geometry
3D geometry
modular arithmetic
trigonometry
AMC
USA(J)MO
USAMO

I'm interested in how AoPSers fare against the almighty awesome demon of the most frightful and fearful thingies called careless errors. So, please participate in the poll.

I suppose I didn't lose any points due to "careless" errors; it was mostly me just being stupid.

Yea.. I knew my answer was wrong, somehow I ended up bubbling it anyway...
must be horrible for whoever lost a LOT of points... if you lost 32 points and scored 118.5, thats still pretty good  :P

No counting problems => no stupid mistakes.

I had a 136.5, made 4 stupid mistakes, got a 112.5.

By the way, mathtyro, you can't get a 118.5 after losing 32 points lol.

Sometimes careless mistake can help people gain points.(think about it; it is possible) :P

you cant lose 32 points period.

I meant 31.5 close enough ok?:p

I shold have had a 132. maybe a 138 if I read the cube-plane problem but then again I did get lucky in bothering to read the long and boringly worded laura-bike problem. or else I'd have gotten a 124.5

Still, losing 1.5 points isnt as bad as it could be I guess..

(when you are saying it in english: "I could have scored one point five points more" it sounds weird and toungetwisting)

Careless errors... Most fearful one is bubbling them one space up or down and losing a lot of points

Evaluate your work, check calculations, assess whether you really are convinced that you have the right answer and that you might not have made a mistake, don't be satisfied with just thinking you probably have the right answer if your gut says otherwise. For example, I almost missed 6 and 18 this year, but I had a feeling that I might have messed up my exponents on 6 and decided to check my answer, and since it was easy to check anyways; for 18, I almost put down (A)6 but decided that that seemed too easy, and sought to see where I could get another factor of 2 out of the expression. I ended up missing not a single point this year, nor two years ago. Last year I made a bunch (get good sleep, don't panic during the test, go on if you're not making progress, but don't give up early and jump back and forth without making any progress at all. if you're missing low numbered problems that should be easy, go back to them first before moving on to the hard ones, think about how to work smartly and use your time wisely).

Check check counting problems, they suck. Look at the answer choices, if you have really close answer choices, ask yourself why they're there and whether you couldn't have easily made a 1 off error, or a sign error, or a factor of whatever error. There were some people who had the wrong sign on 14, I had paused for a little while to make sure I had the right sign. Remember, it is better to invest a little extra time and get the right answer, then to squeeze more points out by solving another problem and missing questions you could have gotten right. Organize your scratch work.

[quote="leoxnlin"]Let x be the number of points careless errors cost you on the AMC 10/12a. Which of the following is true?[/quote]

[hide="Solution."]
=====
$ 5^k\plus{}1 \equiv 2 \not \equiv 0 \pmod{4} \implies 6\ \text{points}$
$ 1\ \text{misbubble} \implies 6\ \text{points}$
$ \implies x\equal{}6\plus{}6\equal{}6\times 2\equal{}16 \implies \boxed{14<x<20}$
=====
[/hide]

[quote="jvenezuela716"][quote="leoxnlin"]Let x be the number of points careless errors cost you on the AMC 10/12a. Which of the following is true?[/quote]

[hide="Solution."]

$ 6 \times 2 \equal{} 16 ??$ add 6 points for that :P
=====
$ 5^k \plus{} 1 \equiv 2 \not \equiv 0 \pmod{4} \implies 6\ \text{points}$
$ 1\ \text{misbubble} \implies 6\ \text{points}$
$ \implies x \equal{} 6 \plus{} 6 \equal{} 6\times 2 \equal{} 16 \implies \boxed{14 < x < 20}$
=====
[/hide][/quote]

for me
3 * 3 does not = 27 - 4.5 points
14/2 = 7 not 14 - 6 points
got #5 wrong - 6 points
so 16.5 points off which would have brought me to a 122.5- still bad

6*6 = 12 -> 6 points (triangle being divided into 2 parts of equal area)
2^6|64 therefore 2^6 is the largest power that divides 10...064 for any number of 0's
sahcohtoa?  somehow sine=adj/hyp -> 4.5 pts + at least 5 minutes that I could have solved another problem in (another 4.5)

so in all, I supposed this would be 21.  Ok, that's not the fairest guess, the 2^6 wasn't entirely careless it just wasn't thinking the problem all the way through.

Still got 115.5.  Anyone have an idea what USAMO index it will take to qualify?

I missed two questions due to careless errors and left one blank, so either 12 or 16.5 (depending on whether I would have gotten the last one; I'm assuming I wouldn't have gotten it wrong without another careless error.)

I lost about 12 pts.  :(
First mistake:
$ 3\plus{}4\equal{}12$
Second mistake:
$ 5^2\plus{}1\equiv1 \: (mod 2)$

I don't know why, but once I make a mistake like the first one, I do the same thing
while I check over my answers... So, I think the only solution is do it right the first time..
But how...

[quote="jvenezuela716"][quote="leoxnlin"]Let x be the number of points careless errors cost you on the AMC 10/12a. Which of the following is true?[/quote]

[hide="Solution."]
=====
$ 5^k \plus{} 1 \equiv 2 \not \equiv 0 \pmod{4} \implies 6\ \text{points}$
$ 1\ \text{misbubble} \implies 6\ \text{points}$
$ \implies x \equal{} 6 \plus{} 6 \equal{} 6\times 2 \equal{} 16 \implies \boxed{14 < x < 20}$
=====
[/hide][/quote]

uh... 6+6=6*2=12, not 16...

I remember in my class a kid made a mistake, then made another mistake that would have corrected the first, but then he corrected the second mistake and got it wrong.

blah about 14 on the 12, then I bubbled 24. I shouldn't have. :(

But I got 25 right even though I made the worst careless error ever. I chugged some values out then I thought that -2-sqrt(3) and 0 alternated ad infinitum.

The trick is to do lots of official countdowns so you can make those careless errors GO AWAY.

I lost 21 points from 4 careless errors on the AMC 10A...stupid me :mad:

[quote="lotsofmath"]

uh... 6+6=6*2=12, not 16...[/quote]

Yeah, that was a joke.  :roll:

I should have gotten a 121.5. I ended up getting a 99 give or take 6.

0 careless errors, got 111 on the AMC 12. Hopefully I can get 0 careless errors on the B as well, except with better performance.

However, last year on the AIME, careless errors cost me about 6 points, bringing me from a 10 to a 4.

[hide="Last year"]
4 questions wrong $ \implies 24$ points $ \implies 20<x<26$
[/hide]
[hide="This year"]
$ \minus{}1$ questions wrong $ \implies \minus{}6$ points $ \implies \text{none of the answer choices} \implies \text{leave blank}$
[/hide]

yeah I think he got it right? I know of quite a few people who got questions right by accident (cough #24 cough)

um, cough, 12/18 points i missed were careless, and the last, i probly could've gottten, but i literally spent like 30 seconds doing it.

[quote="lordviking"]so you got a question RIGHT from a careless error? Or does the -1 mean something else?[/quote]

Hint: Two wrongs don't make a right, but two sign errors make a right :lol:

But I realized that and fixed both of them, and it turned out I was right in the first place.

[quote="Yongyi781"]Hint: Two wrongs don't make a right, but two sign errors make a right :lol:[/quote]Same thing with two off-by-one errors, as long as they're in opposite directions. :D


[b]The three points (1,2,4),(2,5,6), and (x,y,10) are on the same line. What are x and y? ;) [/b]

[hide]Its easy by use of determinants[/hide]

Hmm looking at the answers, is there some reason that this might work?

[hide]
$\text{Given two points: } (x_1,y_1,z_1) \text{ and } (x_2,y_2,z_2)$

$\text{A point on on the line determined by the aforementioned points can be described as: }$

$(x_1 + a*(x_1-x_2), y_1 + a*(y_1 - y_2), z_1 + a*(z_1 - z_2))$

$\text{Where a is a constant}$[/hide]

This is what happens if we use that method in the given problem:

[hide]$\text{Let's call our point } (x_3,y_3,z_3)$

$\text{Using knowledge from above, we say } z_3 = z_1 + a* (z_1 - z_2) \text{ and substituting } z_1\text{, } z_2\text{, and } z_3 \text{ we get: }$

$a= -3$

$\text{Using this knowledge we can quickly solve for the other two points just by plugging in the known info and we get: }$

$x_3 = 1 + -3*(1-2) = 4 \text{ and } y_3 = 2 + -3*(2-5) = 11$

$\text{therefore our point is: } (4,11,10) \text{ as posted in the post above}$[/hide]

It makes sense intuitively, just wondering where the justification is.. probably somewhere in the slope

I just solved it in like 2 seconds by looking at slope.  The changes the the $x,y,z$ values are $1,3,2$, respectively, from the first point to the second.  Since the $z$ value changes by $4$ from the second to the third, the changes will be $2,6,4$, giving the point $(2+2,5+6,6+4)=(4,11,10)$.

Yeah, Dan, you did it the way I proposed.

As I thought, it has to do with the slope.

Yeah, it's a really simple problem, when you look at it that way...


Before conference swim championships, Suzanne’s best time in the 200 yard butterfly race was 2 minutes, 23 seconds.  In the conference championships, she won 10th place by swimming the 200 yard butterfly in 2 minutes, 15 seconds.  What per cent decrease is that?

A.3.6%	B.5.6%	C.5.9%	D.6.2%	E.36.9%

Convert everything into seconds.
Time before:143 seconds
Time after: 135 seconds
Subtract 135 from 143 and divide this answer by the original, which is 143. We get $ 143-135=8 \Rightarrow \dfrac{8}{143}\approx .0559$ Multiplying this by 100, we get $ .0559 \cdot 100=5.59 \approx 5.6%
$ So the answer is $ \boxed{B)  5.6}$


floor function
logarithms
modular arithmetic
number theory unsolved
number theory

For every positive integer $ n$ let
\[ a_n\equal{}\sum_{k\equal{}n}^{2n}\frac{(2k\plus{}1)^n}{k}\]
Show that there exists no $ n$, for which $ a_n$ is a non-negative integer.

There exists a prime $ p$ st $ n<p<2n$ for Bertrand. The problem follows.

[quote="bboypa"]There exists a prime $ p$ st $ n < p < 2n$ for Bertrand. The problem follows.[/quote]
Sorry, but I don't really understand how this problem follows from Betrand's Postulate  :maybe:
Could you please explain?

[quote="Moonmathpi496"][quote="aadil"] what is bertrand's postulate? [/quote]
Actually it is exactly what bboypa said, For all $ n\geq 2$ there exists a prime $ p$ such that, $ n < p < 2n$

BTW can anyone post a complete solution to this problem? I don't really think that Bertrand postulate is going to solve this problem directly.[/quote]

I think bboypa means:

$ a_n$ is integer $ \Leftrightarrow \sum_{k\equal{}n}^{2n}\frac{1^n}{k}$.
Now Let $ n < p < 2n$ be a prime (Bertrand's postulate).
Then the denominator of $ \sum_{k\equal{}n}^{2n}\frac{1}{k}\minus{} \frac{1}{p}$ in simplest form has factor $ p$ , which is impossible as $ n<p<2n$

I think I can prove it without using Bertrand's postulate. Only with elementary solution :

$A = \sum_{k=n}^{2n}\frac{1}{k}$ must be integer. Assume that $\lfloor \log_2 n \rfloor = m$. So, if we compute the denominator of $A$ (in reduced fraction), the exponent of $2$ in it is $m$. But the parity of  the numerator is $\equiv 1 \pmod{2}$. So it is impossible that $A$ is integer.  :wink:

[i]*(Sorry if my English bad, because english not my mother language)[/i]


trigonometry
trig identities
Law of Cosines

A clock has an hour hand with a length of 6 centimeters and a minute hand with a length of 12 centimeters.  What is the distance between the tip of the hour hand and the tip of the minute hand when the clock reads 12:30?

[quote="matt276eagles"][hide]The angle between the hands is $165^{\circ}$.  By the Law of Cosines, the desired distance is $\sqrt{6^{2}+12^{2}-2\cdot 6\cdot 12\cos 165^{\circ}}=\sqrt{180-144\cos 165^{\circ}}=6\sqrt{5+4\cos 15^{\circ}}$
$=6\sqrt{5+4\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)}=\boxed{6\sqrt{5+\sqrt{6}+\sqrt{2}}}$.[/hide][/quote]

can you generalise the formulae for the angle between the hands?

Generalization:

Let the time be $h: m$.  If $h=12$ then let $h=0$ instead.  Then the angle of the hours hand is $30h+\frac{m}{2}$.  The angle of the minutes hand is $6m$.  Then the angle between the hands is $|30h-\frac{11m}{2}|$.

For example, for $h=12$ and $m=30$, then we let $h=0$.  Then angle between the hands in this case is $|30\cdot 0-\frac{11\cdot 30}{2}|=|-165|=165$.  If the angle measure $m$ is greater than 180, then we take $360-m$ instead.


An ice cream shop offers 6 kinds of ice cream. What is the greatest number of two scoop sundaes that can be made such that each sundae contains two types of ice cream and no two sundaes are the same combination?

There are $ \binom{6}{2} \equal{} \boxed{15}$ ways to choose two different scoops.

[quote=aops008]Does it matter the order in which the scoops are put on the ice cream cone?[/quote]

No, it does not. Combinations deal with the number of committees. If you were to choose a two person committee with persons A, B, and C, AB is the same as BA. However, permutations deal with the order, for example AB is NOT the same as BA

yeah, it's just

Combination - Doesn't matter the order
Permutation - Order matters


Sn denotes the sum of the squares of N consecutive natural nos.Which cud be possible value of Sn,if N=15 ?

a 2425
b 1480
c 3400
d 3655

i am getting quite long solution? can anyone suggest better way...

Is this your long solution?
$ \frac{(n\plus{}15)(2(n\plus{}15)\plus{}1)(n\plus{}16)\minus{}n(n\plus{}1)(2n\plus{}1)}6$
where the result fits one of the answers?
After the subtracting, we would get a quadratic, which should be solvable, but this multiplying out may take some time.


Hi ,
In the following command,
\documentclass[pdftex, 12pt, a4paper]{scrreprt}

What does pdftex do?
and what package should I add to use this command.Can anybodt explain it.
and also what does it mean for this next command,
\usepackage[pdftex,plainpages=false,pdfpagelabels]{hyperref}
I am using both of these but i am getting error.

Thanks
Koreno

pdftex basically means that you are compiling your TeX document directly to a PDF file, rather than to a DVI file (which is TeX's default behavior).  This gives you access to a few special PDF commands; perhaps the most commonly used are \pdfpagewidth and \pdfpageheight, which set the page size of the output PDF file.

As for  the hyperref options, I would check the hyperref's package documentation.  Go to: [url]http://www.ctan.org/info?id=hyperref[/url]

//pdftex basically means that you are compiling your TeX document directly to a PDF file, //rather than to a DVI file (which is TeX's default behavior). This gives you access to a few //special PDF commands; perhaps the most commonly used are \pdfpagewidth and //\pdfpageheight, which set the page size of the output PDF file.

//As for the hyperref options, I would check the hyperref's package documentation. Go to: //http://www.ctan.org/info?id=hyperref

hi DPatrick,

If I want to use this command ,\documentclass[pdftex, 12pt, a4paper]{scrreprt}
which package should i add ,
I have just simply written this command , but it is giving error, please help
thanks
koreno

Since I don't know anything about the scrreprt class that you're using, I can't really help you.  You'll need to check the class's documentation.

[quote="koreno"]hi DPatrick,If I want to use this command ,\documentclass[pdftex, 12pt, a4paper]{scrreprt}
which package should i add ,
I have just simply written this command , but it is giving error, please help
thanks
koreno[/quote]

The problem is probably that you're not importing whichever package has the "scrreprt" documentclass. Try replacing that with "report" and see if it will compile.


group theory
abstract algebra
induction
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ G$ be a group and let $ n$ a positive integer.  If $ N$ is a normal subgruop of index $ n$, it is easy to see that $ g^n\in N$ for all $ g\in G$. Is it remainds true when $ N$ is not normal?
I am interested in the case $ G$ finite.

No.  For example, let $ G \equal{} S_4, n \equal{} 3, N \equal{} S_2 \times S_2, g \equal{} (1234)$.

Are you sure the result is true for normal subgroups?

It is true for normal subgroups, by Lagrange's theorem applied to G/N.

However, for non-normal subgroups it is not true.  For instance, take G=S3, N=<(1,2)>, n=3, g=(2,3), g^n=(2,3).

The previous example had N of index n=6, not n=3.

Whoops.  The example still works, though, if we take $ N$ to be generated by $ (12)$ and $ (34)$, since $ g^6 \equal{} (13)(24)$.

Oh yes, sorry.  It is an example for the non-normal subgroup.  <(1,2),(3,4)> is not normal since it contains (1,2)(3,4), but not its conjugate (1,2,3,4)^2 = (1,3)(2,4).

The K4 subgroup is actually < (1,2)(3,4), (1,3)(2,4) >.  So abstractly it is C2 x C2, but as a permutation group it is the regular permutation representation, not the direct product S2 x S2.

If $ G$ es a nilpotent group then there exist a chain of subgroups $ N\equal{}N_1\lhd N_2\lhd\ldots\lhd N_{r\minus{}1}\lhd N_r\equal{} G$. This is because in a nilpotent group a proper subgroup is a proper subgroup of its normalizer. Then $ g^{[G: N]}\equal{}(\ldots((g^{[G: N_{r\minus{}1}]})^{[N_{r\minus{}1}: N_{r\minus{}2}]})^{\ldots} )^{[N_2: N_1]}$. Then we can show by induction that $ g^{[G: N]}\in N$ because $ g^{[N_i: N_{i\plus{}1}]}\in N_{i\plus{}1}$ if $ g\in N_i$ since $ N_{i\plus{}1}\lhd N_i$.

What can you say about a family bigger that the class of nilpotent group?. For example $ G$ solvable.

Well, it is true for any subnormal subgroup N that g^[G:N] is in N for every g in G.

However, a subgroup need not be subnormal for this to happen, since it might just be very small:  the non-central subgroups of order 2 in the dihedral group of order 12 are not subnormal, but since the dihedral group of order 12 has exponent 6, g^[G:N]=1 is always in N.

Now if a finite group has the property that g^[G:N] in N for every subgroup N of G and every g in G, then G must be nilpotent:  Let N be a Sylow p-subgroup of G, and let K be the intersection of the Sylow p-subgroups of G.  Since H = < { g^[G:N] : g in G } > is a normal subgroup contained in N, and so H is contained in K, and so g^[G:N]=1 mod K for all g in G/K.  Hence the exponent of G/K divides [G:N].  However, [G:N] is relatively prime to p, so G/K contains no elements of order p, and N=K is the unique Sylow p-subgroup of G.

A group $ G$ is polycyclic if there is a chain of subgroups $ 1 \equal{} H_0\lhd H_1\lhd\ldots\lhd H_{r \minus{} 1}\lhd H_n \equal{} G$ such that $ H_i/H_{i \minus{} 1}$ is cyclic. For example, all finitely generated nilpotent group is polycyclic.
It is a theorem that: If $ G$ is polycyclic and every finite image of $ G$ is nilpotent, then $ G$ is nilpotent.

Then you have showed: If $ G$ is a polycyclic group such that $ g^{[G: H]}\in H$ for all finite index subgroup $ H$ and for all $ g\in G$, then $ G$ is nilpotent.


[quote="rockwell"]The archery tournament:

At an archery tournament, eight of the twenty competitors were left-handed and the remainders were right-handed.

1.	Find the probability that an archer chosen at random is left-handed.
2.	If a pair of archers is chosen at random is left-handed. Find the probability that exactly one of the pair is left-handed.

Due to the weather conditions it is more difficult for left-handers to hit the target. Practice sessions indicated that the probability of a left-hander hitting the target was 0.7, while for right-hander it was 0.9.
3.	Find the probability that an archer chosen at random is left-handed and hits the target.
4.	Find the probability that an archer chosen at random hits the target.
5.	If an archer hits the target, find the probability that the person was right-handed.
6.	Fin the probability that two archers chosen at random are the same hand and both hit the target.
7.	Draw up a table which shows the probability for each outcome possible for two randomly chosen arches.[/quote]

[hide="Edited"]
1.	Find the probability that an archer chosen at random is left-handed.

$P(L)=\frac{8}{28} =\frac{2}{7}$.

2.	If a pair of archers is chosen at random is left-handed. Find the probability that exactly one of the pair is left-handed.

$P(\text{exactly one leftie}) =P(A_1 \cap A_2)$
$=P(A_1=R)\cdot P(A_2=L|A_1=R) + P(A_2=R)\cdot P(A_1 = L|A_2 =R)$
$=\frac{20}{28}\cdot\frac{8}{27}+\frac{20}{27}\cdot\frac{8}{28}$
$=\frac{(2)(20)(8)}{(27)(28)}$
$=\frac{80}{189}$

Due to the weather conditions it is more difficult for left-handers to hit the target. Practice sessions indicated that the probability of a left-hander hitting the target was 0.7, while for right-hander it was 0.9.
3.	Find the probability that an archer chosen at random is left-handed and hits the target.

$P(H \cap L) = P(L)\cdot P(H|L) = (\frac{8}{28})(0.7)$

4.	Find the probability that an archer chosen at random hits the target.

$P(H) =P(L)P(H|L) + P(R)P(H|R) = (\frac{8}{28})(0.7) +(\frac{20}{28})(0.9)$

5.	If an archer hits the target, find the probability that the person was right-handed.

$P(R|H)=\frac{P(R)P(H|R)}{P(H)} =\frac{(\frac{20}{28})(0.9)}{\frac{23.6}{28}}$

6.	Fin the probability that two archers chosen at random are the same hand and both hit the target.

*My answer to #6 is wrong. If I figure out why, I'll fix it.

7.	Draw up a table which shows the probability for each outcome possible for two randomly chosen arches.

I hate drawing tables; I'll leave that to you.[/hide]


induction
integration
calculus
combinatorics unsolved
combinatorics

Show that for nonnegative integers m and n:

${\sum_{j=0}^{m}(-1)^{j}\frac{\binom{m}{j}}{n+j+1}}=\sum_{k=0}^{n}(-1)^{k}\frac{\binom{n}{k}}{m+k+1}$

I was stumped

Let $f(x)=x^{n}(1-x)^{m}= \sum_{j=0}^{m}(-1)^{j}\binom{m}{j}x^{j+n}.$
Then the LHS is $A = \int_{0}^{1}f(x)dx$

Let $g(x)=x^{m}(1-x)^{n}= \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}x^{k+m}.$
Then the RHS is $B = \int_{0}^{1}g(u)du$.

Now, using the variable change $u = 1-x$ in $A$, we easily see that $A=B;$

Pierre.

My solution is not as nice as the solution using integration but it is quite straightforward, and since you asked for a solution using induction I am going to sketch it anyway:

Denote the right hand side by $g(n,m)$. Then we can prove by induction over $m$ that for all $n$ we have $g(n,m) = g(m,n)$, and obviously this is what we have to show.

$m=0$ is easy. Just use the binomial identity for $g(n,0)$.

For the inductive step we simply use $\binom{n}{k}=\binom{n+1}{k+1}-\binom{n}{k+1}$
in order to compute $g(n,m+1)$ and after a short calculation we get
\[g(n,m+1) = g(n,m)-g(n+1,m) \quad \text{ i.e. }\quad g(n,m) = g(n+1,m)+g(n,m+1). \]
Interchanging $m$ and $n$ in the last formula gives
\[g(m,n) = g(m+1,n)+g(m,n+1). \]
This establishes the inductive step:
\[g(n,m) = g(m,n), g(n+1,m) = g(m,n+1) \quad \Rightarrow \quad g(n,m+1) = g(m+1,n). \]


Let $ a,b,c \in\mathbb{C}$ so that $ |a|\equal{}|b|\equal{}|c|$ and $ a\plus{}b\equal{}c$ . Find the value of $ a^4\plus{}b^4\plus{}c^4$ .

Put $ a\equal{}uc$, $ b\equal{}vc$. Since $ u\plus{}v\equal{}1$, $ Im(v)\equal{}\minus{}Im(u)$. Moreover $ |u|\equal{}|v|\equal{}1$, so $ Re(u)\equal{}Re(v)$. Since $ Re(u)\plus{}Re(v)\equal{}1$, $ Re(u)\equal{}Re(v)\equal{}1/2$, WLOG $ Im(u)\equal{}\sqrt{3}/2$ and $ Im(v)\equal{}\minus{}\sqrt{3}/2$ (the other possibility is symetric). Since $ a\equal{}(1/2\plus{}\sqrt{3}i/2)c$, $ b\equal{}(1/2\minus{}\sqrt{3}i/2)c$, we have $ a^4\plus{}b^4\plus{}c^4\equal{}0$. Not hard, but nice one :)


[quote="Mashimaru"]Let $ O,I$ be the circumcentre and incentre of $ triangle ABC$ respectively. Let $ D,E,F$ be the points where $ (I)$ touches $ BC,CA,AB$ and $ P$ is the centroid of $ \triangle DEF$. Prove that $ OI\geq 6IP$.[/quote]
It is clearly an application of inversion. After the inversion around I with radius r, O becomes the nine-point center K of triangle DEF.
So, IP=2IK/3. The circle (ABC) becomes nine-point circle of DEF, which implies $ \frac {IO}{IK} \equal{} \frac {R}{\frac {r}{2}} \ge 4$, thus $ OI\ge 4IK \equal{} 6IP$.


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove the inequality
$ \sum\frac {ab}{(a \plus{} b)^2} \ge \frac {3}{4}\cdot \frac {ab \plus{} bc \plus{} ca}{2(a^2 \plus{} b^2 \plus{} c^2) \minus{} ab \minus{} bc \minus{} ca}.$
:)[/quote]
An expanding gives
$ \sum_{sym}(5a^5b^3 \minus{} 5a^4b^4 \plus{} 4a^6bc \plus{} 27a^5b^2c \minus{} 3a^4b^3c \plus{} 5a^4b^2c^2 \minus{} 33a^3b^3c^2)\geq0,$
which is obvious.

The best of this type is
$ \sum \frac {ab}{2a^2 \minus{} ab \plus{} 2b^2} \ge \frac {ab \plus{} bc \plus{} ca}{2(a^2 \plus{} b^2 \plus{} c^2) \minus{} ab \minus{} bc \minus{} ca},$
which can be proved by Cauchy Schwarz:
$ \sum \frac{ab}{2a^2\minus{}ab\plus{}2b^2} \ge \frac{(ab\plus{}bc\plus{}ca)^2}{\sum ab(2a^2\minus{}ab\plus{}2b^2)},$
and
$ \sum ab(2a^2\minus{}ab\plus{}2b^2)\equal{}(ab\plus{}bc\plus{}ca)(2a^2\plus{}2b^2\plus{}2c^2\minus{}ab\minus{}bc\minus{}ca).$
:)

For more general, we can prove that:
If $ a,b,c$ are nonnegative real numbers, then we have
$ \sum \frac{ab}{a^2\plus{}kab\plus{}b^2} \ge \frac{1}{k\plus{}2} \cdot \frac{(u\minus{}2)(a^2\plus{}b^2\plus{}c^2)\plus{}(5\minus{}u)(ab\plus{}bc\plus{}ca)}{u(a^2\plus{}b^2\plus{}c^2)\plus{}(1\minus{}u)(ab\plus{}bc\plus{}ca)},$
where $ 2 \ge k \ge \minus{}\frac{1}{2},$ and $ 2 \ge u \ge \frac{2\minus{}k}{2\plus{}k}.$
:)

[quote="can_hang2007"]For more general, we can prove that:
If $ a,b,c$ are nonnegative real numbers, then we have
$ \sum \frac {ab}{a^2 \plus{} kab \plus{} b^2} \ge \frac {1}{k \plus{} 2} \cdot \frac {(u \minus{} 2)(a^2 \plus{} b^2 \plus{} c^2) \plus{} (5 \minus{} u)(ab \plus{} bc \plus{} ca)}{u(a^2 \plus{} b^2 \plus{} c^2) \plus{} (1 \minus{} u)(ab \plus{} bc \plus{} ca)},$
where $ 2 \ge k \ge \minus{} \frac {1}{2},$ and $ 2 \ge u \ge \frac {2 \minus{} k}{2 \plus{} k}.$
:)[/quote]
I think this is a little nicer.  :wink:

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that

$ (a)\ \ \ \ \  15\sum \frac {ab}{2a^2 \plus{} ab \plus{} 2b^2}\plus{}\frac {8(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca} \ge 17$;

$ (b)\ \ \ \ \  m(\sum \frac {ab}{a^2 \plus{} kab \plus{} b^2}\minus{}\frac 3{k\plus{}2})\plus{}\frac {a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\minus{}1 \ge 0$,

for $ \minus{}2<k\le 2$ and $ m\le \frac {(k\plus{}2)(k\minus{}1\plus{}\sqrt{5\minus{}2k})}{4}$.


Prove that:$\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}\leq \frac{27}{10}$
where $x;y;z$ are positive real numbers satisfying $x+y+z=1$

I think use $\frac{1}{x^{2}+1}\leq-\frac{27}{50}(x-\frac{1}{3})+\frac{9}{10}$, etc...

P.S: Why name of this topic is ''hard''? You need pay attention name of topic.


For what non-zero integer K is it true that 15k/2k+5/6=(15K+6)/(2K+6)?

[hide="the answer"]
15k/2k is just 15/2, so the left hand side would be 50/6
so:
50/6 = (15K+6)/(2K+6)
cross multiplying...
90K+36=100K+300
10K=-264
thus[b] K= -26.4[/b]
thats not an integer though, so can anyonme tell me what i did wrong??
im sortof tired and in a rush so i didnt think very hard... maybe there is some easy trick or something?[/hide]

Basic algebra  :spider: Maybe this should go in Mathcounts or something...

[quote="4everwise"]Yeah. This shouldn't have been in this forum.[/quote]
'prob the basics or classroom math?

Oh right...this is just a basic thing....

maybe the tricky "integer" part of the question makes it a high school question?  :D or maybe we're trying to figure out what was the typo

There's obviously a typo; I got the same answer as sheepwarrior. Also, since it's extremely easy anyway, this should either be moved or locked/deleted...

Most likely the question is $\frac{15k}{2k} + \frac{5}{6} = \frac{15k + 5}{2k+6}$

[quote="break314dancer"]So then $K=-27$, right?[/quote]

Yup :yup:  :10:  :clap2:

Because the $k$s in $15k/2k$ cancel out and then you have $25/3=(15k+5)/(2k+6)$ and you can cross multiply and solve it from there so you get $k=-27$

[quote="Archimedes27"]
\[\frac{15k}{2k}+\frac56 = \frac{15K+6}{2K+6}\]
[/quote]
[quote="sheepwarrior"]
thus[b] K= -26.4[/b]can anyonme tell me what i did wrong??
[/quote]
look at the numerator of the fraction on the right hand side... you did right, but Archimedes27 made a mistake... it should be
\[\frac{15k}{2k}+\frac56 = \frac{15K+5}{2K+6}\]


$1. \text{If we have:} \sin{x}=\frac{a+b}{a-b}, 0<a<b,\text{find:}\sqrt{\cot^2{x}-\cos^2{x}}.$
$2. \text{If we know that:} \tan(\frac{\pi}{4}+x)=\frac12, \text{then find the value of:} \frac{\sin2x-\cos^2{x}}{1+\cos2x}.$

2.[hide]Using the sum formula for tangents on $\tan\left(\frac{\pi}{4}+x\right)=\frac 12$, we get $\frac{1+\tan x}{1-\tan x}=\frac 12$, or $\tan x=-\frac 13$. Writing the desired value as $\frac{2\sin x\cos x-\cos^2 x}{2\cos^2 x}$ and simplifying, it becomes $\tan x-\frac 12$. -1/3-1/2=-5/6, the answer.
[/hide]

I got -4ab/(a^2 - b^2) just as shobber asserted. My method differed slightly; instead of using (1- sinx^2)^2, I took out cosx^2 and continued on. My method is not as efficient, however, for you run into the added difficulty of finding cosx interms of a and b in order to complete the problem

I solved the problem identically. Good job


What command would I type in the command prompt to get latex2html (from MiKTeX) to convert a certain file?

Have you installed Latex2html obtainable from http://www.latex2html.org/ ? It's quite tricky to do so in Windows (see for example http://www.mayer.dial.pipex.com/l2h.htm which may help - though some people still find it doesn't work). There are alternatives for converting to HTML: see http://www.mayer.dial.pipex.com/samples/comparisons.htm for comparisons and http://www.mayer.dial.pipex.com/samples/example.htm for conversion examples.

Please note that although MiKTeX appears to include [url=http://www.cis.ohio-state.edu/~gurari/TeX4ht/mn.html]Tex4ht[/url] as an option, it doesn't include the relevant files so they need to be downloaded separately.


He hasn't worked for AoPS for a while -- he now runs a school for gifted math/science students in Birmingham, AL (MIST Academy).

N00000000000000000000000000000000000000000000000000000000000000000  :( :noo:

I really like his number theory book!
So he's not going to write intermediate number theory ? :noo:

[quote="Poincare"]So he's not going to write intermediate number theory ? :noo:[/quote]

Nope - I believe Mr. Rusczyk will write it (after the calculus book?). I too, am waiting for Inter NT  :lol:

[quote="rrusczyk"]He hasn't worked for AoPS [b]for a while[/b] -- he now runs a school for gifted math/science students in Birmingham, AL (MIST Academy).[/quote]

Does that mean Mr. Crawford will come back?

Youre gonna have a calculus book?!!  :wow:  :wow:  :wow:

I don't think Mathew will be coming back anytime soon. He has different things going on for him at the moment and he is pretty involved with his school.
About the Interm NT book, I doubt we will write one, or, anyhow, it's not on our list of priorities as of this moment.

Dave is writing the calculus book.

[quote="Poincare"]I was reading mr.crawford's blog and i noticed that i hadn't posted anything since early 2007 did he quit or something? I really hope he didn't  :([/quote]

Though he doesn't work for AoPS anymore, he still has a blog. You can read it here [url]http://www.mistacademy.com/blog/[/url]

MIST is currently expanding to my town, Huntsville, and I bet he is really busy with getting it set up and stuff. I know I am taking MIST this year but I am not sure about high school years.

Edit: Oh, yay! He's coming to our high school too :D!! This means I [i]will[/i] have him more than this year! :D

Wow!! You're lucky!!

So I assume he won't be coming back to AoPS for at least a long time?

[quote="isabella2296"]So I assume he won't be coming back to AoPS for at least a long time?[/quote]The chances are indeed very slim. Actually we do hope he doesn't come back, because that will mean his school project will be successful ;)

The MIST Academy project goes well, and I am excited about working one day a week in Huntsville.  There is a chance it becomes two days (or a day-and-a-half really).

I have considered the idea of one day writing an Intermediate Number Theory book, but right now other priorities limit the chances that I will write one very soon.  However, it may be that out of MIST Academy eventually comes enough material to compile one.  Or, perhaps it will be a smaller volume outside of the AoPS philosophy of a complete textbook -- just something to bridge a certain set of concepts for dedicated students.

Mathew: do you think maybe the time has come for you to edit the "location" in your user profile?

I just tried to write "Done."  That message size was too small.  So I'm writing more words to convey the same message.


AMC
AIME
function
probability
calculus
number theory
relatively prime

Solve the problem and post a new one with the same post.

m and n are relatively prime positive integers. The coefficients of x:^2: and x:^3: in the expansion of (mx + n)2000 are equal. Find m + n.

edit: Fixed a typo

Did you mean (mx + n)2000? Spoiler:[hide]

Expand using binomial theorem:

(mx + n)2000 = .... + 2000C3 (mx)3 n1997 + 2000C2 (mx)2 n1998 + ...

So 2000C3 m3 n1997 = 2000C3 m2 n1998

666 m = n

Because (m,n) = 1 the only solution in pos. int's is (m,n) = (1,666).

m + n = 1 + 666 = 667.[/hide]

2 boxes contain both black and white marbles. Altogether, there are 25 marbles. Randomly pick a marble from each box.

Prob(both marbles are black) = 27/50

Prob(both marbles are white) = m/n, m and n relatively prime pos ints

Find m + n.

wyra wrote:2 boxes contain both black and white marbles. Altogether, there are 25 marbles. Randomly pick a marble from each box.

Prob(both marbles are black) = 27/50
Prob(both marbles are white) = m/n, m and n relatively prime pos ints

Find m + n.

[hide]It seems like the easiest way to do this problem is to pick some numbers that work, and evaluate P(both marbles are white) using those numbers.  Say the two boxes contain 5 and 20 marbles, respectively.  Then (#black in box 1)/5 * (#black in box2)/20 = 27/50, so (#black in box1)(#black in box2) = 54.  If #black in box1 = 3, then #black in box2 = 18.  Since there are 5 marbles total in box1, 2 are white; since there are 18 marbles total in box2, 2 are white.  2/20*2/5 = 1/25.  So m + n = 1 + 25 = 26.

I know there's a more rigorous way of doing this, but it would be fine for AIME, since we need only the answer....[/hide]

The function f, defined on the set of ordered pairs of positive integers, satisifes the following properties:

f(x, x) = x

f(x, y) = f(y, x)

(x+y)*f(x,y) = yf(x, x+y).

Find f(14, 54).

mathfanatic wrote:
The function f, defined on the set of ordered pairs of positive integers, satisifes the following properties:

f(x, x) = x
f(x, y) = f(y, x)
(x+y)*f(x,y) = yf(x, x+y).

Find f(14, 54).

Spoiler:[hide] We rewrite the third identity as f(x,x+y)=f(x,y)*(x+y)/y. Then

f(14,40+14)=54/40*f(14,40)=54/26*f(14,26)=54/12*f(14,12)=54/12*f(12,14)=54*14/(12*2)*f(2,12)=54*14/(12*2)*((12/10)*(10/8)*(8/6)*(6/4)*(4/2))*f(2,2)=378 if I didn't make any mistakes.

[/hide]

In a five-team tournament, each team plays one game with every other team.  Each team has a 50% chance of winning any game it plays.  (There are no ties.)  Let m/n be the probability that the tournament will produce neither an undefeated team nor a winless team, where m and n are relatively  prime positive integers.  Find m+n.

--Dan

petra wrote:In a five-team tournament, each team plays one game with every other team.  Each team has a 50% chance of winning any game it plays.  (There are no ties.)  Let m/n be the probability that the tournament will produce neither an undefeated team nor a winless team, where m and n are relatively  prime positive integers.  Find m+n.

[hide]Let us count the ways in which we can lose.  First, let us assume one team wins all 4 of its games.  Then the other matches of the 4 teams can happen in any way we want.  5 teams that can win all their games, C(4, 2) games with 2 outcomes each gives us 5*2C(4, 2) = 320 possibilities.  We have exactly the same number if one team loses all its games.  However, we have double-counted all those possibilities in which we have one team wins all its games and another loses all its games.  We have 5 choices for the first team, 4 for the second, and C(3, 2) other games with 2 outcomes each to determine.  This gives us 5*4*2C(3, 2) = 160 possibilites.  So we have in total 2*320 - 160 = 480 situations in which we win.  There are 10 total games, so 210 = 1024 total outcomes.  Thus, our probability is 480/1024 = 15/32.  However, we wan the compliment of this, which is 17/32, and 17 + 32 = 49.[/hide]

Let R be the reals. The function f : R → R satisfies f(0) = 0 and f(2 + x) = f(2 - x) and f(7 + x) = f(7 - x) for all x. What is the smallest possible number of values x such that |x| ≤ 1000 and f(x) = 0?

JBL wrote:Let R be the reals. The function f : R → R satisfies f(0) = 0 and f(2 + x) = f(2 - x) and f(7 + x) = f(7 - x) for all x. What is the smallest possible number of values x such that |x| ≤ 1000 and f(x) = 0?

[hide]

f(0)=0, so f(2-2)=f(2+2)=f(4)=0, and it's easy to see that all the multiples of 4 are going to be 0. Between -1000 and 1000 inclusive, there are 501 of these.

f(0)=0 so f(7-7)=f(0)=0=f(7+7)=f(14), so all the multiples of 14 are going to be 0. There are 142 of these excluding 0.

However, that's overcounting. lcm(4,14) = 28, so I've counted multiples of 28 twice. So I've counted 70 numbers twice.

But then there are also the numbers that are a multiple of 14 :pm: 4, which will also be zero. There are two of these for every multiple of 14, so there are 284 of them, but we need to subtract some of these too. We need to subtract multiples of 24 and 32 and add back in multiples of 192 and 168. There are 20 of these.

I'm pretty sure I missed some cases, but now I'm sick of this problem and I'm not using paper, so... 501+142-70+284-20 = 837

[/hide]

What is the smallest positive integer with 12 positive even divisors and 6 positive odd divisors?

[quote="confuted"]it's easy to see that all the multiples of 4 are going to be 0[/quote]

I don't see it.  In fact, it isn't true.

oh, whoops, lol... I switched it around to be f(x+2)=f(x-2) ... not paying attention... d'oh

What is the smallest positive integer with 12 positive even divisors and 6 positive odd divisors?

n = 2^a * 3^b * 5^c * 7^d * 11^e * . . .

There are 12 even divisors so that means that a(b+1)(c+1)(d+1)... = 12.  There are 6 odd divisors so (b+1)(c+1)(d+1)... = 6.  A is now certainly 2 after substituting (b+1)(c+1)... into the first equation.  The possibilities of (b, c) so the number will be as small is possible are (1, 2), (2, 1), (5, 0), and (0, 5).  Testing these possibilities, the pair that will give us the smallest integer is (2, 1).   2 :^2: * 3 :^2: * 5 = 180.  180 is the smallest integer with 12 even divisors and 6 odd divisors.

My problem still needs a solution.  And Yannik, you made the part about getting the evens really complicated.  It can be explained much more simply than that.

This round-robin chain thing gets complicated when someone gets a problem wrong but still posts another problem.  Maybe we should make a separate post for each problem...

Oops?  Don't worry I'll get another AIME problem now:

The positive integers A, B, C, D satisfy A :^5:  = B :^4:, C :^3: = D :^2: and C = A + 19.  Find D - B.

Was my solution correct?  How can finding the even divisors be made simpler?

Well, I guess what I mean is it can just be said more nicely:  if a number has o odd factors and is divisible by 2^n, then it has no even factors.

Hehe.  That is, n*o even factors, since you get each odd factor times 2, each odd factor times 4, each odd factor times 8 ...

So you know that it's divisible by 4.  And then you just need to find a number with 6 odd factors.

I guess it's not really particularly simpler -- I just think it sounds nicer.  So my criticism was misplaced.

Anyhow, someone still needs to solve my problem!

Let R be the reals. The function f : R → R satisfies f(0) = 0 and f(2 + x) = f(2 - x) and f(7 + x) = f(7 - x) for all x. What is the smallest possible number of values x such that |x| ≤ 1000 and f(x) = 0?

f(2+x) = f(2-x)
f(2+(x+2)) = f(2-(x+2)) = f(x+4) = f(-x)
f(x) = f(4-x)

and

f(7+x)=f(7-x)
f(7+(x+7) = f(7-(x+7)) = f(x+14)=f(-x)
f(x) = f(14-x)

f(4-x)=f(14-x)
f(4-(4-x))=f(14-(4-x))=f(x)=f(x+10)

If f(0)=0, a possibility for f(x) to be 0 can be x to be a multiple of 10.  There are 201 multiples of 10 between -1000 and 1000 inclusively.  Also f(4)=0 from f(x)=f(4-x) and f(14)=0 from f(x)=f(14-x).  We need to count the integers where x :equiv: 4 (mod 10), there are 200 of those integers.  201 + 200= 401.  401 is the smallest possible number of values of x where f(x) = 0 and |x| ≤ 1000.

Nope, not quite, but you're getting there.  Try using those relations you found to list out the first few values you can get.

I think I fixed it.  I undercounted because I thought I only needed to find the period and count from there on.

You still have one (very very) slight mistake -- you didn't quite add on enough.

Rockin'. And just to repost your problem for the next person to come along:

The positive integers A, B, C, D satisfy A5 = B4 , C3 = D2 and C = A + 19. Find D - B.

i dont think my method is great, but i do think it works:

let A^5 = B^4 = x^20 because to be a 5th and 4th power it must be a 20th power as well

Since C = A+19, that means A+19 must be a square because C^3 has to be a square as well. Therefore, A+19 = x^4+19 = k^2. By trial and error we get x = 3 and k = 10. Therefore:

A^5 = B^4 = 3^20 which means A = 81 and B = 243

C^3 = D^2 = (10^2)^3 which means C = 100 and D = 1000.

Finally, D - B = 757.

NEW PROBLEM:

2a + b + c + d + e = 6
a + 2b + c + d + e = 12
a + b + 2c + d + e = 24
a + b + c + 2d + e = 48
a + b + c + d + 2e = 96

Find 3d + 2e.

(not too hard, but i like the problem)

2a + b + c + d + e = 6
a + 2b + c + d + e = 12
a + b + 2c + d + e = 24
a + b + c + 2d + e = 48
a + b + c + d + 2e = 96

add all the eqns, 6(a+b+c+d+e)= 6(1+2+4+8+16)

a+b+c+d+e=31

Substitude this into 5th and 6th equations to get d=17, e=65.  3d+2e=51+130=181

Find the minimum of [9x :^2: *sin(x) :^2: +4]/(x*sinx) from 0<x<pi

[quote="confuted"][quote="cats..."]Find the minimum of [9x :^2: *sin(x) :^2: +4]/(x*sinx) from 0<x<pi[/quote]
May I use Calculus? Heh. I assume not, since it would make the problem easy (easier?)[/quote]

I dunno, I think the problem is slightly easier to solve without calculus..

Find the minimum of (9x :^2: * sin :^2:x + 4) / (x sin x) from 0<x<pi.

x sin x = u

(9u :^2: + 4) / u = 9u + 4/u.

(9u + 4/u) / 2 :ge: :sqrt:(9u*4/u) = :sqrt: 36 = 6

9u + 4/u :ge: 12

Equality will be attained when u = 2/3 and the minimum of (9x :^2: * sin :^2:x + 4) / (x sin x) is 12.  x sin x will equal 2/3 somewhere between the x values of 0 and :pi: because 0 sin 0 and :pi: sin :pi: equal 0 and :pi:/2 sin :pi:/2 is :pi:/2.  Since x sin x is continuous, there must be an x value for which x sin x = 2/3.

Find the smallest positive integer n such that every digit of 15n is 0 or 8.

WarpedKlown1335 wrote:
Find the smallest positive integer n such that every digit of 15n is 0 or 8.

[hide]

Since 15 and 0 are divisible by 3 but 8 is not, our answer must include at least

three 8s.  There is nothing we can multiply 15 by to get an 8 in the ones place,

so we must have a 0 in the ones place.  Thus, the smallest possible such integer

is 8880.  And 8880=15*592, so we are done.

[/hide]

Determine x^2+y^2+z^2+w^2 if

x^2/(2^2-1^2)+y^2/(2^2-3^2)+z^2/(2^2-5^2)+w^2/(2^2-7^2)=1

x^2/(4^2-1^2)+y^2/(4^2-3^2)+z^2/(4^2-5^2)+w^2/(4^2-7^2)=1

x^2/(6^2-1^2)+y^2/(6^2-3^2)+z^2/(6^2-5^2)+w^2/(6^2-7^2)=1

x^2/(8^2-1^2)+y^2/(8^2-3^2)+z^2/(8^2-5^2)+w^2/(8^2-7^2)=1

--Dan

petra wrote:WarpedKlown1335 wrote:
Find the smallest positive integer n such that every digit of 15n is 0 or 8.

[hide]
Since 15 and 0 are divisible by 3 but 8 is not, our answer must include at least
three 8s.  There is nothing we can multiply 15 by to get an 8 in the ones place,
so we must have a 0 in the ones place.  Thus, the smallest possible such integer
is 8880.  And 8880=15*592, so we are done.
[/hide]

--Dan

It seems more like you have stated what 15n (8880) is, but you never said that the smallest possible value of n was 592.  Good job anyways but just pinpointing that out.

I think maybe I killed the thread by picking a problem that is quite difficult.

If anyone wants to work on this problem but has no idea where to go, here is a hint:

[hide]Define some function, f(t), so that the four equations are the result of evaluating the function at four different values of t.  Then using the function you can argue about x^2+y^2+z^2+w^2.

[/hide]

Hope that gets things moving,

--Dan


limit
advanced fields
advanced fields unsolved

1) Let $ n \equal{} q_1^{e_1}\ldots q_s^{e_s}$ be the prime-power decomposition of a positive integer $ n\geq 2$ and let $ m$ be a positive integer. Show there exists a ring $ R$ such that $ |R| \equal{} n$ and $ |\textrm{Spec}(R)| \equal{} m$ if and only if $ s \leq m \leq \sum e_i$.
2) Prove that $ \lim_{|R|\to \infty} \frac {|\textrm{Spec}(R)|}{|R|}\equal{}0$.

(p.s. This result comes straight from a paper, but since I was so amazed by it, I leave the problem open for those who can come up with another nice solution)

1) "$ \Leftarrow$" we may write $ m \equal{} \sum f_i$ where $ 1 \leq f_i \leq e_i$. if we have found a rings $ R_i$ with $ |R_i| \equal{} q_i^{e_i}$ and $ |Spec(R_i)| \equal{} f_i$, then $ R \equal{} \prod R_i$ satisfies $ |R| \equal{} n$ and $ |Spec(R)| \equal{} m$. now fix some $ i$. to find $ R_i$, choose $ k_j \neq 1$ with $ k_1 \plus{} ... \plus{} k_{f_i} \equal{} e_i$ and let $ R_i$ be the product of the $ \mathbb{F}_{q_i^{k_j}}$.
"$ \Rightarrow$" non-reduced rings make problems :-). considering $ R/Nil(R)$, which is a product of fields, we get $ m \leq \sum e_i$, but we cannot recover $ s \leq m$.

ok it's easy when you know that every finite ring is the direct product of local rings (since they are artinian) and finite local rings have prime power cardinality.


I had a tough time doing this problem:
"The solution of the inequality
1/(x-1)  + 1/(x-2)  + 1/(x-3)  >=  1
is the union of a finite number of disjoint intervals, each of the form a<x<=b (x is greater than a and less than or equal to b).  If the length of the interval a<x<=b is b-a, what is the sum of the lengths of all these intervals?"

Thanks for responding.

hello, you can rewrite your inequality in the form
$ \frac{1}{x\minus{}1}\plus{}\frac{1}{x\minus{}2}\plus{}\frac{1}{x\minus{}3}\minus{}1\equal{}\frac{\minus{}x^3\plus{}9x^2\minus{}23x\plus{}17}{(x\minus{}1)(x\minus{}2)(x\minus{}3)}\geq 0$
Sonnhard.

Hi,

I was just wondering, do you have the answer?

Ok, here's my method.

[hide="My_Solution"]I used Vieta's formulas.

Using the numerator of the equation from Dr. Sonhrad (which isn't too difficult to get, just follow the common denominator rule),

the sum of the roots is 9. The roots are the b parts. So the sum of the b parts is 9.

The non-possibilities are the 1,2,3 (hence the non-equal sign under the < sign), which are the a parts.

So simply do 9-1-2-3 = 3.[/hide]
I hope this helps.


[quote="hoshattack"]A license plate consists of two letters followed by two digits; for example, MP78.  Neither the digits nor the letters may be repeated, and neither the letter "O" nor the digit "0" may be used.  When reading from left to right, the letters must be in alphabetical order, and the digits in increasing order.  How many different license plate combinations are possible?

Please don't solve if you did the problem in that State Competition.[/quote]

[hide]If the letter is A, there are 24 other choices. If B, 23...etc. So 24+23+22+21+...+1=$\frac{(24)(25)}{2}=300$.
For the numbers, if the number is 1, there is 8 choices...until there is 1. 8+7+6...+1=$\frac{(8)(9)}{2}=36$. So total licenses are $300\times 36=\boxed{10800}$?[/hide]

[hide]21600. A has 24 combinations, Y has 1, 1+2+3...+24 is 600.
1 has 8 combinations, 8 has 1 combination, 1+2+3...+8 is 36.
600*36 is 10800.[/hide]

edit: i made an arithmetic error. :(  oops.
i did this problem, but i don't remember it. my memory isn't that good.

[hide]For the letters there are (25*24)/2 possibilities, since O can't be used. For the numbers, likewise, their are (9*8)/2 possib. Multiply: (25*24)/2 * (9*8)/2 = 25*12*9*4 = 300*36 = [b]10800[/b][/hide]

[hide]
There are 24 possible ways for the letters with A (A and O can't be used as second letter), 23 with B, giving you $24+23+22...1$. Apply sum of arithmatic sequence to get $\frac{24(25)}{2}$ or $300$. The reason you don't need to worry about O not being used as the first letter is because the letter P will fill in for that and the letter Q will fill in for how many combinations P should have, etc. Then, there are 8 combinations with 1, 7 with 2, giving you $8+7+6...1$. Apply sum of arithmatic sequence to get $\frac{8(9)}{2}$ or $36$. Finally, multiply to get $10800$.[/hide]

[quote="binonunquineist"][quote="Captain Sugar"][hide]21600. A has 24 combinations, Y has 1, 1+2+3...+24 is 600.
1 has 8 combinations, 8 has 1 combination, 1+2+3...+8 is 36.
[b]600*36 is 10800[/b].[/hide]

edit: i made an arithmetic error. :(  oops.
i did this problem, but i don't remember it. my memory isn't that good.[/quote]

You made another mistake, it should be [b]300*36 = 10800.[/b][/quote]

oh well. you get that i got the right answer. i just typed gibberish. :oops:
i am hyper today. :)

you are happy that u are hyper?   :?

thats a little strange, but we all are somehow...

[hide]25 letters and 9 numbers can be used.
If a is the first letter, then there are 24 choices for the second letter.
If b, then there are 23.
...
there are 24+23+...+1=300 arrangements for letters.

Similarly,
If 0 is the first number, then there are 8 choices for the second letter.
If 1, then there are 7.
...
8+7+...+1=36 arrangments for numbers.

$300*36=\boxed{10800}$[/hide]

[quote="hoshattack"]you are happy that u are hyper?

thats a little strange, but we all are somehow...[/quote]

why wouldn't i be happy that i'm hyper?  :w00t:  :w00tb:  :wow:
look at my user name!


Post your favorite math jokes here.

Here's one I made up:

A mathematician and his brother are driving down the road when they notice they are almost out of gas.  Luckily, there are two gas stations nearby.  One sign reads $\$3.28^\text{ 9}_{10}$ a gallon, and the other reads $\$2.78^{9}$ a gallon.  The mathematician pulls into the first one.
"What are you doing?" his brother says.  "This is much more expensive!"
The mathematician says, "I know what I'm doing."
"What do you mean?  It's more expensive here!"
"That's where you're wrong."  The mathematician smiled and said, "2.78 to the 9th power is almost 10,000.  That's much more than 3.289."

I think you screwed up the Latex Expressions. Please check them.

And the other one is merely resting - it did not die.


category theory
abstract algebra
advanced fields
advanced fields unsolved

Let $ _S B_R$ be an $ S$-$ R$-bimodule, and consider the two covariant functors
\[ F, G \ : \ (R\text{\minus{}Mod})^\circ \  \times  \ (S\text{\minus{}Mod}) \quad  \rightarrow \quad  \mathbb{Z}\text{\minus{}Mod},
\]where
\[ F\equal{} \text{Hom}_R (\minus{}, \text{Hom}_S (B, \minus{})) \qquad \text{and} \qquad G \equal{} \text{Hom}_S (B \otimes_R \minus{}, \minus{}).
\]
Prove that the functors $ F,G$ are naturally equivalent.
[i]i.e.[/i], there exists a natural transfromation
\[ \Phi : F \rightarrow G
\]
such that
\[ \Phi (A, C)
\]
is an isomorphism of abelian groups for all $ (A,C)$ in $ \ldots$.

[b]Note:[/b] $ (R\text{\minus{}Mod})^\circ$ is the category with the same object class as $ R$-Mod, and
\[ \text{Hom}_{(R\text{\minus{}Mod})^\circ} (A, A') \equal{} \text{Hom}_{R\text{\minus{}Mod}} (A, A') \equal{} \text{Hom}_R (A', A).
\]

$ Hom_S(B \otimes_R A,C) \to Hom_R(A,Hom_S(B,C)) , f \mapsto (a \mapsto (b \mapsto f(a \otimes b))$ is an isomorphism which can be checked by using the universal property of the tensor product. it's straight forward to verify its naturality.


Find the ratio of height h of a cushion on a snooker table to the radius r of a ball as shown in the figure, such that when the ball hits the cushion with a pure rolling motion it rebounds with a pure rolling motion. (Assume that the force exerted on the ball by the cushion is horizontal during the impact and that the ball hits the cushion normally)

Assume it goes back with $ v_{1}$ and $ \omega_{1}$
$ v_{1}\equal{} r\omega_{1}$
Just do impulse momentum theorem for the force imparted.
Likewise do for torque and angular momentum.


Ahh... Schools out on the 23rd of June...

haha...mine got out june 7

nowonder there aren't a lot of people posting

i got out June 3rd.  and i voted yes by accident because whed i read the question i thought it meant are you still in school or are you in college.

i feel really lucky...i got out 24 days ago :D

do you go to plublic schools of private?

finished it today......oh yes I'm happy! :D

Finished with school yesterday......might take a while for it all to settle it and I realize that I actually am on summer vacation.

wow, in louisiana school ended april, or may-ish. in california we just ended yesterday. YAY!

In Beijing, most schools end at mid-july. But my school is gonna let us do sth else so we shall end a bit earlier.

oops, i thought it meant that if you are still a student, so take one vote away from yes
summer started on may 19th for me :D

[quote="juicybooty911"]oops, i thought it meant that if you are still a student, so take one vote away from yes[/quote]
i did the same thing.

i got out of school over 20 days ago.

I got out may 25 or so....more than 20 days ago.  :D

Out on 24th of june, 5 school days left.

I have three exams to go. I don't actually go to school except for the exams though.

I'll be done in less than a week.

i got outta school about 20 days ago  :D

We got out of school on the 7th, but we start on August 18th. Grr.

school let out June 10th!!!!! this is to the other people who are still in school:rotfl:

I got out June 2nd. I thought that was late! Well, compared to the private schools here, that is late to get out. (I go to a public school)

I feel bad for all of you that still have school.

Without school you
can:   :icecream:      :football:      :ddr:      :nhl:      :trampoline:       :rotfl:      :love:         :play_ball:      :cleaning:      :evilgrin:      :harhar:      :juggle:     :starwars:      :pilot:                                 :stretcher:                 :surf:

Yeah, you get it.

YOU CAN DO WHATEVER YOU WANT!!!!!!!!!

public school, still have 2 more days of classes:( not that the class will do anything anyway and its not like i ever do anything:) then i also get to look forward to 3 exams, ooo so scary....english(wed), chem(fri) and calc(thurse), ewww what a joke. Yeah you may notice we have not school days off before exams, and some ppl have already done exams even though schools still going on...

[quote="G-UNIT"]I guess my school sucks, we get out on the 30th.[/quote]

Wow, that's REALLY late. I get out on the 23rd, which is already considered later than most schools, even in WA. 4 more days to go...

one school in my county doesn't get out until july 8!!!


geometry rectangle quadratics calculus trigonometry derivative algebra

A soccer player dribbles up the sideline of the field.  Any good player knows that he should shoot whenever the angle on the goal is as wide as possible.  If the dimensions of the field are 150 feet wide by 300 feet long, and the goal (blue in diagram) is 24 feet wide and centered on its side of the field, how far from the corner should our player be when he shoots?  What about for general field and goal sizes?  (You may assume that the field is always longer than it is wide for the general case.)

[hide]THe goal lies in the center. Let ABC the the triangle with BC=24. Let the O and P be the side of the rectangle which the goal lies on.with CP=150-24)/2=63. We want the max of angle BAC, or the max of tan(BAC).

AP=x. tan(CAP)=63/x, tan BAP=87/x

tan(BAC)=(tan(BAP)-tan(CAP))/1+tan(CAP)tan(BAP)

=(24x)/(x^2+(63)(87))=k

rewrite it as

kx^2-24x+(63)(87)=0

quadratic formula we have

x=(24+ :sqrt: (24^2-4k^2(63)(87))/2k

minimum value of k is when (24^2-4k^2(63)(87))=0

=4/ :sqrt:(609)

x=24/(8/ :sqrt:(609))=3:sqrt:(609).[/hide]

Any way I just had a soccer game today.

Oddly enough, this same problem has appeared on both a calculus test I took and a physics placement test I took (except on the latter, they didn't ask you to solve the problem, just to draw a picture!).  Anyway, as beta has already tackled the problem, I'll do the general case.

Suppose that the field has width w, length l, and goal size g.  For simplicity's sake, suppose that w < l and the goal is always on one of the w sides.  The other case is the same, but it's easier to label things this way.

Let's call the angle formed between the bottom edge and the segment drawn to the far post A; similarly, call the angle between the bottom edge and the segment drawn to the near post B.  Then the desired angle is A - B.

Finally, suppose that the player is x units from the endline (the line from corner to corner that passes through the goal).

The length from the near corner to the near post is (w-g)/2.  The length from the near corner to the far post is (w+g)/2.  So

[tex]\displaystyle \tan A = \frac{(w+g)/2}{x} = \frac{w+g}{2x}[/tex]
[tex]\displaystyle \tan B = \frac{(w-g)/2}{x} = \frac{w-g}{2x}[/tex]

By the difference of tangents formula,

[tex]\displaystyle \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}[/tex]
[tex]\displaystyle = \frac{(w+g)/(2x) + (w-g)/(2x)}{1 + (w+g)/(2x) * (w-g)/(2x)}[/tex]
[tex]\displaystyle = \frac{w/x}{1 + (w^2 - g^2)/(4x^2)}[/tex]
[tex]\displaystyle = \frac{4xw}{4x^2 + w^2 - g^2}[/tex]

Since [tex]0 < B < A < \frac{\pi}{2}[/tex], it follows that A-B is in the first quadrant.  So A - B is maximized when tan(A-B) is maximized.

But tan(A-B) is maximized when its derivative is 0.  (Note that w and g are constants, not variables.)

So
[tex]\displaystyle \frac{d}{dx} \frac{4wx}{4x^2 + w^2 - g^2} = 0[/tex]
[tex]\displaystyle \frac{4w(4x^2 + w^2 - g^2) - 4wx(8x)}{(4x^2 + w^2 - g^2)^2} = 0[/tex]
[tex]4w(4x^2 + w^2 - g^2) - 4wx(8x) = 0[/tex]
[tex]16wx^2 + 4w^3 - 4wg^2 - 32wx^2 = 0[/tex]
[tex]16wx^2 = 4w^3 - 4wg^2[/tex]
[tex]x^2 = w^2/4 - g^2/4[/tex]
[tex]x = \frac{\sqrt{w^2 - g^2}}{2}[/tex]

Jenerahlized!

(Check: w = 150, g = 24 yields [tex]x = 3 \sqrt{609}[/tex].  That's comforting.)

[Edit: Fixed TeX problem]

Good job both -- this problem also has a really beautiful geometric solution which does not rely on trigonometry.  I'd love to see if anyone could find it.

A big helping hint towards getting it: [hide]You'll need a circle somewhere.[/hide]

[hide]So what we wanted is a circle with a chord length 24 and tangent to the longer side of the field. The rest we can figure out with Pathygorean Theorem.[/hide]

Correct - how will you justify it though?

[hide]Let the chord be AB. Let point P lies on the perpendicular bisector of AB and on the circle.The closer point P is to AB, the greater angle APB we have. Now put everything into the circle. The smallest circle we can possibly have is when it's tangent to the longer side of the field, let C be the point of tangency. Any inscribed angle will be congruent to angle ACB.  AngleAPB :cong: angleACB. Thus that's the smallest angle possible of APB. [/hide]


Copied directly from AoPSWiki. Can someone explain this problem? Sorry if this has been posted somewhere else...

For all positive integers $ x$, let
$ f(x) \equal{} \begin{cases}1 & \rm{if \ x \equal{} 1} \\
\frac x{10} & \rm{ if \ x \ is \ divisible \ by \ 10} \\
x \plus{} 1 & \rm{otherwise}\end{cases}$
and define a sequence as follows: $ x_1 \equal{} x$ and $ x_{n \plus{} 1} \equal{} f(x_n)$ for all positive integers $ n$. Let $ d(x)$ be the smallest n such that $ x_n \equal{} 1$. (For example, $ d(100) \equal{} 3$ and $ d(87) \equal{} 7$.) Let $ m$ be the number of positive integers $ x$ such that $ d(x) \equal{} 20$. Find the sum of the distinct prime factors of $ m$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=378142#378142

The "Contests" link is there for a reason.


1. A scarlet compound (A) is treated with con nitric acid to give a chocolate

 brown precipitate (B).  The precipitate is filtered and the filtrate is neutralised

with NaOH.  Addition of KI to the resulting solution gives a yellow precipitate

(C).  The precipitate (B) on warming with con nitric acid in the presence of

Mn(NO3)2 produces a pink coloured solution due to the formation of (D).

Explain the reactions.

2. A certain compound X shows the following reactions:  When KI is added to an

aqueous suspension of X containing acetic acid, iodine is liberated.  When CO2

is passed through an aqueous suspension of X the turbidity transforms to a

precipitate.  When a paste of X in water is heated with ethyl alcohol a product

of anaesthetic use is obtained.  Explain.

I have a small suspicion Pardesi. Dont you mind it if it isnt true  :D
Do you have a book that contains exactly these questions?  :D   :D

no i reasearched on the compounds and found out their colours nad then smelt each of them...actually while doing lll these i fell down  :rotfl:  and spilt the substance X on me whne it was heated in ethyl alcohol(actually god told me to heat iit in ethyl alcohol :P ) and bang my wound got perfect...
and then i answered Rakesh's question :huh:

Your answers for the 1. except D are correct and your answer to 2. is also right but
still u should give chances to other people to atlest try those problems.
You shd have hidden ur answers.

some write D as $ \ce{HMnO4}$ or something  like that ...it's actually the hydrated form

i ahve actually seen hundreds(quite magnified) of answers of that

so pardesi you have wrong nved  :D  :D  :D

ofcourse rohit how do u else naswer thses unique questions :wink:

Pls dont degrade urselves by nvying chemistry It is highly logical

yes chemistry(some parts ) are logical...so please tell me how to logically attack question 1 and more specifically 2 :wink:

These 2 questions are tough without knowing the facts i agree  :D   :D   :)

Yeah I accept to a certain extent u shd posses knowledge of colours but
I approach every question by 1-2 hints that I know that can lead me into the answer.
I didnt know anything in the 1. when I glanced it at first but moment PbO2 struck and
with the help of IIT notes I was able to solve.

for ! there is enuf hint with scarlet colour and yellow ppt. with KI


A，B，C将圆O分成三段弧，X是弧AB上的一个动点，$I_1$,$I_2$分别是三角形CAX和三角形CBX的内心。求证：三角形$XI_1I_2$的外接圆交圆O于X之外的一个定点

[quote="shobber"][quote="mecrazywong"]Let D be the midpoint of arc AB [b]containing[/b] C. Then the circumcircle of $\triangle XI_1I_2$ passes through $DI\cap O$.[/quote]
What is $I$?[/quote]
The incenter of $\triangle ABC$.

现在的关键是找出M点是什么！！
M点到底是什么！
这直接影响证明的进度！

I have a solution now.http://www.mathlinks.ro/Forum/viewtopic.php?t=19777

[quote="georgew_bush"]现在的关键是找出M点是什么！！
M点到底是什么！
这直接影响证明的进度！[/quote]
有$M$么?


If you mean the China MO,then find it in the Chinese Community.We've have it in Chinese.


ratio
function
quadratics
geometry
3D geometry
sphere
calculus

Albatross and his friend Frankinfueter are visiting a 560 foot building, than Frankinfueter pushes Albatross off the building. If it takes Albatross 50 seconds to hit the ground, how long will it take falling off a 784 foot buiding?

hehe Frankinfueter is evil

[hide](1/2)*g*t^2 = d is the equation used
where g is the gravitational constant (for earth it is around 9.8 meters/second^2, but its different in this prob...), t is time, and d is distance fallen (in meters)

Plugging in the values of 170.688 m (560 ft) for d and 50 for t lets us find that g = 0.1365504  m/s^2

Now we plug that back into the equation and use d = 238.9632 m (784 ft.) and g = 0.1365504 m/s^2, then solving for t

t = 59.161 sec
[/hide]

[quote="RC-7th"]Kinda hard for getting started... it assumes you know the formula for gravitational pull.[/quote]

I don't think it's hard.  If you know the formula, then it's pretty easy, but if you don't, you won't know what to do.

So was the problem supossed to be set up like this:

$\displaystyle\frac{50}{560} = \displaystyle\frac{x}{784}$

or the way dinoman solved it?

It can't be a simple proportion since the function is a quadratic.[hide]

With a 560-foot building, Albatross' height is given by: $y = 560 - \frac{1}{2}gt^2$
   where $g$ is the acceleration due to gravity (on that planet).

When $t = 50,\;y = 0\;\;\Rightarrow\;\;0 = 560 - \frac{1}{2}g(50^2)\;\;\Rightarrow\;\;g = 0.448\text{ ft/sec}^2$
[It must be an asteroid.]

With a 784-foot building, Albatross' height is: $y = 784 - 0.224t^2$

When $y = 0:\;\;0 = 784 - 0.224t^2\;\;\Rightarrow\;\;t^2 = 3500$

Therefore: $t = \sqrt{3500} \approx 59.16\text{ seconds.}$[/hide]

I'm pretty sure you won't find problems like this on any mathematics competition without them telling you gravitational acceleration. Otherwise, it requires specific previous knowledge that actually encompasses physics.

[quote="Soroban"]It can't be a simple proportion since the function is a quadratic.[/quote]
The discription of how to solve it made it seem like it was -

[quote="A+MATH"]Actually It wasn't suppose to be hard because I gave you how long it took for Albatross to fall off of one building. So you were suppose to use that and find a ratio to find the second time.[/quote]

Sorry about that I forgot every thing I learned in physics class right there. I shouldn't have posted a problem like that on Getting Started. Sorry  :blush:

[quote="A+MATH"]Sorry about that I forgot every thing I learned in physics class right there. I shouldn't have posted a problem like that on Getting Started.[/quote]

Very well then, I'll "promote" this thread to Pre-Olympiad, and see what reasonable assumptions people make about terminal velocity to solve the problem.

what.the.heck?  Why do quadratic equations suddenly become pre-olympiad material?

If limiting velocity is relevant, then air resistance is essentially determined by an experimentally derived constant (and the shape of the body, which is usually assumed to be a sphere), and the equations of motion are governed by a (trivial to derive) differential equation.  In that case, it's still not a pre-olympiad problem.

I think you need to be a lot more careful when moving problems around.

I'm not really sure where to put this thread, but its really more of a physics problem than maths - so I'll move it to other problem solving topics.

Yeah, I guess that admixture of physics in the problem means it belongs here in Other Problem Solving Topics.

But, aside to blahblahblah, what is your conception of a proper math level for Pre-Olympiad Topics challenge problems?

Something like this is a good pre-olympiad problem in my view.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20528

Many of the problems in pre-olympiad currently are a bit on the difficult side, but I think it's better that they're stuck there than in the advanced section, because then the vast majority of posters will never get to look at them.

As for this problem, however, it's either a problem on quadratic equations, which means that it's little more than a glorified getting started problem, or a calculus-based mechanics problem, in which case it's hardly a pre-olympiad problem.


group theory
abstract algebra
superior algebra
superior algebra solved

If $ H$ and $ K$ are normal subgroups of a group $ G$ with $ HK\equal{}G$, prove that

$ G/(H \cap K)\simeq (G/H) \times (G/K)$.

[quote="Fraenkel"]If $ H$ and $ K$ are normal subgroups of a group $ G$ with $ HK \equal{} G$, prove that

$ G/(H \cap K)\simeq (G/H) \times (G/K)$.[/quote]

I believe the way to go is to use the first isomorphism theorem.

It is easy to prove that $ (H \cap K)$ is the kernal for the map $ x\mapsto (xH,xK)$

Now I dont fully understand how to show that $ (G/H) \times (G/K)$ is the image of the map.

By second isomorphism theorem: $ G/H \simeq K/(H \cap K)$ and $ G/K \simeq H/(H \cap K)$

Now, $ K/(H \cap K)$ and $ H/(H \cap K)$ are normal subgroups of $ G/(H \cap K)$ with trivial interesection and they generate $ G/(H \cap K)$. Then $ G/(H \cap K)$ is the inner direct product of those two subgroups.

That is,
$ G/(H \cap K) \simeq K/(H \cap K) \times H/(H \cap K) \simeq G/H \times G/K$

[quote="Fraenkel"]
Now I dont fully understand how to show that $ (G/H) \times (G/K)$ is the image of the map.[/quote]

we have $ HK \equal{} G \equal{} G^{ \minus{} 1} \equal{} KH$. thus, every element in $ G/H \times G/K$ has the form $ (xH,yK)$ with $ x \in K, y \in H$. then $ xy (H \cap K)$ is easily seen to be a preimage.

a generalization: if $ H,K$ are normal subgroups of $ G$, then

$ G/(H \cap K) \cong G/H \times_{G/(HK)} G/K$

[quote="-oo-"]
..with $ x \in K, y \in H$. then $ xy (H \cap K)$ is easily seen to be a preimage.
[/quote]

are you assuming that the group is commutative?

-oo- is skipping a part of the argument here.

If $ x \in H$ and $ y \in K$:
Look at $ xyx^{ \minus{} 1}y^{ \minus{} 1} \equal{} (xyx^{ \minus{} 1})y^{ \minus{} 1} \equal{} x(yx^{ \minus{} 1}y^{ \minus{} 1})$
Because of normality and the way I grouped things, we see that $ xyx^{ \minus{} 1}y^{ \minus{} 1} \in H \cap K$
So indeed $ xy(H \cap K) \equal{} yx(H \cap K)$.

In my proof, this argument is hidden in the fact that I use a special case of this theorem:

If $ H$ and $ K$ are normal subgroups of $ G$, $ H \cap K$ is trivial and $ H,K$ generate $ G$, then $ G$ is the inner direct product of $ H$ and $ K$, that is $ G \simeq H \times K$ under the most natural isomorphism.

And I really think first proving this theorem, and then using my argument is the way that gives most insight to this problem. I don't like using the first isomorphism theorem when you don't get what is going on

[quote="Kalle"]
And I really think first proving this theorem, and then using my argument is the way that gives most insight to this problem. I don't like using the first isomorphism theorem when you don't get what is going on[/quote]

Thank you. I share your opinion, for I did not get what was going on and hated the fact that I would have to simply apply some theorem from the book.

Would this be valid?

Consider the homomorphism

\[ \phi(hk) \equal{} {Hk} \times {Kh}\]

This has kernel $ H\cap K$.


Gently down the stream.

A man in a rowboat takes 5 hours to travel from point A to point B going upstream. The man takes 4 hours to go from point B to point A. Neglecting the current of the water, how long does it take for the man to get from point B to point A?

Is this a trick question?  I got [hide]4 hours[/hide], only becuase the answer is stated in the problem, I think.

h_s_potter2002 wrote:Is this a traick question?  I got [hide]4 hours[/hide], only becuase the answer is stated in the problem, I think.

When he said "The man takes 4 hours to go from point B to point A," it's implied that he will be travelling downstream.  The problem is looking for the rowing speed (without the current).

[hide]The amount of time added by going upstream with the current should be subtracted when going downstream with the current.

x = amount of time added or subtracted by going from point A to point B or point B to point A.

2x = 5-4

2x = 1

x = 0.5

Since you would subtract this number to get the amount of time it would take to get from point B to point A, you have to do the opposite with the number we have now (4).

4+0.5 = 4.5 hours.[/hide]

[color=white]I just took the average of 4 and 5 hours, since the speed is a constant.[/color]

[quote="mdean09"]If you choose a distance at random(sort of) like 80 miles.  Then if you finish in 4hrs your rate is 20 mph.  If you finish in 5 hrs then your rate is 16mph.  Since the current directly affects the rate most equally, you could say the rate without the current is the average of the two, in this case 18mph.

So you travel 80 miles at 18 mph which would take 4 4/9 hours.

Try some other distances and you will find the same time.  Check 60 miles for example.[/quote]

Yes! We have a correct answer. True, speed is constant, but you CANNOT average speed together. Here's an example.

A family is driving to the beach. The first half of the trip has decent traffic, so the overall average speed in that given time is 30 miles per hour. Coming back, there's a bit more traffic, making for an average speed of 20 miles per hour. What was the average speed for the whole trip?

But wasn't this quqstion asking the speed without the current and not the average speed?

Yes, but to figure out the time, we need to know average speed.

[quote="Treething"]Note:

Formula for averaging speeds

a + b / ab[/quote]
You meant harmonic mean right, i.e. $\frac{2ab}{a+b}$?

[quote="Treething"]Note:

Formula for averaging speeds

a + b / ab[/quote]

what is a and b? rate?

edit:whoops... obvious  :blush:

say the distance is 60 miles
that means it took 2 hours there and 3 hours back
that means in 5 hours they traveled 120 miles
that means 24 miles per hour
PS
how do u do the hidden content thing?

When you are posting, you should see a row of buttons above the area in which you write your post. At the end, just highlight what you want to hide and then click the hide button. Either that or type hide in brackets at the beginning of what you want to hide, and then /hide in brackets at the end of what you want to hide.

brackets = square brackets.  [ ].   So type [/hide] at the end and [hide] at the beginning and it will turn out hidden.


find the greatest value of a + 1/(b+1/c) + d + 1/(e+1/f) + g + 1/(h + 1/i), where each letter represents a different non-zero digit.

Do you mean: Find the greatest value of $ a\plus{}\frac{1}{b\plus{}\frac{1}{c}}\plus{}d\plus{}\frac{1}{e\plus{}\frac{1}{f}}\plus{}g\plus{}\frac{1}{h\plus{}\frac{1}{i}}$?

So we have $ a\plus{}\frac{1}{b\plus{}\frac{1}{c}}\plus{}d\plus{}\frac{1}{e\plus{}\frac{1}{f}}\plus{}g\plus{}\frac{1}{h\plus{}\frac{1}{i}}\equal{}a\plus{}\frac{1}{b}\plus{}c\plus{}d\plus{}\frac{1}{e}\plus{}f\plus{}g\plus{}\frac{1}{h}\plus{}i$.  By "digit" I'm assuming you mean integers $ 1\minus{}9$.  So the a, c, d, f, g, and i should be as high as possible.  Likewise, b, e, and h should be as small as possible.  Therefore, our largest total possible will be $ 9\plus{}\frac{1}{1}\plus{}8\plus{}7\plus{}\frac{1}{2}\plus{}6\plus{}5\plus{}\frac{1}{3}\plus{}4\equal{}\boxed{40\frac{5}{6}}$.

[quote="Stank"]$ a \plus{} \frac {1}{b \plus{} \frac {1}{c}} \plus{} d \plus{} \frac {1}{e \plus{} \frac {1}{f}} \plus{} g \plus{} \frac {1}{h \plus{} \frac {1}{i}} \equal{} a \plus{} \frac {1}{b} \plus{} c \plus{} d \plus{} \frac {1}{e} \plus{} f \plus{} g \plus{} \frac {1}{h} \plus{} i$[/quote]
I'm sorry, but this step isn't valid. You can't distribute across the denominator.

We can just simplify that ugly expression to get a much nicer equivalent:

(Just combining the terms of each of the denominators, we get)

$ a \plus{} \frac {1}{\frac {bc \plus{} 1}{c}} \plus{} d \plus{} \frac {1}{\frac {ef \plus{} 1}{f}} \plus{} g \plus{} \frac {1}{\frac {hi \plus{} 1}{i}} \implies a \plus{} \frac {c}{bc \plus{} 1} \plus{} d \plus{} \frac {f}{ef \plus{} 1} \plus{} g \plus{} \frac {i}{hi \plus{} 1}.$

Note that our only choices for digits are $ \{1,2,3,4,5,6,7,8,9\},$ and there are also nine letters, so every digit is used once.

We want $ a,d,$ and $ g$ to be as large as possible. Thus, $ (a,d,g) \equal{} (7,8,9)$ in some order (doesn't matter which). We clearly want to minimize $ b,e,$ and $ h$ to keep the denominators as small as possible. Thus, $ (b,e,h) \equal{} (1,2,3),$ in some order (again, order doesn't matter). So, let $ b \equal{} 1, e \equal{} 2,$ and $ h \equal{} 3.$

We now have
\[ 24 \plus{} \frac {c}{c \plus{} 1} \plus{} \frac {f}{2f \plus{} 1} \plus{} \frac {i}{3i \plus{} 1}.\]
There are only three values left, which are $ 4,5,6.$ We want to maximize each of the fractions as much as possible. $ \text{max}\left(\frac {c}{c \plus{} 1}\right)$ occurs when $ c \equal{} 6,$ at $ \frac {6}{7},$ which is clearly higher than $ 6$ substituted into any other of the fractions. Then we have $ 4$ and $ 5$ left. You can just test those for each and determine what is the largest. And add all the terms up.


geometry
calculus
function
floor function
ARML

Why is it me that keeps posting contest topics...?

Ok that thing was NOT high quality at all. At least 3 problems were faulty (in the Algebra II section). So I got a 45, but wasn't called in the individual prizes  :|

lol algebra 2 was retarded. i ended up getting 39 or 40, but shouldv'e gotten 43 because i didn't put an answer for the faulty questions..

[quote="james4l"]Why is it me that keeps posting contest topics...?[/quote]
Because we are all waiting for you to do it?  :lol:

I wasn't really happy with Discovery Quest. I read through the questions once in the beginning and was like "I'm pretty sure I know how to solve all of these." so the contest just became who can make the fewest errors in their solution.  :|

(For example, one of the questions asked you to prove $ [ABC] \equal{} \frac {abc}{4R}$...)

I believe that the algII test was extremely EXTREMELY screwed up.
look at the 2006 or 2007 sample problems ... some of them are the same as the ones in this years competition.
 :huh:  :huh:
basically.. the test failed.  :mad:

hmm...so do I
those problems were like the exact same as last year for alg 2
I really really should have studied last years test >____<
so basically if you studied last years test and remember how to do everything then you do well
but I still failed :(

It was...quite...disorganized.  :rotfl:

BTW james4l you have 1764 posts! That is a very special number, I forgot why, I just remember it is supposed to be a very special number.

EDIT: Oh, now I remember. It's special because it's square root is special.

It was fun and I'm proud of Paly, but I was definitely very disappointed by the Discovery Quest. I've never really expected high-quality questions from the field day, but making it so easy that they couldn't even decide on a single winner? Sigh. Though maybe I should be happy since if it had been harder Marianna (and everyone else in there that I knew) probably would've owned me.

I took the alg 2 test, and a lot of the first part (1-25) were the same as the last few years... I did a few of the questions just by remembering the answers.
BTW does anyone know when the results will be up? I missed the award ceremony because I left early.

does anyone have the results for the geometry section? I just want like top 5 individual and team, or just part of it if cause i left early to go to my psychologist

calculus exam was just like wtf what is all this stuff with random functions and properties and blah

like, this function: $ \dfrac{d}{dx} \sum_{i=0}^{\abs{\lfloor{x}} i$
Isn't this just 0 for all non-integer x, and undef otherwise?
that makes it even, odd, discrete, periodic, non-decreasing, and a bunch of other stuff.

again, wtf.

Think you meant to have
$ \frac{d}{dx} \sum_{i\equal{}0}^{| \lfloor x \rfloor |} i$

I didn't go  :)
How was this compared to ARML tryouts this year?

This makes ARML tryouts look good... at least the test portion. :roll:

Ehh the first two pages of problems were NOT fun. Other than that, it was ok.

I got a free rubik's magic :)
Now I have 13 -- broken 9.


geometry
ratio
inequalities
analytic geometry
linear algebra
matrix
geometry unsolved

In triangle $ABC$, the medians and bisectors corresponding to sides $BC$, $CA$, $AB$ are $m_a$, $m_b$, $m_c$ and $w_a$, $w_b$, $w_c$ respectively. $P=w_a \cap m_b$, $Q=w_b \cap m_c$, $R=w_c \cap m_a$. Denote the areas of triangle $ABC$ and $PQR$ by $F_1$ and $F_2$ respectively. Find the least positive constant $m$ such that $\frac{F_1}{F_2}<m$ holds for any $\triangle{ABC}$.

[color=indigo]I would guess it's [i]"...find the least $ m$ such that $ \frac{F_2}{F_1} < m$..."[/i] because there is no such $ m$ as it is stated.[/color]

[i]Areal coordinates:[/i] any point on $ w_a$ has coordinates $ \left(\tfrac{x}{x\plus{}b\plus{}c}, \tfrac{b}{x\plus{}b\plus{}c}, \tfrac{c}{x\plus{}b\plus{}c}\right)$, where $ x$ is positive; any point on $ m_b$ has coordinates $ \left(\tfrac{1}{2\plus{}y}, \tfrac{y}{2\plus{}y}, \tfrac{1}{2\plus{}y}\right)$, where $ y$ is positive. Using this we can find that the intersection $ w_a \cap m_b$ has areal coordinates $ \left(\tfrac{c}{b\plus{}2c}, \tfrac{b}{b\plus{}2c}, \tfrac{c}{b\plus{}2c}\right)$. Similarly, $ w_b \cap m_c$ has areal coordinates $ \left(\tfrac{a}{c\plus{}2a}, \tfrac{a}{c\plus{}2a}, \tfrac{c}{c\plus{}2a}\right)$ and $ w_c \cap m_a$ has areal coordinates $ \left(\tfrac{a}{a\plus{}2b}, \tfrac{b}{a\plus{}2b}, \tfrac{b}{a\plus{}2b}\right)$.

Hence
\[ \frac{F_2}{F_1} \equal{} \left|\begin{matrix} \tfrac{c}{b\plus{}2c} & \tfrac{b}{b\plus{}2c} & \tfrac{c}{b\plus{}2c} \\ \tfrac{a}{c\plus{}2a} & \tfrac{a}{c\plus{}2a} & \tfrac{c}{c\plus{}2a} \\ \tfrac{a}{a\plus{}2b} & \tfrac{b}{a\plus{}2b} & \tfrac{b}{a\plus{}2b} \end{matrix}\right| \equal{} \frac{ab^2\plus{}bc^2\plus{}ca^2\minus{}3abc}{(a\plus{}2b)(b\plus{}2c)(c\plus{}2a)}\]
I would guess that the maximum of the above expression occurs when $ a,b\equal{}x, c\to 2x$, which gives $ m \equal{} \tfrac{1}{60}$? Otherwise, it is just tedious taking of derivatives.

Actually, $ a,b=x, c\to 0$ gives $m=\frac{1}{6}$, which turns out to be the maximum.


geometry
analytic geometry
parallelogram
graphing lines
slope

Prove that the points $ (2,-1), (0,2),(2,3) and (4,0)$ are the coordinates of the vertices of a parallelogram and find the angle between its diagonals. It became a problem for me when I got the slope of one diagonal equals to infinity and could not solve it using the formula $ tan(theta)= \frac{m1-m2}{1+m1.m2}$ where m1 and m2 are slopes of diagonals and theta is the angle between diagonals. Its answer is $ -tan\inv{-1}\frac{1}{2} - \frac{pi}{2}$.Please help.Thank you so much.


Given$ a,b,c>0$ and $ abc\equal{}1$ Prove that:
$ \frac{a^2\plus{}2ab\plus{}3}{(a^2\plus{}a\plus{}1)^2}\plus{}\frac{b^2\plus{}2bc\plus{}3}{(b^2\plus{}b\plus{}1)^2}\plus{}\frac{c^2\plus{}2ca\plus{}3}{(c^2\plus{}c\plus{}1)^2} \ge\ 2$ :)

It's obviously trues by Am-Gm because: $ \sum\ \frac{1}{a^2\plus{}a\plus{}1} \geq\ 1$  :wink:
@quykhtn-qa1: [hide]Cach' che' cua? em rat hay nhung qua? thuc. anh ko ko thjk lam` dang. nay` 1 ti' nao`, no ko co nhieu` y nghia~ doi voi nhieu` ng`, va` cach' che' cua? em deu` dua. vao` cac bdt cu~ va`yeu'. Em co the? thay anh da~ tang. em 1 bai` tuong tu, cuc. de~ nhung kho' nghi~ ra hay ko thi` chac' em tu. biet roi` (va` chac em cung~ biet anh co' the? che' vai` nghin` bai` nhu the' nhung chac chan chi? lam` mat thoi gian cua? ng` doc. Anh hi vong. n~ bai` toan lan` sau em "tang." anh se~ ko co' dang. $ abc\equal{}1$ kieu? nay` nua~  :) .[/hide]

[quote="nguoivn"]It's obviously trues by Am-Gm because: $ \sum\ \frac {1}{a^2 \plus{} a \plus{} 1} \geq\ 1$  :wink:
@quykhtn-qa1: [hide]Cach' che' cua? em rat hay nhung qua? thuc. anh ko ko thjk lam` dang. nay` 1 ti' nao`, no ko co nhieu` y nghia~ doi voi nhieu` ng`, va` cach' che' cua? em deu` dua. vao` cac bdt cu~ va`yeu'. Em co the? thay anh da~ tang. em 1 bai` tuong tu, cuc. de~ nhung kho' nghi~ ra hay ko thi` chac' em tu. biet roi` (va` chac em cung~ biet anh co' the? che' vai` nghin` bai` nhu the' nhung chac chan chi? lam` mat thoi gian cua? ng` doc. Anh hi vong. n~ bai` toan lan` sau em "tang." anh se~ ko co' dang. $ abc \equal{} 1$ kieu? nay` nua~  :) .[/hide][/quote]
Thanks.  :blush:


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I have no idea where this belongs to...
http://arxiv.org/pdf/math.GM/0507347

I cant believe what this article says..... Anyway, i havent finished with it but it´s the kind of thing difficult to decide if it is correct or no.

Are you asking me if something which says that mathematics are wrong is true?.... ¡NO!... Somewhere there must be some mistake. But i have no idea where, and i didnt search too much... but most places i read about, people couldnt find any mistake.

 Thats why i post it here, so some clever mind, free of mental slavery can find it.

It's not obvious crankery, but it seems to be following a method that can't possibly work.

Who would have time to read through this densely written document and find out which parts are nonsense?

Wow...

Well, to be honest, I think that all of us at ML could try and analyze it together ;)

7 months later...... what's the conclusion about it?

In a month I shall analise this article with some teachers and researchers from University of Bucharest and Institut of Mathematics. This is a challange. :)

[quote="Cezar Lupu"]In a month I shall analise this article with some teachers and researchers from University of Bucharest nad Institut of Mathematics. This is a challange. :)[/quote]

Please... when you know sth, please let us know

[quote="perfect_radio"]Will you really do that?  :D[/quote]

Of course. After I'll be finished with my exams, I will talk with some strong mathematicians about this attempt of proof. :)


Find $f: \mathbb{R}\to\mathbb{R}$ satisfies:
\[ f(xf(y)+x)=xy+f(x)  \]

molina, as far as I can tell your argument uses at least 2 major tacit assumptions (the invertibility and integrability of functions) which you have no basis for making.  Also, I have no idea where your second line comes from.  And, you're missing at least one solution.  Perhaps a more detailed explanation is in order?

[quote="molina"]I used the fact :

$\large{f(x)\,=\,\int\,f^{\prime}(x)\,dx}$,

but

$\large{f^{\prime}(Cx)\,=\,\frac{f^{-1}(C)}{C}\,=\,k}$

Then

$\large{f(xC)\,=\,\int\,k\,dx\,=\,kx+P}$ .

Is it wrong ?[/quote]

Well it's wrong in the sense that you don't know if the function is differentiable or integrable. You also don't know that it has an inverse, so you can't really use those facts.

I have proved that $f(0) = 0$ , $f$ is one to one and has range the set of all real numbers.Take this as ready , if this could help.I have also proved that

     $f(x+a) = f(x) + a , a = f(1) + 1 , f(a) = a$

  babis


Princeton college Yale Stanford Harvard MIT geometry

why are American Universities on top of all others? ... strange question isn't it ... let's hear some opinions ..

I heard they have a lot of money, they offer more money than any university in the world for students phd/post doc, junior-senior professors, and some others facilities well equiped laboratories...
In one word a lot of advantages.
They also also attract the genius from any country.
Their budgets for research  is very high.

An example Clay Mathematical offer 7 million $ prize for 7 seven problems
http://www.claymath.org/millennium/

what puzzles me is why the UE doesn't have such high research budgets ? I know that Oxford and Cambridge don't offer full paid scholarships to anyone, so most of the good students do not consider them .. :?

US Universities have high tuitions especially private schools.  Private universities and public universities receive endowments from the US Government and from past graduates.  The universities divide up the money through all their programs and their necessities.  Any money left over is for "scholars" coming to the university or concurrently staying at the university.  One has to be well educated and have a good GPA, SAT score, SAT II scores, and good classes to receive scholarships.  They aren't given to everybody you know.

For your information . . . the ivyies (Princeton, Yale, Stanford, Havard, etc . . .) and the techs (MIT, CALTech, Harvey Mudd, etc . . .) do not give any academic scholarships to students (similar to Cambridge university and Oxford University).  The last person I know of that got a scholarship (full ride-> everything paid for) to MIT for academics was a kid that lives in my area named Tomasz Macura.  I won't mention all his accolades but he is a rising "genius" to the public eye.

Here is his website for those that are interested ->

http://userpages.umbc.edu/~tmacur1/

My friend is a "genius-type" also who is at the moment studying at MIT; however, he did not receive any scholarship because you got to understand the difference between him and Tomasz Macura.  Tomasz Macura is completing his undergraduate at just 15 years of age compared to my friend who is just starting his undergraduate at MIT.  Now you know why MIT offered him the scholarship . . . if he can handle undergrad math at age 15 . . . imagine what he can handle when he is 20 etc . . .

my ranting is over . . . I got a big CHEM test tommorow

Well, the basic education in Europe is quite good.

We must see that many scientists are from Europe.

We can say that American basic education is not successful, China is now learning the style of basic education in US, but she has no such an environment like in US, the it is getting worse and worse here now in China. European style is much more suitable.

American education make a great many of enterprisers. US has the best economic. Many scientists moved to US, at least, during and after the world war, many leading European scientistsmoved to US.

Romania is trying to hold on  to its "brain power" (as do other countries), but no country can really dream of competing with the US when it comes to the conditions its universities offer to gifted students.

well if it's all about economics, the EU is almost equal in terms of economy with US. I was wondering why do not they attend more to their students? (grants, etc.)?

Indeed Oxford gives academical scholarship, but they do not give need-base scholarship. MIT, and ivyleagues in general, offer free-tuition, basically giving you aprox. 30,000$ / year, as for Oxford which would give you only 5,000 pounds / year!

maybe they want to work in US?
Does US provides a higher salary?
Or US provides more professor positions?
Does EU still hold the tradition that every department can have at most a few professors?

Well actually in Romania we have a lot of teachers for each department, but most of them have no papers written.

Actually the ex. minister of health had was his univ. professor degree by copying 7 volumes of medicine from French and signing them with his name (!!). Still he is a full-professor, no body has done anything about it. It's a shame for the universitary education in Romania. And this happens a lot.

But I wonder ... is US gives all that money to research, thus keeping them ahead with technological advances, why doesn't EU do that? I mean they too want to be ahead with technology ... but as long as their best students and teachers go to MIT and Harvard ... that's not going to happen, right?

[quote="Valentin Vornicu"]Actually the ex. minister of health had was his univ. professor degree by copying 7 volumes of medicine from French and signing them with his name (!!). Still he is a full-professor, no body has done anything about it. It's a shame for the universitary education in Romania. And this happens a lot. [/quote]
Corruption in Romania..  :D The same thing also happens in China, but not the minister of a government division, but in some universities or academies.
[quote="Valentin Vornicu"]
but as long as their best students and teachers go to MIT and Harvard ... that's not going to happen, right?[/quote]
well, almost the best students go aboard. Many of them go to the US.

There are also in Europe universities which can compete with Harvard,MIT,Princeton etc.:Cambridge,Oxford (as mentioned),Paris
I think that the french universities are the most prestigious in Europe regarding math

France is a land of mathematicians. The years is quite long. It began with Viète, then continued with Descartes, Fermat, Lagrange, Laplace, Galois ... and culminated with one of the greatest mathematicians of all time: Henri Poincar, probably the last to have had complete knowledge of mathematics and their applications.

France has the math tradition.

In the early years, Ecole Polytechnique educated the top-ranking mathematicians and engineers in Europe, no one can bear comparison with the mathematicians there except Gauss. Now, I think, Paris IV and Paris VI are very good at math.

Napoleon has many mathematician friends. It is said that during the time when allies would attack Paris, some Ecole Polytechnique students wanted to join in the army but Napoleon denied their request.

I don't know why there are so many French mathematicians. And in the recent years, many French researchers left France for USA but no mathematicians, conversely, many US mathematicians went to France. Maybe it has something to do with the policy there. Maybe there's a high salary, good work condition and math atmosphere?

[quote] Maybe it has something to do with the policy there. Maybe there's a high salary, good work condition and math atmosphere?[/quote]I know Romanian mathematicians who made Ecole Normale (ex. IMO team members) and decided to go away because those 3 traits you are talking about there do not exist :D

well, the economics of EU stagnates in recent years.
It is said that France cut the budget of sci & tech. Thousands of scientists protest against it.

I don't know much about this, but it looks to me like Romania borrowed a lot from the French education system. First of all, we have something called Bacalaureat (a final exam at the end of high school). Second of all, we have Politehnica, and I've heard some rumors about plans for something called Scoala Normala Superioara (I'm sure you can see the similarities between that and Ecole Normale Superieure :)).

Definately the money in the American University system contributes to their success internationally.  It is easier to get research grants in US universities, and that is probably why they draw a large number of international students.  I attend Cornell University, and I have a lot of international friends, and some of them get money from their governments to go to American schools, while others go because of the budgets.

[quote="zscool"] [..] I attend Cornell University, and I have a lot of international friends, and some of them get money from their governments to go to American schools [..][/quote] incredible. In Romania they don't give you money for staying, I refuse to think that they would give me money to go abroad :)

It is really hard for people outside of US to get money for research due to budget problems and the competition for it is even higher. Last month I was researching for some of the fellowships that they were offering and I bet US has the highest number of fellowships than any other country, not to mention that NSF (National Science Foundation) holds more than 3000 memberships each year.


Ed's calculator is broken. The ``+'' key works correctly, but whenever he tries to compute $ a \minus{} b$, the calculator performs $ b \minus{} a$ instead. What answer does his calculator compute for the expression $ ((8 \minus{} 3) \plus{} 2) \minus{} 4$?

The answer is correct. The problem is ((8 - 3) + 2) - 4, but since his calculator is broken it is really 4-((3-8)+2), or 4-(-3), or 7.

[quote="AIME15"]8-3 becomes 3-8, so -5+2=-3, -3-4 becomes -4-3, which is -7.[/quote]
@AIME15, you forgot that we don't carry the sign of the number $4$, it is the same to the calculator so $(-3)-(4)$ becomes $(4)-(-3)=4+3=\boxed{7}$.

This answer is incorrect.  It is 3-8=-5, -5+2=-3, -3-4=$\boxed{-7}$.  Please fix this answer.


I was doing a USACO contest and one of my programs had a main() function that had a 1000 by 1000 array of shorts. For some reason, everytime I ran the program from my home computer it would give me a signal SIGSEGV but the USACO computers wouldn't be bothered. However, when I ran the program on the USACO computers it worked just fine 0_0. It seemed like everytime I had a program with an array of 1000000 elements in the main() function, my home computer would just keep giving me a signal SIGSEGV. Then today my friend moved these arrays in the main() out of the main() as global variables and the program worked perfectly!

I don't think 1000000 ints is that unreasonable and I don't see why it kept failing on my home computer so could someone please tell me why this happened? I use the DJGPP compiler.

When you make an array global, the array is stored in the heap, but when you make an array inside main (), the array is stored in the stack. (You can probably search for heap and stack online if you want to know more.) Usually the heap is much larger than the stack, and your computer may also have less available space on the stack than the USACO server does, so I think this is what you're seeing.

Thanks for the reply! Do you know how I can check the space available of my computer's stack and heap?

It might depend on the computer, but [url=http://www.delorie.com/djgpp/faq/memory/stack.html]this site[/url] claims that the stack space is defaulted to 256 KB.

But if you just want to find how much stack space is available for your program, you can do a binary search to find the smallest array size that will make the program crash.  :D

As an aside, I believe global variables are usually said to be in static memory, not the heap.  The heap usually refers to dynamically allocated memory, such as that returned from malloc.

If you keep your array inside main, then you can add the word static to the beginning of its declaration.  That is another way of placing it in static memory.

Interesting, I thought that DJGPP doesn't even exists anymore. I would really recommend using GCC instead, it should already be present on your computer if you're running any decent version of Linux. If you're running Windows you can download the MinGW GCC port or better yet, get Dev-Cpp which already comes with the MinGW compiler, graphical IDE and is really easy to install. (And free, of course.)

Anyway, you can tell GCC what stack size you want by setting the appropriate command line option. (Don't know it's name but should be easy to find.)

Regarding algorithm competitions, it's usually a better idea to put big variables in the static memory. (Which is really part of the heap anyway.) Unless you need to reclaim that memory during the lifetime of your program, in which case you should allocate the memory dynamically using malloc or new. (Or with some stack located object that allocates its members on the heap -- vector for example.)


let P chosen (interior) inyhe rectangle ABCD such that <APD+<BPC=180 find
               <DAP+<BCP :(

traslade the triangle APD in such a way that AD coincides with BC, then PBP'C is cyclic and the rest is simple, so finih it by yourselves!

Pascual