AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
AMC 10 A

Will I get 1.5 or 2.5 points for not answering a question?

Billy

[quote="bookaholic"]I think it's changing to 1.5 in 2007. Forgot where I read that, though, so I may be wrong.[/quote]

That's correct. It was in the USAMO qualifier book as well as on the AMC site (I think).

[quote="bookaholic"]I think it's changing to 1.5 in 2007. Forgot where I read that, though, so I may be wrong.[/quote]

if they change it to 1.5, are they changing the minimum score?  :|

er, the minimum score is always going to be 0, pretty much?

According to the 2005 Summary of AMC Results/Awards,

[quote="AMC 12 Chair"]Beginning in 2002 the number of points awarded for a blank response on the AMC12 was increased from 2 to 2.5. As a result, students could qualify for the AIME by solving as few as 11 problems. Analysis of student performances since then reveals that the average number of blank responses by AIME qualifiers has increased from about 8 to 11. This trend, suggesting that many bright students are simply igoring problems of even moderate difficulty, runs counter to our goals. For that reason, beginning in 2007, the reward for a blank response will decrease from 2.5 to 1.5 points. It should be emphasized that in making this change, we are not attempting to reduce the number of AIME qualifiers. Because we are simultaneously working to improve the overall success rate of the first ten problems, we expect that the number of students achieving the minimum qualifying score of 100 will remain roughly constant.[/quote]

BTW I like this change, because 2 blank responses is almost 1 correct answer is... just not right.

Scoring on the 2006 AMC 10 A, 2006 AMC 10 B, 2006 AMC 12 A and the 2006 AMC 12 B will be 6 points for each correctly answered question, 2.5 points for each blank, or unanswered, question and 0 points for an incorrect answer.

See for instance item 4 on the facisimile of the 2006 AMC 10 A cover page which is on our website at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/html/06tm-26-10Acover.html
(or go to the AMC webstie, choose 2006 AMC 10A/ AMC 12 A Teacher's Manual from the What's New column, then using the table of contents, go to page 26).

In 2007, the scoring will change to 6 points for each correctly answered question, 1.5 points for each blank, or unanswered, question and 0 points for an incorrect answer.  This scoring change was announced in the 2005 AMC 101/12 Summary of Results and Awards, will be announced again in the 2006 AMC 10/12 Summary of Results and Awards and will be noted on the advertising brochures for the 2007 contests.  The scoring system will be noted on the cover of the 2007 AMC contests.  The implications of the scoring change will be noted in the 2007 AMC 10/AMC 12 Teachers' Manual.

Qualification for the 2007 AIME from the 2007 AMC 10 will continue to be a total of 120 points, or in the top 1% on the AMC 10.
Qualification for the 2007 AIME from the 2007 AMC 12 will continue to be a total of 100 points or in the top 5% on the AMC 12.

Steve Dunbar
AMC Director

[quote="probability1.01"][quote="JesusFreak197"]I think he means minimum score to qualify, and no, as far as I know, they are not planning to. Why should they?[/quote]

Err.. maybe because most people will have a significantly lower score than before?[/quote]

Not all that much. I'm pretty sure that part of the point is to eliminate those people who qualify for AIME by answering maybe 13 questions on the AMC12 (in case of one wrong) and leaving the rest blank. That shouldn't allow you to qualify. Now, you'll have to answer at least 14 correctly, probably more, and 19 on the AMC10.

Umm... well even if the average borderline qualifying person leaves 8 questions blank, that's still 8 whole points, which disqualifies a lot of people, at least 1000 from a cursory glance at last year's statistics. Obviously I'm ignoring some factors here, but it still strikes me as quite a lot.

But it seems that the AMC people will make the first 10 problems easier.

It gives 2.5 for unanswered Q is becoz it doesn't want us to GUESS
but....now has been changed to 1.5
wut's that mean?

But i figured that it would be a little pointless to change the minimum qualification score because then the qualification would be about the same difficulty, and there's a smaller penalty for guessing.

It shouldn't have much effect on the USAMO qualifying indices (although the USAMO expansion should).  This change is targeted at the borderline AIME qualifiers, who generally do not have much effect on USAMO cutoffs.  People on the borderline for USAMO probably aren't leaving very many questions blank, so they won't lose many points.  I suppose it could still lower the cutoff by a few points, though.

AMC over AIME? Not in this lifetime...  :P

The AIME is the coolest part, no doubt. Plus, if more people demonstrate that they can do better on the AMCs, I'm thinking they will have to make AIMEs harder. Uh-oh. There goes my hopes...

Doesn't matter much to me.
In the past, I've answered every question anyways.  Well, that was on the AMC10.  And, of course, I would make 2 or so minor errors or mistakes.

It would ruin the point of changing it from 2.5 to 1.5 if they also lowered the cutoff for AIME.

[quote="eryaman"]Maybe this will decrease the inflating USAMO qualifying indexes, and place some more emphasis on AMC over AIME.

...on second thought, it probably will only have a small effect on that.[/quote]

Well they will also try to make the AMC's easier(first 10 problems easier, which hmmm does not really help too many people) so yeah your score will stay roughly the same.

But this does hurt a lot if you are leaving like 5-10 blanks (so yeah try not to leave too many blanks that's the intention of the AMC people)

Really good change ... about time. It was ridiculous that 21 right and 4 blank got you almost the same score as 23 right. Generally there isn't a single challenging problem until #21 at the absolute earliest.

Yeah, it was quite ridiculous to qualify for AIME without getting a good number right.

[quote="beta"]Yeah 20 right, 5 blank gets you a higher score than 22 right... that's not fair also. Basically I don't see how the current AMC scores mean anything, like a difference of 0.5 point is a much bigger difference than a difference of 6 points.

Why can't they just make AMC short-answer format(like AIME) instead of multiple choice?[/quote]

It would scare away newbies.

[quote="hockeyadi23"][quote="paladin8"][quote="log0"]I've noticed that on the AMC 10 booklet, it does not actually say that the cutoff is 120 points, just top 1%. So is 120 a generally accepted cutoff score to represent the top 1%, or is it actually part of the competition policy?[/quote]

You might have an old booklet. I believe the policy changed in 2004 to include any 120+ AMC10 scores as qualifying.[/quote]

I noticed that when reading the instructions last year. So basically the baseline is either 120 or the top 1%, whichever is lower.[/quote]

Yeah. It's more like the AMC12 qualifying now, only 20 points higher and 1% instead of 5%.

A kid from our school actually did answer only 12 questions. Checked them over several times and slept for the remaining time. Poor kid bubbled one in wrong...

I think this change is really necessary in order to avoid people from spending time strategizing over the test. In that same time, the person, if preparing, could practice more math, or if it's during the actual test, could be working on math. I personally would prefer an SAT-type system (say, base score 25, +5 for correct answer, -1 for wrong answer)


1.) Prove that for any even $n>1$, $n^{2}-1$ divides $2^{n!}-1$.

2.) Prove that for any $n>2$ the number $1989$ divides $n^{n^{n^{n}}}-n^{n^{n}}$.

What, how can 2) be true? 1989=9*13*17, am I missing something?

Hm these problems look familiar.

Oh wait I've seen these before.

Oh wait that's right I went to AMP :blush:

Yes the problem is right. Amazing isn't it :lol:

[hide="1"] \[n^{2}-1|2^{n!}-1 \iff 2^{n!}\equiv 1\mod{(n^{2}-1)}\]  \[\iff 2^{n!}\equiv 1\mod{n-1}, 1\mod{n+1}\] because n-1 and n+1 are relatively prime, being both odd.

But 2 is also relatively prime to both $n-1$ and $n+1$, and therefore is a unit in both mods.
Thus, $2^{n!}\equiv 1\mod{n-1}, \mod{n+1}\iff \phi(n-1)|n!, \phi(n+1)|n!$

But $\phi(k)<k$ for any natural $k$.  Thus, $\phi(n-1) \le n-2, \phi(n+1) \le n$, and any number less than or equal to $n$ must divide $n!$.  Thus, \[\phi(n-1)|n! \implies 2^{n!}\equiv 1\mod{n-1}\]  \[\phi(n+1)|n! \implies 2^{n!}\equiv 1\mod{n+1}\]  \[\implies 2^{n!}\equiv 1\mod{n^{2}-1}\implies n^{2}-1|2^{n!}+1\] [/hide]

No seriously, can somebody please explain how 2) can possibly be true?

Why not?  Are you sure you aren't thinking the other way?

You have to show \[n^{(n^{[n^{n}]})}\equiv n^{(n^{n})}\mod 9, \mod 13, \mod 17\] Well, if you're looking at that congruence $\mod{x}$, then you look at the first level of exponents $\mod{\phi(x)}$, and the second level $\mod{\phi(\phi(x))}$.  And $\phi(\phi(9))=\phi(6)=2$, $\phi(\phi(13))=\phi(12)=4$, $\phi(\phi(17))=\phi(16)=8$.

There's gotta be a better way then just checking all three mods with exponent reduction...

[quote="TomciO"][quote="K81o7"]
Thus, $2^{n!}\equiv 1\mod{n-1}, \mod{n+1}\iff \phi(n-1)|n!, \phi(n+1)|n!$
[/quote]

That's not true! It's only an implication, not a equivalence! (but of course the implication has a good direction so the solution is correct)[/quote]

Argh...I made the same mistake the same day on a different problem...it should be $\Leftarrow$...


analytic geometry
algorithm
ceiling function
modular arithmetic
number theory
Euclidean algorithm

Source: AMC-12 2000, problem 16

A checkerboard of $13$ rows and $17$ columns has a number written in each squares, beginning in the upper left corner, so that the first row is numbered $1, 2, ..., 17$, the second row $18, 19, ..., 34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1, 2, ..., 13$, the second column, $14, 15, ..., 26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).

$(A)\text{ 222}$
$(B)\text{ 333}$
$(C)\text{ 444}$
$(D)\text{ 555}$
$(E)\text{ 666}$

[I noticed that the AMC-12 2000 only has problems 1-11; I currently have problems 12-25, but is there a reason for the problem absences?]

[hide="my solution... if anyone cares"]

Imagine the numbers on a coordinate plane.

$17(13-y)+x=13x-(y-1)$

Nice diophantine.

$-12x-16y+220=0$ (Using generalized Euclidean Algorithm)
$x=5-4n$
$y=10+3n$

Test all $n$ such that $x,y \ge 0$

$111+56+166+221+1=\boxed{555}$

This problem took me longer than I expected and I'm unhappy with my solution. I know there's some clever way (something to do with 222) to do this... anyone?
[/hide]

[hide]We see that $111$ is in the center of the board under either system.  If $n$ is a solution, then $222-n$ is also a solution.  Then the sum of solutions is $111+222(\text{number of solutions from 1 to 110})$.

Under the first system, $n$ is in row $\left\lceil\frac{n}{17}\right\rceil$ and column $n\pmod{13}$.  Under the second system, $n$ is in row $\left\lceil\frac{n}{13}\right\rceil$ and column $n\pmod{17}$.  Therefore, $\left\lceil\frac{n}{17}\right\rceil\equiv n\pmod{13}$ and $\left\lceil\frac{n}{13}\right\rceil\equiv n\pmod{17}$.

We immediately see that $1$ is a solution.  Once we find that $56$ is another solution, we stop.  The answer is D.[/hide]


Let a be real parameter. Find all a for which the equality $x^{2}+(1-2a)x+2a^{2}+lg(4x^{2}-(8a-1)x+5a^{2})=lg(x^{2}-2(a+1)x-a^{2})$ has only one solution.

[hide=" 2nd Hint"]That equation is equivalent to $f(4x^{2}-(8a-1)x+5a^{2})=f(x^{2}-2(a+1)x-a^{2})$, here $f(x)=\lg{x}+\frac{x}{3}$ is monotone on $(0,\infty)$.[/hide]


inequalities
complex numbers
inequalities proposed

The  triangle  ABC  has  A=30  and  AB=3/4 AC. Find  the  point  P  inside  the triangle  which  minimises  5PA+4PB+3PC

(16 th  vietman  mo , 1978)  thanks!

Here's the best I could do up until now:

We take A to be the origin of the plane, B(3i), C(4w) and D(4) (the things in the brackets are complex numbers; w=cos60+i*sin60). We need to find a z s.t. the sum S=5|z|+4|z-3i|+3|z-4w| is minimised.

5|z|=|4z-3iz|; 4|z-3i|=|12i-4z|; 3|z-4w|=|3iz-12iw|, so
S=|4z-3iz|+|12i-4z|+|3iz-12iw|>=|4z-3iz+12i-4z+3iz-12iw|=|12i-12iw|=12, but I have no idea if this value is reached, or if it's reached for a z inside the triangle ABC. The condition is that the complex numbers z_1=4z-3iz, z_2=12i-4z and z_3=3iz-12iw are collinear and have the same direction: z_3/z_2 and z_1/z_2 are both positive reals. However, I'm a bit scared of the calculations. Can anyone check this? It could work.

I was especially encouraged by the nice-looking result (12) :D It doesn't mean it's correct, though..


Euler
number theory
real analysis
real analysis unsolved

Calculate: $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\plus{}1}}{2n\plus{}1}\cdot\frac{\zeta (2n)}{2^{2n}}.$

There's an obvious thing to try, which is to insert the definition of $ \zeta$ and interchange the order of summation. Doing that, we discover that is it the exact same problem as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218877]this.[/url]

$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}}{2n \plus{} 1}\cdot\frac {\zeta (2n)}{2^{2n}} \equal{} ?$

This is all I got so far.

$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}}{2n \plus{} 1} \equal{} 1 \minus{} \frac {\pi}{4}$

$ \sum_{n \equal{} 1}^{\infty}\frac {1}{2^{2n}} \equal{} \frac {1}{3}$

$ \zeta (2n) \equal{} \frac {2^{2n \minus{} 2}\pi^{2n}|B_{2n}|}{2n!}$ This is some sort of Euler identity from my Intro to Number Theory book. I just can't figure out how to incorporate it into the summation. Maybe someone will be able to telescope some part of it? ^_^ Or not use the derived values for the other parts and be able to cancel stuff. (Just trying to add input >_<) Also $ B$ is a Bernoulli number.

Edited post with regards to Carcul.
P.S -> My apologies :)

Exactly how did you found that $ \sum_{n \equal{} 1}^{\infty}\frac {1}{2^{2^{n}}} \equal{} \frac {1}{3}$?

It's not, of course -- it was a typo for $ \sum_{n \geq 1} \frac{1}{2^{2n}}$.


geometry
circumcircle
analytic geometry
trigonometry
inequalities
calculus
function

Let  $ABC$ is a triangle with the circumcircle $(O)$ and $G$ is the centroid. $AG,BG, CG$ meet $(O)$ at $A', B', C'$ respectively. Prove that $S(A'B'C')\geq\ S(ABC)$ where S(ABC) means the area of the triangle ABC

nice?

no one? I think it's very nice and not so hard

[quote="treegoner"]Notice that $3\frac{S(GB'C')}{S(ABC)} = \frac{S(G'B'C')}{S(GBC)} = \frac{GB'.GC'}{GB.GC} = \frac{(R^2-OG^2)^2}{GB^2.GC^2}$.

According to Leibnitz's theorem, we have

$3R^2-3OG^2 = GA^2+GB^2+GC^2$

Hence $3\frac{S(A'B'C')}{S(ABC)} = \sum3\frac{S(GB'C')}{S(ABC)} = (GA^2+GB^2+GC^2)^2 \sum \frac{1}{GB^2.GC^2} \geq 3$

Thus $S(A'B'C') \geq S(ABC)$.[/quote]

  Your solution is very good . I have to use Kamarta ineq in my solution after proving $(A',B',C')>>(A,B,C)$

treegoner, can you prove that $r_{ABC}\leq\ r_{A'B'C'}$ (r means the inradii)

[quote="treegoner"]That is a very nice question. I did think about it after doing this problem. But up till now, I can't say anything. Are you sure of this result?[/quote]

  I have proved it for twice, but who know? maybe i have made mistakes for twice ???!!!

how it is demonstrated the Leibnitz's theorem ?  :blush:

[quote="treegoner"]I 'm glad that finally I got some free time this morning to think again about this problem. Here is my solution to the problem $r_{ABC} \leq r_{A'B'C'}$.

But first, I need to answer js131090's question. Leibnizt's theorem says that if $(x: y: z)$ is the barycentric coordinate of a point $P$, then for every point $M$ we have the equality $xMA^2 + yMB^2 + zMC^2 = (x+y+z)MP^2 + xPA^2 + yPB^2 + zPC^2$. In particular, if $P$ is the centroid $G$ and $M$ is the circumcenter $O$, Leibnitz's theroem yields that $3R^2 = 3OG^2 + GA^2+GB^2+GC^2$.

Let's get back to the problem. Let $A_1B_1C_1$ be the pedal triangle of $G$ with respect to $ABC$. Then according to sine theorem and Euler's formula for area of pedal triangle, we obtain the sides and area of $A_1B_1C_1$ are respectively $\frac{2}{3}m_a.sinA, \frac{2}{3}m_b.sinB, \frac{2}{3}m_c.sinC, \frac{4S(a^2+b^2+c^2)}{9R^2}(=\frac{1}{4}S(1-\frac{OG^2}{R^2}))$

We have $\frac{B'C'}{BC} = \frac{GC'}{GB} = \frac{GC.GC'}{GB.GC} = \frac{a^2+b^2+c^2}{4m_bm_c}$. Hence $B'C' = \frac{(a^2+b^2+c^2)a}{4m_bm_c}$. Since triangles $A_1B_1C_1$ and $A'B'C'$ are similar, we obtain $\frac{r_{A'B'C'}}{r_{A_1B_1C_1}} = \frac{B'C'}{B_1C_1} = \frac{3(a^2+b^2+c^2)R}{4m_am_bm_c}$. On the other hand, we know that $r_{A_1B_1C_1} = \frac{S_{A_1B_1C_1}}{p_{A_1B_1C_1}} = \frac{S(a^2+b^2+c^2)}{12(\sum m_a.sinA)R^2}$. Hence $r_{A'B'C'} = \frac{S(a^2+b^2+c^2)^2}{8m_am_bm_c(\sum am_a)}$.

We need to show that $\frac{S(a^2+b^2+c^2)^2}{8m_am_bm_c(\sum am_a)} \geq \frac{S}{p}$, which is equivalent to $(a^2+b^2+c^2)^2p \geq 8(\sum am_a)m_am_bm_c$. By applying Tchebychev's inequality and AM-GM inequality, we obtain $\sum am_a \leq \frac{2}{3}p(m_a+m_b+m_c)$ and $m_am_bm_c(m_a+m_b+m_c) \leq \frac{1}{3}(m_a^2+m_b^2+m_c)^2 = \frac{3(a^2+b^2+c^2)^2}{16}$. Therefore, the problem is completely solved.[/quote]

The geometric problem is always should be solve in geometric way but i totally stupid in geometry. Here is my approach tatally calculus, which may not please treegoner:

[b] lemma [b] $(A',B',C')>>(A,B,C) (rad)$.
[b] proof [/b]. WLOG we can assume that $A\geq\ B\geq\ C$ which means $GA\leq\ GB\leq\ GC$. then by the cosin theorem we can proof that $A,B,C\geq\ max(A',B',C')$ and $A,B,C\leq\ min(A',B',C')$ . Then our lemma is proven.

This is the how  to apply our lemma:
the problem <=> $sin\frac{A}{2}\ sin\frac{B}{2}\ sin\frac{C}{2}\leq\sin\frac{A'}{2}\ sin\frac{B'}{2}\ sin\frac{C'}{2}$ <==> $\sum\ ln(sin\frac{A}{2}\ )\leq\sum\ ln(sin\frac{A'}{2}\ )$. (2)
let consider the fucntion $f(x)=ln(sin(x))$ where $0\leq\ x\leq\pi\ /2$. It is easy to see that $f''(x)<0$ Apply the KAMARATA ineq for the function $f(x)$ and 6 numbers $(A',B',C')>>(A,B,C)$ in (2) we have the result.

however thank u treegoner.


function
topology
real analysis
real analysis unsolved

Suppose $ f,g : \mathbb{R} \to \mathbb{C}$ are Borel. Prove that the function $ h(x,y)\equal{}f(x\minus{}y)g(y)$ is also Borel measurable.

Can you prove that $ \tilde h(x,y)\equal{}f(x)g(y)$ is Borel? Because $ h$ is $ \tilde h$ composed with the homeomorphism $ (x,y)\mapsto (x\minus{}y,y)$.


Only x=y=1.
Let $z=2^{x}$. We have $0<z<\sqrt 6$ and $f(z)=\ln{\frac{5-z}{6-z^{2}}}=\frac{\ln 3}{\ln 2}$. Because $f'(z)>0$ for $0<z<\sqrt 6$, we have only one solution z=2.


induction
superior algebra
superior algebra unsolved

Let $ G$ be  finite group and $ \mathcal{K}$ a conjugacy class of $ G$ that generates $ G$. Prove that the following two statements are equivalent:

(1) There exists a positive integer $ m$ such that every element of $ G$ can be written as a product of $ m$ (not necessarily distinct) elements of $ \mathcal{K}$.

(2) $ G$ is equal to its own commutator subgroup.

[i]J. Denes[/i]

From http://mathoverflow.net/questions/109590/the-powers-of-non-empty-subset-of-a-group-that-generate-a-subgroup/109652#109652 we get that $\{e\}\not=N:=( \mathcal{K} )^{|G|}\lhd G$, so we can use Induction to group $G/N$ and it finishes solution.


search
function
number theory unsolved
number theory

Find infinitely many solutions of

 $(a) x^2+y^2+z^2=3xyz$

Davron

What about search function¿ It's very easy to find!

please can you help me with that search function

Just search for "x^2+y^2+z^2=3xyz"... (isn't this the first one one tries¿)


I realize that this is in the teacher's manual, but somehow my school can't quite resolve this issue...
So
When are we allowed to open the official solutions booklet for the AMC 12? Is it when the envelope with the answer sheets has been sealed, or is it when the mail gets picked up, or is it 24 hours?

On the sealed Exan Manager's Envelope itself,  it says in large black block letters:
"Do not open until after the exam has been adminstered and all student answer forms have been mailed."
It also says in smaller print at the bottom:
"The exam may be discussed following its adminstration within the school, but not outside the school until the following day, since there will be schools in different time zones which have not yet given it."
The Teachers' Manual says basically the same thing, on page 5, second column, in Section II, 16. b.

Just to forestall questions about the AIME:  The AIME Solutions pamphlet is NOT mailed with the AIME.  The AIME Solutions Pamphlet is mailed with the AIME RESULTS, not with the AIME itself.  That means your teacher/contest adminstrator does not have the official AIME answers until about 2 weeks after the AIME.

Steve Dunbar
AMC Director


Find 9 odd natural distinct numbers that satisfy
1/(a1) + 1/(a2) + ... + 1/(a9) = 1

a1 means the first number
...
a9 means the ninth number

[quote="bibobeo"]sorry, but what does "perfect number" mean?[/quote]
A number who's factors (besides itself), when added, are equal to itself.

The first few are 6, 28, 496, 8128.

Well, then I don't think perfect numbers can be right.

http://en.wikipedia.org/wiki/Perfect_numbers

Read the minor results section.

Oh, [b]odd.[/b]  This is a slightly different problem.  You might want to try induction instead.

I found one, but I'm not sure that it is unique.

3   5   7   9   11   15   35   45   231

[hide]use the fact that $\frac{1}{a(a+1)}+\frac{1}{a+1}=\frac{1}{a}$
then expand $\frac{1}{1}=\frac{1}{a}$
[/hide]


Write each trig expression as an algebraic expression containing u and v.

1) sin(sin^-1 (u) - cos^-1 (v))

2) sec(tan^-1 (u) + cos^-1 (v))

hello, (1) is equals to $ v\sqrt{1\minus{}u^2}\plus{}u\sqrt{1\minus{}v^2}$ and (2) $ \frac{\sqrt{1\plus{}u^2}}{v\minus{}u\sqrt{1\minus{}v^2}}$
Sonnhard.

I agree with the second one but, unless I've done something silly, for the first one I get:

sin(a-b) = sin a cos b - cos a sin b

=uv - sqrt(1 - u^2)sqrt(1 - v^2).

(Sorry, I can't do LaTeX at the moment.)

$ \sin (\arcsin u \minus{} \arccos v)$
$ \equal{} \sin \left( \frac{\pi}{2} \minus{} \arccos u \minus{} \arccos v \right)$
$ \equal{} \cos (\arccos u \plus{} \arccos v)$
$ \equal{} \cos (\arccos u) \cos (\arccos v) \minus{} \sin (\arccos u) \sin (\arccos v)$
$ \equal{} \boxed{uv \minus{} \sqrt {1 \minus{} u^2} \sqrt {1 \minus{} v^2}}$

[quote="TZF"]$ \sin (\arcsin u \minus{} \arccos v)$ (Equation 1)
$ \equal{} \sin \left( \frac {\pi}{2} \minus{} \arccos u \minus{} \arccos v \right)$ (Equation 2)
$ \equal{} \cos (\arccos u \plus{} \arccos v)$ (Equation 3)
$ \equal{} \cos (\arccos u) \cos (\arccos v) \minus{} \sin (\arccos u) \sin (\arccos v)$ (Equation 4)
$ \equal{} \boxed{uv \minus{} \sqrt {1 \minus{} u^2} \sqrt {1 \minus{} v^2}}$ (Equation 5)[/quote]

I'm sorry, the only part I understand is when you got from Equation 3 to Equation 4. Can you explain all the other parts in more detail please. I'm still learning trig.  :)

Also, can somebody give a solution to number 2?

To help you, here is TZF's solution with the addition of "Equation 1A" and "Equation 2A":

$ \sin (\arcsin u\minus{}\arccos v)$ (Equation 1)
$ \equal{}\sin\left(\left(\frac{\pi}{2}\minus{}\arccos u \right) \minus{}\arccos v\right)$ (Equation 1A)
$ \equal{}\sin\left(\frac{\pi}{2}\minus{}\arccos u\minus{}\arccos v\right)$ (Equation 2)
$ \equal{}\sin\left(\frac{\pi}{2}\minus{} \left( \arccos u \plus{} \arccos v\right) \right)$ (Equation 2A)
$ \equal{}\cos (\arccos u\plus{}\arccos v)$ (Equation 3)
$ \equal{}\cos (\arccos u)\cos (\arccos v)\minus{}\sin (\arccos u)\sin (\arccos v)$ (Equation 4)
$ \equal{}\boxed{uv\minus{}\sqrt{1\minus{}u^{2}}\sqrt{1\minus{}v^{2}}}$ (Equation 5)

We get from Equation 2A to Equation 3 by taking advantage of confunctions . The rule is $ \sin \left(\frac{\pi}{2} \minus{} \theta\right) \equal{} \cos \theta$ (and also $ \cos \left(\frac{\pi}{2} \minus{} \theta\right) \equal{} \sin \theta$). Search for "cofunction identities" for more information.

Since sine and cosine are cofunctions, it follows that $ \arcsin u \plus{} \arccos u \equal{} \frac{\pi}{2}$. Therefore, $ \arcsin u \equal{} \frac{\pi}{2} \minus{} \arccos u$, which was used to get from Equation 1 to Equation 1A.

[Note: You could solve problem (1) without using cofunctions, instead simply applying the angle difference formula for sine to the given expression. U

To get from Equation 4 to Equation 5, you can use the following idea, which I used to solve problem (2) below:

Let $ \arctan u \equal{} \theta$. Then $ \tan \theta \equal{} u \equal{} \frac{u}{1}$. Draw the angle in standard position, draw the corresponding triangle, and label the sides. The opposite side is $ u$ and adjacent side is $ 1$. You can find that the hypotenuse is $ \sqrt{1\plus{}u^2}$ by the Pythagorean theorem. Now you can find any trig function of $ \theta$ using the triangle you have drawn. For example, $ \cos(\arctan u) \equal{} \cos(\theta) \equal{} \frac{1}{\sqrt{1\plus{}u^2}}$. Do the same thing for $ \arccos v$.

Solution to problem (2):

$ \sec(\arctan u \plus{} \arccos v)$

$ \equal{} \frac{1}{\cos(\arctan u \plus{} \arccos v)}$

$ \equal{} \frac{1}{\cos(\arctan u)\cos(\arccos v) \minus{} \sin(\arctan u)\sin(\arccos v)}$

$ \equal{} \frac{1}{  \left( \frac{1}{\sqrt{1\plus{}u^2}} \right) \left( v \right) \minus{} \left( \frac{u}{\sqrt{1\plus{}u^2}} \right) \left( \sqrt{1\minus{}v^2} \right) }$

$ \equal{} \frac{1}{  \left( \frac{v}{\sqrt{1\plus{}u^2}} \right) \minus{} \left( \frac{u \sqrt{1\minus{}v^2}}{\sqrt{1\plus{}u^2}} \right) }$

$ \equal{} \frac{1}{ \frac{v \minus{} u \sqrt{1\minus{}v^2}}{\sqrt{1\plus{}u^2}} }$

$ \equal{} \boxed{\frac{\sqrt{1\plus{}u^2}}{v \minus{} u \sqrt{1\minus{}v^2}}}$

Oh. I get it. Thanks a lot. That explains a lot.  :)


A triangle is formed by joining three points $ A,B$ and $ C$ where $ \overrightarrow a,\overrightarrow b$ and $ \overrightarrow c$ are  the position vectors respectively.

If $ P(\overrightarrow r)$ be a point inside the triangle,then can $ \overrightarrow r$ be expressed as

$ \overrightarrow r\equal{}x\overrightarrow a\plus{}y\overrightarrow b\plus{}z\overrightarrow c$

where $ x,y,z$ are scalars such that $ x\plus{}y\plus{}z\equal{}1,x>0,y>0,z>0$.

I am not able to generate a proof using vector methods.But first of all I would want to know whether above statement is correct or not.

The statement is true. $ P(\overrightarrow r)$be inside the triangle $ ABC$ except boundary.  Use $ z\equal{}1\minus{}x\minus{}y$ to prove.


inequalities
calculus
algebra unsolved
algebra

[u][i]Problem[/i][/u]

   [i] Find all real numbers  $ x$  such that the following inequality holds for all non negative
          numbers $ a , b , c$

 $ [  a^{2} \plus{} b^{2} \plus{} (x\minus{}1)c^{2} ] [ a^{2} \plus{} c^{2} \plus{} (x\minus{}1)b^{2} ] [ b^{2} \plus{} c^{2} \plus{} (x\minus{}1)a^{2} ]$

 $ \leq (a^{2} \plus{} xbc)(b^{2} \plus{} xac)( c^{2} \plus{} xab)$[/i]

    [i]  My proposed problem[/i]

[i] I hope it would be nice problem[/i]

[i]  Scorpius119 , Nayel, ....Let perform [/i]

 [i]  Love An forever :love:
 Huỳnh Võ Phương An[/i]

If we take $ c\to 0$ we will get:
$ (a^2 \plus{} b^2)(a^2 \plus{} (x \minus{} 1)b^2)(b^2 \plus{} (x \minus{} 1)a^2)\leq a^3 b^3 x$
Now put here $ a \equal{} b$ :
$ 2a^6 x^2\leq a^6 x$ or
$ x\left(x \minus{} \frac {1}{2}\right)\leq 0.$
This means that $ x \in (0,\frac {1}{2})$
Maybe this helps... :)

for the post above x>0.
$ c \to 0 \text{ and } b\equal{}ka, a,k \in R^\plus{}$
so we must x such that for all k we have:
$ f(k)\equal{}(k^2x^2)\plus{}x(1\minus{}k^4\minus{}k^2\minus{}k\minus{}\frac{k^3}{k^2\plus{}1})\plus{}(k^2\plus{}k\minus{}1\minus{}k^3) \le 0$
$ \implies f(k)\equal{}(k^6\minus{}2k^3\minus{}k\plus{}1)x\plus{}k^2(k^2\plus{}1)x^2\minus{}(k^2\plus{}1)(k\plus{}1)(k\minus{}1)^2 \le 0$ but $ lim_{k \to \infty}\equal{}\plus{} \infty$, contradiction.

Are you shure?
$ \frac {1}{k^2}x^2 \plus{} x\left(\frac {1}{k^4} \minus{} 1 \minus{} \frac {1}{k^2} \minus{} \frac {1}{k^3} \minus{} \frac {k}{k^2 \plus{} 1}\right) \plus{} \left(\frac {1}{k^2} \plus{} \frac {1}{k^3} \minus{} \frac {1}{k^4} \minus{} \frac {1}{k}\right)\leq 0$
If we take $ k\to \infty$ we will have : $ \minus{} x\leq 0$. This is true if $ x\geq 0$

[quote="Yuriy Solovyov"]Are you shure?
$ \frac {1}{k^2}x^2 \plus{} x\left(\frac {1}{k^4} \minus{} 1 \minus{} \frac {1}{k^2} \minus{} \frac {1}{k^3} \minus{}\frac {1}{k(k^2 \plus{} 1)} \right) \plus{} \left(\frac {1}{k^2} \plus{} \frac {1}{k^3} \minus{} \frac {1}{k^4} \minus{} \frac {1}{k}\right)\leq 0$
If we take $ k\to \infty$ we will have : $ \minus{} x\leq 0$. This is true if $ x\geq 0$[/quote]

however you're right, i'am wrong in calculos, in fact:
$ p(x) \equal{} x^2(k^2(k^2 \plus{} 1)) \plus{} x((k^2 \minus{} 1)^2(k^2 \plus{} 1) \minus{} k^3) \minus{} ((k^2 \minus{} 1)^2(k^2 \plus{} 1)) \le 0$
and we note $ x_1 \le 0 < x_2 \forall k$ and $ min(x_2) \equal{} p(1) \equal{} \frac {1}{2}$

(i've calcolated min{f (x) = ((x3-(((x2)-1)2)(x2+1))+sqrt((x3-(((x2)-1)2)(x2+1))2+4(x2)(x2+1)(x2+1)(x2-1)(x2-1)))/(2(x2)(x2+1))} with http://wims.unice.fr/wims/wims.cgi )

in conclusion $ \forall x \in (0,\frac {1}{2})$
 :roll:


real analysis
advanced fields
advanced fields unsolved

Prove that if $\mathbb{R}^{k}\subset \bigcup_{1}^{\infty}F_{n}$ where each $F_{n}$ is a closed subset of $\mathbb{R}^{k}$, then at least one $F_{n}$ has a nonempty interior.

(using only ideas developed up to chapter 2 in Rudin)

Don't double post, please. This post is nearly identical to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146107[/url].


The more you have of it, the less you see. What is it?

i have made a riddles thing. like it feel free to post riddles and ANSWERS!!!

it has several answers
Cataract,myopia,hypermetropia,................ :rotfl:  :rotfl:

or
KNOWLEDGE :rotfl:

hmmmm knowledge is a good answer, but it's actually darkness. good job though.

next riddle i am posting.
wat's in the middle of nowhere? common riddle.....easy

now[b]h[/b]ere

answer is [b]h[/b]  :rotfl:

good job ur good
what goes up an never comes down? this one shoud have more than one answer but it says it only has one.

a satellite launched with a speed greater than escape velocity :rotfl:  :rotfl:  :rotfl:

now hav a look at this one wat is such that it never degrades with respect to the human body?
(ans=*****)

r u mad or wat.... :mad:  :o
u kno i don't mind using slang :P

and can u prove my ans wrong?? :rotfl:  :rotfl:

@rt and ritu,i am unable to understand you both

even i am not able to understand it................. :rotfl:  :rotfl:  :rotfl:

:rotfl: is telling a different story altogether

common you two can tell it to me :P

tell this one

in a boxing match, non of the men threw a punch ,yet there was a winner.How's so
??????

what is the need of [b]throwing[/b] a punch :rotfl:  :rotfl:

ok
nobody hit a punch.....

but throwing a punch means the same neways :lol:  :D

the other one gave up
 :rotfl:    :rotfl:  :rotfl:

see thiss is a variation of a common ques (produced by myself) :P

General ans is "it was a women's match so no men threw a punch" :rotfl:

But i've twisted the ques  and it says "None [b]of[/b] the men" threw a punch that means men were there in the match!!! :P

Now don't say the referee of a woman's match didn't threw the punch :P


Here is the Weather Forecast for the Penn State site.
[img]http://img98.imageshack.us/img98/7616/pennstateqb8.png[/img]
Source: NOAA.gov, accessed Wed May 28, 9:30pm

Assuming the Day lasts 12 hours and the Night lasts 12 hours, what is the probability it thunders for exactly 36 hours during ARML? (Showers != Thunder)

Well first of all, it can be any real number between 0 and 48...I don't know the techincal terms for this but wouldn't the probability be negligble? Also was your intention to put this in the ARML forum, because this isn't the ARML forum...

Possibly he was assuming that if it storms one night, it will storm the entire night.

Yeah it was, my bad. And yeah tjhance is right.

Are we supposed to solve that or is it just to inform us that it's likely going to rain?

Anyways, answer below

[hide]The probability for it raining 36 hours in a row is the probability that it rains for exactly three (in a row) of the five periods of day and night.

20%*60%*60% + 60%*60%*50% + 60%*50%*40% = 37.2%[/hide]

That doesn't answer the question.  The question asks to find the probability that it thunders exactly 36 hours, not necessarily consecutively, during ARML.

So you're going to have $ \binom{5}{2}$ terms to add up. That's just ugly.

Edit: Errr... never mind.

The forecast for Las Vegas: Friday high 91/low 66, mainly sunny.  Saturday 94/69, sunshine.  Sunday 94/70, abundant sunshine.  What is the probability that it will ever rain in Vegas?  ;)

As far as I can remember, there's been a thunderstorm at Penn State every single Friday night I've gone to ARML.  (That's 7 years, I believe.)  Not that this has anything to do with the subject at hand ....

Not gonna happen at Las Vegas. Forecast now is for sunny both Friday and Saturday, high of 89 on Friday (a little cooler than normal), low of 70 Friday night, high of 95 on Saturday.

Prediction for Penn State: It rains all Friday Night and then is relatively dry and sunny Saturday.
There was absolutely no logical reasoning behind this prediction  :P but that's exactly what happened the last many years...

Chances have been upgraded to 70% / 70% / 60%  :(


Simplify ($ i^2 \equal{} \minus{} 1$):
\[ \frac {5(3 \plus{} i \plus{} 2i^2)^2 \minus{} (1 \plus{} 2i \plus{} 3i^2 \plus{} 4i^3)^3 \minus{} (i \plus{} 2i^2)^4}{1 \plus{} 3i \minus{} 3i^2 \minus{} i^3}
\]
[hide]This question made me laugh when I saw it; I probably worked on it for half of the given time and still got it wrong.  :(

And just now, I typed it into my graphing calculator and it took me half a minute to get the right answer.[/hide]


real analysis
function
real analysis unsolved

Let $\mu$ be a measure that is defined on all Lebesgue measurable subsets of the real numbers. Suppose there is a function $u$ from $[0,\infty)$ into $[0,\infty)$ with 0 and 1 as fixed points, and the property that $\mu((a,b)) = u(b-a)$ for $a < b$ in $\mathbb{R}$. Is it true that $\mu$ is the same as $\lambda_{1}$?

it is a consequence of this: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=measure&t=114694

by "measure" i hope you mean a positive measure, don't you?

if so, the function $u$ has to be monotone, and it has no atoms (since $u(1)=1<+\infty$). so it's continuous.
but $\mu$ is additive, so $u(c-a) = u(c-b)+u(b-a)$ whenever $0\le a\le b\le c$.
now define $\tilde{u}: \mathbb{R}\to \mathbb{R}$ as follows: $\tilde{u}(x) = u(x)$ for $x\ge 0$, and $\tilde{u}(x) =-u(-x)$ for $x<0$.
it's easy to check that this function is still additive (this time on the whole real line, and not just on positive reals), and since it's continuous, and $\tilde{u}(1)=1$, $\tilde{u}$ is the identity, hence the measure is lebesgue measure.


function
integration
abstract algebra
calculus
complex analysis
complex analysis unsolved

How can I show that if $ f \in H^p$ then the harmonic function
$ u_f(z) \equal{} \int_0^{2\pi} P_z(t)|f(e^{it})|^p \, \frac {dt}{2\pi}$
($ P_z(t)$ is the Poisson kernel)
is the least harmonic function that majorates $ |f|^p$.

I can show that it does majorate (by writing $ f$ as a Blaschke product times a zero-free $ g \in H^p$), but I have no idea about the least part.

Do you understand how to do it if $ f$ is smooth up to the boundary?

In this case I can write $ u_f$ as an integral over a circle of smaller radius (with an appropriate poisson kernel), right? So it would follow from the Poisson representation of a harmonic function that extends smoothly to the boundary of a circle.

Yes, that's the idea. Use smaller circles and then go to the limit pointwise.

Ok, thanks.  :blush:  I was convinced this problem was very hard, because it was stated without proof but also not as an exercise. Again the psychological effect of thinking something is easy/hard


i do pop, mild rock, hip hop, maybe some rap if its good

I voted for rock..

You missed tons of genres.. But even if there were 25 options, you would miss sth.. So I guess it`s ok..

I like almost all genres.. But especially rock..

My favorite artists are: Pink Floyd, Radiohead, R.E.M., Air, Pearl Jam, A-Ha, Interpol, The Shins, Husker Du, Pixies, Suede, Vangelis, PJ Harvey, Belle & Sebastian, Bob Marley, Bad Religion, Erykah Badu...

These days I like to listen Goo Goo Dolls, Mike Oldfield, Charlatans UK, Grandaddy..

I frequently find that some people miss the distinction between rock and rock'n'roll.  The latter, I prefer.

I love classical music - especially from the romantic period for those who care - but I'll listen to everything except rap, country, and polka. And I'll even listen to some country.

Bob Dylan is a literary genius, though. I tell you, there's nothing like singing Bob Dylan at the top of your lungs with the car windows down and having people stare at you as if you were a leper...

"But I was so much older then; I'm younger than that now" has to be the most profound line in all of music.

[quote="devilsbrother190"]im into christian rock and punk rock

switchfoot and reliant k are two of my favorite bands[/quote]

Ah, switchfoot. Good stuff.

Oldies, i.e. Rock 'n Roll.  That means the Beatles, the Stones, the Bee Gees, Van Morrison, Kansas, Bob Seger, Jim Croce, Cat Stevens, Billy Joel, the Animals, BTO, the Beach Boys, Boston, CCR, Elvis, the Righteous Brothers, Queen, the Temptations, the Doobies, and so on.

But I like some modern rock (if that's the right genre) musicians - Counting Crows, DMB, Sting, and occasional songs by Matchbox 20, U2, etc.

[quote="mathfanatic"]Oldies, i.e. Rock 'n Roll.  That means the Beatles, the Stones, the Bee Gees, Van Morrison, Kansas, Bob Seger, Jim Croce, Cat Stevens, Billy Joel, the Animals, BTO, the Beach Boys, Boston, CCR, Elvis, the Righteous Brothers, Queen, the Temptations, the Doobies, and so on.
[/quote]

Nice to meet a fellow Oldies fan.

EMINEM! EMINEM!....

Besides him...  Jimmy Eat World, Sum 41, Coheed and Cambria, Dashboard Confessional, Ben Folds Five, John Mayer, Funeral for a Friend, and some DMB...

um...shouldn't u add a poll for "none"? i don't listen to music.

I really enjoy listening to the Beatles. Their songs are really good! My favorite: "The Long and Winding Road"

But I like all kinds of music, even Country! Actually, now that I think about it, I do not like rap. I believe that rap is distatesful and should not be considered music. Of course, that's my opinion.....

so...aren't you going to add a "none" poll?

[quote="Danbert"]I only listen to classical, especially Beethoven and Chopin.  I can't stand other stuff.
No angry replies, please.[/quote]

ROBINHOOD3000 ANGRY...only kidding, of course.

What, no Lizst?

I like classical music and Weird Al music. I HATE rap and country. And I listen to any else. Yay classical is in second place. :)

[quote="A+MATH"]I like classical music and Weird Al music.[/quote]

That's an interesting mix.

And Robinhood3000, same to you.  Although I think we've already discussed this in the Oldies thread.

True enough, but it can't be said enough.

That, and I think my Oldies thread has all but vanished.

I despise rap.
J-pop, J-rock ballads, instrumental jazz, classical, and opera.
Not a real big fan of American pop music.


geometry
circumcircle
incenter
ratio
geometric transformation
reflection
parallelogram

Dear Mathlinkers,
let ABC be a triangle, 1 the circumcircle of ABC, I the incentre of ABC, M a point on BC,
and 2, 3 the Thébault's circles wrt to M.
Prove : the radical axis of 2 and 3 passes through the midpoint of IM.
Any synthetic proof for my the two last problems?
Sincerely
Jean-Louis

In $ AM$ cuts the circumcircle $ (O)$ again at $ M'.$ Let $ P$ be tangency point of the A-mixtilinear incircle with $ (O).$ Let $ AI$ cut $ (O)$ again at $ X$ and let $ (X)$ be a circle with center $ X$ and radius $ XB \equal{} XC \equal{} XI.$ Let the Thebault circles $ (2), (3)$ touch $ BC$ at $ U, V$ and $ (O)$ at $ E, F.$ Inversion in $ (X)$ takes $ BC$ to $ (O)$ and $ U, V$ to $ E, F,$ hence $ \overline{XU} \cdot \overline{XE} \equal{} XI^2 \equal{} \overline{XV} \cdot \overline{XF},$ $ UVFE$ is cyclic and moreover, $ X$ is on the radical axis of $ (2), (3)$ (well known). According to [url]http://www.mathlinks.ro/viewtopic.php?t=233986[/url], $ UVPM'$ is also cyclic. Radical axis $ UV$ of the circles $ \odot(UVFE), \odot(UVPM'),$ radical axis $ EF$ of the circles $ \odot(UVFE), (O)$ and radical axis $ M'P$ of the circles $ \odot(UVPM'), (O)$ meet at their radical center $ Q \in BC.$ $ E, F$ are external similarity centers of the circle pairs $ (O), (2)$ and $ (O), (3),$ hence $ Q \in BC$ is the external similarity center of $ (2), (3).$ By Thebault theorem, $ IQ$ is the common center line of the incircle $ (I)$ and the circles $ (2), (3).$

Parallel to $ AM$ through $ I$ cuts $ BC$ at $ S.$ According to the proof of [url]http://www.mathlinks.ro/viewtopic.php?t=233986[/url], $ XS \perp IQ.$ Let perpendicular to $ IQ$ through $ I$ cut $ AM$ at $ K.$ $ \triangle AIK \sim \triangle IXS$ are centrally similar, having parallel sides. Let the incircle touch $ BC$ at $ D,$ let $ A'$ be midpoint of $ \overline{BC},$ and let parallel $ k \parallel BC$ through $ K$ and perpendicular $ p \perp BC$ through $ I$ intersect at $ T.$ $ T, A'$ are corresponding points of the centrally similar $ \triangle AIK \sim \triangle IXS$ with similarity coefficient  $ \frac {\overline{IT}}{\overline{XA'}} \equal{} \frac {\overline{AI}}{\overline{IX}}.$ Let the angle bisector $ AI$ cut the side $ BC$ at $ X'$ and let $ (I_a)$ be the A-excircle of the $ \triangle ABC$ centered at $ I_a.$ The double ratio $ \frac {\overline{AI}}{\overline{AI_a}} \equal{} \minus{} \frac {\overline{X'I}}{\overline{X'I_a}}$ is harmonic on account of $ A, X'$ being the external and internal similarity centers of $ (I), (I_a).$ In addition, $ X$ is the midpoint of $ \overline{II_a},$ therefore $ \frac {\overline{AI}}{\overline{IX}} \equal{} \minus{} \frac {\overline{IX'}}{\overline{XX'}} \equal{} \minus{} \frac {\overline{ID}}{\overline{XA'}}.$ As a result, $ \overline{IT} \equal{} \minus{} \overline{ID},$ so that the parallel line $ k \equiv KT \parallel BC$ is tangent to the incircle $ (I)$ at $ T.$ Since $ IK \perp IQ$ by definition and $ I$ is on a midparallel $ l$ of $ k \parallel BC,$ the line $ KQ$ is a reflection of both the incircle tangent $ BC$ in $ IQ$ and the incircle tangent $ KT$ in $ IK,$ i.e., a tangent of $ (I).$

Let the midparallel $ l$ of $ k \parallel BC$ cut $ AM$ at $ L.$ As $ IS \parallel KL$ and $ \overline{IS} \equal{} \overline{KL},$ $ KLSI$ is a parallelogram and $ SL \parallel IK$ and $ SL \perp IQ.$ As $ XS \perp IQ$ as well and $ X$ lies on the radical axis of the circles $ (2), (3),$ $ XSL$ is their radical axis. But $ LMSI$ is a parallelogram, its diagonals $ SL, IM$ cutting each other in half.

Though it might be the same as Vladimir's, here is mine: [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=777557517&t=214426[/url] (read posts #2, #3, #5).


calculus
integration
trigonometry
function
calculus computations

Supose you have the integral $ \int_0^\pi\sin(\theta)\,d\theta$
Everyone knows that the result of this integral is 2, but.. what if we tried a variable change like:
$ y \equal{} \sin(\theta)$
for $ \theta\equal{}0$, $ y\equal{}0$
for $ \theta\equal{}\pi$, $ y\equal{}0$
so you get an integral going from 0 to 0 which means the result is zero... where's the flaw?  :oops:  :?:  :ninja:  :huh:

You have to go all the way through with the substitution- what about the $ dy$ and $ d\theta$? We have $ dy \equal{} \cos\theta\,d\theta$, or $ d\theta \equal{} \pm\frac1{\sqrt {1 \minus{} y^2}}\,dy$. That $ \pm$ is trouble, since it changes in the middle of the integral. We could make it into $ \int_0^1 \frac {y}{\sqrt {1 \minus{} y^2}}\,dy \plus{} \int_1^0 \frac { \minus{} y}{\sqrt {1 \minus{} y^2}}\,dy \equal{} 2\int_0^1 \frac {y}{\sqrt {1 \minus{} y^2}}\,dy$, and that form is clearly nonzero.
[Edit- corrected transformed integrals]

Recall what the substitution rule actually states:
Given a continuously differentiable function $ g(y)$ and a continuous function $ f(x)$ it is true that
\[ \int_a^b{f(g(y))g'(y)dy = \int_{g(a)}^{g(b)}{f(x)dx}
}\]
Now in your case $ f(x) = \sin (x)$, but what is your $ g$?  Well, you don't state it, but since you're formally making the substitution $ y = \sin \theta$, you're implicitly setting $ g(y) = \arcsin y$.  The problem with this is that $ \arcsin y = \pi$ has no solution (recall the range of $ \arcsin$ is $ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$), so the substitution simply isn't valid.  Though, as jmerry suggested, there are tweaks you can do to make it work.

The most important thing to get out of this is that the substitution rule is really a very precisely stated theorem -- and all theorems you encounter in mathematics are valid only if the stated conditions are met.  If you ignore the conditions and proceed heuristically, you'll inevitably stumble onto 'contradictions' like this one.

We can always force a substitution through with an inverse- if we're substituting something one-to-one. If we're not, the d_ factor doesn't work uniquely, and we have exactly this sort of problem.

Great answers everyone! thanks a lot.
This is why we, physicists, need mathematicians... to kick us back into doing things the proper way.


HI,
I am solving "Problem solving strategies " by ARthur Engel.I am able to solve all the chapters , save one i.e. Extremal principle.The approaches seem to me very ingenious and hard to guess.Do you know any of the sites or material where this principle is explained in detail ?
thanks
kishore

i think i know what you are talking about.
in ACoPS it is referred to as the extreme principle.
basically what you do is is arrange what you are looking at it some sort of order. then you look at the smallest or largest element. sometimes this is combined w/ proof by contridiction to prove that some statement is true or false.


ugh, don't ask about art class today...

yes, this is actually sorta relevant

hint: shen (hao shen) is in my art class

[quote="illusiat"]i predict that alan from clarke will definitely make it to nats. David Fink also has a good chance, if he actually studies. Shen and Pejin also might make it. I predict that there is 1 random guy who nobody knows who gets to nats. for team, clarke 1st, diamond 2nd, acton ~7th :wink:[/quote]

No. No predictions. They'll be wrong. They are useless. Especially a year beforehand. No. Bad.

Yeah, don't lower your already low self esteem.

He's in 8th grade, he will be in high school next year........I like how I am on the list, it makes me feel special. :lol: Art class? I am missing something?

And you forgot about Jonathan Tidor from Clarke, that was sick this year, and all the incoming maybe beastly 5th graders.

Ok now you need to start lowering your ridiculously high self-esteem.

Can I predict 5 Clarke kids make State Countdown?  Or is that hoping?  I always get hoping and predicting mixed up :D

Yeah, what? Wasn't state like...9 days ago? wtf? But I'll be a hypocrite.

2009 Results:

1. Carl
2. Carl
3. Carl
4. Alan Chou
5. Samlmaml

team:

1. Carl
2. Clarke

Thanks cat nose. So I'm improving 2 places since last year? Well actually, since at least 2 of those carls would be DQed, I'd still make nats. But I'd probably have to go to mexico again like last year...(the reason i didn't go last year)

[quote]i predict that alan from clarke will definitely make it to nats. David Fink also has a good chance, if he actually studies. Shen and Pejin also might make it. I predict that there is 1 random guy who nobody knows who gets to nats. for team, clarke 1st, diamond 2nd, acton ~7th[/quote]

why will acton get 7th next year???
You never know who will appear in RJ Grey next year......

and
maybe I'll stay back in 8th grade ( I don't know why it is possible with the grades I have now... but...)
and I might get first or second in state... yay!!!!   :clap: :clap:

wait.... is mathcounts based on age or grade (I think it's grade)

Yea grade.
But it would be pointless to waste one year of your life just to participate in a Math Competition.

[quote="shentang"]He's in 8th grade, he will be in high school next year........I like how I am on the list, it makes me feel special. :lol: Art class? I am missing something?

And you forgot about Jonathan Tidor from Clarke, [b]who[/b] was sick this year, and all the incoming maybe beastly 5th graders.[/quote]

oh yeah, I forgot, it was during lunch...

Hao, did you just call Jonathan "that"?

Since I think that's enough of my team predicting how they'll do next year when they still haveother things to focus on(AIME, being first in the nation in SIGMA, breaking the record for most points in IMLEM, a little thing called MC nats.....) I'm going to lock this one up.  Wait until summer to start predicting your nonsense again :lol:


Let a,b,c,r,s be real numbers such that ar^2+br+c=0
                                                        -as^2+bs+c=0
  Prove that there exist u is real numbers such that (u-r)(u-s)<=0 and (a/2)u^2+bu+c=0

It is quite simple.  :)

 Note that $ar^2+br+c= 0$ and $-as^2+bs+c= 0$ implies that $a(r^2+s^2)= 0$, by substracting the two equations. Then, $a=0$ or $r^2+s^2=0$.
 If $a=0$, we have that $br+c=bs+c\Rightarrow r=s$. Then, we have

 $(u-r)(u-s)\leq0\Leftrightarrow(u-r)^2\leq0\Leftrightarrow (u-r)^2=0\Leftrightarrow u=r=s$.
So, if $a=0$, we have that $bu+c=0\Rightarrow u=r=s=-c/b$.

Now, if $r^2+s^2=0$, it happens that $r=s=0$. So, we have that
$(u-r)(u-s)\leq0\Leftrightarrow(u-0)^2\leq0\Leftrightarrow u^2\leq0\Leftrightarrow u=r=s=0$.
Then, if a is a real number different form zero, u=0 satisfies all the conditions of the problem.

[quote="carlaperez_20"] Note that $ar^2+br+c= 0$ and $-as^2+bs+c= 0$ implies that $a(r^2+s^2)= 0$, by substracting the two equations.[/quote]

If I'm subtracting correctly, we get $a(r^2+s^2)+b(r-s)=0$

agree with farenhajt that carlaperez_20's solution is false.Subtract 2 equation we get a(r^2+r^2)+b(r-s)=0 not
a(r^2+s^2)=0
   carlaperez_20 can you check your solution???
Anyone has the solution?? Please help me.

[quote="bui_haiha2000"]...and (a/2)u^2+bu+c=0...[/quote]
Check it, please. I think the condition is false

It's easy to see that $r, s$ are the roots of the quadratic equation:     $ax^2+bx+c=0$
By Viète's formula:
   $r+s=\frac{-b}{a}$
   $r.s=\frac{c}{a}$
We need to prove that thre exists a real number u satisfying:
   $(u-s)(u-r) \le 0$
 $\Longleftrightarrow u^2-(r+s)u+rs \le 0$
 $\Longleftrightarrow u^2+\frac{b}{a}.u+\frac{c}{a} \le 0$
 $\Longleftrightarrow au^2+bu+c \le 0$
This is true because $\Delta \ge 0$ (see the function's graph).
Then consider:
 $\frac{a}{2}u^2+bu+c=0$
 $\Longleftrightarrow \frac{a}{2}u^2=au^2+bu+c$
  */if $a>0$
    $\frac{a}{2}u^2=au^2+bu+c \le 0$
    $\Longleftrightarrow u^2 \le 0$
    $\Longleftrightarrow u=0$
  */if $a<0$
    similarly like above...:$u=0$
  This means $r \le 0 \le s$, which is equivalent to $rs=\frac{c}{a}<0$: unsure (not appear in the problems' condition)
 :?:  :!:  :?:


calculus
integration
trigonometry
calculus computations

[quote="kunny"]Calculate
\[\int x\cos (1+x^{2}) dx \]
[/quote]
Set $u=1+x^{2}\Longleftrightarrow du=2xdx$, which yields:

\begin{eqnarray*}\int{x\cos (1+x^{2})dx}&=& \frac{1}{2}\int{\cos udu}\hfill \\ &=& \frac{1}{2}\sin (1+x^{2})+c \hfill \\ \end{eqnarray*}


For any three positive reals $a, b, c$, we have

$\displaystyle \frac{a^3+abc}{\sqrt{b^2+c^2}}+\frac{b^3+abc}{\sqrt{c^2+a^2}}+\frac{c^3+abc}{\sqrt{a^2+b^2}}\ge\frac{\sqrt{2}}{3}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2$

WLOG assume $a \ge b \ge c$. By Chebysev's inequality,
\[\displaystyle\sum \frac{a^3+abc}{\sqrt{b^2+c^2}} \ge \frac{1}{3}(\displaystyle\sum a^3+abc)(\displaystyle\sum \frac{1}{\sqrt{b^2+c^2}}).\]
By Schur's inequality,
\[\displaystyle\sum a^3+abc \ge \displaystyle\sum a(b^2+c^2),\]
so together with Cauchy Schwarz's inequality,
\[\frac{1}{3}(\displaystyle\sum a^3+abc)(\displaystyle\sum \frac{1}{\sqrt{b^2+c^2}}) \ge \frac{1}{3}(\displaystyle\sum a(b^2+c^2))(\displaystyle\sum \frac{1}{\sqrt{b^2+c^2}}) \ge \frac{1}{3}(\displaystyle\sum \sqrt{a\sqrt{b^2+c^2}})^2.\]
Since $\sqrt{\sqrt{b^2+c^2}} \ge \sqrt{\sqrt{2}}\frac{\sqrt{b}+\sqrt{c}}{2}$, so
\[\displaystyle\sum \frac{a^3+abc}{\sqrt{b^2+c^2}} \ge \frac{1}{3}(\displaystyle\sum \sqrt{a\sqrt{b^2+c^2}})^2 \ge \frac{1}{3}(\displaystyle\sum \sqrt{a}(\sqrt{\sqrt{2}}\frac{\sqrt{b}+\sqrt{c}}{2}))^2 =\frac{\sqrt{2}}{3} (\displaystyle\sum \sqrt{ab})^2.\]

Here is my solution:
 $\frac{a^3+abc}{\sqrt{b^2+c^2}}-\sqrt{2}bc=\sqrt{2} \left(\sqrt{\frac{a^2}{b^2+c^2}\frac{1}{2}}(a^2+bc)-bc\right)\geq
\sqrt{2}\left( (a^2+bc) \frac{2}{\frac{1}{\frac{a^2}{b^2+c^2}}+2}-bc\right)=\sqrt{2}\left( \frac{2a^4-bc(b^2+c^2)}{2a^2+b^2+c^2}\right)\geq \sqrt{2}\frac{1}{2}\left(2a^2-b^2-c^2\right)$

So, $\frac{a^3+abc}{\sqrt{b^2+c^2}}+\frac{b^3+abc}{\sqrt{a^2+c^2}}+\frac{c^3+abc}{\sqrt{a^2+b^2}}\geq \sqrt{2}\left(ab+ac+bc\right)\geq \frac{\sqrt{2}}{3}\left( \sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2$  :)


I was having trouble sleeping last night and I tossed and turned well into the night. Our local town hall has a clock which strikes on the hour and also strikes just once on the half hour. During one of my more awake moments I heard the clock strike once, but I could not tell what time it was. Half an hour later it struck once again, but I still could not tell what time it was. Finally, half an hour later it struck once again and I knew what the time was. What time was it?


$ a,b,c>0$ Prove that !
$ (a \plus{} b)(b \plus{} c)(a \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b) \le 8(abc)^2$

[quote="bigbang195"]$ a,b,c > 0$ Prove that !
$ (a \plus{} b)(b \plus{} c)(a \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b) \le 8(abc)^2$[/quote]

$ \Longleftrightarrow \sum(a^3(b \plus{} c)(a \minus{} b)( a\minus{}c)) \plus{} (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{}$ $ 4abc\sum(a(a \minus{} b)(a\minus{}c))\geq 0$

 :lol:


Find the sum of even positive divisors of 10,000.

$ (2\plus{}2^{2}\plus{}2^{3}\plus{}2^{4})(5^{0}\plus{}5\plus{}5^{2}\plus{}5^{3}\plus{}5^{4})$

ps: is this a well known formula or something, it looks beautiful

Well, $ 10^{4}\equal{} (2^{4})(5^{4})$

We can have 0, 1, 2, 3, or 4 factors of 5 in such a divisor, and 1, 2, 3, or 4 factors of 2 (we must have at least 1 because the number should be even).

So, the product $ \left(\sum_{i \equal{} 1}^{4}2^{i}\right)\left(\sum_{j \equal{} 0}^{4}5^{j}\right) \equal{} 23430$ gives the sum of all possible numbers $ 2^{i}5^{j},\; i\in\{ 1,2,3,4\},\; j\in\{0,1,2,3,4\}$. This is exactly the sum that Mohamed wrote.


linear algebra
matrix
geometry
graph theory
advanced fields
advanced fields unsolved

Prove that bipartite graphs has no [url=http://mathworld.wolfram.com/IsospectralGraphs.html]isospectral[/url] graphs among connected graphs.

Let me add a few details: a graph on $n$ vertices $v_{1},\dots,v_{n}$ can be represented by an $n\times n$ matrix, in which the $(i,j)$-entry is the number of edges between $v_{i}$ and $v_{j}$. This [i]adjacency matrix[/i] depends on the enumeration of vertices, but its eigenvalues do not. Since the matrix is symmetric, it has $n$ eigenvalues (counting multiplicities), which form the spectrum of the graph. A natural (and very difficult) question is: which graphs are determined by the spectrum?

Here we consider only [url=http://mathworld.wolfram.com/SimpleGraph.html]simple[/url] graphs (i.e. no multiple edges and no loops starting and ending at the same vertex). The adjacency matrix of such a graph has only $0$s and $1$s, with $0$s on the diagonal. For example, the complete graph on $3$ vertices (a triangle) has
\[A=\begin{pmatrix}0& 1& 1\\ 1& 0 & 1 \\ 1& 1& 0 \end{pmatrix}\]
Let $G$ be a [url=http://mathworld.wolfram.com/BipartiteGraph.html]bipartite graph[/url]. If we enumerate its edges so that $v_{1},\dots,v_{m}$ are in the first part, and $v_{m+1},\dots,v_{n}$ are in the second, then the adjacency matrix will look like this:
\[A_{G}=\begin{pmatrix}0& C^{T}\\ C& 0 \end{pmatrix}\]
where $C$ is a rectangular $m\times(n-m)$ matrix. The claim is that there is no connected graph on $n$ vertices with the same spectrum as $G$.

I don't have a solution, but can offer a simple observation: two $n\times n$ matrices $A_{1}$ and $A_{2}$ have the same spectrum if and only if $A_{1}^{k}$ and $A_{2}^{k}$ have the same trace for all $k\ge 1$ ($1\le k\le n$ is enough). One can try to interpret the trace of $A_{G}^{k}$ in terms of the geometry of $G$...

Thanks for your reply. Let me specialize the question for the moment. Prove that [i]complete[/i] bipartite graphs has no isospectral graphs among connected graphs.
(PS I doubt the general case.)


function
algebra
rational function
algebra unsolved

This is a generalization of
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157768
Let $ a,b,c$ be 3 reals s.t. $ a+b+c<0,a^{2}>bc,b^{2}>ac,c^{2}>ab$.
We seek 3 reals $ x,y,z$ s.t.: $ \sqrt{yz}-x=a,\sqrt{xz}-y=b,\sqrt{xy}-z=c$.
1) Find explicitly a solution which is a rational function of $ a,b,c$.
2) Do there exist other solutions ?

Enough  to consider the equation:
$ x^2\minus{}yz\equal{}a,y^2\minus{}xz\equal{}b,z^2\minus{}xy\equal{}c$
We has :
$ (x^2\minus{}yz)^2\minus{}(y^2\minus{}xz)(z^2\minus{}xy)\equal{}a^2\minus{}bc$
so $ x(x^3\plus{}y^3\plus{}z^3\minus{}3xyz)\equal{}a^2\minus{}bc$
Case 1:$ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}0$
so $ x\plus{}y\plus{}z\equal{}0$ or $ x\equal{}y\equal{}z$
*)$ x\plus{}y\plus{}z\equal{}0$ then
$ a^2\equal{}bc,c^2\equal{}ba,a^2\equal{}bc$
So we can find out $ (x,y,z)$
*)$ x\equal{}y\equal{}z$ same method .
Case 2:$ x^3\plus{}y^3\plus{}z^3\minus{}3xyz$ different from 0.
So $ \frac{x}{y}\equal{}\frac{a^2\minus{}bc}{b^2\minus{}ca}$
     $ \frac{x}{z}\equal{}\frac{a^2\minus{}bc}{c^2\minus{}ab}$
Replace to the first equation we can find out solution of this equations.

Hi TTsphn,
1) you must also prove that, in my notations, necessarily $ x,y,z>0$.
In particular your case $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}0$ is impossible.
2) In my notations the requested solution is :
$ x\equal{}\dfrac{(a^2\minus{}bc)^2}{\minus{}d},y\equal{}\cdots$ where $ d\equal{}a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$.
3) You have proved also that there don't exist any other real solution.

I solve for any $ (a,b,c)\in R$ not necessary your condition.
So can $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}0$


geometry perimeter inequalities trigonometry

A convex quadrilateral $ ABCD$ with area $ 2002$ contains a point $ P$ in its interior such that $ PA \equal{} 24$, $ PB \equal{} 32$, $ PC \equal{} 28$, and $ PD \equal{} 45$.  FInd the perimeter of $ ABCD$.

$ \textbf{(A)}\ 4\sqrt {2002}\qquad \textbf{(B)}\ 2\sqrt {8465}\qquad \textbf{(C)}\ 2\left(48 \plus{} \sqrt {2002}\right)$
$ \textbf{(D)}\ 2\sqrt {8633}\qquad \textbf{(E)}\ 4\left(36 \plus{} \sqrt {113}\right)$

you have to get "lucky" with this question.

as in noticing that 24*32+32*28+28*45+45*24=4004

[hide]Yea, I did it inelegantly too. I noticed that if I made the diagonals perpendicular, the area would be $ 2002$.

Then, the perimeter is $ 40\plus{}4\sqrt{113}\plus{}53\plus{}51\equal{}144\plus{}4\sqrt{113}\equal{}4\left(36\plus{}\sqrt{113}\right) \Rightarrow \boxed{E}$.[/hide]

but seriously, honestly, if this is the "best solution", then i do not find this to be a very good problem for an AMC 12 #24 because its just pure luck that you come up with that.  Also, the solution, once you find it, is pretty trivial, just 1 step using pythagorean theorem.

Then again, if you actually took the 2002 AMC 12B, I would assume that you would memorize that $ 2002 \equal{} 11 \times 13 \times 14$, so the solution wouldn't be so surprising (as in luck-wise).

A complete solution is on the wiki: [[2002 AMC 12B Problems/Problem 24]].

should I add how they found the inequality by using the area formula 1/2 absin C to clear things up?

could someone elaborate on this? (the inequality found in the wiki solution)

[color=darkred][b][u]PP[/u].[/b] A convex quadrilateral $ ABCD$ with area $ 2002$ contains a point $ P$ in its interior such that
$ PA \equal{} 24\ ,\ PB \equal{} 32\ ,\ PC \equal{} 28$ and $ PD \equal{} 45$ .  FInd the perimeter of $ ABCD$ .[/color]

[b]Proof[/b] ([u][b]the future[/b][/u]). Denote the area $[ABCD]=S$ . Therefore, $2S=PA\cdot PB\cdot\sin \widehat {APB}+PB\cdot PC\cdot\sin \widehat {BPC}+$

$PC\cdot PD\cdot\sin \widehat {CPD}+PD\cdot PA\cdot\sin \widehat {DPA}\implies$ $2S\le PA\cdot PB+PB\cdot PC+PC\cdot PD+PD\cdot PA$ .

[u]Observe that[/u] $2S=PA\cdot PB+PB\cdot PC+PC\cdot PD+PD\cdot PA$ . Thus, $\sin \widehat {APB}=\sin \widehat {BPC}=$

$\sin \widehat {CPD}=\sin \widehat {DPA}=1$ , .e. $AC\perp BD$ . In conclusion, the perimeter is $40+4\sqrt {113}+53+51=$ $\boxed{4\left(36+\sqrt {113}\right)}$ .

[b]Solution:[/b] I always struggling on these $PA, PB, PC, PD$ problems but seriously what was this. So it is pretty clear that the diagonals multiply to a number greater than or equal to 4004. [hide=Explanation from AoPS Wiki] (Which is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$)[/hide] Furthermore, notice that the diagonals have to be less than 52 and 77 respectively. However, 52 * 77 = 4004. This implies that the diagonals are straight lines, perpendicular, and that $PAC$ and $PBD$ are degenerate triangles. Use Pythagorean theorem and we are done. I guess the main point of this problem was to realize that there is no way other than bounding the diagonals and then ultimately solving the question.


What is the maximum number of non-overlapping $ 2\times 1$ dominoes that can be placed on a $ 8\times 9$ checkerboard if six of them are placed as shown? Each domino must be placed horizontally or vertically so as to cover two adjacent squares of the board.

We can place at most 34 dominoes.

Consider two parts of the board: $ A$ consists of the squares above and to the left of the six dominoes and also the lower left corner, $ B$ consists of the squares below and to the right of the six dominoes and also the upper right corner.

Color the squares in the usual checkerboard style, so that there are 14 black squares and 16 white squares in A, and there are 14 white squares and 16 black squares in B. Hence there are at most 14 dominoes in each part. Therefore there are at most $ 14\plus{}14\plus{}6\equal{}34$. This is clearly achievable.

Let's prove that the answer is 34.
We color the $ 8x9$ checkerboard like a chessboard. Every domino is maked by a black square and a white square.
The square $ (a,b)$ is the intersection of the row $ a$ and column $ b$.
Let divide the checkerboard into two parts as follow:

We can see that in the red part there are 17 black squares and 13 white squares.
Also we can see that in the blue part there are 13 black squares and 17 white squares.
The parts red and blue only have two intersections.
Then when the maximun number of dominoes are placed, we have to place these two dominoes:
1- (7,1)(8,1) and 2-(1,9)(2,9).
And the answer is 28+6=34 because in the both parts the diference of the numbers of squares of diferents colors is at least 2.

Color the checkerboard in a chessboard fashion, and split it into sections, as shown to below to the left. Notice there are $14$ white squares in the orange section, and that it is not possible to place a domino that has a black square inside the orange region and white square outside the orange region. Since each domino covers a black and white square, there can be at most $14$ dominoes that cover squares in the orange region. Similarly, there are at most $14$  dominoes that cover squares in the blue region. There are $6$ dominos placed in the beginning, so we must have at most $14+14+6=34$ dominoes. A construction with $34$ dominoes is shown below to the left. $\square$

We break the $8\times 9$ grid into two sections: $A$ - including the rightmost column, and every square vertically below one of the placed dominoes and $B$ - including the leftmost column and every square which is vertically above one of the already placed dominoes. Then, we color the square in the usual chessboard style. Then, section $A$ has 14 black squares - so we can place almost 14 dominoes on it. Section $B$ has 14 white squares, so we can also place almost 14 dominoes on it. Thus, we cannot place more than $14+14+6=34$ dominoes on this board. Now, we shall give a construction which places 34 dominoes on such a $8\times 9$ board.


function
induction
algebra proposed
algebra

Find all functions $ f: \mathbb{Z} - \{0\}\to\mathbb{Q}$ which satisfy
\[f\left(\frac {x + y}3\right) = \frac {f(x) + f(y)}2,\]
for all $ x,y\in\mathbb{Z} - \{0\}$ such that $ 3\mid x + y$.

[quote="BaBaK Ghalebi"]you are right pco,the problem is not correctly stated,the original problem is the following one:

[b]find all functions $ f: \mathbb{Z} \minus{} \{0\}\to\mathbb{Q}$ which fulfill the following condition:

$ f\left(\frac {x \plus{} y}3\right) \equal{} \frac {f(x) \plus{} f(y)}2$[/b]

so we must have $ 3\mid x \plus{} y$ and also $ x,y\in\mathbb{Z} \minus{} \{0\}$ for example $ x \equal{} 1,y \equal{} 2$[/quote]

Let $ P(x,y)$ be the property $ f\left(\frac {x \plus{} y}3\right) \equal{} \frac {f(x) \plus{} f(y)}2$

$ P(x,2x)$ implies $ f(2x)\equal{}f(x)$ $ \forall x\neq 0$
$ P(3x,3x)$ implies $ f(2x)\equal{}f(3x)$ $ \forall x\neq 0$ and so $ f(3x)\equal{}f(2x)\equal{}f(x)$ $ \forall x\neq 0$

Assume then $ f(nx)\equal{}f(x)$ $ \forall x\neq 0$ for some integer $ n\neq 0$. Then :

$ P(3nx,3x)$ implies $ f((n\plus{}1)x)\equal{}\frac{f(3nx)\plus{}f(3x)}{2}$ and, since we know $ f(3x)\equal{}f(x)$, $ f((n\plus{}1)x)\equal{}\frac{f(nx)\plus{}f(x)}{2}$, and, since $ f(nx)\equal{}f(x)$, $ f((n\plus{}1)x)\equal{}f(x)$
So $ f(nx)\equal{}f(x)$ $ \forall x\neq 0$ and $ \forall n>0$

Then $ P(6x,\minus{}3x)$ gives $ f(x)\equal{}\frac{f(6x)\plus{}f(\minus{}3x)}{2}$ $ \equal{}\frac{f(x)\plus{}f(\minus{}x)}{2}$ and so $ f(\minus{}x)\equal{}f(x)$ $ \forall x$

And so $ f(nx)\equal{}f(x)$ $ \forall x\neq 0$, $ \forall n\in\mathbb{Z}^*$

And so the solutions are $ f(x)\equal{}c$ for any $ c\in\mathbb{Q}$ (and it is easy to check that these necessary conditions are sufficient).

But I don't understand the interest of $ \mathbb{Z}^*$ instead of $ \mathbb{Z}$ or the interest of $ \mathbb{Q}$ instead of $ \mathbb{R}$

[quote="pco"]But I don't understand the interest $ \mathbb{Z}^{*}$ of  instead of $ \mathbb{Z}$ or the interest of $ \mathbb{Q}$ instead of $ \mathbb{R}$[/quote]

this problem was given in an iranian olympiad about 13 years ago,and the official solution uses induction to prove that $ f$ is constant,but as far as I can see,the official solution didn't use tha fact $ f: \mathbb{Z}^{*}\to\mathbb{Q}$ either (instead of $ f: \mathbb{Z}\to\mathbb{R}$)


algebra
polynomial
absolute value
algebra solved

Let P be a polynomial in R[x] satisfied For all x in the interval [0,1] we have |P(x)|<=1.
Prove that:|P(-1/n)|<=2 n+1 -1.(n is degree of P)

Indeed, an easy, but beautiful problem. I hope Im not wrong with my computations, but here is what I think works: consider the numbers $ x_k=\frac{k}{n} $ with $ k=0,1,,n $. Write the Lagrange interpolation formula for these points and take for x the value $ x=\frac{-1}{n} $. Then, use the fact that $ |P(x_k)| $ is at most 1 and the triangle inequality. The absolute value of the terms that appear in the formula is $\frac{(n+1)!}{(i+1)!(n-i)!} $ and the sum of these things is $ 2^{n+1}-1$. Is it correct?


Consider the inequality
\[(a_1+a_2+\dots+a_n)^2\ge 4(a_1a_2+a_2a_3+\cdots+a_na_1).\]
a) Find all $n\ge 3$ such that the inequality is true for positive reals.
b) Find all $n\ge 3$ such that the inequality is true for reals.

Its surely not that obvious. Does anyone no a full solution?

Hint: Use Abel SUmmation Formula for the RHS.

(a) For $n = 3$, if we let $a_1 = a_2 = a_3 = 1$, then we get $9 \geq 12$, which is clearly false.

We claim that the inequality is true for $n \geq 4$. We use induction on $n$.

For $n = 4$, the inequality can be rearranged to get $(a_1 - a_2 + a_3 - a_4)^2 \geq 0$, which is true for all reals $a_1, a_2, a_3, a_4$.
If the inequality is true for $n$, then we will show it is also true for $n + 1$. Note that since the inequality is cyclic, we can assume that $a_1$ is the largest among all $a_i$'s. Then $(a_1 + \ldots + a_{n-1} + (a_n + a_{n+1}))^2 \geq 4(a_1a_2 + \ldots + a_{n-2}a_{n-1} + a_{n-1}(a_n + a_{n+1}) + (a_n + a_{n+1})a_1) = 4(a_1a_2 + \ldots + a_na_{n+1} + a_{n+1}a_1) + 4(a_{n-1}a_{n+1} + a_n(a_1 - a_{n+1})) \geq 4(a_1a_2 + \ldots + a_na_{n+1} + a_{n+1}a_1),$ which completes the induction.

(b) We have already shown that the inequality is false for $n = 3$ and true for $n = 4$. For $n \geq 5$, let $a_1 = a_2 = 1, a_3 = 0, a_4 = -2, a_5 = \ldots = a_n = 0$. Then we get $0 \geq 4$, which is clearly false. So the inequality is only true for $n = 4$.


geometry
circumcircle
geometric transformation
homothety
geometry proposed

[color=blue]
$ABC$ is a triangle and $B',C'$ are two points on its circumcircle so that $\angle{C'AB}=\angle{B'AC}$.The line trough $C'$ parallel to $AC$ intersects the line trough $B'$ parallel to $AB$ at $A'$.Prove that the points $A,A'$ and $A''$ are collinear,where $A''$ is the point of intersection of the lines $BC'$ and $CB'$.    [/color]

I think that the point $A'',$ must be defined as the intersection point of the segment lines $BB',$ $CC',$ instead of $BC',$ $CB'.$

So, we have that $\angle C'CB = \angle C'AB = \angle B'AC = \angle B'BC$ $\Longrightarrow$ $B'C'\parallel BC$ and now, we see that the sidelines of the triangles $\bigtriangleup ABC,$ $\bigtriangleup A'B'C',$ are parallel each other, one per one.

Hence, these triangles are perspective and so, we conclude that the segment lines $AA',$ $BB',$ $CC',$ are concurrent at one point. That is, the point $A''\equiv BB'\cap CC',$ lies on the segment line $AA'$ $($ also on the midperpendicular of the side segment $BC$ $)$ and the proof is completed.

Kostas Vittas.

Dear Vittas,actually there is a typo but it is not what you mentioned(it will be too easy that way :) ).
There should be $\angle{C'BA}=\angle{B'CA}$,insted of$\angle{C'AB}=\angle{B'AC}$    :roll:

You can also think of point $B',C'$ as two point so that $B'C'$ is perpendicular to the diametre trough the vertex $A$(probably this is a better way to define them because there again can be a confusion if both $B'$ and $C'$ lie on the midlle arc $BC$.)

Thank you dear Tiks for the clarification. Now is not so easy, but not difficult.

We denote as $B'',$ $C'',$ the intersection points of the circle $(O),$ from the segment lines $B'A'\parallel AB,$ $C'A'\parallel AC,$ respectively. Also we denote as $S,$ the intersection point of the segment lines $BB'',$ $CC''.$

So, we have that $\angle B''CB = \angle B'CA = \angle C'BA = \angle C''BC$ and hence, we have, as before, $B''C''\parallel BC.$

Because of the triangles $\bigtriangleup ABC,$ $\bigtriangleup A'B''C'',$ have their sidelines parallel each other, one per one, we conclude that these are perspective $($ also Homothetic $),$ with perpspector the point $S.$ So, the points $A,$ $A',$ $S,$ are collinear.

From the non-convex inscribed hexagon $BC'C''CB'B'',$ by Pascal’s theorem, we have that the points $A',$ $A'',$ $S,$ are collinear.

Hence the points $A,$ $A',$ $A'',$ are collinear and the proof is completed.

Kostas Vittas.

Congratulations with a very nice solution!!!!!!!!!!!!!!!!

Just one more thing:from Dezarg's theorem we can see that this problem is equivalent to the fact that lines $BC,B'C'$ and $B_{1}C_{1}$ are concurent where $B_{1}\in{{B'A'}\cap{AC}}$ and $C_{1}\in{{C'A'}\cap{AB}}$.

Now it is time to see what was mentioned in the description of this topic.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55443[/url]

Take the points $B'$ and $C'$ to be the same and you will get the problem from the link above.

Let $C''\in BC'\cap AC, B''\in CB'\cap AB, S\in BC'\cap A'B', T\in CB'\cap A'C'$.

A little and easy angle-chasing shows that
$C''AB\sim C'A'S$
$BAC\sim SA'T$
$CAB''\sim TA'B'$
$B''AC''\sim B'A'C'$.

Thus, the quadrilaterals $C''BCB''$ and $C'STB'$ are also similar, having the corresponding sides parallel. Hence, there exist a homothety taking $C''BCB''$ to $C'STB'$ and thus $A$ to $A'$. But since $A''\in C''C'\cap B''B'$, $A''$ must be their center of similtude, hence $A'', A, A'$ are collinear.

I think the difficult part of this problem was to come up with the construcation of the additional points $B"$ and $C"$ :wink: .

Anyway,thank you for your solution Yimin! :wink:

Also I am wondering,are you going to participate in the IMO 2007?

[quote="Yimin Ge"][quote="Tiks"]Also I am wondering,are you going to participate in the IMO 2007?[/quote]

Provided that I qualify...yes :)
What about you?[/quote]

Me too :wink: ,if I will qualify :) .

See you!!!!!!


Hello,

May be what I am going to ask is somewhere and I just did not look appropertly for it, but, Is there a way to post a same topic in more than one forum at once. Let me try to explain myself better trying to make an example: Pretty often you can find problems (or topics in general) that fit in more than one of the subfora (just in case, plural of subforum) on MathLinks, for example, some problems that can be solved either with Geometry, or a totally different solution using only algebra, or with number theory or combinatorics, and some of us may want to place it in all the subfora that apply.

For example, this problem

http://www.mathlinks.ro/Forum/viewtopic.php?t=26695

I would like to place it both in Combinatorics and in Geometry, but if I do it placing it once on each fora, probably it turns messy, because it becomes to different discussions, that could fit nice in just one.

Is there a way to do this? Was I clear enough with what I mean?

Best regards,

In general, we prefer to discourage such things, because it takes up space in our database.  Plus it make our forum too easy to spam.

Please don't post the same thing in more than one topic.  Try to pick the most appropriate topic, and post it there.

As David pointed out we don't want that. Try not to double-post a problem, although this happens a lot (given the numbers of problems here, there are roughly more than 50% chances that a problem has been already posted somewhere) and use the search function.

As for problems that fit in two categories, placing them in one category is enough, the subforums for Geometry for example are not restricted to solutions only involving Geometry (etc.)


Let
$ A \equal{} {\begin{pmatrix} 1 & 1 & \minus{}1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{pmatrix}}$
$ D \equal{} {\begin{pmatrix} 0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2 \\
\end{pmatrix}}$

Prove that $ \exists P$ invertible matrix such that

$ A \equal{} P^{ \minus{} 1}DP$

Hint: Why is it enough to show that both matrices have the same eigenvalues assuming that we are working, say, over the reals?

Observe the second and third columns of A: the sum of first elements is zero, while the sum of second, and third elements, is 2. Therefore, 2 is an eigenvalue. Also, the first two columns are equal, therefore 0 is another eigenvalue. Finally, the trace equals 3, therefore the other eigenvalue is 1. Since A has all eigenvalues distinct, it is diagonalizable, which means A is similar to the diagonal matrix of its eigenvalues, which is D. $ \Diamond$


linear algebra
matrix
algebra
polynomial
invariant
absolute value
linear algebra unsolved

Let $A\in M_n(Z)$ be an invertible matrix such that $|Tr(A^k)|\leq n$ for all $k$. Prove that there exists a matrix $Q\in GL_n(Z)$ such that $Q^{-1}AQ$ is an upper triangular matrix by blocks, with some 1 and -1 on the main diagonal and such that all the other blocks on the main diagonal have characteristic polynomials some cyclotomic polynomials.

It is sufficient to prove that the eigenvalues $\alpha_i$ of $A$ are all roots of unity.  We shall prove this under the weaker assumption that $\vert Tr A^n\vert < P(n)$ where $P$ is a polynomial.

Firstly we prove that $\vert\alpha_i\vert \le 1$ for all $\alpha_i$.  We have $Tr A^n = \sum_i \alpha_i^n$.  Consider the power series $F(z) = \sum_{n=0}^\infty (Tr A^n) z^n = \sum_i \frac{1}{1 -\alpha_i z}$.  Since the coefficient of $z^n$ in $F$ is bounded by a polynomial in $n$, we have $F$ convergent for any $z$ with $\vert z \vert < 1$ and so the radius of convergence of $F$ is 1.  Hence $F$ is analytic in the open unit disc.  We conclude that all the $\vert \alpha_i \vert \le 1$, otherwise there would be a pole of $F$ in the open disc.

Now we show that if $f$ is a monic integer polynomial with roots non-zero algebraic integers all of modulus $\le 1$ then the roots of $f$ are all roots of unity. This is well-known, but here's a proof I like.  Consider a recurrent sequence $(x_n)_{n=1}^\infty$ with driving (auxiliary) polynomial $f$ and initial terms $0,0,\ldots,0,1$ (impulse response sequence).  The general term is $x_n = \sum_i A_i \alpha_i^n$ for certain constants $A_i$.  Since all $\vert \alpha_i \vert \le 1$, $x_n$ is bounded and so can take on only finitely many values.  Hence the sequence must repeat from some point on and (since the constant term of  $f$ is $\pm 1$) the sequence is reversible, so the sequence is fully cyclic.  Thus the sequence admits $X^M-1$ as a polynomial for for $M$, and so $f$ divides $x^M-1$.  Hence the roots of $f$ are all roots of unity.

I am sorry, but may I ask you why from the fact that the eigenvalues are roots of unity it follows what I asked? I don't think it is so straightforward... or I am blind.

The characteristic polynomial of $A$ is monic with integer entries, so the eigenvalues are algebraic integers.  $A$ is invertible, so they are all non-zero and indeed their product is $\det A = \pm 1$.  My posting aimed to show that they all have absolute value at most 1, and that algebraic integers with this property which are roots of a monic polynomial with constant term $\pm 1$ (another way of saying that all conjugates are also bounded by 1) must be roots of unity.

Oh, sorry, your post again I've just read.  The quick answer is rational canonical form for the invariant subspaces.

Sorry, but is this theorem true in any ring? Here all matrices are supposed to have integer entries and yet the only version of the theorem you are speaking is working with fields.

Oh, I see your point: I'd read $Q$ for $Z$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group, $ \phi : G \to G$ a group endomorphism.  Is the center of $ G$ necessarily stable under $ \phi$?  Is there a finite counterexample?

What means stable here¿
If $ \phi$ is an automorphism, then the center is mapped to the center because it is a characteristic subgroup (it can be defined by all-quantified relations on $ G$).

If $ \phi$ is just a general homomorphism, then this is no longer true. For example, take $ G\equal{} S_3 \times (\mathbb Z / 3 \mathbb Z)$. It's center is $ \{ 0 \} \times (\mathbb Z / 3 \mathbb Z)$. Now take $ \phi(a,b) \equal{} ( (1,2,3)^b , 0)$. This is a homomorphism and maps the center not to itself.


geometry
circumcircle
geometric transformation
reflection
symmetry
homothety
geometry proposed

Dear Mathlinkers,
let ABC be a triangle, A'B'C' the orthic triangle of ABC, A"B"C" the Euler's triangle of ABC, X the meet point of B"C' and C"B', O the center of the circumcircle of ABC and N the center of the Euler's circle of ABC.
Prove : X, A and N are collinear.
Sincerely
Jean-Louis

Let $ K$ be the circumcenter of $ \triangle AB'C'$ lying on $ AH,$ since rays $ AO,AH$ are isogonal WRT $\angle BAC.$ Let $ K_0$ be the reflection of $ K$ about $ B'C'.$ Note that $ \angle HC'B'' \equal{} \angle HB'C''$ $\Longrightarrow$ Rays $C'X, C'K_0$ and $ B'X,B'K_0$ form equal and oppositely directed angles  with respect to the pairs of rays $ C'B',C'A$ and $B'C',B'A$ $\Longrightarrow$ $X,K_0$ are isogonal conjugates WRT $ \triangle AB'C'$ $\Longrightarrow$ Rays $ AK_0$ and $ AX$ are isogonal WRT $ \angle BAC.$ Since $ B'C'$ is antiparallel to $ BC,$ the axial symmetry about the internal bisector of $ \angle BAC$  takes $ B'C'$ into $ B''C''$ parallel to $ BC,$  $ K$ into $ K''$ on the ray $ AO$ and  $ K_0$ into $ {K_0}'$ on the ray $AX.$ Then homothety with center $ A,$ that takes $ B''C''$ into $ BC,$ sends $ K''$ into $ O$ and carries ${K_0}'$ into the reflection $O_a$ of $ O$ about the sideline $ BC.$ Since $ AH \equal{} OO_a$ $\Longrightarrow$ $ AHO_aO$ is a parallelogram.

Dear Luis,
why do you haven't thought at the Pascal's theorem?
Sincerely
Jean-Louis

Dear Jayme,
Yes Pascal's theorem is very useful if we let $ B_1, C_1$ be the midpoint of $ AC, AB$.


2. Proove that for every x,y,z>0 we have:
sqrt((x+y)/(x+z))+sqrt((x+z)/(x+y)) <=(y+z)/sqrt(yz).

this is funny... just look at M ineq 11 posted by me! :D

cheers!

We see :
            sqrt((x+y)/(x+z)) + sqrt((x+z)/(x+y))  \leq  (y+z)/sqrt(yz)
 <=>    sqrt ((1+y/x)/(1+z/x) + sqrt((1+z/x)/(1+y/x))  \leq  (y/x+z/x)/ sqrt((y/x)*(z/x))
       We denote :
            a = y/x
            b = z/x
       The inequality becomes :
             sqrt((1+a)/(1+b)) + sqrt((1+b)/(1+a))  \leq  (a+b)/sqrt(ab)
  <=>     2ab + a 2 b + ab 2  \leq  a 2  + b 2  + a 3 +b 3
        Which is obvious by Cauchy


Find all natural numbers $ k$ which can be represented as the sum of two relatively prime numbers not equal to 1.

Or even shorter :P
$ 2k \equal{} (k\minus{}1) \plus{} (k\plus{}1)$ and $ 2k\plus{}1 \equal{} (k) \plus{} (k\plus{}1)$

[quote="Akashnil"][quote="Mathias_DK"]Or even shorter :P
$ 2k \equal{} (k \minus{} 1) \plus{} (k \plus{} 1)$[/quote]
k=6 :?:[/quote]
^^ My fault, you have to have three cases.. But it is still shorter :P


geometry
inequalities
analytic geometry
trigonometry
angle bisector
geometry unsolved

ABC is a right-angled triangle with the right angle at A.  Let D be the foot of the perpendicular from A to BC, and let E and F be the intersections of the bisector of B with AD and AC respectively.  Prove that DC>2EF.

Thanks

I believe a co-ordinate approach is very effective when solving problems involving right triangles and their altitudes, bisectors,etc

Consider a co-ordinate system with origin at A.  Let $ B\equiv (u,0), C\equiv (0,v)$ and $ \ell (BC)\equal{}a$

So, we have as follows
Equation of internal angle bisector of B:$ vx\plus{}(u\plus{}a)y\minus{}uv\equal{}0$
Equation of AD:$ ux\minus{}vy\equal{}0$

So, $ E\equiv (\frac {uv^2}{a(a\plus{}u)}, \frac {u^2v}{a(a\plus{}u)})$

$ F\equiv (0,\frac {uv}{u\plus{}a})$

$ \ell (EF)\equal{}\frac {uv}{a(a\plus{}u)} \sqrt {2a^2\minus{}2au}$

$ \ell (CD)\equal{}\frac {v^2}{a}$

So,
$ DC>2EF\Leftrightarrow \frac {v^2}{a}>2\frac {uv}{a(a\plus{}u)} \sqrt {2a^2\minus{}2au}$
$ \Leftrightarrow v^2a^2\plus{}u^2v^2\plus{}2v^2au\plus{}8au^3>8u^2a^2$

That inequality should be true (it has to be if the problem is correct).
Can someone prove that inequality?? as i could'nt.

It very easy by Coordinate - Decarter:
We chose coordinate as picture.
Let $ D(0;0), B(\minus{}1; 0), A(0; a)$. Because $ AB^2\plus{}AC^2\equal{}BC^2$ $ \implies$ $ C(a^2; 0)$.
We have EA/ED = BA/BD --> coordinate of E.
We have FA/EC = BA/BC --> coordinate of F.
So we have EF and DC by a. Very easy --> CD > 2EF.
OK.

here's a simple proof .
first observe that in the right triangle $ \cos^{2} \frac {B}{2} \equal{} \frac {(a \plus{} c)^{2}}{4a^{2}}$
also $ BE \equal{} \frac {2cx \cos \frac {B}{2}}{c \plus{} x}$ and $ BF \equal{} \frac {2ac \cos \frac {B}{2}}{a \plus{} c}$
where $ AD \equal{} x$
hence we are supposed to prove
$ a \minus{} x > 2(\frac {2cx \cos \frac {B}{2}}{c \plus{} x} \plus{} \frac {2ac \cos \frac {B}{2}}{a \plus{} c})$
on simplification this gives
$ \frac {(a \plus{} c)(c \plus{} x)}{4c^{2}} > \cos \frac {B}{2}$
use $ \frac {x}{c} \equal{} \cos B$
hence we need to prove $ \cos \frac {B}{2} > \frac {2c}{a \plus{} c}$
which after putting the value of $ \cos \frac {B}{2}$ simplifies to proving
$ (a \plus{} c)^{4} > 16a^{2}c^{2}$ which follows from AM-GM and since $ a \neq c$ the inequality is strict

[quote="thanhnam2902"]It very easy by Coordinate - Decarter:
We chose coordinate as picture.
Let $ D(0;0), B( \minus{} 1; 0), A(0; a)$. Because $ AB^2 \plus{} AC^2 \equal{} BC^2$ $ \implies$ $ C(a^2; 0)$.
We have EA/ED = BA/BD --> coordinate of E.
We have FA/EC = BA/BC --> coordinate of F.
So we have EF and DC by a. Very easy --> CD > 2EF.
OK.[/quote]

Could you explain what you mean by implies coordinate of E, F.

Also could you explain what you mean EF and DC by a.

And then explain how that implies CD>2Ef

[color=darkred] [size=117][b]Nice problem ![/b][/size]

[/color] [quote="mathgeniuse^ln(x)"][color=darkred]$ ABC$ is a right-angled triangle with the right angle at $ A$.
Let $ D$ be the foot of the perpendicular from $ A$ to $ BC$ and let $ E$ and
$ F$ be the intersections of the bisector of $ B$ with $ AD$ and $ AC$ respectively.
Prove that $ DC > 2\cdot EF$ .[/color] [/quote]
[b][u]Proof.[/u][/b] $ BE^2 \equal{} BD\cdot BA \minus{} ED\cdot EA$ $ \implies$ $ \underline {BE^2 < BD\cdot BA} \equal{} \frac {c^3}{a}$ $ \implies$ $ BE^2 < \frac {c^3}{a}$ .

Prove easily that $ EF \equal{} \frac {a \minus{} c}{c}\cdot BE$ . Thus, $ \boxed {EF^2 < \frac {c(a \minus{} c)^2}{a}}$ . But

$ \boxed {\frac {c(a \minus{} c)^2}{a} < \left(\frac 12\cdot DC\right)^2}$ $ \Longleftrightarrow$ $ \frac {4c(a \minus{} c)^2}{a} < \left(\frac {b^2}{a}\right)^2$ $ \Longleftrightarrow$

$ 4ac(a \minus{} c)^2 < (a \minus{} c)^2(a \plus{} c)^2$ $ \Longleftrightarrow$ $ \underline {4ac < (a \plus{} c)^2}$ , what is truly.

Therefore, $ EF^2 < \frac {c(a \minus{} c)^2}{a} < \left(\frac 12\cdot DC\right)^2$ $ \implies$ $ \boxed {DC > 2\cdot EF}$ .

[quote="armpist"]A good candidate for PWW.

  To improve visibility draw bisector AX of angle DAC
  and midline MN of triangle DAC naturally parallel to DC.
  All 3 are concurrent.  EF is always perpendicular to AX
  while MN is mostly not.

  T.Y.
  M.T.[/quote]

What is PWW?

Is it like some sort of geometry method.


I learned of a new nim-like game in economics, though I don't know what it's called.
o
oo
ooo
oooo
ooooo

You have stacks of pins (stacks of 1, 2, 3, 4, 5 pins in my class) and you can take away 1 or 2 pins at a time.  The object of the game is to force your opponent into taking the last pin.  The catch: if you take 2 pins, they must be adjacent and in the same stack.  Of course, the pins to not collapse together.

I represent any position as a set of positive integers.  For instance, My class's position would be $\{1,2,3,4,5\}$.  Now, overload the '+' operator to combine two sets (scalar multiplication follows from addition), let $[N]=1$ if you can force a win by moving next, and $[N]=0$ if you cannot force a win.

Here are some rules I have gathered (maybe you want to verify them):
1) $[N+\{1,1\}]=[N]$
2) If $A$ is a position in which you can force a lose, and $[B]=0$, then $[A+B]=1$
3) $\{3\}\equiv\{1,2\}, \{3,3\}\equiv\{2,2\}$
where $A\equiv B$ means $A$ and $B$ function the same and can replace eachother.  I couldn't find any more useful equivalences.

In the $\{1,2,3,4,5\}\equiv\{1,2,1,2,4,5\}\equiv\{2,2,4,5\}$ position, I believe it is best to take one off the end from the stack with 5 pins.  Some useful reduced losing positions to corner your opponent into are: $\{4\},\{1,5\},\{2,2\},\{4,5\},\{1,4,4\},\{2,2,5\},\{1,2,2,4\},\{2,2,2,2\},\{2,2,4,4\}$, and $\{2,2,2,5\}$

BTW, does anyone know what a "break-and-take game" is?

[quote="juggle7"]I learned of a new nim-like game in economics, though I don't know what it's called.[/quote]

It's probably called something like Misere k-Token Nim. The reason why I say maybe is because there are so many variations of nim out there, I can't imagine people agreeing on the same name for many variations. Misere refers to the style of play where the player that takes the last token/pin loses. k-Token Nim is the variation where each player may remove up to k-tokens each turn. In this case, k=2.

[quote="juggle7"]I represent any position as a set of positive integers.  For instance, My class's position would be $\{1,2,3,4,5\}$.  Now, overload the '+' operator to combine two sets (scalar multiplication follows from addition), let $[N]=1$ if you can force a win by moving next, and $[N]=0$ if you cannot force a win.[/quote]

It's great that you're coming up with your own notation; it shows that you've really been thinking about the games. In game theory, your $[N]=1$ is an N-position, which means that the next person that moves can win. $[N]=0$ is a P-position, which stands for previous player wins. Often, in game theory, we search for P-positions.

[quote="juggle7"]Here are some rules I have gathered (maybe you want to verify them):
1) $[N+\{1,1\}]=[N]$
2) If $A$ is a position in which you can force a lose, and $[B]=0$, then $[A+B]=1$
3) $\{3\}\equiv\{1,2\}, \{3,3\}\equiv\{2,2\}$
where $A\equiv B$ means $A$ and $B$ function the same and can replace eachother.  I couldn't find any more useful equivalences. [/quote]

Here, you are referring to Bogus-Nim Heaps, and this is a big idea in game theory. You might want to look it up. It looks like you're on the right track so far!!

[quote="juggle7"]In the $\{1,2,3,4,5\}\equiv\{1,2,1,2,4,5\}\equiv\{2,2,4,5\}$ position, I believe it is best to take one off the end from the stack with 5 pins.  Some useful reduced losing positions to corner your opponent into are: $\{4\},\{1,5\},\{2,2\},\{4,5\},\{1,4,4\},\{2,2,5\},\{1,2,2,4\},\{2,2,2,2\},\{2,2,4,4\}$, and $\{2,2,2,5\}$[\quote]

Nice start. Actually, if you investigate it a bit more, nim has a VERY interesting solution for finding all P-positions.

[hide] If you're still interested in exploring the game, try expressing the piles in base-2.[/hide]

[quote="juggle7"]BTW, does anyone know what a "break-and-take game" is?[/quote][/quote]

Yes, a take-and-break game is a game in which a player can remove tokens and/or (depending on the variation) "break" or divide the heaps into several smaller heaps. [/hide]


Suppose that an ordinary chessboard is tiled with dominos, each of which covers two adjacent squares of the chessboard.  Show that the number of horizontal dominos is even.

Generalizing to a $ 2k \times 2k$ board. Let $ D$ be the number of horizontal dominoes in a tiling, and let $ a_1, a_2, \ldots a_{2k}$ be such that $ a_i$ is the number of squares in column $ i$ covered by a horizontal domino.

Observe the following facts:
(1) $ a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots \plus{} a_{2k} \equal{} 2D$.
(2) It is necessary (though not sufficient) for a full tiling to have $ 2k \minus{} a_i$ even for all $ 1 \leq i \leq 2k$, since the remaining squares must be covered by vertical dominos. Equivalently, we require that all $ a_i$ are even.
(3) Placing a horizontal domino between columns $ i$ and $ i\plus{}1$ adds $ 1$ to both $ a_i$ and $ a_{i\plus{}1}$. From this we can find the invariant $ a_1 \plus{} a_3 \plus{} \ldots \plus{} a_{2k\minus{}1} \equal{} a_2 \plus{} a_4 \plus{} \ldots \plus{} a_{2k}$. Using (1), we also determine that both expressions are equal to $ D$.

Suppose $ D$ is odd. Then by (3), $ a_1 \plus{} a_3 \plus{} \ldots \plus{} a_{2k\minus{}1} \equal{} a_2 \plus{} a_4 \plus{} \ldots \plus{} a_{2k} \equal{} D$. For those sums to be odd, the sequences $ a_1, a_3, \ldots a_{2k\minus{}1}$ and $ a_2, a_4, \ldots a_{2k}$ each must contain at least one odd term. This contradicts (2). Therefore $ D$ is even.

Here's a wonderful link: http://www.cs.utexas.edu/users/EWD/indexEWDnums.html

I discovered it by "accident" some time ago. Everybody should check it. Great stuff in there.

This particular problem is 1306a.

So, their problem is more general. For those who don't want to search the link, their problem is the number of horizontal dominoes with a white square on the left equals the number of horizontal dominoes with a white square on the right.

The approach of my solution completely fails to solve this. Time to rethink this.

A nice solution to the original problem is as follows:

Color the columns of the chessboard alternately white and black. Then, a horizontal domino covers one of each color and a vertical domino covers two of one color. Finish with a simple parity argument.

The more general problem eludes this technique as well. (at least as far as I see)

[hide="the 1306a problem"]
Consider any vertical line between two columns and the set $ H$ of all horizontal dominoes crossing the line. Besides these dominoes, each domino is on one side of the line or the other. Say on the left side, there are $ a$ black squares and $ a$ white squares. Then if $ b$ is the number of dominoes in $ H$ covering a black square on the left side and $ w$ is the number of dominoes in $ H$ covering a white square on the left side, the number of uncovered white squares in the left section in $ a\minus{}w$ and the number of uncovered black squares is $ a\minus{}b$. Then $ a\minus{}w\equal{}a\minus{}b\implies b\equal{}w$. We can apply this to all particular vertical lines, so the number of dominoes touching a black square on the left side is the same as the number of dominoes touching a white square on the left side.
[/hide]

[b]Further Generalization:[/b] If you tile an $ nk\times nk$ square using $ 1\times n$ tiles, then the number of horizontal tiles is divisible by $ n$.


probability
number theory proposed
number theory

There exists a partition of the set of positive integers in two classes such that from any three consecutive positive integers at least one is in each class and no class contains an infinite arithmetic progression.

Here's a nonconstructive argument.  Color each odd integer red or blue randomly.  Color each even integer differently from its predecessor.  So we have randomly partitioned the integers into two color classes that definitely meet the "three consecutive" condition.

But there are only countable many arithmetic progressions, and each has probability 0 of being colored one color.  So our partition has the required properties with probability 1.  In particular, there is a good partition.

If you want a constructive argument, I guess we can use a diagonalization argument.  Enumerate each of the arithmetic progressions, and kill each one as we go, maintaining the odd-even pattern.

Very nice construction RaviB. In fact the author indicated a concrete such partition. Put in the first class only those numbers $n$ for which $ n+[\sqrt{n}]$ is even. But these solutions are much more natural and beautiful.

Thanks.  I like that concrete partition too.


Let $ P$ be an interior point of triangle $ ABC$, and let $ x,y,z$ denote the distance from $ P$ to $ BC,AC,$ and $ AB$ respectively. Where should $ P$ be located to maximize the product $ xyz$?

It is not so complicated... :blush:

Put $ a\equal{}BC,b\equal{}CA,c\equal{}AB$ then we have: $ ax\plus{}by\plus{}cz\equal{}2S_{ABC}$.

By AM-GM inequality: $ 2S_{ABC}\equal{}ax\plus{}by\plus{}ca\geq 3\sqrt[3]{abcxyz}\Rightarrow xyz\leq\frac{8S_{ABC}^3}{27abc}$

Equality holds if and only if $ ax\equal{}by\equal{}cz\Leftrightarrow S_{PAB}\equal{}S_{PBC}\equal{}S_{PCA}\Leftrightarrow P$ is the centroid of $ \triangle ABC$


When 15 is appended to a list of integers, the mean is increased by 2.  When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1.  How many integers were in the original list?

[hide]let
A: average
X: number of integers in the original list

(AX+15)/(X+1) = A+2
(AX+15+1)/(X+1+1) = A+2-1

Crossmultiply,

AX+A+2X+2=AX+15
AX+2A+X+2=AX+16

Subtract second from first*2,
3X+2=14

X=4[/hide]