trigonometry
geometry
circumcircle
incenter
trig identities
Law of Cosines
geometry unsolved

In a triangle $ ABC$,$ BC$ í smallest edge.Choose $ M$ and $ N$ on $ AB$ and $ AC$ respectively such that $ BM\equal{}BC$ and $ CN\equal{}CB$
Prove that
$ \frac{MN}{BC}\equal{}\sqrt{3\minus{}2(cosA\plus{}cosB\plus{}cosC)}$

From the isosceles triangle $ BCM$ we get $ \angle BCM = \frac {\pi - B}{2}$ and $ CM = 2BC\cos\frac {\pi - B}{2}$. We also have $ CN = BC$ and $ \angle MCN = C - \frac {\pi - B}{2} = \frac {C - A}{2}$. Thus we deduce from the law of cosines,
\begin{align*}MN^2 & = CM^2 + CN^2 - 2CM\cdot CN\cos\angle MCN \\
& = 4BC^2\sin^2\frac B2 + BC^2 - 4BC^2\sin\frac B2\cos\frac {C - A}{2} \\
& = BC^2\left(4\sin^2\frac B2 + 1 - 2\sin\frac {B + C - A}{2} - 2\sin\frac {A + B - C}{2}\right) \\
& = BC^2\left(3 - 2 + 4\sin^2\frac B2 - 2\sin\frac {\pi - 2A}{2} - 2\sin\frac {\pi - 2C}{2}\right) \\
& = BC^2\left(3 - 2\cos B - 2\cos A - 2\cos C\right).\end{align*}

[b]General problem.[/b] Let $\triangle ABC$ be a scalene triangle and $ BC$ is its shortest side. $ M,N$ are two points on $AC, AB$ such that $ BN \equal{} CM \equal{} BC.$ Then $ MN$ is perpendicular to $OI,$ where $ O , I$ are the circumcenter and incenter of  $ \triangle ABC,$ and $\frac {_{MN}}{^{BC}} \equal{} \frac {_{OI}}{^{R}}.$

[b]Proof:[/b] In the isosceles triangles $ \triangle OAC$ and $ \triangle OAB,$ we have

$ R^2 \minus{} OM^2 \equal{} AM \cdot MC , \ R^2 \minus{} ON^2 \equal{} AN \cdot BN$

$\Longrightarrow ON^2 \minus{} OM^2 \equal{} AM \cdot MC \minus{} AN \cdot BN$

$ \Longrightarrow ON^2 \minus{} OM^2\equal{} (CA \minus{} BC)BC \minus{} (AB \minus{} BC)BC \equal{} BC(CA \minus{} AB)$

Since $\triangle MIC$ and $\triangle MIB$ are isosceles, we have $ IM \equal{} IB,$ $IN \equal{} IC.$ Then

$ IN^2 \minus{} IM^2 \equal{} IC^2 \minus{} IB^2 \equal{} BC(CA \minus{} AB)$

$ \Longrightarrow ON^2 \minus{} OM^2 \equal{} IN^2 \minus{} IM^2 \Longrightarrow \ IO \perp MN$

On the other hand, $ \angle NBM \equal{} \angle OCI \equal{} 90^{\circ} \minus{} \frac {_1}{^2}\angle C \minus{} \angle A.$ If  $ L \equiv{} CI \cap BM,$ then it follows that  $ \angle NMB$ and $\angle OIC$ are supplementary, which implies that $ \triangle IOC$ and  $ \triangle MNB$ are pseudo-similar, hence

$ \frac {MN}{OI} \equal{} \frac {BN}{OC} \Longrightarrow \frac {MN}{BC} \equal{} \frac {OI}{R}.$

[quote="dragon001"]In a triangle $ ABC$,$ BC$ í smallest edge.Choose $ M$ and $ N$ on $ AB$ and $ AC$ respectively such that $ BM \equal{} BC$ and $ CN \equal{} CB$
Prove that
$ \frac {MN}{BC} \equal{} \sqrt {3 \minus{} 2(cosA \plus{} cosB \plus{} cosC)}$[/quote]

This problem is on Mathematical and Youth magazine and it can be solved easily by using vector. Don't need trigonometry.
To dragon001: You are bad  :mad:

[b]Livetolove[/b] [b]is right[/b], this topic must be deleted, the problem that has been proposed in this topic is the problem in [b]Mathematics and Youth magazine[/b] in [b]Vietnam[/b], the latest number [b](May/2009)[/b].


conics
parabola
integration
calculus
calculus computations

use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the x-axis. sketch the region and a typical shell.

x=1+y^2, x=0, y=1, y=2

Solution
[hide]Well I'm not going to sketch it but its a parabola opening to the right with a vertex at $(1,0)$.

$x=1+y^{2}$, $x=0$, $y=1$, $y=2$.

So we're revolving around the x-axis:

$V= 2 \pi \int_{a}^{b}p(x) \cdot h(x) dx$

Where $p(x)$ is the distance from the axis of revolution and $h(x)$ is your representative rectangle.  Now:

$V=2 \pi \int_{1}^{2}y(1+y^{2}) dy$
$V=2 \pi \int_{1}^{2}(y+y^{3}) dy$
$V=2 \pi \left[ \frac{y^{2}}{2}+\frac{y^{4}}{4}\right]_{1}^{2}$
$V=2 \pi \left[ \left( \frac{2^{2}}{2}+\frac{2^{4}}{4}\right)-\left( \frac{1^{2}}{2}+\frac{1^{4}}{4}\right) \right]$
$V=2 \pi \left( \frac{3}{2}+\frac{15}{4}\right)$
$\boxed{V=\frac{21 \pi}{2}}$[/hide]


Let a, b, c be real numbers such that $ \frac{\left(a\minus{}b\right)\left(b\minus{}c\right)\left(c\minus{}a\right)}{\left(a\plus{}b\right)\left(b\plus{}c\right)\left(c\plus{}a\right)}\equal{}\frac{1}{11}$. What is the value of
\[ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}a}
\]


There are 7 people on a quidditch team. If there are 7 people who tried out and will be on the team and,2 of them have the same chances to be seeker, how many ways can the quidditch team be formed?

[hide] is there a way to overcount this?[/hide]

what about the beaters, and all the other jobs that end with eepers. are there restrictions on the other positions?

no not at all. they all are very capable of every position but seeker and then only two of them have the ability to catch the snitch.

i really was just wondering if there was a way to overcount it...
[hide] this is actually about a question i myself have i thought it'd be 2*2*6  :oops:  i'm only sure for it if i had 4 people 2 only seekers..2*2*3=12[/hide]

[hide]Choose the person to be seeker first.  There are two choices for the seeker.  Are the other 6 positions distinct?  If they are, then there are $6!$ ways to choose the others, and if not, only $\binom66=1$ way.  Thus, there are either $2\times6!=1440$ or $2\times1=2$ ways.[/hide]

yeah actually, are the positions idstinct? the beaters are the same i guess and so are the three offensive people, forgot what their name is.

yes 7 people tried out for the team 2 can seek and 5 can't but all 7 can chase,beat,and keep.

just to let you guys know your answers are correct! now i only need my hidden question answered.

I have a different answer.  first, there are 2 ways to chose the seeker.  then, there are $\frac{6!}{3!\cdot2!}=60$ways to scramble the other players.  so isn't it $\boxed{120}$?  :maybe:

I am not sure..the only one of these types of problems i have memorized is if i had 4 people and 2 could be seeker...the answer is 12 which is $2*3!$ so I can't tell you where you went wrong..or for that case if I went wrong but I have more faith in theirs...as that is how I did the problem above. And I know that is the right answer. :read: Hmmm...I'll be thinking about it. Maybe I'll get AoPS 1 out.

ahha! i meant if flint and malfoy could be any player and bletchley,pucey,montague,crabbe and goyle could be any of the remaining ones! so they are distinct people..


[hide]Well the first one is true iff $x=y$ or $x=0$ or $y=0$...
And the second one just makes a circle with radius 2...
I don't know what kind of polygon that can possibly make... it's just a kind of plus sign with a diagnol line through it.l [/hide]

I think I may be missing something here...

Yeah, what you said is correct.

It does make a polygon. Substitute the first equation that you wrote into $x^{2}+y^{2}=4$ and see what you get.

are you sure its regular? seems like a convex hexagon but not regular.

[quote="t0rajir0u"][hide="Interesting"] $x^{2}y = xy^{2}\Leftrightarrow (x-y) xy = 0$

Hence $x = y, x = 0, y = 0$ make up our solutions.  We also want points on a circle of radius $2$.  This gives us six points

$(0, \pm 2)$
$(\pm 2, 0)$
$(\pm \sqrt{2}, \pm \sqrt{2})$

Which want to, but cannot, form either a regular hexagon or a regular octagon.  Odd.  [/hide][/quote]

To be completely clear: the last two points are $\pm(\sqrt{2},\sqrt{2})$ (since $(\pm\sqrt{2},\pm\sqrt{2})$ could be interpreted as $4$ points).

[quote="t0rajir0u"]
Which want to, but cannot, form either a regular hexagon or a regular octagon.  Odd.  [/quote]

What do you mean "want to" make a regular hexagon?

I just mean that they are six of the eight points of a regular octagon.  They do not, by themselves, form either a regular hexagon or a regular octagon.  :P  Sorry if I was confusing.

[quote="13375P34K43V312"]Consider the following system:

$x^{2}y=xy^{2}$
$x^{2}+y^{2}=4$
The real solutions make a convex regular polygon when graphed on the Cartesian plane. Find the area of the polygon.[/quote]

[hide]The first equation can be rewritten as $x^{2}y-xy^{2}=xy(x-y)$.

So, the first graph is the line $y=x$ and the coordinate axes.

The second graph is just a circle with radius 2.

We see that all of the points of intersection of the circle are $(2,0)$, $(0,2)$, $(-2,0)$, $(0,-2)$, $(\sqrt{2},\sqrt{2})$, and $(-\sqrt{2},-\sqrt{2})$. This isn't a regular polygon

One can check that the area is $4*\frac{1}{2}*\sin 45^\circ*4+2*\frac{1}{2}*4=4+4\sqrt{2}$.

$4+4\sqrt{2}$[/hide]

[quote="13375P34K43V312"]Doesn't that make 8 points?[/quote]

No.  We are missing $(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$.


number theory
greatest common divisor
relatively prime

Suppose $a, b, n \in \mathbb{Z} $ with $n > 0$. Suppose $ab \equiv 1 \bmod {n}$. Prove that $a$ and $b$ are relatively prime to $n$.

How can I prove this?

Well then let ab=xn+1.  Therefore, ab and xn are relatively prime.  Thus, ab and n are relatively prime, and the condition follows.

The largest number that divides the LHS of ab - xn = 1 is 1 so the GCD of ab and xn is 1, and GCD(ab,n)=1.

He wants $a$ and $b$ seperately. How do we prove that?

$ab = 1 + nq$

$ab - nq = 1$

The $GCD(a,n)|1$ and $GCD(b,n)|1$ so they both equal one.

OK Now I get it. Just a difference of perspective!


I'm going to post 38 Fantastic problems for you to solve from the MOEMS problem of the day. :D Please give comments.

Someone can make up solutions please...

Why did you take the time to compile those problems if they're all posted in another topic?

[quote="anirudh"]where are they???[/quote]
MOEMS Problem of the Day.

its all in the document, you can print it and answer it anywhere...


That would've been ridiculously easy.  Yay, Google calculator!   :lol:

i think he meant without a calculator but whatever...it's still trivial, just a bunch of arithmetic to do:

$((1000+24)^2)^3$

Yes it has to be done without calculaors.How did you get that iversonfan?

Any more easier methods anyone?

Iversonfan just simplified the $2^{60}$ to get $((1000+24)^2)^3$ because powering 1000's gives you a digital sum of 1.

How about if the expression is written in binary - what is the digital sum? :)

[hide="Gyans Extension"]In binary it's just 100000000000... with 59 0's. So the digital sum is 1.[/hide]

What is digital sum and how does it help to get the answer?

It doesn't really. It's more a joke I guess.

[quote="mna851"]I still don't see how to do this problem.[/quote]

Here is basically how do attack this problem.

$2^{60} = (2^{10})^{6} = (1024)^6 = (1000+24)^6 = ((1000+24)^2)^3$

Now, from here, computing $(1000+24)(1000+24) = 1000^2 + 48000+576$ is not too bad. Write this number as $1000^2+48576$. Now, apply the Binomial Theorem.

$(1000^2+48576)^3 = \binom{3}{0} (1000^2)^3 + \binom{3}{1} (1000^2)^2 \cdot 48576 + \binom{3}{2} (1000^2) \cdot 48576^2 + \binom{3}{3} 48576^3$

Just multiply numbers out. It's not "too" bad.

What Gyan said is another question and actually more of joke because finding $2^{60}$ in base 2 is fairly trivial. That's my Centry said "It's a joke."


The obtuse angle of an isosceles triangle is bisected, and each resulting angle is $ 54$ degrees larger than a base angle. How many degrees are in the measure of the obtuse angle.

Let the base angle be $ x$.

$ \frac{180\minus{}2x}{2}\equal{}90\minus{}x$.

$ 90\minus{}x\minus{}x\equal{}54$
$ 2x\equal{}36$
$ x\equal{}18$.
$ 180\minus{}2x\equal{}144$.


analytic geometry
geometry
circumcircle
radical axis
geometry unsolved

Let $ABC$ be a triangle, $l$ a line and $L,M,N$ the feet of perpendiculars to $BC,CA,AB$ through $L,M,N$ respectively, are concurrent. Their intersection is called the orthopole of the line $l$ and the triangle $ABC$.

You can find a proof of the [url=http://www.cut-the-knot.org/Curriculum/Geometry/Orthopole.shtml]Orthopole[/url]. This have a litle generalization, the [i]x-isopole[/i] (I haven't found any result of it, I have a solution with barycentric coordinates but it's quite long).

They are very interesting results about orthopole:

$\blacksquare$ The orthopole, for a given triangle of a circundiameter of that triangle lies on the nine-point circle of the triangle.
$\blacksquare$ If a line meets the circumcircle of a triangle, the Simson lines of the points of intersection with the circle meet in the orthopole of the line for a triangle.
$\blacksquare$ The [url=http://mathworld.wolfram.com/FeuerbachPoint.html]Feuerbach points[/url] of a triangle are the orthopolres, for that triangle, of the circumdiameters passing throught the tritangent centers (respectly).
$\blacksquare$The orthopole of a line for a triangle is the radical center of the three circles tangente to the given line and havian for centers the vertices of the anticomplementary triangle of the given triangle.


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

[quote]The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value. [/quote]

This is taken from the FAQ on the AMC website.  Is it true that you must take the AMC 10 to take advantage of the floor score cutoff.  What if you are still in tenth grade or less, but choose to take the AMC 12 can you still take advantage of the floor score?

> Is it true that you must take the AMC 10 to take advantage of the floor score cutoff.

No.

> What if you are still in tenth grade or less, but choose to take the AMC 12 can you still take > advantage of the floor score?

Yes.

In practice, at least the last two years, I have selected approximately 80 students in 10th grade and below who scored at or above the "floor AIME score" without regard to whether teh students took the AMC 10 or the AMC 12.  (Note that this was broader than the written algorithm, and worked to no one's disadvantage.)

This rule will be omitted in the 2006 AIME/USAMO Teacher's Manual, which is just about ready to go to the printer.  When the final copy goes to the printer, the amended version with the omitted "rule" will also be posted on our website.

AMC Director
Steve Dunbar

I was watching at what amounted to the birth of this rule in 2002 - its intent was clearly to protect the chances of younger students who took that AMC-12 and not the AMC-10. Among students I knew at the time, the original announcement of USAMO qualification included two. One was a 12th grader who qualified by index; the other was a 10th grader with a 126 AMC-10 and a 6 AIME. (Remember that scores are not constant year-to-year; in particular, AIME scores ran lower in 2002 than in some more recent years.)

But another student I knew, a 9th grader, had a 117 AMC-12 and a 6 AIME. The folks in Lincoln must have been thinking about cases like his, and how there was just no fair way to compare an AMC-12 score to an AMC-10 score. The 9th grader did get a belated invitation to take the USAMO, and, it being the summer of greatly expanded MOP, he got the opportunity to attend MOP.

So the "floor" was invented specifically to avoid comparing the AMC-10 to the AMC-12, and to leave opportunities open for students taking the AMC-12.

Why do you select only 80, shouldn't it be all students in 10th grade and below who score at or above the AIME floor score?  Isn't that the rule?

[quote="mna851"]Ok, perhaps I should rephrase my question.  I am currently a sohpmore.  If I take the AMC-12, and qualify for the AIME, and score the floor score or above on the AIME,  but it's still not enough to qualify by index will I be guranteed USAMO qualification?  Similarly, if I take the AMC-10, qualify for the AIME, and score the floor score or above, but once again still not enough to qualify by index, will I be guranteed USAMO qualification?[/quote]

Yes. Wait but for MOP index(USAMO index + 9*USAMO score) don't we have an advantage of taking AMC-10(not that the few extra points will make much of a difference)?

[quote="JesusFreak197"]I don't think index even matters for 10th or younger. While I could be completely off, my understanding is that we just need to score either $\ge 120$ on the AMC10 or $\ge 100$ on the AMC12, and then whatever the floor is on the AIME; there's no index taken into account until 11th and 12th.[/quote]

Hmm yes there is an index for 9th and 10th graders, at least for people who take AMC12 (I am not sure how index for people who took AMC10 is calculated). Otherwise we are saying that people who are in 10th grade or below cannot place in top 160.

JesusFreak thats what I thought too until I read the quote posted in my first post on this thread.

The only differences if you take the AMC12 instead of the 10 is that you have the ability to set the AIME floor value and your score counts towards your school team's score.

One other question though. If say the index cutoff for USAMO is 210 and the floor value is a 7. what if some 10th grader took the AMC12 and gets 130/7? Is the index lowered to 200 or does it stay at 210? I mean, the cutoff should stay at 210, right?

I will attempt to answer the questions in order.

> Why do you select only 80, shouldn't it be all students in 10th grade and below who score at > or above the AIME floor score? Isn't that the rule?

The answer is that we currently can accept approximately 250 for the USAMO. If we take approximately 160 based on index and then approximately 80 young students based on AIME score above the floor, and then approximately 10 from states not otherwise represented, that makes a total of approximately 250.  When the rules were revised in 2002, those proportions were observed to hold in previous years.  Based on a combination of those observations and a design that attempted to give first consideration to the AMC 12 and the index, the Committee on the American Mathematics Competitions approved what are now the existing rules.

It is my observation from selection in 2002, 2003, 2004, and 2005 that after the approximately 160 are chosen by index, then there are approximately 80 young students above the floor AIME value.  The proportions have worked well.  But it is conceviable that some year, I would observe significantly more than 80 young students above the floor value.  In such a case, I have to reserve an objective criterion that allows me to select approximately 80 young students to preserve the other design goal of approximately 250 students being invited to the USAMO.   In practice for the last four years, it has worked out that ALL the 10th grade and below students who scored at or above the floor AIME score have been invited. it's been around 80 each year.  But it could be that some year, either many more or many fewer than 80 would meet that criterion.

> I am currently a sohpmore. If I take the AMC-12, and qualify for the AIME, and score the > floor score or above on the AIME, but it's still not enough to qualify by index will I be
> guranteed USAMO qualification?

Yes, provided there are not more than 80 10th graders and below in the same situation.  That has been the case in 2002, 2003, 2004, and 2005, but of course, I can't guarantee it for the future.

> Similarly, if I take the AMC-10, qualify for the AIME, and score the floor score or above,
> but once again still not enough to qualify by index, will I be guranteed USAMO >qualification?

Same answer.

> Yes. Wait but for MOP index(USAMO index + 9*USAMO score) don't we have an advantage > of taking AMC-10(not that the few extra points will make much of a difference)?

My observation from those 10th graders and below who take both the AMC 10 and the AMC 12 in a given year and reach that decision point, the difference between the two indices makes no difference.  The weight attached to the AIME and the USAMO are so great that in actual practice the difference between the AMC 10 and the AMC 12 gets washed out.  In short, the few extra points do not make much difference.   Although in some mythical "principle" a 10th grader or below "should" score higher  on the AMC 10 than on the AMC 12, in practice I don't see it among those students above the floor value.  The variations I see at this level appear to be due as much to chance, or "test-taking mood", or "luck" or call it what you will.

> I don't think index even matters for 10th or younger. While I could be completely off, my > understanding is that we just need to score either \ge 120 on the AMC10 or \ge 100 on
> the AMC12, and then whatever the floor is on the AIME; there's no index taken into >account until 11th and 12th.

Yes, that's been the case the last 4 years  but of course, I can't guarantee it for the future.  Some day, I might have to rely on an index to select the approximately 80.  But based on 4 year's experience, I don't expect to.

> The only differences if you take the AMC12 instead of the 10 is that you have the ability to > set the AIME floor value and your score counts towards your school team's score.

That's right, that's a good observation.

> One other question though. If say the index cutoff for USAMO is 210 and the floor value is > a 7. what if some 10th grader took the AMC12 and gets 130/7? Is the index lowered to
> 200

No.

> or does it stay at 210? I mean, the cutoff should stay at 210, right?

Yes, it does.

Steve Dunbar
AMC Director

Then why does it say on the website that students choosing between the AMC 10 and AMC 12 should be advised to take the AMC 12 since it will give them an advantage in USAMO selection?

Does this have something to do with the process if there are more than 80 students above the floor value?

[quote="dasherm"]Then why does it say on the website that students choosing between the AMC 10 and AMC 12 should be advised to take the AMC 12 since it will give them an advantage in USAMO selection?[/quote]I don't think so.

It is because they can qualify through two ways, either a high index, or scoring above the Aime floor.

> Then why does it say on the website that students choosing
> between the AMC 10 and AMC 12 should be advised to take
> the AMC 12 since it will give them an advantage in USAMO
> selection?

We (the AMC and I) say that on our website and every time we are asked on the basis of observed data.  I just checked again, and for the 2005 USAMO, there were 115 students in 10th grade and below who qualified for, and were invited to, the USAMO.  Of those 115, 68 (=58%) qualified after taking the AMC 12, and the other 47 qualified after they took the AMC 10.  This further indicates that about 35 10th graders and below were selected for the USAMO on the basis of a high index using their AMC 12 score.  That is consistent with my notes from the selection process.

Furthermore, for 2004 which is the most recent year for which I have handy data, 1565 students in 9th and 10th grade qualified for AIME from the AMC 12.  On the other hand 1309 qualified for the AIME from the AMC 10.  That is, 54% of the 9th and 10th grade students who qualified for AIME did so by means of the AMC 12.  Since you have to take the AIME to qualify for USAMO, the observed data appears to be roughly consistent from year to year, and across criteria.

If you are in 10th grade or below, you should choose which contest, AMC 10 or AMC 12, you elect to take by using this information and your own self-knowledge of where you believe you will do your best work.  Since we do not have the advantage of having that knowledge about you, the AMC and I must rely on the observed facts to suggest decision-making.  The observed facts suggest taking the AMC 12.

Steve Dunbar
AMC Director


trigonometry
geometry
angle bisector
cyclic quadrilateral
geometry solved

Let $ABC$ be a tringle with Fermat point $F$ and isodynamic point $S$ (isogonal conjugate of $F$) denote $d_a$ is distance from $S$ to internal bisector of $A$ similar define $d_b,d_c$ prove that
$SA+SB+SC=FA+FB+FC+ 2(\frac{d_a^2}{SA}+\frac{d_b^2}{SB}+\frac{d_c^2}{SC})$

More generally:

[color=blue][b]Theorem 1.[/b] Let P and Q be two points inside a triangle ABC. Let x, y, z be the distances from the point Q to the angle bisectors of the angles PAQ, PBQ, PCQ, respectively. Then,

$AQ\cdot\sin\measuredangle BPC+BQ\cdot\sin\measuredangle CPA+CQ\cdot\sin\measuredangle APB$
$=AP\cdot\sin\measuredangle BPC+BP\cdot\sin\measuredangle CPA+CP\cdot\sin\measuredangle APB+2\left(\frac{x^2}{AQ}\cdot\sin\measuredangle BPC+\frac{y^2}{BQ}\cdot\sin\measuredangle CPA+\frac{z^2}{CQ}\cdot\sin\measuredangle APB\right)$.[/color]

Why is this more general than your problem? Just let P = F and Q = S. As we know, the first Fermat point P = F and the first isodynamic point Q = S of triangle ABC are isogonal conjugates; thus, the lines AP and AQ are symmetric to each other with respect to the angle bisector of the angle CAB. Thus, the angle bisector of the angle CAB is also the angle bisector of the angle PAQ. Combining this with the fact that Q = S, we see that the distance x from the point Q to the angle bisector of the angle PAQ is the distance $d_a$ from the point S to the angle bisector of the angle CAB. So we have $x=d_a$, and similarly $y=d_b$ and $z=d_c$.

This, together with P = F and Q = S, transforms the above equation into

$AS\cdot\sin\measuredangle BFC+BS\cdot\sin\measuredangle CFA+CS\cdot\sin\measuredangle AFB$
$=AF\cdot\sin\measuredangle BFC+BF\cdot\sin\measuredangle CFA+CF\cdot\sin\measuredangle AFB+2\left(\frac{d_a^2}{AS}\cdot\sin\measuredangle BFC+\frac{d_b^2}{BS}\cdot\sin\measuredangle CFA+\frac{d_c^2}{CS}\cdot\sin\measuredangle AFB\right)$.

Since F is the first Fermat point of triangle ABC, we have < BFC = < CFA = < AFB = 120°, so this equation becomes

$AS\cdot\sin 120^{\circ}+BS\cdot\sin 120^{\circ}+CS\cdot\sin 120^{\circ}$
$=AF\cdot\sin 120^{\circ}+BF\cdot\sin 120^{\circ}+CF\cdot\sin 120^{\circ}+2\left(\frac{d_a^2}{AS}\cdot\sin 120^{\circ}+\frac{d_b^2}{BS}\cdot\sin 120^{\circ}+\frac{d_c^2}{CS}\cdot\sin 120^{\circ}\right)$, thus
$AS+BS+CS=AF+BF+CF+2\left(\frac{d_a^2}{AS}+\frac{d_b^2}{BS}+\frac{d_c^2}{CS}\right)$,

and the problem is solved. Remains to prove Theorem 1.

[i]Proof of Theorem 1.[/i] Let X, Y, Z be the orthogonal projections of the point Q on the lines AP, BP, CP. Then, since < PXQ = 90°, < PYQ = 90°, < PZQ = 90°, the points X, Y, Z lie on the circle with diameter PQ. Thus, < ZXY = < ZPY, so that $\sin\measuredangle ZXY=\sin\measuredangle ZPY=\sin\left(180^{\circ}-\measuredangle BPC\right)=\sin\measuredangle BPC$. Similarly, $\sin\measuredangle XYZ=\sin\measuredangle CPA$ and $\sin\measuredangle YZX=\sin\measuredangle APB$. Now, by the sine law in triangle XYZ, we have $YZ: ZX: XY=\sin\measuredangle ZXY: \sin\measuredangle XYZ: \sin\measuredangle YZX$, what therefore becomes $YZ: ZX: XY=\sin\measuredangle BPC: \sin\measuredangle CPA: \sin\measuredangle APB$.

Now, the Ptolemy theorem for the cyclic quadrilateral PXZY yields

$PX\cdot YZ+PY\cdot ZX=PZ\cdot XY$.

Since $YZ: ZX: XY=\sin\measuredangle BPC: \sin\measuredangle CPA: \sin\measuredangle APB$, this transforms into

$PX\cdot\sin\measuredangle BPC+PY\cdot\sin\measuredangle CPA=PZ\cdot\sin\measuredangle APB$.

Now, this equation depends on the arrangement of the points X, Y, Z with respect to the segments AP, BP, CP. Using [i]directed[/i] segments on the lines AP, BP, CP, where these lines are directed in such a way that the segments AP, BP, CP are positive, this equation becomes

$PX\cdot\sin\measuredangle BPC+PY\cdot\sin\measuredangle CPA+PZ\cdot\sin\measuredangle APB=0$.

Now, if U is the orthogonal projection of the point Q on the angle bisector of the angle PAQ, then $\measuredangle UAQ=\frac{\measuredangle PAQ}{2}$, and the distance x from the point Q to the angle bisector of the angle PAQ is simply the length QU. In the right-angled triangle AUQ, we have $QU=AQ\cdot\sin\measuredangle UAQ$, what therefore becomes $x=AQ\cdot\sin\frac{\measuredangle PAQ}{2}$. Thus,

$2\frac{x^2}{AQ}=2\frac{\left(AQ\cdot\sin\frac{\measuredangle PAQ}{2}\right)^2}{AQ}=2\sin^2\frac{\measuredangle PAQ}{2}\cdot AQ$
$=\left(1-\cos\measuredangle PAQ\right)\cdot AQ=AQ-AQ\cdot\cos\measuredangle PAQ$.

Now, in the right-angled triangle AXQ, we have $AX=AQ\cdot\cos\measuredangle XAQ=AQ\cdot\cos\measuredangle PAQ$, so this becomes

$2\frac{x^2}{AQ}=AQ-AQ\cdot\cos\measuredangle PAQ=AQ-AX=AQ-\left(AP+PX\right)=AQ-AP-PX$,

so that

$AP+2\frac{x^2}{AQ}=AQ-PX$.

Similarly, $BP+2\frac{y^2}{BQ}=BQ-PY$ and $CP+2\frac{z^2}{CQ}=CQ-PZ$. Hence,

$AP\cdot\sin\measuredangle BPC+BP\cdot\sin\measuredangle CPA+CP\cdot\sin\measuredangle APB+2\left(\frac{x^2}{AQ}\cdot\sin\measuredangle BPC+\frac{y^2}{BQ}\cdot\sin\measuredangle CPA+\frac{z^2}{CQ}\cdot\sin\measuredangle APB\right)$
$=\left(AP+2\frac{x^2}{AQ}\right)\cdot\sin\measuredangle BPC+\left(BP+2\frac{y^2}{BQ}\right)\cdot\sin\measuredangle CPA$
$+\left(CP+2\frac{z^2}{CQ}\right)\cdot\sin\measuredangle APB$
$=\left(AQ-PX\right)\cdot\sin\measuredangle BPC+\left(BQ-PY\right)\cdot\sin\measuredangle CPA+\left(CQ-PZ\right)\cdot\sin\measuredangle APB$
$=\left(AQ\cdot\sin\measuredangle BPC+BQ\cdot\sin\measuredangle CPA+CQ\cdot\sin\measuredangle APB\right)-\left(\underbrace{PX\cdot\sin\measuredangle BPC+PY\cdot\sin\measuredangle CPA+PZ\cdot\sin\measuredangle APB}_{=0}\right)$
$=AQ\cdot\sin\measuredangle BPC+BQ\cdot\sin\measuredangle CPA+CQ\cdot\sin\measuredangle APB$,

and Theorem 1 is proven.

  Darij

[quote="darij grinberg"]More generally:

[color=blue][b]Theorem 1.[/b] Let P and Q be two points inside a triangle ABC. Let x, y, z be the distances from the point Q to the angle bisectors of the angles PAQ, PBQ, PCQ, respectively. Then,

........[/quote]

wow darij, I must say, you truly impress me  sometimes :).

Daniel

Thank you Darij, Your proof is very nice!!!  :)  :)


[b]If x is positive integer greater or equal to 8,
f:x->x,
f(x+y)=f(xy)
for all x>=4
y>=4
if f(8)=9
then what is value of f(9)?[/quote]

[hide="Solution"]
Use:

$ f(9) \equal{} f(4 \plus{} 5) \equal{} f(4 \cdot 5) \equal{} f(20)$
$ f(20) \equal{} f(4 \plus{} 16) \equal{} f(4 \cdot 16) \equal{} f(64)$
$ f(64) \equal{} f(8 \cdot 8) \equal{} f(8 \plus{} 8) \equal{} f(16)$
$ f(16) \equal{} f(4 \cdot 4) \equal{} f(4 \plus{} 4) \equal{} f(8)$

Hence, $ \boxed{f(9) \equal{} f(8) \equal{} 9}$ is a solution.
[/hide]

[b]Follow-up:[/b] Can we assert that $ f(m) \equal{} f(n) \; \forall m,n \in \mathbb{N}^{\geq 8}$? If not, can we assert that $ f$ is indeed a function.


Prove that $ a$ is an upper bound of the roots of $ P(x)$ if the synthetic division of $ P(x)$ by $ a$ yields all positives. Prove that $ a$ is a lower bound if synthetic division yields alternating positives and negatives.

If a synthetic division by a yields a polynomial in which all coefficients are positive.  Then the roots of this polynomial must all be negative.  We couldn't have a positive root since that would yield a positive number since every single term would be positive.  But I don't see how a needs to necessarily be an upper bound... a could be negative just like the other roots.  It works for a > 0 though.

For the part I'm going to assume that the term with the highest degree is positive and the next one is negative and so on.  The other way around is essentially the same thing.  If the highest degree of the polynomial is even then the roots most be positive otherwise the whole polynomial would be positive and thus could not equal zero.  If the highest degree of the polynomial is odd then the roots must be positive otherwise the whole thing would be negative and would not be able to equal zero.  But like the previous one a will only be a lower bound for sure if it has a restriction on it.  Specifically a < 0.

When you say "synthetic division by a," do you mean $ \frac{P(x)}{a}$ or $ \frac{P(x)}{x \minus{} a}$?

oh, sorry, i meant $ \frac{P(x)}{x\minus{}a}$


geometry
3D geometry
search
number theory unsolved
number theory

If in an arithmetic progression of positive integers a square and a cube occur then a sixth power also occur.

It is a problem from shortlist - Russian proposition - don't remember the year but it was posted before - try to search for it

Yes, it's ISL 1997 : http://www.mathlinks.ro/Forum/viewtopic.php?t=658 . :)


What is the greatest integer $ x$ such that $ 2^x$ is a factor of $ 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdots 13^{13}$?

Since odd numbers won't contribute factors of $ 2$, we only need to count the number of 2s in $ 2^2\cdot 4^4\cdot 6^6\cdots 12^{12}$.

From here we just count the number of 2s in each factor:
$ 2^2$ clearly has $ 2$
$ 4^4\equal{}2^8$, so $ 8$
$ 6^6\equal{}2^6\cdot 3^6$, so $ 6$
$ 8^8\equal{}2^{24}$ which gives us $ 24$
$ 10^{10}\equal{}2^{10}\cdot 5^{10}$, so $ 10$
$ 12^{12}\equal{}2^{24}\cdot 3^{12}$, so $ 24$
\[ 2\plus{}8\plus{}6\plus{}24\plus{}10\plus{}24\equal{}\boxed{74}\]


probability
expected value
absolute value
AMC

1. I have 25 pennies, 10 nickels, 5 dimes, and 1 quarter in my pocket.  If I remove a coin from my pocket, and each is equally likely to be chosen, what is the expected value of the coin I take, in cents?

2. Compute the sum of all 5-digit integers whose digits are 1, 3, 5, 7 and 9 in some order.

3. Three boxes are set out on the table.  One contains two black balls, one contains two white balls, and one contains a black and a white ball.  Matt removes a ball at random from box A, and it is black.  If he then removes a ball from box B, compute the probability it is white.

4. If  |x| + x + y = 10  and  |y| + x - y = 12,  compute x + y.

[hide]For #2, there are 5! = 120 different numbers that can be made using the digits 13579.  In 24 cases, the ones digit is 1, in 24 cases, it is 3, in 24 cases it is 5, etc.  Thus the sum of all the ones digits is 24(1) + 24(3) + ... 24(9) = 24(1+3+5+7+9) = 24*25 = 600.  Similarly, the sum of all the tens digits is the same as that of the ones digits, only it's worth 10 times as much, so 6000.  So the hundreds digit is worth 60,000, the thousands digit is worth 600,000, and the ten-thousands digit is worth 6 million. Adding it all together, we get 6,666,600.  Which is a big number.  Don't quote me on that.[/hide]

For #3, I assume the chance Matt picks a particular box to be Box A or Box B is 1/3?  And can box B be the same box as box A?

For number 2, good job.  For number 3, boxes A and B are distinct, and he picks boxes at random, and thus with equal probability at each point.

[hide]

1. The total amount of money in there is 150 cents. The total number of coins is 41. So the expected value is 150/41

3. There are two cases in which this can happen. He can pick the first one from the one with two black balls, and the second from the one with one white and one black, or vice verca. The probability of the first is 1/3*1/2*1/2 = 1/12. The second is also 1/12, so the answer is 1/6.

[/hide]

X is positive, since if x is negative, the |x| and the x in equation 1 cancel and y=10. But in equation 2, 10+x-10=12, x=12, and 12 is positive. So x is positive.
Y is negative, since if y were positive, the |y| and -y in equation 2 would cancel, leaving x= to 12. But equation 1, using this and the fact that x is positive and |x|=x, simplifies to 24+y=10. This means that y=-14, a negative. So |y|=-y.
Using this, we can remove absolute value and make a system.

 2(2x+y=10)                          2x+y=10
     x-2y=12                          -2(x- 2y=12)

    4x+2y=20                           2x+y=10
      x-2y=12                            -2x+4y=-24

          5x=32                              3y=-14
            x=32/5                             y=-14/3

SouthPaw, your number 1 is good, but your number 3 needs a bunch of work.  First of all, you answered a different question that I asked.  Second of all, even if you were answering the same problem, your method is not sound.  (1) Think about what finite probability means, and (2) check what the givens of the problem are -- what you know for sure happens.

Topper -- Good job, although you technically did not answer the question asked.

Ok, i think i got #3 [hide]Matt has a 1/2 chance to choose Box A to be the box with 2 blacks and 1/2 to choose box A with 1 black. This means his probability to get a white will be, half the time 3/4 and half the time 1/2. overall probability is 5/8 (1/2*3/4+1/2*1/2)[/hide]

Your first bit is wrong.  It's not [hide]1/2 and 1/2 like you said.[/hide]

JBL wrote:3. Three boxes are set out on the table.  One contains two black balls, one contains two white balls, and one contains a black and a white ball.  Matt removes a ball at random from box A, and it is black.  If he then removes a ball from box B, compute the probability it is white.

[hide]For simplicity's sake, let us call the box with the two black balls "box1", the box with the two white balls "box2", and the box with one of each "box3".

Consider the probability of getting a black ball from each box.  The probabilities are

black out of box1 = 1

black box2 = 0

black box3 = 1/2.

Now, we take the case where he may have picked any of the boxes to be box A.  The chance of getting each of box1, 2 and 3 is 1/3, so the chance of getting a black is 1/3*1 + 1/3*1/2 = 1/2.  But we know that Matt picked a black, so we need to multiply by two.  Now, the probability that he picked a black from box 1 is 1/3 * 2 = 2/3 and the probability that he picked a black from box3 is 1/6*2 = 1/3.

But this is the probability that he picked box1 versus box 3 to be box A.  Hence, the chance that box A = box1 is 2/3; the chance that box A = box3 is 1/3.

Now, a ball will be white all the time if we pick box2 as box B.  It will be white half the time if we pick box3 as box B.

2/3 of the time, we will pick box1 as box A.  This leaves box2 and box3 open to be picked as boxB.  In this case, the probability that the ball is white is then (1/2)*(1) + (1/2)(1/2) = 3/4.  Since this only happens 2/3 of the time, we multiply by 2/3 to get 1/2.

1/3 of the time, we will pick box3 as box A.  This leaves box1 and box3 to be picked as boxB.  In this case, the probability that the ball is white is then (1/2)(0) + (1/2)(1) = 1/2.  But this only happens 1/3 of the time, so we multiply by 1/3 to get 1/6.

Summing these two possibilities together, we obtain 2/3.  Not bad.[/hide]

Now for the moment of truth: am I right?

To the best of my knowledge.  I do feel that there are simpler ways to do this, however.  Draw three boxes, with two blacks in one, a black and a white in another, and two whites in a third.  Now, we know he picked a black ball and there are 3 of them, so the probability he picked any one of the three is 1/3.  If he picked one of the two in the BB box, the probability of picking a white is 3/4.  If he picked the B in the BW, the probability of getting a white is 1/2.  Thus, his total probability is 2/3*3/4 + 1/3*1/2 = 2/3.

The reason we can leave out the box-choosing step is that it's symmetric.  When we're in the case where we've picked a B from BB, there are 4 marbles left and we'll get each of them with equal probability, so that probability is 1/4.  This wouldn't be true if there were 3 marbles in one box and 1 in another -- but in our situation, it lets us simplify somewhat.

Good job, mathfanatic.

Actually, pardon me.  You made a slight arithmetic (reading) mistake near the end -- just look at your last 3 lines, see where you went wrong.


function
inequalities
algebra unsolved
algebra

1)$ a,b,c,d\in \Re$
2) $ a\neq b\neq c\neq d$
3)$ ac\equal{}bd$
4)$ \frac {a} {b} \plus{}\frac {b} {c} \plus{}\frac {c} {d} \plus{}\frac {d} {a} \equal{}4$

Find the maximum value of
$ \frac {a} {c} \plus{}\frac {b} {d} \plus{}\frac {c} {a} \plus{}\frac {d} {b}$

Daniel

$ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4\implies a^2cd+ab^2d+abc^2+bcd^2=4abcd$
$ \implies bd(a+c)(b+d)=4abcd\implies (a+c)^2(b+d)^2=16abcd$
\begin{eqnarray*}\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}&=& \frac{1}{abcd}(da^2b+ab^2c+bc^2d+cd^2a)\\
&=&\frac{1}{abcd}(cd+ab)(da+bc)\\
&\leq& \frac{1}{4abcd}(ab+bc+cd+da)^2\\
&=&\frac{1}{4abcd}(a+c)^2(b+d)^2\\
&=&4\end{eqnarray*}

I understand what you have done, but you have to prove that it is indeed the maximum...if you don't the problem is not finished
( and I think is not the maximum yet I haven't solved it)
Daniel

Well I will write what I have reached...I can't finish the problem.
Obviously the numbers are nonzero.
Suposse $ a$, $ b$, $ c$ and $ d$ are positive ( if all are negative is the same). Then by  $ AM \minus{} GM$  on $ 4)$ we get that $ \frac {a}{b} \equal{} \frac {b}{c}$. Because of $ 3)$ we get that $ b \equal{} d$ a contradiction. So there is at least 1 negative number ( but there is at least 1 positive number) Because of $ 3)$ we obtain that excatly two of them are positive and two of them negative. If $ a$ and $ c$ are negative then all terms in $ 4)$ are negatives and their sum can't be $ 4$. Then, WLOG let $ a$ and $ b$ negative and $ c$ and $ d$ are positive. So in the expression we're trying to maximize all terms are negative and it is easy to see that by $ AM \minus{} GM$ the expression is $ \le \minus{} 4$. Here I'm completly lost. I don't know if I can attain the value $ \minus{} 4$ or if I have to improve the maximum
Thank's for your help

Daniel

PS: Maybe you have misunderstood $ 1)$ It should have said that the four numbers $ a,b,c,d$ are distinct

[color=green]Let´s see...

$ \frac{d}{a}\equal{}\frac{c}{b}$

$ \frac{c}{d}\equal{}\frac{b}{a}$

so

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{d}\plus{}\frac{d}{a}\equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{b}{a}\plus{}\frac{c}{b}$

i.e. $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{d}\plus{}\frac{d}{a}\equal{}(\frac{a}{b}\plus{}\frac{b}{a})\plus{}(\frac{b}{c}\plus{}\frac{c}{b})$

but... you want

$ (\frac{a}{b}\plus{}\frac{b}{a})\plus{}(\frac{b}{c}\plus{}\frac{c}{b})\equal{}4$

but if $ x$ is not $ 1$

$ (\sqrt{x}\minus{}\frac{1}{\sqrt{x}})^2>0$

wich implies $ x\plus{}\frac{1}{x}>2$

for $ x\equal{}\frac{b}{c}$ and $ x\equal{}\frac{a}{b}$ adding we get

$ (\frac{a}{b}\plus{}\frac{b}{a})\plus{}(\frac{b}{c}\plus{}\frac{c}{b})>4$

I must be missing something here :blush: can you help me?

Life is GREEN! :D

TaUrUs*[/color]

[color=green]uhmm sorry i have just see...

$ x$ must be negative...

TaUrUs*
[/color]

[color=green]well we may supose without lose of generality that $ x\equal{}\frac{a}{b}<0$ and $ \frac{b}{c}>0$

this is because $ \frac{a}{b}\plus{}\frac{b}{a}\plus{}\frac{b}{c}\plus{}\frac{c}{b}\equal{}4$

so... again without lose of generality $ a<0,d<0,b>0,c>0$

and $ \frac{a}{c}\plus{}\frac{b}{d}\plus{}\frac{c}{a}\plus{}\frac{d}{b}<0$

so the maximun can not be $ 4$ :wink:

TaUrUs*[/color]

[color=green]

we have $ |x\plus{}\frac{1}{x}|>2$ so since all terms in

$ \frac{a}{c}\plus{}\frac{b}{d}\plus{}\frac{c}{a}\plus{}\frac{d}{b}$ must be negative... we have the maximun when $ \frac{a}{c}\plus{}\frac{c}{a}\equal{}\minus{}2$ and $ \frac{b}{d}\plus{}\frac{d}{b}\equal{}\minus{}2$
that is when $ a\equal{}\minus{}c$ and $ b\equal{}\minus{}d$ then the maximun is $ \minus{}4$.

Life is GREEN! :D

TaUrUs*[/color]

If $ a\equal{}\minus{}c$ and $ b\equal{}\minus{}d$  then $ \minus{}a^2\equal{}\minus{}b^2$  then $ a\equal{}b$ or $ a\equal{}\minus{}b\equal{}d$ Since the numbers are distinct you reach a contradiction
It is driving me mad :o

 :D Daniel

hello, try it with
$ a\equal{}\minus{}\frac{3\minus{}2\sqrt{2}}{\minus{}3\plus{}2\sqrt{2}}$
$ b\equal{}\minus{}1$
$ c\equal{}\minus{}3\plus{}2\sqrt{2}$
$ d\equal{}3\minus{}2\sqrt{2}$
Sonnhard.

I believe the maximum value doesn't exist.

$ ac \equal{} bd\Longleftrightarrow \frac {a}{b} \equal{} \frac {d}{c} \equal{} k$, yielding $ a \equal{} bk,\ d \equal{} ck$.

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a} \equal{} \frac {c}{b} \plus{} \frac {b}{c} \plus{} k \plus{} \frac {1}{k} \equal{} 4$.

$ \therefore \frac {a}{c} \plus{} \frac {b}{d} \plus{} \frac {c}{a} \plus{} \frac {d}{b} \equal{} \left(k \plus{} \frac {1}{k}\right)\left(\frac {c}{b} \plus{} \frac {b}{c}\right) \equal{} \left(k \plus{} \frac {1}{k}\right)\left\{4 \minus{} \left(k \plus{} \frac {1}{k}\right)\right\}$.

Since $ \left|k \plus{} \frac {1}{k}\right| \equal{} |k| \plus{} \left|\frac {1}{k}\right|\geq 2\sqrt {|k|\cdot \left|\frac {1}{k}\right|} \equal{} 2$, we have $ \frac {a}{c} \plus{} \frac {b}{d} \plus{} \frac {c}{a} \plus{} \frac {d}{b}\leq 4$,

the equality holds when $ k \plus{} \frac {1}{k} \equal{} 2\Longleftrightarrow k \equal{} 1$, i.e. $ a \equal{} b$ and $ d \equal{} c$, which contradicts the assumption. $ \therefore \frac {a}{c} \plus{} \frac {b}{d} \plus{} \frac {c}{a} \plus{} \frac {d}{b} < 4$.

Indeed I have just read a very nice solution and the maximum is $ \minus{}12$
So how did you get that result Sonnhard? It will be nice if you post the complete solution not only the answer :)
Thank you

Daniel

hello Daniel, here is my solution. From $ ac=bd$ we get $ d=\frac{ac}{d}$ for
$ d\not=0$ and so we have $ \frac{a}{b}+\frac{b}{a}+\frac{c}{c}+\frac{c}{b}=4$. Let $ f(a,b,c)=\frac{a}{c}+\frac{c}{a}+\frac{b^2}{ac}+\frac{ac}{b^2}$. From the equation above we have
$ c_1=\frac{-a^2+4ab-b^2-(a-b)\sqrt{a^2-6ab+b^2}}{2a}$
or
$ c_2=\frac{-a^2+4ab-b^2+(a-b)\sqrt{a^2-6ab+b^2}}{2a}$
By inserting $ c_1$ in $ f(a,b,c)$ we have
$ f_1(a,b)=-\frac{(a^2+b^2)(a^2-4ab+b^2)}{a^2b^2}$.
This function has a maximum for $ a=-b$ it is $ -12$. Analogously for the case
when we inserting $ c_2$ in $ f(a,b,c)$ and we get $ f_2(a,b)=-\frac{(a^2+b^2
)(a^2-4ab+b^2}{a^2b^2}$.
Sonnhard.

hello Daniel, a little remark. We can prove the inequality only by simplifying
$ \frac{(a^2\plus{}b^2)(a^2\minus{}4ab\plus{}b^2)}{a^2b^2} \le \minus{}12$ , this is equivalent with
$ \frac{(a\plus{}b)^2(a^2\minus{}6ab\plus{}b^2)}{a^2b^2} \geq0$. This is true because for
real $ c$ must be the inequality $ a^2\minus{}6ab\plus{}b^2 \geq0$ fulfilled.
Sonnhard.


Let $ x,y,z\in\mathbb{R}$ such that $ \minus{}1\leq x\plus{}y\plus{}z\leq 1$, $ \minus{}1\leq x\minus{}y\plus{}z\leq 1$ and $ \minus{}8\leq 4x\plus{}2y\plus{}z\leq 8$, $ \minus{}8\leq 4x\minus{}2y\plus{}z\leq 8$

Prove or disprove $ |x|\plus{}4|y|\leq 7$

hello, one solution of your system is $ \minus{}3<x<\minus{}\frac{7}{3},0<y\le x\plus{}3,z \le 1\minus{}y\minus{}x,\minus{}4x\plus{}2y\minus{}8\le z$. You can take $ x\equal{}\minus{}1,y\equal{}2$ and the condition is fulfilled but
$ |x|\plus{}4|y|>7$.
Sonnhard.


If $ a,b,c$ are positive real numbers such that $ abc\equal{}1$, then

$ \frac 1{(a\plus{}1)(a\plus{}2)}\plus{}\frac 1{(b\plus{}1)(b\plus{}2)}\plus{}\frac 1{(c\plus{}1)(c\plus{}2)} \ge \frac 1{2}$.

let $ a\equal{}\frac{yz}{x^2}$ and $ b\equal{}\frac{xz}{y^2}$ and $ c\equal{}\frac{xy}{z^2}$
so we most prove that

  \[ \sum\frac{x^4}{(yz\plus{}x^2)(yz\plus{}2x^2)}\geq\frac12\]
 after using Titu inequality we have
\[ 2(\sum x^2)^2\geq 2\sum x^4 \plus{}3\sum x^2yz \plus{}\sum y^2 z^2\]
 or
\[ \sum x^2y^2\geq\sum x^yz\]

[quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$, then

$ \frac 1{(a \plus{} 1)(a \plus{} 2)} \plus{} \frac 1{(b \plus{} 1)(b \plus{} 2)} \plus{} \frac 1{(c \plus{} 1)(c \plus{} 2)} \ge \frac 1{2}$.[/quote]

 It's hungkhtn'problem which he posted on VIF :wink:  Here my proof:

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

<=> $ \sum2(a\plus{}1)(a\plus{}2)(b\plus{}1)(b\plus{}2)\ge (a\plus{}1)(b\plus{}1)(c\plus{}1)(a\plus{}2)(b\plus{}2)(c\plus{}2)$

<=> $ 4\sum a^2\plus{}3\sum a\ge 3\sum ab\plus{}12$ which true by AM-GM  :)

We have some nother results: For $ a,b,c>0$ such that $ abc\equal{}1$ then

1/ $ \frac 1{(a \plus{} 1)(b \plus{} 2)} \plus{} \frac 1{(b \plus{} 1)(c \plus{} 2)} \plus{} \frac 1{(c \plus{} 1)(a \plus{} 2)} \le \frac 1{2}$

2/ $ \frac 1{\sqrt{(a \plus{} 1)(b \plus{} 1)}} \plus{} \frac 1{\sqrt{(b \plus{} 1)(c \plus{} 1)}} \plus{} \frac 1{\sqrt{(c \plus{} 1)(a \plus{} 1)}} \le \frac 3{2}$

This statement is also true:

If $ a,b,c,d$ are positive real numbers such that $ abcd \equal{} 1$, then

$ \frac 1{(a \plus{} 3)(2a \plus{} 1)} \plus{} \frac 1{(b \plus{} 3)(2b \plus{} 1)} \plus{} \frac 1{(c \plus{} 3)(2c \plus{} 1)}\plus{} \frac 1{(d \plus{} 3)(2d \plus{} 1)} \ge \frac 1{3}$.

More general:

If $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \equal{} 1$, then

$ \sum\frac 1{(a_1 \plus{} n\minus{}1)((n\minus{}2)a_1 \plus{} 1)}  \ge \frac 1{n\minus{}1}$.

[quote="Vasc"]More general:

If $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \equal{} 1$, then

$ \sum\frac 1{(a_1 \plus{} n \minus{} 1)((n \minus{} 2)a_1 \plus{} 1)} \ge \frac 1{n \minus{} 1}$.[/quote]

I think we can set $ a_i\equal{}\frac{\prod_{k\equal{}1}^{n}{x_k}}{x_{i}^{n}}$ and use Cauchy - Schwarz inequality.

[quote="NguyenDungTN"][quote="Vasc"]More general:

If $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \equal{} 1$, then

$ \sum\frac 1{(a_1 \plus{} n \minus{} 1)((n \minus{} 2)a_1 \plus{} 1)} \ge \frac 1{n \minus{} 1}$.[/quote]

I think we can set $ a_i \equal{} \frac {\prod_{k \equal{} 1}^{n}{x_k}}{x_{i}^{n}}$ and use Cauchy - Schwarz inequality.[/quote]

Try this method for $ n\equal{}4$ and $ n\equal{}5$. :?:

More general:

Let $ a_1,a_2,...,a_n$ be positive real numbers such that $ a_1a_2...a_n = 1$. If $ p,q$ are nonnegative real numbers such that ${ p+q=n-1}$, then

$ \frac 1{1+pa_1+qa_1^2}+\frac 1{1+pa_2+qa_2^2}+...+\frac 1{1+pa_n+qa_n^2} \ge 1$.

[hide]Nice particular cases: $ p=n-1$ , $ q=0$ and $ p=2r$, $ q=r^2$ with $ r=\sqrt{n}-1$.[/hide]

let a,b,c>0 such that abc=1 Prove that
$ (ab\plus{}bc\plus{}ca)^2[(a\plus{}b\plus{}c)^2\plus{}ab\plus{}bc\plus{}ca] \geq 4(a\plus{}b\plus{}c)^3$


vector
geometry
parallelogram
probability
expected value
combinatorics proposed
combinatorics

we have vectors $v_1,..,v_n \in \mathbb{R}^n$ such that $|v_n|=1$. Prove that there exist $\epsilon_i \in \{-1,1\}$ where $i=1,..,n$ such that $|\epsilon_1v_1+..+\epsilon_nv_n|\leq \sqrt{n}$

We apply the parallelogram formula repeatedly ($\|u+t\|^2+\|u-t\|^2=2\|u\|^2+2\|t\|^2$, which holds in all Hilbert spaces), we get $\sum\|\varepsilon_1v_1+\ldots+\varepsilon_nv_n\|^2=2^nn$, where the sum is taken over all $n$-uples $(\varepsilon_1,\ldots,\varepsilon_n)\in\{\pm 1\}^n$. Since the sum has $2^n$ terms, the conclusion follows immediately.

thanks for your solution, this idea is quite similar to the one which i know
let $\epsilon_i$ be $-1,1$ with the same probabilities, then the expected value of $X=|\sum \epsilon_iv_i|^2$ is
$E(X)=E((\sum \epsilon_iv_i)(\sum \epsilon_iv_i))=\sum_{i,j}(v_i \cdot v_j)E(\epsilon_i \epsilon_j)$
but if $i \neq j$ then $E(\epsilon_i \cdot \epsilon_j )=E(\epsilon_i)E(\epsilon_j)=0$ and if $i=j$ then $E(\epsilon_i^2)=1$ so $E(X)=\sum v_i\cdot v_i=n$
so there must be sequence of $\epsilon_i$ which gives what we want


Let n be a positive interger and a,b,c are positive numbers. Prove that
$ (a^n\plus{}b^n)^{\frac 1 n}\plus{}(c^n\plus{}d^n)^{\frac 1 n}\ge ((a\plus{}c)^n\plus{}(b\plus{}d)^n)^{\frac 1 n}$

This is the http://en.wikipedia.org/wiki/Minkowski_inequality


It's pretty easy for positive reals: a+b, b+c, c+a all < a+b+c, so our sum a/(a+b)+.. (call it S1)>(a+b+c)/(a+b+c)=1. The same is true for S2=b/(a+b)+c/(b+c)+a/(a+c). But S1+S2=3, so if S2>1 it means that S1<2, so we proved 1<S1 (and S2)<2.


conics hyperbola calculus integration articles MATHCOUNTS

Similar to the AoPS Daisy Chain, except that your word must have no relation to the word before it. For example, if I posted "donkey", you could post "hyperbola", "Zimmerman Telegram", etc., but NOT "mule", "Democrat", etc.

No explanations necessary in this version.

Mr. Rogers

You know,somwhow or the other oyu can get a connection..like elvis presley wore an underwear,so my thing wouldn't be valid....

Pamela Anderson

Ok I make an addition to this (but r.d. may remove this if he wants) - you have to make a connection between the last two before posting your own word.

Fine - Mrs. P.A. has tennis ball sized you-know-whats.

Emoticons.

there should be an emoticon called chinese fan.
No, this is silly, it will become a nonsense makeup game. Cut it out.

Don Quixote

waiting for pictures to go from camera to computer, and wondering about available memory on the computer (as that is what i'm doing now)

reverses gravity on a fly that is hovering near the ceiling.

I remember saying something about not opening pointless threads...


Putnam
integration
linear algebra
matrix
function
linear algebra unsolved

Let $ \sigma_n$ denote the set of all permutations of numbers $ 1,2,3,...,n$.For $ \sigma \in \sigma_n$
Put $ v(\sigma)$ denote the number of fixed point of $ \sigma$
 $ I(\sigma)$ denote the number of if $ i < j$ then $ \sigma(i) > \sigma(j)$
 $ \pi(\sigma)$ denote the number of if $ i < j$ then $ \sigma(i) < \sigma(j)$
 For $ (i,j) \in \{1,2,...,n\}^2$
Calculate :
1) $ \sum_{\sigma \in \sigma_n}{\frac {( \minus{} 1)^{I(\sigma)}}{I(\sigma) \plus{} \pi(\sigma) \plus{} 1}} \equal{} ?$
[hide="Answer of me"] $ \frac {2}{n \plus{} 1}$[/hide]
2) $ \sum_{\sigma \in \sigma_n}{\frac {( \minus{} 1)^{I(\sigma)}}{(v(\sigma) \plus{} 1)(I(\sigma) \plus{} \pi(\sigma) \plus{} 1)}}$
[hide="Answer of me"] :roll: [/hide]

I'm reminded of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64449]a Putnam problem[/url] although it isn't exactly the same.

Putnam trick works on #1:

Let $ A_x = Jx + (1 - x)I$ where $ J$ is the all-ones matrix.

Then it is easy to see that $ \int_0^1 \det(A_x) \mathrm{d}x$ is the desired summation (for part (a)),
this follows by the same logic as Putnam 2005 B6. We also have that

$\det A_x = \det (Jx + (1 - x)I) = (1 - x)^n \det(I + J(x/(1 - x))) = (1 - x)^{n - 1}(1 - x + nx)$.

Integrating the above from 0 to 1 gives $ \frac {2}{n + 1}$.

Part (b) should follow by computing the determinant of $ Jx + (y-x)I$ and integrating both $ x$ and $ y$ from 0 to 1 -- should be a simple computation.

Yes ,It is correct !.... It a is nice solution in matrix

For Answer (2) you can read with $ m\equal{}n\minus{}1$ http://www.mathlinks.ro/viewtopic.php?p=1498321#1498321

Calculate
Edit mistake of me : :blush: For (1),(2) solution same as [b]Nukular[/b]
 Call $ G(\sigma)$ are set fixed point of $ \sigma$ and $ v(\sigma)$ denote the number of $ G(\sigma)$
$ \pi(\sigma) \equal{} \{\sigma(1),\sigma(2),...,\sigma(n)\} \minus{} G(\sigma)$ are set point remain not $ \subset$ $ G(\sigma)$ of $ \sigma$ and $ I^{'}(\sigma)$ denote the number of $ \pi(\sigma)$
$ I(\sigma)$ denote the number of if $ i < j$ then $ \sigma(i) > \sigma(j)$ ,$ (i,j) \in \{1,2,..,n\}^2$
1)$ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{I^{'}(\sigma) \plus{} 1}} \equal{} ?$
2)$ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{(v(\sigma) \plus{} 1)(I^{'}(\sigma) \plus{} 1)}}$
3)
a)$ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{2^{v(\sigma)}}} \equal{} ?$
[hide="Answer of me"]$ \frac {2n \minus{} 1}{2}(\frac { \minus{} 1}{2})^{n \minus{} 1}$[/hide]
b)$ \sum_{\sigma\in\sigma_{n}}{( \minus{} 1)^{I(\sigma)}(e\minus{}1)^{v(\sigma)}} \equal{} ?$
[hide="Answer of me"][/hide]
4)
$ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{2^{v(\sigma)}(I^{'}(\sigma) \plus{} 1)}}$
[hide="Answer"] :play_ball: [/hide]

Sorry, but can anyone translate the problem into English for me? I am not even able to reverse-engineer it from Nukular's solution... What exactly prevents $ I\left(\sigma\right) \plus{} \pi\left(\sigma\right)$ from being $ \frac {n\left(n \minus{} 1\right)}{2}$ all the time? Or $ 1$, if $ i$ and $ j$ are fixed and $ i<j$ ?

  a seriously confused darij

Sorry I am definition false ! If i am define false again , you can edit for me  :blush:  :oops: Thanks !

For 5)
Calculate :
 $ \sum_{\sigma \in \sigma_n}{(\minus{}1)^{I(\sigma)}(\frac{1}{3})^{n\minus{}v(\sigma)}(\frac{4}{3})^{v(\sigma)}}\equal{}?$
 [hide="Answer of me"]$ \frac{n\plus{}3}{3}$[/hide]

Here $ v(\sigma)$ denote the number   fixed point of $ \sigma$
 $ I(\sigma)$ denote the number of if $ i < j$ then $ \sigma(i) > \sigma(j)$
Then :
1) $ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{n \plus{} 1 \minus{} v(\sigma)}} \equal{} \frac {2}{n \plus{} 1}$
2)$ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{(v(\sigma) \plus{} 1)(n \plus{} 1 \minus{} v(\sigma))}} \equal{}$
 [hide]$ \frac {2 \plus{} ( \minus{} 1)^{n \plus{} 1}n}{(n \plus{} 1)(n \plus{} 2)}$[/hide]
        solution same as Nukular
3) $ \sum_{\sigma\in\sigma_{n}}{\frac {( \minus{} 1)^{I(\sigma)}}{v(\sigma) \plus{} m}} \equal{} \int_{0}^{1}{y^{m \minus{} 1}(y \plus{} n \minus{} 1)(y \minus{} 1)^{n \minus{} 1}dy} \equal{} \frac {( \minus{} 1)^{n \plus{} 1}n!(m \plus{} n \minus{} 1)}{m(m \plus{} 1)...(m \plus{} n)} ,\forall m \in \mathbb{N^{*}}$
Use function $ B$ to calculate or read [b]putnam 2005[/b]  have a diferent solution !
Or use [hide="Hint"]$ x^{m \minus{} 1} \det(A_{x}) \equal{} x^{m \minus{} 1}\det( Jx \plus{} (1 \minus{} x)I) \equal{} \sum_{\sigma \in \sigma_n}{( \minus{} 1)^{I(\sigma)}x^{v(\sigma) \plus{} m \minus{} 1}}$[/hide]
3)$ \sum_{\sigma \in \sigma_n}{\frac {(\minus{}1)^{I(\sigma)}2^{v(\sigma)}}{(v(\sigma) \plus{} 1))(v(\sigma) \plus{} 2)}} \equal{} ?$
[hide="Hint"]$ A_{x \plus{} y} \equal{} J_{x \plus{} y} \plus{} (1 \minus{} (x \plus{} y))I$ ,$ \det(A_{x \plus{} y}) \equal{} \sum_{\sigma \in \sigma_n}{( \minus{} 1)^{I(\sigma)}(x \plus{} y)^{v(\sigma)}}$[/hide]
............. ...
 Sorry everyone  :blush:


If 3 fair dice are tossed, what is the probability that the sum of the faces tossed is greater than 12?

A) $ \frac{7}{72}$

B) $ \frac{1}{6}$

C) $ \frac{1}{3}$

D) $ \frac{7}{27}$

E) $ \frac{3}{8}$

Just use casework to solve it, it doesn't take a lot of time.

-rst2007

Think there is a cube with the edge length 6
A(0,0,0)B(0,6,0)C(6,0,0)D(6,6,0)A'(0,0,6)B'(0,6,6)C'(6,0,6)D'(6,6,6)
Think there is a plane which satisfy the equation x+y+z=12
The number of points above the plane and in the cube is 56
(the integer point in a isosceles right-angle equals to n(n+1)/2 so the sum of them is 6*7/2+5*6/2+4*5/2+3*4/2+2*3/2+1*2/2)
and the number of whole points is 216
so the answer is 56/216=7/27

[/img]

you roll the first dice = 6 ways
roll 2nd dice = 6 ways
roll 3rd dice = 6 ways

therefore total ways = 6x6x6

then you need to find the sum of the 3 dices
diagram may help

[img]http://i43.tinypic.com/212ajqx.jpg[/img]

Go clock wise direction with blue dice rolling first.
eg. roll a 6, roll a 6 with black die and roll 6 with red die = 18
so shade the top right square

roll 6,5,6 so shade in next square etc etc
keep proceeding like this!

but there are only 8 possibilities because rolling 5,5,4 is same as rolling 4,5,5

Assume they're all 6.
Now taking them away, x+y+z=5 You have 3 dividers and 5 of a number, so 8C3=56.

56/216=7/27


modular arithmetic
number theory unsolved
number theory

Show that there are no integer solutions to $ 2^n \plus{} 3^n \equal{} x^p$ with $ p\geq 2$.

(I have shown $ p>1000$ and $ n>10^{200000}$, but I'm a bit stuck.)

[quote="Eigenray"]I have shown $ p > 1000$ and $ n > 10^{200000}$.[/quote]
Please show us how you did it.

I will assume here that $ n$ is odd.  The case $ n$ even is not too difficult.

First note that $ 5|x$, and $ p\geq 2$, so $ 5^2|2^n\plus{}3^n$, which implies $ 5|n$.  Now suppose $ d|n$, and that $ q^\alpha \parallel{} 2^d\plus{}3^d$ with $ q\neq 5$, and $ \alpha$ not divisible by $ p$, say $ \alpha\equal{}kp\minus{}r$, $ 0<r<p$.  Then $ q^{kp} | 2^n\plus{}3^n$.  Let $ s|d$ be the order of $ (\minus{}3/2)$ mod $ q^\alpha$.  Since $ (\minus{}3/2)^d \neq 1 \mod q^{\alpha\plus{}1}$, we have that the order of $ (\minus{}3/2)$ mod $ q^{kp}$ is $ sq^{kp\minus{}\alpha}\equal{}sq^r$.  So we must have $ q^r | n$.

In particular, if $ d|n$ is prime, and $ 2^d\plus{}3^d$ is not 5 times a perfect $ p$-th power, then we can always find a larger prime $ q|n$, since $ q\equiv1\mod d$.  So it would suffice to show that $ 2^q\plus{}3^q \equal{} 5x^p$ has no solutions with $ p,q$ prime.  In fact it looks like $ \frac15(2^q\plus{}3^q)$ is always square-free.

In any case, we can use the above to show that $ n$ is divisible by 5, 11, 35839, etc.  (In fact, $ n$ is divisible by at least the $ (p\minus{}1)$-th power of each of these.)

Then I wrote a short program which does the following: for each prime $ p$, take $ d\equal{}5^\alpha 11^\beta$, and look at primes $ q\equiv1\mod pd$, until I find one such that $ 2^{dk}\plus{}3^{dk}\equiv x^p\mod q$ has no solutions with $ k\not\equiv 0\mod 2p$.  This program ruled out $ p<2003$ before I stopped it.  The first few $ p$s are ruled out as follows:

p=2, q=3, d=1
p=3, q=7, d=1
p=5, q=101, d=1
p=7, q=29, d=1
p=11, q=199, d=1
p=13, q=53, d=1
p=17, q=22441, d=5
p=19, q=1559, d=1
p=23, q=47, d=1
p=29, q=59, d=1

@Eigenray: What do you think? Is it really hard?

I don't know.  I got the problem from someone a while ago who doesn't remember where they got it.

[quote="Eigenray"]I will assume here that $ n$ is odd.  The case $ n$ even is not too difficult.
[/quote]

Why is that so easy?
I have an idea which may help:
$ x^p\equal{}3^n\plus{}2^n \equal{} \dfrac{3^{2n}\minus{}2^{2n}}{3^n\minus{}2^n}$
Then, use the fact that $ 3^{2n}\minus{}2^{2n}$ has a primitive prime divisor which is $ \equiv \pm 1 \pmod{2n}$ (such ideas come from Zsigmondy's theorem or what is sometimes called the Primitive Divisors Theorem for Lucas sequences--which in this case would be overkill).


Find all integer solutions to $x+y+z=xyz-2$.

[hide="Remark"] Ah, yeah, this problem isn't too bad with some bounding arguments.  Let $|z| = 0, 1, 2, ...$ and see when it is too large for a solution to exist. [/hide]


inequalities
trigonometry
geometry unsolved
geometry

Prove that for every triangles $ABC$ we have:
$\cos A+\cos B+\cos C \ge \frac{\cos \frac{B-C}{2}+\cos \frac{C-A}{2}+\cos \frac{A-B}{2}}{2}$

Sorry, the right inequality is:
$\cos A+\cos B+\cos C +\frac{3}{2}\ge \cos \frac{B-C}{2}+\cos \frac{C-A}{2}+\cos \frac{A-B}{2}$

Again, the point is showing that $(\sum cosA)^2 \geq \sum sinBsinC$.
 We will show that $\sum cosA \geq 2sin \frac{B}{2}sin \frac{C}{2}$. If we can, the problem is solved because $\sum sin \frac{A}{2} \leq \frac{3}{2}$.
 By applying Schwarzt 's inequality, we have
 $2sin \frac{B}{2}sin \frac{C}{2} = \sum \sqrt{tg \frac{B}{2}tg \frac{C}{2}}\sqrt{sinBsinC} \leq \sqrt{\sum sinBsinC} \leq \sum cosA$


We have 12  balls, 11 of which are completely equivalent, but one of them is different in weight from the others([b]Note that we dont know its lighter or heavier!!![/b]). We have a scale with 2 sides and we can do the weighting action for atmost 3  times! How can we find that different ball?

Also a nice thing that The Zuton Force posted there:

[quote="The Zuton Force"][b]Generalization 1:[/b] Prove that the most economical approach with $ n \minus{} 1$ light balls and $ 1$ heavy ball (or vice versa; provided it is known) uses $ \left \lceil \frac {\log n}{\log 3} \right \rceil$ moves.

[b]Generalization 2:[/b] Prove that the most economical approach with $ n \minus{} 1$ identical balls and $ 1$ different ball uses $ \left \lceil \frac {\log n}{\log 3} \right \rceil \plus{} 1$ moves.

[b]Follow-up:[/b] You have 2008 balls, 2001 of which have one particular weight and 7 of which all have a different, lighter weight. What is the minimum number of weighings needed to separate the balls into the two weight classes?[/quote]

 :)

P.S.i posted this before in Pre-Olympiad but seems it better be posted here too! lets discuss fellows! :cool:

posted before:
[url]http://www.mathlinks.ro/viewtopic.php?highlight=coin&t=34138[/url]

and here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=72042[/url] (in persian)


Look what i found!
http://web.archive.org/web/20030802061249/www.artofproblemsolving.com/wforum/

wow, i wasn't around back then, this is amazing!

hey, i remember when the forum looked like that! back in the good old days..... :)

Hmmmm, I don't remember that....must have joined too late.  :mad:

[quote="xxreddevilzxx"]lol i remember taht. although i rdont have the id back then anymore.[/quote]

Really? I joined 2 months before you, but I don't remember it... hmm, perhaps my memory escapes me these days; I only remember a simplified/condensed version of the current AoPS (not MathLinks layout, which I use because it looks nicer) forum.

I joined some 6 months after that I think...the layout was pretty much as it is now by then.

After seeing that, I searched through the forum and somehow I feel a connection to the old AoPS days even though I was never there.  It seems as if it was a very small, tight-knit group of people who would spam random things like "Fierytycoon and Rippledance."  Have you read the AoPS saga?  Mathfanatic is not only a math genius, but a literary genius as well.  I wish I could have been there.  Oh well.

By the way, I wonder what happened to those old members: i.e. tare, jojobird, confuted, etc. who are not active anymore.  What made them leave?

[quote="bubala"]After seeing that, I searched through the forum and somehow I feel a connection to the old AoPS days even though I was never there.  It seems as if it was a very small, tight-knit group of people who would spam random things like "Fierytycoon and Rippledance."  Have you read the AoPS saga?  Mathfanatic is not only a math genius, but a literary genius as well.  I wish I could have been there.  Oh well.  [/quote]

yes, the saga is amazing....although.....i was just a lil' weird back then....just a forewarning. :)

[quote="bubala"]By the way, I wonder what happened to those old members: i.e. tare, jojobird, confuted, etc. who are not active anymore.  What made them leave?[/quote]Let's not start this discussion all over again. People come and go as they wish. Sometimes they live because others wish them to do so. You can admire the old AoPS :) but don't go off-topic with questions like that  :mad: :) It'll only make the topic closed ;)

[quote="yif man12"]Hey Valentin, do you have anything from the old Mathlinks?  :)[/quote]Of course, I have all the old versions of the page. I plan to do a "history" page of both AoPS and MathLinks, presenting the beginnings of each of them (including how they looked). Sort of like a museum :)

Obviously today's designs are far more better, modern and nice looking than back in the dark, I mean old ages :)

Even though I am not a great mathematician, I remember these days since I joinced apporixamtely the date of your post, in the beginning it was such a unique feeling that there was such a forum, I cant look at that old stuff for some reason, its not archived ?


Currently these are the people I've found on AoPS who are on CT

algebrainic
brianjg1971
cherrypi
coach sue
codykaz
dbkhoa
edursho
Fedge Matthews
Hcb163
Ignite168
imsocool15
Jhonny Ruiz
jrshoch
JSRosen3
JZF
kidnappedturtle
kine
lotrfan
maniacman384
mintedcoin
Naismith
ngade
pillango
rrodgers
seltang
sfrick
snufflingqwert
trig_is_fun88
worthawholebean
Xtrememath88
yif man12
zabZeno

Ex-CTers:
Phelpedo (Now boarding in MA)

[quote="Phelpedo"]Wow this is an active forum.

There's some sort of connecticut ring headed by yif man, but I don't know how to get to it. Try pming him.[/quote]

Yeah I know, we need to get some mods to clean up the spam here.
Anyway, should I add you to the list even though you board in MA?
I'll PM him even though I think the ring is probably as active as this forum.
Btw: Notice a trend in the four people with posts?

Yeah that's it. There's also hockeyadi(something)

I am a 7th grader and I as well live in Conneceticut  :|

Now we have 9, if your school is out, I listed people under the grade they are going into next year.
Does anyone know which grades cherrypi, algebranic, and rrodgers are going into next year? They all said 7th but that could mean 7th next year or 7th this year.

I live in Fairfield,Connecticut and I am in 6th grade. How about u guys? Who else lives in Fairfield County at least....

Well, I go to Fairfield Woods Middle School

[quote="Phelpedo"]BTW, I'm going into 10th next year...I would say 9th grade currently but I got out of school JUNE 2ND BIATCHES!!![/quote]

Hopkins only ends a week later  :huh: Our school has held the record for earliest going back to school, most school days, and latest getting out of school since it opened. 200 days of school, from 3 weeks before Labor Day to last day of June.

Yeah, I remember hearing from a friend at Hopkins how rain ruined Prize Day. Getting days off for review must be nice although all we did in class the last week was talk, do anything, sign yearbooks, etc. Second to last week of school we did about 15 minutes of work in each class.

Oh and I just took the names of everybody from yif man's group since they all say they're in Connecticut

What's NEML?  :huh:

Oh and, want to tell me how qualification for the CT ARML teams is done?

hey Ignite and everyone  i'm Algerbrainic along with Timecat8

[quote="Ignite168"]

What's NEML?  :huh:

Oh and, want to tell me how qualification for the CT ARML teams is done?[/quote]

CT has n arml team? :maybe:  :roll:

i am guessing neml is
new england math leauge?

I live in Darien, Connecticut, very close to stamford.

[quote=Ultroid999OCPN][quote=gcahilly2023][quote=Ultroid999OCPN]er... [b]/bump[/b]?
@above: have you done MATHCOUNTS before?[/quote]

what part of ct are you from? I am near stamford[/quote]

East Lyme. (the eastern chapter is trash)[/quote]

oh! home of lyme's disease! interesting

 :rotfl:

[quote=yangliardi][quote=gcahilly2023][quote=Ultroid999OCPN]er... [b]/bump[/b]?
@above: have you done MATHCOUNTS before?[/quote]

what part of ct are you from? I am near stamford[/quote]

me too[/quote]

are you gonna do mathcounts this year? if so, let's meet there

[quote=gcahilly2023][quote=Ultroid999OCPN]I'll be there...[/quote]

yay... maybe i'll fail at chapter and then will not be at states but that is unlikely because I made it to states this year when I was absolutely terrible[/quote]

Same. Team round saved me. I did SO much better in sixth grade

Add me: I live in East Lyme and I know Ultroid


Give [tex] x,y,z\geq0 [/tex] satisfying $xy+yz+xz=1$.Prove that:
[tex] x\sqrt{1+x^2}+y\sqrt{1+y^2}+z\sqrt{1+z^2} \geq \sqrt{2(x^2+y^2+z^2+1)} [/tex]
 :)

a rather crude inequality. i like inequalities better if they are homogeneous, so let's write
$x\sqrt{1+x^2}=x\sqrt{(x+y)(x+z)}\geq \frac{2x(x+y)(x+z)}{2x+y+z}$,
the inequality being AM-HM. on the other hand,
$\sqrt{2(x^2+y^2+z^2+1)}=2\sqrt{\frac{(x+y)^2+(x+z)^2+(y+z)^2}{4}(xy+xz+yz)}\leq xy+xz+yz+\frac{1}{4}((x+y)^2+(x+z)^2+(y+z)^2)$,
the inequality being AM-GM. now, by some algebra and muirhead, we immediately get the desired inequality. maybe there's a better argument at the end. but at least it shows that the given inequality is really crude since one could use several elementary inequalities first.

Peter

Well, a boring problem. :(
I know it is really weak, however I wrote it as a gift for treegoner.
I can make it more strict but I afraid of I will make it more ugly. :(
For the last inequality,are you sure it will work.I am not brave enough to check it :maybe:

i am sure. i gave it mathematica to check it again and it said that it is true.

Peter


When a natural number is multiplied by itself, the result is a perfect square.  For example 1, 4, and 9 are perfect squares because $1 \times 1 = 1$, $2 \times 2 = 4$, and $3 \times 3 = 9$.  How many perfect squares are less than 10,000?

Well, knowing that 32 2 is equal to 1024, there are 31 perfect squares less than 1000.

[hide]100, because 100 2 is 10,000.[/hide]

[hide="4everwise"] you did all the squares up to 1,000.  the question asks for all the squares to 10,000,[/hide]

actaully both of you are wrong.  [hide]The question asks for how many squares exist smaller than 10,000.  However, 100^2 equals exactly 10,000 and there are 99 squares smaller than 10,000[/hide]

Yeah, the ultimate tricks.
[hide] :sqrt:10000 = 100, so ones whose squares are [b]less[/b] than 10000 only number 99. [/hide]


I'm sorry to tell you, but there is a fatal mistake in your proof.

I found that for ${p \ge \sqrt{3}-1}$, the inequality holds
$a^2+b^2+c^2+3pabc \le (p+1)(a+b+c)$.

Thus, the best inequality of this type is
$a^2+b^2+c^2+3(\sqrt3-1)abc \le \sqrt3(a+b+c)$.

The ugliest solution:
Homogenalise the inquality(we may do so because the degree of RHS is smaller than the degree of every term of LHS).\[(\sum a^2)\sqrt{\sum a^2}+3(3-\sqrt{3})abc\le\sum a^2\sum a\]Since $(\sum a^2)(\sum a-\sqrt{\sum a^2})=(\sum a^2)\frac{2\sum ab}{\sum a+\sqrt{\sum a^2}}$, it is enough to prove $2\sum a^2\sum ab\ge3(3-\sqrt{3})abc(\sum a+\sqrt{\sum a^2})$. Note that $\sum a^3(b+c)\ge2abc\sum a$, thus, it remains to prove\[\sum a^3(b+c)\ge2abc\sqrt{3\sum a^2}\]Expanding,\[(6,2,0)+(4,4,0)+(6,1,1)+(3,3,2)+2(4,3,1)\ge6(4,2,2)\]which is obvious.

Your solution is not ugly if in inequality
$2\sum a^2\sum ab\ge3(3-\sqrt{3})abc(\sum a+\sqrt{\sum a^2})$
replace $\sum a$ with $\sqrt{3\sum a^2}$, then use AM-GM Inequality. ;)

[quote="leepakhin"]Isn't that $(3, 3, 2)\leq (4, 2, 2)$?[/quote]
you have right. I will make the computations because it is possible that mecrazywong
make a typo.I will see this.


An algebraic expression of the form a + bx has the value of 15 when x = 2 and the value of 3 when x = 5.
Calculate a + b.

That's what I get.[hide]
From the given, we have a sytem of equations.

$ a\plus{}2b\equal{}15$
$ a\plus{}5b\equal{}3$

Subtracting, we find that $ \minus{}3b\equal{}12 \implies b\equal{}\minus{}4$.

Plugging this into either of our expressions, we find that $ a\equal{}23$.

So, $ 23\minus{}4\equal{}\boxed{19}$.[/hide]

Great ! By the way which standard are you in ?

You could set up the equation a+2b=15 and a+5b=3.  Put the first equation in the variable As form which would be A=15-2b.  Plug that in for the second equation to get 15-2b+5b=1. THen get 15+3b=3.  3b=-12 b=-4 plug that in for B and solve.

You should get 19.


This is 2 triangles with base 4 (from $ (3,0)$ to $ (3,4)$).
The left triangle has a height of 3, and the right has a height of 1.
Therefore, your area is:
$ \frac{4*3}{2} \plus{} \frac{4*1}{2} \equal{} \boxed{8}$


inequalities geometry circumcircle inradius perimeter ratio trigonometry

 Let ABC be a right triangle.
prove that R $ \ge$ (1+ $ \sqrt2$)r

Sometimes it's useful to explain the variables, but i guess that $ R$ is the circumradius and $ r$ the inradius of triangle $ ABC$. We have $ R \equal{} \frac {c}{2}$ and $ r \equal{} \frac {A}{s}$, where $ s$ is the semi-perimeter and $ A$ is the area of triangle $ ABC$, repsectively. Moreover, $ c \equal{} \sqrt {a^2 \plus{} b^2}$. We have to show that the ratio $ \frac {R}{r}$ is greater or equal to $ 1 \plus{} \sqrt {2}$. Expressing $ R/r$ with the above formulas, one gets
\[ \frac {R}{r} \equal{} \frac {\sqrt {a^2 \plus{} b^2}(a \plus{} b \plus{} \sqrt {a^2 \plus{} b^2})}{2ab} \equal{} \frac {1}{2} \sqrt {\frac {a}{b} \plus{} \frac {b}{a}} \left( \sqrt {\frac {a}{b}} \plus{} \sqrt {\frac {b}{a}} \plus{} \sqrt {\frac {a}{b} \plus{} \frac {b}{a}} \right) \ge \frac {1}{2} \sqrt {2} ( 2 \plus{} \sqrt {2}) \equal{} 1 \plus{} \sqrt {2},\]
where the last inequality follows from AmGm.

hello, just another solution, using the formula $ r\equal{}\frac{a\plus{}b\minus{}c}{2}$ so our inequality ist equivalent to
$ c\geq(1\plus{}\sqrt{2})(a\plus{}b\minus{}c)$
$ c(2\plus{}\sqrt{2})\geq(1\plus{}\sqrt{2})(a\plus{}b)$
$ \sqrt{a^2\plus{}b^2}(2\plus{}\sqrt{2})\geq(1\plus{}\sqrt{2}(a\plus{}b)$
squaring both sides we get
$ 2(a^2\plus{}b^2)(3\plus{}2\sqrt{2})\geq(3\plus{}2\sqrt{2})(a^2\plus{}b^2\plus{}2ab)$
$ (a\minus{}b)^2\geq0$ which is true.
Sonnhard.

[quote="zool007"][color=darkred]Let $ ABC$ be a right triangle. Prove that $ R\ge \left(1\plus{} \sqrt2\right)r$ .[/color] [/quote]

[color=darkblue][b][u]Proof 1[/u].[/b] If suppose w.l.o.g. that  $ A \equal{} 90^{\circ}$ , then $ \boxed {2R \equal{} a}$ and $ \boxed {2r \equal{} b \plus{} c \minus{} a}$ . Therefore $ p \equal{} r \plus{} 2R$ and $ 4R^2 \equal{} a^2 \equal{}$

$ \boxed {b^2 \plus{} c^2\ge 2bc} \equal{} 4S \equal{} 4pr$ $ \Longleftrightarrow$ $ R^2\ge pr$ $ \Longleftrightarrow$ $ R^2\ge (2R \plus{} r)r$ $ \Longleftrightarrow$ $ R^2 \minus{} 2Rr \minus{} r^2\ge 0$ $ \Longleftrightarrow$ $ \boxed {R\ \ge (1 \plus{} \sqrt 2)r\ }$ .

[b][u]Proof 2[/u].[/b] If suppose w.l.o.g. that  $ A \equal{} 90^{\circ}$ , then $ \boxed {2R \equal{} a}$ and $ \boxed {2r \equal{} b \plus{} c \minus{} a}$ . Therefore

 $ \boxed {\ R\ge \left(1 \plus{} \sqrt 2\right)r\ }\Longleftrightarrow$ $ 2R\ge \left(1 \plus{} \sqrt 2\right)\cdot 2r$ $ \Longleftrightarrow$ $ a\ge \left(1 \plus{} \sqrt 2\right)(b \plus{} c \minus{} a)$ $ \Longleftrightarrow$

$ a\left(2 \plus{} \sqrt 2\right)\ge \left(1 \plus{} \sqrt 2\right)(b \plus{} c)\Longleftrightarrow$ $ a\sqrt 2\ge b \plus{} c\Longleftrightarrow \boxed {\ 2\left(b^2 \plus{} c^2\right)\ge (b \plus{} c)^2\ }$ .

[b][u]Proof 3[/u].[/b] Suppose w.l.o.g. that  $ A \equal{} 90^{\circ}$ . I"ll use the well-known relation $ \boxed {\cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} \frac rR}$ ,

i.e. in this case $ \cos B \plus{} \cos C \equal{} 1 \plus{} \frac rR$ . Thus  $ \boxed {\ R\ge \left(1 \plus{} \sqrt 2\right)r\ }\Longleftrightarrow$ $ \frac {1}{\sqrt 2 \plus{} 1}\ge \frac rR$ $ \Longleftrightarrow \sqrt 2\ge 1 \plus{} \frac rR$ $ \Longleftrightarrow$

$ \sqrt 2\ge \cos B \plus{} \cos C\Longleftrightarrow \sqrt 2\ge 2\cos\frac {B \plus{} C}{2}\cos\frac {B \minus{} C}{2}$ $ \Longleftrightarrow \boxed {\ 1\ge\cos\frac {B \minus{} C}{2}\ }$ because $ B \plus{} C \equal{} 90^{\circ}$ .

[b][u]Proof 4[/u].[/b] Suppose w.l.o.g. that  $ A \equal{} 90^{\circ}$ . I"ll use the well-known relation $ \boxed {4\sin\frac A2\sin\frac B2\sin\frac C2 \equal{} \frac rR}$  ,

i.e. in this case $ \frac rR \equal{} 2\sqrt 2\sin\frac B2\sin\frac C2$ . Thus  $ \boxed {\ R\ge \left(1 \plus{} \sqrt 2\right)r\ }\Longleftrightarrow$ $ \frac {1}{\sqrt 2 \plus{} 1}\ge2\sqrt 2\sin\frac B2\sin\frac C2$ $ \Longleftrightarrow$

$ \sqrt 2 \minus{} 1\ge \sqrt 2\left(\cos\frac {B \minus{} C}{2} \minus{} \cos\frac {B \plus{} C}{2}\right)\Longleftrightarrow$ $ \sqrt 2 \minus{} 1\ge \sqrt 2\cos\frac {B \minus{} C}{2} \minus{} 1$ $ \Longleftrightarrow \boxed {\ 1\ge \cos\frac {B \minus{} C}{2}\ }$ .

[b][u]Proof 5 [/u].[/b] Denote the incentre $ I$ and the second intersection $ D$ of $ AI$ with the circumcircle $ C(O,R)$ . Observe that $ ID \equal{} DB \equal{} R\sqrt 2$ ,

 $ OI^2 \equal{} R^2 \minus{} 2Rr$  and $ \boxed {ID\le OI \plus{} OD}$ $ \Longleftrightarrow$ $ R\sqrt 2\le \sqrt {R^2 \minus{} 2Rr} \plus{} R$ $ \Longleftrightarrow$ $ R\left(\sqrt 2 \minus{} 1\right)\ge r$ $ \Longleftrightarrow$ $ \boxed {\ R\ge\left(1 \plus{} \sqrt 2\right)r\ }$ .

[b][u]Remark[/u].[/b] We have equality if and only if $ B \equal{} C \equal{} 45^{\circ}$ , i.e. $ \frac b1 \equal{} \frac c1 \equal{} \frac {a}{\sqrt 2}$ .[/color]

[quote][color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle with the inradius $ r$ and the circumradius $ R$ . Prove that $ \boxed{\ R\ \ge\ \frac {1 \plus{} \sin\frac A2}{\sin A\cos\frac A2}\cdot r\ }$ .[/color][/quote]

[color=darkblue][b][u]Hint[/u].[/b] You can prove it as with the same idea from the last proof ![/color]


How fast can you add? How many coordinates are you plot in 60 seconds?

[b]So, you think you are the best!?! Prove it![/b]

Compete against other math students in a collection of [u]free[/u] [b]math games[/b] at:

http://www.xpmath.com/

Have fun!

-Hui

It's a series of free math games that tests your speed and accuracy in various math skills.

You can compete with others over the internet for [b]prizes[/b] such as [i]Amazon gift certificates[/i].  :roll:

Take a look and you'll see.[/i][/b]


geometry
3D geometry
tetrahedron
geometric transformation
reflection
vector
geometry proposed

Prove that if the bimedians of a tetrahedron are the common perpendiculars to respective sides then its opposite sides are equal.

The segment connecting the midpoints of two opposite sides of a tetrahedron.

As for a solution,

Let $M_{XY}$ be the midpoint of $XY$, where $X,Y$ are vertices from the set $\{A,B,C,D\}$. We get $M_{XY}U=M_{XY}V$ for all permutations $X,Y,U,V$ of $A,B,C,D$ (call this observation $(*)$). From here the conclusion follows easily. We can see that if the tetrahedron satisfies the conclusion, then $(*)$ holds. On the other hand, given $BCD$, $(*)$ determines the tetrahedron uniquely, so it must satisfy the conclusion.

There are other beautiful things to say about this tetrahedron. For example, it's the only tetrahedron with all faces equivalent (i.e. having the same area), or the only tetrahedron for which the angles around each vertex sum up to $\pi$.

Ok!
Kunny posted this problem some time ago (and I solved it). But I don't know whether this thread disappeared or not?

May be, I've missed something, but it seems obvious :
two opposite sides are reflected of each other with respect to the corresponding bimedian hence they have the same length.
Kind regards. Jean-Pierre

Maybe I am missing something, but I don't understand your idea. :?

The reflection wrt the line joining the midpoints of AB and CD maps A to B and C to D, hence AD = BC and AC = BD
Kind regards. Jean-Pierre

I have an other approach:
Let vectorAB=x, vector AC=y, vector AD=z then vector MN=(y+z-x)/2, where M and N are the midponts of AB and CD. Hence
(y+z-x)/2*x=(y+z-x)/2*(z-y)=0
Then xy+xz-x 2 =0 and
        z 2 -y 2 -xz+xy=0. hence
z 2 =(y-x) 2 and y 2 =(z-x) 2 .Finally AD=BC and AC=BD and the conlusion follows.


IMO Shortlist
3D geometry
packing
combinatorics
Coloring

A staircase-brick with 3 steps of width 2 is made of 12 unit cubes. Determine all integers $ n$ for which it is possible to build a cube of side $ n$ using such bricks.

can someone please find a solution or at least a hint please?

There are 12 unit cubes in each brick, hence $ n$ must be divisible by 2 and 3, i.e., $ 6|n$. Let $ n\equal{}6k$. Two staircase-bricks can be used to build a $ 2\times3\times4$ box. Hence it is easy to construct a cube of side $ 12$. Thus $ n\equal{}6k$ is possible for even $ k$. We claim that if $ k$ is odd, we cannot build a cube of side $ n$, i.e., the number of bricks must be even.

Assume there are $ m$ bricks. Assign them to the Cartesian coordinates such that the center of a corner cube is at (0,0,0). Consider the cubes in $ (x,y,z)$ such that $ x,y,z$ are even. We call such cubes "good". In a brick, there can be either 1 or 3 good cubes. Assume that there are $ p$ bricks that contain 3 good cubes. Hence the total number of good cubes is $ 3p\plus{}m\minus{}p\equal{}2p\plus{}m$. But the number of good cubes is $ \frac{12m}{8}$. Hence $ 2p\plus{}m\equal{}\frac{12m}{8}$, or $ m\equal{}4p$. So $ m$ is even. Q.E.D.

Actually you need $ 4|m$ ($ m$ even does not suffice, since there are $ n^3 \equal{} 12m$ would just imply $ 8|n^3$, which gives nothing).

Or did I miss something?

@above yeah we need 4|m
For this we will use comlex combinatorics Let $\epsilon^4=1$ and $\epsilon \neq 1$ and we will assign to each square with coordinates $(x,y,z)$ with $\epsilon^{x+y+z}$
Then for each staircase-brick with starting unit cube $(x,y,z)$ which $x+y+z$ is minimum Then our sum will be 0(it is easy to calculate).And we know that The general sum is
$(\epsilon^0+......+\epsilon^{m-1})^3$ and it is 0.Then 4|m and then we are done

@elcinmusazade what is the reasoning for $4|m?$ As I understand it you proved that the sum of the staircase brick is $0$ and the total sum is $0$ and any number of staircase bricks should sum to $0$ so there should be no restriction?

jgnr has a good idea and I built on it.

Note that $12|n^3$ so $6|n.$ Let $n=6k.$ Note that two staircase bricks can be put together to form a $2\times3\times4$ prism like in the following picture.
[img width=50]https://media.discordapp.net/attachments/925784397469331477/950060134556663839/Screen_Shot_2022-03-06_at_8.59.17_AM.png?width=1108&height=1170[/img]

We can place a three-dimensional array of $6\times4\times3$ of these $2\times3\times4$ prisms to form a $12\times12\times12$ cube. These cubes can then be made into a big cube of side length divisible by $12,$ i.e. when $k$ is divisible by $2.$

Now, we split the $6k\times 6k\times 6k$ into a $3k\times 3k\times 3k$ grid of $2\times 2\times 2$ cubes and in each such cube we call the bottom-left-back corner unit cube [i]special.[/i] Note that in the two layers of the staircase brick, one of them cannot have any special cubes.

[img width=50]https://media.discordapp.net/attachments/925784397469331477/950073693189586984/Screen_Shot_2022-03-06_at_9.53.22_AM.png?width=1332&height=1169[/img]

In the above picture, observe that differently-colored cubes cannot be both special, and one of the four colors must be a special cube, so the number of special cubes per staircase brick is $1$ or $3.$ Note that the number of staircase bricks is $18k^3$ which is even. Thus, the number of special cubes is even and so the number of total $2\times 2\times 2$ cubes is even. This implies that $27k^3$ is even, so $k$ is also even. Therefore the only solutions are $n=12k$ for all positive (nonnegative if you happen to count $0$) integers $k$.

Notice that the volume of such a cube must be a multiple of $12$ and thus $n$ should be a multiple of $6$. We claim that $n$ can be $\boxed{\text{all multiples of }12}$. Notice that two staircases can be combined to create a $2\times 3\times 4$ box, which can clearly be stacked to fill any cube with side length a multiple of $12$.

It suffices to show that $n=12k-6$ for positive integer $k$ is impossible. Suppose we can fill such a cube completely with staircases. Break up such a cube into unit subcubes and let their centers be $(a, b, c)$ for $1\le a,b,c\le n$. Give each subcube a [i]label[/i] from $0$ to $3$ equivalent to the sum of it's coordinates modulo $4$. Now it is easy to check that all staircases placed inside such a cube must have an odd number of unit cubes of every type. Since there are an even number of staircases in filling up a cube, the cube must have an even number of unit cubes of each type.

Break up the side length $12k-6$ cube into one $12k-4$ cube, three $(12k-4)\times (12k-4) \times 2$ prisms, three $(12k-4)\times 2\times 2$ prisms, and one $2\times 2\times 2$ cube. Everything except the $2\times 2\times 2$ cube has an even number of unit cubes of each type, and the $2\times 2\times 2$ cube has an odd number of unit cubes of each type. This means that the big cube has an odd number of unit cubes of each type, contradiction.

We claim the answer is all positive multiples of $12$. Evidently, if $n$ is an integer that works, we have $12\mid n^3$, so $6\mid n$. Hence, $n=6k$. Now, note that we may place two staircase-bricks together to form a $2\times3\times4$ prism. Since $2,3,4\mid12$, it readily follows that the cube can be built when $k$ is even. Now we prove the cube cannot be built if $k$ is odd. Suppose it can and consider a tiling of a $6k\times6k\times6k$ cube with staircase-bricks. We separate the brick into six $6\times6$ layers. In each layer, the staircase-brick fills either a $2\times1,2\times2$, or $2\times3$. If the staircase-brick fills a $2\times2$, then it must fill a $2\times1$ and $2\times3$ in the above and below layers. Hence, the bottom-most layer has no $2\times2$ filled by staircase-bricks. It follows the layer right on top is filled only by $2\times2$. It is straightforward to see that the only way to tile a $6k\times6k$ layer using $2\times2$ pieces is to lay out the pieces in a $3k\times3k$ grid. Therefore, we use $9k^2$ pieces. Now, look at the bottom layer. It is filled by $x$ $2\times3$ and $y$ $2\times1$. Since it is the bottom layer, these must be part of one of the $9k^2$ pieces above. Hence, $x+y=9k^2$. On the other hand, summing areas we see $6x+2y=36k^2$, or $3x+y=18k^2$. Thus, $2x=9k^2$. However, $k$ is odd, so this is impossible. Hence, the cube cannot be built when $k$ is odd as desired. $\blacksquare$

The answer is all multiples of $12$. Evidently these work, because we can form $2\times 3\times 4$ bricks by combining two staircase-bricks.

Now, for odd multiples of $6$, label each cube with a number from $0$ to $3$, representing its 3-D taxicab distance to some predetermined corner. Note that the number of cubes labeled with each number is odd, since we can "reduce" the dimensions $\bmod\text{ 4}$ to obtain a $2\times 2\times 2$ square, where the counts of all numbers are all odd.

Suppose FTSOC that we can form such a cube with staircase-bricks. Then we must clearly use an even number of staircase-bricks. I claim that the counts of each of the labels is odd; this will give a contradiction, since it implies that the counts of each of the numbers in the cube is even. There are two cases to consider. The first case is where the closest cube to the predetermined corner is a cube on the bottom step of the staircase-brick. In this case, each of the labels from $0$ to $3$ appears exactly three times. The second case is where the closest cube is one of the two "back" corners of the staircase-brick. This case gives one label which appears five times, two which appear three times, and one which appears once.

Then, we have our desired contradiction, so we are done. $\blacksquare$

We claim that the answer is $12|n$.

We can cap the staircase on top of itself to form a 2x3x4 block, which can be combined to build a 12x12x12 block, so all multiples of 12 are possible.

Obviously, $n$ is divisible by 6 by volume. Let $n=6m$. Assign coordinates $(i,j,k)$ with $1\leq i,j,k\leq 6m$ to each unit cube. Denote a unit cube as orz if $i,j,k$ are all even.

Claim: If a staircase-brick is placed anywhere, it will occupy an odd number of orz unit cubes. Note that the staircase-brick can be decomposed into a 2x2x2 cube and two small 1x1x2 blocks. Note that exactly one unit cube in the 2x2x2 is orz. Now, the four remaining unit cubes can be paired into two pairs such that the two unit cubes in each pair are either both orz or both not orz (since each coordinate differs by either 0 or 2), so there are an odd number of orz unit cubes.

Note that the total number of staircase-bricks is $\frac{(6m)^3}{12}=18m^3$, which is even, so the total number of orz cubes must be even since each staircase-brick has an odd number of orz cubes. Therefore, $m$ must be even since there are $27m^3$ total orz cubes, so $n$ must be divisible by 12.


Let a and b be positive integers such that (1 + ab)|(a^2 + b^2). Show that the integer (a^2 + b^2)/(1 + ab) must be a perfect square.

[i]p. 20 An Introduction to the Theory of Numbers, Ivan Niven et al[/i]


arithmetic sequence
algebra unsolved
algebra

let $N$ denote the set of all natural numbers
let it be divided into $k$ disjoint subsets namely $S_{1},S_{2}\dots S_{k}$ each of which has it's elements in $arithmetic$ $progression$ .
also let the common differeences be $\alpha_{1},\alpha_{2}\dots\alpha_{k}$
then prove that

$\sum_{i=1}^{k}\frac{1}{\alpha_{i}}$ $=$ $1$

See here, problem 22, page 19, good luck!  :wink:


function
vector
topology
group theory
abstract algebra
algebra
polynomial

[quote="harazi"]Here's a truly beautiful and I think very very difficult problem:
    Consider $f$ be a $C^1$ class function from $R^n$ into itself such that it vanishes only in 0, it tends to infinity (in norm, of course) when the argument tends to infinity and $df(x)$ is invertible at any point. Then $f$ is a difeomorphisme.[/quote]
You don't need the requirement that $f(x) = 0 \, \implies \, x = 0$.
First we show that $f$ is onto. Let $K \subset \mathbb{R}^n$ be compact, then $f^{-1}(K) = \{x \, | \, f(x) \in K\}$ is closed (since $f$ is continuous) and bounded (since otherwise we can find $x_n \to \infty$, yet $f(x_n) \in K$). In short, [i]$f$ is proper.[/i]

So let $a \in \mathbb{R}^n$, and consider $A = f^{-1}(B_{|a|+1}(0))$ (pre-image of closed ball of radius $|a|+1$). This is a compact set, hence the minimum of $g(x) = |f(x)-a|$ is attained at some $x^* \in A$.

Suppose $f(x^*) = b \neq a$, then there are an open neighborhood $U(x^*)$ and an open ball $B_{\epsilon}(b)$ such that $f(U(x^*)) = B_{\epsilon}(b)$, by the inverse function theorem. Thus there is a point $z \in U(x^*)$ such that $f(z) = b + \delta(a-b)$, and therefore $z \in A$ and also $|f(z) - a| < |b-a|$, contradiction. Hence $f(x^*) = a$, and $f$ is onto.

To show that $f$ is 1-1, assume that $f(a) = f(b)$ yet $a \ne b$. Consider $g(t) = f(a + t(b-a))$. Then $g(b) = g(a)$ and hence $0 = g'(c) = Df(a+c(b-a))(b-a)$ for some $0 < c < 1$. Thus $Df(a+c(b-a))$ has a non-trivial null space, contradicting the assumption that  $Df(x)$ is invertible everywhere.  ;)

Continuing the nonsense....We would need the condition of increasing norms for the fact that $f(B_r)$ would be unbounded and thus cover all of $\mathbb{R}^n$....And the $f(0)=0$...is it necessay? I mean, if we take $f(x)=x+c$, $c$ a constant vector, nothing changes, right?
Sorry for wasting your time  :blush:

[quote="fedja"]Oh, no! You do not have Rolle's theorem in dimensions greater than $1$! The surjectivity is simple, indeed, but the injectivity is much harder  ;)[/quote]Right, I noticed it an hour after posting this nonsense :blush:

[quote="DusT"]Ok, this might look like some nonsense....Because I did not make it rigurous...But...can we do something on the lines of :
Take a point $\alpha \in \mathbb{R}^n$. Then locally, by the Inverse function theorem, $f$ will have an inverse $f_{\alpha}$ in some neighbourhood $U_{\alpha}$.  Now, take a compact ball $B_r$...And by the Heine-Borel (I think) theorem there will be an open cover of it with sets $U_{\alpha}$, finite in number. Then, it is easy to see that they will be coherent in their intersections, as inverses of the same functions, so we would have an inverse in the ball...Can we now extend it to $\mathbb{R}^n$ ?
Sorry for this nonsense... :blush:[/quote]
Here's how this can be made precise:

For each $x \in %Error. "mathnn" is a bad command.
{R}^n$, let $k(x) = \#\{z \, | \, f(x)=f(z)\}$, i.e. the number of points whose image agrees with $f(x)$. Since $f$ is onto, $k(x) \ge 1$ for all $x$, and $k(0) = 1$ by one of the original assumptions. Note that $k(x)$ is finite for all $x$, since pre-images of single points cannot have accumulation points (because $f$ is locally invertible) and are compact (because $f$ is proper). By local invertibility of $f$, $C_m = \{x \, | \, k(x) = m\}$ is open for each $m$. Thus $C_1$ is open, non-empty, and since the complement of $C_1$ is $\cup_{m>1}C_m$ and hence open, $C_1$ is also closed. Thus $C_1 = \mathbb{R}^n$, and $f$ is 1-1.

Actually, we don't need the assumption that $f^{-1}(0)=\{0\}$. It follows from the other hypotheses that $f$ is one-to-one, but we need a little more work. Let's take it from the top:

First of all, $f$ is a local homeomorphism, so it's an open map. Secondly, it's proper, and any proper map from a metric space to itself is also closed (easy to prove: if $f(x_{n})\to y$, the preimage of $\{f(x_{n})\}\cup\{y\}$ is compact, so if $x$ is the limit of a subsequence of $(x_{n})_{n}$, then $f(x)$ is the limit of a subsequence of $(f(x_{n}))_{n}$, i.e. $f(x)=y$). From these two we find that $f(\mathbb R^{n})$ is a non-empty clopen subset of the connected space $\mathbb R^{n}$, so it must be the entire space, i.e. $f$ is onto.

$f: \mathbb R^{n}\to\mathbb R^{n}$ is onto, and our hypotheses imply that it's a covering space. Since $\mathbb R^{n}$ is connected, the preimage of every point has the same (finite, because $f$ is proper) number of elements (this is a basic fact on covering spaces; it's also just a restatement of what hpe did, and I'm sorry I didn't see it before he wrote it down :)). Knowing that $|f^{-1}(0)|=1$ immediately tells us that this common value of all $|f^{-1}(x)|$ is $1$, but as I said, we can reach this conclusion without this hypothesis:

Another basic fact about covering spaces is that if $p: \widetilde X\to X$ is a covering map (for well-behaved spaces $\widetilde X,X$, i.e. which have all the connectedness properties we might need: they're connected, locally connected, path-connected, and locally path-connected), then the common value $|p^{-1}(x)|$ is precisely the index of the subgroup $p_{*}(\pi_{1}(\widetilde X))$ in $\pi_{1}(X)$, where $\pi_{1}(\cdot)$ denotes the fundamental group, and $p_{*}$ is the group homomorphism indiced by $p$. In our case, both $\widetilde X$ and $X$ are simply connected, i.e. this index can only be $1$, and we're done.

In conclusion, we now know that:

A $C^{1}$ diffeomorphism which also happens to be a proper map is actually a diffeomorphism.

Note that we can replace the hypothesis of being proper with that of being a closed map. We saw above that proper implies closed. Conversely, for local homeomorphisms, closed implies proper.

S.I Pinchuk, in "A counterexample to the Real Jacobian conjecture" (Math. Z. 217, 1994), gives an example of a non-injective polynomial mapping from $R^{2}$ into itself of degree(p,q)=(10,25) and whose jacobian determinant is everywhere positive in $R^{2}$.

Where is the mistake?

That mapping is probably not proper (i.e. the preimage of a compact set is not always compact). That can happen with polynomials of more than one variable.