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Hi,

This is best wishes for everyone who will be taking the AMC 8 exam this week, especially first-timers. The official testing period has already begun in a few time zones (although it's pretty early in those places).

Here's a friendly reminder: some test-takers will be dying to discuss the content of the AMC 8 on this forum, which has become active since the last AMC 8 exam was given in 2002, but PLEASE don't type a word about the exam in ANY online forum until after the exam-taking window period is over, which means not until AFTER 25 November 2003. Once all the official exam answer sheets have been put in the mail to Lincoln, Nebraska for scoring, only then will we be permitted to discuss the problems on this year's exam.

http://www.unl.edu/amc/e-exams/e4-amc08/amc8.html

The AMC 8 was a good experience for my test-taking group last year. I hope you enjoy it, and I hope if this is your first math competition experience that you will stay involved for future contests.

Rest easy until you take the exam,

Indeed.  Good luck, MathFiend, Syntax Error, Chinaboy, Tare, and everyone else who's taking the exam.  (Although I'm not sure if all of those people are).

To all AMC 8 competitors:

From the AMC staff in Lincoln, NE:

Best of luck, and happy problem solving to all the AMC 8 contestants, starting on Tuesday, Novmber 18, and going through the testing period until November 25, 2003.  We hope you enjoy the contest we have put together for you.

Do allow me to repeat what TokenAdult said in starting this thread:  I know you will be anxious to discuss the contest with your friends, both on-line and off-line.  But as we say in the AMC 8 Teachers' Manual on page 5, in Section VIII.11.  "...the contest and solutions may not be discussed with anyone outside of their contest administration either orally, via email, www, copier, or media of any type until after November 25."  If you must, discuss the contest with your teacher/contest administrator, but wait a week before discussing here or elsewhere.

Thanks!

Steve Dunbar, AMC Director, for the AMC staff and the AMC 8 Sub-Committee of the MAA

I'll ask this directly to you Mr. Dunbar*: Is there still a way to register for the AMC 8?

I really do want to take it but my school still refuses to start math programs such as AMC 8 or MATHCOUNTS mainly because there's no one who wants to coordinate and I'm wondering if I can join other schools instead for the contests.

*Moderator's note: corrected.

Dear Tare --  Given that the main date for the AMC 8 is tomorrow (although the official contest period extends through November 25), it will be hard.  But give the AMC office a call tomorrow, give us your zip code, and we'll see if we can locate other schools in your area administering the contest.  So best answer is to contact me, or the AMC office early on Tuesday.  Contact info is on our web site.

-- Steve Dunbar for the AMC office.

when they say there is only a handful of people that gets a perfect on the AMC 8. How many exactly is a handful? And what about the honor roll thing?

I counted 126 kids from the U.S., plus 59 from Taiwan, bringing a total of 185 (I can add!)

Oh, wait, it says 126 right on the front page.  Meh.  All that counting wasted.  Let's do some math.  There were 157,778 students taking the exam.  126/157,778 means the top .000798 = 0.0798%.

I'm not sure what the exact standard for "honor roll" is, though.

One other thing: in response to MysticTerminator's (earlier) claim that people who did well on the AMC8 were invited to take the AMC10, it says in the booklet that "high scoring students on the 2003 AMC8 will be invited to participate in the 2004 AMC10A and 10B and/or AMC 12A and 12B which are scheduled for TUESDAY, February 10, 2004 and/or February 25, 2004.

Meaning that last years scorers will be "invited" (i.e. encouraged) to take the AMC10/12 this year.

On a side note, I realized that we finally know the identity of the formerly mysterious "AMCDirector".  (Before, I wasn't sure which director it was: J. Douglas Faires?  Titu Andreescu?  David Hankin? etc...)  Not that it matters.

[quote]Syntax Error Posted: Mon Nov 17, 2003 6:05 pm    Post subject:

--------------------------------------------------------------------------------

wait wait wait. 8th graders can still take the amc10 and 12 and stuff righteous? [/quote]

yes, i took the AMC-10 last year as an 8th grader, but i didn't do as well as i thought... lots of dumb mistakes.

[quote="Leeyuhang"]honor roll is for those who received a 20 or higher.[/quote]

That's probably pretty close, numerically, most years, but I'm pretty sure honor roll on the AMC 8 is simply defined as the top 2 percent of test-takers. The AMC 8 is not an "equated" test in which a particular item content score has the same difficulty year by year, so it is possible that one year the cut-off score for honor roll might be different from what it is in another year.

Over the past few years, a 20 on the AMC 8 has certainly been a very respectable score. The curriculum level of the questions is no higher than prealgebra, but the questions are deep enough to require some thoughtful approaches and MASTERY of that level of math.

Good luck to everyone tomorrow . . . .

Oh, too bad Tare doesn't live here in Minnesota, or I would gladly have let him join our local administration of the test.

http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2002-8a/02stats8.html

lol, scores: 25:99+27=127????

[quote="Syntax Error"]wait wait wait. 8th graders can still take the amc10 and 12 and stuff righteous?[/quote]

This gets back to the thread I opened a while ago about grade-skipping in math contests. My son doesn't have a declared grade skip (I call him a sixth-grader, when I think about that silly issue at all, as that is how he would be placed by age in the local public school). As a homeschooler, I can be vague about what grade my children are in, and flexible about what subjects they cover in what year. But it is my intention to sign up my son for the AMC 10 this year (I have just found another homeschooling parent in Minnesota who is encouraging other parents to sign up) even though he will still be only eleven years old at the time he takes the 2004 AMC 10. I would hardly expect him to ace that test the first year he takes it, or even the second or third [grin], but taking the test is good experience. Taking the AMC 8 last year at fifth-grade age was motivational for him, I think, in learning a lot of math in the last year.

Most intellectual contests in the United States don't seem to have a younger age limit. Ask around, and see if you can crash some other group for taking your preferred test if your school won't help you sign up.

[quote="ml2"]lol, scores: 25:99+27=127????[/quote]

Actually, one of them might have failed to mark their gender (or their response wasn't readable - say they marked both, or scribbled over the space), thus accounting for 127 total perfect scores but only 126 acknowledging one gender or another.

Of course, the official results booklet says 126, so I am inclined to believe this figure instead.

Although I don't know how accurate all of this information is.  When I looked at the stats for individual states, there were no 24s in my state (even though my teacher's official results acknowledged that I got a 24).  I wonder what happened with that.

[quote="Syntax Error"]wait wait wait. 8th graders can still take the amc10 and 12 and stuff righteous?[/quote]
Yes, I already took the AMC 10 last year (but not AMC 8 because my middle-school doesn't offer anything but my high school which I take math at offer everything.) and there was no problem marking off "below 9th grade." In fact I believe if you're 8th grade or below and have access to high schools which offer AMC 10/12 then you could take all three.

I'm wondering who's in charge of Massachusetts? If there's no state director than is Mr. Dunbar in charge for those states?

I'm pretty sure Massachusetts has a state director.  Although the page is gone from the website, I still have my handy results book.

Let's see...state director for massachusetts is Prof. Richard Quindley, Dept. of Mathematics, Bridgewater State College, Bridgewater, MA  02325...no, you won't be able to send him a letter in time...so email is 'rquindley at bridgew.edu', with the "at" replaced by @. (I don't want spam robots to pick the address up.)

OK, I've already mailed Mr. Dunbar so I'll mail him after I get a reply.

Where did you get that handy-dandy results book?

Well, the test is over. I think I did well, but I can't say until my class goes over the test tomorrow. I know, no talking about the answers, and I won't. I'm just wondering how the rest of you think you did...


Prove that in a tournamant (any one, not just transitive ones), you cannot have exactly two kings.

(A king is defined as a vertex whose distance from any vertex is at most 2)
(Distance from some vertex a to a vertex b is defined as the shortest directed path from a to b)

first of all you should prove that there always exists a king...
([i]notation:in the following proof from $ x \to y$ we mean that the direction of edge $ xy$ is from $ x$ to $ y$...[/i])
[hide="proof of the existance of king in every tournament"]take the vertex with maximum outgoing degree and name it $ v$ we claim that $ v$ is the vertex we`re searching for...
assume the contrary,i.e. assume that there exists a vertex $ k$ for which we cant reach $ k$ from $ v$ with a path from length at most two...
so we must have $ k \to v$ now let $ S$ denote the set of vertex $ u$ which $ v \to u$ so for every $ u$ in $ S$ we must have $ k \to u$ otherwise the path $ v \to u \to k$ would be a perfect path from $ v$ to $ k$ which is a contradiction,so for every $ u \in S$ we must have $ k \to u$ now let $ d^{+}_{k}$ denote the oudgoing degree of vertex $ k$ now we hvae:
$ d^{+}_{k}=1+d^{+}_{v}>d^{+}v$ which is a contradiction because we assumed that $ v$ has the maximum outgoing degree,so in every tournament there exists a king...[/hide]
now assume that in the tournament $ T$,there exists exactly two kings,name them $ x,y$ and WLOG assume that $ x \to y$
now denote $ S$ as $ S=\{v \in T | v \to x\}$
now note that $ y$ is a king in $ T$,so there must be a path with length at most $ 2$ from $ y$ to every other vertices,including $ x$,but $ x \to y$ so there must exist vetex $ v$ s.t $ v \to x$ thus $ |S|\geq 1$...
now $ S$,itself is a tournament (its a subset of $ T$) so there exists a king in $ S$,name it $ z$,we claim that $ z$ is a king in $ T$...
[b]proof:[/b]first of all note that $ z \in S$ so $ z \to x$ also $ z$ is a king in $ S$ so we can reach every other vertex in $ S$ by a path from length at most $ 2$ from $ z$...
also we know that for every $ u \in T-S-x$ we have $ x \to u$ (otherwis we would have $ u \in S$)...
now consider the path $ z \to x \to u$ so we can reach every vertex in $ T-S-x$ with a path with length at most two from $ z$...
so every vertex in $ T$ is reachable from $ z$ with a path of length at most $ 2$,so $ z$ is a king too,and the claim is proved as wanted...
contradiction...


we choose n+1 point, x0=0,x1,x2,...,xn=1 in [0,1].we asssume the sequence of the distances between these points. in this sequence each numbers occur at least two times  except 1. prove that all xi are rational.

Let c_1,...,c_t) be a basis of the vector space over Q generated by (x_0,...,x_n), with c_1 = x_n, and for all i > 1 c_i = x_j for some j.
Thus, for i = 1,...,n, there exists an unique t-uple (r_1,...,r_t) of rational numbers such that
x_i = sum{1 \leq j \leq n} r_j*c_j.  (1)

We want to prove that t = 1 (which will leads to the fact that each x_i is collinear to x_n = 1 and therefore rational).

Let M (res. m) be the greatest (res. the lowest) of the r_t which appears in the n decompositions of type (1) when i = 1,...,n. Since c_t = c_t and x_0 = 0, it follows that M  \geq  1 and m  \leq  0. Thus M > m.

Let A (resp. a) be the greatest (resp. the lowest) of the numbers x_i for which c_t = M.
Let B (resp. b) be the greatest (resp. the lowest) of the x_i for which c_t is m.

Clearly a  \leq  A and b  \leq  B.
Suppose, for a contradiction, that |A - b| =/= 1.

- If a < B :
Then, there exist x,y among the x_i with (x,y) =/= (b,A) such that y - x = A - b.
Let p_1,...,p_t be the coefficients r_j which appears in (1) for x, and q_1,...,q_t those for y.
If x < b then from the minimality of b, we have p_t > m, and since q_t  \leq  M, the coefficient of y-x on c_t is q_t - p_t < M - m. But, M-m is also the coefficient of A-b on c_t. Contradiction.
Thus, x  \geq  b. Since (x,y) =/= (b,A), we have x > b and y > A.

If y > A then q_t < M, and since p_t  \geq  m it leads to the same contradiction as above.

Then a  \geq  B.

- If b < A :
Let x'_i = 1 - x_iand we use the reasoning above to get the same contradiction.

Then a  \geq  B and b  \geq  A. Thus a  \leq  A  \leq  b  \leq  B  \leq  a.
Then a = A = b = B, from which it follows that M = m. A contradiction.

Thus |A - b| = 1. Then |A - b| = x_n - x_0 = x_n. Since the coefficient of c_t for A - b is M - m  \geq  0, we deduce that A - b = a_n - a_0, that is A = a_n and b = a_0.
Thus A = a_n = c_1 = sum{1 \leq j \leq t} r_j*c_j, with r_t = M =/= 0. Since the c_i's form a basis it follows that t = 1, and we are done.

Remark.
This result is due to Schinzel, and has been generalised by Strauss.

Reference : E.G.Strauss, "Rational dependence in finite sets of numbers", Acta Arithmetica, XI (1965), p. 203-204.

Pierre.


I drew a cube here...

by using x find out the length between A and B

If x is referring to that edge then the space diagonal is always = to x*sqrt3. This is because when finding the space diagonal inside any rectangular prism, the space diagonal is always equal to the squareroot of a^2 + b^2 + c^2, where a, b, and c are the height, width, and length. In a cube though, since all of the dimensions are the same it is really the squareroot of 3x^2 which reduces to x*sqrt3.

[hide]If x=an edge then $x=AB \sqrt {3}$

something to memorize...

[hide]The longest diagonal of a unit cube is sqrt 3

so $x\sqrt 3$[/hide]

[hide="proof"]
Call the vertex directly under A, C.  And the one to the right of C, D

triangle CDB is right.  Its two legs are both length x.  x^2+x^2=2x^2
BC=xsqrt2

Triangle ABC is right becuase the two planes are perpendicular.
The legs are x and xsqrt2

x^2+2x^2=3x^2

AB=xsqrt3[/hide]


if all the points in R^2 are colored with 3 colors

then does the collection of all regular n-gons with monochromatic diagonals form a basis for a topology on R^2

What do you mean with "n-gon"? The whole surface or just the set of vertices?

I would assume the interiors of these polygons, since finite sets of points are not open. Basically, you need to show that any axis-aligned square in the plane contains some above-described n-gon.

You can't assume that, because the question is whether this could be a base for some topology. The discrete topology is a topology, and finite sets are open in it.
We definitely need clarification.

I think examining by the definition would be helpful. This group of set certainly filled up the space.And the second criterion - that any open sets can be filled up by some polygons, is also apparent for we can choose a local polygon inside of any open sets for a given point. Then we're done. I don't think it's interesting to investigate the vicissitudes of possibilities of various kinds of topology.

I think its not the case...

i used it to prove the coloring question in I.N.M.O  :D

but didnt got selected......  :rotfl:


Prove that $ \frac{1}{25}\displaystyle\sum_{k\equal{}0}^{2001}[\frac{2^k}{25}]$ is a natural number.


abstract algebra
linear algebra
matrix
group theory
vector
algebra
function

Let $ f: \mathbb{Z}^n \rightarrow \mathbb{Z}^n$ be a linear map given by a non-singular matrix $ A$. How does cokernel of $ f$ look like (in terms of $ A$) ?

It's a finite subgroup of size $ \det(A)$, if I understand what you're asking.  The image $ f(\mathbb{Z}^n)$ is a lattice spanned by the image of the basis vectors and the elements of the cokernel can be taken to be points in a fundamental region of this lattice.

Ok, I would like to specify, what I mean by above: could you show that when $ A$ is upper-triangular with $ a_1,...,a_n$ on a diagonal then cokernel is isomorphic to $ \mathbb{Z}_{a_1} \oplus ... \oplus  \mathbb{Z}_{a_n}$ ?

It isn't always. For example take
$ \begin{pmatrix} 2 & 1 \\ 0 & 2\end{pmatrix}$

Are you familiar with the structure theorem?  For finite abelian groups it's equivalent to the claim that the cokernel is isomorphic to $ \mathbb{Z}_{d_1} \oplus ... \oplus \mathbb{Z}_{d_n}$ where $ d_1 | d_2 | ... | d_n$ are the [url=http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain]invariant factors[/url] of $ A$.

Thank you for your responses - they are both helpful.


Let $ (z_{n})$ be a complex series with $ z_{n}\to 0$ .

Does always a series $ (\epsilon_{n})\in\{\minus{}1,1\}^{\Bbb{N}}$ exist so that $ \sum_{n \equal{} 0}^{\infty}\epsilon_{n}z_{n}$ converges?

Nice problem. If $ |z_{n}|$ is [i]non-increasing[/i] then choose $ e_{n}$ so that $ \mbox{sgn}(e_{n}z_{n}) \equal{} (\minus{}1)^{n}$ then this will make the series convergece.

But otherwise, I dont have a counterexample.

[quote="jmerry"]You can't make that choice, sylow_theory; we only have $ \pm 1$ to work with, not the whole unit circle.[/quote]

Yes, your right, I was doing this for $ z_{n}\in\mathbb{R}$, his question is more general. My bad.

You may want to look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61690]this topic[/url]


Express the product of $ 2^{35}$ and $ 5^{38}$ in scientific notation.

Scientific notation is a number in the form $ a \times {10}^n$ where $ a$ is a number $ 1 \le a < 10$. The prime factorization of $ 10$ is $ 2^15^1$. Since the number we want to write in scientific notation only contains $ 2's$ and $ 5's$, it makes what we have to do a bit easier. Obviously the $ n$ in the $ a \times {10}^n$ is the smaller one of the exponents, which is $ 35$. The number can thus be written as $ 5^3 \times {10}^{35}$. $ 5^3 \equal{} 125$, but we have to have the $ a$ in $ a \times {10}^n$ be in the interval $ 1 \le a < 10$. When we take a power of $ 10$ out of $ 125$, we have to add a power of $ 10$ to $ {10}^{35}$. The answer is thus $ \boxed{1.25 \times {10}^{37}}$. (in FTW this would be written as $ 1.25 \cdot {10}^{37}$, or $ 1.25 * {10}^{37}$.)


function
algebra
polynomial
algebra unsolved

Find all  bijective functions $f:S\longrightarrow S$ such that $f(3mn + m + n) = 4 f(m) f(n) + f(m) + f(n)$ for all $m, n\in S$ where $S$ is the set of nonnegative integers.

Let A={3k+1| k is non-negative integer}, B={4k+1| k is non-negative integer}. Consider $h:A\to B$ defined by $h(3k+1)=4f(k)+1$. Then we have $h(xy)=h(x)h(y)$.
But unique factorization does not holds in A and B, so we have to ask what their structures are like.
In A, there are three kinds of irreducibles(elements that cannot be expressed as the product of two elements > 1 in A):
1. $p$, where $p=3k+1$ is a prime.
2. $p^2$, where $p=3k+2$ is a prime.
3. $p_1p_2$, where $p_1=3k+2,p_2=3l+2$ are primes.
Also in B there are three corresponding kinds of irreducibles, just replace 3k+1 by 4k+1, 3k+2 by 4k+3.
The structure between the three kinds of irreducibles are different, so every irreducible in A of the first kind must correspond to an irreducible of the first kind in B, and so on. So we are done. It is easy to check the above conditions are enough.

There is still a small problem here, to prove that such h exists, we have to prove that their is a bijective mapping between the first, second, and third kinds of irreducibles. The latter two are clearly infinite and thus denumerable(infinite and countable), so there must be a bijective mapping. For the first kind, they are also denumerably many. But it is not so easy to show that.

To show primes of the form 3k+1 and 4k+1 are infinite, first we consider the factors of Fermat numbers $2^{2^n}+1$ and get infinite primes of the form 4k+1. Next we consider the polynomial $f(n)=n^2+n+1$ and prove that $p|f(n)$ for some n holds for infinitely many p. Finally we prove that such p must be of the form 3k+1.


Find all functions from R->R with the property

$ f(x^3) \minus{} f(y^3) \equal{} (x^2 \plus{} xy \plus{} y^2)(f(x) \minus{} f(y))$  for all x,y at R

Note that if $ f$ is a solution then so is $ af \plus{} b$ for real constants $ a,b$.
Therefore, we may assume that $ f(0) \equal{} 0$ and let $ f(1) \equal{} a$.

It follows for $ y \equal{} 0$ that $ f(x^3) \equal{} x^2f(x)$ for all $ x$.
Substituing in the initial equation, it leads to $ y(x \plus{} y)f(x) \equal{} x(x \plus{} y)f(y)$.
Now, choose $ y \equal{} 1$, it leads to $ f(x) \equal{}a x$ for all $ x\neq \minus{} 1$.
And for $ y \equal{} 2$, since we know now that $ f(2) \equal{} 2a$, we deduce that $ f( \minus{} 1) \equal{} \minus{} a$.
Thus, $ f(x) \equal{} ax$ for all $ x$.

Then, the solutions are the functions of the form $ f(x) \equal{} ax \plus{} b$ for some constants $ a,b$.

Pierre.

[quote="pbornsztein"]Note that if $ f$ is a solution then so is $ af \plus{} b$ for real constants $ a,b$
Therefore, we may assume that [b]$ f(0) \equal{} 0$[/b] and [b]$ f(1) \equal{} 1$[/b].
[/quote]

Why?

Sorry, but I don't understand why did you can assume that $ f(0) \equal{} 0$ and let $ f(1) \equal{} a$.


function
real analysis
real analysis theorems

[b]Theorem (Dimitrie Pompeiu--[1873--1954],see  [4],1946).[/b]
[i] Suppose that $ -\infty<a<b<\infty $ and $ 0\not\in [a,b]\; .$ If $f:[a,b]\to {\mathbb R} $ is continuous on $[a,b] $ and differentiable on $(a,b),$ then there exists at least a point $\xi,\; \xi \in (a,b) ,$ such that
\[(*)\; \; \; \; \; \; \; \; \; \;
\frac{af(b)-bf(a)}{a-b}=f(\xi)-\xi f^{\prime}(\xi) \; .
\]
[/i] [b]Proof 1 (by Tiberiu Popoviciu--[1906--1975]).[/b] Let us consider the interval $[A,B]$ where $ A:=\displaystyle
\min\left(\frac{1}{a},\frac{1}{b}\right)\; ,\; B:=\displaystyle\max\left(\frac{1}{a},\frac{1}{b}\right)\; .\; $ Define $g:[A,B]\to{\mathbb R}$ by $ H(t)=tf\left(\frac{1}{t}\right) , \; t\in [A,B].$ According to Lagrange mean value theorem there exists  $ c\in (A,B)$   such that $ H(B)-H(A)=(B-A)H^{\prime}(c). $ If we denote $\xi=\displaystyle \frac{1}{c} \in (a,b) $ we find (*).
This  elegant proof , using only Lagrenge mean-value theorem, is given by T.Popoviciu (around 1956) .

[b]Proof 2.[/b] Apply Cauchy mean -value theorem to the functions $F,G:[a,b]\to {\mathbb R}  $ where   $F(x)=\displaystyle \frac{f(x)}{x}\; ,\; G(x)=\displaystyle \frac{g(x)}{x} $. Then put $g(x):=1 .$

For a geometrical interpretation of (*) see [2] .

[b]References.[/b]
[1] T. Boggio ,[i] Sur une proposition de M.Pompeiu,[/i] Mathematica (Cluj) 23 (1947/1948) 101--102.
[2] M. Ivan, [i]A note pn a Pompeiu-type theorem,[/i] in [b]Mathematical Analysis and Approximation Theory[/b], Burg
Verlag,2002,129--134.
[3] D. Pompeiu , [i]Sur le theoreme des accroissements finies (premieres note)[/i], Ann.Sci.de l'Univ de Jassy, 3(1906) 244--250.
[4] D. Pompeiu, [i]Sur une proposition analogue au theoreme des  accroissements finis,[/i] Mathematica (Cluj)  22(1946)143--146.
[5] E. Popoviciu , [b]Teoreme de medie din Analiza matematica si legatura lor cu Teoria interpolarii,[/b](Romanian)  Editura Dacia,Cluj,1972.
[6] I.Stamate , [i]asupra unei proprietati a lui Pompeiu,[/i] Lucrarile stiintifice ,Inst.Politehnic din Cluj, nr. 1  (1958)12--15.
[7] C. Szasz , [i]Asupra unei et6oreme de medie,[/i] Gazeta Matematica si Fizica  (Seria -A), nr. 10 (1961) 526--528.

Dimetrie Pompeiu was working high school or university ?

Same question for Tiberiu Popoviciu ?

Yes,Moubi.Both of them. Actually their research was based on discovering quadrature formulas.Tiberiu Popoviciu
was a teacher at University in Cluj_Napoca. ;)

[quote="Moubinool"]Dimitrie Pompeiu was working high school or university ?
Same question for Tiberiu Popoviciu ?[/quote]
Both were professors at university; also both were  members of Romanian Academy.
Tiberiu Popoviciu was interested in the Theory of Convex functions. For instance he published in 1945 a book on Convex Functions [ see [i]Les functions convexes[/i], Actualit\'es Scientifiques et Industrielles , 992,XVII,Paris,1945] . He has introduced, by means of divided difference, the notion of [b]convex functions of higher order.[/b] He has also importanty contributions in Combinatorics and  in Number Theory. He was  a famous problem solver.


geometry
parallelogram
circumcircle
trigonometry
ratio
geometric transformation
rotation

The bisector of $\angle{BAD}$ in the parallellogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. Prove that the centre of the circle passing through the points $C,\ K$ and $L$ lies on the circle passing through the points $B,\ C$ and $D$.

Denote $a = AB = CD,\ b = BC = DA$, $\alpha = \angle DAB = \angle BCD$, $\beta = \angle ABC = \angle CDA$, $\alpha + \beta = 180^\circ$. Assume $a > b, \alpha < \beta$. The casees $b > a$ and/or $\alpha > \beta$ are similar. The triangles $\triangle DAL \sim \triangle CKL$ are both isosceles and similar, because the following angles are equal:

$\angle ALD = 180^\circ - (\angle LDA + \angle DAL) = 180^\circ - \left(\beta + \dfrac \alpha 2 \right) = \dfrac \alpha 2 = \angle DAL$

$\angle KLC = \angle ALD = \dfrac \alpha 2$ (vertical angles)

$\angle LCK = 180^\circ - \angle BCD = 180^\circ - \alpha = \beta = \angle LDA$

Consequently, DL = DA = b, CL = CK = CD - DL = a - b. Let P be the circumcenter of the triangle $\triangle CKL$. Since the triangle $\triangle CKL$ is isosceles, its circumcenter lies on the bisector of the angle $\angle LCK = \beta$ and $\angle PCL = \dfrac \beta 2$. Consider the triangle $\triangle DLP$. Since PL = PC is the radius of the circle (P), the triangle $\triangle PLC$ is isosceles with $\angle PLC = \angle PCL = \dfrac \beta 2$ and the angle $\angle DLP = 180^\circ - \angle PLC = 180^\circ - \dfrac \beta 2$. The sides the triangle $\triangle DLP$ are equal to

$DL = b,\ \ LP = \dfrac{CL}{2 \cos \dfrac{\angle PLC}{2}} = \dfrac{a - b}{2 \cos \dfrac \beta 2}$

Let M the intersection of the bisector of the angle $\angle DAB$ with the line AB. Again, the triangle $\triangle AMD$ is isosceles, because the following angles are equal:

$\angle AMD = 180^\circ - (\angle DAM + \angle MDA) = 180^\circ - \left(\alpha + \dfrac \beta 2\right) = \dfrac \beta 2 = \angle MDA$

Consequently, AM = AD = b, MB = AB - AM = a - b. Consider the triangle $\triangle DMB$. The angle $\angle DMB = 180^\circ - \angle AMD = 180^\circ - \dfrac \beta 2$. The sides of this triangle are equal to

$DM = 2AM \cos \angle AMD = 2b \cos \dfrac \beta 2,\ \ MB = a - b$

As a result, the triangles $\triangle DMB \sim \triangle DLP$ are similar by SAS, because the angles $\angle PLD = \angle BMD = 180^\circ - \dfrac \beta 2$ are equal and the sides ratios $\dfrac{DM}{DL} = \dfrac{MB}{LP} = 2 \cos \dfrac \beta 2$ are also equal. This means that the angles $\angle BDM = \angle PDL$ are equal.

Consider the quadrilateral BCPD. The angle $\angle BCP = \angle BCD + \angle DCP = \alpha + \dfrac \beta 2$. The opposite angle of this quadrilateral is equal to

$\angle PDB = \angle PDL + \angle LDB = \angle BDM + \angle LDB = \angle LDM = \dfrac \beta 2$

The sum of the opposite angles of the quadrilateral BCPD is

$\angle BCP + \angle PDB = \alpha + \dfrac \beta 2 + \dfrac \beta 2 = 180^\circ$

which means that this quadrilateral is cyclic. QED.

Hello Yetti! I am very happy that you are back now!

A simpler solution:

Let M be the midpoint of the arc BCD of the circumcircle of triangle BCD. Then, MB = MD. Since the point M lies on the circumcircle of triangle BCD, we have < MBC = < MDC, what rewrites as < MBK = < MDC. Since the line AK is the angle bisector of the angle BAD, we have < BAK = < DAK; but since the quadrilateral ABCD is a parallelogram, BC || DA, and thus < DAK = < BKA. Hence, < BAK = < BKA. Thus, the triangle ABK is isosceles, so that BK = AB. Since the quadrilateral ABCD is a parallelogram, AB = DC. Hence, BK = DC.

From MB = MD, BK = DC and < MBK = < MDC, it follows that the triangles MBK and MDC are congruent, and thus MK = MC. Similarly, ML = MC. Hence, MC = MK = ML, so the point M is the center of the circle through the points C, K and L. Since the point M lies on the circumcircle of triangle BCD, we thus see that the center of the circle through the points C, K and L lies on the circumcircle of triangle BCD, and the problem is solved.

  Darij

here is my solution:

    If a center of inversion circle  is on circumcircle of BCD, then circle BCD should

    invert into a straight line.  If this line is parallel to diagonal BD (see the inversion

    triangles), then the inversion circle  is  CKL.

    But I like Darij's solution better, because it brings to light the isosceles triangles

    and also because of it's quintessential simplicity.

    Thank you.

    M.T.

We have
$\angle PKB=\angle PCK=\angle PCD$ $(1)$ and $PK=PC$ $(2)$
$\angle BKA=\angle DAK=\angle BAK$ so triangle $ABK$ is isoceles. Thus $CD=AB=BK$ $(3)$
From (1) (2) (3) triangle $DPC$ and $BPK$ are congrurent
so $\angle PDC=\angle PBC$ so $B,C,P,D$ are cyclic $q.e.d$

All angles in this proof are directed. Let $AB$ hit $(BCD)$ at $G$. $BG || CD$, so $GBCD$ is an isosceles trapezoid. Let $GL$ hit $(BCD)$ again at $P'$. $\angle CP'L =  \angle CP'G = \angle CDG = \angle BCD = \angle BAD = 2 \angle LAD = 2 \angle CKL$. $\angle ALD = \angle 180^{\circ} - \angle LDA - \angle DAL = 180^{\circ} - \angle LDA - \frac{\angle 180 - \angle LDA}{2} = \angle DAL$, so $DA = DL$. But $DA = DG = BC$, so $DG = DL$, so $\angle LCP' = \angle DGL = \angle GLD = \angle P'LC$, so $P'L = P'C$, whence $P'$ is the center of $(KLC)$.

Something similar with Huyen Vu:
Easily notice $\triangle KCL$ is isosceles, with $\widehat{CKL}=\widehat{BAD}\implies \widehat{CPL}=\hat A\ (\ 1\ )\implies\widehat {PLC}=\widehat{PCL}\ (\ 2\ )$ and $PC=PL\ (\ 3\ )$. With $DL=AD=BC\ (\ 4\ )$ we infer that a rotation of $\angle CPL$ about P will map triangles $DLP$ and $BCP$, i.e. $\angle DPB=\angle CPL=\angle BAD=\angle BCD$ and, indeed, $B,C,P,D$ are concyclic.

Best regards,
sunken rock

Dear Mathlinkers,
the converse of this problem has been proposed by Virgil Nicula... I have lost the reference...
Sincerely
Jean-Louis


im new to this so until i can learn how to type these questions can someone help with number 23, 24 or 25 in the AMC 10 2005 A test plz?

thanx

You don't have too, they have already been typed, you can find them [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2005]here[/url].

thanx but i already have the problem i was asking if any 1 could help me with them

Oh ok, well if you want the solutions, you can click on the links next to each problem(which will take you to that post of the problem) from the link I gave.

For the last one, isn't the AMC 10 supposed to be non-trig? I managed to do it in a non-trig way, but it was kinda based on assumptions rather than proof.

well i could help you  with 24

basically you the questionis asking how many primes can you add to 48 to get another prime.

i do not know what to do from ther tho get a solution


geometry
3D geometry
tetrahedron
probability

George and Leo each toss a number tetrahedron with the numbers 1-4 on the faces. They then multiply the resulting pair of numbers. If the product is less than 9, George wins; otherwise, Leo wins. What is the probability that Leo will win? Express your answer as a common fraction.

We have for possible values: 1, 2, 3, 4, 6, 8, 9, 12, 16.

There is one possible way to get 1 (each rolls a 1), 2 possible ways to get a 2 (1 and 2 or 2 and 1), 2 ways to get a 3 (3 and 1, 1 and 3), 3 ways to get a 4 (1 and 4, 2 and 2, 4 and 1), 2 ways to get a 6 (2 and 3, 3 and 2), 2 ways to get an 8 (2 and 4, 4 and 2), 1 way to get a 9 (3 and 3), 2 ways to get a 12 (3 and 4, 4 and 3) and 1 way to get 16 (4 and 4).

So we see that there are 12 ways to get less than 9, out of 16 possible toss combinations.  So Leo's chances of winning are $ 1\minus{}\frac{12}{16}\equal{}\frac{1}{4}$.


what is the best way to teach that lim x-->0+ (2x+1)/x=+inf ?
Ideally would be that the student writes in words that the denominator approaches 1 and the numerator 0 and that the fraction is positive, so the limit is infinity.
 But in the economy of a test what can you write?
 My teacher(long time ago) taught me to write with quotations marks "1/0+" and then, separately,  lim x-->0+ (2x+1)/x=+inf .
Now I have to teach students this subject.
So what do you think?

No matter what you do, you're going to run into the fact that most calculus students have no idea how to write math. There's plenty of time to use a few words on a test, but this won't help with them not understanding how to write an explanation.

:D I think student's must studied it in school ;) Anyway you can write
$ \lim_{x\to 0\plus{}}{\frac{2x\plus{}1}{x}}\equal{}\lim_{x\to 0\plus{}}{\left(2 \plus{} \frac{1}{x}\right)}$. Now when $ x$ is very small then $ \frac{1}{x}$ is very large. And because $ x>0$ then we have $ \plus{}\infty$. ;)

The answer depends on what you really want. This example is quite good to illustrate $ \varepsilon\minus{}\delta$ definition of the limit because it is one of the rare cases where you can explicitly find the best $ \delta>0$ for any given $ \varepsilon>0$. So, if you taught the rigorous definitions, you can demand using them here. An alternative is the phrase you suggested as "ideal". I strongly oppose the idea to teach students some kind of special abbreviated language to use on the tests. Nobody else uses it anywhere outside the classroom and it just creates more confusion. If you think that they'll not have enough time to write the full sentences, it merely means that you should not give as little time or as many problems on the test as you intend. You are supposed to check their knowledge of the subject, not their writing speed. So my advice would be to demand the full explanation if you want to check the understanding of the concept or just accept the statement $ \lim_{x\to 0\plus{}}\frac{2x\plus{}1}x\equal{}\plus{}\infty$ as obvious and not requiring any comments at all if you want to check their routine skills by giving them several problems like this. Anything in between just creates a lot of headache for both you and them.

I agree with fedja's comments. Requiring written explanations other than full $ \epsilon$-$ \delta$ proofs isn't likely to help understanding. But you may be able to explore this territory in an oral class discussion.

I like to use paraphrases of the definition. If we say that $ \lim_{x\to0^\plus{}}f(x)\equal{}\infty,$ we're saying that by making $ x$ [b]arbitrarily[/b] small (but still positive), we can make $ f(x)$ [b]arbitrarily[/b] large. (Any attempt to clarify what I mean by "arbitrarily" would have a good chance of walking into the formal definition.)

But the hard part is the understanding that "dividing by an arbitrarily small positive number" is not a single act of algebra; that denominator isn't a single value. Look at $ \frac{2x\plus{}1}{x}.$ [b]Don't[/b] let $ x\equal{}0$; that's not the question. We're not letting $ x$ be zero, but we're letting it be very close to zero. In that case, the numerator, $ 2x\plus{}1$ is very close to $ 1,$ but the denominator is very small. What happens when we divide nearly $ 1$ by something very small? We get something very big, right? Maybe we should look at few numbers; if $ x\equal{}.01,$ we get about $ 100.$ If $ x\equal{}.0001,$ we get about $ 10000$ (I don't really care if it's $ 9998$ or $ 10002,$ that's not important).

But the point is there; we're not dividing by a single value. That $ x\to0^\plus{}$ doesn't represent a single value; it represents a range of possibilities. And we do have useful intuition about that. "Roughly constant" divided by "arbitrarily small" should be "arbitrarily big" - that's really just sound intuition about numbers. Of course "arbitrarily small" divided by "arbitrarily small" doesn't answer a question but creates new questions: [b]how[/b] small, in comparison?

That's the problem with introducing $ ``\frac{1}{0^\plus{}}\equal{}\infty".$ Do it up front and you're offering to make this a single act of "algebra", which misses out on the idea of the range of possibilities. Once you've talked your way through a number of problems and seen what happens in a number of cases, then $ ``\frac{1}{0^\plus{}}\equal{}\infty"$ can simply be a summary of experience - not something that creates new knowledge, but a shorthand for writing what you already know. And in any case, it's not really an explanation.

[quote="fedja"]... as obvious and not requiring any comments at all if you want to check their routine skills by giving them several problems like this.[/quote]
It should be possible to create, as a quiz, a collection of problems (with different answers, representing some of the different things that can happen) in which having a student get all of the correct answers, even without elaborate explanations, would be reasonably solid evidence that the student understands big and small, at least at an elementary level.


A quadrilaterial $ABCD$ is incribed into a circle with a center of $O$. a circle $O'$ touches side $BD$ and side $AC$ at $P$ and $Q$, respectively. And $\odot O'$ touches $\odot O$. Prove that the incenters of $\triangle ABC$ and $\triangle DBC$ are both on the line $PQ$.

It's so hard for me and I couldn't finish it before 12 o'clock today in the exam. Not lots of people (but still there are some) solved it. Some said it can be solved with Casey Theorem, and I believed that this question have been discussed in the forum (though I didn't find it).

I think by asking this problem, the proposers made the competition is a bit unfair because the result is famous. Please look at http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf for solution.

[quote="e.lopes"]it really can be solved using casey, isn't difficult!

how treegoner says, this result is famous.. the problems that i see this year of the chinese tst are all famous. strange  :maybe:[/quote]

You're right, some people use Casey to solve it!


[quote="daermon"][hide]$\frac{2}{15}$ I'll post my solution if I'm right[/hide][/quote]
No, it is not correct.

[quote="mthiyaga"]Mr. Gardner plants 10 rose bushes in a row. Eight of the bushes are white and two are red, and he plants them in random order. What is the probability that he will consecutively plant seven or more white bushes?[/quote]
[hide]There are 6 cases in which 7 or more white bushes are planted together. The total amount of possibilities is $1+2+...+8+9.$ The answer is $\frac{6}{45},$ or $\frac{2}{15}.$ I don't see any fault in my logic. :huh: [/hide]

yeah thats what I did too what did we do wrong mithyaga  :maybe:

I don't want to give away anything....especially since I'm not 100% on my answer, but are you sure there are only 6 ways to have 7 OR MORE in a row?  Did you consider they wouldn't be next to each other starting from the end?

[quote="i_like_pie"][hide]9 cases for 7 or more white bushes in a row, and 45 total possibilities. The answer is $\frac{9}{45},$ or $\frac{1}{5}.$[/hide][/quote]
As frost13 indicated, considering 7 or more white bushes in a row, we see there are 9 possibilities. So i_like_pie's new answer  $\left (\frac{1}{5}\right )$ is correct!

[hide]45 possibilities for 9 things...
and 6 possibilietes. i kwo thats wrong though...[/hide]

[quote="mthiyaga"]Mr. Gardner plants 10 rose bushes in a row. Eight of the bushes are white and two are red, and he plants them in random order. What is the probability that he will consecutively plant seven or more white bushes?[/quote]
[hide="solution"]
if the 7+ white bushes start on the first,
$\frac{8}{10}\frac{7}9\frac{6}8\frac{5}7\frac{4}6\frac{3}5\frac{2}4=\frac{1}{15}$
if the 7+ white bushes start on the second,
if one white bush is already planted,
then its the scenario as before...
if no white bush is already planted,
$\frac{1}5\frac{8}{9}\frac{7}8\frac{6}7\frac{5}6\frac{4}5\frac{3}4\frac{2}3=\frac{2}{45}$
if the 7+ white bushes start on the third,
if two white bushes are already planted
then its the scenario as startin on the first
if one white bush is already planted and it was on the first
$\frac{4}5\frac{2}9\frac{7}{8}\frac{6}7\frac{5}6\frac{4}5\frac{3}4\frac{2}3\frac{1}2=\frac{1}{45}$
if no white bush is already planted, then all the rest are white
$\frac{1}5\frac{1}9=\frac{1}{45}$
if the 7 white bushes start on the fourth,
if one white bush is already planted, and it was either first or seccond, then all the rest have to be white
$(2)\frac{4}5\frac{2}9\frac{1}8=\frac{2}{45}$
add them
$\frac{3+1+2+1+2}{45}=\frac{9}{45}$
The probability is $\frac{1}{5}$
[/hide]


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

Does any one have the statistics about what the cutoff score was to get to the AIME in the past years? If it is online, I searched and didn't find it.

From my memory...
2009 AMC 10A: 120
2009 AMC 10B: 120
2009 AMC 12A: 97.5
2009 AMC 12B: 100 (this is pretty much 100.5)
2008 AMC 10A: 117 (or 118.5; can't remember)
2008 AMC 10B: 120
2008 AMC 12A: 97.5?
2008 AMC 12B: 97.5 or 99?

Or you could just get 120+ on the 10 or 100+ on the 12...there's usually hardly any differene, if any.

Changed 2009 to 2008.

You repeated AMC12B 2009 twice. I am assuming the last one is supposed to be 2008. Thanks.

2008 12B is 97.5 (heh, I got that score).

Note that 120 on the AMC 10 or 100 on the AMC 12 guarantees AIME qualification.

Yep, you just need 120 on the amc10 or 100 on the amc12 to absolutely guarantee. If your only goal is to make AIME and not USAMO, just focus on the first 20 (on the amc 10) and make sure that you get them right at all costs. (lol, thats not what I did. I got all of the last five right and made stupid mistakes on 5 of the first 15. Barely made aime)

I was just wondering, how hard was this amc10a compared to other (well maybe you cant compare because of the calculator changes) Also, how hard was this years AIME 2 compared to others?

I'd say the 10A was fairly easy compared to others, while the AIME II was average, leaning towards more difficult. At least for me.

[quote]what the cutoff score was to get to the AIME in the past years? If it is online, I searched and didn't find it.[/quote]
Try looking in the AMC 12 Archive, under General Statistics.  The record is complete back to 2001.
http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml

Here's how I found it:  I went to the AMC page, then clicked AMC 12, then clicked Archive, then chose a year to look at the General Statistics.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="simo14"]I'd say the 10A was fairly easy compared to others[/quote]

Well yeah, I just realized that it would probably seem easier because of the calculator change. Because we couldn't use a calculator, they probably gave easier questions, unlike #25 of 2008 which involved trig-bashing

No problem on AMC 10 requires trig/logarithms.

Your right, number 25 on the 2008 amc10a didnt [i]require[/i] the use of the law of cosines, but without it the question would be much harder to do.

Also, I think we pretty much answered the OP's question as much as possible.

I trig bash almost all the geometry problems...

Also note that the cutoffs are not usually lower than 100/120.  And if they are, it's only a few points lower.  The AMC10 cutoff tends to be lowered more often than the AMC12 cutoff.

2001 was the exception to what I said: AMC12 cutoff was 84.  Take the 2001 AMC12 and you'll see why.

[quote]Any other major exceptions, anyone?[/quote]

The AIME cutoff on AMC12 in 2000 was 92.

Also, scoring system changes are relevant in considering cutoffs.  In 2000 and 2001, an entirely blank exam was worth 50 points (since blanks were worth 2 points each.)  From 2002 through 2006, an entirely blank exam was worth 62.5 points (since blanks were worth 2.5 points each.) From 2007 onwards, an entirely blank exam was worth 37.5 points (since blanks were worth 1.5 points each.)

Operationally, this means that qualifying for AIME in 2001 could be done with 9 correct questions and the rest blank or 10 correct, 3 wrong, and 12 blank.   Qualifying for the AIME cutoff of 92 in 2000 could be done with 11 correct, 1 wrong, and 13 blank.

In recent years, it has taken several more correct answers to minimally qualify for AIME, partly because the cutoffs were higher and partly because blank answers were not worth as much.


geometry
circumcircle
trapezoid
cyclic quadrilateral
power of a point
radical axis
geometry proposed

Let circle$(O) ,B,C$ on circle .$A$varry on circle .Let line $\delta$ parallel to $BC (A$ not in $\delta$)and intersect AB,AC at M,N respectively.$J$be intersection of circles $(ABN)$and $(ACM)$ .Prove that $AJ$ through a fixed point not depen on $A$ (on circle$(O)$)

Draw the tangents to the circumcircle of $ABC$ in $B$ and $C$, and call their intersection point $S$.
$SC$ and $SB$ intersect $MN$ in $P$ and $Q$ respectively.
Clearly, $PCBQ$ is a isosceles trapezoid, so it is a cyclic quadrilateral, on the circle $\omega$.

Therefore, $S$ must be the radical centre of $(ABN), (ACM)$ and $\omega$, so $S$ must also lie on the radical axis of $(ABN)$ and $(ACM)$, so $S$ lies on $AJ$, while $A$ is variable. $\mathbb{QED}$

Note: $AJ$ is called the symmedian of triangle $ABC$, it is the isogonal conjugate of the median.

nice sol Jan
In my sol I use equal angle


calculus
integration
analytic geometry
function
calculus computations

Can anybody help me with the following problem:

Find the mass of the region (in cylindrical coordinates) r^3<=z<=4, where the density function is rho(r,theta,z)=1z.

Anybody have an idea? I especially need help with problems 1-3, number 4 look manageable. Thanks for your attention.

I have finished all of the problems except this one. Is there someone who know the answer?

Either calculate

\[ \int \limits_{\theta =0}^{2\pi }\; \int \limits_{z=0}^{4}z\;\int \limits_{r=0}^{z^{1/3}}dr\,dz\,d\theta \]

or

\[ \int \limits_{\theta =0}^{2\pi }\; \int \limits_{r=0}^{4^{1/3}}\;\int \limits_{z=r^3}^{4}z\;dz\,dr\,d\theta . \]


geometry
geometric transformation
homothety
ratio

The diagonals of a convex quadrilateral ABCD are mutually perpendicular. Perpendicular lines from the midpoints of sides AB and AD are dropped to their opposite sides CD and CB, respectively. Prove that these two lines and line AC have a common point.

[hide="Hidden by mod"]for an homothety with center A and ratio 2 it's equivalent to say that AC, the perpendicolar line from B to CD, the perpendicolar line from D to BC concur but they concur on the orthocenter of $ \triangle BCD$[/hide]


Following equation:

  n(n+1)(n+2)(n+3) = m^k

Does  the natural number triple (m,n,k) satisfying that equation exist ?

Sorry, k>=2

[quote="gus135791"]Following equation:

  n(n+1)(n+2)(n+3) = m^k

Does  the natural number triple (m,n,k) satisfying that equation exist ?[/quote]

Yes we can consider trivial case
$ m \equal{} n(n\plus{}1)(n\plus{}2)(n\plus{}3),\ n\in\mathbb N,\  k\equal{}1$

if u want $ k$ greater than $ 1$,then the answer will obviously be negative.

Obvious meaning a very difficult theorem by Erdos and Selfridge, shoki !


AMC
USA(J)MO
USAMO
calculus
integration
number theory
Diophantine equation

Determine all non-negative integral solutions $ (n_1,n_2,\dots,n_{14})$ if any, apart from permutations, of the Diophantine Equation $ n_1^4 \plus{} n_2^4 \plus{} \dots \plus{} n_{14}^4 \equal{} 1599$.

[hide]The quartic residues of 16 are 0 and 1, i.e. all $ n_i^4$ are congruent to zero or one modulo sixteen. 1599 is congruent to 15 modulo 16. However,  $ n_1^4 \plus{} n_2^4 \plus{} \dots \plus{} n_{14}^4$ can only be  be 0-14 modulo 16. Therefore, there are no solutions to this equation.[/hide]

For any $ n_k$, $ n_k$ can be even or odd.

If $ n_k$ is even, $ n_k^4 \equiv 0 (mod 8)$

If $ n_k$ is odd, $ (2x\minus{}1)^4\equal{}16x^4\minus{}32x^3\plus{}24x^2\minus{}8x\plus{}1 \equiv 8x(x\plus{}1)\plus{}1 (mod 16)$.

Since x is odd, $ x\plus{}1 \equiv 0 (mod 2)$.

Thus, $ n_k \equiv 1 (mod 16)$, for $ n_k$ odd
$ n_k \equiv 0 (mod 16)$, for $ n_k$ even.

Since $ 1599 \equiv 1600\minus{}1 \equiv 0\minus{}1 \equiv 15 (mod 16)$, there are no solutions.

I think the solution is good, but you could elaborate a bit on the 0-14 residue.  Your solution is short, so I don't know if you would get full full credit.


linear algebra
matrix
calculus
derivative
limit
linear algebra unsolved

$A=(a_{ij})$ real matrix $M_n(R)$
suppose that for any $t>0$ all the entries of $e^{tA}=b_{ij}$ are positives prove that
$a_{ij}\geq 0$ for any $i \neq j$

While trying to solve [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28482]this problem[/url], I remembered this topic. The same differentiation trick works here as well:

Let $f: [0,\infty)\to\mathcal M_n(\mathbb R),\ f(t)=\exp(tA)$. $f$ is differentiable, and $A=f'(0)=\lim_{t\to 0+}\frac{\exp(tA)-I}t$. It's clear now that the hypothesis implies $a_{ij}\ge 0,\ \forall i\ne j$.


calculus
integration
floor function
limit
probability
real analysis
real analysis unsolved

For $0\leq x<1$ we define $g(x)=\sum_{n=1}^{\infty}\frac{1-(-1)^{[}2^{n}x]}{n!}$ where [y] is the whole part of y. Find a)$g(1)=lim_{x\rightarrow1}g(x)$ b)$\int_{0}^{1}g(x)dx$.

can you rewrite your formula!? because there's not sense on it

Martin is new to $\text{\TeX},$ and may not be all that familiar with the uses of {}. I'm going to guess he means this:

[quote="martin1985"]For $0\leq x<1$ we define $g(x)=\sum_{n=1}^{\infty}\frac{1-(-1)^{[2^{n}x]}}{n!}$ where $[y]$ is the whole part of y. Find a) $g(1)=\lim_{x\to1}g(x)$ b) $\int_{0}^{1}g(x)dx$.[/quote]

a) ${g(x)=\sum_{n=1}^{\infty }\left(\frac{1-(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}\right)}$

${g(1)=\lim_{x\to 1}\, \left(\sum_{n=1}^{\infty }\frac{1-(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}\right)}$

${(n\in \mathbb{Z}) \Rightarrow \left(\lim_{x\to 1}\, \left\lfloor 2^{n}x\right\rfloor =2^{n}\right) \Rightarrow \left((-1)^{2^{n}}=1\right)}$

${g(1)=\sum_{n=1}^{\infty }\left(\lim_{x\to 1}\left(\, \frac{1-(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}\right)\right)=\sum_{n=1}^{\infty }\left(\frac{1-(-1)^{\lim_{x\to 1}\, \left\lfloor 2^{n}x\right\rfloor }}{n!}\right)=0}$

I used the fact that ${\lim_{x\to a}\, \left(\sum_{n=1}^{\infty }g(x,n)\right)\equiv \sum_{n=1}^{\infty }\left(\lim_{x\to a}\, g(x,n)\right)}$

But I'm not sure to define $g$ well. so i made another.

${g(x)=\sum_{n=1}^{\infty }\frac{1-(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}=\sum_{n=1}^{\infty }\frac{1}{n!}-\sum_{n=1}^{\infty }\frac{(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}=(-1+e)-\sum_{n=1}^{\infty }\frac{(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}}$

${n\in \mathbb{Z}}$
${\lim_{x\to 1}\, \left(\sum_{n=1}^{\infty }\frac{(-1)^{\left\lfloor 2^{n}x\right\rfloor }}{n!}\right)=\lim_{x\to 1}\, \left(\sum_{n=1}^{\infty }\frac{(-1)^{\left\lfloor 2^{n}\right\rfloor }}{n!}\right)=\lim_{x\to 1}\, \left(\sum_{n=1}^{\infty }\frac{1}{n!}\right)=(-1+e)}$

The expression ${1-(-1)^{[2^{n}x]}}$ equals $2$ if the $n$-th bit in the binary representation of $x$ equals $1$ and equals $0$ otherwise.

a) Therefore, to compute the limit we need essentially (this is a slightly informal argument) to compute $g$ at point $x = 0.111\dots$ - where all digits are equal to $1$. We get
\[\sum_{n=1}^\infty \frac{2}{n!}= 2\left( \sum_{n=0}^\infty \frac{1}{n!}-1\right) = 2 e-2. \]
b) If pick $x\in(0,1)$ uniformly at random then its $n$-th bit will equal $0$ with probability $1/2$ and equal $1$ with probability $1/2$. Therefore, the integral equals
\[\sum_{n=1}^{\infty}\frac{\left(\frac12 \cdot 0+\frac12 \cdot 2\right)}{n!}= e-1. \]


Is the ans $ \frac {qin} {6\cdot Eo}$ ? Eo stands for epsilon nought

And dont be shocked that i answered a physics question  :D

i suppose it is 0
now the let the flux through the surface s attached to the point charge be $ \Phi_{1}$ and those not attached be $ \Phi_{2}$
flux thru each cube $ \equal{}3\Phi_{1}\plus{}3\Phi_{2}$
u can simply prove it by placing another 7 identical cubes such that q is the centre of the big cube of side 2a
the flux through the big cube is $ \frac{q}{\epsilon_{0}}$
so flux thru each cube
now the flux thru the cube is also equl to $ 24 \Phi_{2}\equal{}\frac{q}{\epsilon_{0}}$
so we have $ \frac{q}{8\epsilon_{0}}$
so we have flux $ \Phi_{1}\equal{}0$
which is also obvious from the definition of flux ....
at anypoint fiel dis along the line joining the point charge to it which i sperpendicular to the perpendicular to the surface

i also think the answer will be zero...cos...like in soild state this corner of the cube will be the center of another big cube .....but considering that pt to be the center no of electic field lines cutting thru the given area will be zero.... i.e. E.dS will be zero as E and dS are perpendicular to each other........

Rohit, I appreciate you trying but your answer is wrong. Yours will be right only if it is inside the cube, simple applications of gauss law hold good only in school sums.

Pardesi: Yeah, right method, but ans wrong.

why???
forget about the mathematics...just use simple physical logic...
take anypoint on those sides attached to the charge the field is along the line joining that point to the charge which is on the plane of the side hence at the point the normal is perpendicular to the side hence the field.
i find no flaw in thsi logic


A function is called odd if $f\left(-x\right)=-f\left(x\right)$.  Prove that $f\left(x\right)=\ln\left(x+\sqrt{x^{2}+1}\right)$ is odd.

$\\f\left(-x\right)=\ln\left(\sqrt{x^{2}+1}-x\right) \\ =\ln\left(\frac{x^{2}+1-x^2}{\sqrt{x^2+1}+x}\right)\\ =-\ln\left(\sqrt{x^2+1}+x\right)=-f(x)$

Masoud Zargar


logarithms
limit
real analysis
calculus
calculus computations

If $ x_n \to \plus{}\infty$ and $ a \in \mathbb{R}$ show that $ \lim_{n \to \infty}\left[\sqrt{\log (x_n\plus{}a)} \minus{}\sqrt{ \log x_n}\right]\equal{}0$

[hide]I arrive at this: $ \sqrt{\log (x_n\plus{}a)} \minus{}\sqrt{ \log x_n}\equal{}\frac{\log{\left(1\plus{}\frac{a}{x_n}\right)}}{\sqrt{\log (x_n\plus{}a)} \plus{}\sqrt{\log x_n}}$ but don't know how to continue. Looking for hints. Thank you[/hide]

[quote="mathmusic"]I arrive at this: $ \sqrt{\log (x_n\plus{}a)} \minus{}\sqrt{ \log x_n}\equal{}\frac{\log{\left(1\plus{}\frac{a}{x_n}\right)}}{\sqrt{\log (x_n\plus{}a)} \plus{}\sqrt{\log x_n}}$ but don't know how to continue. [/quote]

Just determine the limit of the numerator and the denominator and since one of them is finite and the limit in the denominator is not $ 0$, you can use $ \lim\frac{a_{n}}{b_{n}}\equal{}\frac{\lim a_{n}}{\lim b_{n}}$.

Thanks tmrfea! i'm taking real analysis and log was not rigorously defined so i wondered if there was some trick there. But i think there is no other way... i need to use that $ \lim\log n \equal{} \plus{}\infty$

$ \sqrt {\log (x_n \plus{} a)} \minus{} \sqrt {\log x_n} \equal{} \frac {\log{\left(1 \plus{} \frac {a}{x_n}\right)}}{\sqrt {\log (x_n \plus{} a)} \plus{} \sqrt {\log x_n}} \equal{} \left(\frac{0}{0}\right)$ is an indeterminate form, how can you evaluate it?

Rajiv, $ \sqrt {\log (x_n \plus{} a)} \plus{} \sqrt {\log x_n}\rightarrow\plus{}\infty$.

[quote="Ognjen09"]Rajiv, $ \sqrt {\log (x_n \plus{} a)} \plus{} \sqrt {\log x_n}\rightarrow \plus{} \infty$.[/quote]
 :oops:

Yet, $ \left(\frac{0}{\infty}\right)$ is indeterminate....! :D

[quote="JBL"][quote="Rajiv"]$ \left(\frac {0}{\infty}\right)$ is indeterminate....! :D[/quote]  No, it isn't.[/quote]

Yes, it isn't. Sorry.


Let $a,$ $b,$ $c$ and $d$ are non-negative numbers. Prove that
$\sum_{sym}(a^{3}b^{3}-2a^{3}bcd+a^{2}b^{2}c^{2})\geq0.$

What is pvthuan's problem?  :?: Can you talk with me and everybody  ? :roll:

zaizai-hoang, ask pvthuan. If he will tell yes, I"ll post the pvthuan's inequality.

Mine is

Let $a,b,c,d$ be non-negative real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+3(abc+bcd+cda+dab).\]

[quote="pvthuan"]Mine is

Let $a,b,c,d$ be non-negative real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+3(abc+bcd+cda+dab). \]
[/quote]
Are you sure that it's equivalent to
$\sum_{sym}(a^{3}b^{3}-2a^{3}bcd+a^{2}b^{2}c^{2})\geq0$ $?$ :wink:
However, it is formal - yes! :D

[quote="pvthuan"]Mine is

Let $a,b,c,d$ be non-negative real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+3(abc+bcd+cda+dab). \]
[/quote]

But this one is not hard at all.
$L-R = \sum (a^{2}b+b^{2}c+c^{2}a-3abc) \geq 0$.

Gib, mine is the following
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+ac+bd.\]
(the previous inequality is not what I mean)


In  any triangle ABC prove that
[tex]18\sqrt{3}\cdot r^2 \leq am_a+bm_b+cm_c \leq \sqrt{bc}\cdot m_a + \sqrt{ca}\cdot m_b + \sqrt{ab}\cdot\ m_c  [/tex]

Thanks

The first one comes from the fact that:
[tex] abc \geq 8\sqrt{\frac{S^3}{3\sqrt3}} [/tex] and
[tex] m_am_bm_c \geq \sqrt{3\sqrt3 S^3}[/tex] .
Just take [tex] r=\frac{2S}{a+b+c} [/tex]  :)


Recall that y = ln(x) and y = e^x are inverse functions. THe graph of y = ln(x) looks as though it might be approaching a horizontal asymptote. Write an argument based on the graph of y = e^x to explain why it does not.

Since they are inverse functions, every (x,y) pair of ln(x) corresponds to a pair (y,x) of e^x.  If ln(x) approaches a horizontal asymptote, that means that y has a limit.  That means there is a limit on the x values of f(x).  This is ridiculous, so obviously there is no horizontal asymptote for ln(x).


Hi everyone :) .
Let $a,b,c>0$, such that \[\sum{ \frac{a}{b}}=5.\]
a) What is the minimal and maximal possible value of $\frac{max(a,b,c)}{min(a,b,c)}$?
Prove that:
b) \[3\sum{a}>\sum{\frac{ab}{c}}\]
c) \[\sum{\frac{a}{b+c}<\frac{5\sqrt15}{6}}\]

Could you make up some other inequalities related to this condition? Thanks.

I thought this problem would be interesting because it can't be solved easily by Muirhead or other technical theorem. At least I hope so. :D
Maybe I shouldn't have written 'very easy'. I hope someone will give it a try.       C'mon it isn't that ugly!

For a): we may assume that $c=max(a,b,c)$ and we must analyze the cases $a\le{b}\le{c}$ and $b\le{a}\le{c}$. The result should be$2+\sqrt3\le\frac{max(a,b,c)}{min(a,b,c)}\le3+2\sqrt2$ and both extremal values can be attained.

Both b) and c) have simple solutions using classical ineqs.


integration
calculus
calculus computations

In my book, they solved the differential equation $y''+xy'+y=0$ using power series, to get $y=c_0 \left( 1-\frac{x^2}{2}+\frac{x^4}{2\cdot4}-\frac{x^6}{2\cdot4\cdot6}+...\right)+c_1 \left(x-\frac{x^3}{3}+\frac{x^5}{3\cdot5}-\frac{x^7}{3\cdot5\cdot7}+...\right)$, is there a closed form answer for this?

Here's a partial closed form: that first half is $c_0\cdot e^{-x^2/2}$.
You can get the other half by reduction of order.

I saw that just by reading the power series; $2\cdot4\cdot6\cdots2n=2^n\cdot n!$, and that makes the series obviously an exponential power series.

Thanks.
When I did the reduction of order I got $y=e^{-x^2/2} \left(C_1+C_2\int e^{-x^2/2}\ dx\right)$, which I guess is about as closed form as this gets.


Let $ f(x)$ be a function in $ [0,1]$ and

$ f(x)\equal{}2x$ 　$ (0\leqq x\leqq \frac{1}{2})$,  $ f(x)\equal{}\minus{}2x\plus{}2$ 　$ (\frac{1}{2}< x\leqq 1)$,

Let $ \{a_n\}$ be a sequence of numbers defined as

$ a_{n}\equal{}f(a_{n\minus{}1})$  $ (n\equal{}2,3,...)$,and its initial value is $ a_1$. Answer the following questions.

$ (1)$ Determine all number b $ \in [0,1]$ such that $ f(b)\equal{}b$

$ (2)$ Determine all the initial value $ a_1$ such that $ a_4$ is equal to one of $ b$ which you determined in $ (1)$

$ (3)$ Determine all the initial value $ a_1$ satisfying the next condition.

       Condition:There exists a number $ n\geqq 1$ such that $ a_n$ is equal to one of $ b$ which you determined in $ (1)$

$ (4)$ If the initial value $ a_1$ does not satisfy the condition of $ (3)$,show that there exists a number $ n\geqq 1$ such that $ a_n\geqq \frac{3}{4}$

$ (5)$ Find the necessary and sufficient condition of the initial value $ a_1$ such that $ \{a_n\}$ converges.

(1) $ b\equal{}0, \frac{2}{3}$
(2), (3) $ a_1 \equal{} \frac{m}{3}\left(\frac{1}{n}\right)^n \ (n\equal{}0, 1, 2, \cdots; m\equal{}0, 1, 2, \cdots, 3\cdot2^n)$
(4) Contraposition
(5) Condition (3)


[hide=Click for solution]
We have $ \frac{3}{9}\plus{}\frac{3}{x}\equal{}\frac{8}{9} \implies \frac{3}{x}\equal{}\frac{5}{9} \implies x\equal{}\frac{27}{5}$, so in three days he did $ \boxed{\frac{15}{27}}$ of the job.

which can be simplified to $ \boxed{\frac{5}{9}}$

i dont understand. can you break it down a little more?

[quote="motiroti"]Jerry worked for one day on a project that he could have completed alone in nine
days. Bill joined Jerry the next day, and they worked together for exactly three
days to complete the project. How much of the job did Bill do in those three days?
Express your answer as a common fraction.[/quote]

Jerry does a whole job in 9 days, so in one day he does $ \frac{1}{9}$ of the job.

The next day, there are $ \frac{8}{9}$ of the job remaining.

In three days, Jerry did $ \frac{3}{9}\equal{}\frac{1}{3}$ of the job.

Thus, Bill did the rest, or $ \frac{5}{9}$ of the job.

basically, we know jerry worked for four days, completing 1/9 a day. therefore, he completed 4/9. well, who did the rest? bill did! thus, bill did $ 1\minus{}\frac49\equal{}\boxed{\frac59}$


Let $AA_1$, $BB_1$ and $CC_1$ be the bisectors of triangle $ABC$ and $\angle AA_1C=\angle AC_1B_1$. Find the $\angle BCA$.

$\angle BCA=120^{\circ}$. I believe that the problem is a classical one. I wonder if it's not already posted on the forum.


geometry
rectangle
conics
hyperbola
parameterization
geometric transformation
reflection

Suppose KLM is a triangle. We are given point A, which lies on line KL on the opposite side of L as K. Show how to construct a rectangle ABCD, such that the vertices B, C, D lie on KM, KL, LM, respectively.

If we take the variable point $D$ on $ML$ and construct $B\in MK$ s.t. $AB\perp AD$, my dynamic sketch shows that the locus of the midpoint of $BD$ is a hyperbola, so we must find the intersection of this hyperbola with the line $KL$ in order to find the center of $ABCD$.

Can we construct the intersection of a hyperbola and a line with the ruler and the compasses? I've just realized that I don't know for sure, but I don't think so :?

Of course, I'm not even sure the locus is a hyperbola, I'd have to prove that first..

Let me restate that: we are given a conic determined by 5 points (it's well-known that 5 points determine a conic), which we know is a hyperbola and a line. Can we construct the intersections of the two using only the ruler and the compasses?

For a given hyperbola we know focal points and parameter 2a. We have to find
  a point on the given line m that has the difference in distances to foci =2a.

  This point is the center of a circle that passes through   one focus and it's
  reflection in m    and is tangent to a circle centered in the other
  focus with radius 2a.

  Construct a circle O1(F1,2a). Construct a reflection G2 of F2 in m.
  Construct arbitrary circle O2 centered on m  s.t. F2 and G2 are on it
  and it crosses O1 in 2 points. The 2 points is the second radical line
  (F2G2 being the first one). Find 2 tangency points for the solution circles
  ( one internal, the other external).

  The centers are the intersection points of m and hyperbola.

  Pestich.

Hi Pestich, your solution is interesting (looks like many construction problems with conics can be reduced to problems about constructing tangent circles), but I think it solves a different problem: You construct the points of intersection of the hyperbola and the line, given the line, the two focii of the hyperbola and the difference 2a. But I think that in Grobber's original problem, not the two focii and the difference 2a were given, but the given data were five arbitrary points on the hyperbola. With your construction, one has to find the focii and the difference first, what is quite difficult (although it's possible).

Here is my way of constructing the points of intersection of an arbitrary conic K with an arbitrary line m, given five points A, B, C, D, E on the conic K:

We assume that no three of the points A, B, C, D, E are collinear (else, the conic K is degenerate). Hence, the line m can pass through at most two of the points A, B, C, D, E. Therefore, we can WLOG assume that the line m doesn't pass through any of the three points A, B, C.

A [b]circumconic[/b] of a triangle will mean a conic passing through the vertices of the triangle. The most known circumconic of a triangle is, of course, the circumcircle.

Now we consider the triangle ABC. The conic K is then a circumconic of this triangle ABC.

If p, q and r are three arbitrary real numbers, then the [b]reciprocal conjugation[/b] with parameters ( p : q : r ) with respect to triangle ABC is defined as a transformation mapping every point P in the plane of triangle ABC to another point P' after the following rule: If $P\left( x:y:z\right) $ are the homogeneous barycentric coordinates of the point P with respect to triangle ABC, then the point P' has the homogeneous barycentric coordinates $P^{\prime }\left( \frac{p}{x}:\frac{q}{y}:\frac{r}{z}\right) $. Of course, isogonal and isotomic conjugation are particular cases of reciprocal conjugation (isogonal conjugation is obtained for $p:q:r=a^{2}:b^{2}:c^{2}$, and isotomic conjugation for p : q : r = a : b : c). Note that every reciprocal conjugation is an involution, i. e. if P' is the image of a point P in some reciprocal conjugation, then P is the image of P' in this reciprocal conjugation.

Now, a reciprocal conjugation always maps a line (not passing through a vertex of the triangle) to a circumconic of the triangle, and a circumconic of the triangle to a line. Hence, if we apply a reciprocal conjugation, we get our circumconic K mapped to a line, but our line m mapped to a circumconic. This, in general, doesn't simplify the problem of finding the points of intersection of K and m, since we have to intersect a line with a conic again, but... if the circumconic which is the image of the line m is the circumcircle of triangle ABC, then we have reduced our problem to intersecting a line with a circle, what is trivial with compass and ruler, and then we can apply our reciprocal conjugation again and get the points of intersection of K and m. Hence, it remains to find a reciprocal conjugation which maps the line m to the circumcircle of triangle ABC.

Here is a way to do this in barycentric coordinates. There is probably a synthetic way (there are some Euclidean constructions of reciprocal conjugations, but I'm too lazy to try them).

Let m ( u : v : w ) be the homogeneous line barycentrics of the line m with respect to triangle ABC; in other words, the equation of the line m in barycentric coordinates is ux + vy + wz = 0. Then, we take $p=\frac{a^2}{u}$, $q=\frac{b^2}{v}$, $r=\frac{c^2}{w}$. It is easy to see that the image of the line m in the reciprocal conjugation with parameters ( p : q : r ) is a circumconic with equation $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}=0$; but this is the equation of the circumcircle of triangle ABC. Hence, our reciprocal conjugation maps the line m to the circumcircle of triangle ABC.

This closes the problem.

Hi Grobber, concerning your nice remark:

[quote="grobber"]Of course, I'm not even sure the locus is a hyperbola, I'd have to prove that first..[/quote]

This I have proven analytically (the first time I have used Cartesian coordinates for two years); let me add an easy corollary: The asymptotes of this hyperbola are parallel to the lines KL and KM.

  Darij


Let ABCD be a quadrilateral and P , Q , R , S be the midpoints of AB , BC , CD , DA respectively. The lines AB , DC intersect in X and the lines AD , BC intersect in Y . Show that triangles XRP and YSQ have the same orthocentre if and only if ABCD is a cyclic quadrilateral.

it is easy first supose   ABCD   is cyclic . and   PH  is altitude  toCD  as RK  is altitude to  AB  and  (RK,PH)=G

then the vector OG =OP+OR

then calculate  vector OK' as intersection of altittudes of Q&S.
 it is easy to show that  VECTOR OG=OK'.
Ffor the inverse supose  the  fulse and pass a circle from A,B,C.


Based on the following two groups of letters:

Group 1: A, E, F, H

Group 2: B, C, D, G

Place each letter from the following letters in its appropriate group: I, J, K, L

[hide]From what I see, Group one would contain I, K, and L, while group 2 would contain the J

Group one has the letter with straight lines, group two has letters with curved lines. [/hide]

[hide]Group 1: I, L, K
Group2 : J

Group one has only straight lines, group 2 has curves.[/hide]


calculus
integration
function
limit
trigonometry
real analysis
real analysis theorems

I suppose this question is about series and both types of proper integrals, but I'll ask the questions about improper integrals in "infinity":

If we have an integrand $f(x)$ and it turns out that after taylor expansion  $f(x) = O(\frac{1}{x^k})$ for some k > 1 (for large x of course). Is it safe to conclude that $f(x)$ is absolutely convergent seeing as by definition $|O(\frac{1}{x^k})| < C\frac{1}{x^k}$ ?

The reason I ask is because I'm so sure it has to be that way, but it feels sketchy to to talk about the convergence simply based on an Ordo-term we don't really know anything about. Is there something I'm not thinking about? I'd also like to know if there is something that can be said about $O(\frac{1}{x^k})$ when $k \le 1$ (all tests seem to fail. for example if we compare with the function $\frac{1}{x^k}$ we don't know anything because we can't assume $O(1)$ has a limit, right?).

If $f(x)=O(x^{-k})$ for some $k>1$ (as $x\to\infty$) you are perfectly safe concluding that  the integral of $f$ from some point to infinity converges absolutely (as long as there are no local issues). That's simply the comparison test.

You're right to say that $f(x)=O(x^{-k})$ for some $k\le 1$ wouldn't be enough to conclude divergence. On the other hand, you can often say more than that:

Example 1: $\int_0^{\infty}\frac{x\,dx}{x^2+x+1}=\infty$

(Never mind that we can actually compute an explicit antiderivative - I don't want to go to that trouble.)

While it's true that $f(x)=O(x^{-1})$ as $x\to\infty,$ that statement is weaker than the statement we can actually make:
$\lim_{x\to\infty}xf(x)=L,$ where $L>0.$
Then we have that for large enough $x$ and some $c>0,$ $f(x)>cx^{-1}$ That gets us divergence by comparison.

This happens often enough to be worth checking for: when you said $f(x)=O(x^{-k})$ as $x\to\infty$ did you actually have and mean the stronger condition $f(x)$ is asymptotic to some constant times $x^{-k}$? Because if that's what you have, you can act on it.

Example 2: $\int_0^{\infty}\frac{\sin x}x\,dx$ does not converge absolutely.

Here we have that $f(x)=O(x^{-1})$ but we don't have $f(x)$ asymptotic to $cx^{-1}$ as in the previous example. We do, intuitively, have that "$|f(x)|$ is comparable to $x^{-1}$ 'most' of the time." There are a couple of different ways to exploit this. One would be to place a sequence of boxes under the graph and estimate them. Here's another:

$\int_0^{\infty}\frac{|\sin x|}x\,dx=\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}x\,dx$

$>\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{(k+1)\pi}\,dx=\sum_{k=0}^ {\infty}\frac1{(k+1)\pi}\int_0^{\pi}\sin x\,dx=\frac2{\pi}\sum_{k=0}^{\infty}\frac1{k+1 } =\infty.$

Of course, $\int_0^{\infty}\frac{\sin x}x\,dx$ does converge conditionally. An integration by part turns it into

$\int_0^{\infty}\frac{1-\cos x}{x^2}\,dx$

where the removable discontinuity at zero causes no trouble and we can use the fact that this function is $O(x^{-2})$ as $x\to\infty$ to argue its convergence.

Wow. Really nice answer. I wish I could help you out like you help me out, but I doubt that's going to be possible. I did not mean that $f(x)$ was asymptotic; in fact I don't know precisely what that means. I use a definition of big-O that is equivalent to the one you have in the document you have written.

It feels weird that a function that is $O\left(\frac{1}{\sqrt{x}}\right)$ for large $x$ could be divergent, because my intuitive and informal understanding of $O(g(x))$ is that the function behaves approximately like $g(x)$, but I suppose this shows that this is not true (I know there are other reasons why this is not true as well, but it works quite well to think like that most of the time).


$ \begin {cases} 3 \left (x\plus{}\frac{1}{x}\right ) \equal{}4 \left (y\plus{}\frac{1}{y}\right )\equal{}5 \left (z\plus{}\frac{1}{z}\right ) \\xy\plus{}yz\plus{}zx\equal{}1\end {cases}$

[quote="thanhnam2902"]$ \begin {cases} 3 \left (x \plus{} \frac {1}{x}\right ) \equal{} 4 \left (y \plus{} \frac {1}{y}\right ) \equal{} 5 \left (z \plus{} \frac {1}{z}\right ) \\
xy \plus{} yz \plus{} zx \equal{} 1\end {cases}$[/quote]

Denote $ x \equal{} \tan \frac{A}{2},y \equal{} \tan \frac{B}{2},z \equal{} \tan \frac{C}{2}$ with $ A,B,C \in \left( {0,\pi } \right)$ then
\[ \Rightarrow \left\{ \begin{array}{l}
 x \equal{} \frac{1}{3} \\
 y \equal{} \frac{1}{2} \\
 z \equal{} 1 \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 x \equal{}  \minus{} \frac{1}{3} \\
 y \equal{}  \minus{} \frac{1}{2} \\
 z \equal{}  \minus{} 1 \\
 \end{array} \right.
\]

[quote="Honey_S"][quote="thanhnam2902"]$ \begin {cases} 3 \left (x \plus{} \frac {1}{x}\right ) \equal{} 4 \left (y \plus{} \frac {1}{y}\right ) \equal{} 5 \left (z \plus{} \frac {1}{z}\right ) \\
xy \plus{} yz \plus{} zx \equal{} 1\end {cases}$[/quote]

Denote $ x \equal{} \tan \frac {A}{2},y \equal{} \tan \frac {B}{2},z \equal{} \tan \frac {C}{2}$ with $ A,B,C \in \left( {0,\pi } \right)$ then
\[ \Rightarrow \left\{\begin{array}{l} x \equal{} \frac {1}{3} \\
y \equal{} \frac {1}{2} \\
z \equal{} 1 \\
\end{array} \right. \vee \left\{\begin{array}{l} x \equal{} \minus{} \frac {1}{3} \\
y \equal{} \minus{} \frac {1}{2} \\
z \equal{} \minus{} 1 \\
\end{array} \right.
\]
[/quote]
Thank you very much. You are a good friend. Have you got some nice other problem of system same my post. If you have, Can you send to me? Thank you.

Hi Honey_S. I can done this system by use ineq AG-AM. What do you think of it.
$ \begin {cases} x \plus{} y \plus{} z \equal{} z^{2^{2003}} \plus{} 2.z^{2^{2002}} \\
x^4 \plus{} y^4 \plus{} z^4 \equal{} 2.z^{2^{2004}} \\
(x \plus{} y)^{z \minus{} 1} \equal{} (z \plus{} 2004)^{x \minus{} y}\end {cases}$

Thank you very much.


can someone explain nthis solution to me?

[img]http://i42.tinypic.com/ixw6m8.jpg[/img]

That is the definition of the distribution for a uniform random variable.

ok thanks
but im just wondering how they got the function in the middle i.e. $ \frac{x\plus{}0.00000005}{0.000001}$


[quote="jasamper88"]Compute:
$\sum_{k=0}^{n} \cot^4 \frac{k\pi}{2n+1}$[/quote]

$\cot 0 = \infty$

[quote="jasamper88"]Compute:
$\sum_{k=0}^{n} \cot^4 \frac{k\pi}{2n+1}$[/quote]
I believe it is supposed to be $\sum_{k=1}^{n} \cot^4 \frac{k\pi}{2n+1}$.

Masoud Zargar


function
algebra
functional equation
algebra unsolved

Is there exist a function $f: \mathbb{Q}\rightarrow\mathbb{Q}$ satisfying: $f(x+f(y))=f(x)+y$  $\forall x,y\in\mathbb{Q}$

$x=0\Rightarrow f(f(y))=f(0)+y\ \ \forall y\in\mathbb{Q}\Rightarrow f$ is injective.
$x=y=0\Rightarrow f(f(0))=f(0)\Rightarrow f(0)=0$.
$\Rightarrow f(f(y))=y\ \ \forall y\in\mathbb{Q}$ and replacing $y$ by $f(y)$ gives $f(x+y)=f(x)+f(y) \ \ \forall x,y\in\mathbb{Q}$.
So $f$ satisfies Cauchy's functional equation and $f(x)=xf(1)\ \ \forall x\in\mathbb{Q}$.
Then $f(f(y))=y\ \ \forall y\in\mathbb{Q}$ gives $f(1)=\pm 1$,
and therefore $f(x)=x\ \ \forall x\in\mathbb{Q}$ or $f(x)=-x\ \ \forall x\in\mathbb{Q}$.

dsnt exist such function .the same method used by olorin and you will find a contradiction such $f\left(1\right)^{2}=-1$.

[quote="hien"]How about the similar problem?
Is there exist functions: $f: \mathbb{Q}\rightarrow\mathbb{Q}$ satifying:
$f(x+f(y))=f(x)-y$  $\forall x,y\in\mathbb{Q}$[/quote]

we suppose that $f$ exist
$fof(y)=-y$ and $f(0)=0$
so, $f(x-y)=f(x)+f(y)$
$x=y$ => $\forall x\in Q;\ 2f(x)=f(0)=0$ then $\forall y\in Q$ ;$2y=2f(x)-2f(x+f(y))\neq 0$ contradiction

i  can't see   the contradiction :blush:

here you are a similare problem:
1:find all monotonic functions such:
$f: \mathbb{R}\rightarrow\mathbb{R}$ and $f(x+f(y))=f(x)+y$.
2:prove that for every $n\geg{1}$ dosnt exist a monotonic function satifyng:
$f_{l}eft(x+f(y))=f(x)+y^{n}$

[quote="fermat3"]i  can't see   the contradiction :blush:[/quote]
we have $\forall x\in Q, 2f(x)=0$ so, $\forall y\in Q,2y=-2f(x+f(y))+2f(x)=0$ you see now :lol:

[quote="aviateurpilot"]$f(x-y)=f(x)+f(y)$[/quote]
It's $f(x-y)=f(x)-f(y)$ not $f(x-y)=f(x)+f(y)$.
Then $f(y)+f(x-y)=f(x)$ and $x=y+z$ gives $f(y)+f(z)=f(y+z)$.


A strip of sinc metal weighing 2.00 grams is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:

Zn (s) + 2AgN03 (aq) = 2Ag(s) + Zn(N03)2 (aq)

The equals sign should be an arrow. Sorry about the subscripts.

How many grams of Ag will form?

1) Convert the given masses of each reagent in number of moles.

2) Decide which is the limiting reagent.

3) Assuming 100% yield, find the number of moles of silver expected to form.

4) Convert that number of moles in a mass.


Of the first 100 natural numbers, 'k' numbers are randomly chosen, if the sum of the 'k' numbers is even A wins, if it is odd 'B' wins, find the values of 'k' for which the game is fair.

Any takers for this problem, shall I try posting this in the combinatorics section?

[hide="Partial solution"]For $k$ odd, the game is always fair.  We see this by constructing a bijection between winning combinations for $A$ and winning combinations for $B$ -- in any set in which $A$ wins, add 1 to every member (sending 100 to 1).  This changes the parity of the sum, and it is bijective, so we're good.

Even $k$ are not immediately obvious, and I haven't sat down to work it out.  Someone else care to give it a go?[/hide]

For $k=2$, the game is unfair to $A.$ I guess the game is unfair to either one of them for even $k$.

Well, by computer calculation I can verify that it is unfair for all even numbers (depending on whether $k \equiv 0$ or 2 mod 4), but the difference in probability is very small, especially for $k$ relatively close to 50 (we're talking on the order of $10^{-14}$).

Haven't had any good thoughts on how to prove it, though.

By using generating functions, I think we can show that that the difference in probability (for $k$ even) is exactly \[\binom{50}{k/2}/ \binom{100}{k}\,.\] Edit:
Here's a combinatorial interpretation for the case when $k$ is even.

First, let's partition the set 1, 2, ..., 100 into 50 pairs: {1, 2}, {3, 4}, etc.  If the $k$ numbers chosen happen to be $k/2$ of these pairs, then A always wins (if $k/2$ is even) or $B$ always wins (if $k/2$ is odd).  We will show that the remaining cases divide fairly.

The proof is by bijection, similar to Joel's argument.  Suppose we have $k$ numbers from 1, 2, ..., 100 that are not all pairs as above.  Look at the smallest number whose partner was not chosen.  Change that number to its partner.  That changes the parity of the sum.  So we have a bijection from the remaining wins of A to the remaining wins of B.


I am 6 years older than my sister. What was the number of years in my sister's age when I was three times as old as she was?

x= sisters age, x+6= your age

$ 3x \equal{} x \plus{} 6$

$ 2x \equal{} 6$

$ x \equal{} 3$


consider a square $ABCD$ with side length 1.
at every vertext we draw a quarter circle inside the square (for example if the center is A the quareter circle passes through B and D).

what is the area common to all four quarter circles?

[hide="Solution"]Drawing the circles carefully, we see that the region desired is formed by a small square as well as four regions of $\frac{1}{12}$ of each of the four circles, thus $30^\circ$ arc. Thus, by using LoC, we have that the area of the square is $2-\sqrt{3}$ and we know that the area of each of the circular regions is $\frac{\pi}{12}-\frac{1}{2} \cdot 1 \cdot 1 \cdot \sin{30^\circ}=\frac{\pi-3}{12}$, so the total area is $\boxed{\frac{\pi+3-3\sqrt{3}}{3}}$[/hide]


logarithms
limit
real analysis
real analysis unsolved

Let S be the series $ 1 \minus{} 1/3 \plus{} 1/5 \minus{} 1/7 \plus{} 1/9\ldots$. Let $ S_r$ be the rearrengement of $ S$ in blocks of $ 2r$ consecutive terms followed by $ r$ negative terms. For example $ S_1 \equal{} (1 \plus{} 1/5 \minus{} 1/3) \plus{} (1/9 \plus{} 1/13 \minus{} 1/7) \plus{} \ldots$. Find the sum of $ S_r$.

You can use the fact that the sum of $ S$ is $ \frac {\pi}{4}$.

If i'm not mistaken it should be $ \frac{\pi\plus{}\ln (2)}{4}$ independent of $ r$...

Yes, it seems that it does not depends of $ r$, at least that is what I found. Whats your proof?

I just looked at the partial sums and used the limit $ \lim_{n\to \infty} \sum_{i\equal{}n}^{2n}\frac1{i}\equal{}\ln(2)$


Let $ X_1, X_2, ..., X_n$ be independently and identically distributed random variables with:
\[ P(X_i \equal{} j) \equal{} \frac 16 \forall j \equal{} 1,2, ..., 6\]
($ X_i$'s are the value obtained via independent rolls of a fair dice.)
For $ k \leq n$, find:
\[ E\left[ \frac{\sum_{i\equal{}1}^k {X_i}}{\sum_{i\equal{}1}^n {X_i}} \right]\]

\[ E\left[ \frac {\sum_{i \equal{} 1}^k {X_i}}{\sum_{i \equal{} 1}^n {X_i}} \right] \equal{} \dfrac{k}{n}?
\]

Yes, that is correct. Let $ U\equal{}\sum_{i\equal{}1}^kX_i$ and $ V\equal{}\sum_{i\equal{}1}^nX_i.$

If we pick a possible value $ m$ for $ V$ and condition on $ V\equal{}m,$ then each term of the sum will, on average, have contributed equally. We can say that because any permutation of the terms of the sum is equally likely. That means that $ E[X_i\mid V\equal{}m]\equal{}\frac{m}{n}.$

The expectation of a sum is the sum of the expectations, so we can add up that previous result to get that $ E[U\mid V\equal{}m]\equal{}\frac{km}n.$

It follows that $ E\left[U/V\mid V\equal{}m\right]\equal{}\frac{km}n\cdot\frac1m\equal{}\frac{k}n.$

Then the expectation is the expectation of the conditional expectations, but if the conditional expectations are all the same, then that is easy to compute. Hence
$ E\left[\frac{U}{V}\right]\equal{}\frac{k}n.$

[quote="Kent Merryfield"] That means that $ E[X_i\mid V \equal{} m] \equal{} \frac {m}{n}.$ [/quote]

Did you mean using $ E[\sum_{i\equal{}1}^n X_i \mid V \equal{} m] \equal{} m$ and $ E[X_i\mid V \equal{} m] \equal{} E[X_j\mid V \equal{} m] \forall i, j$ ?

Yes, that's the explanation.

Note that the exact distribution of $ X_i$ was immaterial - the argument never used that it was specifically a roll of a die. It could have been anything.

Another derivation could be:
${{ E[\frac{\sum_{i=1}^{k}{X_i}}{\sum_{i=1}^{n}{X_i}}] = \sum_{i=1}^{k} {E[\frac{X_i}{\sum_{j=1}^{n}{X_j}}}]} = k {E[\frac{X_1}{\sum_{i=1}^{n}{X_i}}}]}$ (1).

Plugging $ k=n$, we have: ${{ n E[\frac{X_1}{\sum_{i=1}^{k}{X_i}}}] = E[1] = 1 \Rightarrow E[\frac{X_1}{\sum_{i=1}^{k}{X_i}}}] = \frac{1}{n}$ (2).

With (1) and (2), the conclusion follows.

My New Question: Could we remove the hypothesis of $ X_i$ to be independent? (just keep the "identical" part)

Look at your own latest argument. Where, if anywhere, did  you use independence? The answer is that you never did. (That the expectation of the sum is the sum of the expectations does not require any assumption about how the terms are related or correlated - it's just the linearity of the integral.) And since you never did use independence then yes, you can drop that hypothesis.

Erm... I do not see how just the identical distribution of $ X_i$ implies the identical distribution (or even equality of expectations) of $ \frac{X_j}{\sum_{i\equal{}1}^n X_i}$ for different $ j$. :?

[quote="Kent Merryfield"]Hmm... OK, now I'm not sure. Can you find an example that breaks this?[/quote]
For example, let $ n\equal{}3, \ X_1 \equal{} X_2 \equal{} 7 \minus{} X_3,$ then
\[ E\frac {X_1}{X_1 \plus{} 7}\not \equal{} E\frac{7 \minus{} X_1}{X_1 \plus{} 7}.
\]


HMMT Harvard college ARML Stanford geometry

Does anyone know the exact location of this year's HMMT? Where is the science center? Is there a street address available? I've e-mailed them, and I have not received a response yet.

If you'll look at the other thread about HMMT, you'll see that I was kind enough to go dig up the Harvard interactive map of the area.  Give me a minute and I'll find the link again.

... ah, here it is!

First, the FAQ for HMMT:
[url]http://web.mit.edu/hmmt/www/faq.shtml[/url], which apparently contains a link to a map.

and

[quote="JBL"][url]http://www.map.harvard.edu/[/url] is a map of the Harvard area.
http://www.map.harvard.edu/level2.cfm?mapname=camb_allston&tile=F7&series=N should be a map centered on the Science Center.

It's pretty easy to find -- big concrete building with a layered roof and big black steel beams sticking out of the side making it look like some kind of spider or something.[/quote]

Joel, you know I was trying to find you the whole time.

I bumped into Charles, Antony, Yelena, the various HMMT staff (Will and Jeff went AWOL) and some others, though YOU seem to be missing....

There's ARML, and I have every intention of being at the West Coast site for ARML 2004.

I wouldn't dare to call SMT similar to HMMT. Perhaps the format is somewhat similar, but the event isn't anything like as prestigious. However, having said that, I do remember that at least one team from San Diego went last year. No one seems to want to give me full score reports from contests this year, so I'm not very up-to-date on Bay Area contest results.

[quote="JBL"]No, Joel was unfortunately not there.[/quote]
Aren't you Joel? :?:

Um can someone please split this and lock the original thread or something?

[quote]Aren't you Joel?[/quote]

He is just fond of speaking of himself in the third person.  :mrgreen:

Interesting, so does Simon...is there a connection? :lol:

[quote=KnowingAnt][quote=texas137]>"Is there some west coast math competition?"

There are competitions similar to HMMT at Stanford and at Rice. I think all 3 were this week-end.[/quote]

yeah berkeley math tournament as well[/quote]

Thanks! I'll be sure to check it out

[quote=exp-ipi-1][quote=KnowingAnt][quote=texas137]>"Is there some west coast math competition?"

There are competitions similar to HMMT at Stanford and at Rice. I think all 3 were this week-end.[/quote]

yeah berkeley math tournament as well[/quote]

you do realize that the post was 20 years ago[/quote]

5 years 6 days ago? feb 29 2004 --> feb 29 2008 --> feb 29 2012 --> feb 29 2016 --> feb 29 2020 --> feb 29 2024 + 6 days = march 6 2024

[quote=KnowingAnt][quote=texas137]>"Is there some west coast math competition?"

There are competitions similar to HMMT at Stanford and at Rice. I think all 3 were this week-end.[/quote]

yeah berkeley math tournament as well[/quote]

micheal did you hit your head


geometry
ratio
Asymptote
algebra
system of equations

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point.  The areas of four of these triangles are as indicated.  Find the area of triangle $ABC$.

[asy]
size(200);
pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C);
draw(F--C--A--B--C^^A--D^^B--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("84", centroid(H, C, E), fontsize(9.5));
label("35", centroid(H, B, D), fontsize(9.5));
label("30", centroid(H, F, B), fontsize(9.5));
label("40", centroid(H, A, F), fontsize(9.5));[/asy]

[hide]The three lines are altitudes of the triangle so opposite interior triangles are similar by a a scale factor of 2.8 so multiply 35 by 2.8(=98) and 40 by 2.8(=112) and then add all 6 areas together.  this gives you an area of 399. [/hide][/hide]

[hide="different hint"]
If two triangles share the same base, the ratio of their areas is the ratio of their altitudes.
If two triangles share the same altitude, the ratio of their areas is the ratio of their bases.
[/hide]

EDIT: oops, "there" $\ne$ "their" :blush:

how can mkottman prove his solution is correct? it seems quite flawed to me

[hide]
We set the left blank area to $y$, and set the right blank area to $x$.

Using the fact that triangles with similar altitudes, we see that if we have $AB$ as the base, we get:
$\frac{124+y}{65+x}=\frac{30}{40} \rightarrow y= \frac{4x-112}{3}$.

Using $AC$ as the base, we get:
$\frac{y}{84}=\frac{70+y}{119+x}$.

Solving the system of equations, we get:

$x=70, y=56 \rightarrow Area=56+70+35+30+40+84=315$.
[/hide]

[hide="Here is an interesting proof."] [color=red][b]
A new enunciation.[/b] Let $ABC$ be a triangle and a its inside point $I\ .$ Denote $M\in AI\cap BC$, $N\in BI\cap CA$, $P\in CI\cap AB\ .$ I know the areas $[AIP]=84$, $[BIM]=40$, $[CIM]=30$, $[CIN]=35\ .$ Ascertain the areas $[AIN]$ and $[BIP]\ .$[/color]

[color=blue][b]The first proof.[/b] Denote $[AIN]=x\ ,$ $[BIP]=y\ .$ Apply the [/color][color=red]Menelaus' theorem[/color] [color=blue]to the transversal $\overline {BMC}$ for the triangle $\triangle ANI\ :$ $\frac{BI}{BN}\cdot\frac{CN}{CA}\cdot\frac{MA}{MI}=1$ $\Longrightarrow$ $\frac{70}{105}\cdot \frac{35}{35+x}\cdot\frac{65+x}{30}=1$ $\Longrightarrow$ $\frac{65+x}{9}=$ $\frac{35+x}{7}=$ $\frac{30}{2}=15$ $\Longrightarrow$ $\boxed {\ x=70\ }\ .$ Apply the [/color][color=red]Ceva's theorem[/color] [color=blue]to the point$I\in (ABC)\ :$ $\frac{MB}{MC}\cdot\frac{NC}{NA}\cdot\frac{PA}{PB}=1$ $\Longrightarrow$ $\frac{40}{30}\cdot \frac{35}{70}\cdot\frac{84}{y}=1$ $\Longrightarrow$ $\boxed {\ y=56\ }\ .$ Therefore, $\boxed {\ S=315\ }\ .$[/color]

[color=darkblue][b]The second proof.[/b] [size=150][u]Yetti[/u][/size] remarked that to know
the area $[AIP]$ isn't necesarry ! Indeed, denote $[AIP]=z\ .$
$\frac{IP}{IC}=\frac{[AIP]}{[AIC]}=\frac{[BIP]}{[BIC]}$ $\Longrightarrow$ $\frac{z}{x+35}=\frac{y}{70}\ .$
$\frac{IN}{IB}=\frac{[AIN]}{[AIB]}=\frac{[CIN]}{[CIB]}$ $\Longrightarrow$ $\frac{x}{y+z}=\frac 12\ .$
$\frac{IM}{IA}=\frac{[BIM]}{[BIA]}=\frac{[CIM]}{[CIA]}$ $\Longrightarrow$ $\frac{40}{y+z}=\frac{30}{x+35}\ .$ Therefore,
$70z=y(x+35)\ ;\ 2x=y+z\ ;\ 3(y+z)=4(x+35)$ $\Longrightarrow$
$\boxed {\ x=70\ ;\ y=56\ ;\ z=84\ ;\ S=315\ }\ .$[/color][/hide]

The areas 40, 30, 35 are sufficient to find the remaining 3 areas - the area 84 is not necessary. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39142]www.mathlinks.ro/Forum/viewtopic.php?t=39142[/url].

[color=red] I added a second proof in the my previous reply owing to the [b]Yetti's remark[/b] from the his last reply.[/color]

in zellex method,

the ratio he took is  (124+y)/(75+x)=30/40

but i think it should be (124+y)/(75+x)=40/30

since ,let the common interior point be P and the point from C to AB be M

in triangle ACM area = (1/2)*h*AM
likewise BCM area = (1/2)*h*BM

ratio is of the bases is the ratio of areas=AM/BM

like wise the ratio of areas of AMP and BMP are AM/BM
 can you check this ,is this correct or wrong

[quote="beta9"]Why are the lines all neccessarily altitudes?[/quote]

they're not, they share the same base, so the ratio of their areas is the ratio of their altitudes.

[quote="Zellex"][hide]
We set the left blank area to $ y$, and set the right blank area to $ x$.

Using the fact that triangles with similar altitudes, we see that if we have $ AB$ as the base, we get:
$ \frac {124 \plus{} y}{65 \plus{} x} \equal{} \frac {30}{40} \rightarrow y \equal{} \frac {4x \minus{} 112}{3}$.

Using $ AC$ as the base, we get:
$ \frac {y}{84} \equal{} \frac {70 \plus{} y}{119 \plus{} x}$.

Solving the system of equations, we get:

$ x \equal{} 70, y \equal{} 56 \rightarrow Area \equal{} 56 \plus{} 70 \plus{} 35 \plus{} 30 \plus{} 40 \plus{} 84 \equal{} 315$.
[/hide][/quote]
Just wondering, I got exactly the same thing as Zellex using AB as a base, but then I used BC as a base but after some simplification it turns out that the equation is a duplicate of the first equation using base AB... however, using base AC as Zellex did won't give you a duplicate of the first equation using base AB... is there a reason for it?

That would be because the missing areas are both in one triangle, so we will only x and y terms again.  (x and y are the missing areas).  But if you use AC as a base x and y are in different triangles so we will have xy terms, therefore we can solve the system of equations.

[hide="Solution"]

Denote the point on $AB$ to be $D$, the point on $AC$ to be $E$ and the point on $BC$ to be $F$.

[asy]
/* File unicodetex not found. */

 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.74cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 7.44, ymin = -2.78, ymax = 6.3;  /* image dimensions */

 /* draw figures */
draw((0,0)--(5,0));
draw((5,0)--(2.82,3.42));
draw((2.82,3.42)--(0,0));
draw((2.82,3.42)--(2.78,0));
draw((5,0)--(1.51,1.83));
draw((0,0)--(3.96,1.64));
 /* dots and labels */
dot((0,0),dotstyle);
label("$A$", (0.08,0.12), NE * labelscalefactor);
dot((5,0),dotstyle);
label("$B$", (5.08,0.12), NE * labelscalefactor);
dot((2.82,3.42),dotstyle);
label("$C$", (2.9,3.54), NE * labelscalefactor);
dot((2.78,0),dotstyle);
label("$D$", (2.86,0.12), NE * labelscalefactor);
dot((1.51,1.83),dotstyle);
label("$E$", (1.58,1.94), NE * labelscalefactor);
dot((3.96,1.64),dotstyle);
label("$F$", (4.04,1.76), NE * labelscalefactor);
dot((2.79,1.16),dotstyle);
label("$M$", (2.88,1.28), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

 Let $CD\cap BE\cap AF=M$.  Note that $\frac{[ADM]}{[BDM]}=\frac43=\frac{AD}{BD}$ because of same altitudes.  Using Ceva's on these lines intersecting at $M$ gives us $\frac43*\frac{BF}{CF}*\frac{CE}{AE}=1$.  Using similar logic as before, we have $\frac{BF}{CF}=\frac{[BMF]}{[CMF]}$ and $\frac{CE}{AE}=\frac{[CEM]}{[AEM]}$.  Therefore, we now have $\frac43*\frac{35}{[CMF]}*\frac{84}{[AEM]}=1$.  Denote $[CMF]=a$ and $[AEM]=b$ now as well.  Therefore, we get $ab=28*4*35$.

Now, $\frac{[ACD]}{[CDB]}=\frac43$ or therefore $\frac{124+b}{65+a}=\frac43$.

Hence, now $372+3b=260+4a\implies 112+3b=4a$.  Therefore, $b=\frac{4a-112}{3}$.

Substituting this back into $ab=28*4*35$ gives us $\frac43*a(a-28)=28*4*35$.  This gives us $a(a-28)=28*3*35$.  We list out the factors, which are $2*2*7*3*5*7$.  Therefore, we find $a=2*5*7\implies a=70$.  Next, $b=\frac{280-112}{3}\implies b=56$.

Therefore, the sum of all the areas is $56+70+40+30+35+84=\boxed{315}$.  [/hide]

Mass points + same base same altitude kills this problem! note how the 84 is a red herring:

[hide]$[APF]:BPF]=AF:FB=4:3$ so assign masses of $3$ and $4$ to $A$ and $B$ respectively. Drawing $\overline{PB}$ we see that $[APB]:[BDP]=AP:PD=2:1$ so assign a mass of $6$ to $D$. Since the center of mass at $P$ is $9$ then $C$ has to have a mass of $2$ to balance with $F$; it follows that $[ABC]:[ABD]=3:1$ hence $[ABC]=3(30+35+40)=\boxed{315}$

[img]http://oi46.tinypic.com/2ngd16p.jpg[/img][/hide]

What's a red herring? I remember hearing that term used before...

In short, a red herring is something that will troll noobs to the heavens.

Bumping a five-year-dead thread, lol

[hide=My Solution]
[asy]
size(200);
pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C);
draw(F--C--A--B--C^^A--D^^B--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("84", centroid(H, C, E), fontsize(9.5));
label("35", centroid(H, B, D), fontsize(9.5));
label("30", centroid(H, F, B), fontsize(9.5));
label("40", centroid(H, A, F), fontsize(9.5));
label("$y$", centroid(H, D, C));
label("$x$", centroid(H, A, E));[/asy]
[i]Note: $x$ and $y$ refer to the areas.[/i]

We begin by using the same-height-same-base formula to help establish some ratios $$\frac{105}{84 + x + y} = \frac{35}{y}$$ which yields $$105y = 84 \cdot 35 + 35x + 35y$$ so that $$70y - 35x = 84 \cdot 70$$ ultimately leading to $2y - x = 84$ and thus, $$x = 2y - 84.$$

Note that if we replace $x$ with $2y - 84$, the sum of the areas of the little triangles is $2y - 84 + 84 + 40 + 35 + 30 + y = 3y + 105$ $= 3 \cdot (y + 35)$, which happens to be (if $Z$ is the intersection of the lines) $\triangle ABZ$.  Also, see that by same-height-same-base, we can establish that $$\frac{40}{30} = \frac{124 + x}{65 + y}$$ and that $y + 35 + 30$ is exactly $\frac{3}{7}$ of the large triangle.  The only difference these two regions share is the triangle with area 30.  Thus, if $a$ is the area of the large triangle, $$30 = (\frac{3}{7} - \frac{1}{3}) a.$$ Solving gives us $a = \boxed{315}$, which is the correct answer.
[/hide]

[s]Finding that took an hour.  :P[/s]

Edit: you can also reach the conclusion that $ABZ$ is $\frac{1}{3}$ of $[ABC]$ by area ratios, which is quicker and cleaner than my original algebra.

[hide=an amazing solution] firstly denote point intersection of line from  $A$ and  $CB$ be $F$ using the lemma mentioned in this solution, [url]https://artofproblemsolving.com/community/u650014h509770p25873315[/url]notice: $\frac{AP}{PF}=\frac{AD}{DB}(1+\frac{BF}{BC})$, from base division theorem , $\frac{AD}{DB}=\frac{4}{3}$ and $\frac{AP}{PF}=\frac{2}{1}$. so we have $\frac{BF}{BC}=\frac{1}{2}$,so again from base division theorem $\frac{[ACF]}{[ABF]}=\frac{1}{2}$, so $[ACF]=210$ hence $[ABC]=210+105=\boxed{315}$, interestingly there was no use of triangle which had area $84$[/hide]

Ceva's theorem. We have $$\frac{4}{3}\cdot \frac{35}{a}\cdot \frac{84}{b}=1$$

We also have $$\frac{70}{84+b}=\frac{35}{a}$$. The first equation gives $$ab=35\cdot 112=3920$$

Now, we have $70a-35b=35\cdot 84\to 2a-b=84$. Substitute $b=2a-84$ gives $$a(2a-84)=3920\to a=70$$

Plug back in gives $b=140-84=56$. Thus, the area is $40+30+35+70+84+56=\boxed{315}$.

We have that $\frac{4}{3}\cdot \frac{35}{a}\cdot \frac{84}{b}=1$, and $\frac{35}{a}=\frac{70}{84+b}$, using the first equation, $ab=3920$, and $2a-b=84$, solving, $a=70$, and $b=56$, adding up all the areas, give us $\boxed{315}.$


logarithms
real analysis
real analysis unsolved

bename khoda,,
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one more time : http://pagesperso-orange.fr/gery.huvent/articlespdf/article3.pdf
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