A nicer approach: multiply $x^2+x=2y^2$ by $4$, the left hand side gets $(2x+1)^2-1$ then. Now continue with Pell...
For the Pythagorean triplets: this can be proved the same manner/follows by your rewriting from the above (yes it's true).

but ZetaX it gets really complicated with pyth triplets.
i will try to read Pell equation(i'm just trying to read them for allot of time now
and always ...done smthin else).
thank you very much! :)

ok...i'm trying to continue what i wrote
we have now $s(\beta^2-\alpha^2)=t+1,2s\alpha\beta=t$
but we know $(t,t+1)$=1 so i figgure $s=1$
now we have to solve the system $\beta^2 -\alpha^2=t+1,2\alpha\beta=t$
we actually need to find $f(t)=\alpha$ and $g(t)=\beta$

$\beta=\frac{t}{2\alpha}$ we substitute that in the first relation
we get to $t^2-4\alpha^4=4\alpha^2(t+1)$ now we substitute $t=2\alpha\beta$
we get to
$\alpha^2+(2\beta)\alpha+(1-\beta^2)=0$
we try to get $\alpha$ from here.
if we calculate $\Delta=4(2\beta^4 -1)$
to get integer solutions we have to have $\Delta$ a square.
so when is $2\beta^4 -1$ a square ?

i am really going around in circles.

I'm reading it as proving there are an infinite number of simultaneously triangular and square numbers.

[quote="spx2"]$\beta=\frac{t}{2\alpha}$ we substitute that in the first relation
we get to $t^2-4\alpha^4=4\alpha^2(t+1)$ now we substitute $t=2\alpha\beta$
we get to
$\alpha^2+(2\beta)\alpha+(1-\beta^2)=0$
we try to get $\alpha$ from here.
if we calculate $\Delta=4(2\beta^4 -1)$
to get integer solutions we have to have $\Delta$ a square.
so when is $2\beta^4 -1$ a square ?[/quote]
At first solving one of them for $t$ and setting into the other is a lot easier. At second, $2\beta^4-1=y^2$ has only finetely many solutions as posted recently. At third, this is of no interest because $\Delta = 8 \beta^2-4$, which is more or less the same you started with.

Really, both your problems are totally equivalent to Pell Equations, so you will need to use more or less the same idea.

oh sorry zetax you are right $\Delta=4(\beta^2-1)$
and finding when $\Delta$ is a square brings me back
to the same problem as before.

[quote="spx2"]proove that $x^2+x=2y^2$ has infinite solutions

i seem to not get any of these problems right...i would like some advice if possible.
ill write thoughts about this problem
$x(x+1)=2y^2$
it seems clear that $x(x+1)$ is even
if x is a square $x=k^2$ the question i ask myself is when is $x+1=2p^2$,
can also be written $k^2+1=2p^2$ so we must have k odd.
now $k=2t+1$.
from this we have $p=\pm \sqrt{2t^2+2t+1}$.
now when is $2t^2+2t+1$ a square ?
if we write $2t^2+2t+1=t^2+(t+1)^2$.
so i am asking when is $t^2+(t+1)^2$ is a square.
can there be infinite many pythagorean triples of the form $t,t+1,f(t,t+1)$ ?
if this is so the problem is solved.
i see in my book that pythagorean triples $x,y,z$ that satisfy $x^2+y^2=z^2$ have
the general formula $x=s(\beta ^2 - \alpha^2),y=2s\alpha \beta,z=s(\beta^2 +\alpha^2)$
is it possible to get a good solution following this path ?
im stuck  :(
please help ![/quote]

My opinion two consecutive numbers never been squares.

$3^2+4^2=5^2$
$21^2+20^2=29^2$
$119^2+120^2=169^2$
$697^2+696^2=985^2$
were taken from here
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html
and i'm sure the paires i wrote above are not the only ones

ok...exactly below where those pairs are written
there is also one more interesting observation
i will quote from the site above mentioned

[quote]Kayne Johnston aged 13 has also found the neat pattern to compute this table without using the m and n generators, each row being computed simply from the two rows before:
the next row in the table has a smallest side that is 6 times the previous smallest side minus the smallest side before that plus 2.
For instance, after the first two rows:
a     b=a+1     h=sqrt(a^2+b^2)
3     4             5
21   20            29[/quote]

so this table stretches...allot so we have a rule for generating pythagorean triples
of the kind we needed.

this is more of a non-orthodoxic solution...ZetaX please tell me where should i read
Pell Equations.
Also,ZetaX how could this kid discover this pattern in those triples ?


Let $ n$ be a positive integer and $ a_1,a_2,\ldots,a_n$ be positive reals such that $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$.

$ A \equal{} \text{min}\left\{\frac {a_1}{1 \plus{} a_1},\frac {a_2}{1 \plus{} a_1 \plus{} a_2},\ldots,\frac {a_n}{1 \plus{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n}\right\}$.

Determine the maximum value of $ A$.

Answer: $ 1\minus{}2^{\minus{}\frac{1}{n}}$, right?

That's right kunny :) ,
can you explain how you got it if possble.

Actually, the answer pretty much gives it away.

From the given,
\begin{align*}
A &\le \frac{a_1}{1 + a_1}, \\
A &\le \frac{a_2}{1 + a_1 + a_2}, \\
&\dots, \\
A &\le \frac{a_n}{1 + a_1 + a_2 + \dots + a_n},
\end{align*}
so
\begin{align*}
1 - A &\ge \frac{1}{1 + a_1}, \\
1 - A &\ge \frac{1 + a_1}{1 + a_1 + a_2}, \\
&\dots, \\
1 - A &\ge \frac{1 + a_1 + a_2 + \dots + a_{n - 1}}{1 + a_1 + a_2 + \dots + a_n}.
\end{align*}
Multiplying all these inequalities, we get
\[ (1 - A)^n \ge \frac{1}{1 + a_1 + a_2 + \dots + a_n} = \frac{1}{2} = 2^{-1}.\]
Therefore, $ 1 - A \ge 2^{-1/n}$, or $ A \le 1 - 2^{-1/n}$.  Equality occurs if $ a_k = 2^{(k - 1)/n} (2^{1/n} - 1)$ for all $ 1 \le k \le n$.

Equality should occur when $ \frac {a_k}{1 + a_1 + a_2 + \ldots + a_k}\}$ are all equal with k ranging from 1 to n, but is there a simple way to solve for $ a_k$ explicitly?


Does any one know a good way to find the determinant of a 4 by 4 or a 5 by 5 matrix?

Can someone tell me how to do number one of this NAT? :
http://www.natassessment.com/2007_TS/2003_TS_Test_Excerpt.doc

Be aware that there are plenty of zeroes in your matrix
thus check for the products without them...

http://en.wikipedia.org/wiki/Cofactor_%28linear_algebra%29
The section titled "Cofactor expansion" gives a formula for the determinant, which is the one I use. Note that this works for any row and column; it's easiest to use one with a lot of $ 0$s.

I'm sorry, I still don't quite understand...can someone explain a little more? thank you.

Here are some hints:

If you add or subtract ($ n\cdot$ a row) to another row, the determinant stays the same. Try to eliminate nonzero numbers.

Then use minors.

cofactor method is pretty much the only one i know, as mellowmelon said. if the wiki doesn't help, look up cofactor method on like answers.com.
it's a really helpful method and it can be used for finding the determinants of any nxn matrix.


pigeonhole principle
modular arithmetic
number theory
number theory unsolved

Find, with proof, all positive integers $m$ for which there does [b]not[/b] exist a Fibonacci number $F_{n}\equiv a \bmod m$ for at least one $a$ in the set $\{ 0,1,2, \ldots, m-1 \}$.

This is an interesting problem.  Where did you get it?  So far I've found possibly $m = 4k, k > 1$.

My own observation. :)  I was looking at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153892]this[/url] thread, and noticed that for a few (presumably more) specific moduli have remainders that are skipped. I was thinking that perhaps a pigeonhole argument can be applied here as well, but I'm not quite sure how.

I don't understand the problem... I may be being really stupid, but the third Fibonacci number 2 is congruent to 2 mod 8, 2 being in the set :?:  What am I missing?

There are no Fibonacci numbers $\equiv 6 \bmod 8, 6 \bmod 12$ (note the important phrase [i]at least one[/i] $a$).  I have not tested for higher numbers.  It seems to me that this problem might be very difficult in the general case...

$34 \equiv 6 \mod 28$, but I think there always has to one value of $b$ such that there is no number congruent to $6b \mod 4k$.

Also, $m=13$ satisfies the conditions of the problem.

[quote="t0rajir0u"]It seems to me that this problem might be very difficult in the general case...[/quote]

Then should I post it in the [b]Olympiad Number Theory, Open Questions[/b] forum as well?

I think it would fit better in the [b]number theory unsolved problems[/b] maybe some one (probably harazi or vess or Rust) will answer it sooner than you think...  :D

Some computations:

It seems that the complementary set is much more manageable.  In addition to 1, it contains:

2, 4 (but not 8)
3, 9, 27, 81, 243
5, 25, 125, 625
7 (but not 49)
6, 10, 14, 15, 20, 30, 35, 45, 50, 70, 75, 100, 150, 175, 225

and no other numbers less than 243.  (I have not checked the range 243 to 625.)

Some unsupported conjectures:

1) No prime larger than 7 is on the list.
2) All powers of 3 and 5 are on the list.
3) There is a discernable pattern to the composite numbers on the list, beyond the fact that if $n$ is not on the list, no multiple of $n$ is.

[i]Later edit[/i]:
This exists in the OEIS, sequence [url=http://www.research.att.com/~njas/sequences/A079002]A079002[/url].

[quote="OEIS"]References:  R. L. Graham, D. E. Knuth and O. Patashnick, "Concrete Mathematics", second edition, Addison Wesley, ex. 6.85, p. 318, p. 562
Formula:  Consists of the integers of the form: 5^k, 2*5^k, 4*5^k, 3^j*5^k, 6*5^k, 7*5^k and 14*5^k[/quote]

I'll say some obvious thing:

1) if $\left(\frac{5}{p}\right) =1$ then the Fibonacci sequence is periodic modulo p and its period is p-1, so it can't contain all residues.

2) if $\left(\frac{5}{p}\right) =-1$ and $\left(\frac{-1}{p}\right) =1$ then the Fibonacci sequence is periodic modulo p and the period divides $2(p+1)$, but $F_{n}+F_{n+p+1}\equiv 0$ and $F_{n}\equiv\pm F_{p+1-n}$ and $F_{\frac{p+1}{2}}\equiv 0$ so there are not enough residues.

If $p\equiv 3,7\pmod{20}$ obviously we can't solve the problem this way, I've been trying to show that $X=\{F_{i}\; |\; 2\ge i\ge \frac{p+1}{2}\}$ cannot be a gaussian set, but without any success...


function inequalities calculus derivative algebra domain limit

Show that if a Jensen-convexe(ex: f(m*x+(1-m)*y)<=m*f(x)+(1-m)*f(y),with m in (0,1)) function f defined over an opened interval is bounded then f  is continous.

This is a standard theorem, and can presumably be found in most textbooks which define convex functions.

We can prove it by proving something stronger: $f$ has a left-hand derivative and a right-hand derivative everywhere.

Fix $x$ in the domain of $f$ and define $g_x(h)=\frac{f(x+h)-f(x)}h$. By repeatedly using the definition of convex given above (namely, the graph of the function lies below any secant line segment), we can show the following inequalities:

If $0<h_1<h_2, g_x(h_1)\le g_x(h_2)$
If $h_2<h_1<0, g_x(h_2)\le g_x(h_1)$
If $h>0$ and $k<0, g_x(k)\le g_x(h)$.

Since monotone bounded functions have limits, we can see that both one-sided limits of $g_x(h)$ exists as $h\to 0$, and (by the third inequality), $\lim_{h\to 0^-}g_x(h)\le \lim_{h\to 0^+}g_x(h)$. The fact that these limits exist implies that $\lim_{h\to 0}(f(x+h)-f(x))= 0$ from both sides. Hence, $f$ is continuous at $x$.

A direct consequence of this proof is that for each $x$ there exists a slope $m$ (depending on $x$) such that $f(x)+mh\le f(x+h)$ for all $h$ such that $x+h$ is in the domain of $f$. To show that, we choose any $m$ such that $\lim_{h\to 0^-}g_x(h)\le m\le \lim_{h\to 0^+}g_x(h)$. The "supporting line" in this inequality is then the key to an efficient proof of Jensen's Inequality.

I have a more sticky question: if  $f:[0,1]\rightarrow \R$, and
$2f((x+y)/2)\le f(x)+f(y)$, then $f$ is not necessarily convex. It is convex, if it is either coninious or picewise monotonic; the question is wether Lebesgue-measurability is enough or not.

Consider Hammel basis in $\mathbb{R}$ over $\mathbb{Q}$. Define $f(x)$ as the sum of coefficients of basis's elements in the representation of $x$. It is clear $2\cdot f\left(\frac{x+y}{2}\right)=f(x)+f(y)$. And $f(x)$ is not a convex function.

Note that Myth's example does not directly answer Fedor Petrov's question, since the function in his example is not measurable.

Do you mean Fedor asked us whether Lebesgue-measurable function with property $2f((x+y)/2)\leq f(x)+f(y)$ is convex?

Kent!
Do you have more constructive remark?  ;)

Finally, I positively resolved Fedor's question.

Consider two points $a$ and $b$, $b>a$. Let $I=(a,b)$, $l(x)=f(a)+(x-a)\frac{f(b)-f(a)}{b-a})$. We will prove that $X=\{x\in I\,|\,f(x)\leq l(x)\}=I$ (it means $f(x)$ is convex function).

[b]Remark.[/b] It is clear that $\frac{x+y}{2}\in X$ for all $x,y,\in X$.

[b]Case 1.[/b] $|X|=b-a$. Suppose there is $x\in I$ s.t. $x\not\in X$. Consider  neighborhood $(x-\epsilon,x+\epsilon)\subset I$. Due to remark we have $x+\delta\in X$ implies $x-\delta\not\in X$, $|\delta|<\epsilon$. It follows that $|X\cap (x-\epsilon,x+\epsilon)|\leq \epsilon$ $\Rightarrow$ $|X|\leq b-a-\epsilon$ -- contradiction.

[b]Case 2.[/b] $\mes X<b-a$. In this case there are exists $\epsilon>0$ s.t. $|Y|>0$, where $Y=\{y\in I\,|\,f(y)>l(y)+\epsilon\}.$
Let $c=\frac{a+b}{2}$. We have $|Y\cap (a,c)|\geq |Y|/2$ OR $|Y\cap (c,b)|\geq |Y|/2$.
Suppose WLOG $|Y\cap (a,c)|\geq |Y|/2$. Consider set $Z=2\cdot(Y\cap (a,c))-a\subset I.$ It is easy to show that $f(z)>l(z)+2\epsilon$ for all $z\in Z$, since $f(z)+f(a)\geq 2f(y)$, where $y=\frac{z+a}{2}\in Y$. Note that $|Z|\geq |Y|$ !!! So we can repeat our construction for set $Z$ instead of $Y$ and $2\epsilon$ instead of $\epsilon$, and so on. It follows that $|\{t\in I\,|\,f(t)>l(t)+k\}|\geq |Y|$ for all positive $k$ -- contradiction.

I have just corrected text, since I was sure that there is LaTeX command \mes!


[quote="tdl"]Let $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Find the maximum and minimum value of: $ |a^3 \plus{} b^3 \plus{} c^3 \minus{} abc|$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179316
The minimum equal $ 0.$ ( $ c\equal{}0$ and $ a\equal{}\minus{}b.$ )

Let $ a,b,c\ge0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}1$.Find the maximum and minimum value of:
$ P\equal{}|a\plus{}b\plus{}c\minus{}abc|$


quadratics
arithmetic sequence
number theory
Diophantine equation

Find three different natural numbers in an arithmetic progression whose product is a perfect square.

1,25,49

Just think of easiest solution possible, and try to make it work :P

Suppose we have $ a\minus{}b,a,a\plus{}b$

Their product is $ P\equal{}(a\minus{}b)(a)(a\plus{}b)$

Multiply by each term by $ P$ and you still have an arithmetic progression whose product is $ P^4$

What if they ask for [i]all[/i] solutions?

[hide="Soltion if none in the sequence is a perfect square(may be not)"]Let's fix a number that is not a perfect square to make the sequence.Prime factorize it and make a set with the primes that make this set.then put those primes into two disjoint sets.You can include any two distict squares in those sets.You'll get all sequences with that number in it.(I think)
Oops you have to choose three numbers that create a arithmetyic sequence.
[/hide]

Ok make a diophantine equation with the product of the primes to solve for integer values that needs to be multiplied with those multiplied number such that they create an arithmetic progression.This is easy.I'll show it later,may be tomorrow.

Well, this is stupid but here goes
[hide="Stupid solution"]let A & B be the number found by multiplying the elements of the subsets and A<B.And C be the fixed one.
then the diophantines will be
                                       B*x^2=C+|C-A*y^2|
                                               or
                                      B*x=C+|C-A*x|
Then A*y^2,C,B*x^2 are the wanted progression.[/hide]

I do not understand your solution. How does this solve the problem? We don't know anything about the solvability of that equation...

Well I don't know what conditions a quadratic diophantine equation of this kind needs.But if this equations have no solutions then there's no progression with that number C.I did say that this was stupid. :D


Could someone try to explain cyclic expressions to me?  I know a little bit about them but I want to learn more.

Btw,  here is a problem I recently did that relates to them:

Find all real triples (x, y, z) such that when any one of these numbers is added to the product of the other two, the result is 2

I know the answer but I would like to see what other people think about the problem.  I was also wondering about something that I thought about after doing the problem(look in spoiler):

They are cyclic and all equal 2.  After finding the answers I thought that maybe the fact that they are cyclic and equal to the same thing would imply that the only way the equation could be fulfilled is if all three variables equal the same thing.  Would this general method always work, why or why not?

A function $f\left( x_1, x_2, ..., x_n\right)$ of n variables is called cyclic if we have $f\left( x_1, x_2, ..., x_n\right) = f\left( x_2, x_3, ..., x_n, x_1\right)$ for every n-tuple $\left( x_1;\;x_2;\;x_3;\;...;\;x_n\right)$ (of course, if this n-tuple belongs to the set on which the function f is defined).

[quote="joml88"]Find all real triples (x, y, z) such that when any one of these numbers is added to the product of the other two, the result is 2[/quote]

In fact, this is a quite easy problem. You know that

  yz + x = 2;
  zx + y = 2;
  xy + z = 2.

Hence, yz + x = zx + y, and thus

  0 = (yz + x) - (zx + y) = z (y - x) + (x - y) = -z (x - y) + (x - y) = (1 - z) (x - y).

This leaves you two possibilities: either 1 - z = 0, or x - y = 0.

If 1 - z = 0, then z = 1, and you are left with y + x = 2, x + y = 2 and xy + 1 = 2. In other words, x + y = 2 and xy = 1. From x + y = 2, you infer y = 2 - x, and now, substituting into xy = 1, this gives you the equation x (2 - x) = 1, which is a quadratic equation for x. This equation has only one solution, namely x = 1. Thus, y = 2 - x = 2 - 1 = 1. Hence, you have found only one solution: (x, y, z) = (1, 1, 1).

Now consider the case x - y = 0. Then x = y. Now, just the same way as you obtained 0 = (1 - z) (x - y), you can find 0 = (1 - x) (y - z). Hence, either 1 - x = 0 or y - z = 0. In the case 1 - x = 0, you have x = 1, hence y = x = 1, and by the equation xy + z = 2 you get z = 1, too, so that again you arrive at the only solution triple (x, y, z) = (1, 1, 1). In the case y - z = 0, you have y = z, what, combined with x = y, yields x = y = z. Now, you can rewrite your initial equation xy + z = 2 as $x^2 + x = 2$, what is a quadratic equation again. This quadratic equation has the two solutions x = 1 and x = -2. Hence you get the two solution triples (x, y, z) = (1, 1, 1) and (x, y, z) = (-2, -2, -2).

By the above, we see that the only solutions to your problem are (x, y, z) = (1, 1, 1) and (x, y, z) = (-2, -2, -2). (You can easily check that these two triples do actually satisfy the three equations!)

[quote="joml88"]maybe the fact that they are cyclic and equal to the same thing would imply that the only way the equation could be fulfilled is if all three variables equal the same thing.[/quote]

Hmm, I don't know whether this general rule is true. Actually I wonder why you put a spoiler around it - it's a question you ask, not a solution to a problem you propose, so you don't need to hide it!

  Darij

And on the same note, is a cyclic expression with a cyclic constraint always maximized when all variables are equal?


1. How many lattice points must you have to be certain that 3 of them form a trianlge with an integer area.

2. Find a general formula for n dimensions (n>1)

Then they would have area $0$ which is an integer.

[quote="beta9"]I don't think that any number of lattice points would guarentee such a triangle.  Couldn't all the points just be co-linear.[/quote]

Having 3 collinear points actually does give a triangle, but since it's a line it's called a degenerate triangle. Its area is 0 since it has an altitude of length 0.

Yes 3 collinear points count as a triangle of integer area.
[hide="hint"]
Think about the parity of the coordinates
[/hide]


i think i've read it, how long was it? just check :)

[quote="Rep123max"]Has anybody read China Boy by Gus Lee? Did you like it? I have to read it for summer reading and just wanted to know what people thought about it. Also, has anybody read the Source by Michner (some spelling like that)?[/quote]
What's this?? :?

It's a book.

Any you are reviving dead threads  :noo:


for the first conversion,

1.sulphonate to get -SO3H group at the para position
2.treat with CH3Cl/AlCl3 to get the methylated group at the o-position
3.heat with superheated water to remove the sulphonyl group.

Is my conversion right :wink: ????????????????????????

wont your first step yield multiple results?
And for the [b] $ n^n \plus{} 1$ th time [/b]  :D
[b] [size=200]
YOU CANNOT ALKYLATE A RING CONTAINING NH2 GROUP   :P

[/size][/b]

hey madness............
NH2 is not a deactivating group only NO2

cause AlCl3 will coordinate will NH2 and also it might form NH(CH3)

Yes, coordination of nitrogen with the Lewis acid transform the-NRR' group into a powerfull deactivating group.

[quote="madness"][quote="srinitrs"][quote="Carcul"]How to make the following conversions?

1) Aniline ----> o-toluidine

2) Phenol ----> o-cresol

3) p-Cresol ----> p-toluidine

3) R-COOH ----> 2-R-oxyrane*

4) 2-R-oxyrane ----> R-COOH*

* For example, conversion of benzoic acid to phenyloxyrane, and vice-versa.[/quote]
for the second conversion
acylate the OH group using CH3COCl and you will get C6H5OCOCH3 now add AlCl3/CS2 and do fries rearrangement and you will get C6H4(OH)(COCH3) now add tosyl chloride and protect the OH group and then do wolf kischner reduction to get an ethyl group instead of the acetyl one and then oxidise it to COOH and then add P/HI and then reduce it to CH3  and remove the tosyl goup using H+??? am i right??? :D  :D[/quote]

I think Carcul has already made it twice clear that P/HI is not a method used in synthesis  :D[/quote]
NO PROBLEM THEN YOU CAN ALWAYS CONVERT THE ACID TO KETONE AND DO A WOLF KISCHNER REDUCTION???NOW IS IT RIGHT???

[quote="Srinitrs"]FOR CONVERSION 1
add NaNO2/HCl at O0c and then treat the diazonium salt with Pb(NO3)2 and then convert the benzene using friedel crafts alkylation to toluene and then protect the para position by sulphonating it and then nitrate the ring and the reduce usin Sn/HCl ad remove SO3h using a base am i right???[/quote]

Yes, that's right, although I have never heard of treating a diazonium salt with Pb(NO3)2.

This may work:

3. Start off by protecting the other functional groups on the R-chain, I guess. Then, reduce the carboxylic acid with LiAlH_4 to the alcohol. Tosylate with TsCl and substitute with Br. Add Mg, attack on formaldehyde, and acid quench to get an alcohol with one more carbon (the higher homologue?). Tosylate and eliminate with a bulky base, such as LDA, and let the double bond attack a peroxyacid, such as mCPBA, to form the epoxide (oxirane).

4. Open the epoxide S_N2 with iodide, and then eliminate the iodide with LDA. Hydrogenate the double bond with Lindlar's catalyst, tosylate the hydroxyl, and eliminate that with LDA. Apply ozonolysis with a reductive workup of dimethyl sulfide to give the alcohol version of what we want and formaldehyde. Distill out the formaldehyde (I think it should have a lower boiling point) and oxidize the remaining alcohol to the carboxylic acid.

[quote="MysticTerminator"]3. Start off by protecting the (...) as LDA, and let the double bond attack a peroxyacid, such as mCPBA, to form the epoxide (oxirane).[/quote]

That may work, but its unnecessarily long.

[quote="MysticTerminator"]4. Open the epoxide S_N2 with iodide, and then eliminate the iodide with LDA.[/quote]

What happened to the -OH group?

er, I don't think the OH would do anything at that stage? I mean, I don't think it would leave since it's not a good leaving group unless it's protonated, right?

Right, but how would you eliminate iodide having your molecule an acidic -OH group? Am I missing something?

[quote]MysticTerminator wrote:
4. Open the epoxide S_N2 with iodide, and then eliminate the iodide with LDA.[/quote]
yeah, you would have deprotonation followed by an intramolecular SN2 reaction that would occur and bring you back to the epoxide, rather than elimination.

for conversion 4:

1. reduce the epoxide with LiAlH4 to get the secondary alkoxide
2. oxidize with CrO3 in H2SO4/H2O to get the methylketone
3. react with 3 equivalents of Br2 in NaOH (haloform reaction) and follow with acidic workup to get R-COOH

and for conversion 3:

1. treat the carboxylic acid with 2 equivalents of methyl lithium and follow with acidic workup to get the methyl ketone
2. treat with 1 equivalent of LDA under kinetic control (-78 degrees C) and add Br2 to get R-C(=O)-CH2-Br
3. reduce with sodium borohydride in ethanol and treat with sodium hydroxide to get the epoxide


whos has the lowest rating?

and if it was a glitch then it doesnt count.

[quote="RollandWu"]well i've gotten 1000 and stayed there for a while...
and i also saw a few people with sub-1100s[/quote]

Well, [b]you bombed[/b]. That doesn't count.

When I first started, I was horrible. Back then, I considered 1140 to be [u]good[/u]. I got a low of 1107 without bombing. But with lots of practice and time, plus AoPS Vol. 1 power, I got a lot better. Now my rating is pretty steady from 1320 - 1340, occasionally soaring to the 1360s and 1370s, and occasionally dipping into the low 1300s and high 1200s.

i saw someone (i think) with 1000-something.

One time i think that donkeykong had a rating of -14 or -16.

Apparently, so.  But the lowest rating in this version is 1000.

[quote="mewto55555"]Lowest ever is ROllandWu and I with 1. Lowest w/o glitch was ROllandWU and Brut3Forc3 at 1000. IDK lowest w/o bombing. My lowest w/o bombing/glitch is 1198.[/quote]

i got 0 once, so ha!!!!  :P

but yeah...it got fixed.. :(

i had a 1060 a couple months before! and it wasnt a glitch and i didnt do it on purpose :P

[quote="mewto55555"]Yehah ROlland and I, if you click on our names, our lowest ratiing shows up as 1  :D[/quote]thats nice....

[quote="almath"]I am at 773 now. I'm going at a fast rate. Because i start a cd. leave it. start another one. then leave it. i keep doing that.[/quote]

that doesn't work

trust me

pi when you play a game you play the whole game. Leaving makes it faster. It works. MAybe thats why it took you a long time to get a low rating and and why i'm going at a much faster rate then you were. the strategy does work. you lose more points faster.

There's no need for this here.  If you want to post updates, post 'em in FF.

[quote="PI-Dimension"]well ding tried to make me gain points once so i left the game started a cd with a guest (cough me cough) and I didn't lose anything so hah I win[/quote]

when you play with a guest, you do not gain or lose rating points

also, if the rating difference is too big, you will not gain or lose rating points either

How large does the rating difference have to be?

About 700~800 points from what I've seen, and around 600 points it gets difficult.

it depends
a person with 2000 rating playing a person with 1200 rating is different from a 1000 playing a 200-quoted from levans on FTW

bottom line: there is no exact number

I'm at 1826, and I gain +1ish from 1300s, so about 500 if you get to 1800.

i tried to leave CDs, but for some reason my rating went up???

there's a duke tipper with a rating of 923. obviously not hacked. did they revise the rating methods??

yes
levans, changed the rating system to fix the bug

I think not a lot of people are bombign anymore. And i am not reviving a thread, it has to be a month old. it isn't yet.

it doesn't have to be a month old
there is no rule that sets the time at a month (not that i have seen anyway)
plz just don't post things in threads that have died out


what are the last three digits of
$2005^{2005}$

[hide="Hints"]Chinese Remainder Theorem and Euler's Totient function[/hide]
A pretty standard problem!

[hide]We can take this $\pmod{125}$ (trivial) and $\pmod{8}$ and then use CRT.

${2005}^{2005}\equiv 5^{2005}\pmod{8}$
and by Euler's Totient theorem,

$5^{2005}\equiv 5 \pmod{8}$

So by CRT, the last three digits are $125$.[/hide]


calculus
derivative
trigonometry
geometry
geometric transformation
LaTeX
limit

ok...so i have an equation which i can't get into a form suitable for differentiation..could anyone help me?

y=(x 2  over 2 :pi:) + 2sinx
and i can't get the 2:pi: to the top of the fraction

oh ...and another thing.. sin4 :pi: over 3= sin( :pi:  (+4 :pi:  over 3))=-sin 7 :pi:  over 3 which then =
- :sqrt: 3/2....but why does it = - :sqrt: 3/2    ?????

thanks alot for anyone who can help!! sorry bout the use of emoticans...im new!

So, let me re-write what I think you meant, and you tell me if it's right:

$y = \frac{x^2}{2\pi}+2\sin x$
and
$\sin{\frac{4\pi}{3}}=\sin{\left(\pi + \frac{4\pi}{3}\right)} =- \sin{\frac{7\pi}{3}} = -\frac{\sqrt{3}}{2}$

Are those the equations you were trying to write?

[quote="Dee!"]ok...so i have an equation which i can't get into a form suitable for differentiation..could anyone help me?

y=(x 2  over 2 :pi:) + 2sinx
and i can't get the 2:pi: to the top of the fraction[/quote]If all you want to find is $\frac{dy}{dx}$ the answer is simply $\frac{x}{\pi}-2\cos{x}$. I'm not sure what the problem is there.[quote]oh ...and another thing.. sin4 :pi: over 3= sin( :pi:  (+4 :pi:  over 3))=-sin 7 :pi:  over 3 which then =
- :sqrt: 3/2....but why does it = - :sqrt: 3/2    ?????[/quote]Here is my attempted translation of what you said:

$\sin\frac{4\pi}{3}=\sin\left(\pi+\frac{4\pi}{3}\right)=-\sin\frac{7\pi}{3}=-\frac{\sqrt{3}}{2}$

You understand everything there except why it is that $-\sin\frac{7\pi}{3}=-\frac{\sqrt{3}}{2}$, or $\sin\frac{7\pi}{3}=\frac{\sqrt{3}}{2}$. The quick answer is to refer to the unit circle. Note that the angle $\frac{7\pi}{3}$ is equal to $60$ degrees, so what you have there is that the sine of 60 degrees is $\frac{\sqrt{3}}{2}$. Let us know if that doesn't satisfy you.

yea joel...thats wut i was trying to say...how did you get it to look like tht???
oh and antonio how does
 x 2 / 2 :pi: +2sinx
=
x 2 / :pi: -2sin x...or wutever u said

and how do u figure that  7 :pi:/ 3 =60 degrees...? cuz i got it to equal 420 degrees

btw..told u i was stupid

[quote="Dee!"]
oh and antonio how does
 x 2 / 2 :pi: +2sinx
=
x 2 / :pi: -2sin x...or wutever u said
[/quote]

It doesn't equal that. Nor differentiate to that. But anyway, if you want to differentiate $x^2$, you get $2x$. If you want to differentiate $\sin x$, you get $\cos x$. In both cases you just have a constant multiplied on which stays there - so $\frac{1}{2\pi} x^2$ differentiates to $\frac{1}{2\pi}\cdot 2x$, and $2\sin x$ differentiates to $2\cos x$.

[quote="Dee!"]
oh and antonio how does
 x 2 / 2 :pi: +2sinx
=
x 2 / :pi: -2sin x...or wutever u said[/quote]What I said was that $\frac{d}{dx}\left(\frac{x^2}{2\pi}+2\sin{x}\right)=\frac{x}{\pi}+2\cos{x}$ (I actually had [i]minus[/i] $2\cos{x}$ but that was just a silly mistake). If you want to know why that is here's the "long solution"

First, $\frac{d}{dx}\left(f(x)+g(x)\right)=\frac{d}{dx}f(x)+\frac{d}{dx}g(x),$ so we can say that $\frac{d}{dx}\left(\frac{x^2}{2\pi}+2\sin{x}\right)=\frac{d}{dx}\frac{x^2}{2\pi}+\frac{d}{dx}2\sin{x}.$

Next, $\frac{d}{dx}kf(x)=k\frac{d}{dx}f(x),$ so we have $\frac{d}{dx}\frac{x^2}{2\pi}+\frac{d}{dx}2\sin{x}=\frac{1}{2\pi}\frac{d}{dx}{x^2}+2\frac{d}{dx}\sin{x}.$

Finally we use the power rule, $\frac{d}{dx}{x^n}=nx^{n-1}$ and the fact that $\frac{d}{dx}\sin{x}=\cos{x}$ to see that $\frac{1}{2\pi}\frac{d}{dx}{x^2}+2\frac{d}{dx}\sin{x}=\frac{1}{2\pi}2x+2\cos{x}=\frac{x}{\pi}+2\cos{x}.$

I hope that helps.

[quote]and how do u figure that  7 :pi:/ 3 =60 degrees...? cuz i got it to equal 420 degrees[/quote]The angle $\theta$ is always equivalent to the angle $\theta+2k\pi$ for any integer $k$. This is because $2\pi$ is a full revolution, which leaves the angle unchanged. For example, say you run three quarters of the way around a circular track. Now if you run all the way around twice more (two full revolutions, or $2$ times $2\pi$), you're once again three quarters of the way around the track. Notice that if $\theta=\frac{7\pi}{3}$ then $\theta-2\pi=\frac{\pi}{3},$ or $60$ degrees.

\frac{a}{b} becomes $\frac{a}{b}$.  (This forum uses Tex -- if you know how to use the syntax, you surround your math text in dollar signs [code]$ TEX HERE $[/code] and it gets beautifully formatted.

[quote="Dee!"]goodness me u guys are confusing...twenty lines explanation which could have been replaced with one...only joking..thanks alot for putting in the effort...i finally understand..but i still don't know how to put the over sign in....i have been told im a fast learning...wut were they thinking![/quote]I thought that might be a problem; I'm glad you understand it now though.

As for the "over sign" and the math formatting we use in the forums, you can go to http://www.artofproblemsolving.com and click on LaTeX to learn the syntax. Here are the basics:

(on the left is the text without dollar signs around it, on the right is with the dollar signs)

a+b $a+b$
a^b $a^b$
a^bc $a^bc$
a^{bc} $a^{bc}$
\frac{a}{b} $\frac{a}{b}$
\sin{x} $\sin{x}$
\sqrt{x} $\sqrt{x}$
a_1,a_2^2,a_3^3,\ldots,a_n^n,a_{n+1}^{n+1},\ldots $a_1,a_2^2,a_3^3,\ldots,a_n^n,a_{n+1}^{n+1},\ldots$
\{1,2,3,\ldots\} $\{1,2,3,\ldots\}$
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$
\sum_{k=0}^n\frac{1}{k!} $\sum_{k=0}^n\frac{1}{k!}$
N=\frac{N}{2}+\frac{N}{4}+\frac{N}{8}+\frac{N}{16}+\cdots $N=\frac{N}{2}+\frac{N}{4}+\frac{N}{8}+\frac{N}{16}+\cdots$
f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)(x-a)^n}{n!} $f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)(x-a)^n}{n!}$

..and so forth.


[quote="rateesquad"]Hey,

I get this sort of a problem studying for GMAT. Sad, I know.

150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?

Set up a venn diagram with the following:
(all three) = 3
(basketball and baseball) = x
(basketball and soccer) = y
(soccer and baseball) = z

So $ x+y+z = 27$

(none) = n

$ 63-x-z+42-x-y+39-y-z+x+y+z+3+n= 150$
$ 147-x-y-z+n= 150$
$ 147-(x+y+z)+n = 150$
$ 147-(27)+n = 150$
$ n = 30$

Is this right or should it be 60?[/quote]

$ x+y+z$ is NOT equal to 27. Be careful, you overcounted. By adding $ x$, $ y$, and $ z$ together, you counted the center of the Venn Diagram (all three sports) and extra two times. So you have to subtract $ 2 \cdot 3$ for overcounting.

[quote="i_like_pie"][hide="PIE"]$ 66+45+42-27+3+n=150$

$ n=\boxed{21}$[/hide][/quote]
i think it should be $ ....-27-6$ instead of $ ......-27+3$ because they play three sports and your counting them 3 times instead of once becase they count for people who play all three so you have to subtract 2 times that. so i think $ n=28$. btw, whats up with the penguin avatars

I'm pretty sure that abacadaea is right and the answer is actually $ 153-27-6=150-n \to  n=30$

[hide="I think"] Let the ones who play baseball+basketball be x, basketball+soccer be y, and baseball+soccer be z. n=people who play none
Then, 150=66-x-y-3+45-y-z-3+42-x-z-3+x+y+z+3+n
or $ n=30$

BTW, x+y+z is 27 because it says 27 play [b]exactly [/b]2 sports[/hide]

Using variables makes it more complicated than it needs to be.

Too bad. I know inclusion-exlusion works too, I just did it that way to make sure.


Find all functions $ f \colon \mathbb{Z} \mapsto \mathbb{R}$ which satisfy:
i) $ f(x)f(y) \equal{} f(x \plus{} y) \plus{} f(x \minus{} y)$ for all integers $ x$, $ y$
ii) $ f(0) \neq 0$
iii) $ f(1) \equal{} \frac {5}{2}$

[quote="Brut3Forc3"][hide="Solution"]Let $ x \equal{} y \equal{} 0$. We have $ f(0)^2 \equal{} 2f(0) \Rightarrow f(0) \equal{} 2$, since $ f(0)\ne0$. Now plug in $ \minus{} y$ for $ y$, giving $ f(x)f( \minus{} y) \equal{} f(x \minus{} y) \plus{} f(x \plus{} y) \equal{} f(x)f(y) \Rightarrow f(y) \equal{} f( \minus{} y)$. Thus, we only need to consider $ f(x)$ for nonnegative integer $ x$. Now let $ y \equal{} 1$, giving $ \frac {5}{2}f(x) \equal{} f(x \plus{} 1) \plus{} f(x \minus{} 1) \iff f(x \plus{} 1) \equal{} \frac {5}{2}f(x) \minus{} f(x \minus{} 1)$. Since we have fixed values for $ f(0), f(1)$, and the function value for any consecutive two integers uniquely determines the next, we know that the function must be unique. Letting $ f(n) \equal{} a_n$ (since $ n\in\mathbb{N}_0$), we have $ a_{n \plus{} 1} \equal{} \frac {5}{2}a_n \minus{} a_{n \minus{} 1}$, and solving for this linear recurrence gives $ f(x) \equal{} 2^x \plus{} 2^{ \minus{} x}$.[/hide][/quote]

Awesome solution. Thanks


calculus
integration
trigonometry
logarithms
complex analysis
real analysis
real analysis unsolved

$ (i)$ is a famous sum first showed by ramanujan to hardy. i know how to show it, so i won't spoil it for others who may be able to as well. however,  i haven't encountered $ (ii)$ before, so i will try and solve that now.

OK, i have had received numerous PMs and e-mails asking for the solution to the first question, and instead of me giving away the answer (which by the way, is not at all hard to arrive at), i will give  hints and hopefully, you will be able to work your way. if not,  i will post the full solution.

first, notice that a generalization can be done,  which states

$ \sum_{n\equal{}1}^{\infty}{\frac{n^{4p\plus{}1}}{e^{2\pi\, n}\minus{}1}}\equal{}\frac{\textbf B_{4p\plus{}2}}{4(2p\plus{}1)}\qquad\qquad \ldots  (1)$

( where $ \textbf B_m$ is the $ m^{th}$ bernoulli number)

 to prove $ (1)$, recall  ( for example, via complex analysis),

$ \int_0^\infty \;\frac{x^{2p\minus{}1}}{e^x\minus{}1}\;\textbf dx\;\equal{}\;\Gamma(2p)\cdot\zeta(2p)\;\equal{}\;\Gamma(2p)\;\frac{(\minus{}1)^{p\plus{}1}\;\textbf B_{2p}\;(2\pi)^{2p}}{2\,(2p)!}$

$ \Rightarrow \qquad \int_0^\infty \;\frac{x^{4p\plus{}1}}{e^{2\pi\,x}\minus{}1}\;\textbf dx\;\equal{}\;\frac{\textbf B_{4p\plus{}2}}{4(2p\plus{}1)}$

now look at the integral and look at the summation on the LHS in $ (1)$..... can you take it from here ?  :) some rigorous  argument will be necessary to prove  the [i]equivalence[/i] of the integral and the  summation.

finally, put $ p\equal{}3\quad\Rightarrow\quad \textbf B_{14}\equal{}\frac{7}{6}$ to get what we want.

[ i have no clue how ramanujan solved it, but i doubt it would have been the same way.  :maybe: he had  peculiar ways of solving sums and integrals, so i have always been curious as to how he solved this one, as was hardy when he first received the letter from ramanujan that stated  the formula. ]

[color=green]Long unnecessary quote removed by moderator[/color]

Thank very much.

Misan, can you please give us the proof for the equivalence of the integral and the summation.

And can someone prove the following three series?

$ (1) \quad \sum_{k\in\Bbb{Z}} \frac {1}{\cosh k\pi } \equal{} \frac {1}{\sqrt {2\pi}}\, \frac {\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}$

$ (2) \quad \sum_{k \equal{} 1}^\infty \frac {( \minus{} 1)^k}{k\sinh k\pi} \equal{} \frac {\pi}{12} \minus{} \frac {\ln 2}{2}$

$ (3) \quad \sum_{k \equal{} 1}^\infty \frac {( \minus{} 1)^k}{k^3\sinh k\pi } \equal{} \minus{} \frac {\pi^3}{360}$

Yes misan we can easily show that $ \int_{0}^{\infty}{\frac {x^p}{e^x \minus{} 1}dx} \equal{} \Gamma(p \plus{} 1)\zeta (p \plus{} 1)$ even without complex analys:

$ \frac {x^p}{e^x \minus{} 1} \equal{} \frac {x^p e^{ \minus{} x}}{1 \minus{} e^{ \minus{} x}} \equal{} x^p e^{ \minus{} x} \sum_{k \equal{} 0}^{\infty}{e^{ \minus{} kx}} \equal{} x^p\sum_{k \equal{} 1}^{\infty}{e^{ \minus{} kx}}$

And now $ \int_{0}^{\infty}{\frac {x^p}{e^x \minus{} 1}dx} \equal{} \sum_{k \equal{} 1}^{\infty}{\int_{0}^{\infty}{x^p e^{ \minus{} kx}dx}} \equal{} \Gamma (p \plus{} 1)\sum_{k \equal{} 1}^{\infty}{\frac {1}{k^{p \plus{} 1}}} \equal{} \Gamma (p \plus{} 1) \zeta (p \plus{} 1)$

Alsow $ \int_{0}^{\infty}{\frac {x^p}{e^{2\pi x} \minus{} 1}dx} \equal{} \frac {1}{(2\pi) ^{p \plus{} 1}}\Gamma (p \plus{} 1)\zeta (p \plus{} 1)$ and if $ p$ is odd then

$ \int_{0}^{\infty}{\frac {x^{2r \minus{} 1}}{e^{2\pi x} \minus{} 1}dx} \equal{} \frac {1}{(2\pi) ^{2r}}\Gamma (2r)\cdot \frac {1}{2}|B_{2r}|\frac {(2\pi)^{2r}}{(2r)!} \equal{} \frac {|B_{2r}|}{4r}$

For example $ \int_{0}^{\infty}{\frac {x}{e^{2\pi x} \minus{} 1}dx} \equal{} \frac {1/6}{4\cdot 1} \equal{} \frac {1}{24} \equal{} \frac {7/6}{4\cdot 7} \equal{} \int_{0}^{\infty}{\frac {x^{13}}{e^{2\pi x} \minus{} 1}dx}$. Interesting fact :)

As I can see using computer, your generalization is true, if $ p \equal{} 4r \plus{} 1\geq 5$. Only in this case $ \int_{0}^{\infty}{\frac {x^{p}}{e^{2\pi x} \minus{} 1}dx} \equal{} \sum_{k \equal{} 1}^{\infty}{\frac {k^{p}}{e^{2\pi k} \minus{} 1}}$

Hm... I think that equivalence between sum and integral maybe we can get from Euler's summation fromula.


Sam is moving to a colder climate.  Right now his car's 20 quart radiator contains a mixture of 20% antifreeze and 80% water.  At his new location, he will need the mixture to be 50% antifreeze.  How many quarts of his current mixture does he need to change for pure antifreeze to rezch the required 50% concentration?

[hide]He wants 10 and 10 his current is 16 and 4 so the answer must be 6.[/hide]

Oooh oops duh!
[hide]He needs to take out 6 quarts of water total and every quart is 4/5 water so he must take out 7.5 quarts of the entire thing. Then he can replace that with 7.5 of pure antifreeze for a 50-50 mixture.[/hide]

I also fell for the same trap Tarquin did..this was a question on an exam I took, and I got it wrong.... :(

I'm lucky that I didn't guess on that question, because I would've gotten it wrong, but the answer is

[hide="The answer"] 7.5 [/hide]

yeah... first i also made the same mistake as tarquin, and suddenly im like ahhh! you can't remove only one ingredient of the mixture! My science teacher wouldn't have been very happy with me if she saw me there....

All the questions I got wrong on that test was because I wasn't sure of.  If I had not answered all the ones I wasn't sure of I would've increased my score my 5 points...

I got a 71 and you got a 75, Secret Asian.

[quote="mathgeek2006"]I got 6, then couldn't figure it out(It was a multiple choice test).  There was a choice, "none of these", but I wasn't willing to take the risk.  I feel very lucky.[/quote]

Sorry if i get a little personal, but i was just wondering which test did you guys take?

The Utah State Math Exam.  It's for Utah students only.

Oooooh, Utah students only....well, aren't [i]you[/i] special! Get to take your Utah students' [i]exam[/i]...

Naw, I'm just kidding. Don't lynch me.

I've searched for a bunch of State Math Exams(For other states), and it appears that many have them.  But on Google, the two that come up most are Utah and North Carolina.  It's usually one of the main colleges in the state that hosts it.

Well, most/all states have some sort of government-run proficiency test for public schools, right?

Um..the Utah State Math Exam isn't a proficeintcy test.  It's really a competition.  I took 3rd and Mathgeek took 6th.

Oh - sorry. That was a little unclear to me. My bad.

Probably since it's an exam, it seems unclear.  If it were called Utah State Math Contest, it would probably be more clear.

Is the Utah test hard? I live in Washington, and I think that our state math tests are fairly easy. So far, the major ones that we have are Math Is Cool and the Washington State Math Championship. I took both this year a grade up and still they seemed easy.

If you don't mind, could you please show a question from the test? Just curious :lol:

thx

Wasn't the thread started on a question that came from the exam? That would be one question.

ohh.... i never thought of that  :blush:  :P
That would be around a medium difficulty problem on the average of all 8th grade tests (see how easy they are now?). However, this years test was really easy compared to even the other years. For me, the hardest problem was:

If the largest possible circle is cut from an equilateral triangle, to the nearest whole percent how much of the triangle is part of the circle?

a. 69% b. 57% c. 66% d. 63% e. 60%

FYI: we have thirty minutes to do thirty questions here

I could just tell that there were a million tiebreaks here because the first 10 q's are counted towards our team score and the team member's score (but not name) is shown on the little spreadsheet we get after the competition.

Even on the teams as low as 27th place, I could still see 10s

Lol. Good job! You took sixth place in the 8th grade division at the Utah. That's especially awesome since you go to Churchilljrhigh! Remember MG, me, and O Master would kick even more (@ $\$ $) as sophomores or froshes.

Remember, if you want the link for the USME, just go to my blog.

are claculators allowed on the Utah State Math Exam?? How about the Washington contests?

There are no calculators allowed on the USME (That crosses off my brain) ;)


inequalities
algebra
polynomial
number theory proposed
number theory

Let $ p$ be a prime and let $ k \ge 2$ be a positive integer. Suppose that $ q$ is the smallest prime that divides $ k$. Prove that $ p^k \minus{} p^{\frac {k}{q}} \ge \phi(p^k \minus{} 1)$.

Let $ k/q\equal{}m$, then $ (p^m\minus{}1,\frac{p^k\minus{}1}{p^m\minus{}1})\equal{}(p^m\minus{}1,q)$.
If $ (p^m\minus{}1,q)\equal{}1$ then $ \phi (p^k\minus{}1)\equal{}\phi(p^m\minus{}1)\phi (\frac{p^k\minus{}1}{p^m\minus{}1})\le (p^m\minus{}2)\frac{p^k\minus{}p^m}{p^m\minus{}1}\equal{}$
$ \equal{}p^k\minus{}p^{(q\minus{}1)m}\minus{}...\minus{}p^{2m}\minus{}2p^m<p^k(1\minus{}p^{\minus{}m})<p^k\minus{}p^m.$
If $ q|p^m\minus{}1$, then $ \phi(p^k\minus{}1)<(q\minus{}1)/q (p^k\minus{}p^{(q\minus{}1)m}\minus{}...\minus{}2p^m<p^k(1\minus{}p^{\minus{}m})$.

Nice job  :lol:  There is a very elegant but non-elementary solution to this problem, and I was curious if someone could find an elementary one. If someone would like I will post this.

EDIT: Nice job, t0rajir0u. You found exactly the solution I was thinking of.

$ \varphi(p^k \minus{} 1)$ is the number of primitive roots in $ \mathbb{F}_{p^k}$, so equivalently what we want to prove is that at least $ p^{ \frac{k}{q} }$ elements of $ \mathbb{F}_{p^k}$ are not primitive roots.  But this is true of all solutions to the polynomial $ x^{ p^{ \frac{k}{q} } } \equal{} x$.


hi
i think its cool that those math jams are free for every community member. and i'd really like to just try it out. but there is a problem. time. i am from germany and over there is german time. i looked at the schedule and at the moment all of them are starting either 2 am or 5 am my time which is both not the best time to do math, when u ask me. are there gonna be any math jams so they can be used internationally? maybe around 1pm ET?
peeta

The problem with having them at the time that you suggest is that our core audience -- North American middle and high school students -- is in school at that time.  We've picked 7:30 p.m. Eastern / 4:30 p.m. Pacific as the default time for classes and Math Jams since it seems to best suit our US and Canada students (not too late for students on the East Coast, not too early for students on the West Coast).  Unfortunately, we realize that this time is not so great for our European members.

Fortunately, all of our Math Jams are [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]transcripted[/url], so even if you can't attend a Math Jam live, you can still read the transcript.

and whats about weekends? and i know some schools already have summer break. (we still have like 2 months till break here) wouldnt that enable the main audience to join math jams around midday already?
peeta

ps: thx for the tip about the transcripts, im gonna take a look into them

On weekends we teach the San Diego Math Circle, some classes and Independent Study Office Hours and we aren't particularly interested in adding to our schedule in that regard.  Hopefully as time goes on we will come up with creative ideas to leverage our capacity such that Europeans and others are better able to gain full benefit from the site and our educational materials.


How many ways are there to arrange 6 people around a circular table with 7 seats?

Fix the empty chair.  Then, there are $ 6\cdot 5\cdot 4 \cdot 3\cdot 2 \cdot 1 \equal{} \boxed{720}$ ways to arrange the people around the table

[quote=pifinity][font=TImes new roman][hide=Even more simple]Having an empty seat means that each person has an extra seat to choose from. Since at the end, there is only one place for the empty seat, this problem is the same as just arranging 7 people into a round table with 7 seats. Therefore, the answer is $\boxed{\tfrac{7!}{7}=720}.$ $$\definecolor{color}{rgb}{0.7, 0.9, 0.4} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$[/quote]

Wait, is it just me or does the "A pifinity solution" at the end get easier and easier to see as you stare longer and longer at it (the green rectangle starts appearing darker)? Then when you look at something else, like opening a new tab, staring at your screen, then looking back at the green rectangle, the text becomes blurry again? The pixels aren't changing either, you can check by taking a screenshot and looking at that. It can't be just me?

You are correct! I believe it's a phenomenon called "Retinal Fatigue."

Wait nevermind, I realized it was because my display was changing brightnesses. Thanks for fooling me, Windows 10!

With an extra seat is just like having the 7 people so  $\frac{7!}{7} = \boxed 6! $

[hide=sol]Evidently there will be an empty seat. To avoid having to account fro rotations, we say the head of the table is an empty seat. So, we have $6!=\boxed{720}$ ways.[/hide]

[hide=HowToSeeIt]Simply treat the empty seat as a person, and now you have 7 people to sit around 7 seats.
This method will not work for other problems (Well, unless your seating $x-1$ people around a table meant for $x$).


The point $ A(\minus{}7, 4)$ is reflected across the $ x$-axis onto point $ B$.
Point $ B$ is reflected over the $ y$-axis onto point $ C$. What is the
sum of the coordinates of point $ C$?

When the coordinate is reflected across the $ x$-axis, the $ y$-coordinate is changed to it's additive inverse.  So point $ B$ is $ (\minus{}7,\minus{}4)$.  Same goes for when it is reflected over the $ y$-axis, but instead, the $ x$-coordinate is changed.  So point $ C$ is $ (7, \minus{}4)$.  So our answer is $ 7\plus{}\minus{}4\equal{}\boxed {3}$


Tons [url=http://www.agmath.com/media//DIR_57727/NumberTheory.pdf]here[/url].  Sorry, no solutions... but this is a good starting point.

Hmm..

For something more (quite a bit more) challenging, you could try PEN (which is in the olympiad section of this forum)

[quote="boohsunil"]can anyone give me problems on number theory?[/quote]

Intro to Number Theory is a great book for number theory problems. It talks a lot about strategies how to solve complex number theory problems and knowledge and imagination to go beyond depts.
AOPs vol.1 does well, but does not have that much material devoted to just number theory.
I have not taken the intro. to number theory class, but it depends if you're struggling, then do it, otherwise just focus on the book, like I did.

Preparing for AMC 10/12 for number theory questions: Use the book [i]First steps for Math Olympians[/i] or practice old tests.

Be careful about trying to memorize too many theorems. You will do better if you learn by doing problems, and pick up the relevant theorems as you go.

Maybe you can go for books like Challenge and thrill of Pre college mathemeatics ,
104 problems in Number theory , Number theory -structures,examples ,problems.

[quote="Vithulan MV"]Hey
I need Thearoms from basic to higher level.......... Where should I get them?  :blush:[/quote]

Theroms? Well, don't find theroms, they will be confusing, if you don't understand them, derive them, and then apply them. If you want theroms, just go through the Intro/Intermediate books from AOPs. They have great explanations how to derive those theroms.


LaTeX
number theory unsolved
number theory

find all naturals x,y such that (2^x)-5 = (3^y)

Find all natural numbers $ x$ and $ y$ such that $ 2^{x}\minus{} 5 \equal{} 3^{y}$

thank you but before that try to solve the problem (if you can)!!!

I'm new to using this LaTeX thing so I might come back and edit it later, like when I'm not at work, but basically:

3^y = (2^x) - 5 ; take the ln of both sides

y*ln(3) = ln( (2^x) - 5 ) ; divide by ln(3)

y = ln ( (2^x) - 5 ) / ln(3) ; rewrite as a log base 3

y = log_3 ( (2^x) - 5 )

From here it's pretty clear that the solutions in natural numbers are anything where (2^x) - 5 is 0 (mod 3), which is equivalent to finding when (2^x) - 2 is 0 (mod 3), so we just look for when 2^x = 2 (mod 3)

2^0 = 1 (mod 3)
2^1 = 2 (mod 3) (but gives a negative answer)
2^2 = 1 (mod 3)
2^3 = 2 (mod 3)
etc., etc., clearly it keeps repeating 1,2,1,2,...

So x = 2^(2k+1) for k>0 works.  And then I suppose you would just say that y is the integer of log_3 (x).

Sorry about the formatting.

also we can say that x and y are odd and we can write if (2^x)-5 = (3^y) then (2^x)-2 = (3^y)+3.

Political Silence's solution is incorrect, because he has only found a necessary condition, not a sufficient one. Does $ 2^7\minus{}5\equal{}3^y$ have any solutions?

Actually, this problem has already been posted. It can be solved by the technique of considering it modulo $ 5440$; see [url=http://www.mathlinks.ro/viewtopic.php?t=271932]here[/url]. :) In general, to solve problems of this form, you must [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=48431]consider[/url] the cycles that the powers form modulo some number $ n$--for a well-chosen $ n$.

now maybe this solution may not appear appealing..
but by graphical approach we can prove that no. of intersections possible is 2..
the graph of $ 2^x \minus{} 5$ shall be negative initially before $ x\geq\ 2.$ something and then positive. So it shall intersect $ 3^y$ at exactly 2 points...
which are obviously $ x\equal{}3,5$


function
limit
logarithms
number theory unsolved
number theory

Maybe this is the right section. I don't know how to prove that $ \lim_{x\to\infty}\frac{\sum_{n\le x}(\theta(n\plus{}h)\minus{}\theta(n))}{hx}\equal{}0$, where $ x,h\in\mathbb{R}$ and $ \theta(x)\equal{}\sum_{p\le x}\log p$, $ p$ primes.

My guess is that is wrong, since we know that $ \theta(x)\equal{}\Theta(x)$, that is, there exist $ c,d\in\mathbb{R}$ such that $ cx\le\theta(x)\le dx$, but I was assured that it's correct :maybe:

[quote="Stewie"]Maybe this is the right section. I don't know how to prove that $ \lim_{x\to\infty}\frac{\sum_{n\le x}(\theta(n\plus{}h)\minus{}\theta(n))}{hx}\equal{} 0$, where $ x,h\in\mathbb{R}$ and $ \theta(x) \equal{}\sum_{p\le x}\log p$, $ p$ primes.

My guess is that is wrong, since we know that $ \theta(x) \equal{}\Theta(x)$, that is, there exist $ c,d\in\mathbb{R}$ such that $ cx\le\theta(x)\le dx$, but I was assured that it's correct :maybe:[/quote]
It is not correct. Let h is natural, then $ \frac{\sum_{n\le x }(\theta(n\plus{}h)\minus{}\theta(n))}{hx}\equal{}\frac{\sum_{i\equal{}1}^{h}(\theta([x]\plus{}i)\minus{}\theta(i))}{hx}\equal{}1\plus{}o(1)$.

[quote="Rust"]It is not correct. Let h is natural, then $ \frac{\sum_{n\le x }(\theta(n\plus{}h)\minus{}\theta(n))}{hx}\equal{}\frac{\sum_{i \equal{} 1}^{h}(\theta([x]\plus{}i)\minus{}\theta(i))}{hx}\equal{} 1\plus{}o(1)$.[/quote]
Hmm, can you explain in detail the last two equalities?

$ \sum_{n\le x }(\theta(n\plus{}h)\minus{}\theta(n \equal{}\sum_{i\equal{}h\plus{}1}^{[x]\plus{}h}\theta(i)\minus{}\sum_{i\equal{}1}^{[n]}\theta(i)\equal{}\sum_{i \equal{} 1}^{h}(\theta([x]\plus{}i)\minus{}\theta(i))$.
Use $ \theta(x)\equal{}x(1\plus{}o(1))$.


Let the $x_i 1 \le i \le n$ be real numbers (not neccessery positive ) such that

$\displaystyle \sum_{i=1}^n x_i^2=1$.

Find the minimum and maxmimum of the following statement:

$\displaystyle  \sum_{1 \le i < j \le n} x_ix_j +\sum_{i=1}^n x_i$

Here, the maximum :

$\sum_{i=1}^n x_i \leq \sqrt{n} * \sqrt{\sum_{i=1}^n {x_i}^2} = \sqrt{n}$ (AM-QM)
$\sum_{1 \leq i \leq j \leq n} x_i*x_j= \frac{(\sum_{i=1}^n x_i)^2-\sum_{i=1}^n {x_i}^2}{2}=\frac{(\sum_{i=1}^n x_i)^2-1}{2} \leq \frac{n-1}{2}$

The maximum is thus  $\sqrt{n} + \frac{n-1}{2}$ and the equality occurs when $x_1=x_2=...=x_n=\sqrt{\frac{1}{n}}$

Nice try for using the means but we have that means are just for the positive numbers .So we can not have this, Can we ?
$\frac{\sum_{i=1}^n x_i}{n} \le \sqrt{\sum_{i=1}^n x_i^2} /\sqrt{n}$ ,
But we can finish it by this:
as you have said :

$\sum_{1 \le i <j \le n} x_ix_j =((\sum_{i=1}^n x_i)^2 -\sum_{i=1}^n x_i^2) /2=
 \frac{(\sum_{i=1}^n x_i)^2 }{2}-\frac{1}{2}$

So we shall just find the minimum and maximim of the :
$\sum_{i=1}^n x_i$ when $\sum_{i=1}^n x_i^2=1$

After Editing:Read the follwoing posts please, this one is wrong.

Really , I dont think so   cause of this trivail  counterexample

$x=y=-1$

$(\sqrt{xy}=1)\ge \frac{x+y}{2}=-2$

[quote="lomos_lupin"]Really , I dont think so ;)  cause of this trivail (indeed) counterexample

$x=y=-1$

$(\sqrt{xy}=1)\ge \frac{x+y}{2}=-2$[/quote]

That's AM-GM... AM-QM or AM-RMS does I think.

paladin8 is correct. AM-QM or AM-RMS holds for all real numbers because

$\text{AM(reals)}\leq\text{AM(positive)}\leq\text{RMS}$

lomos_lupin, your counterexample applies to AM-GM, we are doing AM-QM not AM-GM

Well ,quite right , My counterexample is absoulotly wrong .I missread the words :blush:

and then min should be -1, since if we denote $\sum x_i = x$, then the given thing is equivalent to $(1/2)(x^2+2x-1)=(1/2)(x+1)^2-1$


geometry
probability
calculus
integration
geometry solved

I'm not even sure if geometry is the right area for this, because I don't know how to do it.  I don't think the answer can turn out nice, but I don't know...

It’s raining!  Each raindrop can hit the ground anywhere within a fixed n x n region on a flat plane.  The plane starts out completely dry.  On impact, each raindrop leaves a wet circle of radius r, with the impact point as its center.  After t raindrops have fallen, what is the area of the part of the plane that will end up wet, on average?

Thanks for your help in advance, people!

First than all, you are right, geometry is not the best place to post this problem. It is more a probability kind of problem. You have a random variable and you want to know the expectation value of it.

The true is that I don't have a good idea right now about how to solve it without a sumatory of integrals, and as you posted that here and in the High-School Forum, I suppose you haven't seen it yet. I will think about it, and if I find some nice solution, I will post it.

Usually you shouldn't post a problem twice in the forum. You posted the same problem here and in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=192438]this other thread[/url].


search
function
number theory proposed
number theory

Does there exist a solution $(x,y,z,u,v)$ in integers greater than $1998$ to the equation $x^{2}+y^{2}+z^{2}+u^{2}+v^{2}=xyzuv-65$?


If $ a,b,c$ be natural numbers such that $ gcd(a,b)\equal{}(b,c)\equal{}(c,a)\equal{}1$, then prove that $ 2abc\minus{}ab\minus{}bc\minus{}ca$ is the largest integer which cannot be expressed in the form $ xbc\plus{}yca\plus{}zab$, where $ x,y,z$ are non-negative integers.


1. Prove that the number p^p-1 has a prime factor of the form pk+1, where p is a prime.

2. Prove that there are infinitely many primes of the form 2kp+1 where p is a prime.

3. Prove that there are infinitely many primes ending in a 9.

Don't use Dirichlet's thm. I think I have solns for 1 and 2, but i'm still working on 3.

Isn't #2 (and maybe also #3) incredibly hard? The only way Euler could prove that there are infinitely many 4n+1 primes was by considering the sum of the reciprocals of the 4n+1 primes minus the sum of the reciprocals of the 4n+3 primes and then doing magic. He ended up showing somehow that this difference of sums converged, and that implies that there are infinitely many 4n+1 primes because the sum of the reciprocals of the primes diverges.

I don't know about #3, but #2 is not very hard using #1.  In fact it's possible to use a generalization of #1 involving cyclotomic polynomials to prove that there are infinitely many primes that are 1 mod n for any integer n.  (This was in one of Reid's number theory handouts at MOP.)

#3 of course is equivalent to showing that there are infinitely primes that are -1 mod 5, which I'm not sure how to do.

I would however believe you about Euler, since finding that proof without the hint requires some cleverness.

[quote="ComplexZeta"]Isn't #2 (and maybe also #3) incredibly hard? [/quote]

Here's what I did for 1 (the idea can be used for 2):

It's easy to show that p-1 and p^(p-1)+..+p+1 are coprime. We then take a prime q|p^(p-1)+..+p+1. q doesn't divide p-1, so q|p^p-1 and p|p^(q-1)-1, but since a_n = p^n-1 is a Mersenne sequence, we get q|p^(q-1,p) - 1. (q-1,p) =/=1 because q doesn't divide p-1, so p|q-1 (p is prime), meaning that q=pk+1.

Hope it's correct.

Hm.. that looks vaguely similar to IMO 2003 #6 :)

I think I have a solution for the third problem (even if it is a bit weird for me to answer my own post  :D ). It can be found here:[url]

http://www.mathlinks.ro/phpBB/viewtopic.php?t=1545[/url]

Oh, and about that thing with primes of the form 4k+1:

It's well-known that if a prime of form 4k+3 divides a^2+b^2 then it divides a and b, so any prime which divides a^2+1 is of the form 4k+1. If we assume there are only finitely many primes of this kind, let those be p1,..,pn. We then take a s.t. pi|a for all i and then a^2+1 MUST have a prime factor (of the form 4k+1) different from p1,..,pn. his means that there are infinitely many primes of form 4k+1. It can generally be proven that if p divides a^(2^t)+b^(2^t) and p doesn't divide a (or b, of course), then p has form
2^(t+1)*k+1, and this can be used to prove that there are infinitely many primes of the form 2^k*t+1 for any natural k.


[quote="pascal_1623"]A merchant has determined that if the selling price of a peach is 15 cents, he will sell 400 each day. He has also determined that for each cent the price is increased, he will sell 16 fewer peaches each day. At what price, in cents, should he sell each peach in order to maximize his daily income from peachs.
This is my most favorite problem :D.[/quote]

[hide]$x=$ number of cents raised.
Income is $(400-16x)(15+x)$. Expanding, $-16x^2+160x+6000$ and the cents raised to maximize profit is $\frac{160}{32}=5$, adding to 15, we get $\boxed{20}$. [/hide]

[hide]Let the cost of the peaches in cents be $x$.  Then we have the number of peaches sold to be $400-16(x-15)$.  Now, we have an equation for the number of cents he will receive.

$x(400-16x+240)=640x-16x^2$

The maximum point of a parabola can be found at $-\frac{B}{2A}$ (this gives the $x$ coordinate), where $Ax^2+Bx+C=0$ is the equation.
Plug in to get the equation to get $-\frac{640}{-32}=\boxed{20\text{cents}}$[/hide]


Julie begins counting backward from 1000 by 2's, and, at the same time, Tony begins counting forward from 100 be 3's. If they cout at the same rate, what number will they say at the same time?

[hide="Solution"] $1000-2x=100+3x$, $5x=900$, $x=180$. So the answer is $100+180\times3=\boxed{640}$ :D I think this is right.[/hide]


Find $\lim_{x\to\infty}\sqrt{9x^2+x} -3x$

The first limit I got from my math class.  :D


integration
trigonometry
calculus
function
calculus computations

[color=darkblue]Find $ I \equal{} \int_{0}^{\frac {\pi}{2}} \frac {\cos{x} \minus{} \sin{x}}{(1 \plus{} \cos{x} \plus{} \sin{x})^2} dx$

[b][u]question[/u][/b]:

$ I \equal{} 0$ is right or wrong?[/color]

hello, let $ t \equal{} 1 \plus{} \cos(x) \plus{} \sin(x)$ then we have $ dt \equal{} (\cos(x) \minus{} \sin(x))\,dx$ and our integral is now
$ \int_{2}^2\frac {1}{t^2}\,dt \equal{} 0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, let $ t \equal{} 1 \plus{} \cos(x) \plus{} \sin(x)$ then we have $ dt \equal{} (\cos(x) \minus{} \sin(x))\,dx$ and our integral is now
$ \int_{2}^2\frac {1}{t}\,dt \equal{} 0$.
Sonnhard.[/quote]
Nope. It is $ \int_{2}^2\frac {1}{t^2}\,dt \equal{} 0$.

hello, yes you have right, it was a typo it must be $ \int_{2}^2\frac{1}{t^2}\,dt\equal{}0$. Thank you for your hint.
Sonnhard.

Alternatively, making the substitution $ u \equal{} \frac {\pi}{2} \minus{} x$ interchanges sines and cosines.  This shows that $ I \equal{} \minus{} I$.  Or (slightly more complicated) put $ x \equal{} u \plus{} \frac{\pi}{4}$ to see that we have the integral of an odd function over a symmetric interval.


inequalities
LaTeX
inequalities proposed

$\LaTeX$ed:

[quote="Yulia"]So we should prove
$a^{2}c+b^{2}a+c^{2}b\ge ab+bc+ac$
By AM-GM
$\frac{a^{2}c}{3}+\frac{b^{2}a}{3}+\frac{b^{2}a}{3}\ge ab$, we used that $c=\frac{1}{ab}$

$\frac{b^{2}a}{3}+\frac{c^{2}b}{3}+\frac{c^{2}b}{3}\ge bc$

$\frac{c^{2}b}{3}+\frac{a^{2}c}{3}+\frac{a^{2}c}{3}\ge ac$[/quote]

nice solution! :)  yes this is quite easy. actually i created it from another one i created. i'm trying to create a harder inequality from it at the moment. i'll post it as soon as i'm done. :wink:

[quote="nayel"]Sorry if the following inequality already belongs to someone else, but i also created it. :)

Let $a,b,c>0$ such that $abc=1$. Prove that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\]
[/quote]
Very nice proof, Yulia!  :lol: See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88272


geometry
3D geometry
calculus
calculus computations

5.  Find the point closest to the origin on the curve of the intersection of the plane $ 2y\plus{}4z\equal{}5$ and the cone $ z\equal{}4x^2\plus{}4y^2$.

With this one, I'm clueless as to what to do.

Minimize $ g(x,y,z)\equal{}x^2\plus{}y^2\plus{}z^2$ subject to these two constraints.


What is the number of degrees in the acute angle formed by the hands of a clock at 6:44 P.M.?

[hide="solution"] The hour hand will be $\frac{1}{2}$ of the way around the clock PLUS $\frac{44}{60}\cdot \frac{1}{12}= \frac{11}{120}$ for the additional chunk added on by the extra minutes, for a total of  $\frac{71}{120}$. Converting into degrees from the top, we get 213 degrees.

The minute hand will be $\frac{44}{60}= \frac{11}{15}$ of the way around, which is 264 degrees.

Taking the positive difference, we get $264-213=\boxed{51}$. [/hide]

Your minute hand is right but your hour hand is wrong.
[hide]
6/12 * 360 = 180 degrees. In 60 minutes it moves an additional 360/12 = 30 degrees, so in 44 minutes it moves 44/60 * 30 = 22 degrees, so the hour hand is a total of 202 degrees. 264-202 is $62 \text{degrees}$.[/hide]

hahahaha there was a similar problem on the vestavia test.

Also about angles at another time i think  :lol:

Don't you mean #9?
I almost missed that one.


pigeonhole principle
geometry
circumcircle
rhombus
perpendicular bisector
combinatorics proposed
combinatorics

Let $P$ and $Q$ be two distinct points in the plane. Let us denote by $m(PQ)$ the segment bisector of $PQ$. Let $S$ be a finite subset of the plane, with more than one element, that satisfies the following properties:
(i) If $P$ and $Q$ are in $S$, then $m(PQ)$ intersects $S$.
(ii) If $P_1Q_1, P_2Q_2, P_3Q_3$ are three diferent segments such that its endpoints are points of $S$, then, there is non point in $S$ such that it intersects the three lines $m(P_1Q_1)$, $m(P_2Q_2)$, and $m(P_3Q_3)$.
Find the number of points that $S$ may contain.

No solution after over a year?  Let us see if this one works...

Assume that $ S$ contains exactly $ n$ points.  We may then define $ \binom{n}{2}\equal{}\frac{n(n\minus{}1)}{2}$ segments.  Assume $ n > 5$.  Then, there are more than $ 2n$ segments, or more than $ 2n$ perpendicular bisectors, passing through the $ n$ points of $ S$.  By Dirichlet's pigeonhole principle, there is at least one point such that 3 perpendicular bisectors pass through it, contradiction.  Or $ n\leq5$.

For $ n \equal{} 5$, take $ S$ as the set of vertices of a regular pentagon.  Through each vertex pass exactly two perpendicular bisectors, those of the opposite side and the diagonal parallel to the latter.  For $ n \equal{} 3$, take $ S$ as the set of vertices of an equilateral triangle.  Through each vertex pass exactly one perpendicular bisector, that of the opposite side.  So $ S$ may contain 3 or 5 points.

$ S$ may not contain 2 points, since there is no point left in $ S$, for the perpendicular bisector of the segment they determine, to intersect.  Assume a set $ S$ may be found with exactly 4 points, and call them $ A,B,C,D$.  Assume wlog that the perpendicular bisector of $ AB$ passes through $ C$.  Then, $ ABC$ is isosceles in $ C$.  Assume that $ ABC$ is not equilateral, then the perpendicular bisectors of $ BC$, $ CA$ cannot pass through $ A$, $ B$, respectively, or they both pass through $ D$, which is then the circumcenter of $ ABC$, and the perpendicular bisector of $ AB$ passes also through it, contradiction.  $ ABC$ is then equilateral.  Consider now the perpendicular bisector of $ CD$.  Since it passes through $ A$ wlog, $ CDA$ is also equilateral, or $ ABCD$ is a rhombus with lesser angle $ \pi/3$ and greater angle $ 2\pi/3$, or the perpendicular bisector of $ BD$ passes through both $ A$ and $ C$.  So, as before, $ ABD$ and $ BCD$ are equilateral, which is absurd.

So $ S$ may contain only 3 or 5 points.


Does anyone know what anthropology is?

I have heard of it before, and I have looked up the word. However, the definition, according to http://www.dictionary.com, is not really concrete. What does an anthropology do? Are there any famous anthropologists in history, or is anthropology a relatively new discipline?

Why do people use dictionaries on the internet?  Why not just look it up in Merriam Webster or something like that?  I looked in Merriam and it said the study of humans and their ancestors.

[quote="tetrahedr0n"]Thats pretty much what dictionary.com says... i don't see the problem with their definitions. they are pretty concrete.

[quote="dictionary.com"]1. The scientific study of the origin, the behavior, and the physical, social, and cultural development of humans.

2. That part of Christian theology concerning the genesis, nature, and future of humans, especially as contrasted with the nature of God: changing the church's anthropology to include more positive images of women (Priscilla Hart).
[/quote][/quote]

I don't understand what is the "physical, social, and cultural development of humans." Can you give me an example?

The reason that I'm asking is that my college interviewer is an anthropologist, and I just wanted to know what on earth an anthropologist is.

Part of the reason http://www.merriam-webster.com (also reachable via http://www.m-w.com) is better than http://www.dictionary.com is that the Merriam-Webster Online Dictionary, if you're doing debate or something similar, sounds a little more authoritative than "dictionary dot com."  Another reason is that Merriam Webster has a more easily navigated website, at least in my experience.  There's no reason to choose dictionary.com over Merriam Webster, except maybe for the online translator that is linked from their website, but that's about it.

I meant Merriam Webster as in the hard back copy, the actual book.  I don't see why people look things up on the computer.  It probably takes longer than jut looking it up, unless you have a whole bunch of words.

Eh, if you're going to use a hard copy Dictionary, you might as well go Oxford English or American Heritage.  I dunno, I suppose they strike me as being particularly authoritative.  Or perhaps it's just the hypnotic effect of a cool binding.  Oooh, navy blue Pleather...

And as for taking more time, that really isn't the case for me, unless you've got a super-slow Internet or a mess of computer maintenance on your "To-Do" List.  You type in the URL, type in the word, and it's done.

Unless you're my sister, who types at five WPM...


Let $ ABC$ be a triangle, B-mixtilinear incircle touch circumcircle $ (O)$ at $ A',P$ ia an arbitrary point on line $ BC$, the circle diameter $ PA$ cuts $ (O)$ at $ E, AE\cap BC \equal{} N$, line $ PA'\cap BC \equal{} N$ prove that $ MN$ is always through a fix point.

if P is an arbitrary point on line BC, how PA' meet BC at N??I don't understand  :(
Could you draw us a figure?

oh sorry  :oops: here is correct problem

Let $ ABC$ be a triangle, B-mixtilinear incircle touch circumcircle $ (O)$ at $ A',P$ ia an arbitrary point on line $ BC$, the circle diameter $ PA$ cuts $ (O)$ at $ E, AE\cap BC \equal{} N$, line $ PA'\cap (O) \equal{} N$ prove that $ MN$ is always through a fix point.

Your problem can be generalized:

Let $ ABC$ be a triangle, $ C\left(O\right)$ is the circumcircle, the point $ X$ is on the side $ BC$ and
consider the circle tangent to $ AX$, $ BX$ and intouche $ C\left(O\right)$ (at $ A'$) simultanously.
$ P$ is an arbitrary point on the line $ BC$, the circle with diameter $ AP$ cuts $ C\left(O\right)$ at $ E$,
$ \left\{M\right\} \equal{} AE\cap BC$, $ \left\{N\right\} \equal{} A'P\cap C\left(O\right)$. Prove that $ MN$ is always through a fix point ($ F$).

If $ AD\bot BC$, $ D\in BC$ and $ A'L\parallel{}BC$, $ L\in C\left(O\right)$ then the fix point $ F$ is the intersection
point of the line $ LD$ with the circumcircle $ C\left(O\right)$.  :)

[quote="Petry"]Hello!

Let $ ABC$ be a triangle, $ C\left(O\right)$ is the circumcircle and $ A'\in C\left(O\right)$.
$ P$ is an arbitrary point on the line $ BC$, the circle with diameter $ AP$ cuts $ C\left(O\right)$ at $ E$,
$ \left\{M\right\} \equal{} AE\cap BC$, $ \left\{N\right\} \equal{} A'P\cap C\left(O\right)$. Prove that $ MN$ is always through a fix point $ \left(F\right)$.

Let $ K\left(O'\right)$ be the circle with diameter $ AP$.
Construct $ A'L\parallel{}BC$, $ L\in C\left(O\right)$ and $ AD\bot BC$, $ D\in BC$.
$ \left\{F\right\} \equal{} LD\cap C\left(O\right)$. Let's prove $ F\in MN$.
$ \left\{M'\right\} \equal{} NF\cap BC$ and $ \left\{E'\right\} \equal{} M'A\cap C\left(O\right)$.
Let's prove that $ E' \equal{} E$.
$ \angle FNP \equal{} \angle DLA' \equal{} \angle LDC\Rightarrow$ the quadrilateral $ DPNF$ is cyclic
$ M'F\cdot M'N \equal{} M'D\cdot M'P$ and $ M'F\cdot M'N \equal{} M'E'\cdot M'A$ $ \Rightarrow M'D\cdot M'P \equal{} M'E'\cdot M'A\Rightarrow$
$ \Rightarrow$ the quadrilateral $ DPAE'$ is cyclic $ \Rightarrow E'\in K\left(O'\right)$.
So $ C\left(O\right)\cap K\left(O'\right) \equal{} \left\{A,E'\right\}\Rightarrow E' \equal{} E\Rightarrow M' \equal{} M\Rightarrow$
$ \Rightarrow F\in MN$.

P.S. The circle tangent to $ AX$, $ BX$ and intouche $ C\left(O\right)$ (at $ A'$) simultanously is not necessary![/quote]

Very nice, thanks, here is my solution, actually the problem is as you Petry I inspire it from a problem when $ A'\equiv A$, here we need $ A'$ fix point only  :)

Let $ MN\cap (O) \equal{} \{N,F\}$ we will prove that $ F$ is fix point, indeed we easy to show

$ MD\cdot MP \equal{} ME\cdot MA \equal{} MF\cdot MN\Rightarrow FDPN$ is cyclic, let $ FD\cap (O) \equal{} \{F,L\}$ from $ FDPN$ is cyclic $ \Rightarrow \angle MDF \equal{} \angle FNP \equal{} \angle FLA'$ so $ A'L\parallel BC\Rightarrow L$ is fix point, so $ \{F\} \equal{} DL\cap BC$ it is fix point.


I am having trouble understanding about 13 :blush: solutions from Intro to Algebra. Here they are:

11.4.3: I am having trouble understanding everything.

11.4.4: I am having trouble understanding pretty much all of it :oops:.

11.5.6: $ (a)$: I'm having trouble understanding everything.

11.45: I didn't understand any of it.

11.47: I understood none of it.

11.49 $ (d)$: zippo understanding.

This is sorta embarrassing...

11.51: Nothing registered.

11.52: No understanding at all.

It's really embarrassing that I have to ask so much, but it's really important that I don't leave anything without clarifying. So please: HEEELP!!

Don't be embarrassed at all. Most of the people on AoPS are some of the best problem solvers. Don't be intimidated. You'll definitely get a lot better with practice.

For those of us who don't have the book but still want to help you, could you post a couple of them?

And as 123456789 said, don't be embarrassed. :)


Ok, I announced that I was going to post some self - made problems. Some are intermediate, some are "advanced". This is the first one :

Show that the numbers 16, 1156, 111556, ... are perfect squares.

(In general, such a number consists of k + 1 ones, k fives and a six.)

[color=cyan]Nice problem.  A bit tough, but doable.[/color]

Arne wrote:Ok, I announced that I was going to post some self - made problems. Some are intermediate, some are "advanced". This is the first one :

Show that the numbers 16, 1156, 111556, ... are perfect squares.

(In general, such a numlber consists of k + 1 ones, k fives and a six.)

[hide]i see this, and i think, maybe it would help to find a general formula. note that the numbers are 15+1, 1155+1, 111555+1, etc. The general formula for n ones followed by n fives is going to be the number that consists of n ones multiplied by (10^n+5). The number that is n ones is (10^n-1)/9. So we multiply and add 1 to get (10^n+5)(10^n-1)/9+1. We factor the first term as ((10^n+5)/3)((10^n-1)/3) (we know both of those are divisible by 3; look at the mods). Then we know the difference between them is 6/3=2. If you multiply 2 numbers that are 2 apart and add 1, the result will be a square ((a-1)(a-1)+1=a:^2:) [/hide]

[color=cyan]Ooh, that was really clean.  When I did it it was much messier.[/color]

Here's an intuitive proof...I'm not sure if it qualifies as a valid one...

Suppose we have a number with k+1 digits, where the last digit is 4 and the first k digits are all 3s. We can let k=0 if we want. We can multiply the number by itself like below:

                 33..........334
               [u]x33..........334[/u]
               133..........336
             100..........002
           100..........002           <--k+1 numbers to add up
         100..........002
etc, etc, etc.
Imagine those numbers indented like you were multiplying 33.......334 x 33.......334 out on paper by hand.

The last digit will clearly be 6. Since, in doing the multiplication, we have to add the results from multiplying in each individual decimal place, we have k+1 numbers to add, as shown above. We then have k+1 ones at the front of the resulting square number. Then, in the first number at the top of the addition, 133..........336, there are k 3s, each of which has a 2 below it along with a bunch of zeroes. When we add, those will become k 5s, and the last digit is, again, 6. Since k can be any non-negative integer, all numbers composed of k+1 ones, k fives, and a six (i.e. 16, 1156, 111556, etc.) are perfect squares of the numbers composed of k threes and a units digit of 4. I know I didn't explain this proof too well, so I'll give an example...

                33334
              x[u]33334[/u]
              133336
            100002
          100002
        100002
      [u]100002        [/u]
      1111155556
Again, imagine as if the numbers above were indented as if you were multiplying 33334 x 33334 out on paper by hand.

Hope this wasn't too confusing...

Charles[code][/code]

Mystic, I think u  meant (a-1)(a+1)+1=a^2, not (a-1)(a-1)+1=a^2.
But, i agree, it is a nice proof
i was wondering, did u have those formulas memorized or did u derive them or something?

ANd Joel, what was ur method?

[color=#00B0B0]I also looked for a parametric form for the number in terms of the number of digits.  I just did it a really ugly way, and he did it a very nice way.[/color]

yea
but still wanna kno how mystic can memorize all those formulas like dat

[color=orange]Not memorized, worked out and then written down . . . you can do the algebra yourself, it isn't tricky . . . 11111111....11111  is 9999....9999 divided by 9, which is (100000....00000 - 1)/9  which is (10^n - 1)/9.  See?[/color]

yeah you're right, that's what I meant to say, darkquantum.

I didn't memorize all those formulas, I just worked them out.


A =log2

What is the shortest interval of A ? (shortest distance)

[quote="telemat"]B<A<C      B=?  C=?[/quote]

There is no shortest distance, but you can get arbitrarily close to 0. If you allow equality, the minimum is 0 with $ B\equal{}C\equal{}\log_2 A.$

I'm sorry, but this question makes exactly no sense. As $ \log 2$ is a real number there are rational series that converge to it from below and above, so the length of the interval converges to 0 :huh:

[quote="bertram"]As $ \log 2$ is a real number[/quote]

You mean an irrational number.  In particular, the even and odd convergents of the continued fraction of $ \log 2$ converge to it from above and below, respectively.

Unless I'm badly mistaken every real number is defined as a limit of a rational sequence. And by some obscure logic we also can make a series that is strictly less and one that is strictly more than the number... So for $ \log 2$ one could use $ 0.3,0.30,0.301,\ldots$ and $ 0.4,0.31,0.302,\ldots$ and the difference of both is converging to 0.... which was what I tried to explain with my last post  :maybe:

Sorry, I was thinking in terms too strict.  Of course a rational number can also be approximated by an infinite rational sequence  :oops:


Mixing up notation there a bit, aren't we?

Anyway, $ {1\over{n! \plus{} (n\plus{}1)!}}\equal{}{1\over{(n\plus{}1)!}}\minus{}{1\over{(n\plus{}2)!}}$, which should help a little bit. :)


What should I do if my school does not sponsor (administer) the mathcounts and AMC tests?

For AMC you could find another place that sponsors it. Where I live, the major libraries usually are testing centers, and you can take the test there if you can't take it in school. For MATHCOUNTS, I'm afriad you have to have your school sponser it in order to participate.


Deterimine the area enclosed by the graph of the following function: $ 3x^2 \minus{} 24x\plus{}3y^2\plus{}6y\equal{}0$.

[quote="jeez123"]Deterimine the area enclosed by the graph of the following function: $ 3x^2 \minus{} 24x \plus{} 3y^2 \plus{} 6y \equal{} 0$.[/quote]

[hide]Divide by 3, and we have $ x^2\minus{}8x\plus{}y^2\plus{}2y\equal{}0\Rightarrow x^2\minus{}8x\plus{}16\plus{}y^2\plus{}2y\plus{}1\equal{}17\Rightarrow(x\minus{}4)^2\plus{}(y\plus{}1)^2\equal{}17$.

This is a circle with radius $ \sqrt{17}$, so our area is $ \pi\cdot(\sqrt{17})^2\equal{}17\pi$[/hide]


geometry
3D geometry
tetrahedron
AoPSwiki
geometry unsolved

Where can I get more info on how to solve high school level tetrahedron problems?

EDIT: also, how do I work with dihedral angles to find distances within the interior of a tetrahedron?

Hopefully this has nothing to do with any ongoing contest ..  :roll:

At any rate, try AoPSWiki - [[Special:WhatLinksHere/Tetrahedron]].


Surely $k<p$?

Part (a) is true by definition (and is true regardless of the parity of $p$ or $k$).

As for part b:
[hide]
$(-1)^{k-1}=-1$, as k is even, and:
$(p-k)!(p-k+1)...(p-1)=(p-1)!$, hence the problem is reduced to:
$[(p-k)!(k-1)!]+(p-1)!$

If $p$ is composite, then this is clearly divisible by p (as $p=ab$, for some $a$ and $b$, and $a$ and $b$ are smaller than $p-1$, hence they both divide $(p-1)!$ and $[(p-k)!(k-1)!]$).

Now all we have to do is prove that it's true when $p$ is prime. By Wilson, $(p-1)! \equiv-1 \mod{p}$. Hence, if we can prove that $[(p-k)!(k-1)!] \equiv 1 \mod{p}$, we are done. However, I haven't been able to prove this.  :blush:
[/hide]

[quote="rjt"]Hence, if we can prove that $[(p-k)!(k-1)!] \equiv 1 \mod{p}$, we are done. However, I haven't been able to prove this.  :blush:
[/quote]
$(p-k)!(k-1)! \equiv (-1)^{k-1}(p-k)! (-1)(-2)\ldots(-k+1) \equiv$
$\equiv(-1)(p-k)!(p-1)(p-2)\ldots(p-k+1) =-(p-1)! \equiv 1(mod\ p)$
(we use the fact that $k$ is even so (-1)^{k-1}=-1 and also, in the last step, Wilson's theorem)

EDIT: I deleted some nonsense (see Krysztof's post below). Sorry for not reading the rjt's post properly.

We just don't have to split the cases of p prime and composite. The conguence from Krysztof's proof
$(p-k)!(k-1)!\equiv(-1)(p-k)!(p-1)(p-2)\ldots(p-k+1) (mod\ p)$ is exactly the given one.

[quote="Kondr"]So we suppose p is a prime? .[/quote]
No, rjt has already solved the case when p is composite, so I only wrote the remaining part of his proof.


[quote="riddler"]Can you reply in a forum which isnt you're nationality?[/quote]
if you can understand the language.

if you are learning a language at school, and don't know it well, could you start a post in that froum for people who are learning that language?

In general just how free are these national forums

I have a huge interest in about every nation on the world and its people.

Am I even allowed to have done this post here not being a moderator? :ninja:

Of course you are allowed :) If you were not allowed you couldn't have posted in the first place :D

You can obviously interact with any members of the site on any open forums, as long as you will keep the discussions in the proper decency limits.

i've been visiting all the forums of different countries as long as i can understand the language.


function
calculus
derivative
analytic geometry
graphing lines
slope

Can someone help out with this. How do I tackle this question?

Find the equation of the normal to the curve $ y \equal{} 3{x^2} \minus{} 2x \minus{} 1$ which is parallel to the line $ y \equal{} x \minus{} 3$.

I have found the gradient function $ \frac{dy}{dx} \equal{} 6x \minus{} 2$. But what do I do next?

Thanks

[quote="pizza1512"]Can someone help out with this. How do I tackle this question?

Find the equation of the normal to the curve $ y \equal{} 3{x^2} \minus{} 2x \minus{} 1$ which is parallel to the line $ y \equal{} x \minus{} 3$.

I have found the gradient function $ \frac {dy}{dx} \equal{} 6x \minus{} 2$. But what do I do next?

Thanks[/quote]

Wrong forum; should be in Calculus forum.

Anyways, just set the derivative to be $ \minus{}1$ (that's the slope of the perpendicular line) to obtain $ x \equal{} 1/6$. Then the line has slope $ 1$ and passes through $ (1/6, f(1/6)) \equal{} (1/6, \minus{}5/4)$. So $ y\equal{}x\minus{}1/6\minus{}5/4\equal{}x\minus{}17/12$.

EDIT: Thought it said tangent.

not quite, it needs to be normal to the cure, so we want the derivative to be $ \minus{}1$.

[hide="My solution"]Normals are perpendicular to their curves, so we are looking for $ \frac {dy}{dx} \equal{} 6x \minus{} 2 \equal{} \minus{} 1$ as the normal at this point will be the negative reciprocal, or $ 1$, which will be parallel to $ y \equal{} x \minus{} 3$ as desired. This slightly twisty logic is the tricky part, for me anyway, after this it is simple enough algebra.

Solving for the equation $ 6x \minus{} 2 \equal{} \minus{} 1$ we get $ x \equal{} \frac {1}{6}$. Substitution back into $ 3x^{2} \minus{} 2x \minus{} 1$ gives us a y value of $ \frac {3}{36} \minus{} \frac {2}{6} \minus{} 1$ = -$ \frac {5}{4}$. The equation of the line that goes through $ (\frac {1}{6}, \minus{} \frac {5}{4})$ with gradient $ 1$ is:

$ y \minus{} \frac { \minus{} 5}{4} \equal{} 1(x \minus{} \frac {1}{6})$
$ y \equal{} x \minus{} \frac {1}{6} \minus{} \frac {5}{4}$
$ y \equal{} x \minus{} \frac {2}{12} \minus{} \frac {15}{12}$
$ y \equal{} x \minus{} \frac {17}{12}$

And we are done. At least, I certainly hope so.

[b]EDIT:[/b] Whoops, forgot to $ \minus{} 1$ from the fraction the first time around.[/hide]


geometry
trigonometry
circumcircle
rhombus
geometric transformation
reflection
angle bisector

We have to postpone question-specific comments for tomorrow, but how did people do generally?

I did atrociously badly :( (probably about 11 or 12 marks). I'm just praying they take other criteria (like my performance in the Bath, BMO1, Hungary exams) into account for Trinity selection, but I have a feeling I might not be there.

How depressing... and I was having such a good year.

I found it hard too.
I think my mark will be either 12ish or 21ish.
Dominic

I thought it was harder than last years, though I managed all 4.
11/12 was sufficient for Trinity last year, IIRC, so don't give up hope.
Sorry you didn't do as well as you might have expected,

Jack

edit @ zibi:  people will still be sitting this all day, so we can't give anything question-specific away until tomorrow

[quote]zibi: people will still be sitting this all day, so we can't give anything question-specific away until tomorrow[/quote]
I don't understand. Is this contest still in progress or ... ?? http://www.mathlinks.ro/Forum/viewtopic.php?t=130981 ???

That's from two years ago :)
The contest happens at a time of your choice during the course of today.

Jack

I suppose I can take solace in the fact that it wasn't just me who found it difficult; I almost certainly got 0. For the last couple of papers I managed to complete two questions completely, and on all the past papers I've done at least one question without help. Oh well - hopefully next year it'll be a nicer paper!

I did two questions although im not totally sure of the second one.  It will be down to the generosity of my marker.  :)  More wil be posted on this subject later.
Sam Siebert

To make matters worse, it appears that I didn't read the geometry question properly, completely missing an important condition regarding an angle being fixed rather than any acute angle (thereby demonstrating how my 3 hours of work was actually in vain from the start). I don't know whether to be happy about this, or to cry at my babbling incompetence.

I found it moderately hard, but not too bad. I think I solved two of the problems, so hoping for the 18-21 mark bracket. (That said, I thought I'd solved two last year and only got 14).

Oddly enough, I wasted almost as long as Tom missing the geometry condition, though fortunately enough I rediscovered it half an hour from the end, in time to solve it.

@ zibi (and others who want to know the questions) I may be still up at midnight (90 minutes from now) if in which case I'll post the problems then.

Ok, some of the rest of my work done and it now gone midnight, here are the problems:

[b]1.[/b] Triangle $ABC$ has integer-length sides, and $AC=2007$. The internal bisector of $\angle BAC$ meets $BC$ at $D$. Given that $AB=CD$, determine $AB$ and $BC$.

[b]2.[/b] Show that there are infinitely many pairs of positive integers $(m,n)$ such that

$\frac{m+1}{n}+\frac{n+1}{m}$

is a positive integer.

[b]3.[/b] Let $ABC$ be an acute-angled triangle with $AB>AC$ and $BAC=60^{\circ}$. Denote the circumcentre by $O$ and the orthocentre by $H$ and let $OH$ meet $AB$ at $P$ and $AC$ at $Q$. Prove that $PO=HQ$.

[b]4.[/b] In the land of Hexagonia, the six cities are connected by a rail network such that there is a direct rail line connecting each pair of cities. On Sundays, some lines may be closed for repair. The passengers' rail charter stipulates that any city must be accessible by rail from any other (not necessarily directly) at all times. In how many different ways can some of the lines be closed subject to this condition?

I guess someone could post these in Pre-Olympiad or something; I won't right now.

I gather from Sloane's dictionary of integer sequences that the answer to question 4 was actually 26704.

26704 - are you sure?  If so thats down to one question for me.  (Yes it was Q1).  I got 26--3, but I don't remember what the blanks were.
Surely it should be 26703, however, as I assumed keeping all lines open didn't count (from the wording of the question).
Sam Siebert

Yes, 26704 (or, as Sam says, 26703)
Question 2 was very nice, and question 3 unfair - knowing something was a huge advantage (i.e. making the problem trivial).

Jack

I agree on Q3 to an extent, but to another extent it's ok. I think for BMO2 it is okay to put in some more squad-friendly questions, and when I looked at that question I thought "this looks like it's here to check on squad geometry" (anyone who had to read the definitions of orthocentre and circumcentre would probably have been intimidated away from the question anyhow).

This is surely more of an issue on Q1, where the angle bisector theorem practically trivialised it?

I liked this year's paper more than I did last year's - I managed 1 and 3. Question 2 was really annoying, in the sense that I ended up with something Pell-like and just stated it had infinite solutions (couldn't prove it) - but could probably have found the recurrence using an earlier part of my method.

I agree with Jack about Q3, you do have to be pretty competent with triangle centres, which untrained people may not be.

Is there not going to be a squad discussion via the mailing list? I've just seen scattered posts between here and nrich.

Alison

I disagree with Tom - I think it's fair to expect those sitting BMO2 to know the angle bisector theorem, and I contest that it's trivial even from there.  On the other hand, it's not fair to set a question which is [i]actually[/i] trivialised by AH=2RcosA.

Jack

I wouldn't say it's trivialised any more so than Q1. You still have to know about or prove the isogonal angle <OBC = A', and the angles at the top, unless your method is more direct than mine. I do agree with you that knowing AH=2OL makes the problem much much easier.

Incidentally, rather interestingly, this implies that the powers of P and Q wrt the Nine Point Circle are equal (this was one of my many attempted lines of attack on the false general case), so (PF)(PN)=(QE)(QM).

My method was to say BAO=CAH, which if you know the formula for AH you will certainly know, and then it is clear  that AOP and AHQ are congruent.

Jack

Oh yeah... ok you win. I was using the more roundabout method Alison hinted at when she pointed out the condition to me as I sent her my geogebra screens and said "look! they aren't equal! it's false!", I guess because I had got used to working with A rather than $\pi/3$

It looks like everyone found this quite hard which is quite comforing as I felt awfull about it. I reckon i gave a complete answer to Q2 although that was just about it - I dont think i've ever solved a geometry problem in something like this. I cant imagine I'll have qualified for any further events but this is the first year I even tried the SMC so I'm happy to have made it here.

I think question 1 was good as an easier question and although that the angle bisector theorem makes it much simpler, proving the angle bisector theorem isn't too hard, and some people might reasonably have deduced it in the exam.

Proving AH=2RcosA would be much harder, though, particularly if you didn't know it in advance, which probably not many people did (lucky I had learned that far through the trig drill by then). I also wondered at the time (before I realised that $\angle A=60^{\circ}$) if there was a proof based on the fact that N is the midpoint of OH: does anyone known of any alternative - based on this fact or not - proofs to Q3?

This fact reduces (using trig knowledge), via powers of a point, to showing that OH bisects the angle A at the orthocentre, which I'm not sure is that easy to prove without using some facts.

If the angle bisector of A meets the circumcircle at M, then BOHC is cyclic centre M, and it can then be shown that AOMH is a rhombus, providing an alternative proof that AH=AO.

Jack

I don't follow. M lies outside the convex hull of ABHO (if it is convex, which I don't think it is), so surely this can't be true?

Oooh, so this is a rather interesting special case, because M is the reflection of O in BC, so this is the case in BMO2 2006 Q3 where P coincides with the circumcentre :).

I managed one question and I'm pretty pleased with that. I spent a while on Q2 and Q4 but didn't really get anywhere - could anyone give an outline solution of these two?

For Q2 try writing out some simple examples, and guessing at a recurrence which allows you to turn a solution into a bigger one. If you get the right one, it's a simple case of showing it works, which isn't too hard.

For Q4, consider the ways of adding a vertex (and some edges) to 5-graphs to create a connected 6-graph. This is a mammoth casebash, but it does provide the correct answer, and in about 30 minutes of work, which is ok. Does anyone else know if there is a nice way?

It seemed the setters of this year's BMO were making a point - read Bradley and Gardiner. That formula that Jack was mentioning appears in there and it quite a standard result when dealing with orthocentres.

I still wish they'd set my question though.

Q1, Q3 - Geometry, both killable
Q2 - Nice number theory question
Q4 - Yuk! Counting...(sigh so glad I didn't have to waste countless sheets of A4)

Are you allowed to reveal what your question was?

Afraid not. It's still on the list for next year.