quadratics
modular arithmetic
number theory
number theory proposed

$ k\equal{}2,3$ are the only solutions, which can be checked easily. Let $ k\ge 4.$If $ \phi(k)\plus{}1<k$ then we can choose $ n\equal{}1,m\equal{}\phi(k)\plus{}1$ which are clearly different, and we are done. Otherwise $ k$ is prime and greater than $ 3$ so we can choose $ n\equal{}k\minus{}1,m\equal{}\frac{k\plus{}1}{2}$ or $ n\equal{}1,m\equal{}\frac{k\plus{}1}{2}$, and we are done by FLT.

I agree for the case where $ k$ is prime and $ > 3$.
But if $ m \equal{} \phi(k) \plus{} 1$ then $ m^{m\minus{}1} \equiv 1 \bmod{k}$ is not true if $ \gcd(k,  \phi(k)\plus{}1) > 1$
This can easily happen e.g $ k \equal{} 15 \implies \phi(k)\plus{}1 \equal{} 9$

[quote="FelixD"]Determine all integers $ k\ge 2$ such that for all pairs $ (m$, $ n)$ of different positive integers not greater than $ k$, the number $ n^{n \minus{} 1} \minus{} m^{m \minus{} 1}$ is not divisible by $ k$.[/quote]
$ k \equal{} 2,3$ are  solutions of this problem . We divide problem into two cases :
If $ k$ is even and is greater than 2 . It is easy to see that $ (m,n) \equal{} (k \minus{} 1,1 )$ then $ k$ is a divisor of number $ n^{n \minus{} 1} \minus{} m^{m \minus{} 1}$ .Therefore if $ k > 2$ then exist $ (m,n)$ satisfies condition .
Remain case is $ k$ is odd . We will show that $ k$ must have form $ p^a$ where $ p$ is a prime . If not then exists $ p,q$ such that $ n \equal{} pq$ where $ p,q > 1$ and $ \gcd(p,q) \equal{} 1$ . By the chinese remainder theorem , exist n such that :
\[ n\equiv 1 (\mod p)\]

\[ n\equiv \minus{} 1 (\mod q )\]
It is easy to show that $ n$ is different from $ 1$ and $ k \minus{} 1$ and $ n^{n \minus{} 1}\equiv 1 , \minus{} 1 (\mod k )$ . If this congruence is 1 then we choose $ m \equal{} 1$ and the other we choose $ m \equal{} k \minus{} 1$ . So we only need to solve this problem when $ n \equal{} p^a$ . But is is quite easy .
If $ a > 2$ then we choose $ m \equal{} \varphi(p^a) \plus{} 1$ and $ n \equal{} 1$ then $ p^a |n^{n \minus{} 1} \minus{} m^{m \minus{} 1}$
If $ a \equal{} 1$ (it is equivalent to the fact that $ k$ is a prime number . Suppose it does not exist pair of $ (m,n)$ satisfies condition then $ n^{n \minus{} 1}$ is a complete reside system mod p . Because the is exactly $ \frac {p \minus{} 1}{2}$ numbers are quadratic residue mod p and if n is odd then $ n^{n \minus{} 1}$ is a quadratic residue mod p so if n is even then $ n^{n \minus{} 1}$ is not a quadratic residue mod p . Applying property of Jacobi's symbol we have $ n$ is also a nonquadartic residue mod p when $ n$ is even . If $ p > 3$ then $ 4$ is an even number and a quadratic residue mod p and we will have contradiction . Thus $ p$ must be equal 3 and it is easy to check that $ 3$ is a suitable number .
Hence ,all numbers satisfy condition are $ k \equal{} 2$ and $ k \equal{} 3$

This seems to be a hard problem unless I'm missing an easy trick.

We try to rule out all cases $ k > 3$

For primes we can find $ (m,n)$ as described by Ahwingsecretagent above.

For numbers that have a square factor greater than one $ x^2 | k$ then we take $ m \equal{} \frac{k}{x}$ and  $ n \equal{} k$

For numbers that are products of two primes $ k \equal{} pq, p < q$ consider two cases:
  1st case: $ p \nmid q\minus{}1$:
  solve $ m \equal{} ap \equiv 1 \bmod{q}$ for $ 0 < a < q$ and
  solve $ n \equal{} bp \equiv 1 \bmod{q\minus{}1}$ for $ 0 < a < q$
  these have solutions because $ (p,q) \equal{} (p,q\minus{}1) \equal{} 1$

  $ m^{m\minus{}1} \equiv 0 \bmod{p}$ and $ n^{n\minus{}1} \equiv 0 \bmod{p}$
  $ m^{m\minus{}1} \equiv 1 \bmod{q}$ and $ n^{n\minus{}1} \equiv 1 \bmod{p}$

  so $ m^{m\minus{}1} \equiv n^{n\minus{}1} \bmod{pq}$ as required

  2nd case $ p \mid q\minus{}1$:
  then take $ m\equal{}1$ and $ n \equal{} \phi(k) \plus{}1 \equal{} (p\minus{}1)(q\minus{}1) \plus{} 1$ which works because $ (k,n) \equal{} 1$

This leaves cases where $ k$ is the product of 3 or more distinct primes

TTsphn i don't understand something
if  p | n-1
 and q | n+1 then pq | n^2-1 or n | n^2-1

what have I misunderstood

just for record only three guys got more then one point on this task  (8,5,4) i got one point on proving its impossible for even k(>2)

It's just a typo. Where he has $ n \equal{} pq$ it should be $ k \equal{} pq$. $ n$ is a different number.

And it's not corect. For example:

$ p\equal{}3$ and $ q\equal{}5$.
$ 4\equal{}1$ mod $ 3$ and $ 4\equal{}\minus{}1$ mod $ 5$ but $ 4^3\equal{}4$ mod $ 15$ but $ 4\neq 1,\minus{}1$ mod $ 15$

[b]The correction of the part why an odd $ k$ can have only one distinct prime devisor should be:[/b]
by chinese reminder theorem we have $ n_1\equal{}1$ mod $ p$, $ n_1\equal{}\minus{}1$ mod $ q$ and $ n_2\equal{}\minus{}1$ mod $ p$, $ n_2\equal{}1$ mod $ q$. Obviously $ n_1,n_2\neq 1,\minus{}1$ and $ n_1\neq n_2$ mod $ k$ ($ k\equal{}pq$).

$ n_1\plus{}n_2\equal{}0$ mod $ p$ and $ n_1\plus{}n_2\equal{}0$ mod $ q$ therefore $ n_1\plus{}n_2\equal{}0$ mod $ k$. but since $ 1\leq n_1,n_2\leq k$ we have $ n_1\plus{}n_2\equal{}k$

Since $ k$ is odd one of $ n_1,n_2$ is odd. wlog let it be $ n_1$. Than $ n_1^{n_1\minus{}1}\equal{}1^{n_1\minus{}1}\equal{}1$ mod $ p$ and $ n_1^{n_1\minus{}1}\equal{}(\minus{}1)^{n_1\minus{}1}\equal{}1$ mod $ q$ (because $ n_1\minus{}1$ is even). therefore $ n_1^{n_1\minus{}1}\equal{}1$ mod $ k$ by chinese reminder theorem.

[b]idea of alternative solution:[/b]
if $ n$ is odd than $ n^{n\minus{}1}$ is a quadratic residue mod $ k$. The numbers $ 1^0,3^2,5^4..$ are all quadratic residues mod $ k$. The number $ 4^3\equal{}8^2$ is also a quadratic residue. But there are roughly $ \frac{k}{2}$ distinct quadratic residues for odd $ k>3$...

I agree that the solution of TTsphn is incorrect.

However the quadratic residue method proposed by Jure the frEEEk can easily be made to work.

The number of distinct quadratic residues $ \bmod{k}$ is at most $ [\frac{k}{2}] \plus{} 1$.
That is because $ (\minus{}x)^2 \equiv x^2 \bmod{k}$ and $ x \equiv \minus{}x \bmod{k}$ has one or two solutions $ \bmod{k}$ according to whether $ k$ is odd or even.

$ n^{n\minus{}1}$ is a quadratic residue $ \bmod{k}$ if $ n$ is odd or if $ n$ is a square or if $ n\equal{}k$. For $ k \ge 5$ this gives more quadratic residues than the number of distinct residues $ \bmod{k}$ so there must be some that are $ \equiv \bmod{k}$

[quote="PhilG"]I agree that the solution of TTsphn is incorrect.

However the quadratic residue method proposed by Jure the frEEEk can easily be made to work.

The number of distinct quadratic residues $ \bmod{k}$ is at most $ [\frac {k}{2}] \plus{} 1$.
That is because $ ( \minus{} x)^2 \equiv x^2 \bmod{k}$ and $ x \equiv \minus{} x \bmod{k}$ has one or two solutions $ \bmod{k}$ according to whether $ k$ is odd or even.

$ n^{n \minus{} 1}$ is a quadratic residue $ \bmod{k}$ if $ n$ is odd or if $ n$ is a square or if $ n \equal{} k$. For $ k \ge 5$ this gives more quadratic residues than the number of distinct residues $ \bmod{k}$ so there must be some that are $ \equiv \bmod{k}$[/quote]
Excuse me ! . This is also my idea when I see this problem  again . Indeed ,we can find the number of quadratic residue mod $ n$ but in this case ,your idea is enough to solve this problem .

[quote="FelixD"]Determine all integers $ k\ge 2$ such that for all pairs $ (m$, $ n)$ of different positive integers not greater than $ k$, the number $ n^{n\minus{}1}\minus{}m^{m\minus{}1}$ is not divisible by $ k$.[/quote]

[hide="Solution"]Let $k$ be an integer satisfying the condition. Note that the condition is equivalent to $S_k=\{n^{n-1}:n=1,2,\dots,k\}$ being a complete residue system modulo $k$.

Suppose that $k$ is not squarefree, i.e. there exists a prime $p$ such that $p^2|k$. Write $k=p^t k_1$, where $p\not | k_1$, and choose $n=k$, $m=\frac kp=p^{t-1} k_1$. Then of course $k|n^{n-1}$. We will also have $k|m^{m-1}$ if $m\ge 3$, because $m^{m-1}$ is trivially divisible by $k_1$, and also is divisible by $p^t$ because the exponent of $p$ in $m^{m-1}$ is $(t-1)(m-1)\ge 2(t-1)=t+(t-2)\ge t$, which leads to a contradiction. Notice that $m\ge p$ forces $p=2$, but if $n\ge 6$ then $m=\frac n2\ge 3$. This leaves us with $k=4$, which is not a solution because $1^0\equiv 3^2\pmod 4$.

So $k$ is squarefree. Suppose that there is a prime $p\ge 5$ that divides $k$, set $k=pk_1$, and consider the following set of $2k_1+1$ numbers:
\[B_k=\left\{n^{n-1}:\, n=1,p+1,\dots,(k_1-1)p+1,\quad p-1,2p-1,\dots,k_1p-1,\quad \frac{p+1}2\right\}.\]
($B_k$ indeed has $2k_1+1$ different elements: the first $2k_1$ are clearly pairwise different, and also $1<\frac{p+1}2<p-1$ if $p>3$.)
Our key observation is that every element of $B_k\subset S_k$ is $\equiv \pm 1\pmod p$. This is obviously true for the first $2k_1$ elements, moreover putting $x=\left(\frac{p+1}2\right)^{(p-1)/2}$, we have $x^2\equiv 1\pmod p$ by Fermat's Little Theorem, hence $p|x^2-1=(x-1)(x+1)\implies x\equiv \pm1\pmod p$. But if $S_k$ were a complete residue system modulo $k$, then it would contain exactly $2\cdot \frac kp=2k_1$ elements that are $\pm 1$ modulo $p$, contradiction.

This leaves us with squarefree $k$ divisible with no prime $p\ge 5$, these are $2$, $3$ and $6$. It is easy to check that $2$ and $3$ works, but $k=6$ is no solution due to $5^4\equiv (-1)^4=1^0\pmod 6$. Therefore the solutions: $k=2$ and $k=3$.[/hide]

[quote=Ahwingsecretag]Otherwise $ k$ is prime and greater than $ 3$ so we can choose $ n=k-1,m=\frac{k+1}{2}$ or $ n=1,m=\frac{k+1}{2}$, and we are done by FLT.[/quote]

can anyone explain this please. thanks


Hi,

Let $ A,B,C,D$ and $ E$ be five points in space. Assume that : $ AB\equal{}BC\equal{}CD\equal{}DE\equal{}EA$ and $ \angle EAB\equal{}
\angle ABC\equal{}\angle BCD\equal{}\angle CDE\equal{}\angle DEA$. Prove that these five points are coplanar.

do you have some ideas how to attack this hard problem ? :help:

Thanks


$x^2+y^2+z^2=3,x\geq0,y\geq0,z\geq0.$
Prove that $8+xyz\geq3(x+y+z).$

[quote="barra"]It seem for me that the inequality is so thick. Am I wrong somewhere?
$(x+y+z)^2=1+2(xy+xz+yz)\leq 3$ So $3(x+y+z)\leq 3\sqrt{3}<8$[/quote]
Sorry, $x^2+y^2+z^2=3.$ Thank you.

We have:
$ 3(x\plus{}y\plus{}z)\minus{}xyz\equal{}3(x\plus{}y)\plus{}z(3\minus{}xy) \leq \sqrt{(x^2\plus{}y^2\plus{}2xy\plus{}z^2)(9\plus{}9\minus{}6xy\plus{}x^2y^2)} \equal{}\sqrt{(3\plus{}2xy)(18\minus{}6xy\plus{}x^2y^2)}$
We need prove that
$ (3\plus{}2xy)(18\minus{}6xy\plus{}x^2y^2) \leq 64$
I think this ineq is true!

[quote="chien than"]
We need prove that
$ (3 \plus{} 2xy)(18 \minus{} 6xy \plus{} x^2y^2) \leq 64$
I think this ineq is true![/quote]
Try $ xy\equal{}1.$ :wink:


$26659 .$ Pentru $x \in \mathbb{R}$ fie $f(x) = \sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}$ . Calculati partea intreaga din $f(x)$ .

                                    [b] Romanta Ghita si Ioan Ghita , Blaj , Alba[/b]

eu cred ca e $0$, se foloseste: $\left\lfloor{x}\right\rfloor-\left\lfloor{y}\right\rfloor-1\leq \left\lfloor{x-y}\right\rfloor\leq \left\lfloor{x}\right\rfloor-\left\lfloor{y}\right\rfloor$

trebuie sa ai prea mare rabdare sa postezi solutia completa aici, ca e lung semnu ala pt partea intreaga  :wink:

Faza finală 1978, problema 1:
Să se afle mulţimea valorilor posibile ale expresiei $\sqrt{a^{2}+a+1}-\sqrt{a^{2}-a+1}$ dacă $a\in \mathbb{R}$, propusă de Sorin Rădulescu. Răspuns: intervalul $(-1,1).$  :)

a, n-am efectuat calculele pana la capat, dar am vazut ca iese asa pana la urma. am uitat un caz probabil.

da.. :) cam usoara ... luand cateva cazuri particulare intuesti imediat ca este cuprinsa in $(-1,1)$ ... lucru care se poate demonstra muncitoreste .. daca e pana acolo  :) .

Problema nu e chiar uşoară...şi există şi soluţii ingenioase, de exemplu folosind o interpretare geometrică  :)

Cred ca am facut demonstratia asta la clasa.

Fie $A \left( x-\frac12, 0 \right), \, B \left( x+\frac12, 0 \right), \, C \left( 0, \frac{\sqrt 3}2 \right)$. Mai ramane doar de aplicat inegalitatea triunghiului
($AC+AB \geq BC, \, BC+AB \geq AC$) si de verificat faptul ca egalitatea nu se atinge.


linear algebra
matrix
combinatorics solved
combinatorics

Let X be a set ,|X|=n (n>=5) and Y the set of all subsets of X such
that |Y|=n+2 and for all T  \in Y then |T|=3 .Prove that
 exist A,B \in  Y such that |A \cap B|=1

For all A,B in Y we have |A \cap B| =0,1 or 2.  (Can't be 3 because they are all different). Assume that for each A and B, their intersection contains an even number of elements. Let Y = {A1,A2...,A(n+2)} and build a matrix X with n+2 lines and n columns such that on the i-th line there is a 1 on the j,k and l columns if A_i={j,k,l} and 0 otherwise. It's easy to see that in Z_2, X * X^t = I_(n+2). So n+2 = rank(X*X^t) <= rank(X) <= n, a contradiction. Doesn't this work with |Y|=n+1 as well ?


Let $ ABC$ be an acute-angled triangle with $ \overline{BC}\equal{}5$ . Let $ E$ be the foot of height on the side $ \overline{AC}$ and $ F$ the midpoint of side $ \overline{AB}$. If $ \overline{BE}\equal{}\overline{CF}\equal{}4$, calculate the area of triangle $ ABC$.

Portuguese question:
Seja ABC um triângulo acutângulo com BC = 5. Seja E o pé da altura relativa ao lado AC e F o ponto médio do lado AB. Se BE = CF = 4, calcule a área do triângulo ABC.

Anyone ?  :D
OBS: All of the angles of this triangle are littler than 90 degrees.

Applying STEWART in triangles ABC  for height and median we find the area as 8*3^.5-6

Another solution :
paralell to BE through F cut AC on G.So FG=2 and CG=12^.5 and AG=12^.5-3
Area triangle AFG=12^.5-3
Area triangle FGC=12^.5
Area triangle AFC=2*12^.5-3
Area triangle ABC=2*AFC=4*12^.5-6=8*^.5-6

The answer correct is $ 8\sqrt{3}\minus{}6$

$ 8\sqrt{3}\minus{}6$ = 8*^.5-6 ?


algorithm
modular arithmetic
number theory unsolved
number theory

Yes.  For example, sequence [url=http://www.research.att.com/~njas/sequences/A019444]A019444[/url] in the OEIS.  (This is one time when the greedy algorithm actually works nicely.)  See also the following posts:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1642[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30381[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84828[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158438[/url]

[quote="SUPERMAN2"]Does there exist an infinite sequence of positive integers,containing every positive integer exactly once,such that the sum of the $ n$ first term is divisible by $ n$ for every $ n$ ?[/quote]
I found this problem very nice and did not look at the references given by nsato.

[hide="My solution"]
We can build such a sequence $ a_n$ with the following algorithm :
Let $ S_n \equal{} \sum_{k \equal{} 1}^na_k$
Let $ p_n \equal{} \min\{x\in\mathbb N$ such that $ x\neq a_k$ $ \forall k\in[1,n]\}$
Let $ a_1 \equal{} 1$, $ S_1 \equal{} 1$ and $ p_1 \equal{} 2$

Consider that the sequence is built up to n. Then :
If $ S_n \plus{} p_n \equal{} 0\pmod {n \plus{} 1}$ then take $ a_{n \plus{} 1} \equal{} p_n$. As a consequence $ p_{n \plus{} 1} > p_n$
If $ S_n \plus{} p_n\neq 0\pmod {n \plus{} 1}$ :
Choose an integer $ k$ such that $ A \equal{} k(n \plus{} 2)(n \plus{} 1) \minus{} S_n \plus{} p_n(n \plus{} 1)$ be a positive integer not yet used. (it's obviously always possible).

Let $ a_{n \plus{} 1} \equal{} A$ and $ a_{n \plus{} 2} \equal{} p_n$

$ S_{n \plus{} 1} \equal{} S_n \plus{} A \equal{} k(n \plus{} 2)(n \plus{} 1) \plus{} p_n(n \plus{} 1) \equal{} 0\pmod {n \plus{} 1}$
$ S_{n \plus{} 2} \equal{} S_{n \plus{} 1} \plus{} p_n \equal{} k(n \plus{} 2)(n \plus{} 1) \plus{} p_n(n \plus{} 2) \equal{} 0\pmod {n \plus{} 2}$

As a consequence $ p_{n \plus{} 1} \equal{} p_n$ but $ p_{n \plus{} 2} > p_n$

And so the sequence $ p_{2k}$ is strictly increasing and so all integers will be exactly once in the sequence $ a_i$
Q.E.D.

[/hide]

Thanks for these nice problems :)


geometry
geometric transformation
reflection
inequalities
trigonometry
rotation
function

Given a unit circle O with points A, B, and C on O equidistant from each other.  A point P is chosen inside ABC.  If a = AP, b = BP, and c = CP, prove that a triangle can be formed with sides a, b, and c.  Further prove that the area of this triangle is dependant only on the distance from P to O.

Note: edited to reflect the fact that the radius of circle O is supposed to be 1.  Sorry about that.

The first part of this problem is called Pompeiu's theorem (Pompeiu was a Romanian mathematician). We just use Ptolemy's inequality (it also works for concave quadrilaterals):

PA*BC <= PB*AC+PC*AB, but, since AB=BC=CA, we can simplify this length to get PA <= PB+PC. Equality is reached in Ptolemy iff ABCP is a cyclic quadrilateral, but it's obviously impossible, since P is inside ABC.

Haven't thought abt the second part yet.

For the first part, I just said that each length clearly has to be less than the side of the triangle, whereas the sum of the other two is always greater than the side of the triangle, by the triangle inequality.

The way I did the second part was definitely valid (unless I made some silly mistake) and also definitely much too ugly -- I know there has to have been a more elegant method.

Here's what I did for the second part:

Let OA, OB, OC=R (in our case it's 1). Let D, E, F be the projections of P on sides BC, CA, AB. From sine thm in PEF we have EF=PA*sin60 (and the analogous ones for FD, DE). This means that we have to prove the same statement about the triangles DEF, where D, E, F re projections of some pt P on the sides of the trinagle.

We have PA^2+PB^2+PC^2=3PO^2+3R^2 (I'm always forgetting how this thm is called; maybe it's Leibniz, but I'm not sure; it can be proven by using dot product and the fact that PA+PB+PC=3PO as vectors). This means that DE^2+EF^2+FD^2=(9/4)*(PO^2+R^2), so for constant PO DE^2+EF^2+FD^2 is constant. We use cosine thm in PEF to get EF^2=PE^2+PF^2+PE*PF (and the analogous ones from triangles PFD, PDE). This means that for constant PO the value of 2(PD^2+PE^2+PF^2)+PD*PE+PE*PF+PF*PD is constant. We also know that PD+PE+PF is always constant. By combining these 2 we get the fact that PD*PE+PE*PF+PF*PD is constant for constant PO, but the area of DEF is (PD*PE+PE*PF+PF*PD)*sin(120)/2, so it's constant for constant PO, which is exactly what we want.

I know it looks long, but it really isn't.

I had a nice solution but it's hard to illustrate without a diagram.

For each vertex of the triangle, rotate the triangle and point P 60 degrees clockwise around that vertex to produce a new triangle.  This gives us a bigger triangle with area 4 times that of the original equilateral triangle (call that area T).  But this triangle can also be dissected into regions with total area 2T, three equilateral triangles whose side lengths are PA, PB, and PC respectively, and three copies of a triangle with sides PA, PB, and PC.  From that you know that the area of the desired triangle is a function of PA^2+PB^2+PC^2 which in turn depends only on PO.

Apart from minor config issues it's pretty nice.

This is indeed very nice. This is what I was looking for: a beautiful proof involving nice geometric constructions, but I got impatient and I only came up with what I have already posted :) .


HI PEOPLE! My name is Sara. I am 14 years old, from the Chicago suburbs in Illinois... I can't wait for the 13th!! =DD

Hi Sara!  I am looking forward to seeing you at the camp.  We have a big group of people this year.  We will have a lot of fun!

AH! I can't wait! Only 1 more day till SUNDAY! YAY!!! =DDD

Hi Sara! It's Stephanie...if you can remember. I bet you had a great time, cause I did! YAY!!

i remember you! you sat right next to anna! I asked where you got your b-ball shorts from!


What is the absolute value of the difference between the squares of 101 and 99?


How many natural numbers are between $ \sqrt{10}$ and $ \sqrt{90}$?

The ceiling of $ \sqrt{10}$ is 4, and the floor of $ \sqrt{90}$ is 9. Thus, there are 9-4+1=6 natural numbers in between $ \sqrt{10}$ and $ \sqrt{90}$.

[quote=LiBoy]Yes, everyone in my FTW game got 6.
Is this a glitch?[/quote]

[quote=isabella2296]Strangely, on the review, it said the answer was $ 7$. Perhaps a bug?[/quote]

$\text{I think they might've fixed it. It says }6\text{ on FTW.}$


Actually, it is not an olympiad problem. Does anybody know proper section for such problems?

The "Other Problem Solving Topics" forum in the High School section?

[quote="mustyoldjoe"]
4, 7, 6, 2, 3, 7, 2, ?, ?.[/quote]
differences: 3, 1, 4, 1, 4, 5, ?, ?
3+1=4, 1+4=5, then its 4+5=9 because its the two digits counted from the right side of the equation.  This makes the missing differences 4 and 5, with a difference of 9 coming after those two.
now for the signs, its +, -, -, +, +, -, ?, ?
seems to me the next two will be - and then +
so its -4, then +5
making it: 4, 7, 6, 2, 3, 7, 2, -2, 3


How would I go about typesetting the funny looking P used in the statement of problem 5?

Thanks.


Let $ A,B,C \in S_{n}(R)$ such that : [b]AB=BA,BC=CB,CA=AC .[/b]
Slove the system of matrix equations  :
    $ \begin{cases} A^{3}\minus{}9B^{2}\plus{}27B\minus{}27I_{n}\equal{}O_{n}  \\ B^{3}\minus{}9C^{2}\plus{}27C\minus{}27I_{n}\equal{}O_{n} \\ C^{3}\minus{}9A^{2}\plus{}27A\minus{}27I_{n}\equal{}O_{n}
\end {cases}$
 Here $ I_{n}$ is identity matrix .
 $ S_{n}(R)$  are set real symmetric nxn matrices .
 :wink:  :wink:

[hide="Hint"]Reduce to the case that $ A,B,C$ are diagonal.[/hide]


integration
trigonometry
calculus
calculus computations

interesting itegral :?:
\[\int\frac{\sin^{2}x}{(x\cos x-\sin x)}dx \]


probability
function
probability and stats

Good afternoon.

Could anyone please help me out with the following problem?

Suppose $ T: (\Omega_{1}, B_{1})\rightarrow(\Omega_{2}, B_{2})$ is a measurable mapping and X is a random variable on $ \Omega_{1}$.  If $ X\in\sigma(T)/B(R)$, then there exists a random variable Y on $ (\Omega_{2}, B_{2})$ such that $ X(\omega_{1}) \equal{} Y(T(\omega_1))$ for all $ \omega_{1}\in\Omega_{1}$.

I think there is one very natural way to define Y, but the problem is proving Y is a random variable i.e. $ Y\in B_{2}/B(R)$.

Thank you for your efforts.

suppose that $ X$ is a simple function $ X \equal{} \sum \alpha_i 1_{A_i}$. Because $ X$ is $ \sigma(T)$ measurable, $ A_i \equal{} T^{\minus{}1}(F_i)$. Hence $ X(w) \equal{} g(T(w))$ where $ g \equal{} \sum \alpha_i 1_{F_i}$.
Now, for the general case, remember that any function $ X: \Omega \to \mathbb{R}$ is the pointwise limite of simple functions $ g_n$. You should be able to conclude ..

Hi and thank you for your reply.

I am not sure that X(w) = g(T(w)) for all w in $ \Omega_{1}$.

I will assume that your Ai and Fi are disjoint.  What happens when for example

$ w_{1}\in A_{1}$
$ w_{2}\in A_{2}$
$ T(w_{1}) \equal{} T(w_{2})\in F_{1}$

Then $ X(w_{2}) \equal{} \alpha_{2} \neq g(T(w_2)) \equal{} \alpha_{1}.$

Also, this question was posed before the introduction to the approximation by simple functions, and I am not sure I am supposed to use that approach.

Regards.


This is the thread to name an actor or actress or character and describe him or her if you would like.

Spongebob.
That dude is the most amazing actor that ever existed.....Who can say/do things THAT STUPID without cracking up?

Gary the pet snail.

lives with a sponge that talks.

spongebob's best friend patrick

or eric cartman is the best actor


algebra
polynomial
inequalities
number theory proposed
number theory

Find all polynomials $f,g$ with integer coefficients such that any odd prime can be represented either in the form $ f(n)$ (with positive integer n) or in the form $ g(n)$ with positive integer n.  For 3 or 4 or 5 polynomials I think it becomes a little bit more difficult to solve.

If both have degree greater than one, their rate of growing will be to big to reach all primes.

WLOG assume $\deg(f) \geq \deg(g)$:

If $g$ is constant $=c$, $f$ has to reach exactly the primes different from $c$, so $f$ has to be linear, $f(x) = ax+b$. If $|a|>2$, there are always two different prime residue classes $\mod a$ and every of these contains infinity many primes, so we get that $|a|>3$. Simple checking gives now that $b \mod 2$ is necessary when $|a|=2$ (and also sufficient). But now $c$ cannot be an odd prime, since then there are two representations of $c$ ( I hope 'either ... or' is meant exclusive  :?  ).

If $g$ is linear, $g(x)=ax+b$, and $f$ has degree greater than one, by the same arguments as above you get that $f$ doesn't reach any prime, $|a|<3$ and the same restrictions on $b$ again iff $|a|=2$.

If $f,g$ are both linear, $f(x)=ax+b, g(x) =cx+d$, you also have three possibilities:
- only one of them (WLOG $g$) reaches primes, you get again the same coefficients for $g$ as before, and $\gcd (a,b)>1$ and not $prime$
- if (WLOG $f$) reaches exactly one prime $p\neq 2$, you get a contradiction, because also $g$ would reach that prime
- if both reach at least two different primes each, you get that $a=\pm c$ and that $b,d$ represent two different residue classes $\mod a$, and they have to represent all of them, so $\phi (a) = 2$.

Is there a solution not using the strong version of dirichlet's theorem on prime numbers¿

I also hope that all is correct...

EDIT: forget that $n$ has to be positive... well, this gives only some restrictions to the linear polynomials occuring in it (some inequalities on $a,b,c,d$)

One of them must be linear (else they take to little number of values), if one of them is $f(x)=\pm x+K$ or $f(x)=\pm 2x+2K+1$, then other may be arbitrary, if not, the other also must be linear, and they are $\pm 4x+1+4K$ and $\pm 4x+3+4L$ or $p_1x+1+p_1M$ and $p_2x+2+p_2N$, where $p_{1,2}=\pm 3$ or $\pm 6$.

Am I right? I think, it may be done for any number of polynomials by finite time.


What?

The Answer's 1 based on your own wording since only one bag can hold 190 marbles... -__-

[hide="beasted"]Take a bag with one marble, put it inside another bag that has one marble, keep doing it until you have 190 bags.[/hide]

wouldnt it be 19 because you need to have a different number if marbles in each bag, so the n(n+1)/=190 so n must equal 19?

EDIT:never mind, wow that one really forces u to think.

You put one marble in a bag. Then you put another marble AND the first bag in a second bag. And so on.
Not mathematically messy, but it requires creativity.

Yea, that is really cool...
I sort of agree with the NATS answer

[quote="BOGTRO"][quote="ajain"][quote="BOGTRO"]Uh... Duh?

Each bag holds 1 so 190?

Duh?

Duh?[/quote]

But each bag can't have the same number of marbles!!![/quote]

Yeah, so bag #1 goes into Bag #2, Bag #2 goes into Bag #3, etc.[/quote]

There was a similar problem in an old MOEMS Elementary division problem (but with a smaller number). The official answer book assumed the bags weren't put inside each other, but I assume that enough people solved it the "trick" way so as to get it right after a protest.

[quote="nimskat"]19 because since the #of marbles can't be the same, you start with one then two, and so on.
so 1+2+3+4......+19=190

because sum of first n natural numbers is n(n+1)/2
so 19(20)/2
380/2= 190

 :rotfl:
cant  see why you other ppl didn't get it!
 :rotfl:[/quote]

you know everything?

cause i dont know anybody who does.  :|

[quote="nimskat"]19 because since the #of marbles can't be the same, you start with one then two, and so on.
so 1+2+3+4......+19=190

because sum of first n natural numbers is n(n+1)/2
so 19(20)/2
380/2= 190

 :rotfl:
cant  see why you other ppl didn't get it!
 :rotfl:[/quote]
OH SHNAP.
That's wrong.

i agree it's 190 but..
I think I've seen a problem like this on a State/National test...or something like this. was this problem taken directly from a MC test? because I don't believe they would give trick questions like this :maybe:

it was- just read my post higher up on the page.

[quote="Ihatepie"][quote="illusiat"]what is the maximum number of bags that can be used to hold 190 marbles, if each bag must contain at least 1 marble and no bag can contain the same amount of marbles?

ANSWER
[hide]190[/hide][/quote]

this was actually a mathcounts nats problem, where the mathcounts board got 19 as the answer, but the answer became controversial when the answer that some earlier people stated was said. They eventually accepted both answers.[/quote]

Lol, did you get that from the Greatest Mathcounts Problems book?


ComplexZeta and Peter VDD. Look at the moderators for the overall forums; those are also the moderators for the sub-forums under each main forum.

Complex zeta is one and peterdd is another, i dont know of any other "supreme" mods

[quote="jhollenbeck"]i dont know of any other "supreme" mods[/quote]

That would be the Admin.

Actually admin[b]s[/b], which are the yellow-named people, those being rrusczyk, vRusczyk, Valentin Vornicu, MCrawford, DPatrick, and nsato.

so admins are moderators that can go to any forum and moderate :?

admins can do whatever they want. they can moderate all forums, as well, as alter the site, as add a forum, make people moderators, change your username and other stuff.

go to http://www.phpbbforfree.com and sign up, then u be an admin on your forum  :) . Then you can open it up in a frame in your website  like [url=http://4aunlimited.co.nr]mine[/url]


I'm pleased to announce that the Math Kangaroo will now also take place in Toronto and Edmonton. Everybody is of course welcome to participate!

[i]Summary:[/i]
The Math Kangaroo is a game-contest for students in grades 3-12 and will take place on Sunday, March 26, 2006 in the regions around Ottawa, Toronto and Edmonton.
The test consists of 24 multiple-choice questions and the working time is 60 min. The test is divided into 3 parts. A correct answer for part 1 (questions 1-8) is worth 3 points, 4 points for each correct answer in part 2( 9-16) and 5 points for each correct answer in part 3(questions 17-24). A blank is worth 0 points.
There is a penalty of one point for each incorrect answer, but you start off with 24 so the min. score is 0 :D

Some info about the contest can be found on

[url=http://www.math.concordia.ab.ca/~marinova/kangaroo/index.html]Canadian Headquarters[/url]

and

[url=http://www.mathkang.org/]European Headquarters (in French)[/url]

It's hard for me to comment on the difficulty of the problems, since each country is allowed to make changes to a few questions (make them harder) so I think the contest will have some interesting problems. :ninja:

Dimitri Dziabenko

How many of you guys are participating in this.

I remember writing it back in Russia :P
I actually won something in grade 5 and 6. But the thing is pretty easy as far as i remember it.

The problems and their structure are similar to Gauss, Pascal etc. from Waterloo

i think it's much easier than fermat cayley etc

[quote="lena_siz"]@Miu:
I've never taken either Fermat or Cayley. If its from the pascal series, which i tried unofficially about a year later, I really don't think so. I've taken Math Kangaroo this year and barely solidly gone through half the test. So I'm realyl not sure.[/quote]

Hmm...which grade are you in? (I mean, which division did you take for Kangaroo)

I took the gr.12 Kangaroo, and I don't think there are any hard problems in it, compared to Fermat, which has a solid hard question (No.25). That's why I think Kangaroo is easier ^^"

I'm in Gr. 9, and I took gr 9 - 10 division this year for the first time (just found out about it recently). I never actually took Pascal (Cayley, Fermat) officially, just as practice at home. But from those I have done already, I think Pascal requires more knowledge of formulas and special approaches, while Math Kangaroo is more of a time - wise contest which requires creative and very fast thinking. I don't think problems in MK were hard, but just not enough time to think.


calculus
derivative
trigonometry
calculus computations

If $ z\equal{}f(x, y)$, where $ x \equal{} r\cos \theta$ and $ y \equal{} r\sin \theta$, show that

$ \frac{\partial^{2} z}{\partial x^{2}} \plus{} \frac{\partial^{2} z}{\partial y^{2}} \equal{} \frac{\partial^{2} z}{\partial r^{2}} \plus{} \frac{1}{r^{2}} \frac{\partial^{2} z}{\partial \theta^{2}} \plus{} \frac{1}{r}\frac{\partial z}{\partial r}$.

So how can you show that this is true starting from $ \frac{\partial^{2} z}{\partial x^{2}} \plus{} \frac{\partial^{2} z}{\partial y^{2}}$? I think I can do it backwards, but it doesn't seem right.

First write the Chain Rule (it does look backwards, but that's the way things are):
\[ \begin{cases}f_r & \equal{} f_x\cos\theta \plus{} f_y\sin\theta \\
f_{\theta} & \equal{} \minus{} f_x r\sin\theta \plus{} f_y r\cos\theta\end{cases}
\]
Solve this system for the derivatives that you want:
\[ \begin{cases}f_x & \equal{} f_r\cos\theta \minus{} \frac {1}{r}f_{\theta}\sin\theta \\
f_y & \equal{} f_r\sin\theta \plus{} \frac {1}{r}f_{\theta}\cos\theta\end{cases}
\] Rinse and repeat.

OK, so

$ \frac{\partial z}{\partial \theta}\equal{}\minus{}\frac{\partial z}{\partial x}r\sin \theta \plus{} \frac{\partial z}{\partial y}r \cos \theta$

For the next second partial derivative, I got,

$ \frac{\partial^{2} z}{\partial \theta^{2}} \equal{} \frac{\partial (\frac{\partial z}{\partial \theta})}{\partial \theta} \equal{} \frac{\partial (\frac{\partial z}{\partial \theta})}{\partial x}\frac{\partial x}{\partial \theta} \plus{} \frac{\partial (\frac{\partial z}{\partial \theta})}{\partial y}\frac{\partial y}{\partial \theta}.$

Which simplifies to

$ \minus{}\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial \theta}\right) r \sin \theta \plus{} \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial \theta}\right) r \cos \theta.$

But the book got:
$ \frac{\partial^{2} z}{\partial \theta^{2}} \equal{} \frac{\partial (\frac{\partial z}{\partial \theta})}{\partial \theta} \equal{} \frac{\partial (\frac{\partial z}{\partial \theta})}{\partial x}\frac{\partial x}{\partial \theta} \plus{} \frac{\partial (\frac{\partial z}{\partial \theta})}{\partial y}\frac{\partial y}{\partial \theta} \minus{} r\frac{\partial z}{\partial r}.$

Which simplifies to

$ \minus{}\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial \theta}\right) r \sin \theta \plus{} \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial \theta}\right) r \cos \theta \minus{} r\frac{\partial z}{\partial r}.$

What did I do wrong? Why is there a $ \minus{}r\frac{\partial z}{\partial r}$?

Since the formula $ \frac{\partial z}{\partial \theta}$ involves multiplication, you will need the product rule when taking the second derivative.
\[ \frac{\partial^2 z}{\partial \theta^2}\equal{}\minus{}\frac{\partial }{\partial \theta}\left(\frac{\partial z}{\partial x}\right)r\sin \theta \minus{} \frac{\partial z}{\partial x}r\cos \theta \plus{} \dots\]


Let $ a,b,c\in\mathbb{R}$ (not real positive) such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
$ \frac{1\minus{}2ab}{c}\plus{}\frac{1\minus{}2bc}{a}\plus{}\frac{1\minus{}2ca}{b}\geq 7$

[quote="bambang"]Let $ a,b,c\in\mathbb{R}$ (not real positive) such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \frac {1 \minus{} 2ab}{c} \plus{} \frac {1 \minus{} 2bc}{a} \plus{} \frac {1 \minus{} 2ca}{b}\geq 7$[/quote]
Try $ a\rightarrow0^\minus{}.$ :wink:

Sorry I'm wrong :blush: That's real positive

[quote="bambang"]Let $ a,b,c\in\mathbb{R}$ (not real positive) such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \frac {1 \minus{} 2ab}{c} \plus{} \frac {1 \minus{} 2bc}{a} \plus{} \frac {1 \minus{} 2ca}{b}\geq 7$[/quote]
$ \frac {1 \minus{} 2ab}{c} \plus{} \frac {1 \minus{} 2bc}{a} \plus{} \frac {1 \minus{} 2ca}{b} \equal{} \sum \frac {(a \plus{} b \plus{} c)^2 \minus{} 2ab}{c}$
$ \equal{} \sum \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 2c(a \plus{} b)}{c} \equal{} (a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} 4(a \plus{} b \plus{} c) \ge \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c} \plus{} 4 \ge 3(a \plus{} b \plus{} c) \plus{} 4 \equal{} 7$
(Because by Cauchy-Schwarz: $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge \frac {9}{a \plus{} b \plus{} c}$ and $ 3(a^2 \plus{} b^2 \plus{} c^2) \ge (a \plus{} b \plus{} c)^2$ )

________
Editted!
Thanks, bambak!

[quote="NguyenDungTN"]
$ \equal{} \sum \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 2c(a \plus{} b)}{c} \equal{} (a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} 2(a \plus{} b \plus{} c)$[/quote]
it should be $ 4(a\plus{}b\plus{}c)$

Since $ 1 \minus{} 2ab \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2ab > 0$, similarly $ 1 \minus{} 2bc > 0,\ 1 \minus{} 2ca > 0$, we have

$ \frac {1 \minus{} 2ab}{c} \plus{} \frac {1 \minus{} 2bc}{a} \plus{} \frac {1 \minus{} 2ca}{b}\geq 7$

$ \Longleftrightarrow ab(1 \minus{} 2ab) \plus{} bc(1 \minus{} 2bc) \plus{} ca(1 \minus{} 2ca)\geq 7abc$

$ \Longleftrightarrow ab \plus{} bc \plus{} ca \minus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\geq 7abc$

$ \Longleftrightarrow ab \plus{} bc \plus{} ca \minus{} 2\{(ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(a \plus{} b \plus{} c)\}\geq 7abc$

$ \Longleftrightarrow ab \plus{} bc \plus{} ca \minus{} 2(ab \plus{} bc \plus{} ca)^2\geq 3abc.$

This inequality can be proved as follows for $ a \plus{} b \plus{} c \equal{} 1,\ a > 0,\ b > 0,\ c > 0.$

Now $ (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ca)\Longleftrightarrow 1\geq 3(ab \plus{} bc \plus{} ca) > 0$, yielding $ 0 < ab \plus{} bc \plus{} ca\leq \frac {1}{3}.$ therefore

$ ab \plus{} bc \plus{} ca \minus{} 2(ab \plus{} bc \plus{} ca)^2\geq \frac {1}{3}(ab \plus{} bc \plus{} ca)\geq \frac {1}{3}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\geq 3abc$ by A.M.-G.M. , which completes the proof.


algebra
polynomial
number theory unsolved
number theory

for every integer x polynomial P(x) with second degree  is a square of a natural number. Prove that P(x) is a square of another polynomial.

 :oops:


The incircle of a triangle ABC touches BC at D and AC at E. Let K be a point on CB with CK=3.BD and L be a point on AC with AE=CL. Lines AK and BL meets at P. If Q is the midpoint of BC, I is the incenter and G is the centroid of ABC prove that:

a)IQ and AK are parallel.

b)the triangles AIG and QPG have equal area.

It must be "$ CK\equal{}BD$"
(a) Let $ ID$ cut $ (I)$ at $ N$. This result is well-known: $ A,N,K$ are collinear.
Then $ IQ//AK$ because $ IQ$ is the midline of triangle $ NDK$.
(b) Let IG extended twice its length to the point $ P'$. We will show that $ P'$ is the [b]Nagel[/b] point of triangle $ ABC$.
By [b]Thales[/b]'s theorem: $ AP'//IQ$ then applying result $ (a)$, $ P'$ lies on $ AK$.
Similarly, we can prove that $ P'$ lies on $ BL$ and $ P'\equiv P$
$ S_{AIG}\equal{}2S_{IGQ}\equal{}S_{GPQ}$


modular arithmetic
quadratics
number theory unsolved
number theory

From INTERMEDIATE Forum:

Prove that $n^3+23$ can never be a perfect square.

The folks in the forum were not getting anywhere, and after several brief failed attempts (trying residues (mod 4)) a friend suggested posting this in here.

This may be trivial, if I am overlooking the obvious  :blush:

I think you are referring to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20032]this[/url] topic.  If you check out the message right before the last one, you'll see a really big hint.

The first part is a $\pmod 4$ approach: if $n$ is even, then that number is $3\pmod 4$, and if $n$ is $3\pmod 4$, then the number is $2\pmod 4$, and these can't be perfect squares. This means that $n\equiv 1\pmod 4$.

We now have, as blahblahblah said, $y^2+4=(n+3)(n^2-3n+9)$ (actually, he wrote $(n+3)(n^2-3n+27)$, but that's a typo). Since $n\equiv 1\pmod 4$, we find that $n^2-3n+9\equiv 3\pmod 4$, so $n^2-3n+9$ has a prime factor of the form $4k+3$, which must then also divide $y^2+4$. However, it's well-known that if a prime of the form $4k+3$ divides a sum of squares, then it divides both squares, meanind that our prime divides $2$, giving us a contradiction.

Why does a prime of the form 4k+3 divide both squares in a sum of squares?

If we have $p|a^2+b^2$ and assume that $a,b\not\equiv 0\pmod p$, then $\left(\frac {a^2} p\right)=-\left(\frac {b^2} p\right)$, so $-1$ is a quadratic residue $\pmod p$. This is impossible if $p=4k+3$ because $\left(\frac {-1} p\right)\equiv (-1)^{\frac{p-1}2}\equiv -1\pmod p$.

[$\left(\frac ap\right)$ is the Legendre symbol]


It is given triangle ABC with points M, N, P on its sides (internally). U, V, W are intersection points of AN and CM, BP and AN, CM and BP. Triangle UVW is equaliteral. If AM=BN=CP is it true that ABC is equaliteral?

(The problem is edited because of valuable vittasko's remark (b) was too easy. Thank you, vittasko!).

P.S. It is a new problem composed by me today. I believe you will like it, especially vittasko.

I've an idea for solution. But I'm not good enough with the trygonometry.
1. Denote:
a) angles AMC, BNA, CPB with M, N, P respectively;
b) AM=BN=CP=x;
c) UV=VW=WU=y;.
2. Look at the triangles AMU, BNV and BMW - using Sine Law 3 times and fact that sum in angles in angles in triangle is 180 degrees we get a relation between x, y, M, N.
3. By similar way we may get a relations between x, y, N, P and x, y, P, M.
4. By excluding x, y from these relations we will get an equation only for M, N, P and I suppose it will gave as with some beautiful application of the trygonometry that M=N=P ... and from here it is easy.

I've no enough skills in trygonometry and please if someone with such skills to write the right steps for the solution or only hints.

Comment:
If I'm right it is very probably to have lots of interesting problems in the following that every math competition may be proud with and with differences in their solutions that are not routine:
Suppose M, N, P are points on AB, BC, CA respectively (internally). U, V, W are intersection points of CM and AN; AN and BP; BP and CM; respectively. I think that if we have two triples
a) segment1=segment2=sement3 and segment11=segment22=sement33;
b) segment1=segment2=sement3 and angle1=angle2=angle3;
c) angle1=angle2=angle3 and angle11=angle22=angle33;
(In all triple we have "symmetric" angles/segments). I will not ask you to solve all such problems but I'll be very happy if you compose such a problem, find a correct solution and it appears in some competition or magazine who knows they may even appears at IMO...


Princeton college Harvard Yale Stanford MIT Columbia

What are #admitted and #applied for each 50 states within the United States for Princeton, Harvard, Yale, Stanford, MIT, CalTech, etc for early processes (decision and action)

I had hard time locating this information; is it currently available anywhere?

Sorry I don't have such information. How are you going to use it anyway?  :maybe:

PTynan89 how can you apply to more than one university EA? isn't that illegal?
BTW i applied to Princeton and i got deferred :( Can anyone tell me what counts most in the process :maybe: I had good (not excellent) scores and excellent academic achievements...
I would really appreciate if someone could give me some hints (i am an international student) :)

[quote="ciprian"]how can you apply to more than one university EA?[/quote]

That depends on the rules of the most restrictive university to which you apply. For example, MIT and Caltech both allow simultaneous early action applications to their college and to other colleges, so many math-liking young people submit early action ("EA") applications to both MIT and Caltech. But Harvard (this year), Stanford, and Yale allow only an early action application to their own college, not to any other college at the same time, a system that is called "restrictive early action" or "single-choice early action" ("SCEA"). So if you decide to apply early to Stanford you can't apply early to MIT at the same time--by Stanford's rules, not by MIT's.

A different system is early decision ("ED"), in which the applicant promises to enroll, for sure, at the college that admits him in the early round. It would be inconsistent to apply to more than one college on that basis.

Next year, Harvard will drop SCEA and Princeton will drop ED and both of those colleges will have a single admission deadline for all applicants, at the regular action time of year.

[quote="PTynan89"]Well, I'll tell you about my school, since DE is a very small state, and we have probably over 50% of the ivy acceptances alone ach year.

3 other people and I applied to MIT EA, all deferred.
I applied to Caltech EA as well, and was deferred again.
Harvard: 2 ppl, both deferred (although I heard that 2 ppl outside of my school got in EA)
Princeton: 1 that I know of, and he got deferred
Yale: 2 applied early, not sure but I think both got deferred (IMO one was borderline good enough to get in, then other borderline rejection)
Stanford: 1 EA that I know of, not sure of the result (had a feeling she'd get in, but this year just seems really negative overall, so idk...)

The only acceptances to an ivy were 1 girl to Columbia, (was black, if that's important to you) and 2 to UPenn.

So no males EA/ED at all to any of those schools or any of the other ivies, which was rather strange. (the 2 outside of my school that got into harvard were both girls, btw, and also, all of the unsure, possibly got in were girls)

SIDE NOTE: This was not a sexist post, and I am in no way trying to say that person X got in because she was a female.

Last year it was pretty evenly divided for EA/ED between genders, so I was merely noting the apparent peculiarity this year.

As for the other 49 states... sorry[/quote]

Oh yeah, that girl [i]did[/i] get into Stanford btw.

So there were 5 ivy acceptances and 1 to Stanford so far in all of Delaware (and I'd be willing to bet money there aren't more than about 2 additional ones at most)

Here's a brief overview of the MIT Class of 2011 EA:
http://jzzsxm.googlepages.com/MITData.xls

Compiled by a friend of mine who got in, so thank him, not me.


One square foot of farm land requires four seeds. At this rate, how many seeds will be required to plant ten square yards?

[hide]10 square yards is 90 square feet, or 360 seeds.[/hide]

[hide]
10 square yards is 90 square feet,
So the answer is 90*4=360 seeds[/hide]


Given a $ 4\times 4$ squares table. How many ways that we can fill the table with $ \{0,1\}$ such that two neighbor squares (have one common side) have product which is equal to $ 0$?

Here's the boring way to do it:

Let $ a_n$ be the number of "good" ways to fill a $ 4 \times n$ table so that the leftmost edge (of length 4) is filled with all 1s.  Let $ b_n$ be the number of good ways if the leftmost edge has a 1 in the top corner.  Let $ c_n$ be the number of good ways with the leftmost edge having exactly one 1, in the second position (adjacent to the top corner).  Let $ d_n$ be the number of ways with the leftmost edge having two 1s, one of which is in the top corner and the other of which is in the third position, and let $ e_n$ be the number of ways if the leftmost edge has two 1s, one in each corner.  Then $ a_1 \equal{} b_1 \equal{} c_1 \equal{} d_1 \equal{} e_1 \equal{} 1$, and for $ n \geq 1$ we can express each $ x_{n \plus{} 1}$ as some combination of the $ x_n$.  Then it's just a matter of making a couple dozen additions to reach the answer, which is 1234.

Can you provide more detail in expressing the $ x_i$'s?  :maybe:

For example, $ c_{n \plus{} 1} \equal{} a_n \plus{} 2b_n \plus{} c_n \plus{} d_n \plus{} e_n$ -- it's possible to fill the $ n \plus{} 1$th column as $ (0, 1, 0, 0)$ if and only if the $ n$th column is filled with $ (0, 0, 0, 0)$ (counted by $ a_n$, despite a typo in my previous post), $ (1, 0, 0, 0)$ (counted by $ b_n$), $ (0, 0, 1, 0)$ (counted by $ c_n$), etc.  Similarly $ a_{n \plus{} 1} \equal{} a_n \plus{} 2b_n \plus{} 2c_n \plus{} 2d_n \plus{} e_n$, and similar formulas for each other sequence.

The resulting sequence is [url=http://www.research.att.com/~njas/sequences/A051737]A051737[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  In addition, $ 1 \times n$ matrices with these properties give Fibonacci numbers, $ 2 \times n$ matrices give powers of 3, and $ 3 \times n$ matrices give a modestly simpler recursion than the one above (sequence [url=http://www.research.att.com/~njas/sequences/A051736]A051736[/url]).


let a,b,c  positive real.
prove that
 $ \sqrt{\frac{a}{b\plus{}c}} \plus{}\sqrt{\frac{b}{c\plus{}a}} \plus{}\sqrt{\frac{c}{a\plus{}b}} >2$
  i prove :
$ f(x,y,z) > f(x,y\plus{}z/2,z\plus{}y/2$but very long

It's very old problem! I solve it by using cauchy conversely: square root a.(b+c)<=a+b+c/2. And you will  easily get the result by doing similar to the others!

[quote="Hong Quy"]let a,b,c  positive real.
prove that
 $ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} > 2$
  i prove :
$ f(x,y,z) > f(x,y \plus{} z/2,z \plus{} y/2$but very long[/quote]
$ \sum_{cyc} \sqrt {\frac {a}{b \plus{} c}} \equal{} \frac {a}{\sqrt {a(b \plus{} c)}} > \sum_{cyc} \frac {2a}{a \plus{} b \plus{} c} \equal{} 2$

Let $ a;b;c$ be positive real numbers.Prove that:
$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}} \geq \sqrt{4\plus{}\frac{9(a\minus{}b)^2}{4(a\plus{}b\plus{}c)^2}}$

[quote="chien than"]Let $ a;b;c$ be positive real numbers.Prove that:
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \geq \sqrt {4 \plus{} \frac {9(a \minus{} b)^2}{4(a \plus{} b \plus{} c)^2}}$[/quote]
Try $ b\equal{}c\equal{}1$ and $ a\rightarrow0^\plus{}.$ :wink:


If $p$ is prime , show that we can split all the numbers $2$ through $p-2$ into pairs such that the product of the two members of each pair is congruent to $1$ mod $p$.

And with this lemma, we easily come up with  Wilson's theorem:
$(p-1)!\equiv-1 \mod p$ for any prime $p$.

[quote="rjt"][hide]
First note that if $ab \equiv ac \mod p$, $a,b,c<p$, then $b=c$. As: $ab=kp+1, ac=k'p+1$
$a(b-c)=p(k-k')$
So $p$ divides $b-c$, but this is impossible, so both sides must equal $0$, and $b=c$.

Now, $(1)(1)=1$, and $(-1)(-1)=1$, so there exists no number such that $(-1)(a) \equiv 1 \mod p$ or $(1)(a) \equiv \mod p$, $a \neq-1,1$.

Now, by Euclid's Algorithm, there exists $n$ such that $mn \equiv 1 \mod p$, for every $m$, where $(m,p)=1$. But notice that $(q-p)m+(r+m)p=1$. We can continue to subtract $p$ from the $m$ coefficient and add $m$ to the $p$ coefficient until the coefficient of $m$ is less than $p$. So, for every $a$, we can find a number $b$ such that $ab \equiv 1 \mod p$, where $a,b<p$. Also $b \neq-1,1$ if $a \neq-1,1$, and each pair is unique. Hence we can divide up all of the numbers between $2,p-2$ into such pairs.
[/hide][/quote]
In other words $Z_{p}$ is a field, polinom $x^{2}-1$ has two evident roots but can't have any more because of degree. Hence there is no element other than $\pm 1$ which inverse to itself. Then trivialy we have Wilson's theorem.

By the way [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Wilson_John.html.]Wilson[/url] knew this result from his teacher [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Waring.html]Waring[/url] who published it without proof but it was known to [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Leibniz.html]Leibnitz[/url] about 100 years before. The 1st proof was published by [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Lagrange.html]Lagrange[/url] 10 years after Waring.
It was Euler who had attributed the result to Wilson. It is known that  Euler disliked Leibnitz but perhaps he disliked Lagrange too?


geometry
3D geometry
modular arithmetic
number theory unsolved
number theory

Find all primes p in order to have integers m and n satisfying the conditions.
$ p \equal{} m^{2}\plus{}n^{2}$ e $ p|(m^{3}\plus{}n^{3}\minus{}4)$.

Posted before $ p\equal{}2$ or $ p\equal{}5$.

[quote="danilorj"]I didn't find the topic where it has been posted. And p=13 also satisfies the conditions.[/quote]
What is $ m,n$ for $ p\equal{}13$?
Solution $ 2m^{3}\plus{}2n^{3}\minus{}8\equal{}3(m\plus{}n)(m^{2}\plus{}n^{2})\minus{}(m\plus{}n)^{3}\minus{}8$, therefore $ p|(m\plus{}n)^{3}\plus{}8\equal{}(m\plus{}n\plus{}2)(m^{2}\plus{}n^{2}\plus{}2mn\minus{}2m\minus{}2n\plus{}4)$
If $ p\equal{}m^{2}\plus{}n^{2}>5$, then $ p>m\plus{}n\plus{}2\longrightarrow p|(mn\minus{}m\minus{}n\plus{}2)$, but in these case $ |(m\minus{}1)(n\minus{}1)\plus{}1|<m^{2}\plus{}n^{2}\equal{}p$.
Therefore only $ p\equal{}2,p\equal{}5$ works.

[quote="danilorj"]why $ p|((m\plus{}n)^{3}\plus{}8)$ ?[/quote]

As $ 0\equiv 2m^{3}\plus{}2n^{3}\minus{}8\equal{}3(m\plus{}n)(m^{2}\plus{}n^{2})\minus{}(m\plus{}n)^{3}\minus{}8\equal{}3(m\plus{}n)p\minus{}(m\plus{}n)^{3}\minus{}8$ $ \equiv (m\plus{}n)^{3}\plus{}8\pmod p$.

if you set $ m\equal{}\minus{}2$ and $ n\equal{}\minus{}3$ the condition is satisfied. Hence p=13.


At a new job, Bob's starting monthly salary is $ \$1000$, and he gets a $ \$20$ raise each succeeding month. In total, how much does Bob earn in his first year at this job?

He gets a base of $ \$1000$ for each of the $ 12$ months in that year,

and a total of $ \$0 \plus{} \$20 \plus{} \$40 \plus{} \cdots \plus{} \$220 \equal{} \$20(1 \plus{} 2 \plus{} \cdots \plus{} 11)$

$ \equal{} \$20\cdot11\cdot12\cdot\frac12 \equal{} \$1320$

for a final answer of $ \$1000\cdot12 \plus{} \$1320 \equal{} \boxed{\$13320}$.


[url]http://www.nytimes.com/2008/10/10/education/10math.html[/url]

Many of the observations have been true in my experience as well.

Thoughts?

Hmm, I agree with most of the points here, there definetely is a social stigma problem, the question is what we can do about it.  Unfortunately, I have a feeling it's more likely our negative view of math spreads to Eastern Europe and Asian than for a positive view to begin/come here.

Edit: I can't fix the link, that <a> keeps showing up, argh

[quote="worthawholebean"]http://www.nytimes.com/2008/10/10/education/10math.html/

Many of the observations have been true in my experience as well.

Thoughts?[/quote]

[b]Lesson[/b]: Don't use url tags when you are directly referencing a website. :)

Official word on the bug:
[quote="Valentin Vornicu"]Ok, clearly I've created that by fixing the other stuff. I'll look into it Monday. The code is good, it's just displayed badly in the post.[/quote]


conics
ellipse
trigonometry
analytic geometry
graphing lines
slope
ratio

Here are some fairly difficult ellipse qns for people to practice on. If posting solutions please use the hide function. If mistake is found within the question the post reply with correction.

[b]1.[/b] $ P(a\cos\theta, b\sin\theta\)$  lie on the ellipse $ \frac{x^{2}}{a^{2}}$+ $ \frac{y^{2}}{b^{2}}$ $ = 1$. The tangent at P cuts the x-axis at A and the y-axis at B. T and T' are the feet of the perpendiculars from the foci S and S' respectively to this tangent. X and Y are the feet of the perpendiculars From P to the x- and y-axis respectively
 Show that
[b](a)[/b] $ OX.OA=a^{2}$ [b](b)[/b]$ OY.OB=b^{2}$ [b](c)[/b]$ ST.S'T'=b^{2}$

[b]3.[/b] $ P(a\cos\theta, b\sin\theta)$  lie on the ellipse $ \frac{x^{2}}{a^{2}}$+ $ \frac{y^{2}}{b^{2}}$ $ = 1$. The normal at P cuts the x-axis at G, and N is the foot of the perpendicular from P to the x-axis. Show that
[b](a)[/b]$ OG=e^{2}ON$ [b](b)[/b]$ SG=eSP$ and $ S'G=eS'P$

[b]4.[/b] $ P(a\cos\theta, b\sin\theta)$  lies at an extremity of a latus rectum through one focus S of the ellipse $ \frac{x^{2}}{a^{2}}$+ $ \frac{y^{2}}{b^{2}}$ $ = 1$. The tangent at P cuts the tangent at A(a,0) at R. Show that OR is $ //$ to A'P, where A' is the point (-a,0)

[b]5.[/b]Find the equations and the coordinates of the points of contact of the tangents to $ x^{2}+2y^{2}= 19$which are // to $ x+6y=5$

[b]6.[/b]Find the equations and the coordinates of the points of contact of the tangents to $ 8x^{2}+3y^{2}=35$ from $ (\frac{5}{4},5)$

[b]7.[/b]$ P(a\cos\theta, b\sin\theta)$ and $ Q(a\cos\phi, b\sin\phi)$  lie on the ellipse $ \frac{x^{2}}{a^{2}}$+ $ \frac{y^{2}}{b^{2}}$ $ = 1$. Find the equation of the chord PQ. Hence show that PQ subtends a right angle at the point A(a,0) the PQ passes through a fixed point on the x-axis.

[b]8.[/b]$ P(a\cos\theta, b\sin\theta)$ and lie on the ellipse $ \frac{x^{2}}{a^{2}}$+ $ \frac{y^{2}}{b^{2}}$ $ = 1$ with focis S and S'. The normal at $ P$ meets $ SS'$ at $ G$. Show that $ PG^{2}$ = $ (1-e^{2})PS.PS'$

1a) We can consider a horizontal shrink/stretch that maps the ellipse into a circle. We note that this preserves length ratios....i.e. we are considering the ratio $ \frac{OX}{OV}*\frac{OA}{OV}$ where $ V$ is a vertex. Anyway, in our transformed diagram, we have $ OP^{2}=OX*OA$ since we have that $ \angle OPA=\angle OXP=\frac{\pi}{2}$.

b) This is the same as a)


$ (x^2\plus{}y^2\plus{}z^2)^2 \ge 3(x^3 y\plus{}y^3 z\plus{}z^3 x)$

I know this inequality is very well known but I can't solve it..
Please give me the answer as fast as you can.

thanks :lol:

[quote="limsk1"]$ (x^2 \plus{} y^2 \plus{} z^2)^2 \ge 3(x^3 y \plus{} y^3 z \plus{} z^3 x)$

I know this inequality is very well known but I can't solve it..
Please give me the answer as fast as you can.

thanks :lol:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6026


inequalities
trigonometry
inequalities proposed

Let $ ABC$ be an acute-angled triangle.
Let $ m_a$ be the length of the median of triangle $ ABC$ from the vertex $ A$. Similarly define $ m_b$ and $ m_c$.
Let $ r_a$ be the exradius of triangle $ ABC$ opposite to the vertex $ A$. Similarly define $ r_b$ and $ r_c$.
Let $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$ be the semiperimeter of triangle $ ABC$.
Prove that
$ m_ar_a \plus{} m_br_b \plus{} m_cr_c \leq p^2$.

I have a solution...  I think it can be simplified because i think the solution does not really make use of the substitution.
First let \[
b^2  + c^2  - a^2  = x,c^2  + a^2  - b^2  = y,a^2  + b^2  - c^2  = z
\], then
\[
a = \sqrt {\frac{{y + z}}{2}} ,b = \sqrt {\frac{{x + z}}{2}} ,c = \sqrt {\frac{{x + y}}{2}}
\]
Then it is equivalent to prove:
\[
\begin{array}{l}
 \sum\limits_{cyclic} {\sqrt {4x + y + z} \sqrt {\frac{{\sqrt {(x + y)(x + z)}  - x}}{{\sqrt {(x + y)(x + z)}  + x}}} }  \le \sqrt {x + y}  + \sqrt {y + z}  + \sqrt {z + x}  \\
  \Leftrightarrow \sum\limits_{cyclic} {\sqrt {4x + y + z} \sqrt {1 - \frac{{2x}}{{\sqrt {(x + y)(x + z)}  + x}}} }  \le \sqrt {x + y}  + \sqrt {y + z}  + \sqrt {z + x}  \\
 \end{array}
\]
By am-gm,
\[
\begin{array}{l}
 \sqrt {4x + y + z} \sqrt {1 - \frac{{2x}}{{\sqrt {(x + y)(x + z)}  + x}}}  \\
  \le \sqrt {4x + y + z} \sqrt {1 - \frac{{2x}}{{\frac{{x + y + x + z}}{2} + x}}}  \\
  = \sqrt {4x + y + z} \sqrt {\frac{{y + z}}{{4x + y + z}}}  \\
  = \sqrt {y + z}  \\
 \end{array}
\]
so it is true.

It is easy to prove that
ra=p*tan(A/2);
ma=sqrt(p(p-a));
tan(A/2)=sqrt((p-b)(p-c)/(p(p-a)));
So that,we have
p*(sqrt((p-b)(p-c))+sqrt((p-b)(p-c))+sqrt((p-c)(p-a))<=p*p.
If x=sqrt(p-a);y=sqrt(p-b);z=sqrt(p-c) then we have
p(xy+yz+zx)<=p*(x*x+y*y+z*z).
So the inequality is
ma*ra+mb*rb+mc*rc<=p*p

Example:
Take the right-angled triangle with side 3,4,5. We have
ma=2.5
ra=6
mb=sqrt(13)
rb=3
mc=sqrt(18.25)
rc=2
and p=6
and thus
ma*ra+mb*rb+mc*rc <=p*p

[quote="phuchung"]ma=sqrt(p(p-a));[/quote]..You are wrong.
Why $ m_a \equal{} \sqrt {p(p \minus{} a)},m_a\ge \sqrt{p(p \minus{} a)}$

I have just found an almost geometric proof of the inequality.

[color=blue][b]Initial problem.[/b] For any acute triangle $ ABC$ with medians $ m_a$, $ m_b$, $ m_c$, exradii $ r_a$, $ r_b$, $ r_c$ and semiperimeter $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$, prove the inequality $ m_a\cdot r_a \plus{} m_b\cdot r_b \plus{} m_c\cdot r_c \leq p^2$.[/color]

I rename $ p$ into $ s$:

[color=blue][b]Problem.[/b] For any acute-angled triangle $ ABC$ with medians $ m_a$, $ m_b$, $ m_c$, exradii $ r_a$, $ r_b$, $ r_c$ and semiperimeter $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$, prove the inequality $ m_a\cdot r_a \plus{} m_b\cdot r_b \plus{} m_c\cdot r_c \leq s^2$.[/color]

[i]Solution.[/i] Since the triangle ABC is acute-angled, we have $ m_a\leq\frac {a}{2}\cdot\cot\frac {A}{2}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=21462 post #16). In other words, $ m_a\leq\frac {a}{2}: \tan\frac {A}{2}$. But by the half-angle theorem, $ \tan\frac {A}{2} \equal{} \frac {r_a}{s}$; thus, $ m_a\leq\frac {a}{2}: \frac {r_a}{s} \equal{} \frac {as}{2r_a}$. Multiplication with $ r_a$ yields $ m_a\cdot r_a\leq\frac {as}{2}$. Similarly, $ m_b\cdot r_b\leq\frac {bs}{2}$ and $ m_c\cdot r_c\leq\frac {cs}{2}$. Hence,

$ m_a\cdot r_a \plus{} m_b\cdot r_b \plus{} m_c\cdot r_c \leq\frac {as}{2} \plus{} \frac {bs}{2} \plus{} \frac {cs}{2} \equal{} \frac {a \plus{} b \plus{} c}{2}\cdot s \equal{} s\cdot s \equal{} s^2$,

and the problem is solved.

Note that the condition that triangle $ ABC$ be acute-angled is essential. For obtuse-angled triangles $ ABC$, we have sometimes $ m_a\cdot r_a \plus{} m_b\cdot r_b \plus{} m_c\cdot r_c \leq s^2$ and sometimes $ m_a\cdot r_a \plus{} m_b\cdot r_b \plus{} m_c\cdot r_c \geq s^2$.

  Darij

[color=purple] Nice solution, Darij I have the following weaker but it's true for arbitrary triangle, hope you are interested in  :) : $ m_{a}r_{a}\plus{}m_{b}r_{b}\plus{}m_{c}r_{c}\le \frac{R}{2r}s^2$
  And here is lower boundary : $ (\frac{1}{2}\plus{}\frac{r}{R})s^2\le m_{a}r_{a}\plus{}m_{b}r_{b}\plus{}m_{c}r_{c}$
   Can you give me the better estimation Darij ? Thank you  :) [/color]


(a) Let $A,B,C \in \mathcal M_n \left( \mathbb C \right)$ s.t. $ABC = I_n$. Prove that if $I_n+A+AB,I_n+B+BC,I_n+C+CA$ are invertible, then the sum of their inverses is $I_n$.

(b) Prove that there are no $A,B,C \in \mathcal M_3 \left( \mathbb R \right)$ s.t. \[ AB=BC=CA= \left( \begin{array}{*{20}c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{array} \right) .  \]

For (b) part:

Since $A,B,C \in {\mathbb R}^{3 \times 3}\Rightarrow detA,detB,detC \in {\mathbb R}$.
Let now $detA=x,detB=y,detC=z$. We obtain the following sistem:$xy=-3,yz=-3,zx=-3 \Rightarrow  x^2y^2z^2 = -27<0$ which is absurd!


inequalities
algebra
polynomial
inequalities proposed

I have worked it out,arqady.The method is similar.And here is the problem and the solution.

gb2124, you can not use
 $ b\sqrt{(a^2\plus{}kbc)(c^2\plus{}kac)}\plus{}abc\plus{}\frac{kb^2(a\plus{}c)}{2}\leq$
$ \leq a\sqrt{(b^2\plus{}kac)(c^2\plus{}kab)}\plus{}abc\plus{}\frac{ka^2(b\plus{}c)}{2}$ because $ 4a\plus{}4c\minus{}kb$ can be negative. :wink:
But you can end your reasoning.

Thanks,arqady :blush: .I checked the mistake but I still solved it in a similar way. :oops:

I think you made the same mistake. :wink:

I don't think I have made the same mistake,beacuse we assume that 4a+4c-kb<=0,So we can made the denominator less to make the LHS smaller.Can you tell me what mistake you have found.
Oh,sorry I wrote 4a+4c-kb>=0 however what I really do is the case 4a+4c-kb<=0,

To make it clear,
I will post the whole solution I have corrected to the inequality.
Here it is :lol:

gb2124, now your proof seems true. Beautiful idea! My congratulations!

[quote="gb2124"]Arqady,Can you post your solution to this inequality,I think I proved it so complicated. :(
I wonder if there is some easy proof???[/quote]
Let $ a,$ $ b$ and $ c$ are non-negative numbers and $ 0\leq k\leq2.5 \plus{} \sqrt6.$ Prove that:
\[ \sqrt {a^2 \plus{} kbc} \plus{} \sqrt {b^2 \plus{} kac} \plus{} \sqrt {c^2 \plus{} kab}\geq\sqrt {3(1 \plus{} k)(ab \plus{} ac \plus{} bc)}
\]
Here is my proof for this one ( very similar to your proof and by using of your idea).
If $ c \equal{} 0$ then our inequality is obvious. Hence, we can assume that $ a,$ $ b$ and $ c$ are positive numbers.
$ \sum_{cyc}\sqrt {a^2 \plus{} kbc}\geq\sqrt {3(1 \plus{} k)(ab \plus{} ac \plus{} bc)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2 \minus{} ab) \plus{} 2\sum_{cyc}\sqrt {(a^2 \plus{} kbc)(b^2 \plus{} kac)}\geq(2k \plus{} 2)(ab \plus{} ac \plus{} bc)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2 \plus{} 4\sum_{cyc}\left(\sqrt {(a^2 \plus{} kbc)(b^2 \plus{} kac)} \minus{} ab \minus{} \frac {kc(a \plus{} b)}{2}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2\left(1 \plus{} \frac {4kc(a \plus{} b) \minus{} k^2c^2}{\sqrt {(a^2 \plus{} kbc)(b^2 \plus{} kac)} \plus{} ab \plus{} \frac {kc(a \plus{} b)}{2}}\right)\geq0.$
Let $ a\geq b\geq c.$ Hence, $ S_c\geq0$ because $ 4(a \plus{} b)\geq kc$ ( given  $ 0\leq k\leq2.5 \plus{} \sqrt6$ ),
$ S_b\geq0$ because $ \sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2} \plus{} 4kb(a \plus{} c) \minus{} k^2b^2\geq$
$ \geq\sqrt {ka^3b} \plus{} 4.5kab \minus{} k^2b^2\geq b^2\sqrt k(1 \plus{} 4.5\sqrt k \minus{} \sqrt {k^3})\geq0$ because
for polynomial $ t^3 \minus{} 4.5t \minus{} 1$ there is  only one  positive root :wink: and it's $ \sqrt {2.5 \plus{} \sqrt6}.$
Hence, it remains to prove that $ a^2S_b \plus{} b^2S_a\geq0.$
But $ a^2S_b \plus{} b^2S_a \equal{} a^2 \plus{} b^2 \plus{} \frac {4ka^2b(a \plus{} c) \minus{} k^2a^2b^2}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{}$
$ \plus{} \frac {4kb^2a(b \plus{} c) \minus{} k^2a^2b^2}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\geq$
$ \geq a^2 \plus{} b^2 \plus{} 4kab\left(\frac {a^2}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{} \frac {b^2}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\right) \minus{}$
$ \minus{} k^2a^2b^2\left(\frac {1}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{} \frac {1}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\right)\geq$
$ \geq a^2 \plus{} b^2 \plus{} 4ka^2b^2\left(\frac {1}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{} \frac {1}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\right) \minus{}$
$ \minus{} k^2a^2b^2\left(\frac {1}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{} \frac {1}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\right).$
Thuse, it remains to prove that
$ a^2 \plus{} b^2\geq(k^2 \minus{} 4k)a^2b^2\left(\frac {1}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{} \frac {1}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\right),$
which is obvious for $ 0\leq k\leq4.$
Hence, we need to understand what is happening for $ 4\leq k\leq2.5 \plus{} \sqrt6.$
But in this case
$ (k^2 \minus{} 4k)a^2b^2\left(\frac {1}{\sqrt {(a^2 \plus{} kbc)(c^2 \plus{} kab)} \plus{} ac \plus{} \frac {kb(a \plus{} c)}{2}} \plus{} \frac {1}{\sqrt {(b^2 \plus{} kab)(c^2 \plus{} kab)} \plus{} bc \plus{} \frac {ka(b \plus{} c)}{2}}\right)\leq$
$ \leq(k^2 \minus{} 4k)a^2b^2\left(\frac {1}{\sqrt {ka^3b} \plus{} \frac {kab}{2}} \plus{} \frac {1}{\sqrt {kb^3a} \plus{} \frac {kab}{2}}\right) \equal{}$
$ \equal{} \sqrt {k}(k \minus{} 4)ab\left(\frac {\sqrt b}{\sqrt a \plus{} \frac {\sqrt {kb}}{2}} \plus{} \frac {\sqrt a}{\sqrt b \plus{} \frac {\sqrt {ka}}{2}}\right).$
Id est, we need to prove that $ \frac {a}{b} \plus{} \frac {b}{a}\geq\sqrt {k}(k \minus{} 4)\left(\frac {\sqrt b}{\sqrt a \plus{} \frac {\sqrt {kb}}{2}} \plus{} \frac {\sqrt a}{\sqrt b \plus{} \frac {\sqrt {ka}}{2}}\right),$ which is true for $ 0\leq k\leq2.5 \plus{} \sqrt6.$
For example, if $ \sqrt{\frac{a}{b}}\plus{}\sqrt{\frac{b}{a}}\equal{}u$ then, $ \frac {a}{b} \plus{} \frac {b}{a}\geq\sqrt {k}(k \minus{} 4)\left(\frac {\sqrt b}{\sqrt a \plus{} \frac {\sqrt {kb}}{2}} \plus{} \frac {\sqrt a}{\sqrt b \plus{} \frac {\sqrt {ka}}{2}}\right)\Leftrightarrow$
$ \Leftrightarrow(u\minus{}2)(2\sqrt ku^2\plus{}(2\plus{}\sqrt k)^2u\plus{}8\plus{}2k\plus{}20\sqrt k\minus{}4k\sqrt k)\geq0,$ which is obvious.
Our inequality is proven.  :)


You're not allowed to ask this type of question. Read the problem again.

As a reminder, you are not allowed to ask another person about the problems.  This includes posting on any forum.  We will not answer any questions about the problems while the round is in progress.


[quote="The Zuton Force"]Let $ S$ be the set of ordered triples $ (i,j,k)$ of positive integers such that $ ijk \equal{} 432$.

Let $ N \equal{} \sum_{(i,j,k) \in S} i \plus{} j \plus{} k$.

Find the remainder when $ N$ is divided by $ 1000$.[/quote]

[hide]
Let's first find S.
$ 432 \equal{} 2^4cdot 3^3$
It is a set of 20 members.
Say a member is $ 2^n*3^m$
Then the number of times it is in a set is $ (4\minus{}n\plus{}1)(3\minus{}m\plus{}1)$
So it's an array of numbers:
20 15 10 5
16 12 8 4
12 9 6 3
8 6 4 2
4 3 2 1

Sum the items to get 150?

That can't be right.  No matter what.
[/hide]

OK - revising solution now.  Thanks Zuton for the tip.

Unfortunately, that is incorrect. The likely issue is that you assumed that all factors of 432 are used an equal number of times, whereas this is not the case (the answer does not end in a 0).

[hide="Solution (poorly explained, sorry)."]
Since all the factors of 432 will only be powers of 2 and 3, you can think of them separately.  You know that the powers of 2 have to add up to 4 and those of 3 have to add up to 3, as $ 432 \equal{} 2^43^3$.  So for powers of 2, you can have the following arrangements (the first number being the largest power of 2 found in $ i$, the second number being the same thing for $ j$, and the third being for $ k$):

$ 4 0 0$ (3)
$ 3 1 0$ (6)
$ 2 2 0$ (3)
$ 2 1 1$ (3)

The numbers in parentheses are the number of distinct ways in which the numbers can be rearranged.  There is only one way to place 4 in a given place, there are two ways to place 3 in a given place (for example, $ 3 1 0$ and $ 3 0 1$), three ways to place 2 in a given place (two rearrangements of $ 2 2 0$ and $ 2 1 1$), four ways to place 1 in a given place, and five ways to place 0 in a given place.  You can do the same thing for the powers of 3.

$ 3 0 0$ (3)
$ 2 1 0$ (6)
$ 1 1 1$ (1)

The same pattern exists: there is 1 way to place 3, 2 ways for 2, 3 ways for 1, and 4 ways for 0.  Now you can express this using summation notation, multiplying by 3 because any given number can be any of the three possible positions, not just the first one.

$ 3\sum_{a\equal{}0}^{4} \sum_{b\equal{}0}^{3} 2^a3^b(5\minus{}a)(4\minus{}b) \equal{} 174\sum_{a\equal{}0}^{4}2^a(5\minus{}a) \equal{} 9918$.

The remainder when divided by 1000 is $ \boxed{918}$. [/hide]

[hide="Discussion"] Replace $ 432$ with $ n$.

Let $ d$ be a divisor of $ n$.  In the expression for $ N$, $ d$ appears as the value of $ i$ for every pair $ (j, k)$ such that $ jk \equal{} \frac {n}{d}$, so it appears $ \sigma_0 \left( \frac {n}{d} \right)$ times.  Applying the same logic to $ j$ and $ k$, our desired summation is

$ N(n) \equal{} 3 \sum_{d | n} d \cdot \sigma_0 \left( \frac {n}{d} \right)$.

This is a [url=http://en.wikipedia.org/wiki/Dirichlet_convolution]Dirichlet convolution[/url], which is therefore an arithmetic function.  It suffices to evaluate $ N(n)$ for prime powers. [/hide]

$ \sigma_k(n) \equal{} \sum_{d | n} d^k$.  In particular, $ \sigma_0(n)$ is the number of divisors of $ n$ and $ \sigma_1(n)$ is the sum of the divisors of $ n$.

Begone, your answer is correct :) And thanks for the theory, t0rajir0u

If I may, how would you guys rate the difficulty of this problem, and how time-consuming was the computation?

[i]-- Thanks![/i]

In comparison to everybody else here, I did terribly on the AIME (6/15), so my estimation of its difficulty will probably be higher than anybody else's; however, I'd place it roughly as #12 on an AIME.  As for computation time, I found the correct solution using a computer and then spent an hour working out various manual approaches until the answer fit (if only I had that luxury on an actual test).  Again, that's definitely not representative of everybody else.  The actual working out of the correct solution once you get the formula is quite quick, though -- it takes less than a minute.

[quote="t0rajir0u"]This is a [url=http://en.wikipedia.org/wiki/Dirichlet_convolution]Dirichlet convolution[/url], which is therefore an arithmetic function.  It suffices to evaluate $ N(n)$ for prime powers.[/quote]

Is an arithmetic function the same as a multiplicative function?  That seems to be implied by your last sentence, but I wasn't sure if it was true.

Two types of functions are meant by "multiplicative," so I wanted to be unambiguous.  An arithmetic function is a function $ f : \mathbb{N} \to \mathbb{C}$ such that $ \gcd(a, b) \equal{} 1 \implies f(ab) \equal{} f(a) f(b)$.  A "multiplicative" function can mean a [i]totally multiplicative function[/i] such that $ f(ab) \equal{} f(a) f(b)$ for all $ a, b$.


[color=orange]The Fibonacci sequence is the sequence 1, 1, 2, 3, 5,... where each term is the sum of the previous two terms.  What is the remaider when the 100th term of the sequence is divided by 4?

See here [color=blue]Forum index » Olympiad Section » Geometry » Geometry Proposed & Own Problems[/color]   :P

For your problem: Put $r_{n}=F_{n}\mod{4}(n=1,2,3,...)$. Then $r_{1},r_{2},r_{3},...$ write as $1,1,2,3,1,0,1,1,...$. By induction on $n$ easy prove $r_{n}=r_{n+6}\forall n$ and therefore $r_{100}=r_{6\cdot 16+4}=r_{4}=3$.

[hide="solution"]Let's find a pattern mod 4:

1,1
1,1
2,2
3,3
5,1
8,0
13,1
21,1
34,2
55,3

It looks like the fibbonacci sequence all over again, with a period of 6.

So 100-6-6-6-6-6-...-6=4

The fourth fibbonacci number is 3.[/hide]


geometry
circumcircle
geometric transformation
reflection
incenter
rectangle
vector

$ I$ and $ I_a$ are Inscribed circle and Excircle of △$ ABC$ respectively，$ M$ is the midpoint of $ IF$，$ I_aS//OA$
Then show that ：The circumcircles of △$ KDL$、△$ AOM$、△$ ISI_a$ are Coaxial

assume $ \angle B \ge \angle C$.denote by $ V$ the intersection $ KI$ with $ (O)$.also let $ \frac{\angle B}{2}\minus{}\frac{\angle C}{2}\equal{}\beta$.
we have $ \angle FAK\equal{}\angle KVF\equal{}90^\circ$
so we have $ \angle IVF\equal{}90^\circ$
and since $ M$ is the midpoint of $ IF$ we get $ MI\equal{}MF\equal{}MV$
by $ MV\equal{}MF$ and $ OV\equal{}OF$ we get $ \angle MVO\equal{}\angle MFO\equal{}\angle AFK\equal{}\beta$
so we have $ \angle OVM\equal{}\beta$
also we have $ \angle OAM\equal{}\frac{\angle B}{2}\minus{}\frac{\angle C}{2}\equal{}\beta$ and so we get $ \angle OAM\equal{}\angle OVM\equal{}\beta$.
so the quadrilateral $ OAVM$ is cyclic.
Let $ T$ denote the projection of $ I$ on $ FK$.
then we have $ \angle OTM\equal{}\angle FTM\equal{}\angle TFM\equal{}\beta\equal{}\angle OAM$
so we have that $ OTAVM$ is cyclic.
also we have $ \angle ITK\equal{}\angle IAK\equal{}90^\circ$
so the quadrilateral $ AITK$ is cyclic.
so we have $ \angle TAI\equal{}\angle TKI\equal{}\angle FKV\equal{}\angle FAV\equal{}\angle MAV$.
but it is very well-known that $ A,T,E$ are collinear.so we get $ \angle EAF\equal{}\angle FAV$.let $ \alpha\equal{}\angle EAF$.
obviously $ \angle EI_aA\equal{}\beta\equal{}\angle OAI_a\equal{}\angle AI_aS$ and so $ S$ is the reflection of $ E$ wrt line $ I_aA$.
so we have $ \Delta I_aAS\equal{}\Delta I_aEA$
and so we get $ \angle I_aAS\equal{}\angle  I_aAE\equal{}\angle FAE\equal{}\alpha\equal{}\angle FAV$.
so we get the collinearity of $ A,V,S$.
but we also have $ \angle IVA\equal{}\angle KVA\equal{}\angle KFA\equal{}\beta\equal{}\angle AI_aS$.
so we obtain that $ IVSI_a$ is cyclic.
now we want to prove the collinearity of $ V,D,T$.
for this,we prove that $ \angle ITV\equal{}\angle ITD$ and so we obtain the collinearity.
at first we try to determine $ \angle VTI$.
we prove that $ I$ is the incenter of $ \Delta AVT$.
we already know that $ \angle VAI\equal{}\angle IAT\equal{}\alpha$.
so we must prove that $ \angle IVA\equal{}\beta\equal{}\angle IVT$.
since $ \angle IVF\equal{}\angle ITF\equal{}90^\circ$ we obtain $ \angle IVT\equal{}\angle IFT\equal{}\beta$.
so we r done for this part.
now we have
$ \angle ITV\equal{}\frac{\angle ATV}{2}\equal{}\frac{\angle AOV}{2}\equal{}\angle C\plus{}\angle BAV\equal{}\angle C\plus{}\frac{\angle A}{2}\minus{}\alpha$.
so we get $ \angle ITV\equal{}\angle C\plus{}\frac{\angle A}{2}\minus{}\alpha$.
we know that $ ITA'D$ is a rectangle ($ A'$ is the midpoint of $ BC$).
but we have $ \angle ITD\equal{}\angle IA'D$ since $ IA' \parallel{} AE$ we get $ \angle IA'D\equal{}\angle AEA'\equal{}\angle C\plus{}\frac{\angle A}{2}\minus{}\alpha$.
so we obtain $ \angle ITD\equal{}\angle ITV$ and so $ T,V,D$ are collinear.
so we get $ \angle KVD\equal{}\angle KVT\equal{}\angle IVT\equal{}\beta\equal{}\angle KLD$.
so the quadrilateral $ KDVL$ is cyclic.
so until now we have proved that the circles $ (O_1),(O_2),(O_3)$ all passes through point $ V$ which is on $ (O)$.
Let $ Q$ be the intersection of $ TI$ with $ (O_1)$.we have proved that $ I$ is the incenter of $ \Delta ATV$.as u can see $ AQ$ is tangent to $ (O)$ since $ \angle OAQ\equal{}90^\circ$.
so $ TQ$ is the internal angle bisector of $ \angle ATV$ and since $ TA'$ is the external angle bisector so we have the perpendicualrity of $ TQ$ with $ FK$ and so it is parallel with $ BC$.
now let $ U'$ be the second point of intersection of $ (O_1)$ and $ (O_2)$.
we know that $ TQ$ is parallel to $ DL$.
also we know that $ D,T,V$ are collinear.
also we know that $ D,L$ are on $ (O_2)$ and $ T,Q$ are on $ (O_1)$.
so by Reim's theorem(see the link i mentioned at the end of the proof) $ L,Q,U'$ are collinear.
now let $ U$ be the second point of intersection of $ (O_1)$ and $ (O_3)$
we know that $ AO$ is parallel to $ I_aS$.
we know that $ A,V,S$ are collinear.
also we know that $ A,O$ are on $ (O_1)$ and $ I_a,S$ are on $ (O_3)$.
so again by Reim's theorem $ I_a,O,U$ are collinear.
now we want to prove that $ U'\equal{}U$.
for this we have just to show that $ I_aO$ and $ LQ$ intersect each other on $ (O_1)$
or we have to prove that they are perpendicular since if they are perpendicular then owing to the fact that $ OQ$ is the diameter of $ (O_1)$ the intersection point will lie on $ (O_1)$.
but we know that $ \angle I_aAL\equal{}90^\circ$
so if we let the intersection of $ I_aO$ and $ LQ$ be $ R$ then we must prove that $ I_aLAR$ is cyclic.and then we'll have $ R\equal{}U\equal{}U'$.
now the rest can be done using vectors and proving that $ \vec{LQ}.\vec{OI_a}\equal{}0$ and using the equalities :
$ \vec{QL}\equal{}\vec{QI}\plus{}\vec{ID}\plus{}\vec{DL}$ and $ \vec{OI_a}\equal{}\vec{OA'}\plus{}\vec{A'E}\plus{}\vec{EI_a}$
(note that $ QI\equal{}QA$ and $ \angle AIQ\equal{}\angle C\plus{}\frac{\angle A}{2}$)
or we can try this method...
we'll prove that $ \angle QLA\equal{}\angle AI_aO$ then obviously we'll be done since $ I_aLAR$ will be cyclic.
let $ P$ be the intersection of $ IT$ with $ AL$.obviously since $ IT$ and $ KA$ (or $ KL$) are the altitudes of $ \Delta IFK$ so the point $ P$ is the orthocenter of $ \Delta IFK$ and it's collinear with $ F,V$ since $ FV$ is the other altitude.
since $ (O_1)$ is the nine-point circle of $ \Delta FIK$(because it passes through the midpoint of $ FK(O),FI(M)$ and also the point $ T$ which is the projection of $ I$ on $ FK$). so $ Q$ must be the midpoint of $ PI$.
now we prove that $ \Delta I_aFK \sim \Delta ILP$.then the rest is trivial since if they are similar then we'll have
$ \angle FI_aK\equal{}\angle ILP$ and since $ I_aO$ and $ LQ$ are two corresponding medians then we'll have
$ \angle QLP\equal{}\angle FI_aO$ and so we'll be done.
so let's prove that they are similar..
one can easily see that $ \angle IPL\equal{}\angle KFI_a\equal{}180^\circ\minus{}\beta$.
now it remains to prove that $ \angle ILP\equal{}\angle FI_aK$.
but if they are equal then $ LAWI_a$ is cyclic where $ W\equal{}I_aK \cap LI$.
so this equality is in fact equivalent to the perpendicularity of $ I_aK$ and $ LI$.
so we have just to prove that $ I_aK \perp LI$.
for this we prove that $ K$ is the orthocenter of $ \Delta LII_a$.
we know that $ LK \perp I_aA$ or $ LK \perp I_aI$.
so it remains to prove that $ KI \perp LI_a$.
we have $ \angle IKL\equal{}\angle IKA\equal{}\angle ATI\equal{}\angle AEA'\equal{}\angle C\plus{}\frac{\angle A}{2}\minus{}\alpha$.
also we have $ \angle I_aLK\equal{}\angle I_aLA\equal{}\angle ALE\plus{}\angle ELI_a\equal{}\beta\plus{}\angle EAI_a\equal{}\beta \plus{}\alpha$.
(note tat $ \angle ELI_a\equal{}\angle I_aAE$ because $ I_aEAL$ is cyclic(since $ \angle I_aEL\equal{}\angle I_aAL\equal{}90^\circ$)
so we have
 $ \angle I_aLK\plus{}\angle LKI\equal{}\beta \plus{}\alpha \plus{} \angle C\plus{}\frac{\angle A}{2}\minus{}\alpha\equal{}\beta\plus{}\angle C\plus{}\frac{\angle A}{2}\equal{}\frac{\angle B}{2}\minus{}\frac{\angle C}{2}\plus{}\angle C\plus{}\frac{\angle A}{2}\equal{}\frac{\angle A}{2}\plus{}\frac{\angle B}{2}\plus{}\frac{\angle C}{2}\equal{}90^\circ$
so we get $ KI \perp LI_a$.
so $ K$ is the ortho center of $ \Delta ILI_a$.
and so we have $ LI \perp I_aK$.
so we r DONE and the proof is completed !!

[hide="some other properties which can easily be obtained by using angles and power of point"]
Denote by $ O_4$ the midpoint of $ KI$ which is in fact the intersection of $ VI$ with $ (O_1)$.(in fact,u can easily prove that $ F,K$ are the excenters for $ \Delta AVT$)
as we know the intersection of $ (O_1),(O_2),(O_3)$ are $ V,U$.
then $ O_4,E,U$ are collinear.
let $ V'$ be the second intersection of $ (O_3)$ with $ (O)$.then $ V,A'',V'$ are collinear($ A''\equal{}AI \cap BC$).
also $ F,E,V'$ are collinear.
also $ V,S,F,A'',D,(FK,ES)$ lie on a circle.
[/hide]

[url=http://pagesperso-orange.fr/jl.ayme/apropos.html]Reim's theorem(L'équivalence de Reim,L'équivalence 0)[/url]

Thanks for your proof.
a little long，my english is poor，tomorrow I will have enough time to see your proof
We can use Casey Theorem to prove,$ O1$ is the midpoint of $ O2O3$,this way maybe easy.

yes,and also we can prove that they have this property (after proving that they are coaxial!)
we can use angles to prove that $ O_3,Q,M$ are collinear.
we have $ O_3V\equal{}O_3I$ and $ MV\equal{}MI$ which implies that $ MO_3 \perp VI$ and since $ MQ \perp IV$ so we get the collinearity of $ M,Q,O_3$.
we can easily prove the colinearity of $ O,N,O_2$ where $ N\equal{}AK \cap (O_1)$ by using angles.
and since $ ON \perp VI$ so we get that $ O_3Q \parallel{} OO_2$ and again since $ MN$ and $ OQ$ are diameters passing through $ O_1$ and $ O_2O_3$ passes through $ O_1$ so the quadrilateral $ O_3QO_2O$ is a parallelogram and we r done.

Lemma:
You can see it here (at the end) :
http://jl.ayme.pagesperso-orange.fr/Docs/A%20generalization%20of%20IMO%202009.pdf
•A generalization of the problem 2 (IMO 2009) ; an unexpected proof with the midcircle
by Jean-Louis AYME
Let point X on (CBA) and angle BAX = EAC
Let point A' is Simmetry of point A wrt FK , easy to see that AE -> DA' after this Simmetry , so X is on DA' , so angle DXK = KFA = DLK , so DXKL is cyclic
Let XA intersect Excircle of CBA at point P
Easy that EA = PA , triangle CXA ~ EBA , so CA / BA = XA / EA = XA / PA , so XA*PA = CA*BA = I_aA*IA , so PXI_aI is cyclic
(XIA) is tangent to OA (use Lemm) , so XMOA is cyclic
LA*AK = CA*BA = XA*PA , let line thru secont intersect point W of circles (XI_aI) and (LXK) intersects them at points P and Q respectively then easy to see that AW*AP = AW*AQ , so AP = AQ
Easy to see that FXM ~ A"XO were A'' is intersection point of (CBA and OA)
So XM = FM = MI , so angle IXF = 90 , so I is on XK
Let T is midpoint of IK , easy to see that angle XTA = 2*XKA = XOA , so XMOTA is cyclic , use Lemma for (XI_aI) (DLK) , so problem done


geometry
geometric transformation
homothety
search
trigonometry
angle bisector
geometry proposed

Let $ P$be the intersection of line $ AD$and circle$ ABC$. Denote by$ Y,Z$ the orthogonal projections of $ I$on $ AB,AC$. then $ M,Y,Z$ lies on simson line. Since $ IE//PY$, triangle $ IEF,PYZ$ is homothety and homothety center is $ A$. Therefore $ IN$ is parallel to $ PM$. So $ IN$ is perpendicular to $ BC$.

i was hoping to find a solution in which we can prove that $ \angle NIF\equal{}\angle ACB$ then  if $ NI\cap BC\equal{}H$ it woyld be proved that $ CFIH$ and $ EIHB$ are cyclic and the problem solved!

I think it is problem

 http://www.mathlinks.ro/viewtopic.php?search_id=845163698&t=140884

 but I'm not sure( I have a slow conection this moment to check it).

 babis

[quote="April"]Let be given a triangle $ ABC$. Let $ I$ be a mobile point lies on the angle bisector $ AD$ of $ \angle BAC$. Denote by $ E,\,F$ the orthogonal projections of $ I$ on $ AB,\,AC$, respectively. Let $ N$ the intersection of $ EF$ with median $ AM$ of triangle $ ABC$. Prove that $ IN\perp BC$.[/quote]
lets change the problem a bit!!
let $ I$ be a mobile point on $ AD$# and let $ E,F,K$ be the projections of point $ I$ on $ AB,AC,BC$ respectively,now let $ IK$ intersect $ EF$ at $ N$ we will show that $ N$ lies on the median $ AM$ of triagnel $ ABC$...
its sufficient to show that:

$ \frac{\sin EAN}{\sin FAN}\equal{}\frac{AC}{AB}$

which is true because first we know that $ IE \equal{} IF$ and also $ AE \equal{} AF$,then note that by $ \sin$ law in $ \triangle AEN,\triangle AFN$ we get that:

$ \frac{\sin EAN}{\sin FAN}\equal{}\frac{EN}{FN}\equal{}\frac{\sin EIN}{\sin FIN}$

$ \Rightarrow\frac{\sin EAN}{\sin FAN}\equal{}\frac{\sin EIN}{\sin FIN}(*)$

but in the other hand we know that $ \angle IKB \equal{}\angle IEB \equal{} 90$ so $ EIKB$ is cyclic,similary we can show that $ IKCF$ is cyclic thus we get that $ \angle B \equal{}\angle EIN,\angle C \equal{}\angle FIN$
thus by $ (*)$ we get that:

$ \Rightarrow\frac{\sin EAN}{\sin FAN}\equal{}\frac{\sin B}{\sin C}\equal{}\frac{AC}{AB}$ as wanted...

let a perpendicular line from I to BC intersect EF at G . Draw a parallel line from G to BC and let this line intersect AB and AC at K and L. we have in cyclic quadrilaterals GELI and GIFK : $ \angle GEI\equal{}\angle GLI\ ,\ \angle GKI\equal{}\angle GFI\Longrightarrow\angle KLI\equal{}\angle LKI$ hence G is midpoint of KL and AG passthrough mid of BC and $ G\equiv N$