articles geometry inequalities Princeton college AMC AIME

If I very thouroughly read AoPs Volume 1, and solve all the problems, what score can I expect on the AMC 12  (Please give a range here)?  Thanks!

I'd say 100-120, but that depends on a lot of things - mainly, your creativity and insight, and perhaps your background in math.  I will say that if you read AoPS, your score will improve.

But scores aren't everything, and the real benefit of reading the AoPS and other such books is to have fun and recognize the beauty of math.

i'd say anywhere from 85-120 depending on the difficulty of the amc, everything mathfanatic said, how comfortable you are with the material, and how many careless errors you make.

[quote="samus"]Well, I got 105ish withot even cracking a study book last year. New projections?:[/quote]

I think the AoPS books are wonderful, and want to work through them both myself after nagging my son to get through them, but it's just so hard to say what they will do for your score. The online article "[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop Making Stupid Mistakes[/url]," also by Richard Rusczyk, seems to be at least as needed by some competitors as more problem sets.

Did you get last year's score under actual competition conditions? That's the main thing: to get the score that counts under conditions that count, if you're keeping score. And, yeah, I agree with what has already been said in this thread about pursuing math for its fun and beauty and not obsessing about scores.

[quote="tokenadult"]The online article "[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop Making Stupid Mistakes[/url]," also by Richard Rusczyk, seems to be at least as needed by some competitors as more problem sets. [/quote]

Hehe; tokenadult, do you have that article's URL memorized yet?  Granted, it is a helpful article.

Samus, don't worry about scores.  I would guess that a 120 is attainable next year after reading AoPS1, but that's up to you - will you truly think as hard as you can before finally giving in and looking at the solutions book?  (I'll admit that sometimes I just give up myself.)  Regardless, if you read the first volume, then you should (and probably will want to) read the second one, too, which talks more about complex numbers, analytic geometry, inequalities, and other useful (and mroe importantly, interesting) topics.  After reading that, you should be well on your way to, say, a 150.

But the AoPS books aren't just study books - that's the sort of thing Princeton Review and REA dumb down and grind up.  The AoPS books are the real deal.  Or as close as you can get, anyway.

jeez.
105 without any preparations?

wow.
that's pretty impressive

The AMC 12 is not the SAT - there is no real studying or preparing for it. You can do lots of problems - but no guarantee anything on it will even remotely resemble what you've seen in practice. You practice the test format - but this doesn't mean anything unless you can solve the problems.

That having been said, over half my school's AIME qualifiers take the exam without any sort of practice for it. Unfortunately they are always in the 100-110 bracket. So I don't think Samus' score is extraordinary, although it is good. I do think that an ability to do a majority of the AOPS I problems can be quite helpful.

Yes, I did the AMC 12A the real setting, getting 105.5 (AMC10-118). The thing that really killed me was geometry (around 6-7 wrong, nearly all), especially circles.  I think I solved 13 (knew 10 for sure, guessed on 3), got one wrong, and left 11 blank. Will the books teach me the "tricks" needed to solve certain problems? And Tetrahedr0n, while I know my score isn't extraordinary, I'd like to point out I had never heard about the AMC until the day I took the AMC 10 and AMC 12 last year. I have never studied competition math until about a month ago. Just more info :lol: for you guys.

P.S: Tetrahedr0n, I'm guessing you are from either Troy High School, Livonia Sci-Tech, Andover High School (Bloomfiled), or Bloomfield Marian High School. Those schools are proven to be great in competition math. However, my school in Rochester, had only 2 AIME Qualifiers, and the highest score in the MMPC was only 24 Part 1 (me!).

I wonder how much Vol.2 will help me on AMC and AIME if I go through it.

Volume 1 helped me a lot. But, volume 2 helped more (Although I am not done with in yet). So far, it's really good. I'm hoping for around 120 on the AMC 12. It will be my first year taking it, and also around 130-150 on the AMC 10 - first year also. I don't know why I didn't bother to take them both last year, but that's just silly planning on my part.

[quote]And Tetrahedr0n, while I know my score isn't extraordinary, I'd like to point out I had never heard about the AMC until the day I took the AMC 10 and AMC 12 last year. I have never studied competition math until about a month ago. Just more info  for you guys.
[/quote]

Sorry if my comment came out as an insulting one, I do think your score is quite good, especially if you haven't been involved in competitive math for a while! In my opinion, the AoPS books teach a lot of techniques that are nearly invaluable from problem solving, so yeah,  your scores will definately go up from reading and using it.

[quote]P.S: Tetrahedr0n, I'm guessing you are from either Troy High School, Livonia Sci-Tech, Andover High School (Bloomfiled), or Bloomfield Marian High School. Those schools are proven to be great in competition math. However, my school in Rochester, had only 2 AIME Qualifiers, and the highest score in the MMPC was only 24 Part 1 (me!).[/quote]

You guessed pretty well - I am from Troy High School, altough lately, I wouldn't call our school good in competition math, not after a few kids graduated a year and a half ago. Great to meet someone else from MI, especially from this area on AoPS! Seeing as how MMPC Part I is coming up soon, best of luck to you! 2 mistakes, 2 points on part I last year cost me a silver medal, so I'm going to need luck as well.

I've probably seen you in Mathcounts if you're a 10th grader or less (though, I didn't do to well, once again :) ;no prep at all). Yeah, Troy should be very dominant this year.... Richard Tao, Sunil Agarwal, Dawson Zhou among others (probably you!). Boulan Park was dominant at Mathcounts. Do you go to IAE? About the MMPC.... :blush:....... maybe I should start preparing for it. Last year was embarrasing. What are you doing in preparation? See ya!

[quote="tetrahedr0n"]Haha I'm not one of the people you mentioned! I did beat them in MMPC, but last year sucked for us in the AIME. Dawson, Richard and I got 6s. I'm not really preparing anything for the moment, but I will eventually do some serious practice MMPC part Is. Last year I got 34 and that really cost me...[/quote]

I'm aiming for 30/10 in the MMPC (last year 24/didn't take) in order to get a bronze. Need a lot of work, though: on the practice tests from previous years, I'm getting ~27/5.  How did Sunil do on the AMC/AIME? I still remember him from Mathcounts....... quick mind.  The dynasty of Agarwals/Guptas has faded though, since the last few years. What math classes are you planning to take? See ya!

Sunil scored 103 on the AMC 12 and 4 on the AIME, which is not quite as good as mathcounts, where he was clearly the state champion every year! As for MMPC, 30/10 isn't too hard to get, so I'm sure you'll achieve it if you practice a bit. Last year I got 34/15, so this year I'm just hoping to improve my part two score to 25 or so.

If i get through Art of Problem Solving: Volume 2, and I'm currently in Calc BC, what are my chances of qualifying for the USAMO? I got through about the first 10 chapters of Volume 1 since I was borrowing it from a friend. Any feedback will be appreciated!

Haha..I used to be obsessed with that..I had my whole little plan to get to the USAMO and whatnot...but of course, I had to come to this site, and see that everyone is 100x smarter than me.  :-P  So I guess the real..um, response to that question is, "QUIT DAYDREAMING ABOUT 'MAKING IT IN THE MATH WORLD' AND PICK UP THE AoPS BOOKS!"  As long as you're having fun with mathematics, who cares how far you go. :-P  Just don't get too full of yourself. :P (I'm having a hard time taking my advice, so don't feel bad.   :lol: )

(Four smiley faces in one reply, I rock.)

[quote="xqpx"][quote="tetrahedr0n"]I have no idea how to answer your question, but certainly taking calculus won't really help you.[/quote]
I agree. It is not only because that AMC doesn't have calculus problems (I could be wrong), but also because that doing cookbook exercises (for a course) is a lot easier than and very different from problem solving (e.g. solving AMC and AoPS type of  problems). I took calculus and DE courses before I took AMC8 last year. I didn't even get to all the problems in time. :blush:[/quote]

No calculus problems on the AMC, although calculus could be useful in certain cases.

Having taken calculus BC is only useful in the sense that it's an exercise in certain aspects of problem solving, such as determining a quantity in several different ways or choosing the best angle to approach from. (For example, in finding volumes of rotations, one must choose between using washers and shells; understanding how the two methods are derived and which cases they are useful requires a certain type of thinking that will be helpful on math contests).

[quote="xqpx"]I took calculus and DE courses before I took AMC8 last year.[/quote]

Wow! Differential equations before losing eligibility for AMC 8 is a good bit of math acceleration, even by international standards.

But your point if very well taken. Course exercises, even exercises for hard "college-level" courses, are not the same as competition problems--usually the competition problems are harder, even if their curriculum level is much lower. That's why Richard Rusczyk advises not falling into "[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]The Calculus Trap[/url]" by ignoring opportunities for challenging math activities outside the standard curriculum. (Note, he does say that students who are able to do it should be studying calculus at ages that are "early," even by international standards, but they should also be learning the precalculus subjects in much greater than usual depth, and learning problem-solving techniques in general.)

We're trying to do it "both-and" here, accelerating the standard curriculum (although differential equations will be AFTER his AMC 8 eligibility is past in my son's case) and getting heavily into competition culture too. Both have their benefits. I have no prediction how well my son will do on AMC tests this school year, except that it will most likely be better than last year. Year-on-year growth in understanding is the thing to look for from the AMC tests.

Good luck, everyone.

At some point in college, you have to start thinking to solve the problems. In all my classes last year (some upper division undergraduate classes and some graduate classes), we were only asked to solve ``real'' problems (i.e. they should take you more than a few minutes to solve even if you have a very strong grasp on the material).

[quote="SingTheSorrow"]Just don't get too full of yourself....Four smiley faces in one reply, I rock.[/quote]

Hypocrite  :P

P.S.
 :P  :P  :P  :P  :P  :P  :P


well if you add them all together you will get
$x_1+x_2+x_3+\dots+x_n=1$.
that all i have now...ill try and finish it later. :)


inequalities
geometry
circumcircle
inequalities proposed

For a triangle with sides $ a,b,c$ and the circumradii $ R$, prove that
\[ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\frac{abc}{\sqrt{3}R}\geq 4(abc)^{\frac{2}{3}}.\]

[quote="xavi"]If you prove that

$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}12r^{2}\geq 4\left( abc\right)^{\frac{2}{3}}$,

you are done. (I have no proof of that inequality yet)[/quote]

so is it just a conjecture or verified result?

anyhow, i gotta edit the inequality i posted:

For a triangle [b]whose angles do not exceed $ \frac{2\pi}{3}$[/b], which has sides $ a,b,c$ and the circumradius $ R$, prove that
\[ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\frac{abc}{\sqrt{3}R}\geq 4(abc)^{\frac{2}{3}}.\]

ok.. for such triangle, there exists the Fermat point inside the triangle. Let the Fermat point be $ F$ and let $ FA \equal{} x,FB \equal{} y,FC \equal{} z$. Then after some calculations, the original inequality is equivalent to $ 8(x^{2}\plus{}xy\plus{}y^{2})(y^{2}\plus{}yz\plus{}z^{2})(z^{2}\plus{}zx\plus{}x^{2})\leq (x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}xy\plus{}yz\plus{}zx)^{3}$, which can be checked by an expansion. I think this inequality on $ x,y,z$ is good too.

thank you Kim, it is a very nice solution.

[quote="xavi"]If you prove that

$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}12r^{2}\geq 4\left( abc\right)^{\frac{2}{3}}$,

you are done. (I have no proof of that inequality yet)[/quote]

[quote="xavi"]It was just a conjecture, but now I see that if triangle have an angle near to $ \pi$, it is not true. Maybe it is true also for triangles whose angles do not exceed $ \frac{2\pi}{3}$.
Could you post your proof of the inequality?[/quote]

Your inequality is false in your condition, see the fig however I think the condition must be triangle is acute, but I haven't solve!

@Sung-yoon Kim I known about a problem like you
\[ (x^{2}\plus{}xy\plus{}y^{2})(y^{2}\plus{}yz\plus{}z^{2})(z^{2}\plus{}zx\plus{}x^{2})\leq\frac{3}{64}(x\plus{}y\plus{}z)^{6}\]
Maybe it is stronger than yours but equal $ x \equal{} y,z \equal{} 0$ and other permutions.


I have seen people like GoogleBot, MSNbot, Yahoo Slurp, and i have noticed they have black names instead of the regular blue.  I realize they are robots, but why do they get special colors.  I want special colors too. :P

nevermind, i see now.  But why do we need robots? Are they ever going to post?

Those are the crawlers from Google that were always of the site. Now they are automatically logged in, there's nothing special about that.

There is no way to change your color (for now). There will be one in the future.

[quote="Inspired By Nature"]What is their purpose :?:[/quote]

It's probably some mix of site indexing and taking over the world.


number theory
greatest common divisor
number theory unsolved

Find all $a,b,c$ integers such that
$a|(b-c)$ and $b|(c-a)$ and $c|(a-b)$

[quote="Fabian May"]Let $t_{1}$ be $gcd(a,b)$.
Since c $\mid$ (a-b) , we know  $c \mid t_{1}$.
As well: $b-c \equiv 0 (mod \ a)$
So: $t_{1}\mid c$. So finally c = $t_{1}$.
WE can do the same with a,b and get: a=b=c, which is obviously true.[/quote]
It is not correct. Let m, n  are integers, suth that $2|mn, n\not =2$, then $a=m,c=\frac{mn}{2},b=a-c$ is solution.

ok, have definitely been writing nonsense.
well, one thing was right at least:$gcd(a,b) \mid c$ so we can reduce it to gcd(a,b) = gcd(a,c) = gcd(b,c) = 1.
Case 1: w.l.o.g.: b=c: so $a \equiv 0 (mod \ b)$,
$L_{1}={(t*q/t/t)}$ as $t = \pm gcd(a,b,c)$and q integer.
Case 2: $b \neq c$
w.l.o.g.: |a|>|b|,|c|
Case 2.1.a>0:
$b \equiv c (mod \ a)$
w.l.o.g.: b>0: $\Rightarrow$ c = -a+b, so c negativ.
$II) a \equiv-a+b (mod \ b) \\ 2a \equiv 0 (b)$
Since gcd(a,b)=1, we know: b=1,2
Case 2.1.1. b=1: $L_{2}={(t*q/t/t*(1-q))}$ with q > 1.
Case 2.1.2. b=2: $L_{3}={(t*q/2t/t*(2-q))}$ with q > 2 and odd.
Case 2.2.a<0:
w.l.o.g.: b>0, so c = a+b
$III) a-b \equiv 0 (mod \ (a+b)) \\ 2a \equiv 0 (a+b) \\ so \ (a+b) \mid 2a \\ gcd (a+b,2a) = gcd(a+b,2b) \le gcd(2a,2b)=2.$
This leads us to 2 cases.
Case 2.2.1. a+b = c = -1
$L_{4}={(t*q/t*(-1-q)/-t)}$with q < 1.
Case 2.2.2. a+b = c = -2
$L_{5}={(t*q/t*(-2-q)/-2t)}$with q < 2 and odd.


Gauss
geometry
geometric transformation
homothety
ratio
analytic geometry
geometry unsolved

Let $ ABC$ be a triangle. $ D,E,F$ are in $ BC,CA,AB$, respectively such that $ AD,BE,CF$ are concurrent. Denote $ A_1,B_1,C_1$ be midpoints of $ BC,CA,AB$ and $ D_1,E_1,F_1$ be midpoints of $ EF,FD,DE$, respectively.

Show that: $ A_1D_1,B_1E_1,C_1F_1$ are concurrent./.

First, let me restate the problem so that it may fits my solution.

[b]PROBLEM-[/b][b][color=darkblue]Let $ ABC$ be a triangle and $ A_0,B_0,C_0$ respectively be the point such that $ AA_0,BB_0,CC_0$ are concurrent at $ J$. Let $ M_a,M_b,M_c$ respectively be the midpoints of $ BC,CA,AB$; $ P_a,P_b,P_c$ respectively are the midpoints of $ B_0C_0,C_0A_0,A_0B_0$. Prove that $ M_aP_a,M_bP_b,M_cP_c$ are concurrent.[/color][/b]

[i]Solution[/i]
Let $ N_a,N_b,N_c$ respectively be the midpoints of $ JA,JB,JC$. Let $ H_1,H_2$ respectively be the orthocenters of $ \triangle C_0BJ$, $ \triangle B_0CJ$. Let $ C_1,J_1,B_1$ respetively be the projections of $ H_1$ onto $ BJ,BC_0,JC_0$. Let $ B_2,C_2,J_2$ be the projections of $ H_2$ onto $ CJ,JB_0,CB_0$. It is followed then $ \overline {H_1C_1}.\overline {H_1C_0} \equal{}$ $ \overline {H_1B_1}.\overline {H_1B} \equal{}$ $ \overline {H_1J_1}.\overline {H_1J}$ and $ \overline {H_2C_2}.\overline {H_1C} \equal{}$ $ \overline {H_2B_2}.\overline {H_2B_0} \equal{}$ $ \overline {H_2J_2}.\overline {H_2J}$, which implies that $ \mathcal {P}_{H_1/(M_a)}$ $ \equal{} \mathcal {P}_{H_1/(N_a)}$ $ \equal{} \mathcal {P}_{H_1/(P_a)}$, where $ (M_a),(N_a),(P_a)$ repsetively are the circles with diameters $ BC,AJ,B_0C_0$. Simliarly for $ H_2$. Thus $ H_1H_2$ are the radical axes of $ (M_a),(N_a),(P_a)$ $ \Longrightarrow M_a,N_a,P_a$ are collinear. With the same argument, we also have $ (M_b,N_b,P_b)$, $ (M_c,N_c,P_c)$ are collinear. Let $ d_a$ be the line which contains $ M_a,N_a,P_a$, define the same for $ d_b,d_c$.

Consider the homothety with center $ J$, ratio $ k_1 \equal{} 2$. We obtain $ \mathcal {H}(J,k_1): N_a\mapsto A$, $ N_b\mapsto B$, $ N_c\mapsto C$. Therefore $ \mathcal {H}(J,k_1): \triangle N_aN_cN_c\mapsto \triangle ABC$. Now consider the homothety with center $ G$- the centroid of $ \triangle ABC$, ratio $ k_2 \equal{} \minus{} \frac {1}{2}$, we obtain $ \mathcal {H}(G,k_2): A\mapsto M_a$, $ B\mapsto M_b$, $ C\mapsto M_c$. Hence $ \mathcal {H}(G,k_2): \triangle ABC\mapsto \triangle M_aM_bM_c$. Therefore $ \mathcal {H}(G,k_2)$ $ \circ$ $ \mathcal {H}(J,k_1):$ $ \triangle N_aN_bN_c\mapsto M_aM_bM_c$. But $ \mathcal {H}(G,k_2)$ $ \circ$ $ \mathcal {H}(J,k_1)$ $ \equal{} \mathcal {H}(G', \minus{} 1)$. Hence $ M_aN_a,M_bN_b,M_cN_c$ are concurrent $ \Longrightarrow d_a,d_b,d_c$ are concurrent at $ G'$, which implies to the result of the problem. $ \square$

There is another interesting result about this configuration.

[b]Proposition.[/b] If $ P$ is the concurrency point of of the cevians $ AD,BE,CF,$ then the lines $ AD_1,BE_1,CF_1$ concur in the center of the inconic $ \mathcal{K}$ with perspector $ P.$

[b]Proof.[/b] The concurrency of $ AD_1,BE_1,CF_1$ follows immediatly by the Cevian Nest Theorem. Using barycentric coordinates with respect $ \triangle ABC,$ we have

$ P \ (u: v: w) \ , \ D \ (0: v: w) \ , \ E \ (u: 0: w) \ , \ F \ (u: v: 0)$

$ D_1   \left(\frac {2u \plus{} w \plus{} v}{wv}: \frac {1}{u} \plus{} \frac {1}{w}: \frac {1}{u} \plus{} \frac {1}{v} \right)$

$ E_1  \left(\frac {1}{v} \plus{} \frac {1}{w}: \frac {u \plus{} 2v \plus{} w}{uw}: \frac {1}{u} \plus{} \frac {1}{v}\right)$

$ F_1  \left(\frac {1}{v} \plus{} \frac {1}{w}: \frac {1}{u} \plus{} \frac {1}{w}: \frac {u \plus{} v \plus{} 2w}{uv} \right)$

Therefore, triangles $ \triangle ABC$ and $ \triangle D_1E_1F_1$ are perspective through

$ U  \left(\frac {1}{v} \plus{} \frac {1}{w}: \frac {1}{u} \plus{} \frac {1}{w}: \frac {1}{u} \plus{} \frac {1}{v} \right)$

Complement of the isotomic conjugate of $ P,$ i.e. center of the inconic $ \mathcal{K}$ with perspector $ P.$

[quote="livetolove212"]First, I will change the names of points such that available to my solution :D
Let $ T$ be the intersection of $ AA_2,BB_2,CC_2, G$ be the centroid of triangle ABC.
Applying [b]Gauss[/b]'s line, we have $ M,A_3,A_1$ are collinear.
Applying [b]Menelaus[/b]'s theorem for triangle $ ATG$ with line $ (MVA_1)$ we get:
$ \frac {VG}{VT}.\frac {MT}{MA}.\frac {A_1A}{A_1G} \equal{} 1$
Then $ \frac {VG}{VT} \equal{} \frac {1}{3}$
Similarly we obtain $ B_1B_3, C_1C_3$ also pass through $ V$.
And we are done.[/quote]
It's not your problem. It has been posted at http://www.mathlinks.ro/viewtopic.php?t=214231 .

[quote="vittasko"]See also at the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150123]Concurrent lines II[/url], of [b][size=100]April[/size][/b].
Kostas Vittas.[/quote]

[quote="vittasko"]An interesting result in this configuration, is that the points $ P,$ $ Q,$ $ G,$ where $ G$ is the centroid of $ \bigtriangleup ABC,$ are collinear such that $ PQ \equal{} 3(QG),$ mentioned by [b][size=100]Little Gauss,[/size][/b] in a [b][size=100]gemath's[/size][/b] problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=132528
Kostas Vittas.[/quote]

The above property that you mentioned above is obvious in my proof for this problem at this topic. Since $ \mathcal {H}(G,k_2)\circ \mathcal {H}(J,k_1)\equal{}\mathcal {H}(G',\minus{}1)$, where $ G'$ is the concurrent point of $ d_a,d_b,d_c$. Hence $ G',J,G$ are concurrent

[quote="thaithuan_GC"]
It's not your problem. It has been posted at http://www.mathlinks.ro/viewtopic.php?t=214231 .[/quote]
Sorry I don't know it was posted on Mathlinks before. Last week I discovered it  and then my friend [b]ll931110[/b] post at here :blush:

[quote="livetolove212"]More interesting result: Denote $ G_a, G_b,G_c$ the centroid of $ TBC, TCA, TAB.$ Prove that $ AG_a, $ $BG_b,$ $ CG_c,$ $ A_1A_3,$ $B_1B_3, $ $C_1C_3$ are concurrent.[/quote]
Using the same notation of your figure above. The centroid $ G_a$ of  $ \triangle TBC$ has barycentric coordinates $ G_a  (u: u \plus{} 2v \plus{} w: u \plus{} v \plus{} 2w).$ Thus $ AG_a$ passes through  $ V \ (2u \plus{} v \plus{} w: u \plus{} 2v \plus{} w: u \plus{} v \plus{} 2w)$ and this point satisfies the equation of the Newton line of the quadrangle $ TB_2AC_2,$ namely

$ \mathcal{N}_a \equiv (w \minus{} v)x \plus{} (2u \plus{} v \plus{} w)(y \minus{} z) \equal{} 0$

Thus, $ AG_a,BG_b,CG_c,\mathcal{N}_a,\mathcal{N}_b,\mathcal{N}_c$ concur at the complement of the complement of  $ T.$

[quote="livetolove212"]More interesting result: Denote $ G_a, G_b,G_c$ the centroid of $ TBC, TCA, TAB$. Prove that $ AG_a, BG_b, CG_c, A_1A_3, B_1B_3, C_1C_3$ are concurrent. :)[/quote]
$ \bullet$ Because of $ TG_{b} \equal{} 2G_{b}B_{1}$ and $ TG_{c} \equal{} 2G_{c}C_{1},$ we conclude that the lines through $ G_{b},\ G_{c}$ and parallel to $ AC,\ AB$ respectively, intersect the line segment $ AT,$ at the same point so be it $ A',$ such that $ TA' \equal{} 2A'A.$

It is easy to show now, that the points $ B'\equiv TB\cap A'G_{c}$ and $ G_{a}$ and $ C'\equiv TC\cap A'G_{b},$ are collinear and $ B'C'\parallel BC.$

Because of $ \bigtriangleup G_{a}G_{b}G_{c}$ is the medial triangle of $ \bigtriangleup A'B'C',$ we have that $ G_{b}G_{c}\parallel B'C'$ $ \Longrightarrow$ $ G_{b}G_{c}\parallel BC$ and similarly, $ G_{a}G_{b}\parallel AB$ and $ G_{a}G_{c}\parallel AC.$

So, the triangles $ \bigtriangleup G_{a}G_{b}G_{c},\ \bigtriangleup ABC$ are perspective and based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the line segments $ AG_{a},\ BG_{b},\ CG_{c}$ are concurrent at one point so be it $ P.$

$ \bullet$ We will prove now, that the line segment $ A_{1}A_{3},$ where $ A_{1},\ A_{3}$ are the midpoints of the segments $ BC,\ B_{2}C_{2}$ respectively, passes through the point $ P.$

Because of $ G_{a}G_{b}\parallel AB,$ based on the [b][size=100]Thales theorem[/size][/b], we have that $ \frac{PA}{PG_{a}} \equal{} \frac{AB}{G_{a}G_{b}} \equal{} 3$ $ ,(1)$ because of $ AB \equal{} \frac{3}{2}A'B'$ and $ G_{a}G_{b} \equal{} \frac{1}{2}A'B'$

Let be the point $ M\equiv AT\cap A_{1}P$ and applying the [b][size=100]Menelaos theorem[/size][/b], in triangle $ \bigtriangleup ATG_{a}$ with transversal $ A_{1}PM,$ we have that $ \frac{MA}{MT}\cdot \frac{A_{1}T}{A_{1}G_{a}}\cdot \frac{PG_{a}}{PA} \equal{} 1$ $ \Longrightarrow$ $ MA \equal{} MT$ $ ,(2)$

$ \bullet$ Because of now, the complete quadrilateral $ AB_{2}TC_{2}BC,$ we conclude that the line segment $ A_{1}P\equiv A_{1}M,$ connecting the midpoints $ A_{1},\ M$ of $ BC,\ AT$ respectively, based on the [b][size=100]Newton theorem[/size][/b], passes through the midpoint $ A_{3}$ of $ B_{2}C_{2}.$

That is, the line segment $ A_{1}A_{3}$ as the problem states, passes through the point $ P\equiv AG_{a}\cap BG_{b}\cap CG_{c}$ and similarly for the line segments $ B_{1}B_{3}$ and $ C_{1}C_{3}.$

Hence, the line segments $ AG_{a},\ BG_{b},\ CG_{c},\ A_{1}A_{3},\ B_{1}B_{3},\ C_{1}C_{3},$ are concurrent at point $ P$ and the proof is completed.

Kostas Vittas.

Here is my solution:
Let $ M$ be the midpoint of $ BC, G$ be the centroid of triangle $ ABC, AG_a\cap TG\equal{}\{V\}$
Applying [b]Menelaus[/b]'s theorem: $ \frac{G_aT}{G_aM}.\frac{AM}{AG}.\frac{VG}{VT}\equal{}1$
$ \Rightarrow \frac{VG}{VT}\equal{}\frac{1}{3}$ ...


number theory
greatest common divisor
least common multiple
absolute value

Suppose a, b, c and d are integers such that lcm(a, b, c, d)= a + b + c + d.
Prove that 3|abcd, 5|abcd or 15|abcd.

Hmm, why the 15|abcd? Won't this imply the other two anyway?

Dunno ... I thought this might seem clearer (you know, English is not my mother tongue) and I didn't know how to formulate it otherwise.

Can you give a hint? :-)

I've managed to find a couple of sets that work, prove that we can assume that gcd (a,b,c,d) can be 1 and then multiply all solutions (a,b,c,d) by k to get another solution, and show that 2 must be even and 2 odd if the gcd is 1. I've also found a formula for lcm (a,b,c,d) in terms of the product of the gcds of subsets of (a,b,c,d), but otherwise haven't gotten anywhere.

I think I might have solved this once. The best starting idea is this: we assume 0<=a<=b<=c<=d, and because d|a+b+c+d it means that a+b+c+d=2d, 3d, or 4d, and if at least one of them is <0 we take the one with largest absolute value and we do the same thing as we did with d. This might not be right; I'm too lazy to take it further anyway... :D


Let O be a vertex of a 1 x 1 x 2 rectangular block (cuboid). For a point P on the surface of the block, let d(P) be the minimum of the lengths of paths from O to P along the surface of the block. What is the maximum of d(P)?

This $ f(n)$ is badly underdefined.  For example, $ f(3)$ can be any odd number whatsoever.


I think, the formula of Calculus looks more impressive than that of Algebra.  :D

Is this a commercial production and if so will you be offering a fee to whoever supplies you with this prop?  Do your other suppliers work for free?

I guess some figures and complicated formulas are what you are looking for?

Why don't you just take a look at the posts in this Forum? Many of them are exactly what you ask for: pages of written solutions. You may even find a problem you like and a solution you (or your actor) can understand if you look at more elementary sections, which will, probably, help you to have a natural face expression and behavior of a person writing down a solution of a problem rather than that fancy look you get when trying to concentrate on writing some gibberish (of course, a trained actor can, in principle, mimic the outlook of a math. student but it is much easier to have it come the natural way). As to the demand that they should look impressive, in my humble opinion, the most impressive thing you can do in the film is to write something that makes perfect sense. Also, actual proofs usually look much neater on the paper or on the blackboard than an artificially designed senseless array of symbols. And do not forget about pictures: there are very few pictures here because creating and uploading computer graphics is generally a headache, but quite often, when solving a problem, one draws more than writes. And the pictures and diagrams that make sense look very nice too, believe me.

[quote="LisaD"]Hi,
I work for a production company in Australia and for a film we're doing we need a guy writing solutions to maths problems. Only I am not a mathematician and have no clue what to write! Would someone be able to post say a page of maths solutions (doesnt matter if it makes sense), it just has to look impressive.

Thank you very much!

Lisa[/quote]Lisa,
You probably want to have this post moved to the Round Table, or some other general discussion forum. (The one you posted to is for unsolved problems in Complex Analysis -- a particular subfield of mathematics.)

As long as I'm answering this, I should tell you that as a mathematician nothing spoils my suspension of disbelief faster than maths that are obvious nonsense. If you want to do this right, hire an actual mathematician to consult on your film. (In lieu of that, you'll at least want math that has apparent application to the plot.)


Souce: 360 PMC
Let $a$, $b$, $c$ be nonzero real numbers such that $a+b+c=0$ and $a^3+b^3+c^3=a^5+b^5+c^5$. Prove that
$a^2+b^2+c^2=\frac{6}{5}$.

[hide="Hint & Remark"]
Well, I knew that $a^3+b^3+c^3=3abc$ when $a+b+c=0$...
however, I didn't know that $a^5+b^5+c^5=-5abc(ab+ac+bc)$, though I finally derived this using sum of Newtons (i learned this on AoPS v2  :D ). Does anyone have better way of deriving this?
[/hide]

[quote="Adunakhor"]How are you two getting $a^3+b^3+c^3 = 3abc$?

When I expand $(a+b+c)(a^2+b^2+c^2 - a - b -c)$, I get $a^3+b^3+c^3 - a(b^2+c^2) + b(a^2+c^2) + c(a^2+b^2) - 2(ab+bc+ac)$, which doesn't seem to show anything.... :huh: Unless I messed up somewhere.[/quote]

That's a typo probably, actually a useful and well known identity is

$\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2 = a^3 + b^3 + c^3 -3abc$

it for example allows you to prove that the R.H.S is $\geq 0$

Daniel

[quote="calculuslover800"]Question: What is the rest of that identity you gave for $a^5+b^5+c^5$?  Because obviously it was simplified using the fact that $a+b+c=0$.  And how exactly did you derive it?

[hide="using that hint, solution"]
Since $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-a-b-c)$, and $a+b+c=0$,  $a^3 + b^3 + c^3 = 3abc$.

Similarly (somehow or another), $a^5 + b^5 + c^5 = -5abc(ab+ac+bc)$.  Since these two sums are equal, we have:

$3abc = -5abc(ab+ac+bc)$
$ab + ac + bc = -3/5$

$0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$
$0 = a^2 + b^2 + c^2 - 6/5$

Therefore, $a^2 + b^2 + c^2 = 6/5$

[/hide][/quote]
Basically, I defined a function whose roots are $a$, $b$, $c$:

$f(x)=x^3-(a+b+c)x^2+(ab+ac+bc)x-3abc$.
Then I used Newton's sum to find $a^5+b^5+c^5$....

yea newtons sums works

[hide]

you have function $f(x) = x^3 - (a+b+c)x^2 + (ab+ac + bc)x - abc$

and then define $s_n = a^n + b^n + c^n$ and $c_n$ as the coefficient in front of the $x^{3-n}$ term

since $s_1 = 0$, we know that $s_1 + c_1 = 0 \implies c_1 = 0$

then $s_2 + c_1s_1 + 2c_2 = 0 \implies s_2 = -2c_2$

also $s_3 + c_1s_2 + c_2s_1 + 3c_3 = 0 \implies s_3 = -3c_3$

continuing this we see that:

$s_5 = 5c_2c_3 = s_3 \implies c_2 = -3/5$

and so $s_2 = -2c_2 = 6/5$

[/hide]

[quote="kimby_102"][quote="Adunakhor"]How are you two getting $a^3+b^3+c^3 = 3abc$?

When I expand $(a+b+c)(a^2+b^2+c^2 - a - b -c)$, I get $a^3+b^3+c^3 - a(b^2+c^2) + b(a^2+c^2) + c(a^2+b^2) - 2(ab+bc+ac)$, which doesn't seem to show anything.... :huh: Unless I messed up somewhere.[/quote]
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.
And since $a+b+c=0$, the RHS is zero.
Therefore, $a^3+b^3+c^3=3abc$.[/quote]

Oh ok, thanks kimby_102 and dhernandez  :) That is useful thing to know.

Yea, it's really good to know that factoring...
Here's another problem (from Math Olympiad Treasure):

Prove that the number $\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}$ is an rational number.

[hide]
Use identity
$(m+n)^3=m^3+n^3+3mn(m+n)$ to get
\[ \left(\sqrt[3]{45+29\sqrt2}+\sqrt[3]{45-\sqrt2}\right)^3=(45+29\sqrt2)+(45-\sqrt2)+3\sqrt[3]{(45+29\sqrt2)(45-29\sqrt2)}\left(\sqrt[3]{45+29\sqrt2}+\sqrt[3]{45-\sqrt2}\right) \]
(to use less LaTeX, I'll let $t=\sqrt[3]{45+29\sqrt2}+\sqrt[3]{45-\sqrt2}$)
or
\[ t^3=90+21t \]
By the rational roots theorem, any rational root of this equation must be $\pm$ a factor of 90.  $2\sqrt[3]{45+29\sqrt2}<10$ so we only need to look at factors <10, i.e. $\pm1, \pm2, \pm3, \pm5, \pm6, \pm9$.  $t=6$ works.
So the cubic expression is actually 6, which is rational.
[/hide][/hide]


Find, with proof, the number of positive integers whose base-$n$ representation consists of distinct digits with the property that, except for the leftmost digit, every digit differs by $\pm 1$ from some digit further to the left.

[quote="Peter"]Find, with proof, the number of positive integers whose base-$ n$ representation consists of distinct digits with the property that, except for the leftmost digit, every digit differs by $ \pm 1$ from some digit further to the left.[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?p=258568#p258568


Prove that the sum of multiplicities of two singular points of an irreducible curve of degree $ n$ is at most $ n$, and that the sum of multiplicities of any 5 points is at most $ 2n$

I assume you mean plane curve.

Then any two distinct points determine a line, any distinct 5 points determine a conic.

Intersect these curves and apply Bezout for the desired result.


Suppose that $ a_{1}=3$, $ a_{n+1}=a_{n}(a_{n}+2)$ Find a general formula for $ a_{n}$


ratio
geometry
blogs
number theory
relatively prime

Find all right triangles whose sides form a arithmetic progression.

is this from AMC?

This may help: pythagorean triples are of the form $ t(m^2\minus{}n^2)$, $ 2tmn$, $ t(m^2\plus{}n^2)$.

But i'm going to start with pythagorean theorem...

Let the sides be $ a$, $ a\plus{}b$, and $ a\plus{}2b$.  Then we have

$ a^2 \plus{} (a\plus{}b)^2 \equal{} (a\plus{}2b)^2$
$ 2a^2 \plus{} 2ab \plus{} b^2 \equal{} a^2 \plus{} 4ab \plus{} 4b^2$
$ 3b^2 \plus{} 2ab \minus{} a^2  \equal{}0$

which isn't very helpful... anyway, perhaps try the formulas.  idk.

How about
$ (a\minus{}d)^2\plus{}a^2\equal{}(a\plus{}d)^2\\
a^2\minus{}2ad\plus{}d^2\plus{}a^2\equal{}a^2\plus{}2ad\plus{}d^2\\
a^2\minus{}2ad\equal{}2ad\\
a^2\equal{}4ad\\
a\equal{}4d$
which gives the side ratios $ 4d\minus{}d: 4d: 4d\plus{}d$ or $ 3: 4: 5$.

I was just about to suggest that... anyway, nice job.

[quote="facis"]This may help: pythagorean triples are of the form $ t(m^2 \minus{} n^2)$, $ 2tmn$, $ t(m^2 \plus{} n^2)$.[/quote]
Can you prove this fact?

The most elementary proof is as follows:  let $ a^2 \plus{} b^2 \equal{} c^2$ and suppose WLOG that $ a, b, c$ are pairwise relatively prime.  From inspection $ \bmod 4$ it follows that $ c$ is odd and exactly one of $ a, b$ is also odd.  WLOG this is $ b$.  Then

$ \left( \frac {a}{2} \right)^2 \equal{} \left( \frac {c \minus{} b}{2} \right) \left( \frac {c \plus{} b}{2} \right)$

where all the quantities involved are integers.  The integers on the RHS are relatively prime by assumption, so they must both be squares; hence

$ \frac {c \minus{} b}{2} \equal{} n^2$
$ \frac {c \plus{} b}{2} \equal{} m^2$.

The result follows.  There are two less elementary proofs, one using the Gaussian integers and one using some ideas from algebraic geometry; you can find both on my blog (although I think the first one is hidden pretty deep if it's there).


Gilligan is walking along a trail leading to the lagoon at an average rate of 3 miles per hour.  Skipper is bicycling along the same trail in the same direction at an average rate of 9 miles per hour.  At 12 noon, Gilligan is 5 miles ahead of Skipper.  If Gilligan and Skipper keep the same rates, at what time will Skipper OVERTAKE Gilligan?

[hide]Every hour, Skipper gains 6 miles on Gilligan. The distance between them is 5 miles, so it will take Skipper 50 minutes to overtake Gilligan. Noon + 50 min = 12:50.[/hide]


superior algebra
superior algebra unsolved

Show that if $ L$ is a left ideal of the ring $ M_n(\mathbb{F})$, $ F$ any field, then $ L$ consists of all left multiples of some element $ A\in M_n(\mathbb{F})$.

Every semisimple ring is [url=http://en.wikipedia.org/wiki/Von_Neumann_regular_ring]von neumann regular[/url], see section 4 of Lam's [i]A First Course in Noncommutative Rings[/i].

A von neuman ring is a ring where: for every $ a \in R$, there exists an $ x \in R$ such that $ a \equal{} axa$. You can use this to prove that every left ideal is generated by a single idempotent (also in Lam's book, but you can do this yourself).


Can you factor $ f(x)\equal{}x^4\plus{}4x^3\plus{}12x^2\plus{}18x\plus{}24$

Or you help me equetion: $ x^4\plus{}4x^3\plus{}12x^2\plus{}18x\plus{}24\equal{}0$.

If you are only interested in real zeros, this is not too difficult.
Write the function as a sum of squares to show that there is no zero.


graph theory
combinatorics unsolved
combinatorics

In a tournament, every two of the $n$ players played exactly one match with each other (no
draws). Prove that it is possible either
(i) to partition the league in two groups $A$ and $B$ such that everybody in $A$ defeated everybody in $B$; or
(ii) to arrange all the players in a chain $x_1, x_2, . . . , x_n, x_1$ in such a way that each player defeated his successor.

this is rather straightforward. if the first condition is not true, then take a longest directed path $P$. if there exists a vertex $v$ off $P$ then if there exist consecutive vertices $x,y$ of $P$(with $x\rightarrow y$) such that the edges are directed $x\rightarrow v,v\rightarrow y$ then the path can be enlarged and that is a contradiction.similarly the edge from the first vertex of $P$ to $v$ has to be directed towards $v$(or else it is possible to enlarge the path again). since this has to be the state of every vertex off $P$ it follows that every edge from any vertex of $P$ to a vertex $v$ off $P$ has to be directed towards $v$, we have a partition of the type in the first part unless the path includes all the vertices.

In other words, in a tour graph we can either find a hamiltonian circuit, or two sets $A$ and $B$ as described.

First, given two such sets $A,B$, it's clear that if we start with a vertex from $B$ we can never get out of $B$ by following the directions of the arcs, so there's no hamiltonian circuit.

Conversely, suppose there are no hamiltonian circuits. Let $x_1,x_2,\ldots,x_p,x_1$ be the longest possible circuit (for some $p<n$). Clearly, for every $m>p$ all arcs $x_mx_i,\ i\in\overline{1,p}$ have the same orientation (all away from $x_m$ or all towards in $x_m$). It's easy to see that putting those $x_m$ for which $x_mx_i$ are all away from $x_m$ in $A$, the other $x_m,m>p$ in $B$, and $x_1,\ldots,x_p$ wherever we want, as long as we put them all in the same set, gives the required partition.

Edit:

Ooops, looks like I wasn't here first :).

The tournament T is either not strong or strong. In first case there is its vertex decomposition into A and B such that each vertex of A defeats each vertex of B. Otherwise T is strong and has a Hamiltonian cycle (camion's theorem).


[quote="silouan"]Is it possible to put all the numbers from  1 to 13 in a circle such that  the sum of any two <<neighbour>> numbers is a prime number ?
Is it possible the same for  the numbers from 1 to 16 ?[/quote]

[hide]
I would say it's impossible for the numbers 1 to 13 because since there is no way for neighboring numbers to add up to 2, all the primes must be odd.  For all neighboring sums to be odd, the numbers in the circle must be arranged odd, even, odd, even, odd, even... but since there are 13 numbers then there has to be an odd, odd at the end, resulting in an even and therefore non-prime neighboring sum.

If there's some logical hole in my reasoning please make me aware.[/hide]

[hide]For 1 to 16 the following arrangement works... 16, 3, 14, 5, 12, 7, 10, 9, 8, 11, 6, 13, 4, 15, 2, 1[/hide]

What if you want the sum of two neighbouring numbers to be a perfect square? (Very easy, I know :))

Nice solution jrshoch .My solution was the the same for the first part but it took me some time to figure out .Another good question .
  jrshoch  wrote a possible solution for the numbers from 1 to 16 . Could you find another ? ;)

I doubt anyone can solve this without a computer/calculator. There are 2*8! possible arrangements keeping the odd/even scheme, and you would basically have to check all these...

Actually, $\frac{7!8!}2$ possibilities (about 100 million), a factor of 100 or so larger than I care to do with brute force.  However, it may be possible to build up -- I'm working on it now.


Complete the sidebar with the letters A, B, C, D and E, so that there aren't 2 same letters in one row (horizontal) or in the same column (vertical) or a single diagonal (inclined). Does anyone know the answer and explain?

[img]http://img25.imageshack.us/img25/7991/calsd.jpg[/img]

(It's my answer, I don't know if it's correct)

Check this one out-

1.      A   B   C   D   E
2.      C   D   E   A   B
3.      E   A   B   C   D
4.      B   C   D   E   A
5.      D   E   A   B   C

its almost the same thing you sent except that rows 2 to 5 go the other way in yours

Boa tarde Luis,

There is more than one way of doing this. A square table in which each row and column consists of the same elements in some order is called a Latin square. These crop up in the study of groups - a nice introduction is a paperback in the Teach Yourself Series. Its title is  Mathematical Groups and it is by Tony Barnard and Hugh Neill, 1996, ISBN 0-340-67012-6. I don't know of any in Portuguese.

You could have

ABCDE
BCDEA
CDEAB
DEABC
EABCD

which is the multiplication table (Cayley table) for the cyclic group $ C_{5}$, or the Latin square

ABCDE
BAECD
CDAEB
DEBAC
ECDBA

which is not a group as it does not have associativity.

Hope this helps.


linear algebra
matrix
logarithms
function
limit
algebra
polynomial

One of the exercises in my analysis text asks to prove that if $T \in M(n, \mathbb{R})$ with $\|T-I\| < 1$ (the norm being the operator norm), there exists $S \in M(n, \mathbb{R})$ with $e^{S}= T$ but that invertability is not a sufficient condition, i.e. there exists some nonsingular $T$ with no $S$ for which $e^{S}= T$.

I'm assuming the problem is to be done without much algebra as it is, after all, an analysis text, so I doubt any ideas like considering the Jordan form are in the intended solution.
[hide="My attempts"] were to define the matrix logarithm with a power series, but taking the exponential of this didn't really give me back T very obviously, and to try to apply the inverse function theorem by calculating the differential of the exponential map, but $\lim_{t \to 0}\frac{e^{(S+tH)}-e^{S}}{t}$ doesn't really have a very nice form as $S, H$ don't necessarily commute, so the best option seems to just be expanding it back out in the power series as $H+\frac{HS+SH}{2}+\frac{HSS+SHS+SSH}{3!}+\ldots$. It's not really clear how to analyze whether this is $0$ or not. I found something sort of cool in that if we define $\phi(t) = tH+t^{2}\frac{HS+SH}{2}+t^{3}\frac{HSS+SHS+SSH}{3!}+\ldots$, we are of course interested in $\phi(1)$, and we have that $\frac{d}{dt}\phi(t) = e^{S}H+\phi(t)S = He^{S}+S\phi(t)$, but this didn't really seem to yield any information automatically.[/hide]

any hints?

Analysis text or not, I'm ruling that this belongs in the linear algebra forum.

On the second part: Note that $\det(e^{S})=e^{\text{tr}(S)}.$ Thus if $\det(T)<0,$ then $T$ cannot be the exponential of a real matrix.

Back on the first part: I think you're making too much of trying to use the full generality of the inverse function theorem. In other words, non-commutativity may be less of an issue than you think.

A power series in a matrix must actually be a polynomial in that matrix. The (finite dimensional) space of matrices which are polynomials in a given matrix is a set in which all the matrices commute.

Let's put it this way. Suppose $x$ is a real number such that $|x-1|<1.$ Then I claim that the following is true:

$\sum_{n=0}^{\infty}\frac1{n!}\left(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}k(x-1)^{k}\right)^{n}=x.$

You know this because you know that $e^{\ln x}=x.$ But that identity implies a particular statement about the rearrangement of the power series, with the rearrangements being justified by abolute convergence (that's what $|x-1|<1$ is for.)

Now just put a matrix $T,$ with $\|T-I\|<1,$ into that very same power series identity. It should follow that

$\sum_{n=0}^{\infty}\frac1{n!}\left(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}k(T-I)^{k}\right)^{n}=T.$

Note that all of our power series rearranging here involves only powers of the same matrix, $T-I.$ All of those matrices commute.


Let $a,b,c,d,e$ and $f,g,h,i,j$ be two seperate geometric series with real terms.

Given that $a+f = 6$, $b+g = 10$, $c+h = 24$, and $d+i = 62$, find $e+j$.

just make a system of equations defining the sequences $a_0$ and $a_{n+1}=r*a_n$, etc

write the first series as $a, ar, ar^2, ar^3, ar^4$ and the second as $b, bs, bs^2, bs^3, bs^4$. The sum of the first three terms of each is $a(\frac{r^4-1}{r-1}),b(\frac{s^4-1}{s-1})$ which adds to 102. We also know $a+b=6$, $ar+bs=10$, etc.
solution comes quickly

wow, shame on me of thinking of a brute force solution, this is a very clever problem, i assure you there is an elegant solution
[hide="hint"]recursion[/hide]

never mind I won't solve that way.
anyway, write the series again as:
$a,b,c,d,e$ but the second as $6-a,10-b,24-c,62-d,j$. So we just want j in terms of e.
so $a/b = b/c$, $(6-a)/(10-b)=(10-b)/(24-c)$
$b^2 = ac$, $(10-b)^2 = (6-a)(24-c)$.
$100 - 20\sqrt{ac} + ac = 144 -6c - 24a + ac$.
$44 = 6c + 24a - 20\sqrt{ac}$
this can be manipulated since $(\sqrt{a} + \sqrt{c})^2 = a + c + 2\sqrt{ac}$.


Prove for positive integers $ n$ that $ \sum_{i\equal{}1}^{n}i^3\ge \frac{16n^4}{(\sqrt{n}\plus{}1)^4}$.

the LHS is >= n^3, clearly. Then, divide both sides by n^3, and rearrange so that $ (\sqrt{n}\plus{}1)^{4}\ge16n$. since both sides are positive, we can take the fourth root of both sides, and use AM-GM on $ \sqrt{n},1$.

$ \sum_{i \equal{} 1}^{n}i^3 \equal{} \frac {n^2(n \plus{} 1)^2}{4}$

Therefore we must prove that

$ n^2(n \plus{} 1)^2(\sqrt {n} \plus{} 1)^4 \geq 64n^4$

We can expand the $ n^2(n \plus{} 1)^2 \ge 4n^3$

Therefore it remains to show that $ (\sqrt {n} \plus{} 1)^4 \ge 16n$

Expanding we get $ n^2 \plus{} 4n^{\frac {3}{2}} \plus{} 6n \plus{} 4n^{\frac {1}{2}} \plus{} 1 \ge 16  \sqrt [16]{n^2\left(n^{\frac {3}{2}}\right )^4n^6\left(n^{\frac {1}{2}}\right)^4} \equal{} 16n$ as required

[color=green]Edit: Sorry, didn't get a warning that someone else had already responded.[/color]

[hide="intended solution"]
we rearrange and expand the LHS to get $ \frac{n^4\plus{}n^3\plus{}n^3\plus{}n^2}{4}$, then AM HM kills it.[/hide]


calculate 7^99 without using calculator.
 How many triangles in a grid of 15units on each side?

Don't post the same problems in two different places.

For anyone who wants to reply, the original topic is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1022539#1022539]here[/url].


Two discs, each of radius=$ a$ are placed one on top of the other. The disc on the top begins to move with a velocity=$ v$ such that $ v \propto \frac{1}{A'}$, where $ A'$ is the area exposed of the stationary disc underneath. Find time taken by the moving disc to completely expose the disc underneath.

[quote="Rajiv"]Two discs, each of radius=$ a$ are placed one on top of the other. The disc on the top begins to move with a velocity=$ v$ such that $ v \propto \frac {1}{A'}$, where $ A'$ is the area exposed of the stationary disc underneath. Find time taken by the moving disc to completely expose the disc underneath.[/quote]
No help with this one?? Anybody???


What are some current political topics in China that are perhaps related to the 2008 Beijing Olympics?  What are some of the current environmental topics in China that are perhaps related to the 2008 Beijing Olympics?

I'd appreciate any ideas you may have on these questions!  Thanks!

(I'm not sure if this was posted in the correct forum.  Please move this thread if needed.)

China is in an economics transition from a command economy to a mixed (more market oriented).  The 2008 Olympics sets the stage for China to display its development since the Maoist cultural revolution days. Maybe you can focus on how the 2008 Olympic manifests this shift of economies?

In terms of health and environment, statistics had shown that as an economy grow..they become more environmentally aware..which is not China's case right now. So that's something interesting to look at (well China claims to be aware of it, but policy-wise, I don't see any improvements other than the limited short term changes)


geometry
geometric transformation
rotation
3D geometry

Find the volume of the solid formed by rotating the triangle with vertices(-4,1),(-2,4) and (-1,1) about the line $y=-3x-2$.

[hide] Hint: 3 dimensional figures as rotated about an axis. Just find the volume after drawing the picture. Cones, Solids, etc.[/hide]


Is there any way to determine the exact dates of MOP 2008?
If not, does MOP ever run into July?

I am pretty sure it ran a tiny bit into July last year. But I might be making that up.

It ended on June 30 last year, but i think it was originally scheduled to end the next day.

Tentative dates for MOSP 2008 at this time (January 1) are June 11 - July 2, a total of three weeks.  However these dates are subject to change due to other camps on campus, facility availability, instructor schedules and other circumstances I am unable to predict at this time.  MOSP however, will take place over approximately 21 consecutive days within the boundaries of no earlier than June 9, and no later than July 4, possibly inclusive.

Steve Dunbar
MAA Director, American Mathematics Competitions


$ \{a_k\}_{k\equal{}1,...,n}\ge 0$ and $ \frac{a_{i\minus{}1}\plus{}a_{i\plus{}1}}{2}\le a_i$
Prove that:$ \exists c>0$ s.t. for every $ n$
$ (\sum_{k\equal{}1}^{n}a_k)^2 \ge \frac{3n\minus{}c}{4}\sum_{k\equal{}1}^{n}a_k^2$

[quote="can_hang2007"][quote="hxy09"]$ \{a_k\}_{k \equal{} 1,...,n}\ge 0$ and $ \frac {a_{i \minus{} 1} \plus{} a_{i \plus{} 1}}{2}\le a_i$
Prove that:$ \exists c > 0$ s.t. for every $ n$
$ (\sum_{k \equal{} 1}^{n}a_k)^2 \ge \frac {3n \minus{} c}{4}\sum_{k \equal{} 1}^{n}a_k^2$[/quote]
Dear hxy09,

I am interested in seeing your solution. Can you please post it? Thank you very much.

Regards
Can[/quote]

I haven't solved it then :( (I will continue trying)
I only had a proof for $ \frac {n \minus{} 1}{2}$
$ \frac{3}{4}$ is said to be the best constant :maybe:


find the sum of the two fractions

2 / x-3  and  x+5 / 3-x

a) 1 b)  x+3 / 3-x  c) -1  d) x+3 /x-3   e) none of these

We have $ \frac{2}{x\minus{}3}\plus{}\frac{x\plus{}5}{3\minus{}x}\equal{}\frac{\minus{}2}{3\minus{}x}\plus{}\frac{x\plus{}5}{3\minus{}x}\equal{}\frac{x\plus{}3}{3\minus{}x}$. So the answer is B.


summer program
PROMYS
MIT
college
number theory

Because I have relapsed in PROMYS-withdrawl, I'm here to answer any questions about PROMYS.

For those of you who have not heard about PROMYS, visit [url=http://www.promys.org]their website[/url].

okay couple questions:

1. Can first year students do research projects? It seems like on PROMYS website that only returning students can do advanced seminars AND research projects... although on earlier PROMYS threads from last year it said some ppl enjoyed their research projects as a first year students.

2. Do we get to visit anywhere other than MIT/Harvard campus (as mentioned in other threads).... if then where?

[quote="xxreddevilzxx"]okay couple questions:

1. Can first year students do research projects? It seems like on PROMYS website that only returning students can do advanced seminars AND research projects... although on earlier PROMYS threads from last year it said some ppl enjoyed their research projects as a first year students. [/quote]

The typical first-year student does number theory problem sets and participates in a first-year lab group, which is like a research project, but is done in a more controlled environment.

There are some exceptional first years who finish their NT psets quickly and also attend 2nd-year courses (advanced seminars) and/or complete research projects. If you know that you want to do a research project, just talk to your counselor when you get there.

[quote="xxreddevilzxx"]
2. Do we get to visit anywhere other than MIT/Harvard campus (as mentioned in other threads).... if then where?[/quote]

What other campuses are you interested in?

PROMYS activities tend to be rather informal, so there are no official tours arranged, or anything. Campus tours of MIT/Harvard usually consist of an excited and eager counselor (who is also a current student at the institution) that shows you around. My visit to MIT was great because the counselor put a lot of thought into the tour and tried to take us to some of her favorite spots on campus, not just the places that admissions wants you to see.

wait does that mean second yr students on a normal day just stays in when the first year students take their class, and instead take their advanced seminar class in the afternoon??? It sounds like if we can finish the problem set and then able to go to the advanced seminar then it must be held in the afternoon.

A typical weekday goes something like this:

9:00-10:30 or so: NT Lecture (all students)

11ish: 2nd-year seminar

Lunch

1ish: another 2nd-year seminar

2ish: Lab/Research Groups

I don't remember all the exact times, but if my memory serves me well, it's physically possible to attend NT, attend all 2nd-year courses, and participate in a research group. (Not all classes/labs meet every day.)

People work on their psets whenever they can. Some will go out for a run/play frisbee in the morning and work in the afternoon and evenings. Others will work first, play later. It's up to you to make it work.

All students attend NT because Glenn doesn't want people to forget their NT.

[quote="tokenadult"]Do 1st-year students attend the second-year lectures?[/quote]

They may choose to. It's quite a large workload if you do, though. Even second-year students don't always attend all the second-year lectures.

[quote="beta"]For those of us who went to PROMYS last year, are we supposed to receive some kind of second-year invitation?[/quote]

Yeah, I got one last year, but I didn't go again. IIRC, it should have the second-year courses and stuff on it.

[quote="mel"][quote="beta"]For those of us who went to PROMYS last year, are we supposed to receive some kind of second-year invitation?[/quote]

Yeah, I got my letter a few weeks ago.

If you didn't get one, you should contact the office, as the deadline for submitting it is coming up. (It's possible that there was some confusion or perhaps your letter got lost in the mail -- who knows.)[/quote]

Ah I guess they probably sent it to my old address.

[quote="K81o7"]About how long should the "essay" be? (that is question d)
I assumed that they wanted something short, so I made it just about 5-10 sentences (word count tells me 127 words), but I haven't sent it yet, so I want to make sure.  Do they want a full page essay, or just that?[/quote]

I can't speak with much authority on how applications are evaluated, but if it says anything, my written response for (d) was <200 words.

Honestly, I would say put as much effort into the pset as possible. After all, PROMYS isn't about you telling people how much you like math; you actually have to do the math.


Two positive integers $a,b$ are chosen so that $1 \leq a \leq b < 50$.

The value of $S=a+b$ is given to mathematician $S$. The value of $Q=a^{2}+b^{2}$ is given to mathematician $Q$.

Both $S$ and $Q$ are given the information in this problem.

A conservation between them goes as follows:

  [b](1)[/b]  $Q$: I can't tell what the numbers are.
  [b](2)[/b]  $S$: I can't tell what the numbers are.
  [b](3)[/b]  $Q$: I can't tell what the numbers are.
  [b](4)[/b]  $S$: I can't tell what the numbers are.
  [b](5)[/b]  $Q$: I can't tell what the numbers are.
  [b](6)[/b]  $S$: I can't tell what the numbers are.
  [b](7)[/b]  $Q$: Now I can tell what the numbers are.

What are the two numbers $a$ and $b$?   :ninja:

[quote="pontios"]The answer is $\boxed{a=8, \ b=9}$

[hide="Assuming that the solution is unique"][i]and I think it is![/i]

Well, I don't know if there is an easier way to solve this, so I brute forced it

$1\leq a\leq b\leq S-1\Rightarrow 1\leq a\leq [\frac{S}{2}]$

We can prove that $2Q=S^{2}+(S-2a)^{2}$

That is, given $S$ we can find a possible $Q$ by the formula

$\boxed{2Q=S^{2}+t^{2}}$, where $t=\{\begin{array}{ll}0,2,4,..., S-2,& \mbox{if S is even}\\ 1,3,5,..., S-2,& \mbox{if S is odd}\end{array}$

This is very useful because it allows us to make a table:

[img]9394[/img][img]9395[/img]

If there is not a number in the cell $(S,Q)$, it means that there is not a solution.
For example, there is a solution if $(S,Q)=(6,18)$, but $(5,18)$ is not possible

$\star$ For any $k\geq 0$ we define the subset of the positive integers $Q_{k}=\{Q|Q$  can be written as $a^{2}+b^{2}$ in $k$ different ways $\}$

Of course the numbers of the set $Q_{0}= \{1,3,4,6,7,..\}$ are not shown on the table

The first cell of each column shows how many non-empty cells are there in that column.
So, if $Q\in Q_{k}$, then the first cell of that column contains the number $k$

We put this number at any non-empty cell in the column
For example, the column of $Q=65$ has two solutions, so we put '2' to each of the corresponding cells

The first cell of each row is a little different. It shows how many cells in that row are $>1$. Let $\Phi(S)$ be this number

$\hline$

$\blacksquare$ Suppose that $Q\in Q_{1}$,
So, $Q$ is some of the numbers $2,5,8,10,13,...$.

Then the mathematician $Q$ will find the answer at the first step (there is only one pair $(a,b)$ such that $a^{2}+b^{2}=Q$)

But he said "I don't know", at the first step, so we are sure that $Q\not\in Q_{1}$

So, actually the mathematician $S$ doesn't care about the columns which have an '1' at the top
He only cares about the number $\Phi(S)$, which shows how many cells in the row $S$ are $>1$

If $\Phi(S)=0$, then $S$ knows that $Q$ will find the answer at the beginning (at the first step)
If $\Phi(S)=1$ (the green cells) then $Q$ fails to find the number and consequently $S$ finds it

$\blacksquare$ For example, suppose that $(S,Q)=(8,50)$.

$Q$ sees that there are two possible answers, $S=8$ or $S=10$. So he says "I don't know"
$S$ knows that the sum is $S=8$. He looks at the row of $8$ and he finds that there is only one cell $>1$, and this is the cell which reffers to $Q=50$, so he gives the correct answer.

We get similar results if $(S,Q)=(10,50)$

If $Q\not\in Q_{1}$ and $\Phi(S)>1$ then the person $S$ will not be able to find the answer at the 2nd step (when he talks first time)

We'll check "all" the possible cases when $Q\not\in Q_{1}$

$\blacksquare\ Q=65$
[b](1)[/b] $Q$ does not know because $S=9$ or $S=11$
[b](2)[/b]
- If $S=9$ then $S$ ends the game
- If $S$ doesn't know, then $Q$ concludes that $S\neq 9$, so $S=11$, end of the game

The above procedure could be represented by the following diagram:
[img]9401[/img]
(the [b]0[/b] on the arrow means that the person does not know; the [b]1[/b] means that the person knows)

[color=indigo]It is very important to notice that:
$S=11$ and the answers are
[b](1) Q: "I don't know"
(2) S: "I don't know"[/b]
, and at the 3rd step $Q$ ends the game[/color]

$\blacksquare\ Q=125$. Then $S=15$ or $S=13$
But $\Phi(15)=1$ (though it's not shown on the table). So if $S=15$, the person $S$ finds the answer
$\Phi(13)=3$, so if $S=13$ then the person $S$ cannot conclude. Then $Q$ knows that $S\neq 15 \Rightarrow S=13$
[img]9404[/img]

$\blacksquare\ Q=85$
[b](1)[/b] $Q$ doesn't know because $S=11$ or $S=13$ (he also knows that $S$ will answer "I don't know" because $\Phi(11),\Phi(13)>1$)
[b](2)[/b] $S$ doesn't know. This does not give a new information to $Q$, so
[b](3)[/b] $Q$ doesn't know
[b](4).a[/b] If $S=11$ then $S$ finds the answer.
Indeed, if $Q$ was equal to $65$ then the player $Q$ would find the answer at the 3rd step.
Since $Q$ does not know, it means that $Q\neq 65$

[b](4).b[/b] If $S=13$ then $S$ knows that $Q\neq 125$ (if it was then the game would be terminated at the previous step). So $Q=85$ or $Q=145$, so has not yet enough information, and he says "I don't know"

[b](5)[/b] $Q$ thinks that, if $S$ was equal to $11$, then $S$ would terminate the game.
Since the answer of $S$ in [b](4)[/b] was "I don't know", $Q$ concludes that $S=13$
So $Q$ terminates the game
[img]9403[/img]

$\blacksquare\ Q=145$. Then $S=13$ or $S=17$

[img]9405[/img][img]9406[/img]

After the set of the answers $0000$ the mathematician $Q$ still can't conclude if $S=13$ or $S=17$.
So he says "I don't know". The set of the answers now is $00000$

Let's see how $S$ is thinking

$\blacktriangleright$ Suppose that $S=13$. This could mean that $Q=85,125$ or $145$
The mathematician $S$ knows that:

If $Q=125$ then the game should be terminated after a set of answers $001$
If $Q=85$ then the game should be terminated after a set of answers $00001$

But the set of the answers is $00000$, proclaiming that $Q=145$

$\blacktriangleright$ If $S=17$, then the person $S$ says "I don't know", and $Q$ understands that $S\neq 13\Rightarrow S=17$

Now the set of the answers is $0000001$, exactly what the problem asked

So the answer is $S=17, Q=145\Rightarrow \boxed{a=8, \ b=9}$[/hide][/quote]

Well done!! That was a great solution.  :D


[quote="dogcat"]Determine all the solutions where $ x,y,s,t\in\mathbb{Q}$:
\[ t^2 \plus{} (s \plus{} x)^2 \equal{} s^2 \plus{} y^2 \equal{} (y \plus{} t)^2 \plus{} x^2\]
[/quote]

Let $ a \equal{} s \plus{} \frac x2$
Let $ b \equal{} \frac x2$
Let $ c \equal{} \frac {y \plus{} t}2$
Let $ d \equal{} \frac {y \minus{} t}2$

The equalities become :
$ (c \minus{} d)^2 \plus{} (a \plus{} b)^2 \equal{} (a \minus{} b)^2 \plus{} (c \plus{} d)^2 \equal{} 4c^2 \plus{} 4b^2$

The first part $ (c \minus{} d)^2 \plus{} (a \plus{} b)^2 \equal{} (a \minus{} b)^2 \plus{} (c \plus{} d)^2$ implies $ ab \equal{} cd$
The second part $ (a \minus{} b)^2 \plus{} (c \plus{} d)^2 \equal{} 4c^2 \plus{} 4b^2$ implies then $ a^2 \plus{} d^2 \equal{} 3(b^2 \plus{} c^2)$

If $ c \equal{} 0$, then $ ab \equal{} 0$ and either $ a \equal{} 0$ and so $ d^2 \equal{} 3b^2$ and so $ a \equal{} b \equal{} c \equal{} d \equal{} 0$, either $ b \equal{} 0$ and so $ a^2 \plus{} d^2 \equal{} 0$ and so $ a \equal{} b \equal{} c \equal{} d \equal{} 0$

If $ c\neq 0$, $ d \equal{} \frac {ab}c$ and the second equation becomes $ \frac {a^2(b^2 \plus{} c^2)}{c^2} \equal{} 3(b^2 \plus{} c^2)$ and so $ a^2 \equal{} 3c^2$, which is impossible in non zero rationals.

Hence the unique solution $ x \equal{} y \equal{} s \equal{} t \equal{} 0$


Hi AoPS and MathLinks members;

I was just wondering what university people prefer: University of Toronto or University of Waterloo? Kindly tell me how they differ and how they specialize in certain areas. Since this is a math forum, in terms of math, which is better? Knowing ahead of time will help me a lot in making an educated choice to further continue my education.

Please reply back soon,

I don't think that you should really worry about university when you haven't even entered high school yet. By the time you get to your senior grade, your interests could very well change.

In my opinion, one should just apply to both of them and then base the decision on financial aids, scholarships and such. It seems to me that their similarities outweigh their differences.

+1 vote for Waterloo!!!

:D

They have their own math faculty, you can specialize in whatever stream of math you want, and they offer an enormous variety of courses.  The profs kick ass too.

[quote="TheDreamer"]Haha good job for planning ahead, unlike me, who still doesn't know the answer to this question in grade 11 :s[/quote]
:D lol same here, still haven't decide my major yet... o_o;; mm as to the original post, dont worry too much about it! things change a lot, just pick the one best for you. waterloo has a really good coop program i heard.. mmm  :P

They are both good universities.

I attend Waterloo, and I enjoy the quite small city it is placed in and the co-op.  But remember if you choose co-op there is no more summer holidays for five years!

Wait for a few years and research it then when you have made up your mind of what you like to study and what kind of field you like to work in after graduation.

Good luck

The only difference which would probably mean anything to you at this point in time is that U of T offers better scholarships and Waterloo is better for Computer Science.

Beyond that you can make distinctions like Waterloo is more focused on discrete/applied math whereas U of T is more about pure math, but even for 90% of undergrads that won't make a difference.

I've said this before in other topics, but don't underestimate the fact that your lifestyle may be quite different in Toronto (a big city) versus Waterloo (not so much).  Some people would go crazy living in a relatively small place like Waterloo.  Other people would feel overwhelmed living in Toronto.  Still other people wouldn't care too much.

Coop has its downsides too, like no summer break (as someone mentioned), as well as delaying one year.  Just wondering, what kind of coop jobs do they give to university math students anywy?

[quote="TheDreamer"]Haha good job for planning ahead, unlike me, who still doesn't know the answer to this question in grade 11 :s[/quote]
 :rotfl:
Same here. I'm also grade 11 but I haven't figure out where to go.

[quote="d.lam.86"]Good point but still, which would you choose? Are you going to either one of them, Mr. Zhao?[/quote]

Lol, Yufei's too good for both of them. It's the ivy league for Mr. Zhao! :)

I'm also in Gr.11, prolly going to apply to as many as my hand can write applications for. If you are going for med or bio related, go to U of T, if it's more comp and math related, go to waterloo. There is a new program out at waterloo on nanotech that might be extremely interesting since it really combines all three major fields: chem, bio, and physics. Nanotech is also a fairly new field which means lots of research opportunities.

Actually, this is great timing for me -- I'm having a hard time deciding (two weeks left) between the two for studying math and computer science (without co-op).

First, Toronto does seem a much more interesting place to attend college than the small city of Waterloo, though occasional trips from the latter to the former are probably feasible.

For computer science, they both offer a broad range of classes, owing to large departments. The student population at Waterloo has a much larger pool of contest CS people, shown by enormous participation in TopCoder and a great ACM team.

For (pure) mathematics, it seems both have a pretty similar range of classes. From what I know, the olympiad math people are split between the two fairly evenly.

Now, just because some people picked a school doesn't mean it wasn't the wrong one. What gets me about Waterloo is that their program (in math) is too structured: you have to put yourself into one field of study (applied math, statistics, computer science, pure math, or combinatorics & optimization) or possibly do a double major. There seems to be quite a bit of restriction on which courses you can take and particularly on those outside of your specialization. Toronto, on the other hand, seems to be less restrictive with their specialist/major options in math and CS.

Despite reading the sites of both universities and various emails, I still am not convinced that Waterloo would allow me as much freedom as I would like in taking pure math, computer science, combinatorics & optimization, and non-math courses. (By the way, departments other than math/CS are supposed to be better at Toronto.)

What I would like to know (apart from the above): How exactly do the math/CS departments compare at the schools in terms of strengths, teaching, class size, etc.? Does Toronto allow placement tests for the introductory courses? (Waterloo allows neither those nor Advanced Placement credit.) How much worse is it to be in Waterloo as opposed to the big city of Toronto?

I'm a strong advocate for UWaterloo. I received my B.Math (Applied Math Co-op) from UWaterloo and strongly recommend it. Because it has its own faculty, the math program is highly developed, interdisciplinary, and diverse. You can major in a particular math, avoid taking courses that don't interest you, and UWaterloo has high standards. Professors are very friendly and approachable (UToronto profs can be more "distant" and "political") and the UWaterloo profs actually know who you are. The campus isn't a "big city" like UToronto (which can be good or bad... depending on who you are). UWaterloo also has a good co-op program (which can be a good thing too).

The best thing to do would be to visit UWaterloo and also UToronto (go to the Fields Institute downtown) and talk to profs and students. Other than the city, I think the only main reason why I'd choose UToronto is for it's "Pure Math", but UWaterloo has more "applications/specializations" in math that'll interest you more I think.

Currently I'm in grad school in the USA and, to quote my supervisor, "You're much more advanced than your peers because you know more 'math' at a higher level." I attribute that comment to my education at UWaterloo.

Nevertheless, do your research. Visit the universities, talk to students, grads, profs, etc and go where your heart tells you to go.

Geoff

Wow... I'm in exactly this position as well... Good timing...
I'm deciding between Waterloo math (non co-op)
and Toronto Engineering Science.

Two weeks left.  This discussion has helped.

What math? pure math?

Why no co-op?  I thought that co-op was one of the main strengths of Waterloo.

My bro went to Waterloo and recommends it.

___________________________
[url=http://www.studentdump.com]Homework Help[/url]


Altitude AH and median BM intersect inside triangle ABC and are equal. If angle ACB = 41, find angle MBC.

[hide="Solution"]
Drop a perpendicular $MN$ from $M$ to side $BC$. WLOG $BM=AH=1$, and we have similar triangles $\triangle AHC \sim \triangle MNC$ by AA. Thus, \[\frac{MN}{AH}= \frac{MC}{AC}\implies MN=\frac{1}{2}\implies \angle MBN = \arcsin \left( \frac{1}{2}\right) = \boxed{30^\circ}\]. [/hide]


Am I all alone here? :( I haven't heard from any of ya'll since July...

Billy

i do, but not as often as i wish i could. =\


The problem below was submitted by me and  it has been solved

M_a + M_b +M_c >= (a^2 +b^2 +c^2)/2R

M_a...... are the medians of Triangle ABC and R is the circumradius.

Can someone provide an upper bound of the form

K(a^2 +b^2 +c^2)/2R where K is a positive constant >1. ?

There is no upper bound. Consider the triangle ABC with angle ACB nearly 180. Then M_a+M_b+M_c is nearly (b+a/2)+(a+b/2)+0=3/2(a+b). a^2+b^2+c^2 is nearly a^2+b^2+(a+b)^2=2(a^2+ab+b^2). We need a constant k with

3/2(a+b)<k(a^2+b^2+ab)/R.

But for fixed a and b (=1 for example) R has no upper bound in the considered triangle. We can always go so close to 180 degrees that the right side becomes smaller than the left one.

Misha


Prove that

$ p^2\plus{}q^2\equal{}r^2\plus{}13$

doesn't have solutions in prime numbers.

[quote="NikolayKaz"]Prove that

$ p^2 \plus{} q^2 \equal{} r^2 \plus{} 13$

doesn't have solutions in prime numbers.[/quote]
If $ r\equal{}2$ we have $ p^2\plus{}q^2\equal{}17$ which doesn't have solutions
Else we have $ r^2 \plus{} 13 \equiv 1\plus{}13 \equiv 6 \bmod 8$
But $ p^2\plus{}q^2 \equiv 6 \bmod 8$ doesn't have any solutions.


a,b,c are the three sides of a triangle,
ra,rb,rc for escribed circle radius,ta,tb,tc for angular bisector
prove that: $ \frac {ta}{ra}\plus{}\frac {tb}{rb}\plus{}\frac {tc}{rc}\geq3$

[quote="xiaohui"]a,b,c are the three sides of a triangle,
ra,rb,rc for escribed circle radius,ta,tb,tc for angular bisector
prove that: $ \frac {ta}{ra} \plus{} \frac {tb}{rb} \plus{} \frac {tc}{rc}\geq3$[/quote]
Let $ h_a , h_b, h_c$ - altitudes of $ \triangle ABC$ , $ s \minus{}$ semiperimeter of  $ \triangle ABC$
$ h_a\equal{} \frac{2S_{ABC}}{a}$ , $ r_a\equal{} \frac{S_{ABC}}{s\minus{}a}$
Then : $ \frac {t_a}{r_a} \plus{} \frac {t_b}{r_b} \plus{} \frac {t_c}{r_c} \ge \frac {h_a}{r_a} \plus{} \frac {h_b}{r_b} \plus{} \frac {h_c}{r_c}$ $ \equal{} 2(\frac{s\minus{}a}{a}\plus{}\frac{s\minus{}b}{b}\plus{}\frac{s\minus{}c}{c})\equal{}(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\minus{}6 \ge 3;$ :)


Find all polynomials $ p(x)$ with real coefficients  such that

$ p(x)p(x\plus{}1)\equal{}p(x^2\plus{}x\plus{}1)$

Which numbers can be roots of $ p(x)$?  :wink:

[hide]
First we show that $ P(x)$ has no real roots
Let $ a$ be a real root of $ P(x)$
$ 0 \equal{} P(a)P(a \plus{} 1) \equal{} P(a^2 \plus{} a \plus{} 1)$ and $ 0 \equal{} P(a \minus{} 1)P(a) \equal{} P(a^2 \minus{} a \plus{} 1)$

So if $ f(x) \equal{} x^2 \plus{} x \plus{} 1$ and $ a$ is a root of $ P(x)$ then so is $ f(a), f^2(a)...f^n(a)$

Since $ a^2 \plus{} a \plus{} 1 > 0 \Longrightarrow f^{n \plus{} 1}(a) > f^n(a)$ and there are infinitely many roots of $ P(x)$ - contradiction.
Therefore $ P(x)$ has no real roots

Let $ x \equal{} 0 \Longrightarrow P(0)P(1) \equal{} P(1) \Longrightarrow P(0) \equal{} 1 (\because P(1)\not \equal{} 0)$

$ P(0) \equal{} 1$ implies that the product of all roots (complex of course) is $ 1$
So if $ b$ is a complex root then $ \|b\| \equal{} 1, \|b^2 \plus{} b \plus{} 1\| \equal{} 1, \|b^2 \minus{} b \plus{} 1\| \equal{} 1$

However of one of $ b^2\pm b \plus{} 1$ we will have $ 1 \equal{} \|b^2 \pm b \plus{} 1\| \equal{} \|b\| \plus{} \|b^2 \plus{} 1\|$

Therefore $ \|b^2 \plus{} 1\| \equal{} 0 \Longrightarrow b^2 \plus{} 1 \equal{} 0$

So we find the only polynomial $ P(x) \equal{} (x^2 \plus{} 1)^n$

Also
$ P(x^2 \plus{} x \plus{} 1) \equal{} ((x^2 \plus{} x \plus{} 1)^2 \plus{} 1)^n \equal{} ((x^2 \plus{} 1)(x^2 \plus{} 2x \plus{} 2))^n \equal{} (x^2 \plus{} 1)^n((x \plus{} 1)^2 \plus{} 1)^n \equal{} P(x)P(x \plus{} 1)$

[/hide]

Edit: fail

[quote="ocha"][hide]$ P(0) \equal{} 1$ implies that the product of all roots (complex of course) is $ 1$
So if $ b$ is a complex root then $ \|b\| \equal{} 1, \|b^2 \plus{} b \plus{} 1\| \equal{} 1, \|b^2 \minus{} b \plus{} 1\| \equal{} 1$[/hide][/quote]
[hide]Just because the product of all the roots is $ 1$, does not mean that each root needs to have a magnitude of $ 1$.

For instance, consider the polynomial:
$ f(x) \equal{} x^4 \plus{} 2x^3 \plus{} \frac {17}{4}x^2 \plus{} \frac {1}{2}x \plus{} 1 \equal{} (x^2 \plus{} 2x \plus{} 4)(x^2 \plus{} \frac {1}{4}) \equal{}$ $ (x \minus{} (1 \plus{} i\sqrt {3}))(x \plus{} (1 \plus{} i\sqrt {3}))(x \minus{} \frac {1}{2}i)(x \plus{} \frac {1}{2}i)$.
The product of the roots is $ 1$, yet not all the roots have magnitude $ 1$.

Also, you missed one trivial polynomial solution.
[/hide]
[hide]
If $ P(x) \equal{} c$ is a constant polynomial, then $ c \cdot c \equal{} c \Rightarrow c \equal{} 0,1$. Thus, $ P(x) \equal{} 0$ and $ P(x) \equal{} 1$ are the only constant solutions.

Now assume that $ P(x)$ is non-constant. Then $ P(x)$ has a complex zero: $ z$.

$ x \equal{} z$ yields: $ P(z^2 \plus{} z \plus{} 1) \equal{} P(z)P(z \plus{} 1) \equal{} 0$, so $ z^2 \plus{} z \plus{} 1$ is also a zero.

$ x \equal{} z \minus{} 1$ yields: $ P(z^2 \minus{} z \plus{} 1) \equal{} P(z \minus{} 1)P(z) \equal{} 0$, so $ z^2 \minus{} z \plus{} 1$ is also a zero.

Lemma:
If $ Re(z) > 0$, then $ |z^2 \plus{} z \plus{} 1| > |z|$.
If $ Re(z) < 0$, then $ |z^2 \minus{} z \plus{} 1| > |z|$.
If $ Re(z) \equal{} 0$, then $ |z^2 \plus{} z \plus{} 1|,|z^2 \minus{} z \plus{} 1| \ge |z|$ with equality iff $ z \equal{} \pm i$.
[hide="Proof:"]Let $ z \equal{} a \plus{} bi$.

$ z^2 \plus{} z \plus{} 1 \equal{} (a^2 \plus{} a \plus{} 1 \minus{} b^2) \plus{} (2ab \plus{} b)i$
$ |z^2 \plus{} z \plus{} 1| > |z| \Longleftrightarrow (a^2 \plus{} a \plus{} 1 \minus{} b^2)^2 \plus{} (2ab \plus{} b)^2 > a^2 \plus{} b^2$ $ \Longleftrightarrow (a^4 \plus{} 2a^3 \plus{} 2a^2 \plus{} 2a) \plus{} (2a^2 \plus{} 2a)b^2 \plus{} (b^2 \minus{} 1)^2 > 0$
which is true for all $ Re(z) \equal{} a > 0$.

$ z^2 \minus{} z \plus{} 1 \equal{} (a^2 \minus{} a \plus{} 1 \minus{} b^2) \plus{} (2ab \minus{} b)i$
$ |z^2 \minus{} z \plus{} 1| > |z| \Longleftrightarrow (a^2 \minus{} a \plus{} 1 \minus{} b^2)^2 \plus{} (2ab \minus{} b)^2 > a^2 \plus{} b^2$ $ \Longleftrightarrow (a^4 \minus{} 2a^3 \plus{} 2a^2 \minus{} 2a) \plus{} (2a^2 \minus{} 2a)b^2 \plus{} (b^2 \minus{} 1)^2 > 0$
which is true for all $ Re(z) \equal{} a < 0$.

If $ Re(z) \equal{} a \equal{} 0$ then $ z \equal{} bi$, $ |z| \equal{} b$ and:
$ z^2 \pm z \plus{} 1 \equal{} (1 \minus{} b^2) \pm (b)i$.
It is easy to see that $ |z^2 \pm z \plus{} 1| \ge |z|$ with equality when $ b \equal{} \pm1 \Rightarrow z \equal{} \pm i$.

End Lemma [/hide]
Thus, if we can find a zero $ z_0$, then we can construct an infinite sequence of roots:
$ \{ z_n \}$ such that $ z_{n \plus{} 1} \equal{} z^2_n \pm z_n \plus{} 1$ and $ |z_{n}| \ge |z_{n \minus{} 1}|$ for all $ n \in \mathbb{N}$.

If $ z_n \neq \pm i$ for all $ n \in \mathbb{N}_0$, then $ \{ z_n \}$ is an infinite sequence whose magnitude is strictly increasing, and thus, $ P(x)$ will have an infinite number of distinct roots, a contradiction.

So, there exists $ n \in \mathbb{N}_0$ such that $ z_n \equal{} \pm i$.

Suppose that $ z_{n \minus{} 1} \neq \pm i$. Then we must have $ z^2_{n \minus{} 1} \pm z_{n \minus{} 1} \plus{} 1 \equal{} \pm i$ Solving yields $ z_{n \minus{} 1} \equal{} \pm(1 \plus{} i)$ (Since $ z \neq \pm i$).

However, then $ |z_{n \minus{} 1}| > |z_{n}|$ which is a contradiction.

Thus, the only possible sequence of roots with a finite number of distinct values is $ z_n \equal{} \pm i$ for all $ n \in \mathbb{N}_0$.

Therefore, the only possible roots are $ z \equal{} \pm i$.

So, $ P(x) \equal{} K(x \minus{} i)^{n_1}(x \plus{} i)^{n_2}$ for some $ n_1,n_2 \in \mathbb{N}_0$ and $ K \in \mathbb{R}$ is non-zero.

Since $ P(x)$ must be a real polynomial, $ n_1 \equal{} n_2 \equal{} n$.

Therefore, $ P(x) \equal{} K(x^2 \plus{} 1)^n$. Plugging this yields $ K \equal{} 1$ as the only non-zero solution.

Thus, the solution is $ P(x) \equal{} (x^2 \plus{} 1)^n$. [/hide]


algorithm
algebra
polynomial
number theory
Euclidean algorithm
number theory proposed

I just can't believe that there was this problem on the [i]IMO[/i].

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for any positive integer n.

Well, MithsApprentice posted the following solution:

Let K be a positive common divisor of the numerator and denominator.  Let $21n+4=Ak$ and $14n+3=Bk$.  THen, $k(3B-2A)=3Bk-2Ak=(42n+9)-(42n+8)=1$ thus it is irreducible.

Well, you could use the Euclidean Algorithm:

$(21n+4, 14n+3)=(14n+3, 7n+1)=(7n+1, 7n+2)=1$.   :idea:  :!:

The two solutions are basically equivalent.  A third easy solution is by polynomial division.  The IMO was much smaller and easier back then.


In how many ways can 8 people be seated in a row if

a) there are no constraints as to their seating arrangement?

b) persons X and Y must sit next to one another?

c) there are 4 women and 4 men and no 2 men or 2

d)women can sit next to each other?

e)there are 4 married couples and each couple must sit together?

f)Î there are 4 men and they must sit next to eachother?

[quote="DiscreetFourierTransform"][hide="1"]
$ 8!$
[/hide]
[hide="2"]
Treat them as a "superperson"
There are $ 2!$ ways to arrange the superperson, and then there are 7! ways to put the superperson in the group.
$ 7! * 2!$
[/hide]
[hide="3"]
I'm a little unclear on this one.
We can either have MWMWMWMW
or WMWMWMWM

There are 4! arrangements for the men in each case, and 4! arrangements of the women.
For two cases, that's $ 2(4! * 4!) = 1152$
[/hide]
[hide="4"]
The married couples can be treated as super people, as well.
There are 4! ways to arrange the couples, times 2! ways to arrange each couple together:
$ 4! * 2! = 48$ ways.
[/hide]
[hide="5"]
Let the block of four men be M and the women be w.
We can have
Mwwww
wMwww
wwMww
wwwMw
wwwwM

that's five cases.
In each case, the men can arrange themselves in 4! ways, so we have the amount of ways is equal to $ 5 * 4! = 120$
[/hide][/quote]

Your number 4 (about 4 married couples ) is wrong, its $ 4!*2^{4}$, there are 2 ways within each paring
Your number 5 is wrong also, i think you misread it. you never counted for the other 4 people.

[hide="Solution to 5"] The solution to number 5 should be $ 5*4!*4!$, which equals $ 2880$.[/hide]

this is just like the impure problem #2 i solved these the same way


prove that for $ \forall n\in\mathbb{N}$
1)$ \lim_{n\to\infty}\frac {2n^2 \minus{} 1}{n^2 \plus{} 1} \equal{} 1$

2)$ \lim_{n\to\infty}\frac {2n^3 \minus{} 4}{n^3 \plus{} 1} \equal{} 2$

3)$ \lim_{n\to\infty}\frac {2^n \plus{} 1}{2^n} \equal{} 1$

4)$ \lim_{n\to\infty}\left(\dfrac{(n \minus{} 1)! \plus{} (n \minus{} 2)!}{n!}\right) \equal{} 0$

5)$ \lim_{n\to\infty}\left(\dfrac{\ln(n^{10})}{n}\right) \equal{} 0$

L'hopital's. Or, squeeze play if you're feeling perky, I guess.

Well, I'm very curious how would you like to apply de l'Hospital's rule to these examples :wink:
All these limits are just working thrugh dividing the numerators and denominators :wink:

Number one does equal $ 2$:

$ \lim_{n\to\infty}\frac{2n^2\minus{}1}{n^2\plus{}1}$

Now after simplifying you have: $ \lim_{n\to\infty}\frac{n^2(2\minus{}\displaystyle\frac{1}{n^2})}{n^2(1\plus{}\displaystyle\frac{1}{n^2})}$

You can now eliminate the $ n^2$ terms leaving: $ \lim_{n\to\infty}\frac{2\minus{}\displaystyle\frac{1}{n^2}}{1\plus{}\displaystyle\frac{1}{n^2}}$

Now by using the fact: If $ r$ is a positive rational number and $ c$ is any real number then, $ \lim_{x\to\infty}\frac{c}{x^r}\equal{}0$

We can now simplify the limit further: $ \lim_{n\to\infty}\frac{2\minus{}0}{1\plus{}0}\equal{}2$

So number one is false.

If the question says "prove", are you not meant to use the $ \varepsilon$ definitions? Or can you use limit rules etc?

What beans said is why I didn't just divide things out.

[quote="JZambrano201"]Number one does equal $ 2$:

$ \lim_{n\to\infty}\frac {2n^2 \minus{} 1}{n^2 \plus{} 1}$

Now after simplifying you have: $ \lim_{n\to\infty}\frac {n^2(2 \minus{} \displaystyle\frac {1}{n^2})}{n^2(1 \plus{} \displaystyle\frac {1}{n^2})}$

You can now eliminate the $ n^2$ terms leaving: $ \lim_{n\to\infty}\frac {2 \minus{} \displaystyle\frac {1}{n^2}}{1 \plus{} \displaystyle\frac {1}{n^2}}$

Now by using the fact: If $ r$ is a positive rational number and $ c$ is any real number then, $ \lim_{x\to\infty}\frac {c}{x^r} \equal{} 0$

We can now simplify the limit further: $ \lim_{n\to\infty}\frac {2 \minus{} 0}{1 \plus{} 0} \equal{} 2$

So number one is false.[/quote]

Strange rearrangement

I would have done

$ \lim_{n\to\infty}\frac {2n^2 \minus{} 1}{n^2 \plus{} 1} \equal{} \lim_{n\to\infty}\frac {2(n^2 \plus{} 1) \minus{} 3}{n^2 \plus{} 1} \equal{} \lim_{n\to\infty} (2 \plus{} \frac { \minus{} 3}{n^2 \plus{} 1}) \equal{} 2$

Still...

Actually I saw this method in a set of notes on calc from a prof. at lamar university . You can find it at this url: http://tutorial.math.lamar.edu/

ur method is absolutely right, JZambrano201...

and i don find any weirdness in this proof,SImonM

um sorry to sound stupid but what the upside down A thing??!!??!!

$ \forall$ means "for all."  $ \forall n \in \mathbb{N}$ means "for all positive integers $ n$," but that's not really relevant to the limit notation.

None of these are particularly hard, in fact 1-3 can be done by factoring the highest power of x and abusing the rule Jzambrano pointed out.

4 just requires that you split your fractions and re-write $ n!$ as $ n(n\minus{}1)!$ or $ n(n\minus{}1)(n\minus{}2)!$ and cancel appropriately. 5 is comparing rates of growth. Basically;

$ \lim_{n \rightarrow \infty} 10 \left(\frac{\ln(n)}{n}\right)$. Since $ n$ grows faster than $ \ln(n)$, the denominator wins so to speak, thus the answer is 0. You could also just use L'H's, which gives us $ \lim_{n \rightarrow \infty} \frac{1/n}{1}$, which complies with the answer I just gave.

[quote="Prime factorization"] You could also just use L'H's, which gives us [/quote]

Well, I said before that de l'Hospitals rule [b]cannot[/b] be applied here. If you really want to do that in a similar way try Stolz's theorem. :wink:


if log 14=x, log 15=y, and log 16=z, then determine the number of elements in S= {log 1, log 2,..., log 100} which can be written in the form ax+by+cz+d for rational numbers a,b,c, and d. Looks tough

[hide]$ \log N \equal{} ax\plus{}by\plus{}cz\plus{}d$
$ \equal{}a\log(14)\plus{}b\log(15)\plus{}c\log(16)\plus{}d\log(10)$
$ \equal{}\log(14^a\cdot15^b\cdot16^c\cdot10^d)$
$ \equal{}\log(2^{a\plus{}4c\plus{}d}\cdot3^{b}\cdot5^{b\plus{}d}\cdot7^{a})$
$ N\equal{}2^{a\plus{}4c\plus{}d}\cdot3^{b}\cdot5^{b\plus{}d}\cdot7^{a}$

If $ N\equal{}2^{k_1}\cdot3^{k_2}\cdot5^{k_3}\cdot7^{k_4}$, then $ a\equal{}k_4$, $ b\equal{}k_2$, $ d\equal{}k_3\minus{}k_2$, and $ c\equal{}\frac14(k_1\plus{}k_2\minus{}k_3\minus{}k_4)$ yields $ N\equal{}2^{a\plus{}4c\plus{}d}\cdot3^{b}\cdot5^{b\plus{}d}\cdot7^{a}$.

Thus, only integers in the form $ N\equal{}2^{k_1}\cdot3^{k_2}\cdot5^{k_3}\cdot7^{k_4}$ for rational $ k_1,k_2,k_3,k_4$ can satisfy $ \log(N)\equal{}ax\plus{}by\plus{}cz\plus{}d$. Since $ N$ is an integer, $ k_1,k_2,k_3,k_4$ are also integers. Thus, $ N$ can only have $ 2,3,5,7$ as prime factors.

Of the integers $ 1$ to $ 100$, $ 50$ don't have $ 2$ as a prime factor.
Of these $ 50$ integers, $ 33$ also don't have $ 3$ as a prime factor.
Of these $ 33$ integers, $ 26$ also don't have $ 5$ as a prime factor.
Of these $ 26$ integers, $ 22$ also don't have $ 7$ as a prime factor.

All of these $ 22$ integers, except for $ 1$, have a prime factor other than $ 2,3,5,7$. Thus, there are $ 100\minus{}22\plus{}1\equal{}79$ integers expressible in the form $ N\equal{}2^{k_1}\cdot3^{k_2}\cdot5^{k_3}\cdot7^{k_4}$, and thus $ 79$ elements of $ \{\log(N)|N\equal{}\overline{1,100}\}$ can be expressed as $ ax\plus{}by\plus{}cz\plus{}d$.[/hide]

[hide="Initial Thoughts"]Naming log 14=x, log 15=y, and log 16=z is totally pointless, and actually hinders you from solving the problem.  We also want to make ax+by+cz+d into a logarithm because then we only have to look inside the log and compare it to an integer.[/hide]
[hide="Solution"]We have
$ a\log(14)\plus{}b\log(15)\plus{}c\log(16)\plus{}d$
$ \equal{}a\log(14)\plus{}b\log(15)\plus{}c\log(16)\plus{}d\log(10)$
$ \equal{}\log(14^a*15^b*16^c*10^d)$
$ \equal{}\log(2^{a\plus{}4c\plus{}d}*3^{b}*5^{b\plus{}d}*7^{a})$

This means that we have to find all integers n such that
$ 1\leq n\leq 100 \text{ and } n\equal{}2^{a\plus{}4c\plus{}d}*3^{b}*5^{b\plus{}d}*7^{a}$

We know that $ 11^2>100$ and therefore the only n that don't satisfy that equation are those who are prime and greater than 7.  We have 25 primes below 100, 4 of which aren't greater than 7.  Therefore there are 21 numbers that don't satisfy $ n\equal{}2^{a\plus{}4c\plus{}d}*3^{b}*5^{b\plus{}d}*7^{a}$.
This means that $ 100\minus{}21\equal{}\boxed{79}$ do satisfy the equation.[/hide]


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

We've got a poll for AIME scores, not Index scores.

Index is sum of AMC 10/12 score and 10 times AIME score.

Feel free to post both scores in a message, but try to keep it free of "I feel stupid, should have gotten an 11" etc. so as to make for quicker tabulation.

My scores:
AMC 12 B: 132.0
AIME I: 6
Index: 192.0

People are obviously gonna vote for whatever score they got :P , unless its like rediculously high or low. I got 207.5.

So 141 + 110 = 251. (Wow... I can't add... anyway)

Ubemaya, you know the index doesnt' apply to you, right? And 8 is definitely enough.

Me, 112.5 + 50 = 162.5. Not all that great.

[quote="Ubemaya"]You mean 251?

I got 150 on AMC10A and 8 on AIME I, for 230. Although I got confused when filling it out and thought that AMC 10 doesn't count, so I put my 124.5 AMC 12B score down, making my index 204.5... I probably didn't make USAMO :( .[/quote]

Ubemaya, if things are done the same this year as previous, then your coach will receive a report with both of your scores listed and the corresponding USAMO score.  I believe that the highest USAMO score will be the one that counts, even if you put the lower score down on the scantron sheet.

[quote="Aryth"]AMC 10A + AIME = 214  :(
To Ubemaya: don't worry; I'm pretty sure that the score you put for the box is only used for identification purposes.[/quote]

So you got 144/7?

[quote="probability1.01"]Dude what I am not a liar@[/quote]

Dude, no way... who would call you that! Lemme at 'em. :bruce:

Haha...

What's your real index, like 280? Or something?

yeah... sadsad.
Oh well. Four more years? lol.

Wait, can we use AMC 10 for our index if we didn't take AMC 12 at all?

Indices: 208, 214 (138/7 and 144/7). Obviously I used my 214 index. I feel that I have a really good chance at making USAMO because last year's cutoff was only like 217, and with the harder AMC and harder AIME, I'm pretty sure it'll drop. (Yes, I know that I can qualify with the floor, but there's the chance that nobody got 145.5/7 or 150/7, which would mean that I'd be setting the floor. That's why I took AMC12).

120+70=190

all these round numbers
hopefully 7 will be enough for the floor.

[quote="white_horse_king88"]So 141 + 110 = 251. (Wow... I can't add... anyway)[/quote]

Lol, neither can I, or else I would have had an 8 on the AIME rather than a 3...

AMC 12A: 112.5
AIME I: 3 (Grrr)
Index: 142.5

I better make USAMO next year or things might get nasty...

[quote="arcticmac"]My scores:
AMC 12 B: 132.0
AIME I: 6
Index: 192.0[/quote]
 :!: same as me, except I took AMC12A.  :lol:

what would you say the chances of us getting in are?

My index is 241 (141 AMC 12B, 10 AIME).

We could use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78302&vote=viewresult]this poll about USAMO indices last year[/url], (when the qualifying index was 217) to predict this year's qualifying index.  But then you have to also have to add the effect of the USAMO expansion.

According to that the index will be about a 195. But that would make the floor a 5, so I misdoubt me that it'll be a 195.

AMC 12B: 114
AIME I: 8
Index: 194.0

:(

need to make usamo...can i use the aime floor? hehe...

i don't mean to complain, but seeing as the cutoff will most probably be somewhere between 210 and 225, we should probably split that into 4 or so categories. not many people who got 240+ have doubts about qualifying...

Let's see...

AMC-10: 144
AIME: 6

Index: 204

I didn't think AMC-10 applied to index?  I thought if you took the 10 it was just the floor...

Yeah, I don't think the AMC 10 really counts.
You could multiply your AMC 10 score by $\frac{5}{6}$... :P

Haha... right!  yeppp, thanks, I can't add either...

AMC: 126
AIME: 6
Index: 186

I know, I rule.

<('.'<)

(>'.')>

<('.'<)

(>'.')>

Another 40 or 50 points would be cool...


Determine all finite consecutive natural number sequences that sum $ 1000$ .

For $ 0\leq k<n$ we have: $ \sum_{t\equal{}k\plus{}1}^nt\equal{}1000$ is equivalent to $ (n\minus{}k)(n\plus{}k\plus{}1)\equal{}2^45^3$. Obviously one of the LHS factors is odd, the other even. This implies for some $ a\in\{0,1,2,3\}$ that (1) $ n\minus{}k\equal{}5^{3\minus{}a}2^4,n\plus{}k\plus{}1\equal{}5^a$ or (2) $ n\minus{}k\equal{}5^a,n\plus{}k\plus{}1\equal{}5^{3\minus{}a}2^4$ holds. Since $ n\plus{}k\plus{}1>n\minus{}k$ we get in case (1) $ a\equal{}3$ and in case (2) $ a<3$. Case (1) yields the solution $ k\equal{}54,n\equal{}70$ and case (2) provides the solutions $ (k,n)\in\{(999,1000),(197,202),(27,52)\}$ (remember the sum of consecutive numbers starts at $ k\plus{}1$ and ends at $ n$)


Could anybody post the problems from the 2007 Mediterranean Mathematics Olympiad?
I have searched MathLinks and I have googled for it, but did not manage to find anything.

http://www.mathlinks.ro/viewtopic.php?t=186242

This is first problem (Also look at my post in this topic)


vector
linear algebra
linear algebra unsolved

of cours it will work for every x.

Thanks.

[quote="Bert"]Why is this a vectorspace?
$\{ \alpha+\beta sin(x) | \alpha \ , \beta \in R \}$

Does the zero vector exist? And so Why?

Thanks.[/quote]
Check the definition of a vector space....


I have been presented with a list of numbers and challanged to find the next number in the sequence. i've tried myself and have been shot down and told "your only seeing it in one dimension". i hope someone can help me out.

1
11
21
1211
111221
312211
13112221
111321211
31131211131221
13211311123113112211
...

well...i solved it...and i'm want to punch myself.
go ahead and submit answers if you think you got it too.

1) This is patterning rather than Number Theory.
2) If you solved it, this should not be in Number Theory Unsolved.
3) A moderator please move this.
4) [hide="Answer"]11131221133112132113212221[/hide]

This sequence is known as [i]die Gleichniszahlen-Reihe[/i].

[b]Follow-up:[/b]  Prove that no digit greater than $3$ appears in the sequence.

[hide]Assume it did.  WLOG let the digit be 1.  There are two cases:

CASE I: the first 1 is the $(2n)$th digit.
The sequence is ... x 1 1 1 1 ...  So there are x 1's, followed by one 1.  However, this would be written as ... x+1 1 ..., and we have a contradiction.

CASE II: the first 1 is the $(2n+1)$th digit.
The sequence is ... 1 1 1 1 ...  So there is one 1, followed by one 1.  However, this would be written as ... 2 1 ..., and a contradiction occurs.[/hide]

man...this problem is like ok...have you seen it...or have you not...i guess that i don't think in this way...but i just read the solution in acops

Yay! Audioactive chemistry! After a period of time (bounded above by something smaller than 50 that I forget), any starting sequence decays into just 92 elements. Note the similarity to the (approximately) 92 elements occurring in nature.
[url]http://mathworld.wolfram.com/CosmologicalTheorem.html[/url]

The ratio of sizes of these terms approaches a certain constant which is a root of a 71st degree polynomial.
[url]http://mathworld.wolfram.com/ConwaysConstant.html[/url]


function
complex analysis
complex analysis unsolved

Let $ f$ be an analytic function from $ \{z: |z|<1\}$ to the complex plane  such that $ |f(z)|<1/|z|$ for $ z$, except zero. Prove that $ |f(z)|\le 1$ for all $ |z|<1$

Hint: Maximum modulus theorem on subdiscs of the unit disc.


我自己做了这份试卷，但无参考答案，不知对与错，故向诸位求教。

第三大题我做的是130，不过过程有点问题。

I am interested in your method of Problem 10. (Does it involve circumcentre?)

这道题曾出在两年前的上海市高中数学竞赛上，可以解三角方程，也可以用旋转和对称，不过那次的答案好象解法有问题（答案没问题）。

Problem 10:
$T(\overrightarrow{PO})$, then $P \to P'$. Thus $BQ=BP'=BC$ hence $B$ is the circumcentre of $\triangle{QP'C}$.  Let $\angle{A}=2x$, then $\angle{ACB}=\frac{180^o-2x}{2}$. Since $BQ=BC$, so $\angle{BCQ}=\frac{90^o+x}{2}$. $PP' || AB \Longrightarrow \angle{PP'C}=2x$. So $\angle{PCP'}=\angle{PP'C}=2x$. $\angle{QP'C}=\frac{1}{2}\angle{QBP'}=x$, since:
\[ \angle{ACB}=\angle{BCQ}+\angle{QCP'}+\angle{P'CP} \]
\[ \frac{180^o-2x}{2}=\frac{90^o+x}{2}+x+2x \]
\[ x=10^o \]
Hence: $\angle{A}=2x=20^o$.

I love this problem. :)


function
trigonometry
calculus
calculus computations

Prove that for all $ a$ function
$ y(x)\equal{}cos(cos(x))\plus{}a\cdot sin(sin(x))$ is a periodic function
and find it`s smallest period...

$ H(x)\equal{}\cos (\cos (x))\equal{}\cos (\cos (x\plus{}\pi ))$

$ M(x)\equal{}\sin (\sin (x))\equal{}\sin (\sin (x\plus{}2 \pi ))$

With that I prooved that $ H$ and $ M$ are periodic with period $ \pi$ and $ 2 \pi$

The sum of 2 period function is a periodic function such as the period is , in this case, $ 2 \pi$

this is simply, can you see why?

[quote="EulerIntegral"]$ H(x) \equal{} \cos (\cos (x)) \equal{} \cos (\cos (x \plus{} \pi ))$

$ M(x) \equal{} \sin (\sin (x)) \equal{} \sin (\sin (x \plus{} 2 \pi ))$

With that I prooved that $ H$ and $ M$ are periodic with period $ \pi$ and $ 2 \pi$

The sum of 2 period function is a periodic function such as the period is , in this case, $ 2 \pi$

this is simply, can you see why?[/quote]
 :)  :roll:  it`s just a training...

A more challenge problem would be.

Proove or Disproove that there is $ \alpha_n$ non-integer such as the function below is periodic (with different coefficientes, the case where all alpha are equal is simply)

$ \cos \left(\sin \left(\cos \left(\sin (x) \plus{} x \alpha _1\right) \plus{} x \alpha _2\right) \plus{} \ldots x \alpha _n \right)$

[hide="Proof"]
Suppose $ \alpha$ is rational and $ miD$ is a operator "minimum integer denominator"

$ miD\left( \frac{972}{314}\right)\equal{}miD\left( \frac{486}{157}\right) \equal{}157$

$ miD(\alpha_1)miD(\alpha_2)\ldots miD(\alpha_n) Pi$ is a period of the function.

With a little bit more can be prooved that for irrational $ \alpha$ the period of the function is infinity, or, doesnt exist

[/hide]