign in at least one coordinate: remove some $\langle{a,21-a}\rangle$, $\langle{b,-(21-b)}\rangle$ with $a,b>0$. This time, we take $\langle{-(a-5),27-a}\rangle$ (so $a>5$; note that doing something with $-(27-b)$ instead is analogous). Our negative $x$-coordinates are limited to $-6$ (and thus $-5$, by parity), so $a<11$, which leaves us $a=8,10$ (as $a\perp 21$). Also, since our negative $x$-coordinate sum of $-6$ must be ``split'' between two of the three vectors, we must have $\langle{a+b+6,-(16-a-b)}\rangle$, and thus $\langle{-(11-a),-(11+a)}\rangle$.

[b]Comment.[/b] We have $S_m\approx \frac{3}{\pi^2}m^2 + O(m\log{m})$ and $T_m\approx \frac{2}{\pi^2}m^3 + O(m^2\log{m})$ using Mobius inversion, where $\frac{2}{\pi^2}\approx\frac15$. (Note $\frac{21^3}{5} < 2003 < \frac{22^3}{5}$.) So in general, the answer is $\Theta(N^{2/3})$ in an $N\times N$ lattice grid (we have an obvious construction for $n=4S_m$ when $N=T_m+1$).

Also see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=98971]China TST 2005[/url] for a problem using this asymptotic bound.[/hide]

[hide="constructions generated above, up to obvious symmetries (swapping $x,y$ coordinates, negating all $x$ or all $y$ coordinates)"]
---
remove (8,13), (1,-20); add (-3,19), (15,-7), (-3,-19)
remove (8,13), (5,-16); add (-3,19), (19,-3), (-3,-19)
---
remove (10,11), (1,-20); add (-5,17), (17,-5), (-1,-21)
remove (10,11), (5,-16); add (-5,17), (21,-1), (-1,-21)[/hide]


[hide]Each side has one more white tile than a black tile, so a 15-by-15 would have 8 white tiles on each edge. The white tiles form a square, and 8*8=[b](C) 64[/b][/hide]

[quote=zhenghua][hide="Solution"]
===============================

We notice that each side has one more while tile than a black one so:
$$ \left(\frac{15}{2}+\frac{1}{2}\right)^2=(8)^2=\boxed{64} $$
Thus the answer is $\boxed{C) \ 64}$

===============================[/quote]

Stop bumping all these 10+ years old threads


Please suggest some books for Chemistry olympiad.

U.S. Chemistry Olympiad:

Study from Peter Atkins' Chemical Principles!!! <---essential

uh...besides that...
some books used at study camp:
Peter Atkins' Physical Chemistry   :lol:
Skoog, etc. Introduction to Analytical Chemistry
Francis Carey's Organic Chemistry
Miessler and Tarr's Inorganic Chemistry
some biochem book...

I think you mean Chemistry Olympiad in India
For the first level NSEC (objective), you just need the JEE Chem knowledge + some knowledge on biomolecules (more than sufficient)
For InChO, you JEE Chem knowledge + spectroscopy + biomolecules (im writing these because they might not be covered in jee classes before InChO) and a few other topics is required.
See [url=http://olympiads.hbcse.tifr.res.in/subjects/chemistry/stages/stages]this[/url]

If u get selected for the TC ask me for further info

cool , thanks guys . @ Madness , the link's not working.  @ Chemrock , thanks for the best wishes. Which books are good for problems . MTG ? RCM ? OPA ? are they sufficient?

Links to NSEC and INCHO papers will be very helpful or please upload them here if you can.

Strange, it was working for me yesterday when I searched for the HBCSE site from google and went to the olympiads page.... I will try to post that information here later. NSEC papers, well, i think some samples will be available in your school if its a centre for the exam. Otherwise refer to [url=http://joigroup.freezoka.com/viewtopic.php?f=56&t=102&p=351&hilit=InChO#p351]InChO 2007[/url] or [url=http://joigroup.freezoka.com/viewtopic.php?f=56&t=101&p=350&hilit=InChO+2005#p350]InChO 2005[/url]. I dont think NSEC papers are availlable on the net. Instead, you could buy the MTG Chemistry Olympiad book of various papers to get some NSEC and InChO papers (not all will be there too) and some IChO and other countries chem olympiad papers.


It's not really an error but in my PDF, I made my own solution form (for some reason, the form that the USAMTS site has doesn't work with my $\LaTeX$).

Anyway, I put all the information needed (Name, ID #, Year #, Round #, Problem #). But I have little problem. Because I'm putting all my solutions in one PDF, my number 2's solution doesn't start at page 1. It starts in page, say 4.

Is there anyway I can fix the page number? Or should I just put each solution in different PDF?

The one on USAMTS site should work! Make sure you have fancyhdr.sty in the same directory as your solutions file, and make sure you have the usamts include file in the same directory as well. What's the error message? :?

It just doesn't work!

Argh.. I don't have time to try to fix the error in USAMTS. I already tried this long time ago and it didn't work that time neither.

Can anyone actually answer my question? Thanks.

[quote="Silverfalcon"]Never mind.

I'm just going to send different PDF for each solution with title like "USAMTS Solution 2005 Round 1 Problem #."[/quote]Please do [b]not[/b] do this!  Send us only 1 file.

We'd rather have one file with consecutive page numbers.

You can reset page numbers in LaTeX by the command \setcounter{page}{1}


Two linear operators $ T$ and $ U$  on a finite-dimensional vector space $ V$ are called simultaneously diagonalizable if there exists and ordered basid $ \beta$ for $ V$ such that both $ [T]_{\beta}$ and $ [U]_{\beta}$ are diagonal matrices.  Similarly, matrices $ A,B\in M_{n\times n}(F)$ are called simultaneously diagonalizable if there exists an invertible matrix $ Q\in M_{n\times n}$ such that both $ Q^{\minus{}1}AQ$ and $ Q^{\minus{}1}BQ$ are diagonal matrices.

(a) Prove that if both $ T$ and $ U$ are simultaneously diagonalizable linear operators on a finite-dimensional vector space $ V$, then the matrices $ [T]_{\beta}$ and $ [U]_{\beta}$ are simultaneously diagonalizable for any ordered basis $ \beta$
(b) Prove that if $ A$ and $ B$ are simultaneously diagonalizable matrices, then $ L_{A}$ and $ L_{B}$ are simultaneously diagonalizable linear operators.
(c) Let $ T$ be a diagonalizable linear operator on a finite-dimensional vector space, and let $ m$ be any positive integer.  Prove that $ T$ and $ T^{m}$ are simultaneously diagonalizable.

a)  think of the change of basis matrix;  if $ [T]_B \equal{} P[T]_{B'}P^{\minus{}1}$ then $ Q^{\minus{}1}[T]_BQ \equal{} (P^{\minus{}1}Q)^{\minus{}1}[T]_{B'}(P^{\minus{}1}Q)$ and similarly for $ U$...

b)  I'm not sure what $ L_A$ and $ L_B$ denote.

c)  If $ T \equal{} PDP^{\minus{}1}$ where $ D\equal{}\text{diag}(\lambda_1,\ldots,\lambda_n)$, then $ T^m \equal{} PD^mP^{\minus{}1}$, where $ D^m\equal{}\text{diag}(\lambda_1^m,\ldots,\lambda_n^m)$

$ L_{A}$ and $ L_{B}$ denote the left-hand transformation.

I get what you did for the first part but I'm not completely sure how why the third part is right.  Could you clarify this?

When you take a basis B so that [T]_B is a diagonal matrix, you see that [T^m]_B = ([T]_B)^m is also diagonal.

So in this example, instead of $ T$ and $ U$ as our transformations, we're just using $ T$ and $ T^{m}$.  If that is the case, then it makes sense.  Thank you.

In part 3 you are asked to show that T and T^m are simultaneously diagonalizable, so you are not working with U in this case.

BTW for part 2, if you choose a right basis then L_A and L_B can be represented as A and B in matrix. If you know that A,B are simultaneously diagonalizable, that means after using possibly another basis, L_A and L_B can be represented as a conjugate of A and B respectively, which can be chosen as the diagonal one.


Find all pairs (y,z) of integers that satisfy the equation

 (y^2).(z^2)-4.(z^2)+4.(y^2)+8.y.z+3.z=0

 İs anybody can solve this?I have been waiting for weeks...

I'll solve half of the problem, I'll leave the solution of the second half to you.

(y^2-4)z^2 + (8y+3)z + 4y^2=0

since y,z are integers, therefore the discriminant is the square of an integer

(8y+3)^2 - 4(y^2-4)(4y^2) = x^2 (x is an integer)

-16y^4+128y^2+48y+9 = x^2

|y| cannot be bigger than 3, otherwise if |y|>=4, then

x^2 = -16y^4+128y^2+48y+9 <= -256y^2+128y^2+48y+9 < 0 <x^2 a contradiction

therefore the possible values of y are -3,-2,-1,0,1,2,3

substitue these values (some of them maybe wrong) into the original equation to find the possible values of z, I think you can solve the rest of the problem yourself


function
superior algebra
superior algebra unsolved

Let G be a finite group, f is a function on G, and k an integer, denote by \Phi^k(f) the function s\mapsto f(s^k).
Show that R(G) is stable under the operators \Phi^k,k\in Z.
Help me!

[quote="thanhluyen"]Let $ G$ be a finite group, $ f$ is a function on $ G$, and $ k$ an integer, denote by $ \Phi^k(f)$ the function $ s\mapsto f(s^k)$.
Show that $ R(G)$ is stable under the operators $ \Phi^k,k\in Z$.
Help me![/quote]

^--- applied dollar signs.

What do you mean by $ R(G)$?  Do you mean the set of all representations as functions on $ G$?  If so, I think that what you're saying is a trivial consequence of the fact that representations are homomorphisms, but it's quite likely that I'm misinterpreting...

$ R(G)\equal{}Z\chi_1\oplus \cdots\ \oplus Z\chi_h$, with $ \chi_i$ be distinct irreducible characters of $ G$.
I don't think that it is trivial


Tamara knows that the arithmetic mean of her five quiz scores is 95$ \%$. However, she has misplaced one of these quizzes. The ones she can find have scores of 100$ \%$, 100$ \%$, 99$ \%$ and 98$ \%$. What is her score on the misplaced quiz?

The sum of her scores should be $ 95*5\equal{}475$ We now subtract the scores we know from $ 475$ and we get $ \boxed{78}$

[hide="This solution MAY be a little bit faster:"]
Call the test she misplaced $ x$.  We can set up this equation:
\[ \frac{100\plus{}100\plus{}99\plus{}98\plus{}x}{5}\equal{}95\]
We can change this into:
\[ \frac{(95\plus{}95\plus{}95\plus{}95)\plus{}(5\plus{}5\plus{}4\plus{}3\plus{}x)}{5}\equal{}\frac{(4\cdot95)\plus{}(95)}5\equal{}95\]
For that equation to be true, we know that $ 5\plus{}5\plus{}4\plus{}3\plus{}x\equal{}95$.  So we solve, and get $ x\equal{}\boxed{78}$.  But since this is countdown, instead of setting up all of those equations, you would be able to see that $ 100\minus{}95\plus{}100\minus{}95\plus{}99\minus{}95\plus{}98\minus{}95\plus{}x\equal{}5\plus{}5\plus{}4\plus{}3\plus{}x\equal{}95$ fairly quickly.
[/hide]


Two 3-digit numbers each have the same hundreds digit as units digit (but the numbers may have different units digits).  The product of the numbers is 436995.  What are the numbers?

[hide]Since the product is divisible by 5, and given the palindrome requirement, one of the numbers must be 5(something)5. Then I just tested every tens digit by dividing 436995 by 505, 515, etc., until I got numbers 585 and 747.[/hide]

Took a while, but finally solved it. Solved with GeronimoStilton.

Note that $9 \mid 436995$. Thus, we seek to factorize $9 \cdot 48555$. Note that $9 \mid 48555$, so we seek to factorize $3^4 \cdot 5395$. Note that $5 \mid 5395$, so we seek to factorize $3^4 \cdot 5 \cdot 1079$. We can see that $1079=13 \cdot 83$, because $1079=1001+78$. So now we want to rearrange the factors of $3^4 \cdot 5 \cdot 13 \cdot 83$ in a good way. The key observation is that $83$ is a big number, so we should check multiples of $83$ that are between $100$ and $1000$. Since $83 \approx 1000/12$, there are $11$ such multiples.

This is the part that actually took a while.

Now, $166$ doesn’t work, $249$ doesn’t work, $332$ doesn’t work, $415$ doesn’t work, $498$ doesn’t work, $581$ doesn’t work, $664$ doesn’t work, $747$ [b]does[/b] work, $830$ doesn’t work, $913$ doesn’t work, and $996$ doesn’t work. Thus, one of the desired three-digit integers is $747$.

Finally, the other three-digit integer must be $3^2 \cdot 5 \cdot 13 = 585$, which is indeed a palindrome.

Thus, the desired factorization is $585 \cdot 747$.

We seek factors of $436995$ that are palindromes: $3^{2} \cdot 5 \cdot 13 = 585,$ so our other number is $3^{2} \cdot 83 = 747$


trigonometry algebra polynomial function ratio

I don't understand what you are saying. Could you be more specific?

Not really, all I know is that there seems to be a relationship between the roots of certain polynomials and trig expressions.

Any help would be appreciated.

maybe this might (somewhat) help you?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=269942&highlight=#269942

well, it's not that hard actually...

just consider a trig expression which you know can be represented as the sum/product of finite root expressions
let's take $\tan\frac\pi8=\sqrt2-1$ as an example:

put $x=\sqrt2-1$
$\Rightarrow x+1=\sqrt2$
Squaring both sides gives us $x^2+2x-1=0$, which has the solution $\tan\frac\pi8$

That's basically all you have to do: squaring till you get rid of the $\sqrt$ signs...
However, everytime you square both sides, you double the amount of solutions.
What's more: the solutions you add are linked to the original solution.
In this case, you add the solution $x+1=-\sqrt2$, which is $-\cot\frac\pi8$

[quote="Jimmy"]Can somebody explain how certain trigonometric expressions (such as $\sin \frac{\pi}{7}$) can be expressed as the roots of a polynomial? (Or something like that, I'm not really sure how it works.)

Thanks.[/quote]

Jimmy,

Trig functions such as $\sin x , \cos x , \tan x$ etc. may be written as sums of infinite power series. Therefore, if $\sin \frac{\pi}{7}$ were expressable as the root of a polynomial, said sum would equal $0$. For $x = \frac{\pi}{7}$, $\sin x \neq 0 \therefore \sin \frac{\pi}{7}$ cannot be expressed as the roots of a polynomial.

I believe what you're referring to is the fact that sine and cosine functions yield ratios of lengths of the sides of a right triangle; specifically if the length of the hypotenuse is equal to the square root of a perfect square, e.g. $\text{hyp}= \sqrt{4}$, and the length of the opposite side is not equal to the square root of a perfect square, such as, $\text{opp}=\sqrt{3}$ then $\sin x =\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}$.

In this case, $x = \arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.

[quote="Jimmy"][quote="vidyamanohar"][quote="Jimmy"]Can somebody explain how certain trigonometric expressions (such as $\sin \frac{\pi}{7}$) can be expressed as the roots of a polynomial? (Or something like that, I'm not really sure how it works.)

Thanks.[/quote]

$let\,\,x = \,\frac{{2\pi }}{7}$

$ \Rightarrow {\rm  7x =  2}\pi$
$ \Rightarrow {\rm  4x =  2}\pi-3x$

$\Rightarrow $ Sin4x=sin( ${2}\pi$-3x)=-Sin3x

$\Rightarrow $2Sin2x Cos2x= 4 $Sin^3 x$ -3 Sinx

$\Rightarrow $ 4Cosx(2 $Cos^2 x$ -1)= 4(1-$Cos^2 x$) –3, canceling Sinx term both sides is possible as sinx in nonzero for given value of x.

Calling Cosx as t

We have $8t^3+4t^2-4t-1=0$ whose root is Cosx i.e., $\cos\frac{2\pi}{7}$
Writing $t=1-2y^2$, where y represents $\sin \frac{\pi}{7}$ , we can write the equation $8t^3+4t^2-4t-1=0$ as
$64y^6-112y^4+56y^2-7=0$

Hence $\sin \frac{\pi}{7}$  is a root of $64y^6-112y^4+56y^2-7=0$.

I have skipped certain steps here.  Hope that will not be a problem to understand the process.[/quote]

That's exactly what I needed, thanks.[/quote]

A similar problem was answered at
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=278815#278815


There is a sequence of number
$1$
$11$
$21$
$1211$
$111221$
$312211$
.
.
.
.
could you find the next number ?or you can find the  formula of general term ?

Just claim each term of the sequence and you will find the general pattern :
Start from '1', which can be read 'One 1' so that it has to be written as '11'.
Thus tou have 'two ones' which gives '21', and so on...

Then the next term is 'one 3, one 1, two 2, two 1'
which is 13112221.

Pierre.

This sequence was given in an Informatics (Computer Studies) contest in Romania a few years ago! :)

In my opinion this sequence of  number is differ from others.'caz It seems that this sequence of number was introduce itself  and you cannt find the formula of general term .You can only use another language to tell other person what the sequence of number was said but you cannt use math language ,don't you ?

The most important question is the difference between "natural language"and "math language ".Can you only use "math language "in math question ?

That 's true :it is a very  interesting sequence of number .

As far I as I know, this sequence has been much studied, and there are some results about some of its properties that I do not remember now. Things about the the fact that no number gretare than 5 (or something like that) will never appear in it, and so on.

Probably Google will give some links....(sorry to be not more precise).

Pierre.

May be google cannt give me some links,caz I don't know the sequence's name by English.  :maybe:

But someone said this sequence of number can be depicted in function??????
$supernatural$ :D  :D  :D

you may start research from here :
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi

Pierre.

hey, Pierre Bornsztein.
the website which you posted may have been cancelled.

Try this one
[url]http://www.research.att.com/~njas/sequences/index.html[/url]

It is an artificial enquence?or you can find something like this enquence in the $nature$?

[quote]As far I as I know, this sequence has been much studied, and there are some results about some of its properties that I do not remember now. Things about the the fact that no number gretare than 5 (or something like that) will never appear in it, and so on.
[/quote]

it's easy to see no number larger than 3 can appear there: look for 1st such n>3 in the sequence, then it must be the "counting" entry, but not the "counted" (in the natural sense), therefore there must be n consecutive identical entries /leq 3, i.e. 1111, 2222 or 3333, each giving something like "k 2s, two 2s", which cannot occur by construction


Find the product when
4-Methyl 2,7octa-dione reacts with $ \ce{OH}^{\minus{}}$ .......
the temp.,concentration of base,solvent etc. are [b]not[/b] given

[hide="Answer"]I would vote on 1-acetyl-2,4-dimethylcyclopent-1-ene, but other products can also form.[/hide]


[color=blue]Sean $a, b, c$ reales positivos. Probar que:
\[\frac{2a}{a^{2}+bc}+\frac{2b}{b^{2}+ca}+\frac{2c}{c^{2}+ab}\leq \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\][/color]
Carlos Bravo

Lima  -PERU


1) if AB+BA=0, A^2 * B^3=B^3*A^2.

2) iF A^3 =0, A -I is singular.
hint : A^3 -I^3 =(A-I) (...)

sorry i really have no idea how to start these two qustions. :(

Thank you.

1: $ AABBB\equal{}\minus{}ABABB\equal{}ABBAB\equal{}\minus{}ABBBA\equal{}BABBA\equal{}\minus{}BBABA\equal{}BBBAA$
2: Your statement is false. Did you mean nonsingular?

yes, nonsingular.
sorry for the typo.

And thank you .

The second question's hint is a good one.  Note that $ A^3 \minus{} I^3 \equal{} \minus{}I$ and that a matrix $ A$ is nonsingular iff there exists another matrix $ B$ such that $ AB \equal{} I$.

[quote="HAPPYSTUDENT"]A^3 -I^3 = -I how did you get this one?[/quote]
have you tried to use the hint?

$ A^3 \equal{} 0$ and $ I^3 \equal{} I$ by definition.


AMC
AIME
percent
USA(J)MO
USAMO
MATHCOUNTS
AMC 12

As another poster noted, the grand statistics from the 2004 AMC exams are now on the web.  This includes both state and national levels, comparisons by gender, grade, etc.

So without further adieu,
[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2004-12a/04AMC1012Aanswers-pict.html]here it is[/url].

I feel retarded, such a high percentage(19%) made the AIME on 12 B.

So weird I get a better score on A than B(off by 3 pts to qualify on A), the cutoff percent for 12A is like 8%

I am going to redeem myself next year.

I feel more retarded than you Beta because I was .5 from getting into the AIME  due to some stupid eye problem (I missread the problem).

Another stat for USAMO hopefuls:

The students tied for the 160th best score have a score of 133.5 on the AMC12.

Do any veteran AIME takers know how soon after the test the cutoff is reported?
Also, were solution packets sent out to the teachers along with the tests? Or do we need to wait until the answers are posted on the website?
Thanks.

I'm fairly certain that (at least in the past) they did not send solutions with the AIME.

one question... when they say

"National Awards are given to a minimum of 10 high scoring students and 5 schools (based on the team score) in each of these
regions."

does it mean at least 10 awards for each of 10A,10B,12A,12B or just the 10 highest regardless of test?

JBL is right. The package with the exams does not include an answer key. I think it was posted on the website pretty quickly last year (like the next day).

Last year the AIME cut-off was up in the air until only a week or so before the USAMO.

ThAzN1 - where are you getting that?

i got it from the amc10/12 teacher's manual...

[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04tm12/1012pdf/12Atm04-text.pdf[/url]

its near the bottom...

Looking at the summary book from last year, only AMC 12 scores count for those awards. They don't distinguish btwn A and B dates. It looks like a school or individual that did both dates would just be recognized once, for the higher score.

That's different from the honor/merit rolls and list of perfect scorers. Those do have separate lists for 10A, 10B, 12A, 12B.

[quote="ThAzN1"]hmmm... wierd cuz there was a 10b winner from missouri last year...

[url]http://www.unl.edu/amc/e-exams/e1-archives/2002-2003/0203StateWinnersPic.html[/url][/quote]

I think that was an award issued by the state AMC organization.

Just curious, about what is the school winner score in your schools?

I read from this year's 10B stats that the overall school winning score was 108.66.  I was one point away from school winner though...our school winner score for 10B was 117 (very low), but I had lower...116...lol.

the highest AMC 10B score at my school suprised everybody. some seventh grader who wasn't a leading mathcounts member or did extremely well on the AMC 8 got a 133.5 I believe.

[quote="ThAzN1"]hmmm... wierd cuz there was a 10b winner from missouri last year...

[url]http://www.unl.edu/amc/e-exams/e1-archives/2002-2003/0203StateWinnersPic.html[/url][/quote]

Different award.

The school and student awards by region, which you originally asked about, are awarded by AMC national office and are based only on AMC12 (with A and B lumped together).  There is also recognition for top individuals by state, with separate lists for the 4 exams.

Ow ow, the finally got my studly pic (taken at USABO ironically enough) up on their site.

Now for the AIME. Eep.

[quote="rlee"]Tiankai Liu got perfect AMC12 scores on BOTH A AND B?

*faints*[/quote]

He did that last year too.

Cool, Alison is up on the perfect scorers page too.

Tiankai Liu is the shiznit, but I mean, why bother taking the B if you already have a perfect on the A. I would have taken the B had my school offered it, and had I been in the mood to take yet ANOTHER test. But yeah props to the dude for perfecting both, though I don't really see the point other than to flaunt your skillz. He's a math-skank. jkjk

I wonder how good that kid was when he was 15...

Hey, I'm the only person from New Hampshire on that page not from Phillips Exeter. Makes me feel kinda strange, since usually all NH perfect scorers are from Exeter :mrgreen:

Well my friend got a perfect on 10B, Nimish Ramanlal.  He's a sophomore.  I may have been going to his high school, Seminole, too if I didn't go to my zoned school, Lake Howell.

[quote="Idiot_without_a_village"]Ow ow, the finally got my studly pic (taken at USABO ironically enough) up on their site.[/quote]

The picture next to you was taken at the computing olympiad. I guess it's good to get all the bang possible out of having to dress up in a suit and tie.


calculus
integration
function
geometry
trapezoid
real analysis
calculus computations

[size=150]find[/size]
$ I \equal{} \int_{ \minus{} 1}^{1} \exp\left({\frac { \minus{} 1}{1 \minus{} x^2}}\right) dx$

[color=green]Edited to look better- Moderator[/color]

hello, this integral does not converge on the interval $ [\minus{}1,1]$.
Sonnhard.

No, you're wrong. The integrand is a smooth function bounded above by $ 1$. It certainly does converge.

This is also a very old thread, but it's interesting enough.

I see no reason to expect this to have a closed-form answer. Maybe it does, but I'm not even going to try.

You'd think with it being $ C^{\infty}$ that numerical methods would work especially well, but some of my usual stuff doesn't work as well as it might. I think the best method is just right-endpoint Riemann sums on $ [\minus{}1,1].$ My best estimate is that $ I\approxeq .44399382.$

A standard prescription for creating a smooth approximation of the identity on $ \mathbb{R}$ with some nice properties is this:

Let $ \varphi(x)\equal{}\begin{cases}\frac1I\exp\left(\frac{\minus{}1}{1\minus{}x^2}\right),&\minus{}1<x<1\\0,&\text{otherwise}\end{cases},$ and let $ \varphi_{\varepsilon}(x)\equal{}\frac1{\varepsilon}\varphi\left(\frac{x}{\varepsilon}\right).$

Then $ \varphi_{\varepsilon}$ is $ C^{\infty},$ supported on $ [\minus{}\varepsilon,\varepsilon],$ and has integral equal to $ 1.$

It's not just $ C^{\infty}$- it extends to be $ C^{\infty}$ and periodic, with $ [\minus{}1,1]$ a full period. That's why the simple Riemann sums are unusually good.

look we have
$ I \equal{}\int_{\minus{}1}^{1}\exp\left({\frac{\minus{}1}{1\minus{}x^{2}}}\right) dx$
or
 $ I\equal{}2\int_{0}^{1}\exp\left({\frac{\minus{}1}{1\minus{}x^{2}}}\right) dx$

in other words its value at 0 is 1/e and its value at 1 is 0
Using the basic concept we can assume it to be triangle and the answer is 1/e

That's a numerical method (the trapezoid rule), and a very crude one, with a rather large error. I could use that as the very first step of a many-step process (Romberg integration), but I would like to improve beyond that.

Answer (involves Bessel functions)...

[hide]$ {{\rm e}^{\minus{}1/2}} \left( {{\rm K}_1(1/2)}\minus{}
{{\rm K}_0(1/2)} \right)$[/hide]

hello, yes i was wrong now i have got $ \int_{\minus{}1}^1e^{\minus{}\frac{1}{1\minus{}x^2}}\,dx\approx 0.443993816168079437823048921170$
Sonnhard.


function
algebra
domain
functional equation

Can someone explain to me why you can "solve" functions?

Where does it all come from? (i.e how can I just substitute y = 0 into a functional identity and expect that the result will be an answer; after all, I could do anything to the functional equation)

Can anybody justify to me why you can manipulate functional identities that way?

I am very confused.

Thanks

when you are solving a functional identity problem, your domain is all single-variable functions, so you can expect your answer to be f(x) = something with x.

you can manipulate your function any way you like, as long as the result always holds true for the original identity. try a simple one:

f(x+y) + f(x-y) = 2x^2 + 2y^2

substiute y=0. we can do this since it doesnt change the problem. the functional equationsm must be true for all x and y, including y=0. we get:

2f(x) = 2x^2 ,
f(x) = x^2.

check it:

(x+y)^2 + (x-y)^2 = 2x^2 + 2y^2,
so it works.

in summary, when you are solving a functional equations, you can make any substitutions which fit the functions' domain.

How do we know that f(x) = x^2 is the only solution? After all, there are an endless number of manipulations we could do to the problem.

For the following example,

(paraphrased from AoPS II)

yf(x) = xf(y)

f(x)/x = f(y)/y

Define g(t) = f(t)/t

g(x) = g(y)

Clearly, g(x) is a constant, say c.

f(x)/x = c

f(x) = cs is the family of solutions.

How do we know that g(x) is "clearly" a constant?

This all seems so awkward to me-- it feels like I am pulling things out of nowhere.

Okay, I've always had trouble with functional equations as well.  The key principle is this:  if you are given a functional equation, it must hold true [i]for every possible choice of value for the variables involved[/i].  So, if you are given the functional equation f(x + y) = x + y, you know that that has to be true [i]for every value of x and y[/i].  In particular, that means that it must be true when y = 0, so that means right away the only possible function is f(x) = x, and we can check to see that it does in fact work.  Now, for the problem you gave, we have
y*f(x) = x*f(y).  This must work no matter what x and y are.  In particular it must work when y = 0 and x = 1, for example, which gives us 0*f(1) = 1*f(0),  f(0) = 0.  Okay, now we know something about the function -- it has to be 0 at 0, no matter what.  That's not great information, but it is something.
Now, lets see what else we can find out.  If we take two values for x and y that aren't 0, we have
y*f(x) = x*f(y), which means that
f(x) / x = f(y) / y.  You might want to do this for several reasons.  One good one is that it brings all the x's to one side and all the y's to another, which makes substitutions easier (since there aren't any x's and y's mixing together).  It also just kind of looks nice.
Now, as your solution suggests, let g(t) = f(t)/t when t is not 0.  Then we have g(x) = g(y), [i]for any choices of x and y[/i] (as long as neither is 0).  This is the definition of being constant -- it's always the same.  So, f(x) / x = c for some c.  f(x) = c*x for some c, for every x not 0, and f(0) = 0.  We can restate this more simply as f(x) = c*x.  So, all solutions must satisfy that property.  Then, you go back to check that any function with that property works.  And they do.  So you're done.

Was that more helpful?


limit
integration
trigonometry
calculus
calculus computations

Calculate:
$ \displaystyle\lim_{\lambda\to\ \plus{}\infty}\sqrt {\lambda}\displaystyle\int^1_0 e^{i\lambda x^2}\,dx$

$ e^{i\lambda x^2} \equal{} \cos (\lambda x^2) \plus{} i\cdot\sin (\lambda x^2)$

$ \int_{0}^{1}\cos (\lambda x^2)\;dx \equal{} \frac {1}{\sqrt {\lambda}}\int_{0}^{\sqrt {\lambda}}\cos (x^2)\;dx$

$ \int_{0}^{1}\sin (\lambda x^2)\;dx \equal{} \frac {1}{\sqrt {\lambda}}\int_{0}^{\sqrt {\lambda}}\sin (x^2)\;dx$

and we get:

$ \lim_{\lambda\to\ \plus{} \infty}\sqrt {\lambda}\displaystyle\int^1_0 e^{i\lambda x^2}\,dx \equal{} \lim_{\lambda\to\ \plus{} \infty}\sqrt {\lambda}\cdot\left(\frac {1}{\sqrt {\lambda}}\int_{0}^{\sqrt {\lambda}}\cos (x^2)\;dx \plus{} i\cdot\frac {1}{\sqrt {\lambda}}\int_{0}^{\sqrt {\lambda}}\sin (x^2)\;dx\right) \equal{}$

$ \equal{} \int_{0}^{\infty}\cos (x^2)\;dx \plus{} i\cdot \int_{0}^{\infty}\sin (x^2)\;dx \equal{} \frac {\sqrt {2\pi}}{4} \plus{} i\cdot \frac {\sqrt {2\pi}}{4}$

[url=http://en.wikipedia.org/wiki/Fresnel_integral]Fresnel integral[/url]


Post your guess on how much Spiderman 2 will make this saturday.

farthest guess to actual gets gmail account. Post your answer to closest hundredth of a million.

NO EDITS, if you edit ur post, the post is not counted (if you want you can post a new prediction before deadline and just write on your other post INVALID). if you don't do this invalid thing, i will take ur first post.

deadline is noon saturday

My Guess is..

$84.50 Million US Dollars

85.397 million (pi * e * 10 million :) )

Are we talking about for just Saturday or the total from opening day to saturday?

You're talking just Saturday, right? I guess $51.21 million.

Oh...just saturday...my post above invalid then

$43.25 US Million dollars

My guess would be

$69.5 million

By the way, I saw it and it was awesome

According to [url]http://www.boxofficeprophets.com/column/index.cfm?columnID=8418[/url], the Satuday gross for Spider-Man 2 was $33.8 million.

Unlike (ahem) certain other participants, JSRosen3 was timely in posting his nearest-to-actual guess, so I congratulate him. I also rue not making my guess a good bit lower.  :)

[quote="JSRosen3"][quote="joml88"]And confuted wins because everyone else overbid.  O, wait we're not on The Price is Right.[/quote]

Yeah, hopefully ml2 didn't mean closest without going over...[/quote]
If he did, I'll pass the invitation along to whoever was actually closest, since I already have a gmail account and I was joking.

[quote="ml2"]Post your guess on how much Spiderman 2 will make this saturday.

[b]farthest[/b] guess to actual gets gmail account. Post your answer to closest hundredth of a million.

NO EDITS, if you edit ur post, the post is not counted (if you want you can post a new prediction before deadline and just write on your other post INVALID). if you don't do this invalid thing, i will take ur first post.

deadline is noon saturday[/quote]

Did anyone read my first post? Farthest is Demon, and thus he wins.

????????

You edited your original post after the contest was over...originally it said "closest guess to actual".

jeez, I was just kidding!  anyways, JSRosen, PM me your email

argh...lost by 2 million...and i wanted one...=(

Yeah like how many 14 year bumps do I have?


[quote="mhuarancca"]por siacaso alguien sabe los resultados de la APMO en peru , por q creoq es requisito para el selectivo de la imo ???? :blush:[/quote]

los resultados ya están, preguntale a algún peruano de MathLinks que capaz que sabe


geometry
perimeter
AMC
AIME
number theory
relatively prime
AIME I

The two squares shown share the same center $O$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$
[asy]
real alpha = 25;
pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin;
pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z;
draw(W--X--Y--Z--cycle^^w--x--y--z--cycle);
pair A=intersectionpoint(Y--Z, y--z),
C=intersectionpoint(Y--X, y--x),
E=intersectionpoint(W--X, w--x),
G=intersectionpoint(W--Z, w--z),
B=intersectionpoint(Y--Z, y--x),
D=intersectionpoint(Y--X, w--x),
F=intersectionpoint(W--X, w--z),
H=intersectionpoint(W--Z, y--z);
dot(O);
label("$O$", O, SE);
label("$A$", A, dir(O--A));
label("$B$", B, dir(O--B));
label("$C$", C, dir(O--C));
label("$D$", D, dir(O--D));
label("$E$", E, dir(O--E));
label("$F$", F, dir(O--F));
label("$G$", G, dir(O--G));
label("$H$", H, dir(O--H));[/asy]

[quote="PenguinIntegral"]Kalva says that the longer side of the triangle is equal to AB ("It is clear from the diagram that the perimeter of the triangle is the side of the square 1"), however this is obviously untrue. I am confused...[/quote]

[hide="No, thats right..."]
Draw $BF$ and $AE$. The figure is symmetric about both, so the perimeter of the triangle is equal to the side of the triangle. Call the sides of the triangle $x$ and $y$; we then have:
$x^{2}+y^{2}= \left(\frac{43}{99}\right)^{2}$ and $x+y+\frac{43}{99}=1$.
Solve this system; then the area of each triangle is equal to $\frac{x \cdot y}{2}= \frac{13}{396}$. Thus the area of the octagon is $1-4 \left( \frac{13}{396}\right) = \frac{86}{99}\Longrightarrow 86+99 = \boxed{185}$[/hide]

[quote="sapphyre571"]Let XAB be the right triangle with longest side AB (X is the corner of one of the squares). If A and B lie on segment JK such that J, A, B, K lie on a line in that order and J and K are corners of the square that does not have X as a corner, then JA=AX and XB=BK so the perimeter of XAB is JK, which is true because the triangles XAB and JAH are similar by angle-angle-angle, and the circles are concentric, so they are congruent. same principle applies for triangles XAB and KBC.[/quote]
But if you have JA=AB then it follows AB=AX and it's not a right triangle...

probably a typo - JA=AX, XB=BK -> AX+AB+XB=JA+AB+BK=1

well the answer is correct (according to kalva anyway)

[quote="4everwise"]The two squares shown share the same center $O$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[img]6599[/img][/quote]

This question was on an AIME I remember.

solution:

[hide]
We know that the area of each square is $1$, So we can find the area of the octagon by subtracting the area of the trianble by $4$ times the area of the triangles that are on the corners of the square.

First, since AB= $43/99$ and the squares have sides of length 1, we know that the sum of the two segements of the side of the square A and B are on (the one to the left and the one to the right of AB) is $1-43/99=56/99$.

Lets call these two segments $a$ and $b$.

Now, we also know that the length of the sides of the legs of the right triangle that contain AB (extended outside of the side of the upright square) is also $a$ and $b$ (whose sum is $56/99$).

Now we can set up two equations:

$a+b=\frac{56}{99}$ (sum of sides equal $1-43/99=56/99$.)

and $a^{2}+b^{2}=\left(\frac{43}{99}\right)^{2}$ (right triangle with AB as hypoteneus).

Next, we know that the area of each triangle on the four corners of the square is $\left(\frac{1}{2}\right)ab$ and four times that is $2ab$. So we're trying to find $1-2ab$.

From the two equations above, we square the first one and subtract the second one to obtain $2ab=\frac{13}{99}$
so the area of the octagon is
$1-\frac{13}{99}\=\frac{m}{n}\=\frac{86}{99}$ so $m+n=86+99=\boxed{185}$

[/hide]

[quote="PenguinIntegral"]JAH and AXB are congruent, right? But XA=AJ, right? Hence AB=AX and it isn't right...[/quote]

Oh I didnt put the congruence statement in order (it should be JAH=XAB)
JA=AX
JH=XB
AB=AH

I did not say JA=AB, because it's obviously not true.

[quote="Debdut"]How can you say man all the right triangles are conguent?[/quote]

Because  the 2 squares share the same center, all of their vertices are points on the same circle.
Suppose we have right triangles AXB and AJH on the diagram.
Then all their angles are equal by vertical angles.
We also know that OX=OJ because the radii of a circle are equal (the squares are concentric).
Therefore, AXB and AJH are congruent by ASA.
The same logic applies to any other pair of the outer right triangles.

@gqiao What is the side you used in the ASA?

I'm not exactly sure about gqiao's method, but I know that you can draw line $HB$.

This gives $\triangle XBH$ congruent to $\triangle JHB$.

From there we conclude that $HJ$ congruent to $XB$.

Hello, I found the problem a simple algebraic substitution, but I did make an assumption to begin with.

I assumed that all the side triangles are congruent without a rigorous proof.

All I have is that the angles are same by vertical angles, but that only shows that they are similar. How do we further prove that they are indeed congruent? Which side shall I introduce?

Please help, so that I can be sure of congruences in the future.

Since the figure has a 90 degree rotational symmetry, it’s sufficient to prove that two adjacent triangles arte congruent.
$\triangle BOK\cong \triangle  BXO \Rightarrow BK=BX$
By $90º$ counterclockwise rotation  $BK=HJ \Rightarrow BX=HJ$.
Remembering that they are rectangular and using vertical angles, follows
$\triangle BXA \cong \triangle AJH$.

[quote=skipiano]Maybe connect all the points with a circle with centre O?[/quote]

@skipiano, how does that help at all?

Let $X_1X_2X_3X_4$ and $Y_1Y_2Y_3Y_3$ be squares as shown in the figure.

Let $\angle Y_1OX_1 = \alpha$. If we rotate $Y_1Y_2Y_3Y_4$ at $O$ with $\alpha$, we will get a square on the circumcircle of the square. So $X_1X_2X_3X_4$ is $\alpha - $ rotated of $Y_1Y_2Y_3Y_4$.
Since $\angle OY_1H = \angle HX_1O = 45^\circ$ and $OY_1 = OX_1$, then $X_1H = Y_1H$.
Similarly $Y_1G = X_4G$ and so on.
This makes all small right triangles congruent to $\triangle X_1AH$.
So $AB=BC=CD=DE=EF=FG=GH$.
Draw perpendicular to $AB$ from $O$. Let this line meets $AB$ at $X$ and $EF$ at $Y$.
$XY=1$ and $OX=OY=\dfrac 12$
Area of octagon is $8 \cdot [OAB] = 8 \cdot \dfrac 12 \cdot AB \cdot OX = 2\cdot AB = \dfrac {86}{99}$.


I have found using google big pdf document with ~200 olympiad inequalities. Now, I can't find it. Does anyone have it? Can someone give me a link or send it by email?

It's not book. It's pdf document with many inqualities from different olympiads, without solutions.

may be this [url]http://www.imo.org.yu/index.php?options=gl|Journ&p=00c32mm4[/url]

I don't know who is author.

It starts with some well-known theorems (cauchy, ....) and then there is
a long list of inequalities from different olympiads (IMO, IMO Shortlist, Other IO s, National Olympiads,  ...)
most of them are for 3 variables
a,b,c > 0 , g(a,b,c) = d
prove that f(a,b,c) ? e

I have found it by typing "Olympiad inequalities" on google, but it's disappeared.

It is free on Internet, and I can upload it :D

That one doesn't quite fit his description. I wonder if he meant http://web.mit.edu/~tmildorf/www/Inequalities.pdf

No, that's not what I need.

Once more how it looks:

Theorems
1. AM-GM
2.. ..
3...
...
10....

Problems

1. (IMO) a,b,c > 0  prove that ....
2 (IMO Shortlist) . a,b,c > 0 prove that ...
3. (...)  x,y,z > 0 prove that ....
......
187. ...

I have downloaded it several times but now I can't find it.
I have it elsewhere on disk, but I thought every user from "MathLinks" had it and  would help.

Yes, it's that. :)

Thank you very much.

thanx, Can you give me a link again?  :lol:

[quote="kdano"]Or maybe you're looking for i2006.pdf at http://www.mathlinks.ro/Forum/viewtopic.php?t=140815[/quote]
thanks for the link.. perfect.. :)