geometry
trapezoid
trigonometry
trig identities
Law of Sines
geometry proposed

Given a trapezoid $ ABCD$ with $ AB\bot BC,\ BC\bot DC.$ Let $ N$ be the point which divides the side length $ AD$ internally such that $ AB : DC$. If $ BN\bot CN$, then prove that $ AB \plus{} DC \equal{} AD.$

Because of $ BN\perp CN,$ we have that the point $ N$ lies on the circle $ (K)$ with diameter $ BC,$ which tangents to $ AB,\ CD,$ at points $ B,\ C,$ respectively and let be the point $ E\equiv CD\cap BN.$

Because of $ CE\parallel AB,$ applying the [b][size=100]Thales theorem[/size][/b], we have that $ \frac {AB}{DE} \equal{} \frac {AN}{ND} \equal{} \frac {AB}{CD}$ $ ,(1)$

From $ (1)$ $ \Longrightarrow$ $ CD \equal{} DE$ $ \Longrightarrow$ $ CD \equal{} DE \equal{} ND$ $ ,(2)$ ( because of $ \angle CNE \equal{} 90^{o}$ ).

From $ (2),$ we conclude that $ \angle DCN \equal{} \angle NCD$ $ ,(3)$ and so, the line segment $ AD,$ tangents to $ (K),$ at point $ N.$

Hence, we conclude that $ AD \equal{} AN \plus{} ND \equal{} AB \plus{} CD$ and the proof is completed.

Kostas Vittas.

PS. The circle $ (K)$ and the line segment $ AD,$ may have only one common point, because $ N$ is a unique point between $ A,\ D,$ such that $ \frac {AN}{ND} \equal{} r$ ( Here, $ r \equal{} \frac {AB}{CD},$ ).

Thank you for the synthetic solution, vittasko. :lol:

My solution is using vector.

[quote="Kunny & Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b] Given a trapezoid $ ABCD$ with $ AB\parallel CD$ .  Consider the point

$ N\in (AD)$ so that $ \frac {NA}{ND} \equal{} \frac {AB}{DC}$ . Prove that $ NB\perp NC\ \Longleftrightarrow\ AB \plus{} DC \equal{} AD$ .[/color] [/quote]

[color=darkblue][b][u]Proof 1[/u].[/b] Denote $ L\in AD\cap BC$ and $ P\in BC$ for which $ NP\parallel AB$ . Thus, $ \frac {NA}{ND} \equal{} \frac {AB}{DC} \equal{} \frac {LA}{LD}$    $ \Longleftrightarrow$ $ \frac {NA}{ND} \equal{} \frac {LA}{LD}$ $ \Longleftrightarrow$

the division $ (A,D\ ;N,L)$ is harmonically $ \Longleftrightarrow$ the division $ (B,C\ ;P,L)$ is harmonically. Therefore, $ NB\perp NC$ $ \Longleftrightarrow$

 $ \widehat {BNL}\equiv\widehat {BNP}$ $ \Longleftrightarrow$  $ \widehat {BNA}\equiv\widehat {ABN}$ $ \Longleftrightarrow$ $ AN \equal{} AB\ \ \wedge\ \ DN \equal{} DC$ $ \Longleftrightarrow$ $ AD \equal{} AB \plus{} DC$ .[/color]

[color=darkblue][b][u]Proof 2[/u].[/b] Denote $ E\in CN\cap AB$ , $ F\in BN\cap CD$ and suppose $ NB\perp NC$ . Observe that $ BA\parallel DF$ $ \implies$

$ ANB\sim DNF$ $ \implies$ $ \frac {ND}{CD}\equal{}\frac {NA}{AB}\equal{}\frac {ND}{DF}$ $ \implies$ $ CD\equal{}DF\equal{}ND$ $ \implies$ $ BA\equal{}AE\equal{}AN$ $ \implies$ $ AD\equal{}BA\plus{}CD$ .[/color]

Thank you, Virgil Nicula.

[url=http://www004.upp.so-net.ne.jp/s_honma/urawaza/angle.htm]This is my Memo.[/url]

Let $ \angle ANB\equal{}x$, then $ \angle CNB\equal{}90\minus{}x$. Let $ \angle ABN\equal{}y$, then $ \angle NCD\equal{}90\minus{}y$.
By the law of sines,
$ \frac{sin x}{AB}\equal{}\frac{sin y}{AN}$ and $ \frac{cos x}{CD}\equal{}\frac{cos y}{ND}$
Dividing the two, and using that $ \frac{AB}{CD}\equal{}\frac{AN}{ND}$, we have $ tanx\equal{}tany$, so $ x\equal{}y$, so $ \triangle ANB$ and $ \triangle CND$ are isoceles, so $ AB\plus{}CD\equal{}AD$.

What exactly is Thales' Theorem?  I mean, I don't really see what Thales' theorem is.

Thales' theorem isn't really a theorem, in my opinion. It's just similar triangles. In other words, for any two parallel segments AB and CD, if AC and BD intersect at E, then $ \frac{EC}{EA}\equal{}\frac{ED}{EB}$

[quote="dgreenb801"]Thales' theorem isn't really a theorem, in my opinion. It's just similar triangles.[/quote]
Correction ,[b]Thales theorem[/b] in fact is a theorem , and actually the triangle similarity criterias come from Thales theorem  :)
Put in other way : Thales theorem was first

[quote="dgreenb801"]Thales' theorem isn't really a theorem, in my opinion. ...[/quote]  :(   :mad:
A good exersize is to prove that $ \frac{AN}{ND} \equal{} \frac{AB}{DE}$ ( see the schema [b]t=299612[/b] ), applying only the [b][size=100]Thales theorem[/size][/b], without any reference in similarity.

As [b][size=100]Tiger100[/size][/b] said, the [b][size=100]Thales theorem[/size][/b] was first and it is well known.

Kostas Vittas.

I'm sorry, I had read a post before on mathlinks saying that you wouldn't really think of it as a theorem, now I see it is wrong  :blush:

[b]A last commentary.[/b] I think that is nicerly the following equivalent enunciation.
Thus the proposed problem becomes very simply and a nice property in a trapezoid.

[quote][color=darkred]Let $ BCFE$ be a trapezoid , where $ BE\parallel CF$ and $ BF\perp CE$ . Denote  the midpoints

$ A$ , $ D$ of the parallel sides $ BE$ , $ CF$ respectively. Prove that $ 2\cdot AD\equal{}BE\plus{}CF$ .[/color][/quote]


Hi,

I need to know the exact date all AP tests end for this year (for COP B). Can anyone tell when it exactly end?

Thanks a lot.

(I heard talks of upcoming Friday and others on 13th I want to know which one's exact.)

Next Friday, May 13th is the last day (except for alternate testers, which should be relatively few people)... whoa, Friday the 13th. Didn't even notice that before. Good luck to anyone who has an AP test that day...

yea.... for the alternate test takers it is 19th i believe.


The consecutive angles of a particular trapezoid form an arithmetic sequence.  If the largest angle measures $ 120^{\circ}$, what is the measure of the smallest angle?

lets say 4 angles are x, y, z, and 120.

y+z=180, and x+120=180 which x is the smallest angle.

so x=60

answer : 60


group theory
abstract algebra
complex numbers
real analysis
real analysis unsolved

let $G=(0,\infty)$, so its a group under the multiplication operation. what are the subgroups of G that are compact subsets of $G$?

Suppose $H$ is a subset of $G.$ If $r\in H$ and $r\ne1,$ then becuase $H$ is closed under multiplication, $\{r^{n}: n\in\mathbb{Z}\}\subset H.$

That's not a compact set and no set containing it can be compact. Hence the only compact subgroup of $G$ is the trivial subgroup $\{1\}.$

If you were to expand this to ask about the group (under multiplication) of nonzero real numbers, then you would have as compact subgroups both the trivial group and the two-element group $\{1,-1\}.$

Going further, the multiplicative group of nonzero complex numbers has a rich supply of compact subgroups.

@dr kent

 for the complex case,i found that the unit circle is compact and i couldnt find anything else. could it be possible that these are the only ones? i just thought about it now and i may be wrong. let us  consider the homomorphism from any compact subgroup of $C^{*}$ to $(0,\infty)$ which sends $z$ to $modz$ . this map is continuous and the image should be compact and this is clearly just ${1}$, so for any z in this compact subgroup mod z =1. am i correct?

For any $n\in \mathbb{N},$ the set $\{e^{2\pi i k/n}|k=0,1\dots n-1\}$ of $n$th roots of unity is a compact subgroup of $\mathbb{C}^{*}.$

kouboy: your argument is correct and it shows that any compact subgroup of $\mathbb C^{*}$ is contained in $\mathbb T=\{z\in\mathbb C\colon |z|=1\}$. In fact, $\mathbb T$ and the cyclic groups mentioned by Kent Merryfield are the only compact subgroups of $\mathbb C^{*}$.

thanks for the help dr kent and mlok  :)


Express [(root(x+h)) - (roo(x))] / h  as  1 / [(root(x+h)) + (root(x))]  by manipulating the first equation.

$\frac{\sqrt{x+h}- \sqrt{x}}{h}$ as$ \frac{1}{ \sqrt{x+h} + \sqrt{x}}$
do you know about rationalizing the bottom parts of fractions? the idea is that if you have square roots,
$\sqrt{a} + \sqrt{b}$, then if you multiply that by $\sqrt{a} - \sqrt{b}$, you will get $a - b$

generally, this is a property you [b]need[/b] to learn:
$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$ or $(a + b)(a-b) = a^2 - b^2$

if you are not convinced that it is true, try multiplying it out yourself (FOIL)
so if you don't recognize this in your problem, it asks you to simplify the top into 1, so we first want to get rid of the square roots, and using the property that i showed earlier,
$\frac{\sqrt{x+h}- \sqrt{x}}{h}* \frac{\sqrt{x+h}+ \sqrt{x}}{\sqrt{x+h}+ \sqrt{x}}=$
$\frac{x+ h - x}{h*\sqrt{x+h}+ \sqrt{x}}= \frac{1}{ \sqrt{x+h} + \sqrt{x}}$
and the x's cancel out, then the h's also cancel, giving your desired answer
generally, you should remember how to get rid of the square roots by the property that i showed you (or move the square roots to the other side of the fraction)


geometry AMC AIME trapezoid rectangle probability ratio

A circle radius r is inscribed in ABCD. It touches AB at P and CD at Q. AP = 19, PB = 26, CQ = 37, QD = 23. Find r.

ABC is a triangle with angle B = 90o. AD is an angle bisector. E lies on the side AB with AE = 3, EB = 9, and F lies on the side AC with AF = 10, FC = 27. EF meets AD at G. Find the nearest integer to area GDCF.

ABC is a triangle with sides 3, 4, 5 and DEFG is a 6 x 7 rectangle. A line divides ABC into a triangle T1 and a trapezoid R1. Another line divides the rectangle into a triangle T2 and a trapezoid R2, so that T1 and T2 are similar, and R1 and R2 are similar. Find the smallest possible value of area T1.

A circle radius 1 is randomly placed so that it lies entirely inside a 15 x 36 rectangle ABCD. Find the probability that it does not meet the diagonal AC.

some of the problmes i still couldnt get after looking at the official solution.(the last two are from last year)

[quote="xxreddevilzxx"]ABC is a triangle with sides 3, 4, 5 and DEFG is a 6 x 7 rectangle. A line divides ABC into a triangle T1 and a trapezoid R1. Another line divides the rectangle into a triangle T2 and a trapezoid R2, so that T1 and T2 are similar, and R1 and R2 are similar. Find the smallest possible value of area T1.
[/quote]

Yes, I remember this. I solved it on last year's AIME. Yay!

[hide]

You have four possible arrangements: 2 for the triangle and 2 for the rectangle. In the triangle you can cut it parallel to side 3 or side 4. In the rectangle, you can cut it so that the line intercepts a side of 6 or a side of 7 (so the triangle formed has one fixed side length of 7 or 6, respectively). So try each of these cases... in each one you set the triangle so that its sides are in a 3:4:5 ratio and then solve for how long the short base of the trapezoid has to be. For example, if one side of the triangle is 7, then 4:3 = 7:21/4, so the short base is 6-21/4 = 3/4. We also know the long base is 6. So the length of the cut in the triangle has ratio 3:6 = 1:2 of the short base which is 3/8. Then the height of T1 is 1/2. So the area is (1/2)(1/2)(3/8) = 3/32. I'm pretty sure this was the answer too (3+32=35). Do this for each case and compare to find out.

Yeah, that was pretty hard to understand. Hopefully you can piece it out. [/hide]

[quote="xxreddevilzxx"]A circle radius 1 is randomly placed so that it lies entirely inside a 15 x 36 rectangle ABCD. Find the probability that it does not meet the diagonal AC.[/quote]

I didn't solve this one on last year, though :(.

[hide]
You've probably figured out by now that you can draw a triangle inside each half of the rectangle (divided by the diagonal) such that is is distance 1 from each AB, BC, and AC and find the areas. Finding the area is the hard part. Let A' be the vertex of the triangle closest to A. Let X be the point where a line perpendicular to AB passing through A' intersects AB. We want to find AX. Consider angle CAB. If we can find $\tan{\frac{CAB}{2}}$ we're set. We know $\tan{CAB} = 12/5$. So using the tangent double angle formula, we get $\tan{2\left(\frac{CAB}{2}\right)} = \frac{2\tan{\frac{CAB}{2}}}{1-\tan^2{\frac{CAB}{2}}} = 12/5$. Solving this, we find $\tan{\frac{CAB}{2}} = 2/3$. Then $A'X/AX = 1/AX = 2/3$. So $AX = 3/2$. We also know that $B'C'$ is distance 1 from $BC$. So $A'B' = 15-3/2-1 = 25/2$. Then $C'B' = (25/2)(12/5) = 30$. So the area of $A'B'C'$ is $(1/2)(25/2)(30) = 375/2$. But since there are two of them as I mentioned in the first sentence, we multiply that by two to get $375$.

EDIT: forgot the little bit about probability, its the area of the triangles over the area of the rectangle distance 1 from each side of the big rectangle = 375/(13*34) = 375/442.
[/hide]

Sorry for the poor TeXing, it's late at night =P.

[quote="xxreddevilzxx"]A circle radius 1 is randomly placed so that it lies entirely inside a 15 x 36 rectangle ABCD. Find the probability that it does not meet the diagonal AC. [/quote]

[hide]circles that lie within rectangle form a 13*34 rectangle
of that, we have to eliminate the circles that touch the diagonal
draw in AC, and draw a circle that is tangent to AC and AD.  Then draw in E on AD and F on AC such that EF is parallel to DC and tangent to the circle.  AE=5x, EF=12x, and AF=13x.  5x+12x=13x+2.  x=.5.  the location of the center is 15-5/2+1=27/2 away from DC.
repeat with a circle tangent to AD and DC, and a circle tangent to DC and AC.  we get a triangle that has sides 27/2-1 and 31-1, so the area is 375/2.  we have two such triangles so the area is 375.  the probability is 375/(13*34) = 375/442[/hide]

[quote="xxreddevilzxx"]ABC is a triangle with angle B = 90o. AD is an angle bisector. E lies on the side AB with AE = 3, EB = 9, and F lies on the side AC with AF = 10, FC = 27. EF meets AD at G. Find the nearest integer to area GDCF. [/quote]

[hide]
This must be the ugliest solution ever.
Pythagorean Thm => BC = BD+CD = 35. Angle Bisector Thm => 12/BD = 37/CD => BD = 60/7. Since cosA = 12/37, we can use Law of Cosines to find EF^2 = 109-60(12/37) = 3313/37 = about 89 so EF = about 9.4. Then Angle Bisector Thm => 3/EG = 10/GF => GF = (10/13)(9.5) = about 7.25. Law of Sines on AEF gives us 9.4/sin(FAE) = 3/sin(AFE) so sin(AFE) = (3/9.4)(35/37) = about 0.3. Using the ab(sinC)/2 formula on AGF, we get that it has an area of (7.25)(10)(0.3)/2 = about 10.9. We can also find the area of ABD to be (60/7)(12)(1/2) = 360/7 = about 51.4. So the area of GDCF = (35)(12)/2 - 51.4 - 10.9 = about 148.

I hope none of my estimates were too far off :D [/hide]

[quote="Mildorf"][quote="xxreddevilzxx"]A circle radius r is inscribed in ABCD. It touches AB at P and CD at Q. AP = 19, PB = 26, CQ = 37, QD = 23. Find r. [/quote]

(That is the Kalva interpretation of AIME 2000 II, Problem #10.)

My solution:
[hide]Let $O$ be the center of incircle $\omega$, and suppose $\overline{BC}$ and $\overline{DA}$ are  tangent to $\omega$ at $R$ and $S$ respectively. Let $\alpha = m\angle{OAP} = m\angle{OSP}$, $\beta = m\angle{OBP} = m\angle{ORP}$, $\gamma = m\angle{OCQ} = m\angle{ORQ}$, and $\delta = m\angle{ODQ} = m\angle{OSQ}$. Then we have

$\frac{\frac{r}{19} + \frac{r}{23}}{1 - \frac{r^2}{19\cdot 23}} = \tan{(\alpha + \delta)} = \tan{\angle{PSQ}} = -\tan{\angle{PRQ}} = -\tan{(\beta + \gamma)} = -\frac{\frac{r}{26} + \frac{r}{37}}{1 - \frac{r^2}{26\cdot 37}}$

which reduces considerably to

$\frac{19+23}{19\cdot23 - r^2} = -\frac{26+37}{26\cdot37 - r^2}$

which can then be solved for $r^2 = 647$ and $r = \sqrt{647}$. (647 is prime.) I don't find this solution particularly inspiring, but it is the best that I have seen to this problem.
[/hide][/quote]

Ahh... that's a pretty good solution. Should've seen it myself. Hmm.


topology
function
advanced fields
advanced fields unsolved

Let $E,F$ be two Banach spaces, and let $f: E\to F$ be a local homeomorphism such that for every $x\in E$ there is a neighborhood $U$ of $x$ such that $\|f(z)-f(t)\|\ge k\|z-t\|,\ \forall z,t\in U$ ($k>0$ is a fixed constant; such a function is also called a local $k$-extension, hence the title).

Prove that $f$ is, in fact, a homeomorphism.

Note that this solves quite a few problems on the forum (after some work, of course :)):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=injective&t=65847[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hadamard&t=69579[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=diffeomorphism&t=98564[/url]

It suffices to prove the following lemma:

if $H: [0;1] \times [0;1] \to F$ is a continuous function such that
$H(s,0)=y_{0}=f(x_{0})$ for every $s \in [0;1]$ then there exists a unique continuous function
$G: [0;1] \times [0;1] \to E$ such that
$H=f \circ G$.

From here the surjectivity is clear and the injectivity can be proved as follows:
suppose that $f(x_{1})=f(x_{2})=Y$ for $x_{1}\neq x_{2}$. Then one consider the function:
$H(s,t)=(1-s)f(t x_{1})+s f(t x_{2})$ and its unique continuous lifting $G(s,t)$.
Then $G(0,1)=x_{1}$ and $G(1,1)=x_{2}$ are in the rank of the continuous function
$s \to G(s,1)$. But $I=G([0;1],1)$ must be a connected subset of $F$, wich is false
because $x_{1}$ and $x_{2}$ are in different connected component of $I$. This is because $f$ is a local homeomorphism
so in a neighboorhood of $x_{1}$, $x_{1}$ is the only solution of $f(x)=Y$.

To prove the lemma on can begin by the easier following result:
if $H: [0;1] \to F$ is a continuous function such that
$H(0)=y_{0}=f(x_{0})$ then there exists a unique continuous function
$G: [0;1] \to E$ such that $H=f \circ G$ and $G(0)=x_{0}$

One can construct $G$ step by step. In a small neighborhood of $H(0)$ on can
define $G=f^{-1}\circ H$ because $f$ is a local homeomorphisme between a neighborhood
of $x_{0}$ and a neighborhood of $f(x_{0})=y_{0}$. Then if one calls $T$ the set of $t \in [0;1]$
such that there exists a continuous function $G$ verifying
$H=f \circ G$
for all $0 \leq x \leq t$, one can easily prove that $T$ is a clopen subset of $[0;1]$.
It is open because $f$ is a local homeomorphism and closed because of the additional assumption
on $f$ and the fact that $E,F$ are complete (the existence of $k$ such that etc ..).

One can prove the initial lemma by the same kind of arguments.


geometry
rectangle
pigeonhole principle
combinatorics proposed
combinatorics

$ 100$ points are enough, and the square may even be enlarged to $ 8.5 \times 8.5$.

[hide]Consider the regular hexagonal lattice of side $ 1$. One can snugly nestle a $ 8.5 \times 5\sqrt{3}$ rectangle (containing a $ 8.5 \times 8.5$ square), such that only $ 33$ hexagons contribute to partition it.
By pigeonhole principle, since $ 100 \equal{} 3\cdot 33 \plus{} 1$, there is (at least) one region of the partition containing (at least) four of the points. But each hexagon is inscribed in a circle of radius $ 1$.[/hide]

I have a strong suspicion this is not the best possible!  :(

What was the original problem?
I don't think it's a good idea to just delete posts...

The original problem posted by N.N.Trung was

[color=blue]
If $ 101$ points are taken arbitrarily inside a $ 8 \times 8$ square, prove there exists a radius $ 1$ disk covering (at leat) $ 4$ of these points.
[/color]


HOw do you do this without actually multiplying everything out:

$f(x)=(x-1)+(x-2)^2+(x-3)^3.....(x-9)^9+(x-10)^{10}$

What is the sum of all ten roots of $f(x)$

well by vieta, the sum $s$ would be $f(x)=x^{10}-sx^9 + ...$ since this function is obviously $f(x)=(x^{10}-10\cdot x^9\cdot 10^1+...)+(x^9-...)+...=x^{10}-99x^9+...$ the sum is $99$


1. Do there exist functions f:R->R and g:R->R , g being periodic, such that  x^3=f([x])+g(x) for all real x?

2. Find all finite sets M subset R containing at least two elements such that (2a/3-b^2) subset M for any two different elements a, b subset M.

3. Any of 6 gossips has her own news. from time to time one of them makes a telephone call to some other gossip and they discuss all the news they know. What the minimum number of the calls in necessary so as (for) all of them to know all the news.

4. Given the integers a nad b satisfy the equility
a^2-(b^2-4b+1)a-(b^4-2b^3)=0, prove that b^2+a is a square of an integer.

b) Do there exist an infinitely many of pairs (a,b) satisfying (*)?

5. Let S(n) be the sum of all different natural divisors of odd natural number n>1 (including  n and 1).

Prove that (S(n))^3<n^4.

6. Let ABCDEF be a convex hexagon such that AB+BC, CD+DE, EF+FA, prove that

(BC/BE)+(DE/DA)+(FA/FC)>=3/2.  When does equility occur?

7.  Let p be a prime number and let f(x) be a polynomial of degree d with integer coefficients such that:

i) f(0)=0, f(1)=1;
ii) for every positive integer n , the reminder of the dividsion of f(n) by p is either 0 or 1.
Prove that d>=p-1.

8. Given any positive real numbers a,b,c prove that

(a/b)+(b/c)+(c/a)>=(a+b/b+c)+(b+c/a+b)+1.

[quote="nurul_ashida"]6. Let ABCDEF be a convex hexagon such that AB+BC, CD+DE, EF+FA, prove that

(BC/BE)+(DE/DA)+(FA/FC)>=3/2.  When does equility occur

8. Given any positive real numbers a,b,c prove that

(a/b)+(b/c)+(c/a)>=(a+b/b+c)+(b+c/a+b)+1.[/quote]
Are you sure that you don't have any typos? In 6. it is obvious that you meant $AB=BC$ etc. but what about 8?

[quote="nurul_ashida"]8. Given any positive real numbers a,b,c prove that

(a/b)+(b/c)+(c/a)>=(a+b/b+c)+(b+c/a+b)+1.[/quote]
If it is $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$, see here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=107534[/url].


I just wanted to know if the invitations had been sent out yet, because I did not get one yet this year (yet?), so I wondered if any of you guys got yours...

Thanks.  :maybe:  :maybe:

I got mine today! I have 6 exams during that week, but hopefully I can persuade my school to excuse me from those exams...I really want to go this year as I was invited last year too but couldn't go =(

Oh, one question, how selective is this seminar? how competitive is it to get invited?

I'm not sure how selective it is... but I heard that waterloo uses some sort of random selection method in determining people to go...

By the way, bluesky, what were your scores on various contests (if you dont mind me asking)? -_-

it also depends on where you live and your gender and age.

for example, people with 136 cayleys from small cities get selected but i know a 99 euclid who didn't.

Got mine yesterday. Scores were:

COMC: 67
Fermat: 114
Euclid: 76
Hypatia: 38

I'm from a small town, so that might be why.

Here were my scores:

COMC: 66 (but made CMO)
Fermat: 130 (I think)
Euclid: 85
Hypatia: 40

But I am from Mississauga... so maybe my chances are very low...  :(

Fermat: 134
Open: high sixties
Euclid: 76
Hypatia: 39

I am from Toronto too

How many ppl are chosen to go each year? 60 for each grade or in total?

I got a invite too.

COMC: 79
Cayley: 150
Galois: 30
Euclid: 100

[quote="Camshaft"]So, to somebody who has gone before, how high level are the topics covered there?[/quote]

I would say the topics are early Olympiad level- ish, slightly less high and less strict than that which is found at the Winter camp. But you learn alot from your peers about Olympiad stuff, you meet alot of smart people, and you get motivated to work harder. So go, and enjoy.

Here is an in-depth analysis of invitations. All of my statements are based on experience of the past two years. It is possible, but unlikely, that the process has changed significantly.

The surest way to get an invitation to the Seminar is to get Honourable Mention or better on CMO, or be on the IMO team. No other Grade 12s will be invited. Fermat provincial winners may also automatically get an invitation. There is no other way to guarantee an invitation. For example, if you got a perfect score on the Cayley, or a Euclid gold medal, or a $ \$$25,000 scholarship from Waterloo, you might not be invited.

For anyone not in one of the above three categories, there are three main factors in getting an invitation, besides contest results:
i) Grade. The Seminar is largely a recruiting event for Waterloo. Hence the vast majority of invitees are in Grades 10 or 11, that is, those who are considering where to apply for University. This also explains why as few Grades 12s as possible are invited. (I personally would have expected the Waterloo organizers to invite some of their large scholarship winners in Grade 12, hoping that these students will impress the younger students. This did not happen last year. I don't have an explanation for it.) A few students in Grades 9 and below will be invited.

ii) Gender. About 40% of invitees are female. This proportion is much larger than, for instance, that found among the top 100 participants in any of the contests. Hence the contests results of female invitees is, on average, significantly lower than that of male invitees.

iii) Location. Again, Waterloo wants to invite prospective students. Students in Ontario are much more likely to come to Waterloo than those in other provinces. For example, if you are in BC, you will probably not get invited unless you are a Fermat winner. It is also possible that the organizers invite disproportionately more Ontarians with the foresight that many out-of-province invitees will decline the invitation, because of transportation costs.

A word on the difficulty of the mathematics covered at the seminar: you shouldn't have any problems unless you, say, got below 60 on the Euclid or above 20 on the CMO. Students are divided into 6 groups based on ability, so a wide range of material is covered.

A final word: The Seminar is a lot of fun. Enjoy the beautiful weather and the rare chance to do lots of geeky things.

Yeah carpo summarized pretty much all of the "invitation theory".
Unlike many other math camps, Seminar is more focused on having fun rather than hardcore training.
Every day there are usually two lectures and a competition, where students get to solve several problems in small groups, and then must write one of them up to hand in to be graded. On the day before departures there are no lectures, and there are several competition rounds. The difficulty of lecture material and competition problems ranges widely from moderate Fermat-difficulty to pre-Olympiad. That's about it for the math part.
There is a huge amount of free time (in comparison to other camps). There are a lot of really fun activities organized, including a movie night, karaoke, and a trip to an amusement park. Plus you will have quite a lot of time to do other activities.
So you are sure to have A LOT of fun, make lots of new friends, and learn tonnes of interesting mathematical and non-mathematical stuff at Seminar! :D

[quote="bluesky2525"]mathgeniuse^ln(x), you are going? (by the way, are you from UTS?)[/quote]

Yeah, but they invited me, so I thought, might as well go.

[quote="Kileel"]Counterexample!  I am in grade 12 and got an invitation to the seminar without qualifying for the IMO Team or making honorable mention or higher on the CMO.[/quote]

I think he said "guarantee an invitation."

[quote="Kileel"]Counterexample!  I am in grade 12 and got an invitation to the seminar without qualifying for the IMO Team or making honorable mention or higher on the CMO.[/quote]
That's good! I'm surprised. My statement was based on last year when Alan Guo and I were the only Grade 12s there. Have fun.

[quote="mathgeniuse^ln(x)"] I think he said "guarantee an invitation." [/quote]
Oh?
[quote="carpo"] No other Grade 12s will be invited.  [/quote]
That's right.

So what were your scores anyways Kileel?

COMC: 77
Euclid: 98

My being from New Brunswick may have had more to do with it though.

Two years ago, Will Ma got invited in grade 12 because he got 100 on Euclid.

I think seminar invitations are pretty random. At seminar, the girls don't appear to be worse at math than the guys - this must mean that the guys invited are not actually the top 30 grade 10s and 11s in Canada, but rather a random selection of the top 100 (or even more).

Also, I think the seminar committee consistently invites exactly one person from Massey, and between one and two people from Waterloo CI. These are usually the top Fermat scorers in these schools. There may be similar patterns in other schools.

[quote="bluesky2525"]Sorry if this is off topic, but how is everyone who's going dealing with exams for school? My school still makes me write them at Waterloo, as the school will fax them there. A question for the people who attended last year, will I have time to study there? Is there a lot of free time and how late can we stay up at night as I am a total night person? Thanks.[/quote]

There is quite a bit of free time. Classes end around 4pm, and the night activities are usually optional. Curfew is at 11pm, but your don probably won't mind if you stay up later, as long as you're in your own room. However, your roommate might want to go to bed. If you really make an effort, I'm sure you can find time to study. If you're discreet, the instructors probably won't notice you studying in class. It's better than sleeping. In fact, last year one student was spotted reading his economics textbook at the second banquet.

All that said, I strongly recommend not planning to study during the Seminar. You probably won't feel like it at all, and the heart of the seminar is spending time with other people. Studying would be a waste.

P.S. Moose room

Thank you carpo. I am going to do as much studying as possible before the seminar, but it's reassuring to know that there will be time to study if I do need to as I have quite a few to write. I agree, spending time with other students is really crucial.

Moose room? So are we staying at Renison? Cool.

Oh, another question, what's the usual style of dressing? i.e. casual, business casual, or formal? Thanks.

Moose Room!!! :ninja:

I think I hold the record for receiving the seminar invitation the latest -- only 5 days before seminar :!: (even though the invitation letter is dated May 15)

Only 2 days left till Seminar! :D

Could someone please post the senior problem sets (from the team competitions?)... I went last year (but not this year), and I would like to do them.

Thanks. :D


[quote="nktp"]More general problem:
Find $ a,b,c,k$ be natural integer satisfy:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{abc} \equal{} k$
Anwser is: $ k \equal{} 3,c \equal{} 1,...,$
Sorry if it have a mistake![/quote]
You are wrong.

May be I wrong but I sure that the solution of problem of delegat is $ k\equal{}3$

[quote="delegat"]Find all positive integers $ a$, $ b$ and $ k$ that satisfy equation:

$ \frac {a^{2} \plus{} b^{2} \plus{} 1}{ab} \equal{} k$[/quote]
The only answer is $ k\equal{}3$,you can find it in [url=http://www.mathlinks.ro/viewtopic.php?t=150373]PEN[/url]  :wink:

the answer to your question is $ k<\equal{}9$ except $ k\equal{}7$

[quote="BaBaK Ghalebi"]the answer to your question is $ k < \equal{} 9$ except $ k \equal{} 7$[/quote]
Why?Could you please write some examples of $ a,b$,such that $ k\leq 9,k\neq 3,7$?

[quote="nktp"][quote="BaBaK Ghalebi"]no I mean the problem with $ a,b,c$,[b]nktp[/b]'s problem...[/quote]
Do you prove your idea?[/quote]
 :oops:
sorry,I have misunderstood the problem.the problem that I had in mind was the following:
$ \frac {(a \plus{} b \plus{} c)^2}{abc} \equal{} k$
anyway,this one has been discussed here if you're interested:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=163252[/url]
maybe we can use the same idea to prove this one,hopefully... :maybe:


function
linear algebra
matrix
advanced fields
advanced fields unsolved

How to show that a topological group $ GL(n,\mathbb{R})$ has exactly two components: $ \{A:\operatorname{det}A>0\}$ and $ \{A:\operatorname{det}A<0\}$?

One remembers the elementary transformations and extends them to smooth path functions , e.g., a right multiplication for $ i\not\equal{} j$ given by $ \phi_{ij,t}: [0,1]\times GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})\,,\,\phi_{ij,t}(x,M)\equal{}M\left(I_{n}\plus{}xtE_{ij}\right)$, where $ t\in\mathbb{R}$ and $ E_{ij}$ is the matrix, which has only zeros except for the (i,j)-entry equal to 1. This connects $ M$ with $ M\left(I_{n}\plus{}tE_{ij}\right)$. Similarly, you define left multiplications. Applying them you find that each matrix can be connected to a diagonal matrix with the same determinant as the original matrix. Any diagonal matrix can obviously be connected to a diagonal matrix with $ \pm 1$-entries preserving the sign of each diagonal entry and hence of the determinant. Each potential pair of $ \minus{}1$-entries can be "resolved" using the elementary transformations above, locally we do for instance $ \left(\begin{array}{cc}\minus{}1 & 0\\0 &\minus{}1\end{array}\right)\rightarrow\left(\begin{array}{cc}\minus{}1 & 0\\2 &\minus{}1\end{array}\right)\rightarrow\left(\begin{array}{cc}1 &\minus{}1\\2 &\minus{}1\end{array}\right)\rightarrow\left(\begin{array}{cc}1 & 0\\2 & 1\end{array}\right)\rightarrow\left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right)$. In complete analogy we can "transpose" by connecting $ \left(\begin{array}{cc}1 & 0\\0 &\minus{}1\end{array}\right)$ with $ \left(\begin{array}{cc}\minus{}1 & 0\\0 & 1\end{array}\right)$, i.e. we can ultimately connect to either the unit matrix, if the determinant was positive or to the unit matrix with the last diagonal element replaced by $ \minus{}1$, if the determinant was negative. This shows that we have at most two components.

Since $ A\rightarrow\det A$ is a continous function the two stated components cannot be connected (any path would smoothly go through some matrix with determinant zero, which is not a valid path in $ GL(n,\mathbb{R})$).


I'm speaking primarily in reference to the MATHCOUNTS Forum, which has lots of young/new users.  But similar comments apply to other problem-solving forums.

When posting a new topic, there's a "Subject" field and a "Subject description" field.  When viewing the topic, I see a "Subject" and a "Source" field.

The inconsistency is a "bug".  More importantly, I think users would be more likely to list the source of their problems if they knew what field they should be using.

Yes. Plesae do change that.

Also, on the first page, could it be possible for the description in small to have in bold, SOURCE, behind it?


[hide=sol that is just so much easier to look at]$\dbinom{16}{5}=\frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}=\\ 2 \times 14 \times 13 \times 12 \text{ (using simplifying) }= \boxed{4368}$[/hide]

[tip=for people who want to yell at me for bumping][quote=AoPSsheriff]You may bump alcumus threads, as they are aattached to problems and are always accessible.[/quote][/tip]


geometry
geometric transformation
rotation
geometry solved

Does there exist a bijection f of

(a) a plane with itself

(b) three-dimensional space with itself

such that for any distinct points A,B line AB and line f(A)f(B) are perpendicular.

a) Yes, the 90deg rotation around a fixed point O.


geometry
3D geometry
sphere
tetrahedron
circumcircle
symmetry
angle bisector

Given a trihedral angle Sxyz. A plane (P) not through S cuts Sx,Sy,Sz respectively at A,B,C. On the plane (P), outside triangle ABC, construct triangles DAB,EBC,FCA which are confruent to the triangles SAB,SBC,SCA respectively. Let (T) be the sphere lying inside Sxyz, but not inside the tetrahedron SABC, toucheing the planes containing the faces of SABC. Prove that (T) touches the plane (P) at the circumcenter of triangle DEF.

I don't have a complete solution, but I believe that this is almost a complete proof.

Let $O$ be the center of $T$. Let $\alpha$ be the plane that contains $SAB$ and let $Q$ be the point where $T$ touches $\alpha$.

Suppose $R$ is the intersection of $AB$ and $SR$. Let $U$ be the point where $T$ touches $P$.

We know that $DR = SR$ because $ABD$ and $SAB$ are congruent. We also know that $RU =  RQ$ because they're both tangent lines to the sphere.

I [i]believe[/i] (but cannot prove) that $U,R,D$ are collinear, so that $UD = UR + RD = RQ + SR = SQ$.

Similarly, if we label $Q' and Q''$ as the points where $T$ touches the planes containing $SBC$ and $SCA$, then $UE = SQ'$ and $UF = SQ''$. But $SQ = SQ' = SQ''$ because they're all tangent lines, so $UD = UE = UF$ which means $U$ is the circumcenter of $DEF$.

Now, for the proof that $U,R,D$ are collinear, I don't have the proof myself, but from many2 times visualizing, I almost believe that the following is correct:
1. $SR \perp AB$
2. If $\beta$ is the plane containing $O, U, Q$ then $\alpha \perp \beta$ and $\beta \perp P$.
3. $S \in \beta$

If we can prove any of the three points above, then it's not hard to establish that $U,R,D$ are collinear.

Denote by $\delta,\alpha,\beta,\gamma$  the planes containing the the faces of the tetrahedron $SABC$ opposite to $S,A,B,C.$ Since $\triangle SBC \cong \triangle EBC,$ it follows by obvious specular symmetry that  the angle bisector (plane) of the dihedral angle $\alpha,\delta,$ which pass through the S-excenter $T$ of  $SABC,$ is also the perpendicular bisector of $\overline{ES}.$ Likewise, the angle bisector of the dihedral angle $\gamma,\delta$  is the perpendicular bisector of $\overline{DS}$ $\Longrightarrow$ $TE=TS=TD$ $\Longrightarrow$ $T$ lies on the perpendicular bisector $m$ of $\overline{DE}.$  Therefore, orthogonal  projection $P$ of $T$ on $\delta$ lies on the intersection  $m \cap \delta,$  i.e. the perpendicular bisector (line) of $DE$ in $\delta$ $\Longrightarrow$ $PE=PD.$ Similarly, $PE=PF$ $\Longrightarrow$ $P$ is the circumcenter of $\triangle DEF.$


ABC triangle. Prove that:
 \sum (a^2+b^2)/4mc  \geq p

One-line argumentation
a2+b2 \geq c \sqrt(2a2+2b2-c2) = c.4mc

Didn't you mean c*2mc instead of c*4mc? (I know it's not that important) :D

My solution:
 \sum (a 2 +b 2 )/mc  \geq  4p
 \sum2 (a 2 +b 2 )/ \sqrt (2(a 2 +b 2 )-c 2 ) \geq 4p
 \sum 2(a 2 +b 2 )/ \sqrt (a 2 +b 2 +2abcosC) \geq 4p
 \sum 2(a 2 +b 2 )/ \sqrt (a+b) 2  \geq 4p
 \sum (a+b) \geq 4p


algebra
polynomial
Vieta
algebra proposed

Prove that polynomial $ f\equal{}X^4\plus{}2X^3\plus{}3X^2\plus{}aX\plus{}b$ can't have all real roots, for any  $ a,b \in R$.

Suppose the contrary: that $ f$ has all real roots for some $ a,b$ and call these roots $ x_1,x_2,x_3,x_4$
According to Vieta, $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}\minus{}2$ and also $ x_1 x_2\plus{}x_1 x_3\plus{}x_1 x_4\plus{}x_2 x_3\plus{}x_2 x_4\plus{}x_3 x_4\equal{}3$
Then, $ (x_1\plus{}x_2\plus{}x_3\plus{}x_4)^2\equal{}(\minus{}2)^2\equal{}>$
$ x_1^2\plus{}x_2^2\plus{}x_3^2\plus{}x_4^2\plus{}2(x_1 x_2\plus{}x_1 x_3\plus{}x_1 x_4\plus{}x_2 x_3\plus{}x_2 x_4\plus{}x_3 x_4)\equal{}4\equal{}>$
$ x_1^2\plus{}x_2^2\plus{}x_3^2\plus{}x_4^2\plus{}2*3\equal{}4\equal{}>x_1^2\plus{}x_2^2\plus{}x_3^2\plus{}x_4^2\equal{}\minus{}2$ which is a contradiction, since we assumed that $ x_1,x_2,x_3,x_4$ were real.


[quote="trippy"]V = pie times r^2 times h solve for h how would i do that?[/quote]
$V=\pi r^{2}h$

$h=\frac{V}{\pi r^{2}}$

Is that what you want?

[quote="trippy"]V = pie times r^2 times h solve for h how would i do that?[/quote]

just do this:

$h=\frac{v}{\pi r^{2}}$


[hide]
The sum of the perpendiculars from a point inside or on the sides of a triangle is always the same and, if the point is on the triangle, equal to the altitude from that side(correct me if I'm wrong).

I know that made no sense, but it gives $4+5=9$ as the height, and so $81$ as the area.
I don't really have much else to go on.[/hide]

I checked out CircleSquared's conjecture in GeoGebra, and it doesn't appear to be true in general.  Unfortunately, I'm not sure how to solve this one either.

Any other thoughts?  Or is the previous answer correct, but does the theorem only works in certain cases?

Hmmm... I don't think that the previous answer is correct...
Here's the way I solved it:
[hide=" my solution"]
Since we see that we have created 2 right triangles - one 3-4-5 triangle, and one 5-12-13 triangle, we note that if we drop an altitude down from B, that hits AC at E, the 2 resulting right triangles will have the same ratios...

Now we see that if the ratio for the new 3-4-5 triangle (created by the altitude) is k, then side length AE will be equal to 3k, and if the ratio for the 5-12-13 triangle is n, then EC will be equal to 12n. Thus, we have $3k+12n = 18$. Also, the length BE will be 4k, but will also be equal to 5n (From the other side). Thus, we have: $4k=5n$ Therefore,$k=\frac{5n}{4}$ Substituting this into the first equation, we see that
$3(\frac{5n}{4})+12n = 18$
$15n+48 n = 72$
$63n = 72$
$n = \frac{8}{7}$
Subbing this in gives $\frac{40}{7}$ as the altitude, and thus, the area is $\frac{360}{7}$[/hide]
Hopefully that's right....

[quote="Nawahd Werdna"]Now we see that if the ratio for the new 3-4-5 triangle (created by the altitude) is k, then side length AE will be equal to 3k, and if the ratio for the 5-1-13 triangle is n, then EC will be equal to 12n.[/quote]sorry, but i don't understand what you mean by the "ratio" a triangle here...   :|
[quote="CircleSquared"]The sum of the perpendiculars from a point inside or on the sides of a triangle is always the same and, if the point is on the triangle, equal to the altitude from that side(correct me if I'm wrong).[/quote]Let $a=BC$, $b=AC$, $c=AB$, $x=PX$, $y=PY$, $z=PZ$.
[asy]size(200); pair A=(0,0), C=(18,0), B=(3,8), P=(5,2); draw(A--B--C--cycle); draw(P--foot(P,A,B)); draw(P--foot(P,B,C)); draw(P--foot(P,A,C)); draw(P--A); draw(P--B); draw(P--C); label("P",P,SE); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("X",foot(P,B,C),NE); label("Y",foot(P,A,C),S); label("Z",foot(P,A,B),NW);[/asy]
so the area of triangle ABC = $[ABC]=\frac{ax+by+cz}{2}$.
the only changing variables in this equation (for a given point $P$) are x, y and z. (the equation is also a plane in the xyz plane  :P )
so the value $x+y+z$ is not constant....
in fact, the value $x+y+z$ (such that x, y, z satisfy $[ABC]=\frac{ax+by+cz}{2}$) is constant if and only if $a=b=c$ (ABC is equilateral.)
let's say we substitute $x'=x+k$ and $y'=y-k$ for $x$ and $y$, so that the value $x'+y'+z=x+y+z$ and
\begin{eqnarray*}[ABC]&=&\frac{ax'+by'+cz}{2}\\ &=&\frac{a(x+k)+b(y-k)+cz}{2}\\ &=&\frac{ax+by+cz}{2}+\frac{ak-bk}{2}\\ &=&\frac{ax+by+cz}{2}\end{eqnarray*}
so $ak-bk=0$ and $a=b$. similarly $b=c$ so $a=b=c$.

[quote]
In triangle ABC, AC=18, and D is the point on line segment AC for which AD=5. Perpendiculars drawn from D to line segment AB and line segment CB have lengths of four and five respectively. What is the area of triangle ABC?
[/quote]
Here's a straightforward solution involving trigonometry:
[hide]
We can use the cosine law to determine angle CAB

4^2 = 5^2 + 3^2 -2(5)(3)cos CAB
CAB = 53.13 degrees

To find the altitude of the triangle that goes through B:

altitude = 18*sin 53.13 = 14.4

Now we use the cosine law again to get angle ABC

5^2 = 13^2 + 12^2 - 2(13)(12)cos ABC
ABC = 22.62 degrees

Finally, angle BCA = 180 - CAB - ABC = 104.25 degrees

The sine law will give us side AC:

18/sin 104.25 = AC/sin 22.62
AC = 7.14

The area of the triangle then is

1/2(base)(altitude) = 1/2(7.14)(14.4) = 51.4
[/hide]
(Next time I'll learn LATEX)

prett straightbforward cept the [i]caculator part[/i]

@minsoens: take the point at each vertex, then d(A,BC)=d(B, CA), etc, then heights equal=> sides equal [think area], nice asy though...i need to learn how to do that better...

original problem:
hmm...so we have two angles and the included side...there is a formula for this...like you can use triangle similarity, but it is essentially the same as the trig method...


inequalities
function
analytic geometry
graphing lines
slope
inequalities unsolved

Prove that the following inequality holds for positive real numbers $ x,y$ in $ [a,b]$
\[ \frac{x^2\plus{}y^2}{xy} \leq \frac{a^2\plus{}b^2}{ab}
\]

[quote="Johan Gunardi"]Prove that the following inequality holds for positive real numbers $ x,y$ in $ [a,b]$
\[ \frac {x^2 \plus{} y^2}{xy} \leq \frac {a^2 \plus{} b^2}{ab}
\]
[/quote]
It's wrong if $ a<0$ and $ b>0.$ :wink:
If $ a>0$ it's true because if $ f(x,y)\equal{}\frac {x^2 \plus{} y^2}{xy}$ then  $ \frac{\partial^2f}{\partial x^2}>0$ and $ \frac{\partial^2f}{\partial y^2}>0.$
Id est, $ \max f\equal{}\max\{f(a,a),f(a,b)\}\equal{}\frac {a^2 \plus{} b^2}{ab}.$

For $ x > 0,\ y > 0$, let $ t = \frac {y}{x}$, consider the graph of the function $ u = t + \frac {1}{t}\equiv f(t)$, for $ t > 0$, if $ 0 < a < b$, then

$ \frac {a}{b} < 1 < \frac {b}{a}$, so we have the maximum value is $ Max\left\{f\left(\frac {a}{b}\right),f\left(\frac {b}{a}\right)\right\} = \frac {a^2 + b^2}{ab}$.

Alternative Solution:

If $ 0 < a < b$, by $ 0 < a\leq x\leq b,\ 0 < a\leq y\leq b$, we have $ \frac {a}{b}\leq \frac {y}{x}\leq \frac {b}{a}$.

$ \therefore \left(\frac {a}{b} - \frac {y}{x}\right)\left(\frac {b}{a} - \frac {y}{x}\right)\leq 0$

$ \Longleftrightarrow (ax - by)(bx - ay)\leq 0$

$ \Longleftrightarrow ab(x^2 + y^2)\leq (a^2 + b^2)xy$

$ \Longleftrightarrow \frac {x^2 + y^2}{xy}\leq \frac {a^2 + b^2}{ab}$. Q.E.D.

Geometric Solution:

Let $ P\in\{(x,\ y)| a\leq x\leq b,\ a\leq y\leq b\}$,and let $ A(a,\ a),\ B(b,\ a),\ C(b,\ b),\ D(a,\ b)$,

${ \boxed{\frac {y}{x} = \frac {y - 0}{x - 0}}\ \cdots the\ slope\ of\ OP}$, yielding the slope of $ OB\leq \frac {y}{x}\leq$ the slope of $ OD$ for $ a > 0,\ b > 0$. Thus we have $ \frac {a}{b}\leq \frac {y}{x}\leq \frac {b}{a}$.

$ \Longleftrightarrow \left(\frac {y}{x} - \frac {a}{b}\right)\left(\frac {y}{x} - \frac {b}{a}\right) \leq 0$.


How many natural numbers are between the square root of 10 and the square root of 90.

I got 6, but the answer says 7???

I counted 4,5,6,7,8, and 9
am I doing something wrong? :?:

so x is between $ 3.1$ and $ 9.4$ so this is $ 4,5,6,7,8,9$. Yeah, it should be $ 6$. Maybe there's an error...

Yep, it's six and the answer key must be wrong.


superior algebra
superior algebra unsolved

Let $ G$ be a set with a closed, associative operation such that
[list]
[*]There exists an $ e \in G$ such that $ ae \equal{} a$ for all $ a \in G$.
[*]For all $ a \in G$, there exists a $ a^{\minus{}1} \in G$ such that $ aa^{\minus{}1} \equal{} e$.[/list]
Prove that $ G$ is a group.

This is from one of the first sections of the book Topics in Algebra by Herstein, and almost all of the problems around here are trivial. Yet I couldn't figure this one out after 45-60 minutes. A solution is nice, but a hint would be nicer.

We must show, that e is a left neutral element and that every $ a^{ \minus{} 1}$ is also left inverse to a:

So let $ a\in G$
$ a^{ \minus{} 1}a \equal{} (a^{ \minus{} 1}e)a \equal{} (a^{ \minus{} 1}(aa^{ \minus{} 1}))a \equal{} (a^{ \minus{} 1}a)(a^{ \minus{} 1}a)$
So if we set $ b: \equal{} a^{ \minus{} 1}a$ we have
$ b \equal{} bb\Rightarrow e \equal{} bb^{ \minus{} 1} \equal{} (bb)b^{ \minus{} 1} \equal{} b(bb^{ \minus{} 1}) \equal{} be \equal{} b$
So every $ a^{ \minus{} 1}$ is a left inverse to a.

To proof that e is left neutral is easy now:
$ ea \equal{} (aa^{ \minus{} 1})a \equal{} a(a^{ \minus{} 1}a) \equal{} ae \equal{} a$

EDIT:
Sorry, but to give a hint is not that easy here. Either you find the answer yourself by manipulating the given axioms again and again or you don't find it, but in this case the hint "just manipulate the axioms" is not that helpfull  :)

Thanks. :D I know what you mean by the edit. I did derive $ a^{\minus{}1}a \equal{} (a^{\minus{}1}a)^2$, but I didn't think of taking advantage of that substitution.


Let $ ABCD$ be a trapezoid with bases $ AD$ and $ BC$ and $ AB \equal{} CD$. An arbitrary circle tangent to SEGMENTS $ AB$ and $ AC$ intersects side $ BC$ at $ M$ and $ N$. Let $ X$ and $ Y$ be the intersections closer to $ D$ of the incircle of triangle $ BCD$ with segments $ DM$ and $ DN$, respectively. Show that line $ XY$ is parallel to line $ AD$.


floor function
inequalities
number theory unsolved
number theory

prove that the sequence [n*squrt(2)] have infinite numbers of the form 2^k
where k natural.

aviateurpilot, the general term is $ \lfloor n \sqrt{2} \rfloor$, not $ \lfloor n \sqrt{n} \rfloor$.

[quote="nsato"]aviateurpilot, the general term is $ \lfloor n \sqrt {2} \rfloor$, not $ \lfloor n \sqrt {n} \rfloor$.[/quote]
sorry :blush:
$ \exists n\in\mathbb{N}: \ \lfloor n \sqrt {2} \rfloor \equal{} 2^{k}\iff\ ([2^{k \minus{} \frac {1}{2}}] \plus{} 1)^2\le 2^{2k \minus{} 1} \plus{} 2^k$

why don't you elaborate? for the benefit of those who didn't have the honor of watching you work...

[hide]Suppose that only finitely many powers of 2 are attained by the sequence. Then all powers of 2 starting with $ 2^m$ (for some $ m$) are skipped.

If $ j\geq m$, then $ 2^j$ is skipped, so there exists a positive integer $ a$ such that
\[ a\sqrt {2} < 2^j,(a \plus{} 1)\sqrt {2} > 2^j \plus{} 1
\]
(the right inequality is strict since $ \sqrt {2}$ is irrational) implying
\[ \frac {2^j \plus{} 1 \minus{} \sqrt {2}}{\sqrt {2}} < a < \frac {2^j}{\sqrt {2}}
\]
Then we have
\[ \frac {2^{j \plus{} 1} \plus{} 2 \minus{} 2\sqrt {2}}{\sqrt {2}} < 2a < \frac {2^{j \plus{} 1}}{\sqrt {2}}
\]
But there also exists $ b$ such that
\[ \frac {2^{j \plus{} 1} \plus{} 1 \minus{} \sqrt {2}}{\sqrt {2}} < b < \frac {2^{j \plus{} 1}}{\sqrt {2}}
\]
We can show that $ 2a \minus{} 1 < b < 2a \plus{} 1$ so that $ b \equal{} 2a$.

As a result, if $ t$ is an integer such that $ t\sqrt {2} < 2^m,(t \plus{} 1)\sqrt {2} > 2^m \plus{} 1$, then we must have $ (2^r\cdot t \plus{} 1)\sqrt {2} > 2^{m \plus{} r} \plus{} 1$ for all nonnegative integers $ r$. This gives
\[ \frac {2^m}{\sqrt {2}} \minus{} \frac {\sqrt {2} \minus{} 1}{2^r\sqrt {2}} < t < \frac {2^m}{\sqrt {2}}
\]
for all nonnegative integers $ r$, a contradiction.[/hide]


trigonometry
geometry
cyclic quadrilateral
geometry unsolved

In triangle $ABC$, $\angle{B} = 70^{\circ}$ and $\angle{C} = 50^{\circ}$. A point $M$ is taken on the side $AB$ such that $\angle{MCB} = 40^{\circ}$ and a point $N$ is taken on the side $AC$ such that $\angle{NBC} = 50^{\circ}$. Find $\angle{NMC}$.

Does anyone have a synthetic solution? :D I'm desperate :)

Using trigonometry I' m sure that you have found the answer. Which is it  ,please ? This will help me a little , so I do not need to make an exact  figure with  geometer sketch - pad to calculate(measure) it or to make the same solution with trigonometry.

 Babis

[i]My dear Arne[/i], I think that you tried with trigonometry and you didn't succed. If you succeded, then you know $\angle NMC$ and now you may be prove synthetically ! [b]Answer:[/b] $m(\widehat {NMC})=30^{\circ}.$

[b]A trigonometrical solution.[/b] I note $m(\angle NMC)=x$. Thus, in the quadrilateral $BCNM$ we have:

$\sin x\sin 20\sin 40\sin 80=\sin (90-x)\sin 70\sin 50\sin 10\Longleftrightarrow\sin x\sin 20\sin 40\cos 10$ $=\cos x\cos 20\cos 40\sin 10\Longleftrightarrow$
$2\sin x\sin ^2 20\cos 10=\cos x\cos 40\sin 10\Longleftrightarrow(1-\cos 40)\cos 10\sin x=\cos 40\sin 10\cos x\Longleftrightarrow$
$\cos 10\sin x=\cos 40(\cos 10\sin x+\sin 10\cos x)\Longleftrightarrow\cos 10\sin x=$ $\cos 40\sin (x+10)\Longleftrightarrow$
$\sin (10+x)-\sin (10-x)=\sin (50+x)-\sin (30-x)\Longleftrightarrow\sin (10+x)-$ $[\sin (10-x)+\sin (50+x)]+\sin (30-x)=0\Longleftrightarrow$
$\sin (10+x)-\cos (20+x)+\sin (30-x)=0\Longleftrightarrow\sin (10+x)-$ $\sin (70-x)+\sin (30-x)\Longleftrightarrow$
$2\sin (x-30)\cos 40+\sin (30-x)=0\Longleftrightarrow(1-2\cos 40)\sin (30-x)=0\Longleftrightarrow$ $x=30\Longleftrightarrow m(\angle NMC)=30^{\circ}.$

[b]A shorter trigonometrical solution.[/b] We observe that the triangles $BCM,\ NBC$ are isosceles $(CB=CM=x,\ NB=NC=y)$.
I note $m(\widehat {NMC})=u$. From the Sinus's law in the triangles $CMN,CBN$ results:

$\frac xy=\frac{\sin (u+10)}{\sin u},\ \frac xy=\frac{\sin 80}{\sin 50}\Longrightarrow\sin u\cos 10=\cos 40\sin (u+10)\Longrightarrow$

$\sin (u+10)+\sin (u-10)=\sin (u+50)+\sin (u-30)\Longrightarrow\sin (u+10)+\sin (30-u)$ $=\sin (u+50)+\sin (10-u)\Longrightarrow$

$\sin (u+10)+\sin (30-u)=2\sin 30\cos (u+20)\Longrightarrow\sin (30-u)=$ $\sin (70-u)-\sin (u+10)\Longrightarrow$

$\sin (30-u)=2\sin (30-u)\cos 40\Longrightarrow (1-2\cos 40)\sin (30-u)=0\Longrightarrow u=30^{\circ}.$

[b]A syntetical solution.[/b] As soon as possible ! I went in a deadlock with syntetical solution. Now we have a occupation !

You are wrong levi :)
I did solve it with trig and I got $ 30^{\circ} $ but I can't find a synthetic proof for this.

[i]Dear Arne,[/i] the your proposed problem is equivalently with the following problem:

$ \triangle ABC,\ AB=AC,\ A=80^{\circ},\ P\in (ABC),\ m(\widehat {PAC})=m(\widehat {PCA})= $ $ 10^{\circ}\Longrightarrow m(\widehat {ABP})=30^{\circ}. $

[u]Remark.[/u] $ (ABC) $ - the interior of the triangle $ ABC $. This problem I solved syntetically. See $ ABC\equiv NBC\Longrightarrow AMNP $ cyclic quadrilateral !

Well , on segment  $ CM $ we take point $ P $ , such that angle $ CNP = 10 $. Then $ MBPN $ is inscribed. We 'll prove that angle $ NBP = 30 $.  We construct equilateral triangle $ BNE $. Then $ P $ is on the bisector of angle $ CNE $  of isosceles triangle $ NEC $ , so triangles $ PBN $ and $ PBE $ are equal and angle $ PBN $ = 1/2  angle $ NBE = 30 $ , and we are done.

  Babis

[i] (This solution was made by  Vagelis Yialos , Kostas Siotos and me in their office.)[/i]

Why $\triangle PBN\equiv \triangle PBE$ ?

Sorry , I was not in my house , when I posted the solution. Please , see the following:

a)	Take $E$ and  $N$  on opposite sides   wrt$BC$
b)	You will find that $\angle PNE = 10 = \angle RNC$  , so $P$ is on the bisector of $\angle ENC$ of isosceles triangle  $\triangle NEC$
c)	 We have $PN =PC = PE$
d)	  We have $BN = BE$, since $\triangle BNE$ has equal sides
e)	It will be $\triangle BPN = \triangle BPE$.

 I hope , I have helped you.

   Babis
 [i]( I thank Levi for giving me the answer of the problem)[/i]

Sorry, I drew wrong the figure. It is O.K. I am the other solution for the last problem, equivalence with the initial problem (see my previous reply).

Dear Ph - An , please forgive me!

  I thought that it was Levi the other solver , becouse I saw only the flag(!).  I own to you the answer 30  for  the angle(see post 2). Really sorry.

      Your trig. solutions are very good. In the past,  I had  solved many problems from this kind using synthetical way. In Romanian I think I have seen many problems of  this kind  in  << Patrascu >> and in Greek  in  << kontogiannis , olympiad geometry >>. Some of them need good techniques.

    Babis

Is it too late for an answer??
See that $BN\bot CM$, so $P$, the reflection of $N$ over $CM$ lies onto $BN$, and construct the equilateral $\triangle PCQ$, $Q,N$ on the same side of $CP$.
See easily $CN=CP=BN=CQ$ and $\triangle MCQ\cong\triangle BCN$, so $QM=QP,\angle MQP=\angle NCP=20^\circ$, $\triangle PNM$ is equilateral and $\angle CMN=30^\circ$.

Best regards,
sunken rock

So far we have $MP=NP$. What about $MN$ ?


linear algebra
matrix
inequalities proposed
inequalities
Poland

In each cell of a matrix $ n\times n$ a number from a set $ \{1,2,\ldots,n^2\}$ is written --- in the first row numbers $ 1,2,\ldots,n$, in the second $ n\plus{}1,n\plus{}2,\ldots,2n$ and so on. Exactly $ n$ of them have been chosen, no two from the same row or the same column. Let us denote by $ a_i$ a number chosen from row number $ i$. Show that:

\[ \frac{1^2}{a_1}\plus{}\frac{2^2}{a_2}\plus{}\ldots \plus{}\frac{n^2}{a_n}\geq \frac{n\plus{}2}{2}\minus{}\frac{1}{n^2\plus{}1}\]

$ a_i \equal{} (i \minus{} 1)\cdot n \plus{} \pi(i)$ where $ \pi$ is a permutation of the set $ \{1,2,...,n\}$, so sum $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n$ is constant, and its $ \frac {n(n \plus{} 1)}2 \plus{} n\frac {n(n \minus{} 1)}2$. Using Caushy gives \[ LHS \geq \frac {(1 \plus{} 2 \plus{} ... \plus{} n)^2}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n} \equal{} \frac {n(n \plus{} 1)^2}{2(n^2 \plus{} 1)} \equal{} \frac {n \plus{} 2}2 \minus{} \frac 1{n^2 \plus{} 1}\].

Note that condition that $ a_i$ is from $ i$th row wasn't neccessery at all...

Could you explain the $ a_i \equal{} (i\minus{}1)n\plus{}\pi(i)$ more? I don't understand permutations much.

a_i=(i-1)n+p(i)
It is better to call it because it is clear that the p (i) s are also different from each other and the idea of some permutation of (1,2, ..., n) comes from.
a_1+a_2+...+a_n does not change even when we select


function
algebra
polynomial
ratio
linear equation
algebra solved

hey
this one is proving hard.plz help
suppose f(x) and g(x) are polynomials with real coefficients such that from infinitely many rational numbers f(x)/g(x) is a rational number.prove that
f(x)/g(x) can be xpressed as a ratio of two polynomials with rational coefficients.
need help fast
cheers

i'm waiting  :D fot the soln of this one

Let a = largest exponent of x in f(x)
let b = largest exponent of x in g(x)
We can express f(x) as the sum of a+1 terms, where each term looks
like a real coefficient, multiplied by (x raised to a power)
If some exponents of x are not representd in f, we can simply insert
it with a zero coefficient.
Analogously, g(x) as b+1 terms
let n = a + b + 2
From our above defined infinite set, let x1, x2, x3, ... xn be n
rational values of x such that f(x)/g(x) is rational
Define function F:
F(c1,c2,c3,...c_a,x) = polynomial in x that looks just like f(x),
except all coefficients in f have been replaced with variables
c1,c2,c3....c_a
Analogously, define function G:
G(d1,d2,d3,...d_b,x) = polynomial in x that look like g(x)
Now, let us set up a system of n equations in n variables as follows:
For i = 1,2,...n define
Equation i: F(c1,c2,c3,...c_a,x_i)/G(d1,d2,d3,...d_b,x_i) = f(x_i)/g
(x_i)
Note that in each equation, x_i is a constant, not a variable.
Also, note that f(x_i)/g(x_i) is always rational
In general, when presented with n equations, in n unknowns, you three
possibilities:
1) no solution
2) exactly 1 solution
3) infinite solutions
You can cross out case 1 right off, because one solution to the above
is to substitute each variable (c1,c2,...c_a,d1,d2,...d_b) with the
coefficients found in f and g.
Furthermore, notice that in each equation, you can multiply left and
right side by G(d1,d2,d3,...d_b,x_i) and wind up with a linear
equation in each variable, making the whole system solvable by
Gaussian elimination. Also, note that each equation consists of only
rational coefficients. (This is so because each term looks like a
variable multiplied by x_i raised to a power. x_i raised to a power
is always rational.) Also, because of the fact we can use Gaussian
elimination on a system linear equations with only rational
coefficients, we can conclude that there is at least one solution
where each c_i and d_i is rational.
Consider one such solution. Define function p,q as follows:
p(x) looks just like f(x) except each coefficient in f has been
replace by the corresponding c_i in our above considered solution.
Analagously, q(x) looks just like g(x) except we substitue in the d_i
for coefficients.
I conjecture that algebraically, p(x)/q(x) = f(x)/g(x)
First, note that p(x)/q(x) = f(x)/g(x) whenever x is one of
x1,x2,x3,...x_n
That means p(x)/q(x) - f(x)/g(x) = 0 for n values of x
In otherwords:
[p(x)g(x) - f(x)q(x)]/[q(x)g(x)] = 0 for n values of x
Therefore the numerator must be 0. That is,
p(x)g(x) - f(x)q(x) = 0 for n values of x
Recall that the highest exponent in x found in these four polynomials
is as follow:
p: a
g: b
f: a
q: b
Therefore, p(x)g(x) - f(x)q(x) is a polynomial whose largest exponent
in x cannot be greater than a+b = n-2
But any polynomial of degree <= n with n or more zero evaluations
must be the zero polynomial. Therefore,
p(x)g(x) - f(x)q(x) = 0 for *all* real x, not just our x_i
Hence p(x)/q(x) - f(x)/g(x) = 0 for all real x
Therefore, f(x)/g(x) is algebraically equivalent to p(x)/q(x)
Since p(x) and q(x) have only rational coefficient, this problem is
proved.

cheers!!! :D

Yes , good work ! I think it's correct .


advanced fields
advanced fields unsolved

Let $ \alpha$ be a projective plane and $ c$ a closed polygon on $ \alpha$. Prove that $ \alpha$ will be decomposed into two regions by $ c$ if and only if there exists a straight line $ g$ in $ \alpha$ which has an even number of points in common with $ c$.

The statement is false.

Take the "triangle" in the projective plane formed for the $X$ axis from $(-\infty,0)$ to $(-1,0)$, the segment from $(-1,0)$ to $(0,1)$, the segment from $(0,1)$ to $(1,0)$ and the remaining of the $X$ axis from $(1,0)$ to $(+\infty,0)$.
It is obvious that the preceding triangle does not separate the projective plane, and the straight line passing for its two points $(0,1)$, $(2,0)$ has exactly this two points in common with the triangle.

Conclusion: we can reformulate the problem saying that the straight line has an even number of points in common with $c$ counting multiplicities (in which case we will count the point $(0,1)$ above with multiplicity two or zero) or else that the straight line $g$ does not pass for any of the vertices of the closed polygon.


Find all integers $ n \ge 3$ such that there are $ n$ points in space, with no three on a line and no four on a circle, such that all the circles pass through three points between them are congruent.

For $n=4$, we consider regular tetrahedron
If $n\geq 5$
[hide=Case 1]There are $4$ point lie on a plane
Since there are $2$ circle with fixed radius pass through $2$ fixed points, called $A,B$
If at least one of them contain other $2$ point, contradiction
So there are exactly $1$ point on each circle, let it be $C,D$, then we apply same argument with $2$ fixed points $B,C$ give a contradiction[/hide]
[hide=Case 2]No $4$ point lie on a plane
For any $3$ vertices, called $A,B,C$
There are other $2$ points, called $D,E$ in space, we apply [hide=Lemma]For any tetrahedron $ABCD$ such that circumradius of $\triangle{ABC},\triangle{ADC},\triangle{ABD},\triangle{DBC}$ is equal
Then $AB=CD,AD=BD,AD=BC$[/hide] to get that distance between $D,E$ to $A,B,C$ are fixed
So $D,E$ are unique determine from $A,B,C$
Then we apply same argument with $A,B,E$, where $C,D$ lie on the same side of plane pass that $3$ points
We get that $D=C$, contradiction[/hide]


[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)(a \plus{} b)(b \plus{} c)(c \plus{} a)\ge 4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$

Equality occurs for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or $ a \equal{} b \equal{} 0,c > 0$ or any cyclic permutation. :)[/quote]
Similar one of Vasc : $ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\ge 4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$

with equality occurs if and only if $ a\equal{}b,c\equal{}0$ or any cyclic permutation.  :)

[quote="nguoivn"]It's equivalent to:
$ \sum\ ab(a^2 \plus{} b^2 \plus{} c^2 \plus{} 3ab \minus{} ca \minus{} cb)(a \minus{} b)^2 \geq\ 0$
obviously trues because $ Sa,Sb,Sc \geq\ 0$  :wink:[/quote]

Yes, but as you see, the calculation is quite complex, so I would like to see a nicer solution for my problem. :)

[quote=dduclam]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)(a \plus{} b)(b \plus{} c)(c \plus{} a)\ge 4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$

Equality occurs for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or $ a \equal{} b \equal{} 0,c > 0$ or any cyclic permutation. :)[/quote]
$LHS-RHS=[\sum_{cyc}a(a-b)(a-c)](a+b)(b+c)(c+a) +(a-b)^2 (b-c)^2 (c-a)^2 \geq 0$
q.e.d

[quote=dduclam]
Similar one of Vasc : $ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\ge 4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$

with equality occurs if and only if $ a\equal{}b,c\equal{}0$ or any cyclic permutation.  :)[/quote]
$LHS-RHS = 4abc(a+b)(b+c)(c+a) +(a-b)^2 (b-c)^2 (c-a)^2 \geq 0$


logarithms
calculus
integration
derivative
algebra
polynomial
binomial theorem

From $ \sum_{n=1}^{\infty}x^{n-1}= \frac{1}{1-x}$ (geometrical series with $ \frac{a_{n+1}}{a_{n}}= x$) we have by integration

$ \sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\ln|1-x|$.

For x = 1/2 we get $ \sum_{n=1}^{\infty}\frac{1}{n \cdot 2^{n}}=-\ln(1-1/2) = \ln 2$.

Use this series

$ \ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$, $ |x|<1$.

when $ x=\frac{1}{2}$ you get your first series.

Now, play with this formula and after some calculations you get that

$ -\int_{0}^{1/2}\frac{\ln(1-x)}{x}dx=\sum_{k=1}^{\infty}\frac{1}{2^{k}k^{2}}$

and

if my calculations are correct

$ \ln \frac{1}{2}\int_{0}^{1/2}\frac{\ln (1-x)}{x}dx-\int_{0}^{1/2}\frac{\ln x \ln(1-x)}{x}dx=-\sum_{k=1}^{\infty}\frac{1}{2^{k}k^{3}}$.

As for the last three sums they are particular cases of the following series:

$ \sum_{k=1}^{\infty}\frac{k^{n}}{a^{k}}$, where  $ a>0$.

For calculating it use the following technique:

$ \frac{k^{n}}{a^{k}}=\frac{P_{n}(k-1)}{a^{k-1}}-\frac{P_{n}(k)}{a^{k}}$ where $ P_{n}$ is a polynomial of degree $ n$ which can be determined by calculations. The sum telescopes and you get that your sum equals $ P(0)$.

For example when $ n=1$ then $ P(x)=x+2$ and hence $ \sum_{k=1}^{\infty}\frac{k}{2^{k}}=P(0)=2$.

[quote="didilica"]and

if my calculations are correct

$ \ln \frac{1}{2}\int_{0}^{1/2}\frac{\ln (1-x)}{x}dx-\int_{0}^{1/2}\frac{\ln x \ln(1-x)}{x}dx=-\sum_{k=1}^{\infty}\frac{1}{2^{k}k^{3}}$.[/quote]
I also got this after interchanging the integration order.
However, I'm having troubles computing the second integral. Could you give a hint?

[quote="Perfect_radio"]However, I'm having troubles computing the second integral. Could you give a hint?[/quote]

$ -\ln|1-x| = \sum_{n=1}^{\infty}\frac{x^{n}}{n}$ for x < 1.

I think it can be calculated by integration by parts as follows

$ \int_{0}^{1}\frac{\ln(x)\ln(1-x)}{x}dx=\frac{1}{2}\ln^{3}\frac{1}{2}+\frac{1}{2}\int_{0}^{1/2}\frac{\ln^{2}(x)}{1-x}dx$

the last integral can be computed by expanding the denominator

$ \int_{0}^{1/2}\frac{\ln^{2}(x)}{1-x}dx=\sum_{k=0}^{\infty}\int_{0}^{1/2}\ln^{2}x x^{k}dx$
and of course that this integral reduces to expressions of the form $ 1/2^{k}k$ and $ 1/2^{k}(k+1)^{2}$ whose sums have been calculated before.

When we put all these calculations together in the integral formula of $ \sum_{k=1}^{\infty}\frac{1}{2^{k}k^{3}}$ we get the desired result.


I need help with this one:

How many zeroes are at the end of 1029!

Is there a simple way to do it?

Yes there is. Zeroes at the end are formed by a 5 and a 2, but there will be lots more 2s than 5s in the expansion. So you need to count the number of times that the powers of 5 go into 1029.
(And you can only go up to 625 since 5 :^5: is greater than 1029.

5 goes into 1029 205 times.
25 goes into 1029 41 times.
125 goes into 1029 8 times.
625 goes into 1029 1 time.

So there are 205+41+1+8=255 zeroes at the end.

The highest power of a prime number $p$ in $n!$ is given by

$\sum_{k=1}^\infty \left \lfloor \frac{n}{p^k} \right \rfloor$

Now you can find it easily.

The highest power of $5$ will obviously be lesser than the highest power of $2$.
$\therefore$ we need to consider only the power of $5$ in $1029!$

The answer is $\sum_{k=1}^\infty \left \lfloor \frac{1029}{5^k}\right \rfloor=205+41+8+1=255$

[quote="arkanm"]I was 7 when this thread was made.[/quote]

lol :rotfl:


Find the sum of all positive integers $a=2^{n}3^{m},$ where $n$ and $m$ are non-negative integers, for which $a^{6}$ is not a divisor of $6^{a}.$

[hide="really not sure"]
expanding, $a^{6}=2^{6n}3^{6m}$, and $6^{a}=2^{a}3^{a}$.
if $a^{6}$ is not a divisor of $6^{a}$, then either $6n>a$ or $6m>a$.
letting $m=0$, $a={2,4,8,16}$ satisfy $6n>a$ if $a=2^{n}$
letting $n=0$, $a={3,9}$ satisfy $6m>a$.
some experimentation leads me to believe that if both $m$ and $n$ are greater than 1, then $a^{6}$ is a divisor of $6^{a}$
so I think the answer is 2+4+8+16+3+9=42?

[/hide]

uhm, according to kalva and the aime archive,  this answer is incorrect.

can someone give me a correct answer with a COMPLETE solution?

thx

Hate to bump such an old topic, but that answer is definitely correct, according to Kalva (and me).[/hide]

In the solution provided above, why is it only that $ 6m > a,$ OR $ 6n > a,$ why can't they both be greater than $ a$ at the same time. That satisfies the conditions, doesn't it?

why is it that if m and n are greater than 1 then it is a divisor despite the fact that one of the exponents of a power of two or three is greater on lhs then rhs

[hide]$2^{6n} \cdot 3^{6n} \nmid 2^{(2^n \cdot 3^m)} \cdot 3^{(2^n \cdot 3^m)}$ means that $6n > 2^n \cdot 3^m$ or $6m > 2^n \cdot 3^m$ (or both).

For $6n > 2^n \cdot 3^m$: rearrange this to $\frac{6}{3^m} n > 2^n$. Now it is obvious that $m < 2$ and testing $m=0$ and $m=1$ we find that the only solutions for this case are $(m,n) = (0,1), (0,2), (0,3), (0,4)$.

Now, do the same with the other case and find that the only other solutions are $(m,n) = (1,0), (2,0)$ so the answer is $2+4+8+16+3+9 = \boxed{42}$.

Also, it is impossible for both to be true, because we have found all possible ways for at least one of the inequalities to be true, but none of those pairs $(m,n)$ satisfy both inequalities.[/hide]

@above: actually, if they are greater than one then the exponents on the lhs are less than those on the rhs...not greater.


limit
real analysis
real analysis unsolved

Let $ f(y,\ a_n) \equal{} (y \plus{} a_1)(y \plus{} a_2)(y \plus{} a_3)\cdots (y \plus{} a_n)$, we have
$ f(y,\ a_n) \equal{} y^n \plus{} \left(\sum_{i \equal{} 1}^n a_i\right)y^{n \minus{} 1} \plus{} \left(\sum_{i\neq j}a_ia_j\right)y^{n \minus{} 2} \plus{} \cdots a_1a_2\cdots a_n$.

$ \therefore f(1, nx) \equal{} (1 \plus{} x)(1 \plus{} 2x)\cdots (1 \plus{} nx) \minus{} 1$
$ \equal{} 1^n \plus{} (1 \plus{} 2 \plus{} \cdots \plus{} n)x\cdot 1^{n \minus{} 1} \plus{} \cdots \plus{} n!x^n \minus{} 1$, yielding the limit is $ \frac {n(n \plus{} 1)}{2}$.

with L'hopital rule it is really easy to get the result.

You are right, but Japanese high school students don't usually study l'Hopital therorem, which is often said by me, that's why I didn't dare to use the theorem.

Japanese high school students must be really stunted educationally

Thank you very much for your post and good ideas :P .

Another, elementary, method to solve this problem is the following one.

Let

 $ L_{n}\equal{}\lim\limits_{x\rightarrow 0}\frac {(1 \plus{} x)(1 \plus{} 2x)...(1 \plus{} nx) \minus{} 1}{x}$.

We have that

$ L_{n}\equal{}\lim\limits_{x\rightarrow 0}\frac {(1 \plus{} x)(1 \plus{} 2x)...(1 \plus{} (n\minus{}1)x) \minus{} 1}{x}$

$ \plus{}\lim\limits_{x\rightarrow 0}\frac{(1\plus{}x)(1\plus{}2x)\cdots(1\plus{}(n\minus{}1)x)nx}{x}.$

It follows that $ L_{n}\equal{}L_{n\minus{}1}\plus{}n$, and the problem is solved.

[quote="didilica"]Another, elementary, method to solve this problem is the following one.

Let

 $ L_{n} \equal{} \lim\limits_{x\rightarrow 0}\frac {(1 \plus{} x)(1 \plus{} 2x)...(1 \plus{} nx) \minus{} 1}{x}$.

We have that

$ L_{n} \equal{} \lim\limits_{x\rightarrow 0}\frac {(1 \plus{} x)(1 \plus{} 2x)...(1 \plus{} (n \minus{} 1)x) \minus{} 1}{x}$

$ \plus{} \lim\limits_{x\rightarrow 0}\frac {(1 \plus{} x)(1 \plus{} 2x)\cdots(1 \plus{} (n \minus{} 1)x)nx}{x}.$

It follows that $ L_{n} \equal{} L_{n \minus{} 1} \plus{} n$, and the problem is solved.[/quote]

Thank you!  :P


Prove that if a,b,y are the three angles of a triangle, then     (tan a/2)^2+(tan b/2)^2+(tan y/2)^2 is le 1.

I think you flipped a sign there.

It should be $\sum{\tan^{2}{\frac{\alpha}{2}}}\geq 1$.

The proof is easy with Jensen on convex function $y=\tan{x}^{2}$

yes, it's $\geq$, and it follows easily from
[hide]
\[\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1\]
[/hide]

My mistake, it was definetely greater than or equal to.


linear algebra
matrix
vector
function
inequalities proposed
inequalities

Let $ {x,y,z,p,q,r}\ge{0},x \plus{} y \plus{} z \equal{} 1,$prove that

\[ \sum_{cyc(x \minus{} > y \minus{} > z)}{\sqrt {px \plus{} qy \plus{} rz}}\geq{\sqrt {p} \plus{} \sqrt {q } \plus{} \sqrt {r}}.\]

We have the general propositional :

suppose $ {x_i,p_i}\ge{0},x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1,$then
\[ \sqrt {n(p_1 \plus{} p_2 \plus{} ... \plus{} p_n)}\geq{\sum_{cyc(x_1 \minus{} > x_2 \minus{} > ... \minus{} > x_n)} {\sqrt {p_1x_1 \plus{} p_2x_2 \plus{} ... \plus{} p_nx_n}}} \geq{\sqrt{p_1}\plus{}\sqrt{p_2}\plus{}...\sqrt{p_n}}.\]

[quote="fjwxcsl"]Let $ {x,y,z,p,q,r}\ge{0},x \plus{} y \plus{} z \equal{} 1,$prove that
\[ \sqrt{3(p\plus{}q\plus{}r)}\geq
{\sum_{cyc(x \minus{} > y \minus{} > z)}{\sqrt {px \plus{} qy \plus{} rz}}}\geq{\sqrt {p } \plus{} \sqrt {q } \plus{} \sqrt {r }}.\]
[/quote]

Nice, but easy with Karamata. See http://www.mathlinks.ro/Forum/viewtopic.php?p=122482. As the circulant matrix made up of the weights $ x_1,...,x_n$ is double stochastic, we see that  $ (p_1,...,p_n)$ majorizes the 'weighted' vector

$ \vec v: \equal{} (p_{1}x_{1} \plus{} p_{2}x_{2} \plus{} ... \plus{} p_{n}x_{n}, p_1x_2 \plus{} ...,\ ...)$.  As $ \sqrt{.}$ is a concave function, this yields the right ineq.

And trivially, $ \vec v\succ (\frac sn,...,\frac sn)$ where $ s: \equal{} p_1 \plus{} ... \plus{} p_n$ which yields the left ineq.
 :)


bueno alguien me puede decir cuanto sale el siguiente problema ?

determine el numero de maneras de permutar $n$ elementos sobre un circulo de modo que $k$ de estos son identicos  :maybe:

hola,$k$ tal aca va la solucion.
mira primero tomemos q todas son iguales, entonces el numero de formas de poner los$n$elementos, esto se puede hacer de $n!$ maneras si estubieran en una fila, pero como estan en una circunferencia abra repeticiones, las cuales seran$n, \Rightarrow $ el numero sera $(n-1)!$, ahora si existen $k$ iguales, kiere decir que al intercambiar estos elementos de lugar no interesa, osea que se repiten$k!$, por lo tanto el numero de formas sera:
                    $(n-1)!/k!$

que pasaria si los elementos repetidos son los n elementos , no se cumpliria la formula :D


calculus
integration
trigonometry
logarithms
calculus computations

$ \int \frac{arcsinx}{x^2} dx$
$ \int \frac{x^2dx}{(1\plus{}x^2)^2}$

[hide]Integrating by parts,

$ P \equal{} \int\frac{\arcsin x}{x^2}\,dx \equal{} \minus{}\frac{\arcsin x}{x} \plus{} \int\frac{1}{x \sqrt{1\minus{}x^2}}\,dx$.

On the last integral make substitution $ x \equal{} \sin t$, $ dx \equal{} \cos t\, dt$, and so

$ \int\frac{1}{x \sqrt{1\minus{}x^2}}\,dx \equal{} \int\csc t\,dt \equal{} \ln\left|\frac{1 \minus{} \cos t}{\sin t}\right| \equal{} \ln\left|\frac{1 \minus{} \sqrt{1\minus{}x^2}}{x}\right|$

and so

$ P \equal{} \minus{}\frac{\arcsin x}{x} \plus{} \ln\left|\frac{1 \minus{} \sqrt{1\minus{}x^2}}{x}\right| \plus{} C$.[/hide]

[hide]$ P' \equal{} \int\frac{x^2}{(1\plus{}x^2)^2}\,dx \equal{} \int\frac{1 \plus{} x^2 \minus{} 1}{(1\plus{}x^2)^2}\,dx \equal{} \int\frac{1}{1\plus{}x^2}\,dx \minus{} \int\frac{1}{(1\plus{}x^2)^2}\,dx$

$ \equal{} \arctan x \minus{} \int\frac{\sec^2t}{(1 \plus{} \tan^2t)^2}\,dt \equal{} \arctan x \minus{} \int\cos^2t\,dt$

and now you can finish by yourself.[/hide]


algebra
polynomial
number theory unsolved
number theory

Let $p$ be a prime number. How many triples $(x,y,z)$ are there, with $0\leq{x}\leq{y}\leq{z}\leq{p-1}$, and such that $(x+y+z)^{2}-xyz$ is divisible by $p$?

Kunterbunt, can you post your solution in detail?  :)

Sorry, I hoped that someone would give an easier proof after seeing the nice solution which is in fact
\[ \frac 1p \binom{p+2}3,  \]
which means that 1 in $p$ of the triples is a good one.

Claim:
The number of triples $0\le x,y,z \le p-1$ with $(x+y+z)^2\equiv xyz \mod p$ is $\frac{(p+2)(p+1)}{6}$ for any prime $p\ge 5$.

I did it the following way

1) There are $(p-1)^2-(p-1)+1$ triples $(x,y,z)$ with $1\le x,y,z \le p-1$ with $(x+y+z)^2\equiv xyz \mod p$.

[hide]
Take a primitive root $\zeta$ for the prime $p$ and look at the $p-1$ triples $(x\zeta^k,y\zeta^k, z \zeta ^k)$ for $k=0,\dots,p-2$. The equation becomes $\zeta^{2k} (x+y+z)^2 \equiv \zeta^{3k} xyz \mod p$ which is equivalent to $(x+y+z)^2(xyz)^{-1}\equiv \zeta^{k}\mod p$ . Obviously exactly one power of $\zeta$ will be equal to this expression unless $x+y+z=0$ when there are no solutions (There are $(p-1)^2-(p-1)$ possibilities for that to happen.)

Therefore we have exactly $\frac{(p-1)^3-(p-1)^2+(p-1)}{p-1}=(p-1)^2-(p-1)+1$ solutions. To get rid of the permutations of the $x,y,z$ we have to essentially divide by $3!=6$, but we have to take care of the cases where some variables are equal. [/hide]

2) There are $p-2$ solutions $(x,x,z)$ with $1\le x,z \le p-1$.

[hide]
We want to solve $(2x+z)^2\equiv x^2z \mod p$. Obviously $z$ has to be a square, so let $z=u^2$ for $u=1,\dots,p-1$. We get
$(2x+u^2)^2\equiv x^2 u^2 \mod p$. Therefore, $2x+u^2\equiv x u \mod p$ (since $-u$ turns up anyway as a possible value of $u$, we can take the plus sign).  So, we get $x \equiv\frac{u^2}{u-2} \mod p$. (For $u=2$, we simply get $4\equiv 0 \mod p$ which only holds for $p=2$, but then this value of $u$ is not allowed.)
[/hide]

3) There is 1 solution $(x,x,x)$ with $1\le x \le p-1$. ($p\not= 3$)

[hide]
We want to solve $9x^2\equiv x^3 \mod p$ which gives $x \equiv 9 \mod p$ (which explains why the formula doesn't hold for $p=3$.)
[/hide]

4) The number of solutions with $1\le x \le y \le z \le p-1$ equals  $\frac16 ((p-1)^2-(p-1)+1+3(p-2) +2)$.

[hide]
We want to divide the number of all triples by $3!=6$, but triples with two equal numbers only occur three times instead of six, so we have to add them three times. Now the triples with three equal numbers have first been counted once and then added three times and have to be added two more times. After that we can divide by 6.
[/hide]

4) There are $\frac{p+1}2$ solutions for $x=0$ and $0\le y\le z \le p-1$. ($p\not=2$)

[hide]
Clearly, we have to solve $y+z \equiv 0 \mod p$ which has $p$ solutions. Since we want $y$ smaller than $z$, we have to divide by two after adding the one solution of $2y\equiv 0 \mod p$. (Here, we acquire the condition $p\not=2$.)
[/hide]

5) In summary we have $\frac16 ((p-1)^2-(p-1)+1+3(p-2) +2)+\frac{p+1}2=\frac{ (p+1)(p+2)} 6=$ for $p\ge 5$.

6) Comment: It's hard to do this without mistakes, but as soon as you see that all relevant numbers are polynomials of degree $\le 2$ in $p$, you know that the result will be a polynomial of degree 2 and you can interpolate with the values for $p=5,7,11$.


But by Pham Kim Hung, not Pham King Hung :rotfl:

For the Lhs:
Since $ f(x)\equal{}\frac{1}{\sqrt{x}}$ is convex and $ a\plus{}b\plus{}c\equal{}1$,then
$ \sum\frac{a}{\sqrt{a\plus{}2b}}\geq\frac{1}{a\plus{}b\plus{}c}$
Could you post your solution rchlch??? :)

[quote="rchlch"]Prove that for all non-negative reals a,b,c with $ a\plus{}b\plus{}c \equal{} 1$,
$ 1\leq\frac{a}{\sqrt{a\plus{}2b}}\plus{}\frac{b}{\sqrt{b\plus{}2c}}\plus{}\frac{c}{\sqrt{c\plus{}2a}}\leq\sqrt{\frac{3}{2}}$

comment: the LHS is not hard, but the right side is rather hard.[/quote]

For the right side, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160719]here[/url]. :wink:


The cost of the daily school lunch increased from $ \$$1.50 to
$ \$$1.95. What was the percent increase?

1.50=100% of original price
Divide by 10, so each 15 cent increase is 10%
1.95-1.5=45 cents, so 45 cents=3*15=3*10% or[b]30%[/b]


A capacitor consists of two coaxial cylindrical plates of radii $a$ and $b$, $b > a$. The capacitor is placed so that its axis is horizontal and is filled with a dielectric fluid (relative permitivity $\varepsilon_{r}$) so that exactly half of its volume is full. Find the capacity of such a capacitor.

if my understanding of the question is right thenn exactly half the volume between the two cylinders is filled with the dielectric medium
let's call the cylinders $C_{a}$ and $C_{b}$ where the radius of each cylinder is same it's subscript in it's name.let's say (later we find both the cases are the same ) that the spae just  after $C_{b}$ is filled with the  dielectric medium

let my gaussian surface be  a cylinder of radius $r$,$r$ $<\frac{b+a}{2}$ and length $x$
now by gauss's law
$\oint_{C_{r}}\vec{E}.\vec{ds}=2%Error. "pirxE" is a bad command.
=\frac{Qx}{\epsilon_{0}l}$
where $Q$ is the charge and $l$ is the length of $C_{a,b}$
this gives $E=\frac{Q}{2\pi r \epsilon_{0}l}$(i assumed cylinders are long enough) now potential diffrence between  a point just near just near the boundary of  the medium and $C_{a}$ is
$\frac{Q\ln\frac{\frac{b+a}{2}}{a}}{2\pi\epsilon_{0}l}$
now consider a similar gaussian surface but this time within the medium
now the medim gets polarised say the polarised charges have magnitude $Q_{p}$
now doing  take a  cylidrical coaxial gaussian surface($C_{t}$) with radius $t$,$t$ $>$ $\frac{b+a}{2}$ and length $x$
by same symmetry arguemnts field  anywhere is outward so we have

$\oint_{C_{t}}\vec{E}\dot\vec{ds}=2\pi t x E=\frac{(Q-Q_{p})x}{\epsilon_{0}l}$ $=$ $\frac{Qx}{\epsilon_{0}\epsilon_{r}l}$
 here i assumed  $C_{a}$ is positively charged.

this gives $E=\frac{Q}{2\pit\epsilon_{0}\epsilon_{r}l}$
this gives potential difference between a pont on $C_{b}$ and a point just inside the dielectric medium as
$\frac{Q\ln\frac{b}{\frac{b+a}{2}}}{2\pi\epsilon_{0}\epsilon_{r}l}$

so potential difference between $C_{a}$ $C_{b}$ assuming $C_{a}$
is positively charged is

$\frac{Q\ln\frac{\frac{b+a}{2}}{a}}{2\pi\epsilon_{0}l}+\frac{Q\ln\frac{b}{\frac{b+a}{2}}}{2\pi\epsilon_{0}\epsilon_{r}l}$
so the capacitance $C$ $=$ $\frac{\ln\frac{\frac{b+a}{2}}{a}}{2\pi\epsilon_{0}l}+\frac{\ln\frac{b}{\frac{b+a}{2}}}{2\pi\epsilon_{0}\epsilon_{r}l}$

No, it's not like that, the surface of the dielectric is horizontal, it occupies the entire lower half of the horizontal capacitor.

sorry i misunderstood ur question about how the dielectric fills it probably a diagram or so could have made things easier... :(

i don't think that the heart of the solution will be changed in a significant way

Sorry for the inconvenience, here's a poorly sketched diagram of the cross section of the capacitor, the grey stuff is the dielectric. And I think that the heart of the solution has changed a bit, since you don't have the symmetry argument which you used to apply Gauss' law :)

here is what I got:
$C=\frac{\pi \varepsilon_{0}(\varepsilon_{r}+1)L}{ln\frac{b}{a}}$
Where $L$ is the length of the cyllinder.

Yes, that's correct :) Will you post your solution? Btw, when will you post the solution of the problem in the marathon?

Gauss's law:
$\varepsilon_{r}\varepsilon_{0}E r\pi L+\varepsilon_{0}E r\pi L=Q$
so the potential-differance between the two cyllinders:
$\Delta \Phi =\int_{a}^{b}Edr=\int_{a}^{b}\frac{Qdr}{\pi \varepsilon_{0}rL(\varepsilon_{r}+1)}=\frac{Q ln\frac{b}{a}}{\pi \varepsilon_{0}L(\varepsilon_{r}+1)}$
So the capacity:
$C=\frac{Q}{\Delta \Phi}=\frac{\pi \varepsilon_{0}L(\varepsilon_{r}+1)}{ln\frac{b}{a}}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

sir
  1.In every cyclic groups, right and left cosets are equal?
  2.Suppose that H is a subgroup of G such that whenever Ha not= Hb then aH not=bH. Prove that gHg' properly contained  in H for all g belongs to G

 My argument is ,the left and right cosets are same ,ie aH=Ha? is it correct?

[quote="barathisid"]  1.In every cyclic groups, right and left cosets are equal?[/quote]
Right and left cosets are equal if and only if the considered subgroup is normal. Since every subgroup of a cyclic group is normal, the statement is correct.
[quote]  2.Suppose that H is a subgroup of G such that whenever Ha not= Hb then aH not=bH. Prove that gHg' properly contained  in H for all g belongs to G
 My argument is ,the left and right cosets are same ,ie aH=Ha? is it correct?[/quote]Could you work that argument out? It is not really a rigorous proof ;) (btw, by g' do you mean $ g^{ \minus{} 1}$ (the inverse of $ g$)?)

For the second question, you may want to take a look [url=http://www.mathlinks.ro/viewtopic.php?t=237015]here[/url].


integration
geometry
vector
3D geometry
sphere
calculus
symmetry

If  $ \int\limits_S \vec{F} d\vec{A} \equal{} 12$ and S is a flat disk of area $ 4\pi$, then $ div\vec{F} \equal{} 3/\pi$. (FALSE)

 If  $ \vec{F}$ is a vector field in 3-space, then $ grad(div\vec{F}) \equal{} \vec{O})$(FALSE)

 If S is the unit sphere oriented outward and \vec{F} is a vector field satisfying  $ \int\limits_S \vec{F} d\vec{A} \equal{} 0$, then $ div\vec{F} \equal{} 0$at all points inside S. (TRUE)

my book just gives these answers but I don't know the explanation why it's either true/false, can someone please give me a short explanation why it is like that

for 1) since $ \int F\cdot dA =12$ and we are itnegrating over a disk of area $ 4\pi$ using Gauss's theorem we would get $ \int (\nabla \cdot F)\cdot dV = \int \frac{3}{\pi} dV = \frac{3}{\pi} \int dV  = 12$. Is there anything amiss after?

2) Use the definitions of the curl and divergence. $ \nabla \times (\nabla \cdot F)$. Do you know how to find the curl as the determinate? How does one find the divergence? ($ F = F_x\^i + F_y\^j + F_z\^k$ then the divergence of $ F$ is $ \nabla \cdot F = (F_x)_x+(F_y)_y+(F_z)_z$
Easier way to deal with this stuff is by using an intuitive approach. What is the divergence physically? What about the curl? Is it always 0? Etc..

3) Use Gauss's again, think about what the intregal of an integrand equal to 0 at all points equals. Ie if $ \vec{a}=0$, what is $ \int \vec{a}\cdot d\vec{X}$.
Since S is oriented so it points outwards (ie. the normal) then if the surface integral equals 0, use Gauss's to reason something about the orientation and $ \int (\nabla \cdot \vec{F}) d\vec{V} = 0$

Well, that's a useless attempt at these.

1: That integral tells you absolutely nothing about the divergence.
2: Just construct an example. $ x^2\cdot i$ is an easy one.
3: What makes you think that the divergence is constant? The book answer is wrong.

O sorry I wasn't trying to make the divergence be constant. Suppose I should have used a capital letter, the integrand is supposed to represent any sort of vector field. Constant or not.
For the $ x^2 \cdot i$, my old prof said that if its meant for 3-space its always considered over the reals...not just for the course.

oh yes, no. 3 is true.. I just mistyped it.. can someone tell me why?

You (FLT) misread what I intended by $ x^2\cdot i$; that $ i$ is one of the standard unit vectors. Of course, almost anything nonlinear works for that problem.
The point on #3: The divergence theorem tells you that the divergence integrates to zero. It doesn't have to be identically zero. If you make $ F$ symmetric around the center, its divergence is antisymmetric and you have an easy counterexample.

In jmerry's example of $ x^2\cdot i,$ that was really $ x^2\,\vec{i}.$ That is, $ \vec{i} \equal{} \langle 1,0,0\rangle,$ the standard unit vector, and not the square root of $ \minus{} 1.$

By the way, exactly that example, $ \vec{F} \equal{} x^2\,\vec{i},$ also works as a counterexample for (3). But I'll make is a tiny bit easier by making it $ \vec{F} \equal{} z^2\,\vec{k}$ instead - essentially the same example but oriented for greater familiarity.

It is true that $ \iint_S\vec{F}\cdot d\vec{A} \equal{} 0$ since the positive flux integral on the upper hemisphere exactly cancels the negative flux integral on the lower hemisphere.

But $ \text{div}\,\vec{F} \equal{} \nabla\cdot\vec{F} \equal{} 2z.$ It is true that the integral of $ 2z$ over the unit ball is zero (again, by symmetry), and that matches the outward flux integral of zero (confirming the divergence theorem). But note that $ 2z$ is certainly not the identically zero function.

So (3) is false.

Edit: or what jmerry said. Same thing.


$ n$ people decide to play a game. There are $ n\minus{}1$ ropes and each of its two ends are in hand of one of the players, in such a way that ropes and players form a tree. (Each person can hold more than rope end.)

At each step a player gives one of the rope ends he is holding to another player. The goal is to make a path of length $ n\minus{}1$ at the end.

But the game regulations change before game starts. Everybody has to give one of his rope ends only two one of his neighbors. Let $ a$ and $ b$ be minimum steps for reaching to goal in these two games. Prove that $ a\equal{}b$ if and only if by removing all players with one rope end (leaves of the tree) the remaining people are on a path. (the remaining graph is a path.)
[img]http://i37.tinypic.com/2l9h1tv.png[/img]

Does anyone have a solution for this problem?

Here's a clearer reformulation of the problem statement.

[hide = Problem]

On a tree, we define an [b] operation [/b] as the following three-step operation. First, select a vertex $v.$ Next, pick a neighbor $w$ of $v$ and remove edge $vw.$ Finally, pick a vertex $t \neq w$ which isn't a neighbor of $v$, and add the edge $vt.$

Define a [b] special operation [/b] as an operation above where $w, t$ are connected.

Define the [b] tastiness [/b] of a tree to be the minimum number of operations we can do to reduce it to a path, and the [b] uniqueness [/b] to be the minimum number of special operations we can do to reduce to a path. Show that a tree has equal tastiness and uniqueness if and only if the subgraph induced by its non-leaf vertices is a path.

[/hide]

The "if" direction is quite easy to check, the tastiness and the uniqueness are both equal to the number of leaves minus two.

Let's now deal with the "only if" direction. Suppose that we have a tree where the subgraph induced by the non-leaves is not a path.

Firstly, note that the tastiness of this graph is still the number of leaves minus two, because whenever there are more than two leaves we can remove an edge connecting to some vertex of degree $>2$ and attach it to a leaf. Let's now show that the uniqueness of this graph is strictly greater than the number of leaves minus two.

The first observation is that any special operation can decrease the number of leaves of the tree by at most one. Right off the bat, this implies that the uniqueness is at least the number of leaves minus two. Here, equality can only be achieved if the number of leaves decreases every time.

We know that there is a vertex $v$ with degree $\ge 3$ in the "defoliated" tree (leaves removed). Clearly, a special operation must result in the "defoliated degree" of $v$ decreasing (number of non-leaf neighbors), as this number is eventually $\le 2.$ This means that some edge was removed and attached to a non-leaf. The number of leaves does not decrease during this special operation, and so we know equality cannot be achieved.

From the previous discussion, the "only if" direction is proven, and so the problem is solved.

$\square$

[hide = Idea]Assume that we have $m$ leaves. We want $2$ leaves, which if we achieve then by handshake lemma we are done. Note that the number of leaves decreases by atmost $1$ in every turn, and infact if it is the first game we can make it decrease by $1$ every turn. However, if a vertex is connected to $3$ or more non-leaves then we cannot necessarily force this. [/hide]


replace every 1 from 1/a , 1/b and 1/c with sum(ab)=1

and you get that:

sum((ab+bc+ca)/a))=2sum(a)+sum(bc/a)>=3sum(a)
why because:

sum(bc/a)>=sum(a) by rearangement!

cheers! hope it's okay!

well what I thought was \sum 1/a = (ab+bc+ca) / abc = (ab+bc+ca)^2 / abc \geq 3abc(a+b+c) / abc = 3(a+b+c) by PM-AM.

You know ineq (a^2/x) +(b^2/y) \geq ((a+b)^2)/(x+y)
You can extend it for n numbers(the demonstration is easy)
So we have 1/a + 1/b + 1/c \geq 9/(a+b+c).
We have to prove that 9/(a+b+c) \geq 3(a+b+c)<=>3 \geq (a+b+c)^2.
Using Cebisev ineq we have 3(ab + bc + ca) \geq (a+b+c)^2 <=>
3 \geq (a+b+c)^2.That's all
I'm right?

isn't first one multiple choice? or the first one of the open questions rounds?

this is the first phase of the olympiad. unfortunately, this was the worse olympiad in years :(.

I can imagine if I look through the questions they've got :?

[quote="Valentin Vornicu"]I've proposed this to the 9th graders:
prove that if a,b,c are positive reals and ab+bc+ca=1 then
1/a+1/b+1/c \geq 3(a+b+c).[/quote]
prove that if $a,b,c$ are positive reals and $ab+bc+ca=1$,  then
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3(a+b+c).\]


Find all the possible values of $k$ porsitive integer,  such that we can find positive integers: $a, b$, such that $k^{a}+1$ can be expressed as $k^{b}+1$, inverting the order of its digits. :?:

oh! i think it not dificult!
first,you slove k<5(note that it have the same of the number   digits)
then,consider with k=1,2,3,4
then if k=3,we have a=3 and b=4

You didn't consider a=b...
This gives k as any power of 10 and a=b. Or, some a,k such that $k^{a}+1$ is a palindrome...


[quote="chien than"][quote="thesunflower"]$ n$ is a positive integer number. Consider $ (2n \plus{} 2)$ real numbers $ a,b,a_1,a_2,...,a_n,b_1,b_2,...,b_n$
such that $ a_i \neq 0, i \equal{} 1,2...,n$ and the function $ F(x) \equal{} \sum\limits_{i \equal{} 1}^n \sqrt {a_ix \plus{} b_i} \minus{} (ax \plus{} b)$ satisfies:
There exists distinct real numbers $ m,n$ such that $ F(m) \equal{} F(p) \equal{} 0$.
Prove that $ m$ and $ p$ are the only real solutions of $ F(x)$[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=170551
To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now[/quote]just wondering, what is M&Y? :huh:


calculus
integration
real analysis
derivative
function
Functional Analysis
real analysis solved

let g(x):R->R s.t :   g(x)=g(x+T),T >0,for each x 's real
and:  int\T_0\g(x)dx =0
also f(x):[a;b]->R s.t int\b_a\f(x)dx  is exist and finite.
Prove that :   int\b_a\f(x)g(nx)dx ->0 as n->inf

[quote="nalpaction"]Let $g(x) : \mathbb{R}\mapsto\mathbb{R}$ s.t $g(x)=g(x+T),T >0,$ for each $x$ real
and: $ \int^T_0 g(x)\,dx =0$
also $f(x) : [a;b]\mapsto\mathbb{R}$ s.t $\int^b_af(x)\,dx$  is exist and finite.
Prove that :  $\int^b_a f(x)g(nx)\,dx \to 0$ as $n\to\infty$[/quote]

This is essentially the Riemann-Lebesgue lemma. I also think we're short a hypothesis: in order to prove it, I had to assume that $g\in\L^{\infty}.$

First, suppose $f\in C^1[a,b].$ Define $G(x)=\int_0^x g(t)\,dt.$ Note that $G$ is bounded on $\mathbb{R}.$ We integrate by parts.

$\int_a^b f(x)g(nx)\,dx=\frac1n\,f(x)G(nx)|_a^b-\frac1n\int_a^b f'(x)G(nx)\,dx.$

Given the assumption that $f$ has a continuous derivative, this whole expression is $O(\frac1n)$ and goes to zero as $n\to\infty.$

This is where I needed the hypothesis that $g$ is bounded. Given $\epsilon>0$, find $f=f_1+f_2$ where $f_1\in C^1$ and $\int_a^b|f_2|<\epsilon.$ Then

$|\int_a^bf_2(x)g(nx)\,dx|\le ||f_2||_1||g||_{\infty}< \epsilon ||g||_{\infty}.$

Did I need to add a hypothesis? Yes. Consider this example:

$g(x)=\frac1{\sqrt{x}}$ for $0<x<\frac12$ and $g(x)=-\frac1{\sqrt{1-x}}$ for $\frac12<x<1.$

Let $f(x)=\frac1{\sqrt{x}}$ on $(0,1).$

Then for all $n$, $\int_0^1f(x)g(nx)\,dx$ diverges.

Or we can first prove that the conclusion holds when $f$ is a step function..then approximate $f$ by step functions...
We need the boundness of $g$ too.

Thanks for pointing out the typo, liyi; it's been fixed now.

The official definition of a separable Banach space is that it contains a countable dense subset. The practical, usable, consequence of the this is that a separable Banach space contains "nice" dense sub[i]spaces[/i]. Often, many different such nice subspaces. The difference between what I did in #3 and what you suggest in #5 lies in different choices for that nice dense subspace of $L^1$: smooth functions on the one hand, step functions on the other.

In either case, to extend the result from the subspace back to the whole Banach space, we need for $g$ to belong to the dual of $L^1$, namely $L^{\infty}.$ The example in #4 was aimed at showing that this is necessary.

i have a solution with the main idea is the same to liyi's.

i meant that f(x) and g(x) are Riemann-integralable so the condition that g(x) is bounded
is concluded  :lol:

Hmm.
So we use the method of Riemann integral.

Consider a partition of $[a,b]$ (The following notations are the classical ones)
Let $m_i$ denote $\inf f$ on each small intervals in the partition.

Since $\int_0^T g(x)dx = 0$, we know that $\int_a^b g(x)dx$ is bounded. Suppose $|g(x)|\leq M$ and $\left|\int_a^b g(x)dx\right|\leq M'$ for all $a,b$.
\begin{eqnarray*}
\left|\int_a^b f(x)g(nx)dx\right| &=& \left|\sum \int_{x_i}^{x_{i+1}} f(x)g(nx)dx\right|\\
& \leq & \left|\sum \int_{x_i}^{x_{i+1}} (f(x)-m_i)g(nx)dx\right| + \left|\sum \int_{x_i}^{x_{i+1}} m_ig(nx)dx\right|\\
&\leq& \sum \int_{x_i}^{x_{i+1}} |(f(x)-m_i)g(nx)|dx\right + \sum |m_i| \left|\int_{x_i}^{x_{i+1}} g(nx)dx\right|\\
&\leq& M  \sum \omega_i \delta_i + \frac{1}{n}M'\sum |m_i|
\end{eqnarray*}

For any given $\epsilon>0$, we can find such a partition such that $\sum \omega_i \delta_i < \frac{\epsilon}{2M}$. Thus the first item on the right side is smaller than $\frac{\epsilon}{2}$.

Once the partition is chosen,  $\sum |m_i|$ is fixed and we can find $n$ sufficiently large s.t. $\frac{1}{n}M'\sum |m_i| < \frac{\epsilon}{2}$.

Thus $\left|\int_a^b f(x)g(nx)dx\right| < \epsilon$ for $n$ sufficiently large.


Hey,The inequality is not homogeneous. :wink:

[quote="pvthuan"]Let $a,b,c,d$ be nonegative real numbers, prove that \[(a^{3}+b^{3}+c^{3}+d^{3}+4abcd)^{2}\leq(a^{2}+b^{2}+c^{2}+d^{2})^{3}.\] [/quote]
And it is wrong. Try $a=b=c=d=1.1.$ :wink:

[quote="arqady"]
And it is wrong. Try $a=b=c=d=1.1.$ :wink:[/quote]
Hey,It is surely wrong because the non-homogenous :)

[quote="zhaobin"]
Hey,It is surely wrong because the non-homogenous :)[/quote]
$(x-1)^{2}\geq0$ is non-homogenous but true. :)

Very sorry. I mean the following:

Let $a,b,c,d$ be nonegative numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=4,\] prove that \[a^{3}+b^{3}+c^{3}+d^{3}+4abcd\leq8.\] It's just an inspiration from the first inequality in the book by Vasc when I read it this early morning. I hope it's true. Forgive for my careless mistake.

The inequality is true. A stronger inequality is the following:

If $a,b,c,d$ are non-negative numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$, then
$a^{3}+b^{3}+c^{3}+d^{3}+abc+bcd+cda+dab \leq 8$.

This is just the last problem of the chapter 5 (problem 34 from page 222, solved at page 264).
It is unsolved in Mathlinks Forum since 3-4 months.
The proof in Algebraic Inequalities is based on EV-Theorem.

[quote="Lovasz"][quote="pvthuan"]You missed one important inequality case, Lovasz.[/quote]
Yes, $a=b=c=d=1$ and $a=b=c=0,\ d=1$.
[quote="zhaobin"][quote="Lovasz"]
Since $f(a,b,c,d)-f(0,0,0,2)=a^{3}+b^{3}+c^{3}+d^{3}-8 \le 0$, [/quote]
What did you do? :?:[/quote]
Do you think it's true?[/quote]
What??
It's obvious wrong.
$f(a,b,c,d)-f(0,0,0,2)=a^{3}+b^{3}+c^{3}+d^{3}-8$
will the equation be true?

[quote="Lovasz"][quote="zhaobin"]What??
It's obvious wrong.
$f(a,b,c,d)-f(0,0,0,2)=a^{3}+b^{3}+c^{3}+d^{3}-8$
will the equation be true?[/quote]
Why is it false, if there was the condition $a^{2}+b^{2}+c^{2}+d^{2}=4$?[/quote]
what is $f(a,b,c,d)$
and what is $f(0,0,0,2)$
I mean the equation.