geometry
ratio
circumcircle
trigonometry
inequalities
geometric inequality
IMO Shortlist

The vertices $D,E,F$ of an equilateral triangle lie on the sides $BC,CA,AB$ respectively of a triangle $ABC.$ If $a,b,c$ are the respective lengths of these sides, and $S$ the area of $ABC,$ prove that

\[ DE \geq \frac{2 \cdot \sqrt{2} \cdot S}{\sqrt{a^2 + b^2 + c^2 + 4 \cdot \sqrt{3} \cdot S}}. \]

does anyone know how to solve this problem??
because i've tried a lot and i couldn't solve it...

The pedal triangle of the 1st isodynamic point Q (the intersection of the 3 Apollonius circles inside the triangle $\triangle ABC$) is equilateral.  For example: Let D, E, F be the feet of normals from Q to BC, CA, AB. The A-Apollonius circle is a locus of points with the ratio of distances $\frac{AB}{AC}$ from the triangle vertices B, C and similarly for the B- and C-Apollonius circles. Hence $\frac{QB}{QC} = \frac{AB}{AC},$ $\frac{QC}{QA} = \frac{BC}{BA},$ $\frac{QA}{QB} = \frac{CA}{CB}.$ The quadrilateral AEQF is cyclic with the right angles at the vertices E, F, hence its circumradius is $R_A = \frac{AQ}{2},$ so that $EF = 2R_A \sin A = AQ \sin A$ and similarly $FD = BQ \sin B,$ $DE = CQ \sin C.$ Consequently,

$\frac{DE}{EF} = \frac{CQ \sin C}{AQ \sin A} = \frac{BC \sin C}{AB \sin A} = 1,\ \ DE = EF$

and similarly EF = FD. From the cyclic quadrilaterals AEQF, BFQD,

$\angle AQB = \angle AQF + \angle BQF = \angle AEF + \angle BDF =$

$= 180^\circ - \angle A - \angle AFE + 180^\circ - \angle B - \angle BFD =$

$= 180^\circ - (\angle A + \angle B) + 180^\circ - (\angle AFE + \angle BFD) = \angle C + 60^\circ$

and similarly, $\angle BQC = \angle A + 60^\circ,$ $\angle CQA = \angle B + 60^\circ.$ Let e = DE = EF = FD be the side length of the equilateral pedal triangle $\triangle DEF$. The area S of the triangle $\triangle ABC$ with circumradius R is

$S = \frac 1 2 [AQ \cdot BQ\ \sin(C + 60^\circ) + BQ \cdot CQ\ \sin(A + 60^\circ) + CQ \cdot AQ\ \sin(B + 60^\circ)] =$

$= \frac{e^2}{2} [\frac{\sin(C + 60^\circ)}{\sin A \sin B} + \frac{\sin(A + 60^\circ)}{\sin B \sin C} + \frac{\sin(B + 60^\circ)}{\sin C \sin A} ] =$

$= \frac{4R^3e^2}{abc} [\sin A \sin(A + 60^\circ) + \sin B \sin(B + 60^\circ) + \sin C \sin(C + 60^\circ)] =$

$= \frac{R^2e^2}{S} [\frac 1 2 (\sin^2 A + \sin^2 B + \sin^2 C) + \frac{\sqrt 3}{2}(\sin A \cos A + \sin B \cos B + \sin C \cos C)] =$

$= \frac{e^2}{8S} [a^2 + b^2 + c^2 + \frac{4 \sqrt 3 R^2}{2}\ (\sin 2A + \sin 2B + \sin 2C)] =$

$= \frac{e^2}{8S} (a^2 + b^2 + c^2 + 4S \sqrt 3)$

$e = \frac{2S \sqrt 2}{\sqrt{a^2 + b^2 + c^2 + 4S \sqrt 3}}$

Thus the expression on the right side of the inequality in question is precisely the side length of the equilateral pedal triangle $\triangle DEF$ of the 1st isodynamic point Q. Any other equilateral triangle $\triangle D'E'F'$ inscribed in the triangle $\triangle ABC,$ so that $D' \in BC,\ E' \in CA,\ F' \in AB$ is obviously obtained from the equilateral pedal triangle $\triangle DEF$ by a spiral similarity with the center Q and similarity coefficient greater than 1, hence its side e' = D'E' is greater than the side e = DE.

if $M$ is the fermat point of $ABC$ then the inequality reduces to :
$\sum |DE| \cdot |MC|\geq 2S$
this is kind of obvious because the product of the diagonals of a quadrilateral is $\geq$ twice its area...

Not true in general, namely if the 1st Fermat point M is outside the triangle $\triangle ABC,$ say, if $\angle C > 120^\circ.$ Then the inequality is reduced to

$|EF| \cdot |MA|+|FD| \cdot |MB|-|DE| \cdot |MC|\geq 2S$

and the "obvious" argument fails.

The 1st isodynamic point Q is then also outside the triangle $\triangle ABC$ (inside its circumcircle) and the angle $\angle AQB,$ the convex one, is equal to

$\angle AQB = \angle AQF+\angle BQF = 180^\circ-\angle AEF+180^\circ-\angle BDF =$

$\angle A+\angle AFE+\angle B+\angle BFD = 180^\circ-\angle C+120^\circ = 360^\circ-(\angle C+60^\circ)$

so that the concave $\angle AQB = \angle C+60^\circ,$ as before. The area S of the triangle $\triangle ABC$ is still

$S = \frac{1}{2}\left[AQ \cdot BQ \sin(C+60^\circ)+BQ \cdot CQ \sin(A+60^\circ)+CQ \cdot AQ \sin(B+60^\circ)\right] =$

$... =\frac{e^{2}}{8S}\left(a^{2}+b^{2}+c^{2}+4S \sqrt 3\right)$

etc.

It is known that among all the equilateral triangles circumscribed in $ \triangle ABC,$ the triangle $ \triangle XYZ$ homothetic to the outer Napoleon triangle of $\triangle ABC$ has the maximum area (side length). Hence, its side is twice the measure of the side $L$ of the outer Napoleon triangle.

$ L \equal{} \sqrt {\frac {_1}{^6}(a^2 \plus{} b^2 \plus{} c^2) \plus{} \frac {2\sqrt {3}}{3}[\triangle ABC]}.$

For every circumcribed equilateral triangle $ \Delta,$ we can associate an equilateral triangle $ \Delta'$  homothetic to $\Delta$ and inscribed in $ \triangle ABC.$ Thus, by Gergonne-Ann theorem, it follows that $ [\triangle ABC]^2 \equal{} [\Delta][\Delta'].$ The area (side L') of  $ \Delta'$ will be minimum if the area (side) of  $ \Delta$ is maximum. In other words, $ [\triangle ABC]^2 \equal{} \frac {_3}{^{16}}(L')^2(4L^2).$

Substituting the value of the side length $ L$ of the Napoleon triangle yields:

$ [\triangle ABC]^2 \equal{} \frac {_1}{^8}(L')^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}[\triangle ABC])$

$\Longrightarrow L' \equal{} \frac {2\sqrt {2}[\triangle ABC]}{\sqrt {a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}[\triangle ABC]}}.$

Here is my solution:
Denote by $ T$ the Toricelly point of triangle $ ABC$.Let $ DE\equal{}EF\equal{}FD\equal{}x$
Appying Ptolemy's theorem
$ S\equal{}[AEFT]\plus{}[CDTE]\plus{}[BDTF]$≤$ \frac{1}{2}(TA.EF\plus{}TC.DE\plus{}TB.FD)\equal{}\frac{x(TA\plus{}TB\plus{}TC)}{2}$
Hence $ x$≥$ \frac{2S}{TA\plus{}TB\plus{}TC}\equal{}\frac{2S \sqrt 2}{\sqrt{a^2\plus{}b^2\plus{}c^2\plus{}4S \sqrt3}}$

[quote="SUPERMAN2"]$ S \equal{} [AEFT] \plus{} [CDTE] \plus{} [BDTF] \ldots$[/quote]
This only works when the Fermat point lies inside ABC. See post #5

Its well known that orthic triangle has least perimeter of all inscribed triangles. Thus $x>\frac{1}{3}\Sigma a cos\alpha$. So if we show $\frac{1}{3}\Sigma a cos\alpha \ge \frac{2S \sqrt 2}{\sqrt{a^2 + b^2 + c^2 + 4S \sqrt 3}}$ Is this inequality true?

My rough solution is exactly same as done in IMO compendium. Assuming that all the angles of the triangle $ABC$ are less than $120^\circ$. Let the point $T$ be the Torricelli point of the triangle. So, $\angle(AT,EF)=\angle(BT,FD)=\angle(CT,DE)=\theta$. So,
$2S= 2(S_{AETF}+S_{BFTD}+S_{CDTE})$
=$(AT+BT+CT)DE\sin\theta\le(AT+BT+CT)DE$

By cosine rule, $AT^2+AT.BT+BT^2=c^2$,
 $BT^2+CT.BT+CT^2=a^2$,
$CT^2+CT.AT+AT^2=b^2$.
And $3(AT.BT+BT.CT+CT.AT) =4\sqrt{3}(S_{ATB}+S_{BTC}+S_{CTA}) =4\sqrt{3}S$

Adding all, we get $2(AT+BT+CT)^2=a^2+b^2+c^2+4\sqrt{3}S$.. This combined with the first inequality, gives the result.

Now let $\angle{C}\ge 120^{\circ}$. Take T to be the point lying on the same side of $AB$ as $C$ such that $\angle BTC=\angle CTA= 60^{\circ}$
Now we have $2S\ge (AT + BT -CT)DE$ and $2(AT+BT-CT)^2=a^2+b^2+c^2+4\sqrt{3}S$.


geometry
3D geometry
number theory proposed
number theory

So let me give this a new thread:

Suppose that $\sum_{\substack{p \text{ prime}\\ p^{2}\mid 2^{p-1}-1}}\frac{1}{\mathrm{ord}_{p}2}< \infty$, where $\mathrm{ord}_{p}2$ denotes the smallest positive integer $k$ with $p \mid 2^{k}-1$. Then show that there is a $c>0$ such that for all $N > 1$, the interval $[1,N]$ contains at least $cN$ integers $n$ such that $2^{n}-1$ is a squarefree number (that is, not divisible by the square of any prime).

Hint. See http://www.mathlinks.ro/Forum/viewtopic.php?t=125372.

(But don't even try to prove--or disprove--that the sum is convergent! It hasn't even been proven whether the sum contains infinitely many terms, or--on the other extreme--whether or not infinitely many primes do not appear in the sum... Naive heuristics suggest that it converges, though).

Needless to say, this can be done in exactly the same way I solved the problem for cube free. One uses your hypothesis and the obvious fact that $\sum_{p}{\frac{1}{p \cdot ord_{2}(p)}}$ converges (the sum is taken over all primes).

Exactly   :P .  By the way, Granville and Soundararajan have shown that under this same hypothesis, $\sum_{p^{2}\mid 2^{p}-2}\frac{1}{\mathrm{ord}_{p}2}< \infty$, one has that there exists an integer $k$ such that almost every odd positive integer is the sum of a squarefree number and $k$ powers of $2$. Moreover, if $\sum_{p^{2}\mid 2^{p}-2}\frac{1}{\mathrm{ord}_{p}2}< 5/8$, one could take $k=1$, i.e. almost all odd integers will be represented as a squarefree and a power of $2$. It is a conjecture of Erd\H{o}s that, in fact, ALL odd positive integers can be so represented (as we know well, we can't replace "squarefree" by "prime").

And by the way, the paper also shows other interesting results along these lines. For example: it may well be that there is no prime $p$ such that $p^{2}$ divides both $2^{p}-2$ and $3^{p}-3$ (no such primes are known, and naive heuristics make us hope/believe that there may well be none -- but again, we are helpless to even prove that there are infinitely many $p$ NOT having this property...) So: assuming this conjecture, another result is that almost every odd integer is of the form $q+2^{a}3^{b}$ with $q$ squarefree.

However, strangely enough, the obvious observation of this topic was missed by in the paper of Granville-Soundararajan (and it is evidently relevant). Here is the paper: http://www.dms.umontreal.ca/~andrew/PDF/wieferich.pdf

Oh, and by the way... The second known Wieferich prime is $3511$  :P .


so,best of luck to all of you giving the exam  :)

give your best and hopefully you all are selected for OCTS

Thanks and All the best to the rest :)

betw -u missed junior science :P
and as far as i know,it used to be OCSC :wink:

I did Junior Science horribly. I wont get through.
Bye guys. All the rest of u will go to Orientation cum Selection Camp this year. congrats.

is it just me or u also felt d chem paper too tough and horribly long.. i cudnt attempt 10 marks and messed up d orgn. problm of nitrogen cmpounds dat was linked. and too many other mistakes mean i wont get more than 50, i think.. any chance??? i think d cutoof wud b 60 arnd. wat do u say???

what will the cutoff be for INChO ? last year what was the cutoff ? any idea anybody . seniors help needed please.

Last yr paper was too easy so the cutoff was somewhere around 6 full questions out of 9 or something like that. :)  But i have no idea how it would change this year. :huh:

Yes last year's InCho was pretty simple.  :)  pf course i didnt even look at the last question which was based on mind-boggling biochemistry :P

If any1 remembers the answer to the follwing question please let me know.Plz.....
In a sample of Na2CO3 and NaHCO3 of mass 0.500g ..............................Find the percentage of Na2CO3 and NaHCO3 in the sample????
Do not remember the entire question.Plz tell me the answer.I'm worried.

Dont remember

i think it was around NAHCO3-- .117 g

and Na2CO3--.28g...

Any idea of cutoff in in INCho and INPho getting 50/80 in physics and 70-80/100 in chem i think ..

That looks quite enough in Chem (the paper was difficult as far as i kno)

But in Physics,we were discussing today--the cut off may very well crosss 70...and 20/35 can be of Kota :(

i asked my frnd after discussion they say that most of getting around 50-60 in Kota from A1 batch

though i didnt hv mre dan 50 sec 4 dis nahco3 ques, i didnt find d data of vol. of hcl used. so i did general calculations and came back... it must hv been my mistake..
pls someone, support me.. tuhin has already depressed me.. am i dat bad in chem... 50 z a bad score i know but z it dat bad.. wats d cutoff going to b lyk???
and wat r ur scores, tuhin_fire and debarun???
or are v dis tym also goin 2 do wat guys did last yr?? pulling each others leg and not telling our own score...

I am not pulling ne body's leg i am getting 70-80/100 but in physics a miserly 50/80.Neither r enuf qualify i s'pose

@einstein2007
Hw much are Pune guys getting in astro senior??

@ fbaggins  i donno abt cutoff last year but i got 65 .
moreover its easier to get selected if u  11th
this time i think 60+ has a gud chance . (aneway i dint attempt nsea this year hoping that i wud clear inpho)

@ aashay....

dunno about others marks....i've got 71.5 n one of ma frnd's got 63.5

[quote="preetham"]moreover its easier to get selected if u  11th [/quote]

He is in 9th as far as i kno :D

full marks to u ritu for i AM in IXth :D !!!! and ritu after seeing marks in junior and senior astro, what do u seriously think shud be cutoff in astro jr and sr??? plz reply immediately becoz tension cutting me in half!!!!

[quote="einstein2007"]i got 72.5 in chem....71.5 in astro...... :(
wat are ma chances??
post ur scores guys....

@nuclearalch....
m prakash n his genius math friends publish their own books for the INMO...
whoever gets thru RMO maharashtra gets to attend a weeklong camp at bhaskaracharya pratishthana in pune, which is a gr8 institute for maths.....esp olympiads...[/quote]
Would it be possible for me to lay my hands on any of those hehe? I guess not....You don't even get standard books like Challenge and Thrill of Pre-College Mathematics in Lucknow...  :huh:
Its honestly PATHETIC! You guys go to Bhaskara whatevers and here we didn't have any TC for INMO by the Govt. Only a private one conducted by my school. No wonder nobody from UP got selected...
[hide]Also Sagar said you know a few good maths books? Can you PM me the ones they teach you at M Prakash and the ones you study otherwise?
Thanks.[/hide]

does anybody know who's the highest or what's the highest score in INChO ? till now we've got einstein2007  with 72.5 .

thought i didn't write Astro, i know some of my friends' scores
76.5
75.5
67

@sid

out of this list which we prepared in the other thread,
[quote]
INChO 2009

vinit (einstein) 72.5
sagar vaze 68
avg 68
lordvoldemort 67
tuhin_fire 71.5
shubham 79
kartik 72
neeraj 72
shivam 74.5
kushagra 72.5
Two chennai boyz 74.5 and 72.5 (avg said)
celestine preetham 55.5
meghna 60
karan 61
sagar chordia 70.5
[/quote]

r u one of those two chennai boys whom avg told about

so max till now is 79 ?

any chances of me getting qualified ?

cutoff kya ho sakta hai ??
any idea ?

[quote="sid24_ss"]hi guys.. am from chennai. I got 74.5 in InChO
Some of the scores from chennai are
74.5 (me)
72.5
68.5
68
66
53
42
etc[/quote]
Hi sid. I am tuhin i got 71.5 do u know anything about andhra . and yes those are the two guys avg talked about. Lord voldemort if andhra doesnot produce out of ordinary results then you will be easily  thru and me 2 :D

@nuclear alchemist...

firstly bhaskar aint a govt inst....its private too....

u can order their books....www.bprim.org....

for maths i havent referred to olympiad stuff since two yrs now.....for iit i like arihant...
for olympiads there's 1.excursions in math-BP
2. Challngs n thrills.......
3. mathematical circles...
4. prob solving strategies-arthur engel
5. hall n knight

btw..u rlly luv chem???i thot such ppl were fictitious.. :rotfl:

Thanks for the site and books Vinit! Though I couldn't find exactly where to order books. And whats that monthly magazine kinda thing? Also there was a summer camp mentioned. Can I attend it? I tried opening that page but looks like it is being updated...
Once again thanks a lot!
I hope you were not referring to me when you said chem lovers?  :o I'll have go and have a bath again if someone laid such an allegation against me! X-(

einstein:-i'll take dat chem blame on me.....
esp, now dat i have got 69 :lol:
i know dats nthng gud, mayb nt even selection bt still... smthing z btr dan nthing.
 i was atually xpcting 56 :o .. my marks prove dat in incho checking was liberal, and step marking was done.
so guys, wat do u think, wat chances do i hv of getting in???????
cnsidering 13-14 guys in dis forum are ahead of me...
wats d scenario lyk??
wats d cutoff gonna b lyk???


integration
trigonometry
calculus
calculus computations

If $ f(x) \equal{} f(a\minus{}x)$, prove that

$ 2\int_{0}^{a} xf(x)\ dx \equal{} a\int_{0}^{a}f(x)\ dx$.

Evaluate

$ \int_{0}^{\frac{\pi}{2}}  \frac{x\sin x\cos x}{\sin^{4} x \plus{} \cos^{4} x}\ dx$.

Performing the substitution $ x \equal{} a \minus{} u$ we have

$ 2 \int_0^a x f(x) \,dx \equal{} 2 \int_0^a (a \minus{} u)f(a \minus{} u)\,du \equal{} 2a \int_0^a f(a \minus{} u)\,du \minus{} 2 \int_0^auf(u)\,du$

$ \equal{} > 4 \int_0^a x f(x) \,dx \equal{} 2a \int_0^a f(u)\,du \equal{} 2a \int_0^a f(x)\,dx$

$ \equal{} > 2 \int_0^a x f(x) \,dx \equal{} a \int_0^a f(x)\,dx$. $ \Diamond$

Now using this result with $ f(x) \equal{} \frac {\sin x \cos x}{\sin^4x \plus{} \cos^4x}$, and noting that $ f(x) \equal{} f \left(\frac {\pi}{2} \minus{} x \right)$, we have

$ 2 \int_0^{\pi/2} x \cdot \frac {\sin x \cos x}{\sin^4x \plus{} \cos^4x}\,dx \equal{} \frac {\pi}{2} \int_0^{\pi/2} \frac {\sin x \cos x}{\sin^4x \plus{} \cos^4x}\,dx$

$ \equal{} \frac {\pi}{4} \int_0^{\pi/2} \frac {2 \tan x \sec^2x}{1 \plus{} (\sec^2x)^2}\,dx$

$ \equal{} \frac {\pi}{4} [\arctan(\sec^2x)]_0^{\pi/2} \equal{} \frac {\pi^2}{16}$.

hello, we can simplify the integrand in to the following form
$ \frac{x\sin(x)\cos(x)}{\sin(x)^4\plus{}\cos(x)^4}\equal{}\frac{2x\sin(2x)}{3\plus{}\cos(4x)}$.
Sonnhard.


I need help with a noncalculator way to do this problem:

Let a, b, c, and d be nonnegative integers. How many solutions exist to

a + 5b + 10c + 25d = 63

[hide="Hint"]
I'm thinking you could do this by cases-obvioiusly d can only be 0, 1, 2 and then this reduces possibilities for c, like if d is 2 then c can only be zero or 1, etc.  That's my first guess just looking at it.[/hide]

Is the answer
[hide]9[/hide]?

If it is, I'll tell you my method.

What is a calculator method to do this problem, anyways?  :?

Random thought: Looks to me like "How many different ways can one make 63 cents using a combination of pennies, nickels, dimes, and quarters?" :)

Towersfreak: Yes, that is certainly one interpretation of the problem. However, this is a reduced version of the problem -- I just need to deal with this.

blahblahblah: I also got 73. The problem, in fact, asked me to basically do what you did via computer, though there was a frill or two. I'm trying to figure out if there is any non-casework way to do this rather terrible problem. BTW, what is are generating functions?

eryaman: The answer is 73, I'm pretty sure.

mkkool: I'm not sure how best to set up the casework...could you elaborate a bit more on your method? Describe it just a bit more? Though what I would really like is a method that does not involve any of that. Does such a method exist?

Thanks for the help guys!

[quote="sran"]
eryaman: The answer is 73, I'm pretty sure.[/quote]Oh, I see what I did wrong now.  I read "positive integers" instead of "nonnegative integers". :blush:

If you regard it as a number theory problem and find the general solution to it, them just stick the conditions on that they most be positive, you get all the solutions. This is the general method for these type of problems, but in this case its a long winded road!

Wait...confused...could you elaborate on this technique a little more?

Errr, its quite long winded, youll find it in any number thoerey book


inequalities
inequalities proposed
algebra


Let a triangle $ABC$  with $\angle{ABC}=45$ and $\angle{ACB}=15$  . Let a point  $D$ at the extension of $BA$ (at the side of A) such that $AD=2AB$ . Find the angle $\angle{BDC}$

See the attached diagram.

$DE\perp AC$

$\angle DAE=60^\circ\implies AE={AD\over 2}=c$

$\triangle ABE$ isosceles $\implies \angle ABE=\frac{180^\circ-120^\circ}{2}=30^\circ\implies\angle EBC=15^\circ\implies EC=EB$

$\angle DAE=60^\circ\implies\angle ADE=30^\circ=\angle ABE\implies ED=EB$

$EC=EB=ED\implies \angle BDC={1\over 2}\angle BEC={1\over 2}(180^\circ-2\cdot 15^\circ)=75^\circ$

(Alternatively: $ED=EC\implies\angle EDC=45^\circ\implies\angle BDC=30^\circ+45^\circ=75^\circ$)

This was problem 47 in the intermidiate geometry marathon. Pilpsie solve it the same was as farenhajt (except for the nice picture :D )

Haven't seen that beforehand. Though I can't think of a much different yet simple solution.

Farenhajt I didn't say that u saw the solution, I just wanted to give credit to Pilpsie. Here is an ugly trigonometrical solution[hide]Let $\angle ACD=x$ From $AD=2AB$ we obtain $\frac{AD}{AC}= \frac{2AB}{AC}$ from Sine theoreme we get $\frac{\sin x}{\sin(x+60}=\frac{2\sin15}{\sin45}$ or $\sin x\sin45=2\sin15\sin(x+60)$
$\frac12(\cos(x-45)-\cos(x+45))=\cos(x+45)-\cos(x+75)$
$\cos(x-45)+\cos(x+75)+\cos(x+75)=3\cos(x+45)$
$\cos(x+15)+\cos(x+75)=3\cos(x+45)$
$2\cos(x+45)\cos30=3\cos(x+45)$ and this shows that $\cos(x+45)=0$ so $x=45$ :wink: [/hide]


What is the remainder of 11 + 22 + 33+ ... + 1010 is divided by 5 ?

Can you explain the pattern? You start with numbers divisible by 11, but the last term $ 1010$ leaves a remainder of $ 9$ when divided by 11.

The pattern is from the sequence $ 1,2,3,.........,10$ in a sense 'double' each integer by writing it again to creat $ 11,22,33,........,1010$ so two one's, two two's,......, two ten's. I think that is the pattern.

The remainder when 11 is divided by 5 is 1
The remainder when 22 is divided by 5 is 2
The remainder when 33 is divided by 5 is 3
The remainder when 44 is divided by 5 is 4
The remainder when 55 is divided by 5 is 0
The same for 66~1010.
So adding the remainders give 2(1+2+3+4)=20
20/5=remainder of 0.


I don't know if this question really belongs here, but oh well.  I am reading a book called Algebra through Problem Solving and there is some symbolism i dont understand.   The bullet ($\bullet$, but actually the bullet was a bit smaller than this) is used quite frequently.  At first I though it was just a way of writing continuation (...) but then I came across statements that look like this:
 $S_6 \bullet F_0 \bullet F_1 \bullet F_2 \bullet F_3 \bullet F_4 \bullet F_5 \bullet F_6
$

      Where $$S_n$ is the the sum of the first n fibonacci numbers, and $F_n$ is the $n^{th}$ fibonacci number.  What does this symbolism mean?  I really have no idea.  The book that I am refering to can be accessed from the links page of this website.

Unless it is just a dot (ie, times), I don't know any standard meaning of that.. the book should define it if it is anything unusual. Do you have a concrete example of when it was used, in what context?

Unfortunately the book does not define the bullet.  It is obvious that it is not used for multiplacation.  The book is can be viewed at [url]http://www.thiel.edu/mathproject/atps/[/url].  The bullet is used in Chapter 2.

I browsed that chapter and saw no such symbol. The only symbols I saw in the expression you mentioned were equality and addition signs. Is it possible that your computer is messed up in some way?

Yeah, I guess the problem was just with my computer. I looked at the chapter on another computer and it uses equal signs and ellipses.Thanks.

Now that I try opening the Chapter 2 PDF file on my other computer I get a cannot find or create the font WPMathA error.  May not be able to desplay or print some characters error.  Does anyone know how to solve this problem?

[quote]I got the same problem too, and my computer didn't have the problem with the web page version. Do you also see bullets in place of ellipses in the PDF version?[/quote]

Yes, the bullets did sometimes replace ellipses as well as + and - signs.

Just do a web search of WPMathA and WPMathB (or whatever text you need) and download the font. Once you do that, install the files and the problem should be fixed.
That's what I did and, afterwards, had my computer had no problem reading any of their symbols since.


Hi, I'm trying to decide how I should arrange my sophomore and junior year schedules.  So I have a few potential options and I'd like some opinions as to which is easiest.  Clearly, I am a math and physics major.  Contemporary Physics Seminar is a 1-credit course that, from what I've been told, isn't too hard.

1) Sophomore: Special Relativity and Waves, Contemporary Physics Seminar, Methods of Complex Analysis, Abstract Algebra I
Junior: Quantum Mechanics I, Advanced Electricity and Magnetism, Honors Analysis I

2) Sophomore: Special Relativity and Waves, Contemporary Physics Seminar, Methods of Complex Analysis, Honors Analysis I
Junior: Quantum Mechanics I, Advanced Electricity and Magnetism, Abstract Algebra I

3) Sophomore: Special Relativity and Waves, Contemporary Physics Seminar, Honors Analysis I, Abstract Algebra I
Junior: Quantum Mechanics I, Advanced Electricity and Magnetism, some other math course(s)

Let's see. Which is the easiest way to commit suicide? No, there's no major difference among your options. Go consult your academic adviser.

Shouldn't this go http://www.artofproblemsolving.com/Forum/index.php?f=143

Luwei, I think you should find other ways to use your time, like helping other people.


conics
parabola
analytic geometry
graphing lines
slope
quadratics
Putnam

The parabola P has focus a distance m from the directrix. The chord AB is normal to P at A. What is the minimum length for AB?

[color=darkblue][b]Answer :[/b] $3\sqrt 3\cdot m\ .$
[hide="Here is the proof."][b]Proof.[/b] Denote $A\left(\frac{a^{2}}{2m},a\right)$ and $B\left(\frac{b^{2}}{2m},b\right)\ .$ Thus, $y^{2}=2mx\Longrightarrow 2yy'=2m\Longrightarrow y'=\frac{m}{y}\Longrightarrow y'|_{A}=\frac{m}{a}\ .$
The slope of the line $AB$ is $-\frac{a}{m}$ $\Longleftrightarrow$  $\frac{y_{B}-y_{A}}{x_{B}-x_{A}}=-\frac{a}{m}$ $\Longleftrightarrow$ $\frac{2m(b-a)}{b^{2}-a^{2}}+\fac am=0$ $\Longleftrightarrow$ $\boxed{\ 2m^{2}+a^{2}+ab=0\ }\ .$
Therefore, $a-b=\frac{2(m^{2}+a^{2})}{a}$, $a+b=-\frac{2m^{2}}{a}$ and $AB^{2}=(a-b)^{2}+\left(\frac{a^{2}-b^{2}}{2m}\right)^{2}\Longleftrightarrow$ $\boxed{\ AB^{2}=\frac{4(m^{2}+a^{2})^{3}}{a^{4}}\ }\ .$
Denote $a^{2}=t>0$, $f(t)=\frac{(t+m^{2})^{3}}{t^{2}}\ .$ Observe that $f'(t)\ .s.s.^{(*)}\ 3(t+m^{2})^{2}\cdot t^{2}-2t\cdot (t+m^{2})^{3}\ .s.s.\ 3t-2(t+m^{2})=t-2m^{2}\ .$
In conclusion, $t_{\mathrm{min}}=2m^{2}$, i.e. $a_{\mathrm{min}}=m\sqrt 2$ and $A(m,m\sqrt 2)$, $B(4m,-2m\sqrt 2)$, $AB^{2}=9m^{2}+18m^{2}=27m^{2}$ $\Longrightarrow$ $\boxed{\ AB_{\mathrm{min}}=3m\sqrt 3\ }\ .$

$.^{(*)}\ \ X\ .s.s.\ Y\Longleftrightarrow$ $XY>0$ or $X=Y=0\ .$[/color][/hide]

The distance from the focus to the directrix is m, so we may put $x=\frac{y^{2}}{2m}$ for the equation of the parabola. Let the point A have coordinates $\left( \frac{b^{2}}{2m},b\right)$ . Differentiate $x=\frac{y^{2}}{2m}$ implicitly to get $1=\frac{yy^{\prime}}{m}\Rightarrow y^{\prime}=\frac{m}{y}$ so that the slope of the tangent line to the parabola at A is $\frac{m}{b}$ and hence the equation of the normal line to the parabola at A (in point-slope form) is $y-b=-\frac{b}{m}\left( x-\frac{b^{2}}{2m}\right)$ which will also intersect the parabola at point B which has coordinates satisfying $x=\frac{y^{2}}{2m},$ like say $\left( \frac{B^{2}}{2m},B\right) ,$ so substitute these into the equation of the normal line to get

 $B-b=-\frac{b}{m}\left( \frac{B^{2}}{2m}-\frac{b^{2}}{2m}\right)$

which is quadratic in B and has the solutions $B=b,-\frac{b^{2}+2m^{2}}{b},$ the first of which is the point A, and the other is the y-coordinate of point B, so the coordinates of B are $\left( \frac{(b^{2}+2m^{2})^{2}}{2b^{2}m},B\right)$ so the square of the distance from A to B is

$d^{ 2}(A,B)= \left( \frac{b^{2}}{2m}-\frac{(b^{2}+2m^{2})^{2}}{2b^{2}m}\right)^{2}+\left( b+\frac{b^{2}+2m^{2}}{b}\right)^{2}= \frac{4(b^{2}+m^{2})^{3}}{b^{4}}$

and since the minimum of the square of the distance occurs at the same location as the minimum distance, we may deal with$d^{2}(A,B)$instead of $d(A,B)$; differentiating w.r.t. b gives

$\frac{d}{db}[d^{2}(A,B)]=\frac{24(b^{2}+m^{2})^{2}b^{5}-16b^{3}(b^{2}+m^{2})^{3}}{b^{8}}=\frac{8(b^{2}+m^{2})^{2}(b^{2}-2m^{2})}{b^{5}}=0$

solving for b gives $b=\pm i |m|, \pm \sqrt{2}|m|$. Obviously the real roots are the desired values of b, and we know they produce minimum values because the distance would be infinite if b=0 and there can be no other maximum. Plug-in $b=\pm \sqrt{2}|m|$ into the formula for $d^{2}(A,B)$ to get

$min\{ d^{2}(A,B)\}= \frac{4(2m^{2}+m^{2})^{3}}{4m^{4}}= 27m^{2}$

taking the root of both sides gives the desired value, namely

$min\{ d(A,B)\}= 3\sqrt{3}|m|$

   -Ben


Altough mentioned in some threads, there seems to be no thread about that game. If you don't know it: http://kayin.pyoko.org/iwbtg/downloads.php .
I want to share it with those that have too much spare time and enough effort  :wink:
It can sometimes be somewhat difficult, be prepared  :D
(Don't have much time to play recently, but got to the graveyard until now)

This game is awesome. I got to Mike Tyson and now I'm stuck. :(

[quote="1=2"]I have a small feeling that there is no Mac version. :([/quote]
[url=http://wiki.winehq.org/MacOSX]Get WINE?[/url]
EDIT: Apparently I can't spell WINE.  :blush:

Hmm...that's like WINE from linux for mac, right?  It never worked well when i used ubuntu.

I never had any problems with WINE (but on the other side, only used it about 5 times...).
You are missing one game with a very adequate difficulty (not unlike most new games that just "motivita" you by being easy).

[quote="ZetaX"](Don't have much time to play recently, but got to the graveyard until now)[/quote]
So you went the way where the first boss is Mike Tyson?
I just passed the graveyard and now I'm on the boss after it...


Cramer's Rule seems rather elongated with some problems, and it seems to be an easy solution to others. However, when dealing with a four variable system, the math can get complicated. So, has anybody ever encountered a situation (i.e. a math conpetition) where they had to use it?

I think it's safe to say that in a math competition, if you have to solve a system of more than three equations, you should be solving the problem in a different way.

Cramer's Rule isn't really a Getting Started subject, so I'll promote this thread to Intermediate.

Cramer's rule is a method for solving systems of equations w/matrices (determinants).  However, it is imo sortof outdated.  If you have a calculator, the easiest way to solve systems w/matrices is using this method:

3x+2y=-9
x - 6y = 2

[ x ]    [  3  2  ]^-1 [ -9 ]
[ y ] = [  1 -6 ]       [ 2  ]

and evaluate this on your calculator.

Notice I put the coefficents in one matix and the constants in another matrix.

Yes.

Personally, I hate Cramer's Rule. I've used it once and never used it again :D.

I'm pretty sure we discussed it briefly in algebra 2 last year, and there's never been a time outside of math class that I can recall using it.

[quote="NAS"]Sorry ???Cramer's Rule??? could someone enlighten me?[/quote]

Cramer's Rule is mostly used for 2 variable systems:

ax+by=c
dx+ey=f
[tex]D[/tex]= determinant of [[a,b][d,e]]=ae-bd
[tex]D_x[/tex]=determinant of [[c,b][f,e]]=ce-bf
[tex]D_y[/tex]= determinant of [[a,c][d,f]]=af-cd

[tex]
\text{}\\
x=\frac{D_x}{D}=\frac{ce-bf}{ae-bd}\\
y=\frac{D_y}{D}=\frac{af-cd}{ae-bd}[/tex]

For any given problem, there is usually a better method than Cramer's rule.  OTOH, I claim that Cramer's rule is a fine method for solving small systems of equations.

In ancient times BC (before calculators), I used Cramer's rule reasonably often.  It is quite likely I used it on the AHSME, etc. in the early 1970's.

I have also coded Cramer's Rule in software.  The regularity and recursive nature of evaluating a determinant makes for very simple code.

Students who are especially adept at mental computations can often perform a 3-variable Cramer's rule computation in their heads.

Finally, some special matrices lend themselves to manual solution by Cramer's rule.  [Sparse matrices or matrices with just -1, 0, and +1, for example.]

Don't use Cramer's rule for everything, but keep it in your bag of tricks.  If you have an open mind, you might be surprised how often Cramer's rule provides a quick, simple solution.

Thanks I think I remember learning somthing about this last year (Alg II) but I don't think that is what my math teacher called it.

Why does anyone need Cramer's rule when you have a TI-89? :)  This question was not meant to be answered.

[quote="Ragingg"]Why does anyone need Cramer's rule when you have a TI-89? :)  [/quote]

Hear, hear.  The "solve" function has saved so much graphite ;)


let AM ,AH,AD are median ,height and bisector of ABC triangle.consider AK is symmetry of AM toward AD.prove that :

\[ \frac{MH}{AM} . \frac{sin BAK}{sin KAM}\equal{}\frac{AB}{BC}\]

[quote="stvs_f"]let AM ,AH,AD are median ,height and bisector of ABC triangle.consider AK is symmetry of AM toward AD.prove that :
\[ \frac {MH}{AM} . \frac {sin BAK}{sin KAM} \equal{} \frac {AB}{BC}
\]
[/quote]

You already posted this problem !  :wink:
http://www.mathlinks.ro/viewtopic.php?p=1410138#1410138


limit logarithms calculus integration inequalities

$ 1\blacktriangleright$ For any $ n\in N^{*}\ ,\ a_{n}>0$ and $ \lim_{n\to\infty }\frac{a_{n}}{n}=\infty$. Prove that $ \lim_{n\to\infty }\sum_{k=1}^{n}\frac{1}{n+a_{k}}=0$.
$ 2\blacktriangleright$ Is well-known that for the sequence $ a_{n}=\frac{\sqrt [n]{n!}}{n}$, $ \lim_{n\to\infty}a_{n}=\frac{1}{e}$. Prove that $ \lim_{n\to\infty}\frac{n}{\ln n}\cdot\left(a_{n}-\frac{1}{e}\right)=\frac{1}{2e}$.

[quote="Virgil Nicula"]$ 1\blacktriangleright$ For any $ n\in N^{*}\ ,\ a_{n}>0$ and $ \lim_{n\to\infty }\frac{a_{n}}{n}=\infty$. Prove that $ \lim_{n\to\infty }\sum_{k=1}^{n}\frac{1}{n+a_{k}}=0$.[/quote]
[hide="Solution"]
$ \lim_{n\to\infty}\frac{{\sum_{k=1}^{n}\frac{1}{1+\frac{a_{k}}{n}}}}{n}=$ by Cesaro-Stolza lemma:
$ \lim_{n\to\infty}\frac{1}{{1+\frac{a_{n}}{n}}}=\left[\frac{1}{1+\infty}\right]=0$[/hide]

[quote="Virgil Nicula"][quote]$ \lim_{n\to\infty}\frac{{\sum_{k=1}^{n}\frac{1}{1+\frac{a_{k}}{n}}}}{n}=$ by Cesaro-Stolza lemma:
$ \lim_{n\to\infty}\frac{1}{{1+\frac{a_{n}}{n}}}=\left[\frac{1}{1+\infty}\right]=0$[/quote]
Is no correctly how you apply Cesaro-Stolz's lemma ![/quote]
Ofc you're right... I'll think about 1st problem later, but now I have solution to the 2nd problem.
[quote="Virgil Nicula"]Is well-known that for the sequence $ a_{n}=\frac{\sqrt[n]{n!}}{n}$, $ \lim_{n\to\infty}a_{n}=\frac{1}{e}$. Prove that $ \lim_{n\to\infty}\frac{n}{\ln n}\cdot\left(a_{n}-\frac{1}{e}\right)=\frac{1}{2e}$.[/quote]
[hide="Solution"]
$ \lim_{n}\frac{n}{\ln n}\left(a_{n}-\frac{1}{e}\right)=\lim_{n}\frac{n}{e\ln n}\left(ea_{n}-1\right)=\frac{1}{e}\lim_{n}\frac{n}{\ln n}\cdot\frac{e^{\ln ea_{n}}-1}{\ln ea_{n}}\cdot \ln ea_{n}=$
becouse $ \lim_{x\to 0}\frac{e^{x}-1}{x}=1$, we'll compute now:

$ \lim_{n\to\infty}\frac{n\ln e\cdot\frac{\sqrt[n]{n!}}{n}}{\ln n}=\lim_{n\to\infty}\frac{\ln\frac{e^{n}}{n^{n}}n!}{\ln n}=$ by Stirling:

$ \lim_{n\to\infty}\frac{\ln\sqrt{2\pi n}e^{\lambda_{n}}}{\ln n}$ for $ \frac{1}{12n+1}<\lambda_{n}<\frac{1}{12n}$, so:

$ \lim_{n\to\infty}\left(\frac{1}{2}+\frac{\ln\sqrt{2\pi}e^{\lambda_{n}}}{\ln n}\right)=\frac{1}{2}$, so:

$ \lim_{n\to\infty}\frac{n}{\ln n}\left(a_{n}-\frac{1}{e}\right)=\frac{1}{2e}$[/hide]

Ahaa...
Can the problem number one be solved with the Riemann Integral theorem ?
Hmm...
HINT : FOr any n>=n0 there exist some constant c so that an/n>=c GBU.

[quote="LideR_M"] ... [b][u]by Stirling :[/u][/b] $ \lim_{n\to\infty}\frac{\ln\sqrt{2\pi n}e^{\lambda_{n}}}{\ln n}$ for $ \frac{1}{12n+1}<\lambda_{n}<\frac{1}{12n}$, so: $ \lim_{n\to\infty}\left(\frac{1}{2}+\frac{\ln\sqrt{2\pi}e^{\lambda_{n}}}{\ln n}\right)=\frac{1}{2}$, so: $ \lim_{n\to\infty}\frac{n}{\ln n}\left(a_{n}-\frac{1}{e}\right)=\frac{1}{2e}$[/quote]
If it is perhaps, [b]without Stirling [/b]!

[quote="Virgil Nicula"][size=111][color=darkred][b]$ 1\blacktriangleright$ For any $ n\in \mathbb{N}^{\ast}\ ,\ a_n>0$ and $ \lim_{n\to\infty }\frac{a_{n}}{n}=\infty$ , prove that $ \lim_{n\to\infty }\sum_{k=1}^{n}\frac{1}{n+a_{k}}=0$.

$ 2\blacktriangleright$ It is well-known that for the sequence $ a_{n}=\frac{\sqrt [n]{n!}}{n}\ ,\ \lim_{n\to\infty}a_{n}=\frac 1{\text e}$. Prove that : $ \lim_{n\to\infty}\frac{n}{\ln n}\cdot\left(a_{n}-\frac{1}{\text e}\right)=\frac{1}{2\text e}$.[/b][/color][/size][/quote]

[hide="Solution to problem (1)"][color=darkblue] Using the simple inequality $x+y\ge 2\sqrt{xy}$ , $x,y>0$ we obtain : $\sum_{k=1}^n\ \frac 1{n+a_k}<\frac 1{2\sqrt n}\cdot\sum_{k=1}^n\ \frac 1{\sqrt {a_k}}$ . On the other hand,

we have : $\lim_{n\to\infty}\ \frac {\sum_{k=1}^{n+1}\frac 1{\sqrt{a_k}}-\sum_{k=1}^n\frac 1{\sqrt{a_k}}}{\sqrt{n+1}-\sqrt{n}}=$ $\lim_{n\to\infty}\ \frac {\sqrt{n+1}+\sqrt n}{\sqrt{a_{n+1}}}=\lim_{n\to\infty}\ \left[\sqrt{\frac {n+1}{a_{n+1}}}\cdot\left(1+\sqrt{\frac {n}{n+1}}\right)\right]=0$ .

Thus, by the [u][b]Cesaro-Stolz[/b] theorem[/u] we get : $\lim_{n\to\infty}\ \frac 1{\sqrt n}\cdot\sum_{k=1}^n\frac 1{\sqrt{a_k}}=0$ . Then we also have : $\boxed{\ \lim_{n\to\infty}\ \sum_{k=1}^n \frac 1{n+a_k}=0\ }$ .[/color]
[/hide]

[hide="Solution to problem (2)"] [color=darkblue]
$\frac n{\ln n}\cdot\left(a_n-\frac 1{\text e}\right)=\frac 1{\text e}\cdot\frac n{\ln n}\cdot\left(\text{e}\cdot a_n-1\right)=\frac 1{\text e}\cdot\frac n{\ln n}\cdot b_n\cdot\frac {\text{e}^{b_n}-1}{b_n}$ , where $b_n=\ln (\text {e}\cdot a_n)=1+\ln a_n\to 0$ .

Therefore, $\lim_{n\to\infty}\ \frac n{\ln n}\cdot\left(a_n-\frac 1{\text e}\right)=\frac 1{\text e}\cdot \lim_{n\to\infty}\ \frac {n\cdot b_n}{\ln n}$ . On the other hand, $\lim_{n\to\infty}\ \frac {(n+1)\cdot b_{n+1}-n\cdot b_n}{\ln (n+1)-\ln n}=$

$=\lim_{n\to\infty}\ \frac {1-n\cdot\ln\frac {n+1}n}{\ln (n+1)-\ln n}=\lim_{n\to\infty}\ \frac {n\cdot\left(1-n\ln\frac {n+1}n\right)}{\ln\left(1+\frac 1n\right)^n}=\frac 12$ , because $\lim_{n\to\infty}\ n\cdot\left(1-n\ln\frac {n+1}n\right)=\frac 12$

(one can easily prove it by [u][b]l'Hospital[/b]'s rule)[/u] . Consequently, $\lim_{n\to\infty}\ \frac {n\cdot b_n}{\ln n}=\frac 12$ ([u][b]Stolz-Cesaro[/b][/u]) and our conclusion follows.[/color]

[/hide]


Aceasta este o problema compusa de mine. Nu e exclus sa mai fi aparut undeva, dar nu stiu nimic despre asa ceva.

Fie a, b, c lungimile laturilor unui triunghi oarecare
Demonstrati, ca:
R = (ab + bc + ac)/2(ha + hb + hc)
(aici, ha este inaltimea corespunzatoare laturii de lungime a a triunghiului)

Incumetati-va :) nu e grea.

cunoscuta si usoara
$R=\frac{abc}{4S}$...etc


[quote="GacKiem"]Let $a_{1}<a_{2}<...<a_{n}$ be positive real numbers and $a_{1}+a_{2}+...+a_{n}=n$. Prove that
\[a_{1}^{2}+3a_{2}^{2}+...+(2n-1)a_{n}^{2}\le n^{2}. \]
[/quote]

Hey,welcome to Mathlinks :lol: !!!
First one thing,I think the condition in the problem should be the opposit one:that is $a_{1}\geq a_{2}\geq ...\geq a_{n}$,shouldn't it?

Anyway,here is my solution for this form of the problem :wink: .

We have that for any $1\leq k\leq n$
\[(k-1)a_{k}\leq a_{1}+a_{2}+...+a_{k-1}\]
<=>

\[(2k-1){a_{k}}^{2}\leq a_{k}(2a_{1}+...+2a_{k-1}+a_{k})\]
<=>
\[(2k-1){a_{k}}^{2}\leq (a_{1}+...+a_{k})^{2}-(a_{1}+...+a_{k-1})^{2}\]

Summing the last inequality for all $k$s,we get that
\[a_{1}^{2}+3a_{2}^{2}+...+(2n-1)a_{n}^{2}\leq (a_{1}+...+a_{n})^{2}=n^{2}\]
 :wink:

[quote="Crux"]EDIT : oups Tiks posted before...
you reversed the inequality...
just square the equality $\sum_{i=1}^{n}a_{i}=n$ to get
$\sum_{i=1}^{n}a_{i}^{2}+2\sum_{1\leq i < j \leq n}a_{i}a_{j}=n^{2}$. Notice that $\sum_{1\leq i < j \leq n}a_{i}a_{j}= \sum_{j=2}^{n}\sum_{i=1}^{j-1}a_{i}a_{j}< \sum_{j=2}^{n}\sum_{i=1}^{j-1}a_{j}^{2}=\sum_{j=2}^{n}(j-1)a_{j}^{2}.$
Thus $n^{2}<a_{1}^{2}+\sum_{j=2}^{n}[1+2(j-1)]a_{j}^{2}=\sum_{j=1}^{n}(2j-1)a_{j}^{2}$[/quote]

I think you reversed the inequality,didn't you?
I reversed the condition  :D :wink: .

Yes, after I posted this one, I realize that I forget to write $a_{1}\ge a_{2}\ge ... \ge a_{n}$. But I can't edit because on 10 minutes after the first post, I can't edit it. Sorry because after that I must log out and haven't time to edit yet.

I like this inequality because it has $n$ cases of equality, that is
\[a_{1}=a_{2}=...=a_{k}=\frac{n}{k},a_{k+1}=...=a_{n}=0 \]
for all $k=1,2,...,n.$ (so surprising)

My solution is the same as Crux but I wonder there is a direct and more beautiful inequality. I suggest to use Abel formula.

Here is another problem (my friend shows me):

Suppose that $0\le a\le b \le c$ and $a+b+c=3$. Prove that
\[\sqrt{3a^{2}+1}+\sqrt{5a^{2}+3b^{2}+1}+\sqrt{7a^{2}+5b^{2}+3c^{2}+1}\le 9. \]
Let's enjoy!

[quote="GacKiem"]Let $a_{1}\ge a_{2}\ge ...\ge a_{n}$ be positive real numbers and $a_{1}+a_{2}+...+a_{n}=n$. Prove that
\[a_{1}^{2}+3a_{2}^{2}+...+(2n-1)a_{n}^{2}\le n^{2}. \]
[/quote]

see http://www.mathlinks.ro/Forum/viewtopic.php?t=108929 .


A set $M$ consists of $n$ elements. Find the greatest $k$ for which there is a collection of $k$ subsets of $M$ such that for any subsets $A_{1},...,A_{j}$ from the collection, there is an element belonging to an odd number of them

[hide="Solution using linear algebra over the field with 2 elements"]
Assume $ M\equal{}\{1,\ldots, n\}$ and identify each subset $ A$ of $ M$ with the vector $ v\equal{}(x_1,x_2, \ldots, x_n)$ where $ x_i\equal{}1$ if $ i \in A$ and $ x_i\equal{}0$ otherwise. If $ A_1, \ldots, A_k$ are subsets as required, then the corresponding vectors $ v_1, \ldots, v_k$ have the following property: for each subset of $ \{v_1,\ldots, v_k\}$, the sum (modulo $ 2$) of the vectors in this subset is non-zero. The reason is that in adding vectors mod $ 2$, we get $ 0$ in coordinate $ i$ iff $ i$ belongs to an even number of the sets in the sum. Since there is always at least one $ i$ which belongs to an odd number of those sets, the sum is not the zero vector.

It follows that the vectors $ v_1,\ldots, v_k$ are linearly independent over $ \mathbb{Z} / 2\mathbb{Z}$. Since they lie in $ (\mathbb{Z} / 2\mathbb{Z})^n$ which is a vector space of dimension $ n$, we must have $ k \leq n$.

Achieving $ k\equal{}n$ is trivial: just take the sets to be singletons $ A_i \equal{} \{i\}$.

[/hide]


a dartboard has two regions, a 7-point region and a 15-point region. If an unlimited number of darts is  allowed, what is the largest finite score that cannot be attained?

[hide]83[/hide]

the formula for this is ab-a-b, but why does that work?
also i like another form of that formula (a-1)(b-1)-1
if the proof is too complicated, just say so and then this may be locked....

thanks for any help! :lol:

It looks like it has something to do with least common multiples but as for why, I'm not sure.

i heard that the proof is way to complicated for mathcounts.

oh....ok thanks!

 :P

this may be locked...

[quote="alanchou"]there is a method to solve this problem without the formula as long as they are relatively prime though.  And it's still MC level :)[/quote]

Can you please post it? Is there a method for if you have 3 regions? 4? 5? 89? I missed a chapter target 2 years ago b/c i never saw a problem like this, but now i know ab-a-b. But I'm still stuck on the ones with even more regions.

I think 3 regions is definitely harder to generalize, and 4... well, yeah

Let's say you wanted to find the largest possible value that cannot be made using non-negative amounts of 7 and 11.  By the formula given, we find it's 7(11)-7-11=$ \boxed{59}$.  But let's say we didn't know this formula.  Let $ S_{0}$ be the set of all integers that are congruent to 0 mod 7 and are greater than or equal to 0.  Then, all the elements in $ S_{0}$ can be created with just 7s.  Let $ S_{1}$ be the set of all the elements that are exactly 11 more than those in $ S_{0}$.  Since $ 11\equiv 4\text{ (mod 7)}$, this set consists of all integers that are congruent to 4 mod 7 and are greater than or equal 11.  Similarly, we define $ S_{2}$ to be the set of all elements exactly 11 more than those in $ S_{1}$, and those are congruent to 1 mod 7 and greater than or equal to 22.  Keep on going... eventually, you'll have the set $ S_{6}$ which consists of all integers that are congruent to 3 mod 7 and are greater than or equal to 66.  The next set, $ S_{7}$, ends up being a subset of $ S_{0}$ (convince yourself this is true).  So, 66 is the smallest integer that can be made that is in $ S_{6}$, we just go down the list

66=6(11)
65 is an element of $ S_{4}$
64 is an element of $ S_{2}$
63 is an element of $ S_{0}$
62 is an element of $ S_{5}$
61 is an element of $ S_{3}$
60 is an element of $ S_{1}$
59 is not an element of $ S_{6}$ because it's less than 66, so the answer is $ \boxed{59}$

The sets are really unnecessary, and once you get your upper bound of 66, it's pretty easy to see that all you have to do is evaluate 66-7.

I think around here, it's commonly called the Chicken McNugget Theorem (search it!).
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43173
For three numbers, I think it's $ 2abc\minus{}(ab\plus{}bc\plus{}ca)$.


[quote="Jean-Louis"]Relativists deny the existence of universal truths, do they have a good argument when mathematics is concerned ? Are there universal truths in mathematics ?

Roger Penrose stated that mathematical truth is absolute, external, and eternal, and not based on man-made criteria ... mathematical objects have a timeless existence of their own..."

I am looking for examples to put in my essay on philosophy of mathematics.[/quote]

Hmm.. I don't think this is an [u][b]Open Question in Algebra[/b][/u]. And by the way there are no examples because nothing in Mathematics can be proven true without assumptions (axioms). And you should be looking elsewhere for philosophy.


algebra
function
domain
algorithm
number theory unsolved
number theory

x,y,z are natural numbers such that
                             xy=z^2+1
prove that there are integers a,b,c,d such that
         x=a^2+b^2;y=c^2+d^2 and z=ac+bd.
i have solved but with great effort!Could anyone solve and post solution!!?
(^2 means square)

Trivial by knowing that the gaussian integers build up an unique factorisation domain:
Write $xy=z^2+1=(z+i)(z-i)$. Now factor $z+i = p_1\cdot p_2\cdot ...\cdot p_n$ and get ${z-i = \overline{z+i} = \overline{p_1\cdot p_2\cdot ...\cdot p_n} = \overline{p_1} \cdot \overline{ p_2}\cdot ...\cdot \overline{p_n}}$ (the primes $p_i$ are not necessary distinct). Now by uniqueness of factorisation, you WLOG get that $x=(p_1\cdot p_2\cdot ...\cdot p_k)(\overline{p_1}\cdot \overline{p_2}\cdot ...\cdot \overline{p_k})=(a+bi)(a-bi)=a^2+b^2$ and $y=(p_{k+1}\cdot p_{k+2}\cdot ...\cdot p_n)(\overline{p_{k+1}}\cdot \overline{p_{k+2}}\cdot ...\cdot \overline{p_n})=(c-di)(c+di)=c^2+d^2$.
Now $(a+bi)(c-di)=(z+i)$ by definition, which gives $z=ac+bd$.

To get a more elementary solution you could elementarize it or use weaker theorems.

one idea:

if $p$ is a prime such that $p|z^2+1$, we have that ${p}\equiv{1}(mod.4)$.

so, all prime divisors of $x$ and $y$ are in the form $4k+1$.

this implies that $\exists a,b,c,d \in Z$ such that $x = (a^2+b^2), y = (c^2+d^2)$.

so, $xy = (ac+bd)^2 + (ad-bc)^2$.  :)

dear e.lopes
let me tell you where i got confused while solving this problem?
        We know that if p|n^2+1 then p=4k+1.
then,We know that for every p=4k+1 there exists integers x,y such that
                          p=x^2+y^2
From here we get that there exists a,b,c,d such that(by using eucllid algorithm)
              x=a^2+b^2and y=c^2+d^2
but how can we establish  ad-bc=1?

hey guy, see my post.

i said $\mbox{one ideia}$, not $\mbox{one full solution}$.  ;)

my solution is the same of ZetaX, but him posted before me.   :)


Hmm, the links for the pdf's don't seem to work anymore, I tried it on two seperate computers, one was a PC and the other was a Mac. Anyone know what happened? Well anyway if anyone has them, would it be possible for you guys to pm/email the pdf's to me?

Thanks in advance

The links are fixed. You can download the files from our page again.


advanced fields
advanced fields theorems

Can you help to solve this problem?

[Sinx y']' + [Cosx+ lambda Sinx] y = 0 ; 1<x<2; y(1) = y(2) = 0

I need to find the eigenvalues and eigenfunctions.

Thanks

Suggestions:
(1) Use a computer.
(2) Try power series.


In AoPS textbooks, a links page has been mentioned, but I have searched all over the website, but a links page is nowhere to be found. Additionally, I have typed the address given in the address bar, but it gives me Error 404- Not Found. Can anyone help?

There are numerous links under Resources->Internet Guide. There are also quite a few in our Wiki.


I know almost everyone who wants to qualify for the USNCO has to take the local section test first. I am wondering what is the usual cut-off score to qualify the USNCO. I mean, you can miss about 10? 5? or 3? to get in? I know different areas' cut-off scores vary, but what is the mid-50% range? What is usually a good score?
Thanks! :)

I can definitely tell you that there is no one "usual" cut-off score.  You're certainly right, they vary... in my section this year, only two people took the local exam in the first place, so the cut-off was writing your name correctly.  I know of places that the cut-off was 5-10 wrong or so.  That's probably about the worst you will see though.  If it helps, last year I got fourth in Minnesota with 6 wrong on the local.

The USNCO doesn't really publish statistics, so it's difficult to say with any certainty what the mid 50% range is.  I would suggest emailing your local coordinator if you really want to know about the cut-off score.  Good luck.


Let $ ABCD$ be a convex quadrilateral, let $ E$ and $ F$ be the midpoints of $ AD, BC$ respectively. Denote by $ M$ the intersection of $ AF$ and $ BE$, $ N$ the intersection of $ CE$ and $ DF$. Find the minimum of
$ \frac {MA}{MF} \plus{} \frac {MB}{ME} \plus{} \frac {NC}{NE} \plus{} \frac {ND}{NF}$

[quote="outback"]Let $ ABCD$ be a convex quadrilateral, let $ E$ and $ F$ be the midpoints of $ AD, BC$ respectively. Denote by $ M$ the intersection of $ AF$ and $ BE$, $ N$ the intersection of $ CE$ and $ DF$. Find the minimum of
$ P\equal{}\frac {MA}{MF} \plus{} \frac {MB}{ME} \plus{} \frac {NC}{NE} \plus{} \frac {ND}{NF}$[/quote]
$ P\equal{}2(\frac{S[FAD]}{S[EBC]}\plus{}\frac{S[EBC]}{S[FAD]})\ge 4$


number theory
greatest common divisor
inequalities

[quote="ThAzN1"]Let $a_1, a_2, \dots , a_n$ be positive integers such that
\[ \frac{a_1+1}{a_2}+\frac{a_2+1}{a_3}+\cdots +\frac{a_n+1}{a_1} \]
is also an integer.  Show that
\[ \text{gcd}(a_1,a_2,\dots , a_n) \leq \sqrt[n]{ a_1a_2\cdots a_n\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_n}\right) }. \][/quote]

Let $d=\text{gcd}(a_1,a_2,\dots ,a_n)$, and write $a_i=d\cdot x_i$ where $x_i\in \mathbb {Z}^+$. If we put $a_{n+1}=a_1$, we obtain that:

$\sum_{i=1}^n \frac{a_i+1}{a_{i+1}}=\sum_{i=1}^n \frac{dx_i+1}{dx_{i+1}}=\frac{dS_1+S_2}{dx_1x_2\cdots x_n}$

Where $S_1=x_1^2x_3...x_n+x_1x_2^2x_4...x_n+....+x_1x_2...x_{n-1}^2+x_2x_3...x_n^2$ and $S_2=x_1x_2...x_n\sum_{i=1}^n \frac{1}{x_i}$. Clearly $d|dS_1+S_2$ (because by assumption $\frac{dS_1+S_2}{dx_1x_2\cdots x_n}$ is an integer), an this implies that $d|S_2$, further $d\leq x_1x_2...x_n\sum_{i=1}^n \frac{1}{x_i}$. Hence $d^n\leq a_1a_2...a_n\sum_{i=1}^n \frac{1}{a_i}$, as desired.


Find all positive integer solution of $(a,b,c)$ if $a^{2}+1$ and $b^{2}+1$ are prime numbers,and $(a^{2}+1)(b^{2}+1)=c^{2}+1$

WLOG, assume that $a\geq b$, we have
$b^{2}(a^{2}+1)=c^{2}-a^{2}=(c-a)(c+a)$
From $a\geq b$ we have $a^{2}+1\geq \sqrt{(a^{2}+1)(b^{2}+1)}>c-a$, and from the fact that $a^{2}+1$ is a prime,
we have $a^{2}+1 | c+a$. But $c+a<2(a^{2}+1)$, we conclude that $a^{2}+1=c+a$, and, thus, $b^{2}=c-a$
Solving this yields $b=a-1$, so $a^{2}+1, b^{2}+1$ have different parity, but both are primes.
Therefore, $b^{2}+1=2$. That is $(a,b,c)=(2,1,3)$
Hence, the solutions to the problem are $(a,b,c)=(1,2,3),(2,1,3)$

Why you can conclude $c+a < 2(a^{2}+1)$ ?
Thx..


A $ 4\times 4\times 4$ open cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?

the number of cubes that touches side of bottom is...

total number of cubes - (the cubes that are inside + the cubes that are touching top)

the number of cubes that are inside is $ 2^3\equal{}8$ and the number of cubes that are touching top is 4.

so the answer is 64-8-4=52.

answer : 52.


What is the sum $ \frac {25}{72} \plus{} \frac {25}{90} \plus{} \frac {25}{110} \plus{} \frac {25}{132} \plus{} ... \plus{} \frac {25}{9900}$?

I first found out that this sequence can also be written as:
$ \frac {25}{8*9} \plus{} \frac {25}{9*10} \plus{} \frac {25}{10*11} \plus{} \frac {25}{11*2} \plus{} ... \plus{} \frac {25}{99*100}$

Is there a special way to factorize this so that the sum can be found?
And if there is, what is it? Thanks!

hello, use that
$ \frac{1}{n(n\plus{}1)}\equal{}\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$.
Sonnhard.

thank you, sonnhard.
now i can solve the problem :)

hello, i have got $ \frac{23}{8}$ as the result of your sum.
Sonnhard.

hello, yes you can prove it by the following way
$ \frac{1}{n(n\plus{}1)}\equal{}\frac{n\plus{}1\minus{}n}{n(n\plus{}1)}\equal{}\frac{n\plus{}1}{n(n\plus{}1)}\minus{}\frac{n}{n(n\plus{}1)}\equal{}\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$.
Sonnhard.

[quote="the cliu"][quote="Dr Sonnhard Graubner"]hello, use that
$ \frac {1}{n(n \plus{} 1)} \equal{} \frac {1}{n} \minus{} \frac {1}{n \plus{} 1}$.
Sonnhard.[/quote]

is there a way to prove this?[/quote]

To obtain the result yourself, just use partial fraction decomposition. In this case, just solve $ \frac {1}{n(n \plus{} 1)} \equal{} \frac {A}{n} \plus{} \frac {B}{n \plus{} 1}$.

I have some thing :
a/(n(n+a)=1/n-1/(n+a)

[quote="KingSmasher3"]Partial Fraction Decomp. is calculus leveled. I think Sonnhard's way works the best.[/quote]

It doesn't require any calculus at all  :roll:

$ \frac 1{n(n\plus{}1)}\equal{}\frac An\plus{}\frac B{n\plus{}1}$

$ 1\equal{}A(n\plus{}1)\plus{}B(n)$.

$ n\equal{}\minus{}1$ gives $ B\equal{}\minus{}1$

$ n\equal{}0$ gives $ A\equal{}1$

$ \frac 1{n(n\plus{}1)}\equal{}\frac 1n\minus{}\frac 1{n\plus{}1}$.

But I agree, the other way is better.

[quote="Arrange your tan"][quote="Ihatepie"][quote="KingSmasher3"]Partial Fraction Decomp. is calculus leveled. I think Sonnhard's way works the best.[/quote] Isn't that what Sonnhard did?[/quote]

No, Dr Sonnhard Graubner, used a "trick"  (here, meaning an equivalent
expression, but it was a calculated one, as opposed to choices in his
new numerator being from trial and error).

He did not go through the method of having to know what degree each
for the split numerators would have to be, using variables such as $ A$ and $ B$,
or having to solve a pair of equations with two unknowns.[/quote]

Well, the original way to come up with this "trick" is to use partial fraction decomposition.


The area of a sector of a circle is $ 12\pi$ square centimeters. The measure of the central angle of the sector is $ 120^\circ$. What is the number of centimeters in the length of a radius of the circle?

Since the central angle is 120 degrees, the sector takes up $ \frac{120}{360} \equal{} \frac{1}{3}$ of the circle. That means the area of the circle is $ 12\pi \times 3 \equal{} 36\pi$, so the radius is 6.


Solve:

$\frac{1}{i^{3}}$

I looked at this, and I obtained

$\frac{1}{-i}$

But the answer is 0 + i

I don't get it.

Where is the zero coming from?

[quote="Genecks"]If $i^{3}$ = -i

and

1=1

then

$\frac{1}{i^{3}}$=

$\frac{1}{-i}$

yeah.. how is the "i" becoming isolated next to that 0?[/quote]
Rationalize your denominator, multiply the numerator and the denominator by $i$.

[quote="Genecks"]If $i^{3}$ = -i

and

1=1

then

$\frac{1}{i^{3}}$=

$\frac{1}{-i}$

yeah.. how is the "i" becoming isolated next to that 0?[/quote]
$\frac{1}{-i}=\frac{1}{-i}\cdot\frac{i}{i}=\frac{i}{1}=i$.

so, it's primarily about getting a 1 or -1 in the denominator? I sometimes wonder why I have to rationalize certain things. I don't understand why. Some form of simplification?

What's wrong with the way I had it?

Oh,, hmm... perhaps the fact it has to be in "a + bi" form?

[quote="Genecks"]I sometimes wonder why I have to rationalize certain things.

...

Oh,, hmm... perhaps the fact it has to be in "a + bi" form?[/quote]

Yes.


"A loan of $\$10000 is being repaid with payments of $\$1000 at the end of each year for 20 years. If each payment is immediately reinvested at 5% effective, find the effective annual rate of interest earned over the 20-year period."

My solution: The answer key says 6.16%.

Is my accumulated value simply 10000? I cannot write the solution here bec. it`s difficult to write the symbols.

I consider the investment(payment) of 1000 at the end of each period for 20 periods at rate i (my unknown) such that the interest is reinvested at rate j (5%).

I have observed that the loan is fully repaid in 10 years, is it not?

10000 = 1000 { 20 + i[ (1.05)^20 - 1)/.05) - 20]/.05 }

Eventually my i is not .0616. Where`s my mistake?

[quote="justmath11"]I cannot write the solution here bec. it`s difficult to write the symbols. [/quote]
You probably can write those symbols with $\text{\LaTeX}$.
Also, you inadvertently triggered $\text{\LaTeX}$ with your dollar signs- either disable BBCode or use $\$ ((dollar)\(dollar)) if you want to talk about money.
[quote]A loan of $\$10000 is being repaid with payments of $\$1000[/quote]

I just edited that (dollar)\(dollar) into justmath11's post.

Next issue: this isn't "higher algebra". It involves finite sums of geometric progressions - that's probably either pre-olympiad or intermediate. I'll move it.


function
parameterization
trigonometry
calculus
calculus computations

I was wondering if you could assist me with this problem.

Find the first non-zero terms, the general term for the cycloid power series, and the interval of convergence for the cycloid power series

I know that the basics for taylor polynomials. for a cycloid,
I know that f(a)+ f'(a)(x-a)+ f''(a)(x-a)^2... will give me the right
answer but I don't know how to do this in parametric mode or what
really do other than what i've listed below. Thanks!

y=-a+acos(theta)
x=a(theta)-asin(theta)

so, dy/dx=-sin(theta)/1-cos(theta)

I also know that d(dy/dx)/dx=d^2y/dx^2

Your question is not well defined. Power series for which function? $ y$ as a function of $ x?$ $ x$ as a function of $ y?$ $ x$ and $ y$ both separately as functions of the parameter $ \theta?$ (The latter is by far the easier question.) And centered where? You just can't say "find me the power series of such-and-such function" without it being obvious where the center should be.

With the paratmetric representations $ x\equal{}a(\theta\minus{}\sin\theta),\ y\equal{}a(1\minus{}\cos\theta),$ we can use the well-known (I have my students memorize them) power series for the sine and cosine to find the following about $ \theta \equal{} 0:$

$ x\equal{}a\left(\theta\minus{}\theta\plus{}\frac16\theta^3\minus{}\frac1{5!}\theta^5\plus{}
\cdots\right)$

$ x\equal{}a\left(\frac{\theta^3}6\minus{}\frac{\theta^5}{5!}\plus{}\cdots\right)$

Similarly, $ y\equal{}a\left(\frac{\theta^2}{2}\minus{}\frac{\theta^4}{4!}\plus{}\cdots\right).$

For small $ \theta,$ we can approximate $ \theta$ by $ \theta\approx\left(\frac{6x}{a}\right)^{\frac13}.$

Substitute this in the series for $ y,$ and we get that for $ x$ and $ y$ near zero,

$ y\approx\left(\frac{9a}{2}\right)^{\frac13}x^{\frac23}.$

This gives some insight into the nature of the non-smooth cusp the curve has at $ (0,0).$

I'm having trouble understanding your response. If I center it about zero or even pi, is there anyway you can help me a little more? Also, my function was a function of theta

y=-a+acos(theta)
x=a(theta)-asin(theta)

The negative is there because i'm dealing with pendulum motion.

Thanks.


Prove that for $a\in\mathbb{R}\setminus{0},$

\[\sum_{n=1}^{\infty}\frac1{n^2+a^2}=\frac{\pi \coth{(a \pi)}}{2a} - \frac{1}{2a^2}\]

[Continuation of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51675]this topic.[/url]]

This alternate form might be more amenable to some manipulations:

\[\sum_{n=-\infty}^{\infty}\frac1{n^2+a^2}=\frac{\pi\coth(\pi a)}a\]


articles quadratics geometry algebra polynomial function MATHCOUNTS

[quote="Rep123max"]Does anyone have tips for doing good on the AMC 10.[/quote]

Yes, there are some good tips here on this site for doing well on math contests. One of my favorites, which I remind my son about all the time, is Richard Rusczyk's article [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop Making Stupid Mistakes[/url], which should help your score on any of the AMC tests or any other math test worthy of the name.

Practicing on old tests is the way to go for getting a good score on a particular brand name of test. Of course, the AMC 10 is such a new test there aren't very many old AMC 10 tests--but there are a lot of old AMC 12 tests and AMC 8 tests, with similar kinds of problems at different levels of difficulty. Be sure to practice according to the same time limits as the real test, as one important thing to learn is how to pace yourself.

Get plenty of scratch paper, read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Mr. Rusczyk's article[/url], and then take a practice AMC 10 under testing conditions.

By the way, what kind of school looks at AMC 10 scores as part of its admission process?

Good luck.

As a supplement to tokenadult's advice, I think there are several topics one should recognize when taking AMC 10 (only in AMC 10!):
1. sequence and set of numbers that involve mean and mode
2. simple quadratic equation (though knowing Viete's formula saves a lot of time) and some linear equations
3. similar figures ( the most common geometry topics since 2000)
4. basic evaluations (know a little bit about polynomials and exponents)
5. symmetric sums ( knowing this can help you to solve problems in less than a minute or so)
6. speed problems
7. basic functions
8. proportion
9. last but not least, basic number theory and counting
also, knowing topics discussed in AoPS vol. 1 should be very helpful.

I still havent been able to seemingly convince my coach to do mathcounts or anything of that sort. She isn't willing to do anymore than she has to. How do you enter into these things? I dont understand, can you enter individualy in these competitions or do you have to go through a school?

Most of the big contests require you to register as a school.  There is one contest, the [url=http://www.nsa.gov/programs/mepp/usamts.html]USAMTS[/url] that I know of that is easy to do on your own.  It's a contest of a very different sort, however.  It's also aimed at high school students.

You've also probably missed registration deadlines for most of the contests for this year.  (I'm not sure about the AMC.)  Websites with more information are here:
http://www.mathcounts.org (I'm 95% positive it's too late for this one -- you might want to get your school to do it for the people who will come after you, though.)
http://www.unl.edu/amc

I'm sure other people can give you more helpful information.

[quote="paladin8"]a math teacher at my school says that the AMC and "beyond" are good for most college admissions. mathcounts, on the other hand, is a middle school contest, which makes it much less convincing to colleges that you are still that good... on the other hand, getting into the USAMO would probably get you into a lot of good colleges.[/quote]

He's applying to St. Marks which is a private (boarding?) high school, not a college.  I think.

[quote="paladin8"]a math teacher at my school says that the AMC and "beyond" are good for most college admissions. mathcounts, on the other hand, is a middle school contest, which makes it much less convincing to colleges that you are still that good... on the other hand, getting into the USAMO would probably get you into a lot of good colleges.[/quote]I believe Mr. Rusczyk got invited to Princeton after winning the USAMO so yeah it does help a lot if you get that high, but if you're like in the AIME then they might disregard it.

Do any of you know a good website that teaches Viete's? I went to mathworld and I searched AoPS for it but I just can't understand it...grrr...and what is "symmetric sums"? Also is there symmetric figure prblem other than triangles?

I had to take the SSAT, which is similar to the ISEE, when I was applying to Exeter.

I live in Oklahoma, and even I know that St. Marks is supposed to be a really goode school.

i didnt think of this at the time, but there are many st. mark's im sure...  there is a St. Mark's school of texas, which he might be talking about.

Is the st. mark's you're thinking of in  MA?

heh, Rep123Max and I are referring to the St. Marks in Texas.

how should the time allocation be divided? (easy, medium, hard)
and at what pace should i go at?

If you think a problem will take more than 3 minutes, don't attempt it until you have done all the ones that you think will take no more than 3 minutes. Then do the ones you think you can do in under 4 minutes, etc.

Yeah -- there's no advantage to doing hard ones instead of easy ones, so do the easy ones first!  You might look through the test for a brief second in order to see which are easy -- usually, one of the last 5 is not very hard at all.  (In my junior year, it was number 25.  The first time I looked at it, I thought it was hard, so I skipped it and checked everything else and came back to it at the end but ran out of time.  Needless to say, I could have done it in a few seconds if I had just looked at it.  So I should have looked at it harder to begin with and done it sooner.)


[quote="Ignite168"]Solve for primary values of $\theta, 0\le\theta<2\pi:$ $\cot\theta=$ $\frac{\tan(A+\frac{\pi}{2})-\tan(A-\frac{\pi}{6})}{1+\tan(A+\frac{\pi}{2})*\tan(A-\frac{\pi}{6})}$.

Wow that was annoying to type up, and I'm assuming ctn = cot.[/quote]
[hide="hint"]$\tan(a-b)=\frac{\tan a-\tan b}{ 1+\tan a \tan b}$
[/hide]


Let ABCD be a quadrilateral inscribed in a cicrle satisfying AB.CD=
AD.BC .Let H,K be respectively the midpoints of the segments AC,BD.
Prove that AK+CK=BH+DH

Let X be a point on line AD s.t. A is the midpt of DX. It's easy to show that the triangles BAX and BCD are similar, so 2CK=BD*CB/AB. We do the same operation for AK, BH, DH and we get:

2AK+2CK=BD(AD/CD+CB/AB)
2BH+2DH=AC(AB/AD+CD/CB)

It's easy now to prove what we want by using Ptolemy's second thm:

AC/BD=(AB*AD+CB*CD)/(BA*BC+DA*DC).

Hope it's clear enough. I didn't feel like writing every single detail...:D


function
inequalities
inequalities proposed

Find the borders of the function $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, when $(a, b, c)$ described ${(R_+^*)}^3$.

It is a really classical result that the minimum is $\frac 3 2$. The upper bound is $+ \infty$ (take $a=1$ and $b,c$ tending to $0$).

In another hand, I think this problem should be in the Inequalities forum, so I move it there.

Pierre.


On the sides of triangle ABC, construct squares ACMN, BCPQ and ABRS and the lines MP, QR and SN. Prove that the triangles CPM, BRQ and ASN have equal area.

Please help! It's supposed to be easy but I can't figure it out.

Isn't this highly trivial? 2[MCP]=MC*CP*sin(<MCP)=ab*sinC=2[ABC]. So done. :D

dear melissa ,problems can be hard saome times,but its going to be ok.
as u know the formula for area of the triangle ABC is:1/2bc sin(CAB).
now we want to know what is the area of the triangle ANS.its:
1/2AN.AS.sin(NAS)
we have: AN=b  AS=c    sin(CAB)=sin(SAN)
and it means that the area of triangle ABC is equal tothearea of triangle NAS.
u can prove that tha areas of the triangles CMP andQBR are the same as the area of ABC.from this way.
is it ok dear melissa? :D  :D  :D

Yes, dear ashegh, I've solved it like a month ago.


floor function
number theory proposed
number theory

Consider the finite sequence $\left\lfloor \frac{k^2}{1998} \right\rfloor$, for $k=1,2,\ldots, 1997$. How many distinct terms are there in this sequence?

[i]Greece[/i]

Here is a complete solution for this forgotten problem.

First, we study when $\left\lfloor \frac{k^2}{1998}\right\rfloor<\left\lfloor \frac{(k+1)^2}{1998}\right\rfloor$. Note that, $\left\lfloor a \right\rfloor = \left\lfloor b\right\rfloor \implies |a-b|< 1$. Hence, in particular, if,
$$
\frac{(k+1)^2}{1998}-\frac{k^2}{1998}=\frac{2k+1}{1998}\geq 1\iff k\geq 999
$$
holds, we have $a_k<a_{k+1}$. Namely, the set $\{a_{999},a_{1000},\dots,a_{1997}\}$ (of cardinality $999$) contains distinct terms.

Next, we consider $k\leq 999$. Note that, in this range, $0\leq \left\lfloor \frac{k^2}{1998} \right\rfloor\leq 499$. Now, fix an integer $n\in [0,499]$. We claim that this $n$ is achievable. To show, we need to prove that, for every $n\in\mathbb{Z}\cap [0,499]$, there is a $k$ such that,
$$
1998n\leq k^2<1998n+1998.
$$
Now, for $k\leq 999$, the difference between two consecutive perfect squares, namely $d_k=k^2-(k-1)^2 =2k-1<1998$. Hence, if we consider the intervals, $I_n=[1998n,1998n+1998)$, we claim that each one contains at least one perfect square. In order to prove this, assume that this is false, and pick the smallest $n$ for which $I_n$ contains no perfect squares (clearly, $I_0,I_1$ contain, which can be verified manually). Now, take the largest square contained in $I_{n-1}$, and denote it $s_k$. By definition, $s_{k+1}\notin I_{n-1}$. However, since $s_{k+1}-s_k<1998$, $s_{k+1}\notin I_{n+1}$, either. Hence, $s_{k+1}\in I_n$, and therefore, $I_n$ must contain a perfect square.

Hence, the total number of distinct terms is $500+999=1499$.

Let $ f:\mathbb{Z}^{+}\rightarrow \mathbb{N}_{\geq 0}$ be the function defined by $ f( x) =x^{2} /1998$.

Since $ f( x+1) -f( x) =( 2x+1) /1998$, the following (1),(2) are true:
(1)  If $ x\geq 999$,  then $ \lfloor f( x+1)\rfloor  >\lfloor f( x)\rfloor $
(2)  If $ x\leq 998$,  then $ \lfloor f( x+1)\rfloor -\lfloor f( x)\rfloor =0,1$

So the answer is $ 1+\lfloor f( 999)\rfloor +( 1997-1000+1) =\boxed{1498}$


function
inequalities
algebra unsolved
algebra

Function $(y+2)^{x}-y^{x}$ increase if x>1, y>0 and decrease if x<1 y>0. Therefore $a^{x}-(a-2)^{x}=(a-2)^{x}-(a-4)^{x}$ had only one solution x=1 if a>4.

You're too quick, Rust, as many times before :)

$\frac{(a-4)^{x}+a^{x}}{2}=\left(\frac{a-4+a}{2}\right)^{x}$

Let $f(a)=a^{x}$. Then $f''(a)=x(x-1)a^{x-2}$. Hence for $x<0$ or $x>1$ the function is convex and Jensen gives $>$ sign in the above equation, and for $0<x<1$ the function is concave and Jensen gives $<$ sign in the above equation. For $x=0,1$ we have equality, hence those are only solutions.


What are the computer facilities at Mathcamp like?  :D

yeah, do we have internet access there? Is that convenient? When can we go there?

Well, last year there was a computer lab (which was a bit farther away) and a small room with three computers in it in the dorm buildings, and I'm assuming that this year the setup will be at least somewhat similar.  But there will definitely be internet access, most likely accessible during any free time as long as there's some computer open.

We always have a computer lab open at some times (the college does require it to be supervised by staff, so we can't make it constantly open), and we try to make those times convenient for everyone (usually in the evening after classes).

We'll also set up some computers that are primarily for e-mail checking, but can be used for other things if no one's waiting.  Those are open pretty much all the time, and are in the main dorms.

Hmm, what if we bring laptops...is wireless available in the dorms?

When it was at UPS(University of Pudget Sound) in 2003, there was internet access in the dorn rooms, though I am not sure if it was wireless.  :? . I would assume the same is true this year, although don't know for sure  :P

My understanding is that this year there will be Internet in the dorms, but I'm not sure what kind.


Ok, so I've been doing this functional equation for ages, and I don't seem to be getting very far.  Hopefully someone will have an idea:
$f(x-f(y))=f(f(y))+xf(y)+f(x)-1$

The solution appears to be $f(x)=1-x^{2}/2$.
Let $I$ be the image of $f$
If $h(x): =f(x)+x^{2}/2-1$, then $h(x+y_{0}+y_{1}+...+y_{n})=h(x)-\frac{n}{2}h(0)$, where $y_{i}\in I$
This implies that all of the elements of $I$ are linearly independent over the rationals.

This is all I know - lets solve it!

Jack

Sorry, it doesn't quite imply linear independence; if $n_{1}+n_{2}+...+n_{s}=m_{1}+m_{2}+...+m_{t}$, $n_{i},\; m_{i}\in \mathbb{N}$
then $\sum n_{i}y_{i}= \sum m_{i}z_{i}$ is allowed where $y_{i},\; z_{i}\in I$

Jack

Whoa this is pretty fun. Let S be the image of f. Now first note that if $a, b \in S$, then setting x = a and y to some value such that f(y) = b, we have

$f(a-b) = f(a)+ab+f(b)-1 = f(b-a)$ (*).

Now since f is obviously not identically zero, there exists c such that f(c) is not zero, so we set $y = c, x = \frac{1}{f(c)}$, and we have

$f(x-f(y)) = f(f(y))+1+f(x)-1 = f(f(y))+f(x)$.

In other words, there exist $p, q, r \in S$ with p+q = r. Then we have f(-p) = f(q-r) = f(r-q) = f(p) for some $p \in S$. However, by setting x = 0, we have

$f(-f(y)) = f(f(y))+f(0)-1$

so that if we choose y such that f(y) = p, we must have f(0) = 1.

Then setting x = f(y), we have $f(0) = 2f(f(y))+f(y)^{2}-1 \implies f(t) = 1-\frac{t^{2}}{2}$ for $t \in S$ (**). We can use (**) to update (*) in that if $a, b \in S$, then

$f(a-b) = f(a)+ab+f(b)-1 = 1-\frac{(a-b)^{2}}{2}$ (***).

To finish, we use (***) in a tricky way. Set y = 0 in the original equation to get

$f(x-1) = f(1)+x+f(x)-1$
$f(x-1)-f(x) = f(1)+x-1$.

Set z = f(1) + x - 1 for convenience. It is evident that $f(x-1), f(x) \in S$, so that by (***), we have

$f(z) = f(f(x-1)-f(x)) = 1-\frac{(f(x-1)-f(x))^{2}}{2}= 1-\frac{z^{2}}{2}$.

Since we can make z any real value by choosing the appropriate x, we have that $f(t) = 1-\frac{t^{2}}{2}$ for all t.

Just to check, this is $\mathbb{R}\rightarrow \mathbb{R}$ right?

Edit: Lol, looks like I was quite severely beaten. I'm gonna try to solve it myself anyway (haven't read that rather long solution yet).

Nice solution probability0.01 - setting x=1/f(y) is the key step, in my opinion (but that's just because its the one I didn't get).
This is question 3 from the 1999 Bucharest IMO, in which no-one scored above 39 and the gold cutoff was 28.
So its probably quite hard.
edit: have just read the official shortlist solution and it is beautiful.  And much too easy!

Jack


A number randomly selected from the set $ \{1,2,3,4,5,6,7\}$ is multiplied by 7. What is the probability that exactly one of the digits in the product is not 1, 2, 4, 5, 7 or 8? Express your answer as a common fraction.

Wait..it says the answer is 2/7....

Also, I think the answer is $ 1/7$, because the the only digit that is not in the set is 9..and thats 1 out of 7?!?!

[quote="GameBot"]A number randomly selected from the set $ \{1,2,3,4,5,6,7\}$ is multiplied by 7. What is the probability that [b]exactly one of the digits[/b] in the product is not 1, 2, 4, 5, 7 or 8? Express your answer as a common fraction.[/quote]
Maybe Bugi typo'd, but I can't tell.  However, we see that the only numbers that fit these restrictions are 35 and 49. So our probability is indeed $ \boxed{\frac {2}{7}}$


Let $ d(n)$ be the largest odd number which divides a given number n. Suppose that $ D(n)$ and $ T(n)$ are
defined by
$ D(n) = d(1)+d(2)+...+d(n)$,
$ T(n) = 1+2+...+n$.
Prove there exists infinitely many positive numbers n such that $ 3D(n) = 2T(n)$


inequalities
symmetry
inequalities proposed

[quote="Ulanbek_Kyzylorda KTL"]a,b,c are positive reals and a+b+c+2=abc is given prove the inequality:ab+bc+ca>=2(a+b+c)[/quote]
$ a\plus{}b\plus{}c\plus{}2\equal{}abc \iff a \equal{} \frac{b\plus{}c\plus{}2}{bc\minus{}1}$, so we are to prove:
$ \frac{(b\plus{}c\plus{}2)(b\plus{}c)}{bc\minus{}1} \plus{} bc \ge 2b\plus{}2c\plus{}\frac{2b\plus{}2c\plus{}4}{bc\minus{}1} \iff$
$ b^2\plus{}bc\plus{}c^2\plus{}2b\plus{}2c \plus{} b^2c^2 \ge 2b^2c \plus{} 2bc^2\plus{}4 \iff$
$ bc(bc\minus{}2b\minus{}2c\plus{}4) \plus{} (b^2\plus{}c^2\minus{}2bc) \minus{} (bc\minus{}2b\minus{}2c\plus{}4) \ge 0$
$ (bc\minus{}1)(b\minus{}2)(c\minus{}2) \plus{} (b\minus{}c)^2 \ge 0$
But by the pigeon hole principle there are at least two numbers in one of the intervals $ [0;2]$ and $ [2;\plus{}\infty)$. Because of symmetry we can assume that $ b$ and $ c$ are these, and we are done :) (Since $ bc\minus{}1 \ge 0 \iff a \ge 0$, so it is true!)

[quote="Ulanbek_Kyzylorda KTL"]a,b,c are positive reals and$ a\plus{}b\plus{}c\plus{}2\equal{}abc$ is given prove the inequality:$ ab\plus{}bc\plus{}ca \ge 2(a\plus{}b\plus{}c)$ [/quote]
Since $ a\plus{}b\plus{}c\plus{}2\equal{}abc$,put $ a\equal{}\frac{y\plus{}z}{x} ,b\equal{}\frac{z\plus{}x}{y} ,c\equal{}\frac{x\plus{}y}{z}$ with $ x,y,z>0$
This inequality equivalent to:
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \ge x^2(y\plus{}z)\plus{}y^2(z\plus{}x)\plus{}z^2(x\plus{}y)$
Which is obvious true by Schur inequality :wink:  :)


Determine the largest real number $ k$ such that the inequality
\[ (ka \plus{} kb \minus{} c)(kb \plus{} kc \minus{} a)(kc \plus{} ka \minus{} b)\le abc
\]
holds for all nonnegative real numbers $ a$, $ b$ and $ c$.

:)

[quote="ElChapin"]Determine the largest real number $ k$ such that the inequality
\[ (ka \plus{} kb \minus{} c)(kb \plus{} kc \minus{} a)(kc \plus{} ka \minus{} b)\le abc
\]
holds for all real numbers $ a$, $ b$ and $ c$.[/quote]
I think $ a,b,c \ge 0$,then $ k_{max}\equal{}1$

Sorry :blush:  I had writed  for all real numbers but is just for nonnegative. I have corrected it.

mmmm may be I should have put this in Pre-Olympiad... ?

:)

[quote="ElChapin"]Sorry :blush:  I had writed  for all real numbers but is just for nonnegative. I have corrected it.

:)[/quote]
If so,choose $ a\equal{}b\equal{}c$,we have $ 2k\minus{}1 \le 1$ ,then $ k{max}\equal{}1$

Suddenly I remembered why did I posted this inequality here...

I should have posted

[b]Find the greatest real number $ k$ such that the inequality
\[ (ka\plus{}kb\minus{}c)(kb\plus{}kc\minus{}a)(kc\plus{}ka\minus{}b)\le (2k\minus{}1)^3abc\] holds for all positive real numbers $ a$, $ b$ and $ c$[/b]

 :rotfl:  :)

Take $ b \equal{} c \equal{} 1$ and $ a \minus{} \minus{} > 0$

$ (k \minus{} 1)2k(k \minus{} 1)\le 0$

$ (k \minus{} 1)^2*2k \le 0$

So $ k$ cannot be greater than 1. Hence $ k \equal{} 1$ is the largest.


geometry
circumcircle
inequalities
inequalities proposed

Prove that in  every triangle the following ineq holds:

$\sum \sqrt{tg\frac{A}{2}}\leq3 \cdot \sqrt{\frac{4R+r}{3p}}$

where
R is the radious of the circumcircle
r is the radious of the inscribed circle
p is the semiperimeter
 :)

Socrates,excuse me,but your inequality is trivial.It follows from Cauchy-Schwarz: $\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{3(x+y+z)}$.Put $x=tg\frac{A}{2}$ and.... and from $ tg(\frac{A}{2})+tg(\frac{B}{2})+tg(\frac{C}{2})=\frac{4R+r}{p}$, we are done. ;)


calculus
analytic geometry
graphing lines
slope

Find the point on the circle centered at the origin of radius one that is closest to the point (2,4).

[quote="max_tm"]is the answer

[hide]
$(\frac{1}{\sqrt5},\frac{2}{\sqrt5})$?

I tried without calculus but I'm in a rush to go snowboarding and just did this in 30 seconds with derivatives... hopefully I didn't make a calculation mistake.
[/hide]

d'oh! The solution is waaaay easier than what I did originally, but I must go and cannot post it. Somebody else is probably posting a solution right now though  :P.[/quote]
That is right but you might want to rationalize the denominator. And when you get back... post your solution.

[hide]The shortest distance will be on the line which connects the origin and the point (because the tangent at the intersection of the line and circle is perpendicular to that line, making it the shortest distance).  The equation (with slope 2 and y-intercept 0) is

$y=2x$

The equation of the circle is

$x^{2}+y^{2}=1$

Plugging in,

$x^{2}+4x^{2}=1$

Solving for $x$ and $y$, we find the point $(sqrt(5)/5,2sqrt(5)/5)$.[/hide]


The perimeter of a right triangle is 42 and the sum of the squares of its sides is 722. Determine the area of the right triangle.

[hide] Let the sides of the triangle be $ a$, $ b$ and $ c$. So, $ a^2\plus{}b^2\plus{}c^2\equal{}722$. But, $ a^2\plus{}b^2\equal{}c^2$, by the Pythegorean Theorem, so $ 2c^2\equal{}722 \implies c\equal{}19$. Now, we also know that $ a\plus{}b\plus{}c\equal{}42$, so $ a\plus{}b\equal{}23$. Now,

$ (a\plus{}b)^2\equal{}a^2\plus{}b^2\plus{}2ab\equal{}529 \implies 2ab\equal{}168$, since $ a^2\plus{}b^2\equal{}361$.

The area is $ \frac{ab}{2}$, so dividing through by $ 4$ yields $ \boxed{42}$.[/hide]

[hide="Solution"]
Let the sides of the triangle be $ a,b,c$ with $ c$ as the hypotenuse.

We know:
$ a^{2} \plus{} b^{2} \equal{} c^{2}$  $ (1)$
$ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 722$  $ (2)$
$ a \plus{} b \plus{} c \equal{} 42$  $ (3)$

From $ (1)$ and $ (2)$, we can solve for $ c$:
$ 2c^{2} \equal{} 722$
$ c \equal{} 19$

Then, from $ (1)$ and $ (3)$ we have
$ a \plus{} b \equal{} 23$
$ a^{2} \plus{} b^{2} \equal{} 722 \minus{} 19^{2} \equal{} 361$

Subtracting, we get
$ (a \plus{} b)^{2} \minus{} a^{2} \minus{} b^{2} \equal{} 2ab \equal{} 168$
$ ab \equal{} 84$

$ area \equal{} \frac {ab}{2} \equal{} 42$[/hide]

Edit: too slow...


Let R be a ring and let e be a unique element in R such that ex=x for all x in R. Show that xe=x for all x in R.

A try :

[hide]

Let x in R.

For all y in R, (e - (x*e-x))*y = e*y - (x*e)*y + x*y = e*y - x*(e*y) + x*y =y - x*y + x*y = y.

So since e is unique,  e-(x-x*e)=e : x*e=x[/hide]


Show that
\[-\frac{2}{3}\cos^{3}x+\cos x = \frac{\cos x \cos 2x+2\sin x \sin 2x}{3}\]

[hide]$\cos^{3}x = \frac{3\cos x+\cos3x}{4}\Rightarrow $

$-\frac{2}{3}\cos^{3}x+\cos x =$

$-\frac{2}{3}\cdot \frac{3\cos x+\cos3x}{4}+\cos x =$

$-\frac{3\cos x+\cos3x}{6}+\cos x=$

$\frac{3\cos x-\cos3x}{6}$

Now

$\cos x = \cos(2x-x) = \cos 2x\cos x+\sin 2x \sin x$

$\cos 3x = \cos(2x+x) = \cos 2x\cos x-\sin 2x \sin x$

So $3\cos x-\cos3x =3 (\cos 2x\cos x+\sin 2x \sin x)-(\cos 2x\cos x-\sin 2x \sin x)=$

$=2\cos 2x\cos x+4\sin 2x \sin x$ and we are done[/hide]


How many sets {a, b, c } of three different, positive single-digit integers have the
property that a^2 + b^2 + c^2 is a perfect square?

Thank you!

[quote="FullMetalNaruto"]so do you just guess and check for problems like these?[/quote]

I wrote a calculator program to solve it. Unless you could eliminate many of the possibilities, guess and check would not be very efficient, as there are $ 9^{3}\equal{}729$ possibilities.

You know...Mathcounts doesnt exactly provide you a calculator for sprint problems (warm-up = sprint). And no there are not  729 sets cuz they have to be distinct and this is a combination.

First of all, the source is actually 2008 Workout 8, Problem 9, so a calculator program is a legit solution :D

Unfortunately, mismatchtea misread the problem - $ a,b,c$ are distinct, so only $ \boxed{3}$ of the sets work.

Here is a solution that will not get you killed by Zuming (but is not particularly nice...).

[hide="Solution"]
The quadratic residues modulo $ 4$ are $ 0,1$, so we use a maximum of $ 1$ odd.

Thus, we will use either $ 3$ evens or $ 2$ evens and $ 1$ odd.

If we use $ 3$ evens, the possible sums are $ 4\plus{}16\plus{}36\equal{}56$, $ 4\plus{}16\plus{}64\equal{}84$, $ 4\plus{}36\plus{}64\equal{}104$, and $ 16\plus{}36\plus{}64\equal{}126$ - no perfect squares.

If we use $ 2$ evens and $ 1$ odd, the possible sums are $ 4\plus{}16\plus{}1\equal{}21$, $ 4\plus{}16\plus{}9\equal{}29$, $ 4\plus{}16\plus{}25\equal{}45$, $ 4\plus{}16\plus{}49\equal{}69$, $ 4\plus{}16\plus{}81\equal{}101$, $ 4\plus{}36\plus{}1\equal{}41$, $ 4\plus{}36\plus{}9\equal{}49$ (!), $ 4\plus{}36\plus{}25\equal{}65$, $ 4\plus{}36\plus{}49\equal{}89$, $ 4\plus{}36\plus{}81\equal{}121$ (!), $ 4\plus{}64\plus{}1\equal{}69$, $ 4\plus{}64\plus{}9\equal{}77$, $ 4\plus{}64\plus{}25\equal{}93$, $ 4\plus{}64\plus{}49\equal{}117$, $ 4\plus{}64\plus{}81\equal{}149$, $ 16\plus{}36\plus{}1\equal{}53$, $ 16\plus{}36\plus{}9\equal{}61$, $ 16\plus{}36\plus{}25\equal{}77$, $ 16\plus{}36\plus{}49\equal{}101$, $ 16\plus{}36\plus{}81\equal{}133$, $ 16\plus{}64\plus{}1\equal{}81$ (!), $ 16\plus{}64\plus{}9\equal{}89$, $ 16\plus{}64\plus{}25\equal{}105$, $ 16\plus{}64\plus{}49\equal{}129$, $ 16\plus{}64\plus{}81\equal{}161$, $ 36\plus{}64\plus{}1\equal{}101$, $ 36\plus{}64\plus{}9\equal{}109$, $ 36\plus{}64\plus{}25\equal{}125$, $ 36\plus{}64\plus{}49\equal{}149$, $ 36\plus{}64\plus{}81\equal{}181$.

The sums that yield a perfect square are denoted by (!).  There are $ \boxed{3}$ of these.
[/hide]

Sorry for posting the wrong topic!  :blush:

But...what the heck is a "quadratic residues modulo"?

[quote="AIME15"]The quadratic residues modulo $ 4$ are $ 0$,$ 1$...[/quote]

This statement is equivalent to saying "The possible remainders of a perfect square when divided by $ 4$ are $ 0$ and $ 1$."


geometry
integration
analytic geometry
ratio
calculus
absolute value

hi frieds ,

  i m new to this form. I am a programmer.  I need to calculate the area of different polygons which is drawn by the user in the screen. User is allowed to draw any closed shape.

He can draw in two ways.

   One is by droping lines.In that case we can use surveyors formula since we know the vertices ( we can get te vetices easily while user draw using mouse) . Is this method is efficient ...??????

  another way he can draw of his own ( freeform drawing). In this case how to find the area of the closed region..?????
  .there wont be any vertices ( shall we consider each and every point on the surface as vertices...?) .

I have attached some sample irregular regions .

thanks in advance.

kannan

I would try Green's formula: the area bounded by a curve $C$ is equal to $\int_{C}x\,dy$ (same as $-\int_{C}y\,dx$), provided that $C$ is traced counterclockwise. If it's clockwise, just change the sign - or take the absolute value and forget about orientation at all. The formula will not work if the curve crosses itself, but it looks like you don't allow that to happen. It should not be hard to implement: $dy$ is just the difference of the $y$-coordinates of two "consecutive points" on the curve.

I'll remember to describe this example to my Calc 3 students.  :)

hai mlok ,

  thanks for ur reply.

  surveyor's formula is nothing but greens theorem , which i have mentioned in the query already ...?

  is this method efficient ..? else is there any other ways to calculate the  area ...???

try by montecarlo... Take a square that contains the shape. Then choose randomly N points in the square. If there are n points inside the shape, compute the ratio $nA/N$ where $A$ is the area of the square? For N large it gives you something correct...

Crux: how do you determine if a point is "inside"? That's a serious  algorithmic question - your method only works if you can do that.

On the other hand, if you can do that, why put up with the lousy accuracy of a Monte Carlo method (where to get one additional significant figure of accuracy you'd need ten times as many points)? The total number of pixels on a computer screen is not more than about $10^{6}.$ You could just brute force it and count all of the pixels that are "inside."

[quote="mlok"]With the surveyor/Green formula the computation time is linear in the length of the curve, which seems efficient to me. You can even run the computation while the curve is being drawn.[/quote]
I was always fond of the symmetric version $\frac12\oint_{C}x\,dy-y\,dx,$ not that I can see any computational advantages for that. (It takes twice the computations, and I don't see any stability issues.)

The biggest issue is that the formula is only correct for simple closed curves, and the formula itself does not provide any automatic detection of self-intersection problems. (It does detect orientation reversals.)

hello
i found the thread very interesting as i have never understood much
about green's formula or curbilinear integrals(as we call them here,the
integrals with that circle in the middle).
it is a enlightment to me to see that Green's formula comes from something
elementary and i was really not aware of this.
the new name for it surveyor seems to be much more common.
i started to read [url=http://www.maa.org/pubs/Calc_articles/ma063.pdf]this article[/url] explaining through elementary methods
how one derives Green's formula but i got stuck along the way on understanding it
at page 6 where Mean Value Theorem is said to be used.
i don't understand how exactly it is beeing used.
if someone is kind enough to explain i will remain forever indebted.
thank you


If $ p_1, p_2, p_3, p_4, p_5, p_6 > 3 \in \mathbb{P}$, prove that $ \displaystyle \sum_{i \equal{} 1}^{n}{p_i^2}$ is divisible by $ n$ for $ n \equal{} 1, 2, 3, 4, 6$.

[hide]

$ p_k\equiv 1\pmod2 \equiv 1,2\pmod3 \equiv 1,3\pmod4 \equiv 1, 5\pmod6$ (sorry if notation sucks) for $ p_k>3$

Since

$ (1)^2\equiv 1\pmod 2$, $ (1)^2\equiv (2)^2\equiv 1\pmod 3$, $ (1)^2\equiv (3)^2\equiv 1\pmod4$, $ (1)^2\equiv (5)^2\equiv 1\pmod6$

$ \displaystyle\sum^n_{i\equal{}1} p_i^2$ modulo n is $ \displaystyle\sum^n_{i\equal{}1} 1^2\equiv n \equiv 0\pmod n$

[/hide]

It is not true for $ n\equal{}4$, take the case where all the $ p_i$'s are 5.

It's clearly true in the case that the $ p_i$'s are all equal for any $ n$.

EDIT: Oh, I misread it, I thought he meant the sum of any 6 primes is divisible by all those numbers  :oops: .


calculus
integration
logarithms
trigonometry
search
calculus computations

This would help you.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2064187152&t=23498[/url]

Sorry i couldn't actually see how it was useful , imean the link . how do u solve this ??can anyone provide me the solution??. please

They solve that exat integral (with different limits) in that link. How is that not helpful? Did you even look at it? ^_^

The author of the thread number 16 of the post linked here is vijaymenon.

The linked thread only helps us with part of the problem. From the result referenced, we have

$ \int_0^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$
$ \equal{} \int_0^{\pi/4} \ln(1 \plus{} \tan x)\mbox{ }dx \plus{} \int_{\pi/4}^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$
$ \equal{} \frac{\pi}{8}\ln 2 \plus{} \int_{\pi/4}^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$.

The remaining integral is a little different to approach. Use the substitution $ 1 \plus{} \tan x \equal{} \frac{1 \plus{} t}{t}$ ($ t$ will run from $ 1$ to $ 0$, but we'll get a negative sign to reverse this in a moment). Using the identity $ \tan x \equal{} \pm \sqrt{\sec^2 x \minus{} 1}$, we can deduce that $ dx \equal{} \minus{}\frac{dt}{1 \plus{} t^2}$. Hence the integral becomes

$ \int_{\pi/4}^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$
$ \equal{} \minus{}\int_1^0 \frac{\ln \frac{1 \plus{} t}{t}}{1 \plus{} t^2}\mbox{ }dt$
$ \equal{} \int_0^1 \frac{\ln \frac{1 \plus{} t}{t}}{1 \plus{} t^2}\mbox{ }dt$
$ \equal{} \int_0^1 \frac{\ln (1 \plus{} t)}{1 \plus{} t^2}\mbox{ }dt \minus{} \int_0^1 \frac{\ln t}{1 \plus{} t^2}\mbox{ }dt$
$ \equal{} \frac{\pi}{8}\ln 2 \plus{} G$,

where

$ G \equal{} \minus{} \int_0^1 \frac{\ln t}{1 \plus{} t^2}\mbox{ }dt$

is [url=http://en.wikipedia.org/wiki/Catalan_constant]Catalan's constant[/url]. Hence the total integral is

$ \int_0^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$
$ \equal{} \int_0^{\pi/4} \ln(1 \plus{} \tan x)\mbox{ }dx \plus{} \int_{\pi/4}^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$
$ \equal{} \frac{\pi}{8}\ln 2 \plus{} \int_{\pi/4}^{\pi/2} \ln(1 \plus{} \tan x)\mbox{ }dx$
$ \equal{} \frac{\pi}{8}\ln 2 \plus{} \frac{\pi}{8}\ln 2 \plus{} G$
$ \equal{} \frac{\pi}{4}\ln 2 \plus{} G$.


given $ a,b,c$ be real numbers.Prove that:
$ (a^2\plus{}5)(b^2\plus{}5)(c^2\plus{}5) \ge\ 72(a\plus{}b\plus{}c)$  :)

My proof:
$ \prod(a^2\plus{}5)\equal{}a^2b^2c^2\plus{}5\sum a^2b^2\plus{}25\sum a^2\plus{}125$
that $ a^2b^2c^2\plus{}2\geq 2\sum ab\minus{}\sum a^2$
       $ 5\sum(a^2b^2\plus{}1)\geq 10\sum ab$
       $ 18\sum(a^2\plus{}1)\geq 36\sum a$
$ \Rightarrow\prod(a^2\plus{}5)\geq 6\sum a^2\plus{}12\sum ab\plus{}36\sum a\plus{}54\geq 72\sum a\Leftrightarrow 6(\sum a\minus{}3)^2\geq 0,Right$

[quote="karis"]My proof:
$ \prod(a^2 \plus{} 5) \equal{} a^2b^2c^2 \plus{} 5\sum a^2b^2 \plus{} 25\sum a^2 \plus{} 125$
that $ a^2b^2c^2 \plus{} 2\geq 2\sum ab \minus{} \sum a^2$
       $ 5\sum(a^2b^2 \plus{} 1)\geq 10\sum ab$
       $ 18\sum(a^2 \plus{} 1)\geq 36\sum a$
$ \Rightarrow\prod(a^2 \plus{} 5)\geq 6\sum a^2 \plus{} 12\sum ab \plus{} 36\sum a \plus{} 54\geq 72\sum a\Leftrightarrow 6(\sum a \minus{} 3)^2\geq 0,Right$[/quote]
NICE SOLUTION :)

Sorry :blush: , Arqady, but I don't understand why

$ (a^2 \plus{} 5)\left(8 \plus{} (b \plus{} c)^2\right)\geq24(a \plus{} b \plus{} c)$

[quote="arqady"][quote="quykhtn-qa1"]given $ a,b,c$ be real numbers.Prove that:
$ (a^2 \plus{} 5)(b^2 \plus{} 5)(c^2 \plus{} 5) \ge\ 72(a \plus{} b \plus{} c)$  :)[/quote]
It follows from $ (a^2 \plus{} 5)\left(8 \plus{} (b \plus{} c)^2\right)\geq24(a \plus{} b \plus{} c)$ and
$ (b^2 \plus{} 5)(c^2 \plus{} 5)\geq3\left(8 \plus{} (b \plus{} c)^2\right).$[/quote]

Thank you very much. :)

[quote="manlio"] Arqady, but I don't understand why

$ (a^2 \plus{} 5)\left(8 \plus{} (b \plus{} c)^2\right)\geq24(a \plus{} b \plus{} c)$

[/quote]
Because $ (a^2 \plus{} 5)\left(8 \plus{} (b \plus{} c)^2\right)\geq24(a \plus{} b \plus{} c)\Leftrightarrow$
$ \Leftrightarrow(a^2\plus{}5)(b\plus{}c)^2\minus{}24(b\plus{}c)\plus{}8(a^2\minus{}3a\plus{}5)\geq0$ and $ \Delta_{b\plus{}c}(a)\leq0\Leftrightarrow a^4\minus{}3a^3\plus{}10a^2\minus{}15a\plus{}7\geq0\Leftrightarrow(a\minus{}1)^2(a^2\minus{}a\plus{}7)\geq0,$
which is true.

Thank you very much Arqady, I like a lot your proof of

$ (a^2\plus{}5)(b^2\plus{}5)(c^2\plus{}5) \geq 24(a\plus{}b\plus{}c)$

Very ingenious!!! :)

[quote="quykhtn-qa1"]given $ a,b,c$ be real numbers.Prove that:
$ (a^2 \plus{} 5)(b^2 \plus{} 5)(c^2 \plus{} 5) \ge\ 72(a \plus{} b \plus{} c)$  :)[/quote]
Here is my solution:
$ (b^2\plus{}5)(c^2\plus{}5) \ge 6(b^2\plus{}c^2\plus{}4)$
$ (a^2\plus{}1\plus{}1\plus{}3)(1\plus{}b^2\plus{}c^2\plus{}3) \ge (a\plus{}b\plus{}c\plus{}3)^2 \ge 12(a\plus{}b\plus{}c)$  :)

[quote="quykhtn-qa1"][quote="quykhtn-qa1"]given $ a,b,c$ be real numbers.Prove that:
$ (a^2 \plus{} 5)(b^2 \plus{} 5)(c^2 \plus{} 5) \ge\ 72(a \plus{} b \plus{} c)$  :)[/quote]
Here is my solution:
$ (b^2 \plus{} 5)(c^2 \plus{} 5) \ge 6(b^2 \plus{} c^2 \plus{} 4)$
[/quote]
Nice! :lol:

Thank you, manlio but I think the quykhtn-qa1's proof better.

[quote="quykhtn-qa1"]Here is my solution:
$ (b^2 \plus{} 5)(c^2 \plus{} 5) \ge 6(b^2 \plus{} c^2 \plus{} 4)$[/quote]
Why should this be true??
For instance $ b\equal{}1/2$ and $ c\equal{}2$ violate this inequality.

[quote="test20"][quote="quykhtn-qa1"]Here is my solution:
$ (b^2 \plus{} 5)(c^2 \plus{} 5) \ge 6(b^2 \plus{} c^2 \plus{} 4)$[/quote]
Why should this be true??
For instance $ b \equal{} 1/2$ and $ c \equal{} 2$ violate this inequality.[/quote]
Because between all three numbers $ a,$ $ b$ and $ c$  or
exists two of them $ x$ and $ y$  such that  $ |x|\leq1$ and $ |y|\leq1$ or
exists two of them $ x$ and $ y$ such that  $ |x|\geq1$ and $ |y|\geq1.$  :wink:
But we can assume that $ x\equal{}b$ and $ y\equal{}c.$
It gives $ (b^2\minus{}1)(c^2\minus{}1)\geq0$ id est, $ (b^2 \plus{} 5)(c^2 \plus{} 5) \ge 6(b^2 \plus{} c^2 \plus{} 4).$


Give a equation $ |x\minus{}a_1|\plus{}....\plus{}|x\minus{}a_{50}|\equal{}|x\minus{}b_1|\plus{}....\plus{}|x\minus{}b_{50}|$ where $ a_1,a_2,...,a_{50},b_1,b_2,...,b_{50}$are
distinct .Find the maximum value of numbers real root of equation.

If $ a_1\plus{}...\plus{}a_{50}\equal{}b_1\plus{}...\plus{}b_{50}$ then infinetely many solutions $ x>max(a_i,b_i)$ or $ x_i<min(a_i,b_i)$.


lol this is gonna be crazy but this is a fight wit pics....first, i post a picture like a cat. then u post a pic of a dog, then the next a fire, etc.

aight i'll start

would u dare kill a penguin?

[img]http://img3.freeimagehosting.net/uploads/cb8cc3bf35.jpg[/img]

ooh, i found it!
now look at mine :P
DRIZZT WILL PWN YOU ALL

yeah, the penguin too

[quote="daermon"]http://www.forbiddenplanet.co.uk/images/C/C9170.jpg
muhaha[/quote]
but stewy doesn't beat chuck norris

[quote="Thales418"]please

someone ban this guys

because this is not funny

they are spamming the formum

ban they by IP number

lets make a petition[/quote]

ur spamming by posting this. havent u seen this? this is GAMES AND FUN FACTORY

[quote="Thales418"]when i post in G&FF is spam

when you post is a creative and beautiful post

stranger....

stop spamming[/quote]

so basically your acknoledging you spam and screw up G&FF while we are good little people that follow the rules?

jeez ur messed up....no offense

[quote="Thales418"]when i post in G&FF is spam

when you post is a creative and beautiful post

stranger....

stop spamming[/quote]

omg he has soo many enters.....i have to scroll down so much and thales ur missing the point of games and fun factory so just stay out aight?

[quote="randomdragoon"]This does though.

[img]

If you don't get it, you haven't watched enough flash/youtube movies[/quote]
I dont get it

yes

this is a good topic

what is the title?

i forget

hauhau

to funny

woooo

lets make a picture fight

street fighters

or the king of figthers?!

maybe mortal kombat

ahhauhau

or other things

hehe

but

what

is

the

purpose

of

the

topic1?

i dont see that

hm......

err.....

hm...

bye bye

see you on moon

yeah

thanks

baka

[quote="Thales418"]bs[/quote]
now THAT was retarded
is it not possible to get him banned?

this is a fun forum

i was making fun

lets laugth together people

for a better world

i have friends. just for largers $n$'s

lets make a topic

dragoons does not ecxist

hehe

padra que dedo

cool

read a calc book

[quote="randomdragoon"]This does though.

[img]http://wiredforbooks.org/images/FredRogers4.jpg[/img]

If you don't get it, you haven't watched enough flash/youtube movies[/quote]

http://www.bocatinta.com/bt/wp-content/apsm.jpg

muhaha
i win

but drunken toy monkeys are supercool :D

[quote="randomdragoon"]d3m0n, I suggest that you rename the topic title to "Thales is a moron" and then restart this game.[/quote]

am i allowed to do it? if so, i will


Prove or disprove this for positive a,b and c!
$ {a}^{3/4}\plus{}{b}^{3/4}\plus{}{c}^{3/4}\geq \frac{3}{\sqrt{2}}\sqrt{\sqrt{a}b\plus{}\sqrt{a}c}$

[quote="k-boy"]Prove or disprove this for positive a,b and c!
$ {a}^{3/4} \plus{} {b}^{3/4} \plus{} {c}^{3/4}\geq \frac {3}{\sqrt {2}}\sqrt {\sqrt {a}b \plus{} \sqrt {a}c}$[/quote]
$ a\rightarrow 0,b\equal{}10,c\equal{}10$

[quote="k-boy"]Prove or disprove this for positive a,b and c!
$ {a}^{3/4} \plus{} {b}^{3/4} \plus{} {c}^{3/4}\geq \frac {3}{\sqrt {2}}\sqrt {\sqrt {a}b \plus{} \sqrt {a}c}$[/quote]
It's wrong. Try $ a\equal{}1,$ $ b\equal{}\sqrt[3]2$ and $ c\equal{}0.2.$ :wink:


Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$, prove that
\[ (1-x)(1-y)+(1-y)(1-z)+(1-z)(1-x)\leq0. \]