AMC
AIME
USA(J)MO
USAMO
calculus
ARML
AIME I

Well, ladies and gents, I must say that I regret not trusting my first instinct. Lol. The people who talked to me the day of AIME I can testify that I had 217/8 as my original guess for qualifying scores, and although the 217 never really changed, the 8 did (under peer pressure :P ). No but on a more serious note, congratulations to EVERYONE who made it...whether you are a sophomore who slid past with a 101.5 AMC 12 and 8 AIME (guessing #14 right in the last 30 seconds) or a freshman with a perfect index. Qualification to the USAMO is, as far as I'm concerned, the largest milestone in competition math.

That said, I started this thread for two purposes. The first is that because qualifying for the USAMO is something worthy of praise, I invite all of this year's qualifiers to post here noting such. The second is that AoPS is simply an unparalleled treasure trove of information and experience for AMC takers and one that should be taken advantage of. Thus, qualifiers should have a place to share their advice, and, likewise, aspiring qualifiers should have a place to ask questions and get them answered. I hope that this thread serves both of these purposes as well.

So once again...Congratulations, everyone. And good luck with the 9-hour energy drain. :)

[quote="filletwho"]Where can I get the "I survived the 2006 USAMO" t-shirts?[/quote]

You can't. Those were made one year and one year only. Lol. Although...I mean...it was so popular, we should seriously consider making some this year. I'm not sure how the AMC would take that though...(probably not well).

[quote="MithsApprentice"]Well, ladies and gents, I must say that I regret not trusting my first instinct. Lol. The people who talked to me the day of AIME I can testify that I had 217/8 as my original guess for qualifying scores, and although the 217 never really changed, the 8 did (under peer pressure :P ). No but on a more serious note, congratulations to EVERYONE who made it...whether you are a sophomore who slid past with a 101.5 AMC 12 and 8 AIME (guessing #14 right in the last 30 seconds) or a freshman with a perfect index. Qualification to the USAMO is, as far as I'm concerned, the largest milestone in competition math.[/quote]

How do you qualify as a sophmore with a 101.5 on the AMC 12 and an 8 on the AIME? That only gives you a 181.5 index...

[quote="calc rulz"]
How do you qualify as a sophmore with a 101.5 on the AMC 12 and an 8 on the AIME? That only gives you a 181.5 index...[/quote]

I'm guessing it's either metaphorical or a reference to me pulling it off as an 8th grader two years ago with a 7, or maybe to someone else doing the same thing before me.

And if I didn't [i]know[/i] who we were talkin' 'bout, I'd say a freshman with a 300 is way more improbable.

I have another competition on April 18th, supposedly at 8 or 9. I wonder how I am going to jump from Calculus to Olympiad Combinatorics. Not so confident on that one.
I consider qualifying for USAMO the biggest accomplishment I have ever had. Since it is impossible (0%, that is) for me to go to MOSP, I wish everyone good luck! Keep up the good work!

[quote="paladin8"]Yup. This time the USAMO is going down... instead of me.[/quote]

W00t.  (Sidenote: do you realize that ten of the people on our ARML team are USAMO qualifiers?  I'm very excited for ARML :) )

[quote="MithsApprentice"]Thus, qualifiers should have a place to share their advice, and, likewise, aspiring qualifiers should have a place to ask questions and get them answered. [/quote]

Well, uh, if I have any useful advice to give, it would probably be to be really careful about checking work.  The AIME's a very slow test so there's a lot of time to do it, and it helps rather a lot, as I think many of us can testify :)

[quote]W00t. (Sidenote: do you realize that ten of the people on our ARML team are USAMO qualifiers? I'm very excited for ARML [/quote]

In my school, we have USAMO qualifiers that aren't even [i]on[/i] the regular team, at least for today's state meet.

[quote="t0rajir0u"]
W00t.  (Sidenote: do you realize that ten of the people on our ARML team are USAMO qualifiers?  I'm very excited for ARML :) )
[/quote]

That's just scary...well, atleast we have 1

[quote="nat mc"][quote="Fermatprime"]And if I didn't [i]know[/i] who we were talkin' 'bout, I'd say a freshman with a 300 is way more improbable.[/quote]
You haven't seen the other discussions?[/quote]

No, he said if.


I think there should be an AMC 8 Math jam. I am very comfortable with almost all AMC 8 problems, but many people on AoPS would probably find it useful.

[quote="junggi"][quote="bpms"][quote="DPatrick"]There's no good time to schedule one.  Schools are allowed (under some circumstances) to administer the AMC 8 as late as November 21, the Tuesday before Thanksgiving.  Then you've got the holiday.  So the earliest we could schedule it would be the week after Thanksgiving, which is 2 weeks after most people have taken the contest, so there's not a lot of interest left.[/quote]
Will there be mathcounts ones?[/quote]

[quote="DPatrick"][quote]Also, can there be mathjams for mc after county state and nats like there is for AMC 10/12 where the admins go over requested problems from the competition?[/quote]We do not have permission to use current-year MC problems for something like this.[/quote][/quote]
I mean like last year with the game math jams and the past problem ones.

There are some as we near Math Counts Time.

Yeah, I joined AoPS like a day after the last one, but they look like fun (after skimming the transcripts)

math jams sound fun. I joined AoPS the day of the Fall STartup Event to discuss it. IS there going to be a math jam for amc 8. Im taking it this november


abstract algebra
group theory
superior algebra
superior algebra unsolved

[quote="abcdefghi"]show that every finitely generated abelian group is the internal direct product of its torsion subgroup and of a torsion-free subgroup[/quote]

The torsion free subgroup is $\mathbb{Z}\times ... \times \mathbb{Z}$, where the Z's appear as many times as its "Betti number" (up to isomorphism).

The torsion subgroup is $\mathbb{Z}_{k}\times ... \mathbb{Z}_{j}$, because each one when raised to the power of lowest common multiple of all the prime powers.

Their product is this group.

The set of all torsion elements forms a group- that's one side of our direct product, and we can't change it. Let $T$ be the torsion subgroup; the quotient $G/T$ is a finitely generated torsion-free abelian group. By standard results, $G/T$ is isomorphic to some $\mathbb{Z}^{n}$; choose a basis $x_{1},x_{2},\dots,x_{n}$ for it.
Now choose $y_{i}\in T+x_{i}$ for each $i$; the $y_{i}$ are linearly independent elements of infinite order, so they generate a torsion-free subgroup $H$ of $G$. By construction, $H+T=G$. Since $T$ is torsion and $H$ is torsion-free, $H$ and $T$ intersect trivially, making the sum direct.

Let me expound a little bit what jmerry said.

For simplicity in notation define, $\mathcal{Z}=\underbrace{\mathbb{Z}\times ... \times \mathbb{Z}}_{n}$
And let $S= \{a_{1},...,a_{n}\}$ be the generating set.

Define, the epimorphism:
$\phi: \mathcal{Z}\to G$
As, $\phi(m_{1},...,m_{n})=m_{1}a_{1}+...+m_{2}a_{2}$
(Note: it is surjective because by assumption $S$ generates $G$).

By the First Isomorphism Theorem,
$\phi[\mathcal{Z}]=\mathcal{Z}\simeq G/\ker \phi$
But $\ker \phi \simeq T[G]$ (torsion subgroup).
That is,
$\mathcal{Z}\simeq G/T[G]$

So we have that $G$ modulo its commutator subgroup $T[G]$ is in fact a free abelian group. Hopefully this helps somewhat.

In fact, my approach is the standard way existence is proven by the fundamental theorem.


absolute value
real analysis
real analysis unsolved

Anyone has any ideas on this one, I need help on proving that

if (absolute value of a_n) for n=1 to infinty , is summable, then so is (a_n) for n=1 to infinity.

[oops.. I misread the problem]

You can prove it using Cauchy's Criterion.

$\left|\sum_{k=m}^n a_k\right| \leq \sum_{k=m}^n |a_k|$

[quote="nerd"]Thanks for the help.

What is the best way to show that summation of a_n and summation of b_n are convergent, the two series  look complex with the absolute values[/quote]
What do you mean by "look complex with the absolute values"?


calculus articles function vector derivative integration algebra

I'm going to be in 9th grade next year, and my grades in precalc in 8th grade were just good enough to get into BC.  It's probably too late for me to change to BC by now, but which one do you think it would be better to take?

...worried about how I'll do on AP exams...

If you really like math a lot I would say to take BC.  I'm taking BC next year but then again I'll be a junior.  You should do fine on the exam if you work hard.

I just finished precalc in 9th grade and I'm going to take BC next year. I think that if you're good at math, (which you probably are considering you're going into calc as a freshman), I wouldn't waste a year taking AB since you will take BC anyway.

Eheheh...I only got an A- in precalc.  And supposedly AB isn't as fast-paced.

Read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]this[/url], an article by [b]Richard Rusczyk[/b].

The voting is unanimous for BC. Take that into consideration when making your decision.

[quote="JSRosen3"]Usually one takes AB [i]or[/i] BC, not both.

I'd strongly recommend BC--if you're already in Calculus by 9th grade, and have had good grades you far, it makes sense to try it. If it's too hard you can probably switch down to AB.[/quote]

I guess my school is different. Of course many people take Calculus their senior year so they can only take one or the other. My school actually has 3 calculus courses. AB, BC, and BC accelerated. People either take AB and then BC, or they take BC accelerated, which meets for two periods and is worth two credits.

My school doesn't offer BC either but I'm just going to be in the AB class, learn the extra stuff on my own, and then take the BC exam.  I'm not sure how much extra stuff is on the BC but some ppl have told me it is like an extra half year compared to AB.  For the mathematically inclined that shouldn't be too hard to overcome.

[quote="Robert0201"]my school doesnt offer BC? How much stuff is in BC thats not in AB, cause I guess I could just take the BC exam and learn the extra stuff.[/quote]
2*|AB| :approx: |BC|

[quote="ComplexZeta"]I don't know why you would even consider taking the AB class. (For that matter, I'm not sure why anyone would if given an option.) You will cover significantly more material in BC than in AB, and so you will learn more. Isn't that the point of going to school?[/quote]

I totally disagree. There's nothing wrong with taking the Calculus AB exam, and there are good reasons why someone might choose to take the AB class (even if given the option). The most significant of which is that not everyone learns at the same pace. So the AB curiculum accurately fits some student's pace of learning. Exposure to more material (presented more quickly) doesn't necessarily translate into "learning more" for every student.

Also in some places in the South (such as SC and NC), a lot of advanced students often take the Calculus AB exam one year followed by the BC exam the next year. This was originally done for logistical reasons (as a fail-safe back before they gave an AB sub-score on the BC exam). But now it just gives some students a 2 year exposure to calculus, and allows teachers to get into some more advanced, or oft neglected, topics during the BC course. Remember that the name of the class that you take and the name of the exam that you take don't necessarily have to be the same thing (or even entirely related to each other).

I, myself, took the Calculus AB exam my freshman year and then the Calculus BC exam my second year. I feel that that sequence worked very well for me. It gave me an extra year to solidify a lot of my pre-cal topics at the same time that I was learning calculus.

What works best for you fluffy is a decision that you'll have to make for herself. But don't feel that you should take the BC class or exam because there's anything"wrong" with the AB class or exam. You should take the class that best fits your learning ability; and then at the end of the year (regardless of which class you've taken) take the exam that best fits your preparation. If you end up taking the AB exam your freshman year you can always take the BC exam the following year.

As for what's covered on which exam:
[u]AB and BC Exam Topics[/u]
(*'s indicate BC-only exam topics)

I. Functions, Graphs, and Limits
- Analysis of graphs
- Limits of functions (including one-sided limits)
- Asymptotic and unbounded behavior
- Continuity as a property of functions
* Parametric, Polar, and Vector Functions

II. Derivatives
- Concept of the derivative
- Derivative at a point
- Derivative as a function
- Second derivatives
- Applications of derivatives
* Analysis of plane curves given in parametric, polar, or vector form
* Euler's Method for numerical solution of differential equations
* L'Hopital's Rule
- Computation of derivatives

III. Integrals
- Interpretations and properties of definite integrals
- Fundamental Theorem of Calculus
- Techniques of anti-differentiation
* integration by parts
* simple partial fractions
* Improper integrals
- Applications of anti-differentiation
* Solving logistic differential equations and using them in modeling
- Numerical approximations to definite integrals

*IV. Polynomial Approximations and Series
- Concept of series
- Series of constants
- Taylor series

[quote="beta"]What's the point of learning calculus in high school? You don't even need it for IMO, one of the hardest math competition for high schoolers in the world[/quote]

The point of learning math isn't just to do well on the IMO. One of the best reasons to take calculus in high school, is that it makes your understanding of science or economics much much better. Many topics in physics for example, are very difficult to get a full understanding of without a knowledge of calculus.

[quote="beta"]What's the point of learning calculus in high school? You don't even need it for IMO, one of the hardest math competition for high schoolers in the world[/quote]

Nearly every math class you take after calculus with take knowledge of calculus for granted. I find it better for a person not to be hindered in college because that person didn't learn calculus in high school. I think it is better to learn calculus in high school and then be free to take a greater selection of classes in college. I also find that, even though complex analysis wasn't a prerequisite for any of my classes, I would have had a much more difficult time with them had I not taken complex analysis in high school. So in my opinion it's good to learn a lot in high school. It's good to learn number theory and combinatorics as well since it starts to be taken for granted more often as you progress through the classes, even though many people don't know it or haven't taken it.

Oh my, AP calc must be a joke!  I can't believe my school would not let me take AB calc (the only thing my school offers).  I am learning all the stuff on that list in a little less than 2 months.  It's not that hard really.

Beta - calculus is cool, at least in my opinion.  I don't know why it's not tested on math competitions.

EDIT: In response to Simon, he makes a really good point.  You are in trouble if you don't know calculus in college - EVEN if you plan to take the class again.  I didn't want to start college with a major disadvantage so I am taking it over the summer, which is a pretty good idea if you ever get stuck in my situation.

I took at least three different math contests (i.e. not the same one a few years, although I did that as well) in high school that had calculus.

Oh, sorry, the only math competitions I have ever done is RIML and AMC12...plus I was just going on what he said.

[quote="beta"]Economics????

I knew you need it for engineering, which is probably the reason I'd take it.[/quote]

Yes, Calculus can be very helpful in economics and finance. In high school, Econ classes (much like Statistics) aren't taught to much depth because students don't have the Calculus background for it.

This is somewhat off topic but does anybody know if you can find out your ap results on the collegeboard website like you can for the SAT without paying any fees?

I agree with Brenda Fine (Mathcamp mentor) that Calculus is overated. I think Calculus is important, but people should know basic group theory and somoe other important basic concepts before they learn calculus.

Back to the original point: It's good to take calculus in high school so you don't have to take it in college. In college, there are many more class options than in high school, and you have to pay for the classes. So you shouldn't waste money and class opportunities in college by not taking calculus in high school.

The term overrated is always a bit strange, because it depends on who's doing the rating. For students and administrators who think of calculus as the "end of all mathematics" - it is in deed overrated. But in the grand scheme of human history, and for its utility, calculus would be difficult to overrate. It is an extremely important and powerful set of ideas.

Unfortunately mathematics is taught as if it were a ladder, with each step leading logically and sequentially to the next; rather than as a very old tree with many twisting and combining branches emanating from a basic trunk, and with roots hidden underneath.

[url=http://www.math-atlas.org/welcome.html]A Mathematical Atlas[/url]

[quote="JSRosen3"]Usually one takes AB [i]or[/i] BC, not both.[/quote]

Actually, almost everyone I know has taken both AB and BC--but only because their schools forced them to.
I took both, too, because I didn't think it was possible to take an AP test without taking the class. Otherwise, I would definitely have just studied the BC syllabus (I think it's only one extra chapter...?) since my school only offered AB, and then taken the BC test.

I don't think that you are correct that taking both AP exams would result in more credit.  The material on the AB exam is entirely covered by the BC exam; on the BC test, you get an AB sub-score (their guess at what you would have gotten on the AB exam, based off the compenents of the test that were AB level).  What college is going to give you additional credit for taking something you should be getting credit for anyhow?

[quote="ml2"]I am just wondering, If i were to take Calc BC, am i allowed to take both AP AB and AP BC exams in the same year (this would give me more college credits)?[/quote]
You can only take one of these tests each year. But as JBL said, you get an AB subscore when taking the BC test.

[quote="JBL"] What college is going to give you additional credit for taking something you should be getting credit for anyhow?[/quote]

I don't quite understand that sentence, what do you mean by something i should be getting credit for anyhow?

Assume you take French.  You demonstrate proficiency in French at, say, a third-year level.  No college will give you additional credit for also demonstrating proficiency at a second-year level.  Similarly, I highly doubt that any college would offer additional credit for taking the AB test in addition to the BC test.  Plus, if what confuted says is correct, it's not possible anyhow.

Take BC.  My school is weird so we have to take both AB and BC (each a year-long course, the way they teach it).  I just finished BC this year and there was so little to learn because essentially they teach most of the calculus sequence in AB so we just did a bit of review and like three chapters.  We went so slowly and were still done well before the AP exam.  Take advantage of your school's much better program.  It's a good idea to take the BC exam because even if you don't do as well as you would have on the AB you could take it over next year (the exam, not the class).

mirebeka, a question for you: My school is the same as yours, was everything on AB covered in BC class?

Does it look bad on my college application that I got into AB instead of BC?
I got in ab instead of bc because i got an A- in precalc
instead of and A

Also can the student decide what test to take at the end of the year?
I just learned calculus in about three-four months and get very good scores on sample BC tests. So I would like to take the BC test at the end of the year.

[quote="qweretyq"]Also can the student decide what test to take at the end of the year?[/quote]

It OUGHT TO BE up to the student to decide which AP test to take at the end of the school year, although schools can be obnoxious about which tests they host. I know a group of homeschoolers who have to shop around each year to find high schools at which to take the AP tests they are ready for.

I took BC as a senior, without having taken AB as a junior.  It was a little slower than it should have been, because half of the people in the class were taking the AB test, and the other half were taking BC (it's a really small public school).  Nevertheless, I learned the material well enough to get a 5 on the test.  So, I would say that a smart young math student like you would probably be able to handle BC.  However, you're only a freshman, so just remember that you'll have four whole years in high school in which to take Calculus.  I took geometry as a freshman, so you're way ahead of where I was!   :)

On calculus and other classes:  Calculus does really help you understand other topics better, particularly physics and economics.  Knowing where some of the graphs and formulas come from can make physics and econ a lot more intuitive and easy to remember.

Anyway, do what you want!  It'll make you happier.[/i]


A play has three acts that are 80 minutes, 50 minutes, and 45 minutes long, respectively. There are also two 15-minute intermissions. If the play starts at 5:30 p.m., at what time p.m. does it end?

there's total 80+50+45 so 175 minutes of play, and total 30 minutes of break.

so its total 205 minutes.

it's 3 hours + 25 minutes so...

5:30 + 3:25 = 8:55

the answer is 8:55.

There is 175 minutes of play and 30 minutes of intermission (what does that mean? breaks?). The total is 205 minutes or 3 hours and 25 minutes.

5:30 + 3:25 = 8:55

Yes, intermission is a break.

Please DO NOT post the exact same solution!!


Find the smallest number that if divided by 7, 9, 11 leaves a remainder of 1, 2, and 3 respectively

[hide="Yay mods!!"]Let $ x$ be this number we are looking for, then

$ x \equiv 1 \mod 7 \implies x \equal{} 7a \plus{} 1$

$ x \equal{} 7a \plus{} 1 \equiv 2 \mod 9$

$ 7a \equiv 1 \equiv 28 \mod 9$

$ a \equiv 4 \mod 9 \implies a  \equal{} 9 b \plus{} 4 \implies x \equal{} 7(9b \plus{} 4) \plus{} 1 \equal{} 63b \plus{} 29$

$ x \equal{} 63b \plus{} 29 \equiv 3 \mod 11$

$ 63b \equiv 8b \equiv 7 \mod 11$

$ b \equiv 5 \mod 11 \implies b \equal{} 11c \plus{} 5 \implies x \equal{} 63(11c\plus{}5) \plus{} 29 \equal{} 693c \plus{} 344$

Hence, the smallest possible $ x$ is $ \boxed{344}$.[/hide]


limit
induction
ratio
calculus
calculus computations

$ \lim_{n \to \infty} \frac{ (2n)^n}{(2n\minus{}1)!}\equal{}0$

I don't know $ (2n\minus{}1)!\equal{}1.3.5\cdots (2n\minus{}1)$ or $ (2n\minus{}1)!\equal{}1.2.3.4 \cdots (2n\minus{}1)$
But  with which case then Is limit $ 0$???

I tried ue Stirling and received answer $ \infty$

$ (2n\minus{}1)! \equal{} 1 \cdot 2 \cdot 3 \cdot ... \cdot (2n\minus{}1)$

Note that: $ \frac{(2n)^n}{(2n \minus{} 1)!} \equal{} \frac{(2n)^{n\plus{}1}}{(2n)!}$ and then use induction to show the denominator is bigger than the numerator for n big.

What is the radius of convergence of the power series $ \sum_{n\equal{}1}^{\infty}\frac{(2n)^n}{(2n\minus{}1)!}x^n\ ?$

[quote="Hong Quy"]$ \lim_{n \to \infty} \frac { (2n)^n}{(2n \minus{} 1)!} \equal{} 0$

I don't know $ (2n \minus{} 1)! \equal{} 1.3.5\cdots (2n \minus{} 1)$ or $ (2n \minus{} 1)! \equal{} 1.2.3.4 \cdots (2n \minus{} 1)$
But  with which case then Is limit $ 0$???

I tried ue Stirling and received answer $ \infty$[/quote]
let$ a_{n} \equal{} \frac { (2n)^n}{(2n \minus{} 1)!}$
$ \frac {a_{n}}{a_{n \minus{} 1}} \equal{} \frac {2n(\frac {n}{n \minus{} 1})^{n \minus{} 1}}{(2n \minus{} 1)(2n \minus{} 2)}$
$ \lim_{n \to \infty}\frac {a_{n}}{a_{n \minus{} 1}} \equal{} 0$
so $ \lim_{n\to \infty}a_{n} \equal{} 0$

Since the series converges for all reals, then you can just use the fact that $ \lim_{n\to \infty} a_{n} \equal{} 0$ to answer the original question.  It all boils down to applying the ratio test.


Prove that among any four distinct numbers from the interval $ (0,\frac{\pi}{2})$ there are two, say $ x,y,$ such that:

$ 8 \cos x \cos y \cos (x\minus{}y)\plus{}1>4(\cos ^2 x\plus{}\cos ^2 y).$

[hide]Some algebraic manipulation yields:

$ 8\cos(x)\cos(y)\cos(x\minus{}y) \plus{} 1 > 4(\cos^{2}(x)\plus{}\cos^{2}(y))$

$ 8\cos(x)\cos(y)[\cos(x)\cos(y) \plus{} \sin(x)\sin(y)] \plus{} 1 > 4\cos^{2}(x)\plus{}4\cos^{2}(y)$

$ 8\cos^2(x)\cos^2(y)  \minus{} 4\cos^{2}(x) \minus{} 4\cos^{2}(y) \plus{} 2 \plus{} 8\cos(x)\cos(y)\sin(x)\sin(y) > 1$

$ 2(2\cos^2(x)\minus{}1)(2\cos^2(y)\minus{}1) \plus{} 2(2\sin(x)\cos(x))(2\sin(y)\cos(y)) > 1$

$ 2\cos(2x)\cos(2y) \plus{} 2\sin(2x)\sin(2y) > 1$

$ \cos(2x \minus{} 2y) > \dfrac{1}{2} \equal{} \cos\left(\dfrac{\pi}{3}\right)$

Since $ x,y \in \left(0,\dfrac{\pi}{2}\right)$, this is equivalent to:

$ |2x\minus{}2y| < \dfrac{\pi}{3} \Longleftrightarrow |x\minus{}y| < \dfrac{\pi}{6}$

Let the four distinct numbers be $ 0 < a < b < c < d < \dfrac{\pi}{2}$.

Suppose that no two of $ a,b,c,d$ satisfy $ |x\minus{}y| < \dfrac{\pi}{6}$.

Then we have:

$ |b\minus{}a| \ge \dfrac{\pi}{6} \Rightarrow b \ge a \plus{} \dfrac{\pi}{6}$

$ |c\minus{}b| \ge \dfrac{\pi}{6} \Rightarrow c \ge b \plus{} \dfrac{\pi}{6} \ge a \plus{} \dfrac{\pi}{3}$

$ |d\minus{}c| \ge \dfrac{\pi}{6} \Rightarrow d \ge c \plus{} \dfrac{\pi}{6} \ge a \plus{} \dfrac{\pi}{2}$

However, if $ a > 0$, then $ d > \dfrac{\pi}{2}$, a contradiction.

Therefore, two of $ a,b,c,d$ satisfy $ |x\minus{}y| < \dfrac{\pi}{6}$, and thus, the original inequality.[/hide]

[quote="moldovan"]Prove that among any [b]four distinct numbers[/b]...[/quote]


geometry
perimeter
incenter
Euler
analytic geometry
angle bisector
geometry proposed

The line passing through the circumcentr and the orthocenter of the not equilateral triangle $ABC$, divides it’s perimeter and area in the same relation. Find that relation.

Lemma:Given are $\triangle ABC$, a line $d$ divides it’s perimeter and area in the same relation then $d$ pass through incenter of $\triangle ABC$
Proof:
Called $d\cap AB,AC=\{M\},\{N\}$ and angle bisector of $A$ cuts $d$=$I$  let distance from $I$ to line $BC$be $x$ and lines $AC,AB$ be $y$ then
$AMN,AMNB$ are the same perimeter therefore
$AM+AN+MN=BM+MN+CN+BC\Rightarrow AM+AN=BM+CN+BC(1)$
$AMN,AMNB$ are the same area therefore
$S(AMI)+S(ANI)=S(MBI)+S(CNI)+S(BCI)\Rightarrow AM.y+AN.y=BM.y+CN.y+BC.x\Rightarrow AM+AN=BM+CN+BC.\frac{x}{y}(2)$
from $(1),(2)$ we have $x=y\Rightarrow I$ is incenter
From lemma we will have relation of this problem is inceter $I$ belong to Euler line of $\triangle ABC$
ok?

[color=darkblue]$I\in HO\Longleftrightarrow I\equiv N\mathrm{\ (the\ Nagel's\ point)\ }\Longleftrightarrow$ the triangle $ABC$ is equilateral, what is [b][u]falsely[/u][/b] ?????[/color]

Why $I\in HO\Leftrightarrow I\equiv N$ ($N$ is only $\in GI$ then $N$ lies on $HO$, too $\not\Rightarrow N\equiv I$) but we can see when $I\in HO\Leftrightarrow \triangle ABC\text{ Isoceles}$ this is a relation! :)

Sorry, I made a great mistake. It is O.K. I am sleeping now. Is 1;32 AM here (west hour in USA).

[hide="Only for Gemath."][color=darkblue][b]Proposed problem (own).[/b] Denote $XYZ$- the area of the triangle $XYZ$ and $P(x,y,z)$- the point $P$ with the barycentrical coordinates $(x,y,z)$ wr.t. $\triangle ABC$.
Let $d$ be a line so that $E\in (AC)\cap d$, $F\in (AB)\cap d$. Denote $AE+AF=k\cdot 2p$ , $[AEF]=s\cdot [ABC]$, where $\{k,s\}\subset [0,1]$.
Prove that for any point $P$ with the barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$ there is the following equivalence :

$\boxed{\ P\ (\ x\ ,\ y\ ,\ z\ )\in EF\Longleftrightarrow (2pkz-cs)(2pky-bs)+s(cy-bz)^{2}=0\ }\ \ (1)\ .$

[b]This problem is the my present for you, Gemath !
I think that with the above extension I expiated the previous mistake.
I hope you can use this result for future developments.[/b]

[b]Proof.[/b] $A(1,0,0)\ ,\ B(0,1,0)\ ,\ C(0,0,1)\ ;\ E(1-e,0,e)\ ,\ F(1-f,f,0)$, where $\overline{AE}=e\cdot\overline{AC}$ and $\overline{AF}=f\cdot\overline{AB}$.
$P\in EF$ $\Longleftrightarrow$ $\left|\begin{array}{ccc}x & y & z\\\ 1-e & 0 & e\\ 1-f & f & 0\end{array}\right|=0$ $\Longleftrightarrow$ $\boxed{\ y\cdot e+z\cdot f=ef\ }\ .$
$AE+AF=2kp$ $\Longleftrightarrow \boxed{\ c\cdot f+b\cdot e=2kp\ }\ \ \ ;\ \ \ [EAF]=s\cdot [ABC]$ $\Longleftrightarrow$ $\boxed{\ ef=s\ }\ .$
Now eiminate the variables $e$ and $f$ between the boxed relations and we shall obtain the required relation $(1)\ .$ I hope I don't err again ![/color][/hide]

Thank Virgil Nicula for your excellent present  :blush: ,that is a good idea for extend this problem.
Form the condition $(3),(4)$ we can solve $e,f$ and we will have the barycentric equation of line $EF$ then we have the codition for $P\in EF$,  I think condition $(1)$ of you equivalent barycentric equation of a line, but really $(1)$ is nice condition and your way to do this problem is very good I like this present very much, thank you again Virgil  :)


limit
real analysis
real analysis unsolved

Find, if exists, the following limit

$ \lim\limits_{n\rightarrow \infty}\sum\limits_{k=1}^{n}\left(\frac{k}{n}\right)^{k}.$

\[ \sum_{k = 1}^n\left(\frac {k}{n}\right)^k = \sum_{j = 0}^{n - 1}\left(1 - \frac {j}{n}\right)^{n - j} = \sum_{j = 0}^{n - 1}\left(\left(1 + \frac {( - j)}{n}\right)^n\right)^{\frac {n - j}{n}}\approx \sum_{j = 0}^{n - 1}{e^{ - j}\to \boxed{\frac {e}{e - 1}}
}\]
It works as a formal calculation and gives the correct limit, but the '$ \approx$' indicates some estimates for large $ n$ that need to be made precise.  Admittedly, bounding the errors to turn the above into a rigorous argument would be a bit of a mess, even if not conceptually difficult.

Your ''proof'' can be fixed to look like a proof if you first note that  $ \sum\limits_{1}^{[\alpha n]}\left(\frac {k}{n}\right)^k\to 0 \ \forall \ 0 < \alpha < 1$ by Dominated convergence thm.


geometry
rectangle
trigonometry
conics
parabola
similar triangles
cyclic quadrilateral

[color=blue][i]In an old greek geometry book (1975) I met this problem :[/i][/color]

[b] Problem[/b]

In a triangle $ ABC$ it is $ AD \equal{} BC$ wher  $ AD$ is the altitude from $ A$  .If  $ H$ is  the orthocenter of triangle $ ABC$, $ M$ is the midpoint of $ AD$ and $ N$ is

the midpoint of  $ BC$ , prove that $ HM \equal{} HN$.

[color=red]**edited[/color]

 Babis

[quote="mr.danh"][quote="stergiu"][color=blue][i]In an old greek geometry book (1975) I met this problem :[/i][/color]

[b] Problem[/b]

In a triangle $ ABC$ it is $ \angle A \equal{} 45$.If $ AD$ is the altitude from $ A$  , $ H$ is  the orthocenter of triangle $ ABC$, $ M$ is the midpoint of $ AD$ and $ N$ is

the midpoint of  $ BC$ , prove that $ HM \equal{} HN$.

 Babis[/quote]
Are you sure that your problem is true?
All figures I have drawn don't fit it.[/quote]

Mmm! I ' m afraid you are right  :( .I changed the greek letters and I have mistaken.Really sorry. I have edited.

[quote="mr.danh"][quote="stergiu"][color=blue][i]In an old greek geometry book (1975) I met this problem :[/i][/color]

[b] Problem[/b]

In a triangle $ ABC$ it is $ AD \equal{} BC$ where  $ AD$ is the altitude from $ A$  .If  $ H$ is  the orthocenter of triangle $ ABC$, $ M$ is the midpoint of $ AD$ and $ N$ is the midpoint of  $ BC$ , prove that $ HM \equal{} HN$.

[color=red]**edited[/color]

 Babis[/quote]
Easy? AH=2ON=BC=AD, so H and D are coincide, consequencetly, triangle ABC is right and HM=HN.  :huh:[/quote]

Well , in the book there is a solution with similar triangles and pythagorian theorem , but I do not like it.This the reason that I asked for another approach. I tried with your method , but why is $ 2ON \equal{} BC$ ? We only know that $ AD \equal{} BC$ and not that $ AH \equal{} BC$, as I wrote in my first post.
 I will try for another solution too and I will write it, if found.
 Babis

[b][size=100][color=DarkBlue]LEMMA. - A rectangle $ ABCD$ is given and let $ E,\ F$ be, two points on the sideline $ AB,$ such that $ AD \equal{} AE \equal{} AF$ $ ($ suppose that $ AD > AB$ and $ B,$ between $ A,\ E$ $ ).$ We draw the line through the point $ E$ and perpendicular to $ CF,$ which intersects the side-segment $ BC$ of $ ABCD,$ at point so be it $ K.$ Prove that $ DK\perp AC.$[/color][/size][/b]

[b][size=100]PROOF OF THE  LEMMA[/size][/b].

We draw the circle $ (A)$ with diameter the segment $ EF,$ which clearly passes through the point $ P\equiv CF\cap EK$ $ ($ because of $ EK\perp CF$ $ )$ and tangents to $ CD$ at point $ D.$

From the cyclic quadrilateral $ BKPF,$ we have that $ (CK)\cdot (CB) \equal{} (CP)\cdot (CF) \equal{} (CD)^{2}$ $ ,(1)$

Let $ Q$ be, the orthogonal projection of $ D$ on $ AC$ and we denote the point $ K'\equiv CB\cap DQ.$

From the similar right triangles $ \bigtriangleup CDK$ and $ \bigtriangleup BCA$ $ ($ because of $ \angle CDK \equal{} \angle BCA$ from $ CD\perp CB$ and $ DK'\perp CA$ $ ),$ we have that $ \frac {CK'}{AB} \equal{} \frac {CD}{CB}$ $ \Longrightarrow$ $ (CK')\cdot (CB) \equal{} (AB)\cdot (CD) \equal{} (CD)^{2}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ (CK')\cdot (CB) \equal{} (CK)\cdot CB)$ $ \Longrightarrow$ $ CK' \equal{} CK$ $ \Longrightarrow$ $ K'\equiv K$ and we are done.

[b][size=100]PROOF  OF  THE  PROPOSED  PROBLEM[/size][/b].

We draw the circle $ (M),$ with diameter the side-segment $ BC$ of the given triangle $ \bigtriangleup ABC,$ which intersects the line segments $ AC,\ NC,$ at points $ E,\ Z,$ respectively, where $ N$ is the midpoint of the segment $ AD.$

Let be the points $ P\equiv AD\cap BZ$ and $ H\equiv AD\cap BE,$ the orthocenter of $ \bigtriangleup ABC,$

Based on the above [b][size=100]Lemma[/size][/b], we conclude that $ KP\perp MN,$ where $ K$ is the tangency point of $ (M),$ from the line through the point $ N,$ because of $ ND\perp BC$ and $ ND \equal{} MB,$ where $ M$ is the midpoint of the segment $ BC.$

$ \bullet$ Because of $ NA \equal{} ND$ and $ CC'\parallel AD,$ where $ CC'$ is the tangent line of $ (M)$ at point $ C,$ we conclude that the pencil $ C.C'EZB$ is in harmonic conjugation.

Because of now, $ \angle CBE \equal{} C'CE$ and $ \angle EBZ \equal{} \angle ECZ$ and $ ZBB' \equal{} \angle ZCB,$ where $ BB'$ is the tangent line of $ (M)$ at point $ B,$ we conclude that the pencil $ B.CEZB'$ is in harmonic conjugation too.

This pencil, is intersected from the line segment $ AD$ and because of $ AD\parallel BB',$ we conclude that $ HD \equal{} HP$ $ ,(1)$

We draw the line through the point $ M$ and parallel to $ KP,$ which intersects the line segment $ AD,$ at point so be it $ Q$ and so, we have that $ MQ\perp MN.$

It is easy to show that $ DQ \equal{} NP$ $ ,(2)$ from the congruent right triangles $ \bigtriangleup NKP$ and $ \bigtriangleup DMQ.$

So, from $ (1),$ $ (2),$ we have that $ HQ \equal{} HN$ $ ,(3)$

Hence, in the right triangle $ \bigtriangleup MNQ,$ from $ (3)$ we conclude that $ HM \equal{} HQ \equal{} HN$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Silouanos Mprazitikos.[/size][/b]

Kostas Vittas.

Thank you very much for the dedication mr Kostas . I really appreciate it and it is a very nice proof .

I will post a purely metrical and trigonometric approach which is very straightforward .

We have $ HM^2 \equal{} \frac {2HB^2 \plus{} 2HC^2 \minus{} BC^2}{4}$ and $ HN^2 \equal{} (\frac {AD}{2} \minus{} HD)^2 \equal{} (\frac {BC}{2} \minus{} HD)^2$

So it suffices to prove that $ \frac {2HB^2 \plus{} 2HC^2 \minus{} BC^2}{4} \equal{} (\frac {BC}{2} \minus{} HD)^2$ or equivalently

$ HB^2 \plus{} HC^2 \equal{} BC^2 \minus{} 2\cdot BD\cdot HD \plus{} 2HD^2$ or equivalently $ HB^2 \plus{} HC^2 \equal{} AH^2 \plus{} HD^2$ or

$ (2R\cos B)^2 \plus{} (2R\cos C)^2 \equal{} (2R\cos A)^2 \plus{} (2R\cos B\cdot\cos C)^2$ or equivalently

$ 1 \minus{} (\sin B)^2 \plus{} 1 \minus{} (\sin C)^2 \equal{} 1 \minus{} (\sin A)^2 \plus{} (1 \minus{} (\sin B)^2)(1 \minus{} (\sin C)^2)$ or equivalently

$ (\sin A)^2 \equal{} (\sin B)^2(\sin C)^2$ which is true since we have

$ (ABC) \equal{} \frac {a\cdot h_a}{2} \equal{} \frac {abc}{4R}$ and since $ a \equal{} h_a$ we have $ 2Ra \equal{} bc$ or $ 2R\cdot 2R\sin A \equal{} (2R\sin B)(2R\sin C)$

which is the desired and we are done

Old Greek book is a major hint.

    As vertex A moves along a line parallel to BC at a distnce from BC equal BC itself,

    the orthocenter moves on parabola with focus being the  midpoint of BC  and directrix

   the A-midline of ABC.

   M.T.

[quote="armpist"]Old Greek book is a major hint.

   ........................................

   M.T.[/quote]
 The name of the book is [b]Χρήστος Ταβανλής, Γεωμετρία Επίπεδος, page 348.[/b] It has apeared in greek before 1975.I hope that this source will be useful to you (I could send you a copy of the book - there are a few  more of them in sail).The only problem is tha the post is more that 15 euros ) and this is no ironie :( , as the above comment in your post.

  Babis

Great proof armpist !  :lol:  .
Thank you for sharing this with us

YAW.

 There was no irony in my reply earlier.

 'Greek book' steered my train of thought along the  conic section lines.

  To Babis:   Thanks you for the book offer, but it will definitely be Greek to me. :)

   M.T.

[quote="armpist"]YAW.

 There was no irony in my reply earlier.

 'Greek book' steered my train of thought along the  conic section lines.

    M.T.[/quote]

...  OK then :)  !

 Babis


Determine the sequence $ (x_n)_{n \in \mathbb{N}^*}$ which satisfies: $ x_1\equal{}1$ and for any $ n \in \mathbb{N}^*$, $ x_{2n\plus{}1}\equal{}2x_n\plus{}1, x_{3n\plus{}1}\equal{}2x_{n\plus{}1}\plus{}n\minus{}1.$

[quote="moldovan"]$ x_1 \equal{} 1$
$ x_{2n \plus{} 1} \equal{} 2x_n \plus{} 1$
$ x_{3n \plus{} 1} \equal{} 2x_{n \plus{} 1} \plus{} n \minus{} 1.$[/quote]
$ x_{6n\plus{}1}\equal{}x_{2(3n)\plus{}1} \equal{} 2x_{3n} \plus{} 1$
$ x_{6n\plus{}1}\equal{}x_{3(2n)\plus{}1}\equal{}2x_{2n\plus{}1}\plus{}2n\minus{}1\equal{}4x_n\plus{}2n\plus{}1$
then $ x_{3n}\equal{}2x_n\plus{}n$

$ x_{6n\plus{}3}\equal{}x_{2(3n\plus{}1)\plus{}1} \equal{} 2x_{3n\plus{}1} \plus{} 1\equal{}4x_{n\plus{}1}\plus{}2(n\minus{}1)\plus{}1\equal{}4x_{n\plus{}1}\plus{}2n\minus{}1$
$ x_{6n\plus{}3}\equal{}x_{3(2n\plus{}1)}\equal{}2x_{2n\plus{}1}\plus{}2n\plus{}1\equal{}4x_n\plus{}2\plus{}2n\plus{}1\equal{}4x_n\plus{}2n\plus{}3$
then $ 4x_{n\plus{}1}\plus{}2n\minus{}1\equal{}4x_n\plus{}2n\plus{}3$ then $ \boxed{x_{n\plus{}1}\equal{}x_{n}\plus{}1}$

trivially by induction $ x_i\equal{}i$


trigonometry function invariant geometry circumcircle logarithms conics

Okay, I just got back from the state test. I took the alg2/trig test, and thought it was pretty easy (at least relative to last year) since the test was written by Jeff Dodd instead of Zhijan W :wink:.

However, I there were a couple of questions I was unsure of.
Does anyone know the answers to the following questions beyond a shadow of a doubt (considering there is no answer key until 3-4 weeks later)?

#'s:
30
34
38
39
40
43
46
50

Yes, i know, that's a lot; not all of us have a math team class  :mad:

I thought it was fairly easy too, a whole lot easier than the practice test I took from 2007. Sad to say I had to skip all the trig problems since we haven't learned trig yet in class, and then a few more outside of those.

#39 and #50 are both D

lifesapain, I think I may know you. did you go to epic elementary?

Dareaper, which school do you go to? (im trying to figure who in AL is on aops, especially since both of u guys are in my grade)

HINTS (cuz im lazy to write up nicer solutions)

[hide="30"] look up the rational roots theorem...NOTE: this problem is not worded correctly in my opinion, since it should say "what is the number of all possible distinct rational roots of p... the way its worded right now the answer should be 9 since it would be like their product is 7200 and the max number of roots would be 1,-1,2,-2,3,-3,5,-5,8[/hide]
[hide="34"]treat it just as algebra...isolate it...
[hide=" if you dont really know about inverses"] by definition, $ f^ \minus{} 1 (f(x)) \equal{} x$, Basically, if you think about a function as a machine where you plug something in and it spits something out, lets say plug in a to get b, then the inverse of that function spits out b when you plug in a ...for except, the logarithmic function is the inverse of the exponential function, and then of course you have the arcsins and arctan's and stuff (now written $ \sin^ \minus{} 1$ etc)[/hide][/hide]
[hide="38"]
hmmmmm ok so like on this one I did it nonrigorously and like i like f(x)=tan x and g(x)= - x and like i just graphed both these on my paper, and I thought about how f(x) is a continuous monotonically increasing function in each of its periods and like i think you can go from there hopefully...[/hide]
[hide="39"]
ok so like u have 3 fractions, and no fraction by itelf can equal 0 when the denominator is 2[/hide]
[hide="40"]
well like one way is to be like hmmm $ (x \minus{} h)^2 \plus{} (y \minus{} k)^2 \equal{} r^2$, which is the equation of a circle (this is an extremely easy equation if you see it as just the pythagorean theorem....take any (x,y) on the circle and use pythagorean theorem to find distance to center (h,k), which is an invariant, r and its pretty ez to see why this formula works) then u can be like hmmmmmmmm i know 3 points (x,y) and need to solve for 3 variables (h,k,r) so like 3 systems of equations...this may seem a little ugly at first but its not really that bad)

or

you can be like well its pretty ez to see that the locus (or set) of all points equidistant from 2 points is the perp bisector of the 2 points, so like in a triangle ABC the circle O lies on the perp bisectors of all 3 of its sides (try to prove that the circumcenter exists....i.e. the concurrency of the 3 perpendicular bisectors...pretty good exercise...) so like find 2 of the per bisectors and find the intersection[/hide]
[hide="43"]
ok so like when you have something ugly you think "how do i get rid of that" (especially in algebra). this can often be done with [i]clever substitutions[/i] so like in this prob be like hmmm interesting...that x in the exponent is like pretty ugly so be like $ y \equal{} e^x$ and rewrite the equations with that substitution[/hide]
[hide="46"]
once again a clever substitution does the job...assuming you understand logs...[hide="hint 2...does look til youve tried just using hint 1"]let $ a \equal{} \log x$[/hide][/hide]
[hide="50"]
hmmm casework.... or it may be helpful to read up on conditional probability...but if you just think about this one....there's like $ 2^3$ cases, but the condition gets rid of one of the 8 cases...[/hide]

if u wanna check:
[hide="answers"]
30 D
34 C
38 C
39 D
40 D
43 A
46 A
50 D[/hide]

good thing i didnt bother to type up full solutions....cuz that took a long time...oh well hope it helps :)

hmm so naturally i failed and did not solve all that i could have...oh well i think i failed less than last year (at least i hope so)

edit: i am so happy to know that there are still some semi active aopsers in AL in high school

I go to Oak Mountain HS

Also, does anyone from a credible school (preferably vestavia or xinke/eugene)  want to post the answers to ALL of the problems (considering they wont post them on the internet for like 3 weeks)?

So, about what score do you guys think will need to place you in the top 25%?

oh, and i forgot, hurdler, i saw in one of your posts that you asked a question about conics that was exactly in the state test.  It was number 48.

Did you actually figure out how to do that one? I mean what the f(x,y) means and how you get those two equations from that?

(btw the problem was:
What is the smallest possible value of $ x^2 \plus{} 2y^2 \plus{} 2xy \plus{} 6x \plus{} 2y$?)

also, please fully explain the solution to 39 and 30

After checking my answers with Eugene, this is what we think the right answers are:

1b
2e
3e
4a
5b
6d
7a
8b
9c
10a
11c
12c
13e
14c
15b
16e
17a
18c
19d
20a
21b
22e
23b
24e
25a
26e
27d
28b
29d
30d
31e
32c
33b
34c
35c
36c
37a
38c
39d
40d
41e
42b
43a
44c
45c
46a
47e
48c
49b
50d

hopefully i didnt typo

39: what happens when b=0,1,2?

the following is probably wrong:
I fail at algebra and don't know calculus, so I'm not sure how he got the system of $ F_x \equal{} blah, F_y \equal{} bleh$. Someone said that that's what you get when you set the partial derivatives equal to 0 or something like that.... (check my thread about that question). My feeling is that the basic idea behind it is when you set the derivative equal to 0, (i.e. derivative is a fancy word for the slope of the tangent to a graph at a point on that graph), the max and min of the graph occurs only when the derivative is equal to 0, so you can find where the max and min occurs if you can find where on the graph the derivative equals 0. I also get the feeling that the 2 systems come from parametrizing the equation or something like that...idk
        $ P(x)$ is the notation for a polynomial in $ x$. A polynomial is basically like an expression (in this case, in terms of $ x$) with the powers (of $ x$, in this case) of the variable being integers (positive I think...you should check wikipedia for the definition). For some reason, we never learn that polynomials do not have to be in just one variable in public schools....the point is that polynomials can also be in terms of 2 or more variables. I'm pretty sure that $ F(x,y)$ is just a function (which doesn't necessarily need to a polynomial...ok im not sure why I started to talk about polynomials) in 2 variables or something like that.

30:
[hide="intuitive explanation"]
Let's say you wanted to find the roots of $ 110x^3 \plus{} 449x^2 \plus{} 508x \plus{} 273 \equal{} 0$. What you would of course do is try to factor it, and you would get $ (2x \plus{} 3)(5x \plus{} 7)(11x \plus{} 13) \equal{} 0$. From here we see that the roots are like $ x \equal{} \frac { \minus{} 3}{2}, \frac { \minus{} 7}{5}, \frac { \minus{} 13}{11}$. Note that the product of the numerators divides the constant term, and that the product of the denominators divides the lead coefficient (in this case they are actually equal since all the roots are rational). If you think about where the numbers come from when you factor and find the roots, the reason for this should be intuitively obvious. This holds true for all rational roots of a polynomial with integer coefficients; that is, if $ \frac pq$ is a root of  a $ P(x)$ with integer coefficients, then $ p$ divides the constant term of the polynomial, and $ q$ divides the leading coefficient.

so like, for this problem, we have that the numerators of the rational roots can be any factor of the constant term 7200, and the denominators must be $ 1$ or $ \minus{} 1$, since those are the only factors of the leading coefficient. Thus, the rational roots of this must also be integers. So like when you find the prime factorization of 7200, you get that it's $ 7200 \equal{} 2^5 \cdot 3^2 \cdot 5^2$.

Then, you may know the formula for the number of  positive divisors of an integer to be the product of one than the coefficients of all the exponents in the prime factorization, so that the number of positive divisors of 7200 is 6 x 3 x 3 = 54. However, you have to account for negative roots too, so you multiply this by 2 to get the answer of 108[/hide]

[hide="advertise for if you wanna learn stuff like this"]If you want to learn more math than what's being taught in public schools in AL, you should totally come to MIST Academy. Mathew Crawford's in charge and he's really really good at teaching higher level math. (Mr. Crawford used to teach all of the classes on aops!). He's only been teaching here is Alabama for about a year, so don't think it's like late to join or anything. He teaches 4th through 12th graders stuff that appears on local math tournaments, as well as harder stuff. You can see some of the problems we do at:
[url]http://www.mistacademy.com/classes.htm[/url][/hide]

edit: did you understand the clever solution to the minimization problem that doesnt use calculus? (the one that was like $ x^2 \plus{} 2y^2 \plus{} 2xy \plus{} 6x \plus{} 2y \equal{} (x \plus{} y \plus{} 3)^2 \plus{} (y \minus{} 2)^2 \minus{} 13$? Then like you basically note that the first 2 things in parenthesis are squares, and squares of real numbers can't be negative, so the minimum is like 0+0 - 13, which occurs when y=2 and x=-5 (you can get that when you set the things in parenthesis equal to 0))

wow i have too much free time...or i just fail at using my free time wisely...

if i were to come to the MIST thing, which course would i enroll in? also, which course does xinke take?

ok, the ones i missed based on your key are, and what i put is:

16c
28a
44d

oh, and i forgot, are there any good AoPS books to buy to help prepare me for next years comp tests?

did u look in the aops bookstore on the sidebar

no, what i meant was that are there any AoPS books with problems that are in the range of difficulty as the alabama comprehensive tests?

Um, I'm not sure which course you would be put in. You would probably be put in whatever course Mr. Crawford puts you in, which means one that he thinks will be challenging for you. Xinke (thats me  :P ) is in the advanced class. I'm guessing you would be put in the intermediate, but it's very possible maybe also the advanced class if you took his classes for a while. (Not that intro classes are easy - I still learned a TON last summer in the intro class, which was on number theory, cuz we did a lot of problems that were much harder than those you might see on the comprehensive test on state written, if that's what you want to compare by...of course his number theory classes are usually pretty challenging). The point is that what class you're in does not depend on how old you are, but how far along you are as a problem solver. For instance, the advanced class right now currently has a really smart 12 year old (maybe she's 11 actually).

Uhh, as for prepping for the state written next year, the thing is, most math books that I am familiar with that covers stuff beyond the really basic stuff (and covers stuff you would see on the state written) you're taught in public schools covers a lot of extra stuff outside of what you're likely to see on local math tournaments, i.e. aops books, along with many other books that I am too lazy to name right now. Actually, I'm now totally sure what all would be on a comprehensive state written test, but I get the feeling that it's like the same stuff for the most part from year to year, so iif your sole goal is to do better on state written, then your best bet would be to try to get your hands on old state tests (which im not sure how to do). Difficulty wise, the state tests are not that hard. Content wise, the state tests seem mostly to cover math team material (which you seem to unfortunately be lacking). The point is, I'm not sure I know of books that can prep you state written without the vast majority of the book being much harder than the difficulty of state written.

dang i talk too much

for those 3 questions, do you think that my answers are wrong (because the answers that I posted are probably not all correct)? or did you catch your error? or are you not sure how to do them?

i caught my error for 2 of them, however, i still think that your answer to 16 if wrong. actually i am definitely sure it is, since answer A is even. here's a hint: $ f(x)\equal{}f(\minus{}x)$ is even, right?

Ohh, so you're the mighty xinke that everyone talks about.  You placed 1st at vestavia right? Yea i actually remember that. It was you (96), eugene(84), elizabeth (78), and me :lol:  (77)....and the rest doesnt matter... :P

Tell me, which one did you miss on that test? Also, i might start PMing you a lot with questions now that i know who you are.... :rotfl:

How much does the MIST cost?

by the way, the official answers are in!!

http://mcis.jsu.edu/mathcontest/

33 also had a typo: it never said that b,c were real, so there is no possible way to find the other root. I just assumed they meant to say b,c were real, and got it right, since I was pretty sure that you could not solve it any other way.

They need to throw 30 and 33 out...I almost left these 2 blank since there was no way to solve them, but I decided to guess what they meant


It's been discussed and over discussed on the forum. Using [url=http://www.mathlinks.ro/Forum/search.php]the search function[/url] one can easily find for example this proof: http://www.mathlinks.ro/Forum/viewtopic.php?p=111071#p111071 ;)

I looked through the search thing and still couldn't figure it out.  will you help me?


A 4-digit number has the following properties:

a) Digit 1 and Digit 2 are the same. Digit 3 and Digit 4 are the same.

b) The number is a perfect square.

What is the number?

[hide="hidden"]$1000a+100a+10b+b=11(100a+b)=q^{2}$

Then $11|q^{2}\implies 11|q$ so $11|(100a+b)\implies 11|(a+b)$

$1\le a\le 9, 0\le b\le 9\implies 1\le a+b\le 18$ Then the only multiple in that satisfies the inequality is $a+b=11$

Substitute that in

$11(99a+11)=11^{2}(9a+1)=q^{2}$

Then $9a+1=p^{2}$, you could factor, but it is easier to test $a=1,2,...,9$

to find $a=7\implies b=4$

Then the number is $7744$[/hide]

An old and nice problem! :) Won't spoil the fun by writing down the solution.


geometry
circumcircle
inequalities
incenter
national olympiad

First Day 10th December 2005

1.prove that for all positive reals a,b,c,d
(a^4+c^4)^(1/2) +(a^4+d^4)^(1/2)+(b^4+c^4)^(1/2)+(b^4+d^4)^(1/2)>=2.2^(1/2)(ad+bc)   holds.
2.in a triangle ABC which satisfies AB<AC<BC,perpendicular bisectors of AC,BC intersects BC,AC at K,L respectively.O,O_1,O_2 are circumcenters of ABC,CKL,OAB respectively.Prove that OCO_1O_2 is a parallelogram.
3.There are n+1 cities and flights between some of them in a country.A good flight between A and B means  a pair of flights from A to B andB to A.for a city A,d_A is the number of flights from A.for all cities A except the capital d_A<=n and for any A,B except the capital which there is no flight between them d_A+d_B<=n.There is no limitation for the number of flights from the capital which is among the n+1 cities.There is no fliht which doesn't have a return.
Find the largest number of the good flights in one day and find all graph of flights which there are good flights with the largest number.

Second Day 11th December 2005
4.Find all nonnegative integer triples (m,n,k) which satisfies 5^m+7^n=k^3.
5.Prove that the following inequality holds for any triangle with sides a,b,c and incenter r;
1/(a^2)+1/(b^2)+1/(c^2)<=1/(4r^2)
6.for a sequence a_n (n=1,2,...) with integer terms there exists a positive integer N such that for any n>=N  a_n is the number of elements of the set {i|1<=i<n and a_i+i>=n}.Find the maximum number of values which exists infinitely many times in the sequence.

Are you sure about problem 2? The Sketchpad doesn't show that $OCO_1O_2$ is a parallelogram.

i don't think there is a mistake;i solved the problem in the exam,if i am not wrong

[quote="cosinerburc"]
          First Day 10th December 2005

1.prove that for all positive reals $a,b,c,d$
$\sqrt{a^4+c^4} +\sqrt{a^4+d^4}+\sqrt{b^4+c^4}+\sqrt{b^4+d^4}\ge 2\sqrt2(ad+bc)$   holds.

2.in a triangle $ABC$ which satisfies $AB<AC<BC$ ,perpendicular bisectors of $AC,BC$ intersects $BC,AC$ at $K,L$ respectively.$O,O_1,O_2$ are circumcenters of $ABC,CKL,OAB$ respectively.Prove that $OCO_1O_2$ is a parallelogram.

3.There are $n+1$ cities and flights between some of them in a country.A good flight between $A$ and $B$ means  a pair of flights from $A$ to $B$ and $B$ to $A$.for a city $A,d_A$ is the number of flights from $A$.for all cities $A$ except the capital $d_A\leq n$ and for any $A,B$ except the capital which there is no flight between them $d_A+d_B\leq n$.There is no limitation for the number of flights from the capital which is among the $n+1$ cities.There is no fliht which doesn't have a return.
Find the largest number of the good flights in one day and find all graph of flights which there are good flights with the largest number.

Second Day 11th December 2005

4.Find all nonnegative integer triples $(m,n,k)$ which satisfies $5^m+7^n=k^3$.

5.Prove that the following inequality holds for any triangle with sides $a,b,c$ and incenter $r$;
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\leq \frac{1}{4r^2}$

6.for a sequence $a_n (n=1,2,...)$ with integer terms there exists a positive integer $N$ such that for any $n\geq N a_n$ is the number of elements of the set ${i|1\leq i<n}$ and ${a_i+i\geq n}$.Find the maximum number of values which exists infinitely many times in the sequence.[/quote]

who can create separate topics for each question? Thanks in advance.

some of these look nice and simple.  :) I'll post some solutions if they can be put into separate threads.


Typical of late August and early September, the summer of 1996 is no exception to the destruction that is wreaked by wildfires. Large portions of the West are being destroyed by lightning-sparked fires. Bryant Gumbel reported on the Today show that a fire in Mendocino, CA, covered 70,000 acres on its ninth day of burning, and the blaze doubled in size each of the previous days.

How many acres did the fire cover on the first day?

This is about chapter level, I think. :)

[hide]The amount of acres burned on the first day is 70000/2^(9-1) acres.  This equals 273.4375 acres.[/hide]

Chapter level, provided you have a calculator :P .


Prove this identity combinatorially (i.e: don't expand out and simplify):

$\binom{2n}{2}= 2\binom{n}{2} +n^2$

Is it really EASY????  I doubt.
I am not good at proving such identities. :(

Err.  Eureka
I got it

[hide="Solution"]
The left side is the number of ways of selecting 2 objects from 2n given objects.
We have to prove that right side is the same.
Let us divide the 2n objects in to two parts A and B.
selcting two objects means either selectimg the 2 from A or selecting the 2 from B or selecting 1 from each. This the right side.

Hence we are done.[/hide]

Hope it is a good solution. cos it is only the $3^{rd}$ or $4^{th}$ such kind of problem which I have solved. :D

Please give some hint on how such problems are solved.

varun

[hide="to elaborate more"]
If you break the the set of $2n$ into $2$ pieces of equal size (size $n$).

Choice 1:
Choose $2$ from only one set.  As there are two ways to do this, the number of choices is $2 {n\choose 2}{n\choose 0} = 2 {n\choose 2}$.

Choice 2:
Choose $1$ from each set.  The number of choices is ${n\choose 1} {n\choose 1} = n \cdot n = n^2$

Thus,
$\binom{2n}{2}= 2\binom{n}{2} +n^2$[/hide]

$\binom{3n}{3}= 3\binom{n}{3} + 6n\binom{2}{n} + n^3$

How would you generalize this?

I'm pretty sure you mean:

$\binom{3n}{3}= 3\binom{n}{3} + 6n\binom{n}{2} + n^3$
This can be proved by:

[hide="proof"]
Break $3n$ into $3$ groups of $n$.

Case 1: $3$ in one group
            As this can be done for each group: ${3\choose 1} {n\choose 3} = 3 {n\choose 3}$

Case 2: $2$ in one group, $1$ in another
            The total number of such arrangements are ${3\choose 1} {2\choose 1} = 6$.
            Then the total number for each arrangement is ${n\choose 2} {n\choose 1} = {n\choose 2} n$.
            Thus the total for case 2 is: $6n {n\choose 2}$.

Case 3: $1$ in each group
            Obviously, ${n\choose 1}^3 = n^3$
[/hide]

At this time I don't know how to generalize the formula for ${n \cdot x \choose x}$ in this way.


The circle $ (I)$ touches all sides of the quadrilateral $ ABCD$. Prove that four orthocenters of triangles $ IAB,IBC,ICD,IDA$ lie on a line.

dear mr.danh,do you have a solution?it seems nice!

Posted before at [url]http://www.mathlinks.ro/viewtopic.php?t=121204[/url].


geometry trapezoid ratio inequalities invariant

I've often heard the expression WLOG and I know it stands for Without Loss of Generality, but what does it mean and how can we apply it in problem solving?

Explanation much appreciated. Thx in advance!  ;)

That means that in certain situations you can narrow down the set of assumptions, in order to simplify the solving, but in doing so you'll still cover all the possible cases since they are either similar or automatically covered by the explored ones.

For example: Solve the equation $x^2+y^2=5$ in integers.

Since the sum of the two terms is odd, they can't have the same parity, therefore one is odd and one is even. Now, if necessary, you can say "WLOG we assume that $x$ is odd", because the cases where $y$ is odd will be obtained by simple switching of the variable values.

It's used a lot in geometry too.

For example, you might be given the popular contest problem where you have a large triangle ABC thats divided into two trapezoids DEFG and FGBC, and a small triangle ADE. You're given that the lengths AD:DF:FC are in a certain ratio, and you want to find the ratio of the area of ADE:FGBC.

In this case, while you could use variables for the side length, it's much easier just to plug in a number of one of them WLOG, since when you're working with ratios, all ratios are preserved when you shrink or enlarge the shape.

[quote="prototypex24"]I've seen the following example from http://www.math.kth.se/~shapiro/problem.html,

[quote]Problem: Prove that the inequality (x + y)^p < x^p + y^p holds whenever x, y are positive numbers and 0 < p < 1.

Solution: Wlog we may assume x + y = 1 (because the inequality in question remains invariant under the transformation of scaling x → tx, y → ty with t positive). But, if x+y = 1, the inequality is obvious since then x < 1, y < 1 so x < xp and y < yp, etc.
[/quote]

What I don't understand is why does "the inequality remain invariant under the transformation of [b]scaling[/b] x → tx, y → ty with t positive"

Help me plz
 :huh:[/quote]

I guess the issue isn't with the wlog, but with the terminology used. It was a fancy way of saying "the truth of the inequality doesn't change if you multiply by a positive constant". This is the justification that allows the writer to say wlog, and to use the specific case, x+y=1.
It's similar to what halluci was saying about considering specific side lengths in geometry problems involving side ratios.

It's basically saying \[ (x+y)^p<x^p+y^p \implies (tx+ty)^p<(tx)^p+(ty)^p  \] because the whole expression was simply multiplied by $t^p$.
All possible ratios $x: y$ in question are covered in the interval $(0, 1)$.  Thus, we only need to show the inequality for $x+y=1$.  You could also think that any point $(x, y)$ in the plane is simply a stretched version of a point inside $x+y=1$.

Thanks everyone! I understand now :D (Bictor717 was right - it was the terminology that prevented me from grasping the essentials of wlog example. :) )

Could someone post some problems involving wlog for me to try on my own? Thanks in advance


If (x+y+z)^4 is multiplied out and the like terms are combined, what is the coefficient of the x^2yz term?

[hide]Answer: 12[/hide]

[hide="For example"]
$ (x^{2}+y^{2}+z^{2}+2xy+2xz+2yz)(x^{2}+y^{2}+z^{2}+2xy+2xz+2yz)$
And now to be $ x^{2}yz$ we must multiplied:
$ x^{2}\cdot 2yz$ or $ 2yz\cdot x^{2}$ or $ 2xy\cdot 2xz$ or $ 2xz\cdot 2xy$, so:
$ (2+2+4+4)x^{2}yz$[/hide]

[hide]We are trying to choose two $ x$'s, a $ y$, and a $ z$ from $ (x+y+z)(x+y+z)(x+y+z)(x+y+z)$. So, the answer is $ \frac{4!}{2!}=12$.[/hide]

[quote="Temperal"]Don't use the distributive property, that takes too much time.

The Multinomial Theorem is much quicker:

http://en.wikipedia.org/wiki/Multinomial_theorem[/quote]

i looked it up...i don't really understand this theorem. could u explain it?

In my opinion, it's not really important to know the "Multinomial Theorem" if you know how to work these problems.  Just use counting skills, like "how many ways are there to choose 3 $ x$s and 2 $ y$s from this product?"

Yeah, I just used basic counting skills like archimedes1 said.

I dont get how to figure this out using counting skills. How do you choose 2,1,1 out of a product?

you take the factorial of the power of the multinomail(in this case 4) and divided it by the factorial of all the powers in the term you are looking for. so it looks something like this $ \frac{4!}{2!1!1!}$

Well, the product you want is $ x^{2}yz$. How many arrangements are there?

Actually, it should be $ \frac{4!}{2!1!1!}=12$.

I didn't know you meant that by "arrangements", I thought you meant how many ways are there to arrange the different x's and y's and z's. Now that I do know what you meant by arrangements, I don't get how you got that?

Actually, Anon your a little confused you answered a LOT harder question.

the x^2yz coefficient is 12

but the number of terms in the expansion of (x+y+z)^{4} is 15, like you said.

Thanks  :lol: , but I still don't get how to find the coefficient by logic (without the Multinomial thing)

Well, it's just the way things work, like abacadaea said.

$ \frac{\text{(Exponent)!}}{\text{(Exponent of x)}!\text{(Exponent of y)!}\text{(Exponent of z)!}}$

[quote="abacadaea"]i dont know whats so confusing.

1. the answer is 12
2. READ the QUESTION. it asks for the coeficient not the arrangements[/quote]

If you had been following the thread, i didn't understand why the answer was 12, and my answer was due to a misunderstanding of mountain.dew's rephrasing of the question to "How many arrangements are there?", which I answered 15. I was not answering the original question...

I still don't get how to get 12 by logic. In other words, prove the Multinomial Theorem.

Uh, I don't think swearing is very nice.

And AnonAmoz, here's a nice proof: http://planetmath.org/encyclopedia/MultinomialTheoremProof.html


conics
ellipse
calculus
geometry
geometric transformation
reflection
real analysis

I was debating whether I should post this in another forum or not. Since I am not sure whether the answer involves calculus, I decided to just post it here. Please move this if it would go better somewhere else.

I was reading AOPS Volume 2 and it said that if you throw ball from a focus of an ellipse in a straight line and it hits the ellipse, then it will bounce off to the other focus. I guess this means that if $ F_1$, $ F_2$ are the foci and P is a point on the ellipse, then the reflection of $ PF_1$ over the line perpendicular to the tangent of the ellipse at P is $ PF_2$. You could probably prove this by explicitly finding the tangent line at P and doing a lot of messy calculus. But AOPS Volume 2 just said "think about why this must be true" which makes me think there is a way to see this without calculus. Is there?

One way to characterize an ellipse is as a locus of points equidistant from a circle and some point inside the circle (where the foci are then the center of the circle and the other point inside it).  Drawing a picture with this construction in mind leads to the geometric intuition behind your statement.

Hmm...I am a bit puzzled why thats true. I usually think of an ellipse as the set of points such that distance to one focus plus the distance to the other focus was a constant. Can you show me how your way of characterizing an ellipse is equivalent to that?

It follows easily from the fact that the sum of the distances from a point to the circle and the point to the circle's center is constant and equal to the circle's radius.

More detail: Let $ C$ be a circle of radius $ r$ and center $ f_1$, and let $ f_2$ be a point within the circle.  Define $ E \equal{} \left \{p\in \mathbb{R}^2: d(p,f_2) \equal{} d(p,C)\right \}$ where $ d$ is the Euclidean distance.  Clearly $ E$ lies entirely within the circle, since for any $ p$ outside $ C$ we have $ d(p,C) < d(p,f_2)$.  It's also easy to verify that for all $ p$ within the circle we have $ d(p,C) \equal{} r \minus{} d(p,f_1)$, so $ E \equal{} \left \{p\in \mathbb{R}^2: d(p,f_1) \plus{} d(p,f_2) \equal{} r\right \}$; i.e. $ E$ is an ellipse with foci $ f_1,f_2$.

Assuming that the ball follows a path akin to a straight line and you must have heard that when light rays travel to hit a mirror then the angle of incidence is equal to the angle of reflection.

The normal  at any point on the ellilse bisects the angle between the focal radii of that point which are $ PF_{1}$ and $ PF_{2}$.The proof of this statment can be found in any elementary text on Ellipse.

Obviously when ball starts from $ F_{1}$ and meets the ellipse at $ P$ it simply follows the law of reflection and consequently reaches $ F_{2}$.

Here's the argument they're probably looking for:

Reflection satisfies a minimum principle. The path of reflection from a line give the minimum distance between two points on the same side of the line for a path that touches that line.

Use the characterization of the ellipse as the locus for which the sum of the distances to the two foci equals a fixed constant.

A tangent line to the ellipse lies outside the ellipse except at the point of tangency. Any path from one focus to that line to the other focus must be longer than the path from one focus to the point of tangency to the other focus - this is from that characterization of the ellipse. But then the point of reflection off that line coincides with the point of tangency, since that is where the length of the combined path reaches its minimum.


calculus
integration
logarithms
trigonometry
function
limit
complex analysis

Evaluate $\int_{0}^\frac{\pi}{2}\ln(\tan\theta) \ d\theta$

First, make the substitution $u=\tan\theta,\ \theta=\arctan u,\ d\theta=\frac{du}{1+u^{2}}.$ The integral becomes

$\int_{0}^{\infty}\frac{\ln u}{1+u^{2}}\,du$

Now, one more substitution: let $t=\ln u,\ u=e^{t},\ du=e^{t}\,dt.$

The integral becomes

$\int_{-\infty}^{\infty}\frac{te^{t}}{1+e^{2t}}\,dt =\int_{-\infty}^{\infty}\frac{t}{e^{t}+e^{-t}}\,dt.$

First note that by straightforward comparison, this improper integral converges absolutely at both ends. At each step, we clearly had absolute convergence. This enables me to make the next remark: this is the integral of an odd function over a symmetric interval. Hence, the integral is zero.

--

You can also go back to the $du$ form of this integral on $(0,\infty)$ and make the substitution $v=\frac1u.$ That's actually what I did first before deciding I liked the other version better. There should also be complex analysis techniques for doing this but you chose to post it in the real analysis section, so I stayed with ordinary calculus.

Thanks for your reply, anyway the way I came up with this was completely by accident; I was trying to evaluate $\int_{0}^\frac{\pi}{2}\ln(\sin\theta) \ d\theta$.  So I made the substitution $u=\frac{\pi}{2}-\theta$, and realized:

$\int_{0}^\frac{\pi}{2}\ln(\sin\theta) \ d \theta = \int_{0}^\frac{\pi}{2}\ln(\cos\theta) \ d \theta \Rightarrow 0=\int_{0}^\frac{\pi}{2}\ln(\sin\theta) \ d \theta-\int_{0}^\frac{\pi}{2}\ln(\cos\theta) \ d \theta =$$\int_{0}^\frac{\pi}{2}\ln(\tan\theta) \ d \theta$

Also, if you don't mind I'd like to see how you could use complex analysis to solve this.

$I=\int_{0}^{\frac{\pi}{2}}\ln\tan x\ dx=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\ln\tan\left(\frac{\pi}{4}-x\right)\ dx=$ $\int_{0}^{\frac{\pi}{4}}\left[\ln\tan \left(\frac{\pi}{4}-x\right)+\ln\tan\left(\frac{\pi}{4}+x\right)\right]\ dx=0\ .$

Well, on the interval $\theta\in\left(0,\frac{\pi}{2}\right)$, $\tan \theta > 0$, so the absolute value sign is unnecessary.

OOOps ! O.K., I am sorry, I was sleeping ! Think I saw wrong down $-\frac{\pi}{2}$.

Complex variables? Ok, but this is going to get a little clumsy. I'll attack the problem as it was after I made my first substitution above:

$\int_{0}^{\infty}\frac{\ln x}{1+x^{2}}\,dx$

Let $f(z)=\frac{\log z}{1+z^{2}}.$ Choose a branch which is continuous on the upper half plane: we may assume that $\arg(z)\in\left(-\frac{\pi}2,\frac{3\pi}2\right).$ We will integrate this around the following closed curve:

1. The real axis from $\epsilon$ to $R.$
2. A semicircle (counterclockwise) in the upper half plane from $R$ to $-R.$
3. The real axis from $-R$ to $-\epsilon.$
4. A semicircle (clockwise) in the upper half plane from $-\epsilon$ to $\epsilon.$

The integral over #1 is just $\int_{\epsilon}^{R}\frac{\ln x}{1+x^{2}}\,dx.$

The integral over #2 can be estimated as $O\left(\frac{\ln R}{R}\right),$ which goes to zero as $R\to\infty.$

The integral over #3 is

$\int_{-R}^{-\epsilon}\frac{\ln|x|+i\pi}{1+x^{2}}\,dx= \int_{\epsilon}^{R}\frac{\ln x}{1+x^{2}}\,dx+i\pi\int_{\epsilon}^{R}\frac{dx}{1+x^{2}}$

The integral over #4 is ${O(\epsilon|\ln\epsilon}),$ which goes to zero as $\epsilon\to0.$

Take the limit, and the integral around the closed curve tends to

$2\int_{0}^{\infty}\frac{\ln x}{1+x^{2}}\,dx+i\pi\cdot\frac{\pi}2.$

By the residue theorem, this equals

$2\pi i\text{Res}_{z=i}f(z)$

$=2\pi i\lim_{z\to i}\frac{(\log z)(z-i)}{(z+i) (z-i)}=2\pi i\cdot\frac{\frac{i\pi}2}{2i}=\frac{i\pi^{2}}2$

So we have

$2\int_{0}^{\infty}\frac{\ln x}{1+x^{2}}\,dx+\frac{i\pi^{2}}2= \frac{i\pi^{2}}2$

from which $\int_{0}^{\infty}\frac{\ln x}{1+x^{2}}\,dx=0.$

---

Not pretty at all. I much prefer the solution with the odd symmetry.

More exactly, for $0<t<\frac{\pi}{4}$ define $F(t)=\int_{\frac{\pi}{4}-t}^{\frac{\pi}{4}+t}\ln \tan x\ dx\ .$
Thus, $F(t)=\int_{-t}^{t}\ln\tan\left(\frac{\pi}{4}+x\right)\ dx=$ $\int_{0}^{t}\ln\left[\tan\left(\frac{\pi}{4}+x\right)+\tan\left(\frac{\pi}{4}-x\right)\right]\ dx=0\ .$
Therefore,  $\int_{0}^{\infty}\ln\tan x\ dx=\lim_{t\nearrow\frac{\pi}{4}}F(t)=0\ .$


how can u factor the following

5(3-4x)^2 - 8(3-4x)(5x-1)

Notice the common $ 3\minus{}4x$ term and proceed accordingly.


search
group theory
abstract algebra
Euler
function
number theory
totient function

Show that in a finite group with order n,written multiplicatively,the equation x^m=e,(e is the identity) has exactly m solutions x in the group for each positive integer m that divides n.

I think you mean: Show that in a finite group $ G$ of order $ n$, [b]if [/b]the equation $ x^m\equal{}e$ has at most $ m$ solutions for $ m \mid n$, [b]then [/b]$ G$ is cyclic.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1961157375&t=90624 for discussion and solution.

I think he meant what he said.  That's a much harder problem.

That isn't true. Take the Klein 4-group. The equation $ x^2\equal{}e$ has $ 4$ solutions.

Yes,i mean what i said by the title "Cyclic Group".I can't prove the "exactly" m solutions.

Can anyone help?

I believe we can prove that for any finite cyclic group $ G$ of order $ n$ , the equation $ x^m\equal{}e$ has exactly $ gcd(m,n)$ solution in $ G$. This is just a special case when $ gcd(m,n)\equal{}m$

Well... in a cyclic group, for every divisor $ d$ of $ n$, there is only one subgroup of order $ d$. This subgroup $ C_d$ is cyclic and for every elements $ x \in C_d$, $ x^d\equal{}1$.
Let's suppose there exist an elements $ g \in G$ such that $ g^d\equal{}1$ and $ g \notin C_d$. Let's $ o(g)\equal{}s$; from the identity $ g^d\equal{}1$ we can find that $ s|d$. $ C_d$ is cycic and so there exist a cyclic subgroup $ C_s \subset C_d$, but $ |C_s|\equal{}|<g>|$ and a cyclic group have only a subgroup for every divisor of $ n$ and so we have $ C_s\equal{}<g>$ and $ g\in C_d$.
So the solution of the equation $ x^d\equal{}1$ in $ G$ are only the elements of the subgroup $ C_d$ and so they are $ d$.

Anyway I consider more interesting the fact that in a cyclic group the set of elements with order $ d$ with $ d|n$ are exactly $ \varphi(d)$ and so that $ \sum_{d|n} \varphi(d) \equal{} n$ where $ \varphi(d)$ is the Euler totient function of $ d$.

[quote="shyong"]I believe we can prove that for any finite cyclic group $ G$ of order $ n$ , the equation $ x^m \equal{} e$ has exactly $ gcd(m,n)$ solution in $ G$. This is just a special case when $ gcd(m,n) \equal{} m$[/quote]

Well, I think is easier to prove the special case first and then to use it to prove the more general one.

In fact $ x^k\equal{}e$ has only the solution $ x\equal{}e$ if $ gcd(k,m)\equal{}1$. So if $ m\equal{}k\cdot gcd(m,n)$ then $ x^m\equal{}e\ \Rightarrow \ x^{k\cdot gcd(m,n)}\equal{}e\ \Rightarrow \ (x^{gcd(m,n)})^k\equal{}e$ that it's true only when $ x^{gcd(m,n)}\equal{}e$.
$ gcd(m,n)|n$ and so for the special case that I have already proved you have that $ x^m\equal{}e$ has exactly $ gcd(m,n)$ solutions.

we can pick a generator $ a\in G$ and notice that if $ gcd(m,n)\equal{}d$ .Suppose $ a^x\in \left<a\right>$, for some $ 0\leq x\leq n\minus{}1$ satisfy the equation then try to solve for $ a^{xm}\equal{}e\equal{}a^{nk}$ for non negative integer $ k$.

we will see that $ x\equiv 0\left(\mod \frac{n}{d}\right)$


2007 年度　　大阪医科大学

$\boxed 1$
　$a$を定数とする。$0\leq x\leq 1$の範囲の曲線 $y=|x(x-a)|\sqrt{1-x}$ と$x$軸で囲まれる部分を $x$ 軸の周りに回転してできる回転体の体積を $V(a)$ とする。
(1)　$V(a)$ を求めよ。
(2)　$V(a)$ を最小にする $a$ を求めよ。

$\boxed 2$
(1)　$2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta)$ を用いて、$(\cos x+\cos 3x+\cos 5x)\sin x =\frac{1}{2}\sin 6x$ を示せ。

(2)　$f(x)=\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}$ とおく。区間$\left[ 0,\frac{\pi}{2}\right]$ における $f(x)$ の最大値を計算せよ。

$\boxed 3$
　$xyz$ 空間にある１辺の長さが１の立方体(周囲の面およびそれで囲まれる内部)$Q$ の頂点を $O(0,0,0),A(1,0,0),B(1,1,0),C(0,1,0),D(0,0,1),E(1,0,1),F(1,1,1),G(0,1,1)$ とする。辺 $DE$ 上の点 $H(a,0,1)$ と辺$FG$ 上の点 $I$ をとり、3点 $O,H,I$ を通る平面を $p$ とする。
(1)　平面$p$ と直線 $BC$ の交点を $J$ とする。$J$ の座標を求めよ。
(2)　$b<a$ のとき、$p$ による立方体 $Q$ の切り口は平行四辺形にならないことを示せ。

$\boxed 4$
(1)　$a>0$ と $0\leq t\leq 1$ を定数とし、$b\geq a$ の範囲で $b$ を変数として、
　　　　　　　　　$f(b)=\log(a+t(b-a))-\log a-t(\log b-\log a)$
　　とおく。$f(b)\geq 0$ を示せ。
(2)　$0<a<b$ とする。$a\leq x\leq b$ に対して $t=\frac{x-a}{b-a}$ とおき(1)を利用して、
　　　　　　　　　$\frac{\log b-\log a}{b-a}(x-a)+\log a \leq \log x$
　　を示せ。
(3)　自然数 $n(\geq 2)$ に対して、$\sum_{k=1}^{n-1}\frac{1}{2}\{ \log k+\log (k+1) \}\leq \int_{1}^{n}\log x dx$ を示せ。
(4)　自然数 $n(\geq 2)$ に対して、$\log (n!) \leq \left( n+\frac{1}{2}\right)\log n-n+1$ を示せ。

$\boxed 5$
　１から６までの数字を１つずつ記した６枚のカードがある。６枚のカードをよく混ぜて２枚取りだし、取り出したカードに記された数字を２つ記載して、取り出したカードを戻す。この操作を繰り返す。
(1)　２回の操作が済んだとき、リストに記された相異なる数字が２，３，４種類である確率をそれぞれ求めよ。
(2)　３回の操作が済んだとき、リストに記された相異なる数字が４，５種類である確率をそれぞれ求めよ。
(3)　３回の操作が済んだときリストに記された相異なる数字が５種類以下で、かつ４回の操作が済んだとき６種類である確率を求めよ。

…というわけで今日受けてきました!! :lol:
全体的にはいつもより簡単で時間も余るくらいだったので、理科、英語で差がつくんじゃないかと思います。
とか言いながら、速報を見ると５番の(2)の片方と(3)を間違えていたのでたぶん85～90点くらいです… :blush:

porteさん、問題提出ありがとうございます。 :)
やはり、例年のことながら、医学部は高得点争いになりそうですね。
是非、この難関に打ち勝って栄冠を勝ち取ってください。
ご健闘をお祈りいたします。

kunny


I checked this a thousand times and read over the question but I just can't get it

-------------
A circle has radius $ 3.4$cm. Find, correct to the nearest $ mm$, the length of an arc of this circle that subtends an angle at the centre of $ 40^{o}$.

Find, correct to the nearest square metre, the area of a sector of a circle of radius $ 100m$ if the angle at the centre is $ 100^{o}$

k, um... that equals 2.4cm, so... I don't get how you calculated that.

Something wrong with your formulas.  What's the correct answer to number two?

[hide="1"]Arc Length=$ 2\pi r\frac{\theta}{360}$
For $ \theta\equal{}40, r\equal{}3.4$
$ 2\pi 3.4\frac{1}{9}$
Turns out it's around $ 2.37$.
[/hide]

[hide="2"]Sector Area=$ \pi r^{2}\frac{\theta}{360}$
For $ \theta\equal{}100, r\equal{}100$
$ \pi 10000\frac{5}{18}$
This one comes out to something like $ 8722.22$.
[/hide]

ok i finally got the answers

qn 2. answer $ 8727$
Probably a calculator error you made. Your working is correct. Try radians

The formulas are fine, they're just in radians versus degrees.

Maybe its the fact that you're doing 3*4 instead of 3.4? o-0 I don't think I noticed that before. Or... maybe your calculator broke.

(also, if you use 3.14 instead of pi on your calc, you get 8722.222...) nbd, just pointing something out. (I have too much time on my hands)


function
induction
algebra proposed
algebra

Find all functions $f: R\to R$ such that:
                            $f(x^{2}+f(y)-y)=f^{2}(x) \forall x,y \in R$

I can prove that $f(x)=0$ or $f(x)=1$ or $f(x)=x$ under the additional assumption that exists a point $y_{0}$ such that $f$ is continous at $y_{0}$, but i dont know if it follows from conditions.
Here it is: let $A=\{t: \exists y\ t=f(y)-y \}$. Note that for all $t_{1}\ne t_{2}\in A$ $t_{2}-t_{1}$ is a period for $f$. If exists $y_{0}$ such that $f$ is continous at $y_{0}$, $t_{2}-t_{1}$ can be made arbitrarily close to zero, so $f$ must be constant due to continuity at $y_{0}$. From this we get $f(x)=0$ or $f(x)=1$. If $A$ consists of 1 element then $f(x)=x$.

Interesting! But both proofs are flawed in many ways.
However, there is enough for me to complete a solution.
(long and tedious, hope someone can shorten it   :blush: )

(*) $f(x^{2}+f(y)-y)=f(x)^{2}$
Assume $f(x)\not =x$ for some $x\in\mathbb{R}$.

(1) There exists a $t>0$ with $f(x+nt)=f(x)$ for all $x\geq 0, n\in\mathbb{N}$:
If $f(x)-x=c$ constant for all $x\in\mathbb{R}$, then (*) gives
$x^{2}+2c=f(x^{2}+c)=f(x^{2}+f(y)-y)=f(x)^{2}=(x+c)^{2}$ for all $x,y\in\mathbb{R}$
$\Rightarrow 2c(1-x)=c^{2}$ constant for all $x\in\mathbb{R}$
$\Rightarrow c=0$ and $f(x)=x$ for all $x\in\mathbb{R}$.
Contradiction!
So $t_{1}=f(x_{1})-x_{1}<f(x_{2})-x_{2}=t_{2}$ for some $x_{1},x_{2}\in\mathbb{R}$.
Then $f(x^{2}+t_{1})=f(x)^{2}=f(x^{2}+t_{2})$ for all $x\in \mathbb{R}$ using (*)
and $f(x)=f(x+(t_{2}-t_{1}))$ for all $x\geq t_{1}$.
By induction we get $f(x)=f(x+n(t_{2}-t_{1}))$ for all $x\geq t_{1},n\in\mathbb{N}$
and $f(x+n(t_{2}-t_{1}))-(x+n(t_{2}-t_{1}))=f(x)-x-n(t_{2}-t_{1}) \mathop{\longrightarrow}_{n\rightarrow \infty}-\infty$.
So we may choose $t_{1}=f(x_{1})-x_{1}\leq 0$ and $t: =t_{2}-t_{1}>0$,
such that $f(x)=f(x+nt)$ for all $x\geq 0, n\in\mathbb{N}$.

(2) $f(x)\geq 0,\ f(f(x^{2}))=f(x)^{2},\ f(-x)=f(x)$ for all $x\in\mathbb{R}$:
For all $z\in\mathbb{R}$ we have $z\geq f(0)-nt=f(nt)-nt$
and $z=x^{2}+f(nt)-nt$ for some $n\in\mathbb{N},x\in\mathbb{R}$ using (1).
And now (*) gives $f(z)=f(x)^{2}\geq 0$ for all $z\in\mathbb{R}$.
$y=x^{2}$ in (*) gives $f(f(x^{2}))=f(x)^{2}$ for all $x\in\mathbb{R}$
and then $f(-x)^{2}=f(f((-x)^{2}))=f(f(x^{2}))=f(x)^{2}$ and $f(-x)=f(x)$ for all $x\in\mathbb{R}$.

(3) $f(f(x))=f(x)$ and $f(x^{2})=f(x)^{2}$ for all $x\in\mathbb{R}$:
We get for all $x\in\mathbb{R}$ using (*) and (2)
$f(x)^{2}=f(x^{2}+f(-f(x^{2}))-(-f(x^{2})))=f(x^{2}+f(x)^{2}+f(x^{2}))$ and
$f(f(x))^{2}=f(f(x)^{2}+f(-x^{2})-(-x^{2}))=f(f(x)^{2}+f(x^{2})+x^{2})$.
So $f(x)^{2}=f(f(x))^{2},\ f(x)=f(f(x)),\ f(x^{2})=f(f(x^{2}))=f(x)^{2}$ for all $x\in\mathbb{R}$ using (2).

And finally
(4) $f$ is constant:
Let's show $f(a)=f(b)$ for all $a,b\geq 0$. The rest follows from (2).
(1) gives $f(a)=f(a+nt)$ for all $n\in\mathbb{N}$.
So let's assume $a\geq b+t^{2}$ and $y: ={a-b-t^{2}\over 2t}\geq 0$.
Then $a+nt\geq f(y^{2})-y^{2}$ and $a+nt=x^{2}+f(y^{2})-y^{2}$ for some $n\in\mathbb{N},x\in\mathbb{R}$.
But (3) and (1) give $f(y^{2})=f(y)^{2}=f(y+t)^{2}=f((y+t)^{2})$
and then $b+nt=a+nt-(a-b)=x^{2}+f(y^{2})-y^{2}-(2yt+t^{2})=x^{2}+f((y+t)^{2})-(y+t)^{2}$.
So (*) and (1) give $f(a)=f(a+nt)=f(x)^{2}=f(b+nt)=f(b)$.

Now if $f\equiv c$, then (*) gives $c=c^{2}$ and $c\in\{0,1\}$.
So there are $3$ solutions $f\equiv 0, f\equiv 1$ and $f(x)=x$ for all $x\in\mathbb{R}$.

For problem : Find all functions $f: R\to R$ such that:
                            $f(x^{2}+f(y)+y)=f^{2}(x) \forall x,y \in R$

See http://www.mathlinks.ro/Forum/viewtopic.php?p=18289#p18289  :wink:


Do you mind if I ask where you go to school and what type of school it is?

Suuuuuuuuuuuure
I live in NYC and go to Forest Hills High School (that's a public school with approximately 3800 students)

I'm a junior, applying second time
Less than 2 years in the US--English is very bad
As a result, I have awful PSAT, awful SAT II, and awful AP
Which means I dont have a slightest chance of being accepted :)

What about you, may I ask?

Absolutely. I'm a sophomore; I live in Oregon; I attend a public high school which currently has ~1500 people in it. Compared to the other sophomores taking the PSAT I did pretty well, but there's definitely room for improvement. I'm currently taking Biology, Physics, and Precal (having challenged Algebra 2) for math/sci courses. I did an interesting independent study earlier this year that involved computer programming and physics, which should help my app. Overall though, I have no idea whether I'll get in. I've never applied to a summer program before, and have no idea how prepared the other applicants will be.

Hopefully we'll run into each other this summer.

What is SSP? Harvard SSP? http://www.ssp.harvard.edu/

eeer....no
another SSP-summer science program, that's held on two campuses in California and New  Mexico, and....but why am i advertising it here?

take a look at it here

http://www.summerscience.org

I've applied this year as an International student from India.

I got in/probably will go this year, Socorro anyone?

I got rejected first round apps. Being an international student is tough lol!


search
combinatorics proposed
combinatorics

There were $n$ people present at a New Year’s party. Determine the value of n, given the following information:
• $n$ is not a multiple of $11$, and $5\leq  n \leq 30$
• Each pair of strangers had exactly two common acquaintances and each pair of acquaintances had common acquaintances.

[quote="cuopbientcd"]It is posted before
 :lol:[/quote]
Link?

Nothing is coming up in the search, can you give a little more info about how the previous problem was stated?


invariant
function
algebra proposed
algebra

The three roots of $P(x) = x^3 - 2x^2 - x + 1$ are $a>b>c \in \mathbb{R}$. Find the value of $a^2b+b^2c+c^2a$.  :D

It is not correct.
We have $a+b+c=2,ab+bc+ca=-1,abc=-1,c<0<b<a.$ Let $u=a^2b+b^c+c^2a>0,v=ab^2+bc^2+ca^2.$
u and v invariant by $s=(a\to b\to c),s^2$ But simmetric functions $u+v,uv$ invariant for all permutations.
We have $u+v=(a+b+c)(ab+bc+ca)-3abc=1,$ and
$uv=a^3b^3+b^3c^3+a^3c^3+3a^2b^2c^2+abc(a^3+b^3+c^3)$.
$a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3-3abc(u+v)-6(abc)^2=-4\Longrightarrow uv=-4+3-(a^3+b^3+c^3).$
$a^3+b^3+c^3=(a+b+c)^3-3(u+v)-6abc=8-3+6=11,uv=-12$. Therefore u and v roots equation:
$z^2-z-12=(z+1)(z-12)$. Because u>0, we have u=12.
$a^2b+b^2c+c^2a=12.$

Yes right. It's the same as our "official" solution

since $\sum a^2b+\sum ab^2=(ab+bc+ca)(a+b+c)-3abc=1,\sum a^2b-\sum ab^2=(a-b)(b-c)(a-c)\ge 0$
we only have to compute$|\prod (a-b)|.$
notice that $P'(a)=(a-b)(a-c)$,etc.
so$|\prod (a-b)|=\sqrt{|P'(a)P'(b)P'(c)|}=\sqrt{\prod{3a^2-4a-1}}=7$
the last step was achieved by Newton's formula(and brutal force!)
hence the vaule is $4$

actually,we can compute $\sum_{sym}x^{2}y$ and $(\sum_{cyc}x^{2})(\sum_{cyc}xy^{2})$
by Newton theorem

Assume the roots of the equation $P(x)=x^3-2x^2-x+1$ are $\alpha,\beta,\gamma.$ By Vieta's Relation we get, $$\alpha+\beta+\gamma=2$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=-1$$ $$\alpha\beta\gamma=-1$$ Given $\alpha>\beta>\gamma\in \mathbb{R}.$ $$\bigg|(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\bigg|=\bigg|\underbrace{(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)}_{=S_1}-\underbrace{(\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2)}_{=S_2}\bigg|$$
Now Simply we have to find an equation whose roots are $S_1$ and $S_2.$ Therefore $$S_1+S_2=\sum_{cyc}{}\alpha^2(\beta+\gamma)\implies\sum_{cyc}{}\alpha^2(2-\alpha)\implies 2(\sum_{cyc}{}\alpha^2)-(\sum_{cyc}{}\alpha^3)\implies12-11=\boxed{1}$$ $$S_1\cdot S_2=\sum_{cyc}{}\alpha^4\beta\gamma+3(\alpha\beta\gamma)^2+\sum_{cyc}{}\alpha^3\beta^3\implies-6+3+-9=\boxed{-12.}$$
Therefore we get, the equation as $$\boxed{x^2-x-12=0=(x-4)(x+3)}$$
Therefore the values are $4$ and $-3$.
Vieta's Relation!! :love:


How exactly would you like to apply game theory to a particular battle?  I can see the case being made that the lead-up to wars can be analyzed by game-theoretic methods (although I'm not sure how good an analysis you could really expect from such a thing), but I really don't see how something like a particular battle qualifies as a game that is easily approachable by mathematics.

[quote="soulsdragon"]Hello,

I was hoping someone could help me out with an argument i am having with a friend.

Can Game theory be applied to war? If so, how? I believe yes, however, am not so versed in Game thoery itself.

For example, Can game theory be applied to the Battle of Rorke's Drift in 1879 in which 139 British soldiers successfully defended a mission station against over 4000 Zulu warriors.

Help would be greatly apprecited!! Thanks!!![/quote]

Not game theory perhaps, but yes COMMON SENSE (that what you really need, ajajajaj)


topology
advanced fields
advanced fields solved

Is $PGL(n,K)$ connected for
(i) $K = \mathbb{R}$, usual topology
(ii) $K = \mathbb{C}$, usual topology
(iii) $K$ field, Zariski topology

For $\mathbb{C}$, it's always connected; $GL_{n}(\mathbb{C})$ is connected, and the quotient of a connected set is connected. For $\mathbb{R}$, it's connected if $n$ is odd and not if $n$ is even; the issue is whether the components with positive and negative determinant are identified by the quotient map.

In the Zariski topology on $K^{n^{2}}$, all open sets are connected since any two nonempty open sets intersect. The set of invertible matrices is open, so $GL_{n}(K)$ is connected in the Zariski topology, and the quotient $PGL_{n}(K)$ is also connected.

For $iii$, I think you need an extra assumption on the field. For example, it should not be finite.

Yes, we need that. The Zariski topology is the discrete topology over a finite field.


find out [b]ALL[/b] a,b,c,r from following equations:
a^2 + b^2 =c^2    ("^2" square....hope everyone understand :P )
[b]and[/b]
(a+r)^2 + (b+r)^2 = (c+r+1)^2

Can anyone help me solve this, plz?? :blush:

I have an example:
a=5 b=12 c=13 and r=3
5-12-13   --->  8-15-17

$a^{2}+b^{2}=c^{2}$ is the pythaogrean theorem.

i think your looking for the triples?

Some Pythagorean Triples:
$x$ $y$ $z$

$3,4,5$

$5,12,13$

$7,24,25$

$9,40,41$

$11,60,61$

from here you can notice that
$x$ increases by 2 and $y$ increases by $8,12,16,20$ and $z=y+1$

Some other useful Pythag. triples are $20-21-29$, $8-15-17$. Also notice that for any odd number n, the other pythagorean triples can be found of n is a leg. $(N^{2}+1)/2$ and $(N^{2}-1)/2$.

I'll try to guess at what you want:
From the first equation, it's fairly obvious you can rewrite it as:
$c = \sqrt{a^{2}+b^{2}}$
Going into the second equation, make that substitution:
$(a+r)^{2}+(b+r)^{2}= (\sqrt{a^{2}+b^{2}}+r+1)^{2}$
Expand, and collect r terms.

I've solved this in terms of c, and I don't want to explain all the algebra behind it.

$c = \frac{r^{2}-2r+2ar+2br-1}{2r+2}$
I'm not entirely sure what the question is.
See if that helps at all.

[quote="FaBreGas-4"]a^2 + b^2 =c^2    ("^2" square....hope everyone understand :P )
[size=200][b]and[/b][/size]
(a+r)^2 + (b+r)^2 = (c+r+1)^2[/quote]

The [b]and[/b] has been sadly ignored by the first two posters.  :(

DiscreetFourierTransform (did you misspell "discrete" intentionally?), you seem to have made a poor substitution similar to that made by another poster on a similar problem involving squares.

I would suggest instead expanding the second expression and then seeing what you can do to simplify it using the first. :)

Hehe, yeah, It was a mispelling...  :maybe:
Yeah, I see the issue at hand. I'm working on that last bit.
Ok, the expansion works out as
$a^{2}+b^{2}+2ar+2br+2r^{2}= c^{2}+2cr+2c+r^{2}+2r+1$
Then, because we know $a^{2}+b^{2}= c^{2}$, we can remove those and rewrite as
$2ar+2br+r^{2}= 2cr+2c+2r+1$
EDIT: Got it: add ONE to each side, then factor by grouping.

Shoot... I'm STILL stuck.

Let's try finding only the [b]integer[/b] solutions, then.   :D

Edit:  Also, you seem to have written $2cr$ as $cr$.  In any case, blindly applying factoring tricks to only [b]part[/b] of an expression won't get you far, since the rest of it remains unfactored.

Yeah, fixed that. Working on more integer solutions.

Ugg. Nevermind. It's late, and I need to get back to other homework. I'll check back for a solution later. See if what I have is a start.

Hi...I'm the poster....I got another solution....but this is one of them....I have to find out more...

 a   b   c      r
36-77-85     3
||   || ||
39-70-89  (89=85+r+1)

My question is to find out ALL a,b,c,r from those  2 equations....You have identify it.....
and...english is not my native language....i try to explain clearly.. :P

PS: the way of the solution which is above ... I used the following:
a = n^2-m^2     b = 2mn     c = n^2+m^2
where n,m are integer and n>m

But....so far, I just these 2 pairs..... :(


Let a triangle ABC (BC>AC). H is on BC such that $ \angle HAC\equal{}\angle ABC$. The bisector of angle BAH meets BH at E. M is the midpoint of side AB. $ F\equal{}ME\cap AH$. Show that $ CF\parallel AE$.

Let $ BX$ be a cevian such that $ BX,AE,MH$ are concurrent.
Now we know that, $ (F,H,X,A) = - 1$ (becuase these points are harmonic)
That is, $ \frac {XF}{XH} = - \frac {AF}{AH}$
We draw a line Through $ C$ parallel to $ AE$.Assume that, it intersects $ AF$ at $ F'$

So, we just need to prove that,
\[ \frac {XF'}{XH} = - \frac {AF'}{AH}
\]
This can be easily shown after some calculation.
[hide="Calculation:"]
(It is actually too long and boring, so I am not going to show it)
But what I did is,
1.Used the fact, $ \triangle{ABC}\sim \triangle{HAC}$ I got $ AH$ and $ HC$
2.Using Ceva's theorem I got $ \frac {XA}{HX} = \frac {BE}{AH} = \frac {BA}{AH} = \frac {b}{a}$ and from here I got, $ HX$
3.Using the fact, $ \triangle{AEH}\sim \triangle{F'CH}$ I got $ HF'$
4.Using all these results I got, $ XF'$

So at last I showed that, $ XF'\cdot AH = AF' \cdot HX$
That is,
\[ \frac {XF'}{XH} = - \frac {AF'}{AH}
\]
[I expressed every length in terms of $ a,b,c$]
[/hide]

for menelaus on BHA and EM we have $ \frac{\overline{FH}}{\overline{FA}} \equal{} \frac{\overline{HE}}{\overline{EB}} \equal{} \frac{\overline{HA}}{c}$ and so $ \frac{\overline{HA}}{\overline{FH}} \equal{} \frac{c \minus{} \overline{HA}}{\overline{HA}}$ but the triangles $ \triangle AHC$ and $ \triangle BAC$ are similar so $ \frac{\overline{CH}}{b} \equal{} \frac{\overline{HA}}{c}$ and then $ \frac{\overline{HA}}{\overline{FH}} \equal{} \frac{b\minus{}\overline{CH}}{\overline{CH}} \equal{} \frac{\overline{HE}}{\overline{HC}}$.
Then the triangles $ \triangle HAE$ and $ \triangle HFC$ are similar and so $ CF \parallel AE$

Applying the [b][size=100]Menelaus theorem[/size][/b], for the triangle $ \bigtriangleup ABH,$ with transversal $ MEF,$ $ \Longrightarrow$ $ \frac {FH}{FA}\cdot \frac {MA}{MB}\cdot \frac {EB}{EH} \equal{} 1$ $ \Longrightarrow$ $ \frac {FH}{FA} \equal{} \frac {EH}{EB}$ $ ,(1)$

Because of $ AE$ is the bisector of the angle $ \angle BAH,$ we have that $ \frac {EH}{EB} \equal{} \frac {AH}{AB}$ $ ,(2)$

The triangles $ \bigtriangleup HBA,$ $ \bigtriangleup HAC,$ have $ \angle ABH \equal{} \angle HAC$ and $ \angle AHB \plus{} \angle AHC \equal{} 180^{o}$ and so,

we conclude that $ \frac {AH}{CH} \equal{} \frac {AB}{CA}$ $ \Longrightarrow$ $ \frac {AH}{AB} \equal{} \frac {CH}{CA}$ $ ,(3)$

From$ (3)$ $ \Longrightarrow$ $ \frac {AH}{AB} \equal{} \frac {CH}{CE}$ $ ,(4)$ because of $ CA \equal{} CE$ $ ($ from $ \angle EAC \equal{} \angle EAH \plus{} \angle HAC \equal{} \angle EAB \plus{} \angle ABE \equal{} \angle AEC$ $ ).$

From $ (1),$ $ (2),$ $ (4)$ $ \Longrightarrow$ $ \frac {FH}{FA} \equal{} \frac {CH}{CE}$ $ ,(5)$

From $ (5)$ we conclude that $ CF\parallel AE$ and the proof is completed.

Kostas Vittas.

PS. [b][size=100]Gabriel[/size][/b] is a faster man :)

[size=150][color=darkblue][b]Easy and nice problem ![/b][/color][/size]

[quote="mr.danh"] [color=darkred]Let $ ABC$ be a triangle with $ a > b$ . Consider $ H\in (BC)$ such that $ \widehat{HAC}\equiv\widehat {ABC}$ . The bisector of

$ \widehat{BAH}$ meets $ BH$ at $ E$ . Denote the midpoint $ M$ of $ [AB]$ and  $ F \equal{} ME\cap AH$ . Show that $ CF\parallel AE$ .[/color] [/quote]

[b][u]Proof (similar with [color=darkred]Kostas Vittas'[/color]).[/u][/b] $ \left\{\begin{array}{c} \widehat {CAE}\equiv\widehat {CAH} \plus{} \widehat {EAH} \\
 \\
\widehat {CEA}\equiv\widehat {CBA} \plus{} \widehat {EAB} \\
 \\
(\ \widehat {CAH}\equiv\widehat {CBA}\ ,\ \widehat {EAH}\equiv\widehat {EAB}\ )\end{array}\right\|$ $ \implies\ \widehat {CAE}\equiv\widehat {CEA}\ \implies\ \boxed {\ CA \equal{} CE\ }$ .

$ \left\{\begin{array}{ccccc} \left\|\begin{array}{ccc} CAH\sim CBA & \implies & \frac {AH}{BA} \equal{} \frac {CH}{CA} \\
 \\
\widehat {EAB}\equiv\widehat {EAH} & \implies & \frac {AH}{AB} \equal{} \frac {EH}{EB}\end{array}\right\| & \implies & \left\|\begin{array}{c} \frac {CH}{CA} \equal{} \frac {EH}{EB} \\
 \\
(\ CA \equal{} CE\ )\end{array}\right\| & \implies & \frac {CH}{CE} \equal{} \frac {EH}{EB} \\
 \\
\mathrm {Menelaus\ for\ }\overline {MEF}\mathrm {\ and\ }\triangle\ ABH \ : & \implies & \left\|\begin{array}{c} \frac {FH}{FA}\cdot\frac {MA}{MB}\cdot\frac {EB}{EH} \equal{} 1 \\
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(\ MA \equal{} MB\ )\end{array}\right\| & \implies & \frac {FH}{FA} \equal{} \frac {EH}{EB}\end{array}\right\|$ $ \implies$ $ \frac {CH}{CE} \equal{} \frac {FH}{FA}$ $ \implies$ $ CF\parallel AE$ .