[b]Waterloo/CEMC[/b]
Canadian Open Math Challenge (COMC) : Wed, Nov 24, 2004
Pascal/Cayley/Fermat [Gr. 9-11]: Wed, Feb 23, 2004
Euclid: Tues, Apr 19, 2004
Fryer/Galois/Hypatia [Gr. 9-11]: Wed, Apr 20, 2004

[b]Tournament of the Towns (Toronto Area)[/b]
Fall Round, Level O: Sun, Oct 24, 2004
Fall Round, Level A: Sun, Oct 31, 2004
Spring Round, Level O: Sun, Feb 27, 2005
Spring Round, Level A: Sun, Mar 6, 2005
(Details see [url]http://www.math.toronto.edu/oz/turgor[/url])

[b]Canadian National Mathematics League (CNML)[/b]
Six 30 min contests, dates may differ by school
1) Tues, Oct 26, 2004
2) Tues, Nov 30, 2004
3) Tues, Jan 11, 2005
4) Tues, Feb 8, 2005
5) Tues, Mar 8, 2005
6) Tues, Apr 5, 2005

[b]U.S. Contests[/b]
AMC 10/12 A: Tues, Feb 1, 2005
AMC 10/12 B: Wed, Feb 16, 2005
AIME A: Tues, Mar 8, 2005
AIME B: Tues, Mar 22, 2005
USAMO: Tues & Wed, Apr 19 & 20, 2004

(It is worth noting that the dates of Euclid/Hypatia coincides with that of USAMO. If it were the same as last year, their time would not coincide, at least for eastern students. But then, one would have to do up to 7 hours of contest in one day. Great job organizers.)

If you have any more contest dates, in particular that of APMO and CMO, then please post.

I think the CMO is held the last Wednesday of March, which would mean that this year it will be Wednesday March 30, 2005. I may be wrong, but it will definately be close to that date

From what I remember, CMO is always being held early in the school year.

Well the COMC is held Nov. 24 so if they are able to mark the contest and get everything ready for the CMO in a month, they would be forced to wait until past the Christmas break, which would mean that it will definately not be near the beginning of the school year.

heh, there are only a few that should bother themselves with Euclid/Hypatia AND the USAMO...
but I am not sure how they are going to hand this though.  But actually, half of the IMO team for Canada is pretty much made already.  anyone who went to IMO last year and still is eligible will be on the team, since that was the point when the selected the younger kids rather than the grade 12s.  Students like Mike out performed those on the team but he still didnt get selected simply because its his last year.


A sequence with first two terms equal $1$ and $24$ respectively is defined by the following rule: each subsequent term is equal to the smallest positive integer which has not yet occurred in the sequence and is not coprime with the previous term. Prove that all positive integers occur in this sequence.

Let $ a_i$ denote terms in the said sequence ($ a_1 \equal{} 1$, $ a_2 \equal{} 24$), and let $ S_n : \equal{} \{a_1,\ldots,a_n\}$, $ S_\infty : \equal{} \bigcup_{n \in \mathbb{Z^\plus{}}} S_n$ = the set of ultimately $ \textit reachable$ numbers,
and $ U : \equal{} \mathbb{Z^\plus{}} \minus{} S_\infty$ = the set of forever $ \textit unreachable$ numbers.  Suppose not all positive integers occur in the sequence i.e. $ U \not \equal{} \emptyset$.
[b]$ U$ cannot contain any composite number[/b]: Suppose $ u \in U$ is composite, and by the Well-Ordering Principle (WOP), we may assume this is the minimal composite.
Then among $ \Psi (u) \equal{} \{$ numbers $ < u$ having a common factor with $ u\}$, $ \exists v \equal{} a_j \in \Psi (u)$ with $ j$ maximal (last number having common factor with $ u$ to appear).  But the next number $ a_{j \plus{} 1} \equal{} u \in S_\infty$. (contradiction) $ \clubsuit$
[b]$ U$ cannot contain any prime number[/b]: Let $ p$ be the minimal (prime) element of $ U$ (using WOP again).  (Note $ p \not \equal{} 2$ because we have $ a_3 \equal{} 2$).
Since the composite number $ 2p \in S_\infty$, $ \exists j > 3$ such that $ 2p \equal{} a_j$.  But the next number would be $ a_{j \plus{} 1} \equal{} p \in S_\infty$ (contradiction) $ \clubsuit$
([b]And obviously $ 1 \notin U$[/b]).  We conclude that $ U \equal{} \emptyset$ and that $ S_\infty \equal{} \mathbb{Z^\plus{}}$ $ \blacksquare$

I believe the above proof is not correct. If we assume for a moment that 8 is the smallest integer in the set of unreachable numbers, then the last (and largest) number smaller than 8 which has a common factor with 8 is 6, which is the fifth number in the sequence. So, according to YeOldeMan's proof, the next number in the sequence must be 8, but it is 3. I'll post my proof later.

Let $U$ be the set of forever unreachable numbers and $n$ be the smallest member of $U$. If $(a_k, n) > 1$, then $a_{k+1} < n$ and, since $n$ is finite, this can only happen finitely often; $M$ exists such that $(M, n) > 1$ and for all $m \ge M$ with $(m, n) > 1$, we have $m \in U$. In particular, $Mp \in U$ for all primes $p$. If now $p | a_k$ for some prime $p$, then $a_{k+1} < Mp$ and this too, can happen at most finitely often. Thus, $M_p$ exists for which $p$ divides $M_p$ and if $p$ divides $m$ for any $m \ge M_p$, $m \in U$. Now, let $k$ be such that $a_k > \max_{p < M_2} M_p$. That is, all prime divisors of $a_k$ are larger than $M_2$. In particular, let $p_1$ be the smallest prime divisor of $a_k$. Then, since $2p_1 > M_2$, we must have $a_{k+1} < 2p_1$. In other words, $a_{k+1} = p_i$ for some prime $p_i$ dividing $a_k$. But then we have $a_{k+2} = 2p_i > M_2$; contradiction.

Note that it's not necessary that $a_2 = 24$; it could be any positive integer larger than $1$. Furthermore, when $a_2 = 4$, I conjecture the following:

1) If $a_k = p$ ($p$ prime), then $a_{k-1} = 2p, a_{k+1} = 3p$. This only holds for $p$ large enough if $a_2 \neq 4$
2) If $a_k = n$ and $n \neq p,3p$, then $k = n + o(n)$. For $n = p$ we have $k = 2n + o(n)$ and for $n = 3p$ we have $k = 2n/3 + o(n)$. (this might be hard to prove, although I have very little idea)

Also, I can imagine the possibility that the sequence is exactly the same for all large enough $k$, whatever the value of $a_2$ is. Where the meaning 'large enough' does depend on the value of $a_2$, of course. I am hesitant, but tempted to call this a conjecture too. For example, whether $a_2 = 4$, or $a_2 = 8$, we have that the set $\{a_1, a_2, .., a_8\}$ equals the set $\{1,2,3,4,6,8,9,12\}$. And in both cases $a_9 = 10$. So, for $k \ge 9$, the sequences are indentical.

I do have more conjectures/questions concerning these sequences, but the above I believe to be the most important.

See http://www.mimuw.edu.pl/~malcin/ekg.pdf for some serious stuff


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

I have 4 proofs to do about subgroups. I've started 3, am stuck on the last one, and only 70% sure of 1. Here they are:

11) Is the set H = {[1 a
                              0 1] | a is in R} a subgroup of GL (2, R) under matrix multiplication? Prove it is or explain why it isn't.
My work so far:
I used the one-step subgroup test. So I want to prove H is a group, right?
1. The property P that distinguishes the elements of H are (1)(1) - 0(a) = 1 - 0 = 1
2. Prove that the identity has property P. So, the identity I = [1 0
                                                                                0 1] has the determinant = (1)(1) - (0)(0) = 1 so I is in H
3. Next is to assume 2 elements a and b have property P. This is where I got stuck. I don't know what elements to assume and how to word it. Since I have a matrix, do I assume 2 matrices with property P?
4. The last step in proving H is a group is to use the assumption about a and b to show that ab^(-1) [read "a b inverse"] has property P. Since I didn't get 3, I can't get 4 so I am having trouble finishing this proof. :maybe:

12) Is the set {[a b
                        c d] | a, b, c, d are in R, and ad - bc = -1} a subgroup of GL (2, R) under matrix multiplication? Prove that it is or explain why it isn't.

My work: My answer is that the set is NOT a subgroup because the identity is not in the set.
The identity I = [1 0
                       0 1] has the determinant = (1)(1) - (0)(0) = 1, but in the set ad - bc = -1.
So by counterexample I will have explained why the set is not a subgroup of GL (2, R).
Is that correct?

I think it's easier to treat multiplication and inverses separately, rather than rolling them into one property. The "one-step" method isn't very useful in practice.
What's the product of $ \begin{bmatrix}1&a\\0&1\end{bmatrix}$ and $ \begin{bmatrix}1&b\\0&1\end{bmatrix}$?

As for your second question- yes. That's the set of matrices of determinant $ \minus{}1$, and the product of any two has determinant $ 1$.


geometry
circumcircle
trigonometry
conics
ellipse
symmetry
power of a point

Let $ABC$ be isosceles with right angle at $A$. $M$ and $N$ are on $BC$ and angle$MAN$=45. Circumcircle of $AMN$ meets $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that $BP+CQ=PQ$

Denote $BC = a,AB = AC = \frac{{a\sqrt 2 }}{2},\angle AMC = \varphi $. If $H$ is the feed of perpendicular from $A$ to $BC$ we have:

$AN = \frac{{AH}}{{\sin \left( {135^\circ  - \varphi } \right)}} = \frac{a}{{2\sin \left( {135^\circ  - \varphi } \right)}}$

$MN = \frac{{AN\sin 45^\circ }}{{\sin \varphi }} = \frac{{\sqrt 2 a}}{{4\sin \varphi  \cdot \sin \left( {135^\circ  - \varphi } \right)}}$

$BM = BH - MH = \frac{a}{2} - \frac{a}{2}\cot \varphi  = \frac{a}{2}\left( {1 - \cot \varphi } \right)$

$BN = BH + HN = \frac{a}{2} + \frac{a}{2}\cot \left( {135^\circ  - \varphi } \right) = \frac{a}{2}\left[ {1 + \cot \left( {135^\circ  - \varphi } \right)} \right]$

$CN = CH - HN = \frac{a}{2} - \frac{a}{2}\cot \left( {135^\circ  - \varphi } \right) = \frac{a}{2}\left[ {1 - \cot \left( {135^\circ  - \varphi } \right)} \right]$

$CM = CH + HM = \frac{a}{2} + \frac{a}{2}\cot \varphi  = \frac{a}{2}\left( {1 + \cot \varphi } \right)$

From the power of a point theorem follows that

$BP = \frac{{BM \cdot BN}}{{AB}} = \frac{{\sqrt 2 }}{4}a\left( {1 + \cot \varphi } \right)\left[ {1 + \cot \left( {135^\circ  - \varphi } \right)} \right]$

$CQ = \frac{{CN \cdot CM}}{{AC}} = \frac{{\sqrt 2 }}{4}a\left( {1 + \cot \varphi } \right)\left[ {1 - \cot \left( {135^\circ  - \varphi } \right)} \right]$

and, summing the previous relations, we obtain

$BP + CQ = \frac{{\sqrt 2 }}{2}a\left[ {1 - \cot \varphi \cot \left( {135^\circ  - \varphi } \right)} \right] =$

$ = \frac{{\sqrt 2 }}{2}a\frac{{\sin \varphi \sin \left( {135^\circ  - \varphi } \right) - \cos \varphi \cos \left( {135^\circ  - \varphi } \right)}}{{\sin \varphi \sin \left( {135^\circ  - \varphi } \right)}} =$

$ =  - \frac{{\sqrt 2 }}{2}a\frac{{\cos \left( {\varphi  + 135^\circ  - \varphi } \right)}}{{\sin \varphi \sin \left( {135^\circ  - \varphi } \right)}} = \frac{a}{{\sin \varphi \sin \left( {135^\circ  - \varphi } \right)}}$

On the other hand, since $PQ$ is the circumdiameter of $\triangle AMN$, the extended sine law yields

$PQ = \frac{{MN}}{{\sin 45^\circ }} = \frac{a}{{2\sin \varphi  \cdot \sin \left( {135^\circ  - \varphi } \right)}}{\rm{   }} \ \ \  \Rightarrow {\rm{   }} \ \ \ PQ = BP + CQ$

and we are done. $\blacksquare$

I have a synthetic solution:

[color=blue][b]Problem.[/b] Let ABC be an isosceles right triangle with the right angle at the vertex A. Let M and N be two points on the side BC such that < MAN = 45°. Let the circumcircle of triangle AMN meet the lines AB and AC at the points P and Q, respectively (apart from the point A, of course). Prove that BP + CQ = PQ.[/color]

[i]Solution.[/i] Since the triangle ABC is right-angled at A, we have < BAC = 90°. In other words, < PAQ = 90°. Since the points P, N, Q and A lie on one circle, the quadrilateral PNQA is cyclic, and thus < PNQ = 180° - < PAQ; since < PAQ = 90°, this becomes < PNQ = 180° - 90° = 90°.

Since the points P, Q, M and A lie on one circle, we have < MPQ = < MAQ. On the other hand, since the triangle ABC is an isosceles right triangle with the right angle at A, we have < BCA = 45°, what, compared with < MAN = 45°, yields < MAN = < BCA. Thus,

  < MPQ = < MAQ = < MAN + < NAC = < BCA + < NAC = < NCA + < NAC.

By the sum of angles of triangle ANC, we have < NCA + < NAC = 180° - < ANC; but clearly, 180° - < ANC = < MNA. Finally, since the points M, N, A and P lie on one circle, the quadrilateral MNAP is cyclic, and thus < MNA = 180° - < MPA. Thus,

  < MPQ = < NCA + < NAC = 180° - < ANC = < MNA = 180° - < MPA = < BPM.

In other words, the line PM is the angle bisector of the angle QPB. Equivalently, the line PM is the exterior angle bisector of the angle QPA. Similarly, the line QN is the exterior angle bisector of the angle PQA. Thus, the lines PM and QN are the exterior angle bisectors at the vertices P and Q of the triangle APQ, and hence, the point of intersection L of these lines PM and QN is the A-excenter of the triangle APQ.

By the sum of angles in triangle PLQ, we have < PLQ = 180° - < LPQ - < LQP. Now, since the line PM is the angle bisector of the angle QPB, we have $\measuredangle MPQ = \frac{\measuredangle BPQ}{2}$; thus,

$\measuredangle LPQ=\measuredangle MPQ=\frac{\measuredangle BPQ}{2}=\frac{180^{\circ}-\measuredangle APQ}{2}=90^{\circ}-\frac{\measuredangle APQ}{2}$;

similarly, $\measuredangle LQP=90^{\circ}-\frac{\measuredangle AQP}{2}$. And thus,

$\measuredangle PLQ=180^{\circ}-\measuredangle LPQ-\measuredangle LQP$
$=180^{\circ}-\left(90^{\circ}-\frac{\measuredangle APQ}{2}\right)-\left(90^{\circ}-\frac{\measuredangle AQP}{2}\right)=\frac{\measuredangle APQ+\measuredangle AQP}{2}$.

By the sum of angles in triangle APQ, we have < APQ + < AQP = 180° - < PAQ; but since < PAQ = < BAC, and < BAC = 90°, we have < APQ + < AQP = 180° - < BAC = 180° - 90° = 90°, and thus

$\measuredangle PLQ=\frac{\measuredangle APQ+\measuredangle AQP}{2}=\frac{90^{\circ}}{2}=45^{\circ}$.

In other words, < PLN = 45°. On the other hand, < PBN = 45° (this is just a rephrase of the equation < ABC = 45°, which is clear since the triangle ABC is an isosceles right triangle with the right angle at A). Hence, < PLN = < PBN. And thus, the points P, N, L and B lie on one circle. Consequently, the quadrilateral BPNL is cyclic, and thus, < PBL = 180° - < PNL. On the other hand, 180° - < PNL = < PNQ, so that < PBL = < PNQ. Since < PNQ = 90°, we thus have < PBL = 90°. In other words, $BL\perp AP$. Hence, the point B is the orthogonal projection of the point L on the line AP. Since the line AP is a sideline of the triangle APQ, and the point L is the A-excenter of this triangle, the point B therefore is the orthogonal projection of the A-excenter of the triangle APQ on its sideline AP; in other words, the point B is the point of tangency of the A-excircle of triangle APQ with its sideline AP. Hence, by a well-known fact, the length AP equals the semiperimeter of triangle APQ; that is: $AB=\frac{AP+AQ+PQ}{2}$. In other words, $2\cdot AB=AP+AQ+PQ$. Since the triangle ABC is an isosceles right triangle with the right angle at A, we have AB = AC, and thus $2\cdot AB=AB+AC$. Hence, the equation $2\cdot AB=AP+AQ+PQ$ becomes AB + AC = AP + AQ + PQ. In other words,

  PQ = (AB + AC) - (AP + AQ) = (AB - AP) + (AC - AQ) = BP + CQ.

And the problem is solved.

  darij

Another interesting thing is that:
$BM^2+CN^2=MN^2$, which can be proved quite easy!

[quote="Sailor"]$BM^2+CN^2=MN^2$[/quote]

Nice fact! Here is my proof (an ugly one, but I really have no time left to simplify it):

We know that < BCA = 45°; in other words, < MCA = 45°. On the other hand, < MAN = 45°. Thus, < MCA = < MAN. Also, trivially, < CMA = < AMN. Thus, the triangles MCA and MAN are similar. Hence, $\frac{AM}{AN}=\frac{CM}{CA}$. Similarly, $\frac{AN}{AM}=\frac{BN}{BA}$. Multiplying these two equations together, we get

$\frac{AM}{AN}\cdot\frac{AN}{AM}=\frac{CM}{CA}\cdot\frac{BN}{BA}$.

In other words,

$1=\frac{CM\cdot BN}{CA\cdot BA}$.

Thus, $CA\cdot BA=CM\cdot BN$. But since the triangle ABC is an isosceles right triangle with the right angle at A, we have CA = BA and $BC=\sqrt2\cdot CA$, so that $BC^2=2\cdot CA^2$. Thus,

$BC^2=2\cdot CA^2=2\cdot CA\cdot CA=2\cdot CA\cdot BA=2\cdot CM\cdot BN$.

Now denote x = BM, y = MN and z = CN. Then, CM = MN + CN = y + z and BN = MN + BM = y + x. Also, BC = BM + MN + CN = x + y + z. Thus, the equation $BC^2=2\cdot CM\cdot BN$ becomes $\left(x+y+z\right)^2=2\cdot\left(y+z\right)\cdot\left(y+x\right)$. After expanding, this becomes

$x^2+y^2+z^2+2yz+2zx+2xy=2y^2+2xy+2yz+2zx$.

This simplifies to $x^2+z^2=y^2$. Remembering that x = BM, y = MN and z = CN, we thus get $BM^2+CN^2=MN^2$. Proof complete.

  darij

The problem I wrote was proposed at Moldavian olimpiad when I was in $9^{th}$ form.
Here is my solution (suppose that B,N,M,C are collinear in this order)
Let $L$ be a point at the same side of BC as A which satisfies the relations:
 $LC\perp{BC}$ and $LC=BN$.
Then we can easily see that $\triangle{ABN}=\triangle{ACL}$ so $AN=AL$ but $\angle{NAM}=\angle{MAL}=45^\cdot$ , hence $\triangle{NAM}=\triangle{LAM}$, from where we get that $MN=LM$ and the result follows!

Knowing that $BM^2+CN^2=MN^2$ we can easely prove that $BP+CQ=PQ$. Let $BM=x$,$CN=y$,$MN=z$. We have $x*(x+z)=BP*BA$ and $y*(y+z)=CQ*CA$. Adding this we get $x^2+(x+y)z+y^2=(BP+CQ)*BA$ or $z^2+(x+y)z=(BP+CQ)*BA$. This is $z*BC=(BP+CQ)*BA$ or $BP+CQ=z*\sqrt2=PQ$ as $PQ$ is diameter and $\angle MAN=45$

[b]Leon[/b], I have a question... How do you prove the theorem in your picture (display)?  I mean, I have proved this resoult if all of them are circles but I think in your display you use ellipses, am I true? :?  I'd like to know the proof If they are ellipses (I mean if it's correct).

[size=117][color=darkred][b]It is a very nice and old problem. The [u]Sailor[/u]'s proof is also very nicely ![/b][/color] [/size]

[quote="warut_suk"][color=darkred]Let $ ABC$ be an isosceles triangle with $ A = 90^{\circ}$ . Let $ \{M,N\}\subset [BC]$ be two points  so that

$ M\in (BN)$ and $ m(\widehat {MAN}) = 45^{\circ}$ . The circumcircle of $ \triangle AMN$ meets $ (AB)$ , $ (AC)$

at the points $ P$ and $ Q$ respectively. Prove that $ \|\begin{array}{c} BP + CQ = PQ \\
 \\
BM^2 + CN^2 = MN^2\end{array}$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u][/b] (metric). Suppose w.l.o.g. that $ AB = AC = 1$ . Denote $ m(\widehat {BAM}) = \phi$ and $ \tan\phi = t\le 1$ .

Thus $ m(\widehat {CAN}) = 45^{\circ} - \phi$ . Apply the [b]Sinus' theorem[/b] in the triangles $ ABM$ and $ ACN$ :

$ \|\begin{array}{ccc} \frac {BM}{\sin\phi} = \frac {1}{\sin (45^{\circ} + \phi )} & \implies & \|\begin{array}{c} \boxed {\ BM = \frac {t\sqrt 2}{1 + t}\ } \\
 \\
CM = \frac {\sqrt 2}{1 + t}\end{array} \\
 \\
\frac {CN}{\sin (45^{\circ} - \phi)} = \frac {1}{\cos\phi} & \implies & \|\begin{array}{c} \boxed {\ CN = \frac {1 - t}{\sqrt 2}\ } \\
 \\
BN = \frac {1 + t}{\sqrt 2}\end{array}\end{array}\|$ $ \implies$ $ \|\begin{array}{ccc} MN = BN - BM & \implies & \boxed {\ MN = \frac {1 + t^2}{(1 + t)\sqrt 2}\ } \\
 \\
BP\cdot BA = BM\cdot BN & \implies & \|\begin{array}{c} \boxed {\ BP = t\ } \\
 \\
AP = 1 - t\end{array} \\
 \\
CQ\cdot CA = CN\cdot CM & \implies & \|\begin{array}{c} \boxed {\ CQ = \frac {1 - t}{1 + t}\ } \\
 \\
AQ = \frac {2t}{1 + t}\end{array} \\
 \\
PQ^2 = AP^2 + AQ^2 & \implies & \boxed {\ PQ = \frac {1 + t^2}{1 + t}\ }\end{array}\|$

$ \|\begin{array}{cccccc} 1\blacktriangleright & \boxed {\ BP + CQ = PQ\ } & \Longleftrightarrow & t + \frac {1 - t}{1 + t} = \frac {1 + t^2}{1 + t} & \Longleftrightarrow & t(1 + t) + (1 - t) = 1 + t^2 \\
 \\
2\blacktriangleright & \boxed {\ BM^2 + CN^2 = MN^2\ } & \Longleftrightarrow & \frac {2t^2}{(1 + t)^2} + \frac {(1 - t)^2}{2} = \frac {(1 + t^2)}{2(1 + t)^2} & \Longleftrightarrow & 4t^2 + (1 - t^2)^2 = (1 + t^2)^2\end{array}\|\ \ \ \mathrm {O\ .\ K\ .\ !}$

[b][u]Remark.[/u][/b] Prove easily that $ PN\cdot QM = MN\cdot AB$ , $ m(\widehat {APQ}) = 2\phi$ and $ \triangle\ BPM\ \sim\ \triangle\ CNQ$ .[/color]

Hi, everybody

Be4 reading these beatiful solutions, I had the following way to find M and N:
Inside the right angle <BAC draw a ray Ax and note T on Ax| AT=AB. Draw angle bisector of angle <BAT which cuts BC at M, draw angle bisector of the angle <CAT and it cuts BC at N.
From the symmetry we get BM=MT and angle <ATM=<ABC=45, analogously TN=CN and <NTA=<NCA=45, while from the right angled triangle MTN we get BM^2+CN^2=MN^2.

After having read these solutions:
Take A' the symmetrical of A w.r.t. BC, draw the quarter of a circle centered A' and ray A'B, to cut AB, AC at B and C respectively. A tangent to this circle will cut (AB), (AC) at P and Q respectively, while A'P and A'Q cut BC at M, N respectively; then easily PQ=BP+CQ.

Best regards,
sunken rock

A blind man sees better the details


abstract algebra
function
group theory
superior algebra
superior algebra unsolved

For ring $R$ with unit, any free $R$-module is isomorphic to a direct sum of $R$ (considered as regular module).

But what about free modules over non-unitary ring?

Let $A$ be a ring with $1$ and let $R$ be an ideal of $A$ such that $R^2=R$. Such ideals exist. For example $A=\mathbb Z/6\mathbb Z$ and $R=(\hat 4)$.
Then $R$ can be seen (even though for commutative algebraists it is hard to) as a ring.
Every abelian group $G$ has an $R-$module structure given by the trivial action $R\times G\to G: (r,g)\to 0$.
This is one reason why the case of modules over rings without $1$ is not as interesting: the axiom $1\cdot x=x$ misses.

I prove that $R$ is not a free $R$-module of rank $1$. Assume there exists $x\in R$ such that for every function $f: \{x\}\to B$, where $B$ is an $R-$module, there exists $\varphi: R\to B$ an $R-$modules homomorphism such that $\varphi(x)=f(x)$.
This can be restated as: For every $B$ as above and $y\in B$, there exists $\varphi: T\to B$ an $R-$module homomorphism such that $\varphi(x)=y$.

[b]Lemma[/b]:Let $B$ be an abelian group with the trivial $R-$module structure. Then $Hom_R(R,B)=0$.
[b]Proof:[/b]: Let $f\in Hom_R(R,B)$ i.e. an $R-$module homomorphism from $R$ to $B$. For every $r\in R$, we have $f(r^2)=r\cdot f(r)$ by linearity. Because the action of $R$ on $B$ is trivial, $r\cdot f(r)=0$.
Because $R=R^2$, for every $r\in R$, there exists $n\in\mathbb N$ and $s_1,\ldots,s_n\in R$ such that $r=s_1^2+\ldots+s_n^2$. Then $f(r)=f(s_1^2+\ldots+s_n^2)$. By linearity, the last is $f(s_1^2)+\ldots+f(s_n^2)=0$. Therefore $f(r)=0$.
This completes the proof of the lemma.

Let now $y\in B$ be such that $y\neq 0$. Then if $\varphi: R\to B$ is an $R-$module homomorphism, then by the lemma $\varphi=0$, so $y=\varphi(x)=0$ which is a contradiction.


Let $ a,b,c,x,y,z\geq 0$,prove that:

$ \left( {y}^{2}\plus{}{z}^{2} \right) x \left( b\minus{}c \right) ^{2}\plus{} \left( {z}^
{2}\plus{}{x}^{2} \right) y \left( c\minus{}a \right) ^{2}\plus{} \left( {x}^{2}\plus{}{y}^{2}
 \right) z \left( a\minus{}b \right) ^{2}\leq  \left( y\plus{}z \right) ^{3}
 \left( b\minus{}c \right) ^{2}\plus{} \left( z\plus{}x \right) ^{3} \left( c\minus{}a \right) ^
{2}\plus{} \left( x\plus{}y \right) ^{3} \left( a\minus{}b \right) ^{2}$
BQ

$ \left( {y}^{2}\plus{}{z}^{2} \right) x \left( b\minus{}c \right) ^{2\,t}\plus{} \left( {
z}^{2}\plus{}{x}^{2} \right) y \left( c\minus{}a \right) ^{2\,t}\plus{} \left( {x}^{2}\plus{}{y
}^{2} \right) z \left( a\minus{}b \right) ^{2\,t}\leq  \left( y\plus{}z \right) ^{3
} \left( b\minus{}c \right) ^{2\,t}\plus{} \left( z\plus{}x \right) ^{3} \left( c\minus{}a
 \right) ^{2\,t}\plus{} \left( x\plus{}y \right) ^{3} \left( a\minus{}b \right) ^{2\,t}$
$ t \in N.$


[quote="Virgil Nicula"][color=darkred]Let $ w$ be the circumcircle of a given triangle $ ABC$ .

Denote $ XX$ - the tangent in the point $ X\in w$ to the circle $ w$.

For a point $ D\in BC$ denote the points $ \{\begin{array}{c}
M\in AD\cap BB\\\
N\in AD\cap CC\end{array}$ and

$ \{\begin{array}{ccccc}
P\in w& , & P\ne B & , & M\in PP\\\
Q\in w & , & Q\ne C & , & N\in QQ\end{array}$ . Prove that $ D\in PQ$ .[/color][/quote]

This problem is directly from by well known theorem: Let $ A,B,C,D$ be four points on circle. Tangent line at $ A,B$ meet $ P$, tangentline at $ C,D$ meet at $ Q$ and $ AC,BD$ meet at $ R$. Then $ P,Q,R$ are collinear.
In problem, Let $ X\equal{}BC\cap PQ$. Then by lemma, $ N,M,X$ are collinear. Hence $ X\equal{}D$.


geometry
incenter
circumcircle
ratio
trigonometry
area of a triangle
Heron\u0027s formula

Here is a problem with no elegant solution in sight.

http://www.geocities.com/gregm112358/Hard_sudoku.pdf

Many thanks for trying.

Dear Alexb,

   Are you going to package a solution into an applet and

   try to sell it later?

   T.Y.
   M.T.

This would depend on the solution. For example, long trigonometric derivations are not very amenable to Java implementation.

But I wonder, do you have a problem with applets being sold?

That's probably OK with you that, say, Mathforum lives off tax payer's money and the AoPS organization sells their books. I do not charge a penny of site visitors. I do not force them to buy an applet. On the other hand, I do maintain my own server and I do spend of my own time. And, between us, if I could I would spend more time at the site. The trouble is I have to make my living somehow. So, perhaps, if you relent somewhat and purchase your next book from my amazon store, you are sure to make a contribution to a good cause. This is of cause in addition to supporting the book's author and the publishing house.

I often receive requests from non-profit organizations to contribute this or that resource to a worthy cause. I always ask this question: do you fellow receive a salary from your dot org? If you do, would you please pay for the resource you request; for I do not receive a salary.

Does the above answer your question?

Best regards,
Alexander Bogomolny
http://www.cut-the-knot.org

Even the worst sweat-shops in South-East Asia
   using under-aged labor at least pay those
   unfortunate kids something out of their future profits.

   T.Y.
   M.T.

Well, I am not sure about your metaphor. Seems to me it's neither here nor there.

I am not a kid and no one pays me for my efforts. As to you, no one forces you to sweat out a solution and I doubt in a sweat shop (not necessarily of the worst kind) dissension would be tolerated.

[quote="alexb"][b]Problem:[/b] Consider a triangle with vertices $ A, B, C$ and opposite sides $ a, b, c.$ Suppos that vertices $ A$ and $ B$ lie on a circle, with $ C$ lying within the circle. Let $ M$ be the midpoint of $ AB$ and let $ N$ be chosen on the arc $ AB$ such that $ MN$ is perpendicular to $ AB.$ Let $ R_C$ be radius of the circle inscribed in the curvilinear triangle with sides $ A, B,$ and arc $ AB.$ Prove that

$ R_c \equal{} r \plus{} \frac {2 (MN) (s \minus{} a) (s \minus{} b)}{cs},$

where $ s \equal{} \frac {_{a \plus{} b \plus{} c}}{^2}$ is the semiperimeter of $ ABC$ and $ r$ is radius of the incircle of $ ABC.$
[/quote]
Extend $ BC$ or $ AC$ (it does not matter which one, say $ BC$) to cut the given circle $ (O)$ again at a point $ P.$ The result is Thebault theorem setup for the $ \triangle ABP.$ The original notation is not handy for calculations; re-label $ C$ as $ D$ and then $ P$ as $ C.$ Use the following notation:

Sides of the $ \triangle ABC, \triangle ABD, \triangle ADC$: $ a \equal{} BC, b \equal{} CA, c \equal{} AB, d \equal{} AD, u \equal{} BD, v \equal{} DC,$ where $ u \plus{} v \equal{} a.$
Semiperimeters: $ s \equal{} \frac {_{a \plus{} b \plus{} c}}{^2}, s_b \equal{} \frac {_{u \plus{} c \plus{} d}}{^2}, s_c \equal{} \frac {_{v \plus{} b \plus{} d}}{^2},$ where $ s \equal{} s_b \plus{} s_c \minus{} d.$
Areas: $ \triangle, \triangle_b, \triangle_c,$ where $ \triangle \equal{} \triangle_b \plus{} \triangle_c.$

Incenters: $ I, I_b, I_c,$ radii of the incircles $ (I), (I_b), (I_c)$:  $ r, r_b, r_c.$
Tangency points of the incircles $ (I), (I_b), (I_c)$ with $ BC$: $ T, T_b, T_c.$

Incenters of the curvilinear $ \triangle ABD, \triangle ADC$: $ J_b, J_c,$ radii of the incircles $ (J_b), (J_c)$: $ \varrho_b, \varrho_c.$
Tangency points of the incircles $ (J_b), (J_c)$ with $ BC$: $ S_b, S_c$ and with $ AD$: $ Q_b, Q_c.$

Midpoints of the segments $ b \equal{} CA, c \equal{} AB$: $ M_b, M_c.$
Midpoints of the circumcircle arcs $ CA, AB$ opposite to $ B, C,$ respectively: $ N_b, N_c.$
Angles: $ \phi \equal{} \angle ADB, \psi \equal{} \angle ADC,$ where $ \phi \plus{} \psi \equal{} \pi.$

In this notation, you want to show:

$ (?)\ \ \frac {\varrho_b}{r_b} \equal{} 1 \plus{} \frac {2(M_cN_c) (s_b \minus{} u) (s_b \minus{} d)}{c r_b s_b},$

or (using Heron's formula)

$ (?)\ \ \frac {\varrho_b}{r_b} \minus{} 1 \equal{} \frac {2(M_cN_c) (s_b \minus{} u) (s_b \minus{} d)}{c \triangle_b} \equal{} \frac {2(M_cN_c) \triangle_b}{c s_b (s_b \minus{} c)} \equal{} \frac {2(M_cN_c) r_b}{c (s_b \minus{} c)}.$

By Thebault theorem, the incenters $ I, J_b, J_c$ are collinear. An important step in the proof shows that the lines $ S_bQ_b, S_cQ_c$ intersect at the incenter $ I.$ This can be used to calculate the desired ratios $ \frac {\varrho_b}{r_b}, \frac {\varrho_c}{r_c}$ (one ratio is sufficient).

$ \overline{DT_b} \equal{} s_b \minus{} c$

$ \overline{DS_b} \equal{} \overline{DT} \plus{} \overline{TS_b} \equal{} \overline{DB} \minus{} \overline{TB} \plus{} \overline{TS_b} \equal{} u \minus{} (s \minus{} b) \plus{} r \tan \frac {\phi}{2}$

$ \frac {\varrho_b}{r_b} \equal{} \frac {\overline{DS_b}}{\overline{DT_b}} \equal{} \frac {u \minus{} (s \minus{} b) \plus{} r \tan \frac {\phi}{2}}{s_b \minus{} c}$

Using $ u \minus{} (s \minus{} c) \equal{} u \minus{} (s_b \plus{} s_c \minus{} d \minus{} b) \equal{} (s_b \minus{} c) \minus{} (s_c \minus{} b),$

$ \frac {\varrho_b}{r_b} \minus{} 1 \equal{} \frac {r \tan \frac {\phi}{2} \minus{} (s_c \minus{} b)}{s_b \minus{} c}$

We have to show equality of this and the above expressions for $ \frac {\varrho_b}{r_b} \minus{} 1.$

$ (?)\ \ \frac {r \tan \frac {\phi}{2} \minus{} (s_c \minus{} b)}{s_b \minus{} c} \equal{} \frac {2(M_cN_c) r_b}{c (s_b \minus{} c)}$

$ (?)\ \ r \tan \frac {\phi}{2} \equal{} \frac {2(M_cN_c) r_b}{c} \plus{} s_c \minus{} b$

$ \angle M_cAN_c \equiv \angle BAN_c \equal{} \angle BCN_c \equal{} \frac {\angle C}{2}.$ Substituting $ \frac {2(M_cN_c)}{c} \equal{} \frac {M_cN_c}{M_cA} \equal{} \tan \frac {C}{2} \equal{} \frac {r}{s \minus{} c}$ and $ \tan \frac {\phi}{2} \equal{} \frac {r_b}{s_b \minus{} c},$

$ (?)\ \ \frac {r r_b}{s_b \minus{} c} \equal{} \frac {r r_b}{s \minus{} c} \plus{} s_c \minus{} b$

Using $ (s \minus{} c) \minus{} (s_b \minus{} c) \equal{} s \minus{} s_b \equal{} s_c \minus{} d,$

$ (?)\ \ \frac {r r_b (s_c \minus{} d)}{(s_b \minus{} c)(s \minus{} c)} \equal{} s_c \minus{} b$

Using $ \frac {r}{s \minus{} c} \equal{} \tan \frac {C}{2} \equal{} \frac {r_c}{s_c \minus{} d},$

$ (?)\ \ \tan \frac {\phi}{2} \tan \frac {\psi}{2} \equal{} \frac {r_b}{s_b \minus{} c} \cdot \frac {r_c}{s_c \minus{} b} \equal{} 1$

which is an obvious identity, because $ \phi \plus{} \psi \equal{} \pi.$

It's a pleasure getting a relevant response. Thank you.

Dear Mathlinkers,
after this beatifull proof of our beloved Vladimir...

another way is possible...next...

Wher can we found other proofs.

Sincerely
Jean-Louis

Dear MLs,

Please, let me know by e-mail if you have a copy of
Karluchenco book on Sangaku (in Russian)
and the last edition of Akopyn book on
geom figures (in Russian).

Thank you.
M.T.

Dear M. T.
thank very much for your initiative...

Sincerely
Jean-Louis

Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/San%20Gaku%201781.pdf] here [/url]

Sincerely
Jean-Louis


inequalities
logarithms
limit
real analysis
real analysis unsolved

Find the lowest $k>1$ such that the following inequality is true for all $n\geq 61$,
\[(\log\log n)^{2}\geq\frac{0,6482}{(k-1)e^{\gamma}}, \]
where $\gamma=\lim_{n\to\infty}1+\frac{1}{2}+\ldots+\frac{1}{n}-\log n$ and $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}.$

What do you mean? Should we just solve a [b]linear[/b] inequality and write the answer up to the forth decimal point?  :maybe:


number theory
relatively prime
Pythagorean Theorem
geometry

Circles of radii 5, 5, 8, and $m/n$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide="Solution"]By drawing a triangle with the radii of each of the larger circles, we form a triangle with sides of $13,13,10$ and so the altitude must be $12$. In one of the two $5,12,13$ right triangles, we can make another right triangle with sides of $5, 5+x$ and $4-x$, where $x$ is the radius of the smallest circle. Note that I got $4-x$ from the altitude, since the radius of the biggest circle plus the smaller circle, which is $8+x$, so the remaining part must be $4-x$. From Pythagorus, we have $5^2+(4-x)^2=(5+x)^2\Longrightarrow x=\frac{8}{9}$, so the answer is $\boxed{17}$[/hide]

Can somebody post a diagram please? :blush:

[hide="Bit different"]

It's basically same but oh well. Let $r$ be the given radius. This right triangle is made up of center of circle with radius 8, center of circle 5 (either one), and the tangent point of two circles with radius 5. The hypotenuse then is 8+5 = 13. Since the tangent point is 5 units away from the center, one leg is 5 units. The other leg must be 12 by Pythagorean Theorem. Now, to find 12 in terms of $r$, note that we can make another right triangle with hypotenuse $r+5,5,$ and $x$. So, we know that $x = \sqrt{(r+5)^2 - 5^2} = \sqrt{r^2+10r}$. Then the unknown height must consist $x,r$ and 8. So:

$12 = x + r + 8$
$12 = \sqrt{r^2 + 10r} + r + 8$
$4 = \sqrt{r^2 + 10r} + r$
$4 - r = \sqrt{r^2 + 10r}$
$16 - 8r + r^2 = r^2 + 10r$
$r = \frac{8}{9} \rightarrow 8+9 = \fbox{17}$ [/hide]

You can also apply Descartes Theorem..(which I just found>)

Let's call the center of the circle with radius 8 A, the left circle with radius 5 B, the middle circle center C, and the center of the right circle with radius 5 D. Note that the distance from A to D is 13, and thus creates a right triangle from the tangency point between circles B and D(let's call this E), the center of D, and A. Thus the height from the tangency point between A and C(let's call this F) and E is 4. Constructing another right triangle between C, E, and D, we see that $EC= EF-x$, and $CD= 5+x$, so we set up the Pythagorean theorem
$5^2+(4-x)^2=(5+x)^2$
Solving this, we find that $x=8/9$, for an answer of $\fbox{17}$.

EDIT: Oops didn't see the first solution

I don't see what I did wrong. Refer to the diagram [url=https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4]here.[/url]

Drop an altitude from the center of the circle with radius $r$ to the intersection of either of the circles with radius $5$ (doesn't matter which, pick any one you like) with the circle of radius $8$. This gives two right triangles. Let the leg common to both triangles be $x$. Then the dimensions are $5,x,5+r$ and $8,x,8+r$. However once the equations are created the result is $x=r=0$ which is obviously wrong. What happened?

[quote=OlympusHero]
Drop an altitude from the center of the circle with radius $r$ to the intersection of either of the circles with radius $5$ (doesn't matter which, pick any one you like) with the circle of radius $8$.[/quote]
You don't know that's an altitude (and in fact it's not).

Oh, okay! Not sure I thought that was an altitude, but thanks! :P

What is the source of this problem? I've seen it before but cannot remember...
BTW, I've tried searching too :)

I'm fairly certain it's an AIME problem...

yes, it's 1997 AIME #4: [url]https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4[/url]


irrational number
algebra unsolved
algebra

If $ a$ is a irrational number then set $ A \equal{}$ {$ a2^n \minus{} [a2^n],n \in N$} is dense in $ [0,1]$.
Prove or disprove it.

false. if, say, $ a \equal{} 0.101001000100001...$ in base 2, then there are no fractional parts of $ 2^na$ in an interval around 1.


Hey Everyone,

I am managing a startup and we are looking for some interns (developers).  Interns will see all aspects of starting a web company (Business / Server Management / Web Development).

It should be fun and could definitely make for a very interesting college / grad school essay!  (We might also give out equity if performance warrants.)

We currently use the following infrastructure in case you are interested:

Linux => Apache => Mongrel => MySQL => Ruby on Rails

We are looking for people who fit the following criteria:

1)  Have at least 2 - 3 hours a night to work
2)  Know some programming (loops at the very minimum)
3)  extremely bright & motivated

If anyone is interested (or has any questions), feel free to reply to this thread or email me directly at (mail at jackpo dot org).


Maybe they were already discussed...
How can I make a link on a post?
Also, how can I make a .gif file from a .fig (taken on Cabrì)?
Thank you very much!  :)

To add a link to your post, use the "URL" button on the posting page, or use the url tag.

See: [url]http://www.artofproblemsolving.com/Forum/faq.php?mode=bbcode[/url]

Questions such as these belong in this form, not the support forum (which is reserved for reporting bugs or other things wrong with the site).


integration
calculus
limit
real analysis
real analysis unsolved

for $f \in C^2[0,+\infty)$, $\int_0^{+\infty}f^2$ and  $\int_0^{+\infty}(f\")^2$
Prove: $\int_0^{+\infty}(f')^2$

 $\int_0^{+\infty}f^2$ means  $\int_0^{+\infty}f^2(x)dx$
thanks for help :)

I would assume that you left out a few words and what you want is the statement that if the first two integrals converge, then so does the third one.

$\int_0^m(f'(x))^2\,dx=\left.f(x)f'(x)\right|_0^m-\int_0^mf(x)f''(x)\,dx$

Let's assume for now that we will later be able to prove that $\lim_{m\to\infty}f(m)f'(m)=0.$ Then we have

$\int_0^{\infty}(f'(x))^2\,dx=f(0)f'(0)-\int_0^{\infty}f(x)f''(x)\,dx$

But $\left|-\int_0^{\infty}f(x)f''(x)\,dx\right|\le\left(\int_0^{\infty}(f(x))^2\,dx\right)^{\frac12} \left(\int_0^{\infty}(f''(x))^2\,dx\right)^{\frac12}$

So if the integrals of $f^2$ and $(f'')^2$ are finite, then so is the integral of $(f')^2.$

I left out one part of the proof; it's annoying, but true. Someone else can fill it in.

Thanks,Kent Merryfield.
But you do is what I do.
I also want to prove:
$\lim_{m \to +\infty} f(m)f'(m)=0$,It seems strange,I have doubted it.But I think it must be right,If not the oringinal problem is wrong.

Now somebody can help me?

Hi,
I have proved that:
$\lim_{x \to +\infty}\frac{f'(x)}{\sqrt x}=0$
so I need to prove $xf^2(x)$ is bound.
Is it ture.
then we can prove:$\lim_{m \to +\infty} f(m)f'(m)=0$
can somebody help me?

This is an interesting results. It may be usefull this theorem(due to hardy and littlewood):

If $\lim f(x)=0$ and $f''$ is bounded then $f'\to 0$.
If it helps you I can give a proof for it.  :lol:

Thanks.
But sorry,I didn't know how to use the theorem you posted.
Anyway,I also like the proof for the theorem. :lol:

[quote="Kent Merryfield"]Let's assume for now that we will later be able to prove that $\lim_{m\to\infty}f(m)f'(m)=0.$[/quote]

How to prove that?

A proof for the Hardy's theorem:
Define $||f||_{\infty,x}=sup_{t\geq x}|f(t)|$.
$f(x+a)=f(x)+af'(x)+\frac{a^2}{2}f''(c_x)$
$f'(x)=\frac{f(x+a)-f(x)-\frac{a^2}{2}f''(c_x)}{a}$
$|f'(x)| \leq \frac{2||f||_{\infty,x} + \frac{a^2}{2}||f''||_{\infty,x}}{a}$
Let $a=2\sqrt{\frac{||f||_{\infty,x}}{||f''||_{\infty,x}}}$
So we have $|f'(x)| \leq 2 \sqrt{||f||_{\infty,x} \cdot ||f''||_{\infty,x}}$ and conclusion follows...
You can also check this post: http://www.mathlinks.ro/Forum/viewtopic.php?t=20015 which seems to be a "solved" problem.  :)

I think it is a number from $[x,x+a]$. Isn't that Taylor's theorem?


induction inequalities trigonometry algebra polynomial Vieta FTW

What did everyone get for Round 4?

Here is what I got:

1. 4 (assume 251 is at the top)

2. 1715 (error argument)

3. 1/2 (I used induction to get a binomial for (a_2^2+a_3^2)^n and a geometric series, but I suppose an easier way would have been to separate the even and odd terms...)

4. Multiplying the inequalities yielded the impossible inequality w^2+wx+x^2<yz<[(wx)^2]/w^2+wx+x^2, and I think casework also works (thank you samath). I actually though this one was really tricky....

5. I failed to solve this one, but I am now aware of two possibilities: (1) trig bashing, which I tried but didn't work for me :maybe: , and (2) similar right triangles and repeated use of the pythagorean theorem.

I think you're supposed to wait until after the Math Jam (in 5.5 hours as of now).

EDIT: never mind, I am wrong.

However,

[hide]
1. 4
2. 1715
3. 1/2
4. VIETA FTW
5. Similar triangles, then Ptolemy Ptolemy Ptolemy!
[/hide]

I think I did pretty good except, of course, for the last one.

[hide]
1. 4
2. 1715
3. 1/2
4. Well, I proved it.
5. ?? I reduced it to representing a tan*cos as a sum of sines, but was stuck after that.

Hopefully my arguments are rigorous enough. I felt like my wording for #3 was shaky, but I know the idea was good (summing over representations instead of $ a_n$, if that makes sense). For #4 I used Cauchy-Schwarz inequality and some clever rearrangements to get a contradiction. It felt good to solve it, since it was really the first competition-level inequality I proved. In retrospect, for #4 I probably should have realized that there were cyclic quadrilaterals.
[/hide]

Hmm anyway I will post my Vieta solution

Consider the polynomial $ P(t)$ with roots $ \minus{}w,\minus{}x,y,z$

Then by Vieta, all the coefficients of $ P$ are positive

However, $ y$ is positive and a root of $ P$; contradiction.

 :)

[hide="My approach for 4"]I showed that the second inequality (in the order given, with $ w\plus{}x<y\plus{}z$ counting as the first) can be rewritten as $ \frac{1}{w}\plus{}\frac{1}{x}<\frac{1}{y}\plus{}\frac{1}{z}$.[/hide]

I did this entire problem on Monday.

1. Assumed 251*2 was at the top, then casework
2. Weird solution
3. Induction thing
4. Really weird solution
5. Something to do with a lot of rotations

I did 1-3 on Monday. And because of that, I didn't actually find a proof for #2 and almost got the wrong answer for #3. So I basically failed this round.

[hide="For number 4,"]
I just kept using AM-GM (mainly $ (w\plus{}x)^2 \ge 4wx$), because that was the only inequality I could think of to use for sums and products. And then I kept forcing the inequality into $ (w\plus{}x)^2$, first by multiplying the two inequalities, then substituting $ w\plus{}x < y\plus{}z$ into the second inequality, and then a contradiction easily appears.
[/hide]

I spent most for this round trig-bashing #5. That was fun.

[quote="McDutchy"]There was a very nice geometric solution to #5 that I haven't seen very many people discussing. Did anyone else besides my self use this solution?[/quote]

Dude Ptolemy was pretty nice...

Gah, so I did see this problem before ... no wonder the motivation felt so obvious :!:  Guess it wasn't a very well-picked problem? :huh:

[quote="kyyuanmathcount"][quote="not_trig"]Hmm anyway I will post my Vieta solution

Consider the polynomial $ P(t)$ with roots $ \minus{} w, \minus{} x,y,z$

Then by Vieta, all the coefficients of $ P$ are positive

However, $ y$ is positive and a root of $ P$; contradiction.

 :)[/quote]

Teehee, look at Engel Chapter 7 (Inequalities) Problem #22.  I actually had that solution written in the book next to their solution (since I worked through that during the summer).  :D[/quote]

Yeah, I saw that one too, but that used like... squares and stuff, actual inequality-solving tactics. :P

OMG! I can't believe this....I noticed the symmetry for #5 and spent like 30 mins trying to work it out, but I drew the parallelogram in the wrong direction... :o Ah well.

[quote="trivial34"]On problem three I found an algebraic approach that works out quite nicely.
[hide]First substitute the series defenitions and then factor out u(1-u). This leaves you with sums of squares of each term. Call this sum S. Next, seperate all the terms into 1 and even and odd sums. Using substitution and factoring gives ((u^2) + (1-u)^2)S + 1 = S. This can be solved for S and then multiplied by u(1-u) to give 1/2. I was very surprised to get this answer: so skeptical that I wrote a program to simulate the summation  :) [/hide][/quote]

I found this too :)

I was like hmm let's get stuff in terms of stuff

hey a quadratic, yay

I was also skeptical when I got the answer to #3.

[hide="Something I noticed"]The closer $ \mu$ is to $ \frac{1}{2}$, the faster the sum seemed to approach $ \frac{1}{2}$.[/hide]

For #4, I expanded the 2nd equation and then made it so that you had (x-z)(y-w) + (some other stuff) was negative so i said that x < z and y > w or x > z and y < w; the 2nd isn't true obviously

so then i expanded it a different way again and showed that each of y and z are greater than each of w and x. Then i substituted w+a in for y and x+b in for z in the first equation and got a contradiction.

[quote="lingomaniac88"]I was also skeptical when I got the answer to #3.

[hide="Something I noticed"]The closer $ \mu$ is to $ \frac {1}{2}$, the faster the sum seemed to approach $ \frac {1}{2}$.[/hide][/quote]

The reason for that is probably obvious, but here it is anyway.
[hide]$ \sum_{k \equal{} 1}^{\infty}a_{2k}a_{2k \plus{} 1} \equal{} (\mu \minus{} \mu^2)\sum_{k \equal{} 1}^{\infty}a^2_k \equal{} (\mu \minus{} \mu^2)\sum_{n \equal{} 0}^{\infty}\left(1 \minus{} 2\mu \plus{} 2\mu^2\right)^n$ (after a lemma and plenty of simplification).

$ S_k \equal{} (\mu \minus{} \mu^2)\sum_{n \equal{} 0}^{k}\left(1 \minus{} 2\mu \plus{} 2\mu^2\right)^n \equal{} \frac {1 \minus{} (1 \minus{} 2\mu \plus{} 2\mu^2)^{k \plus{} 1}}{2}$.

Since $ 1 \minus{} 2\mu \plus{} 2\mu^2$ decreases as $ \mu \to \frac {1}{2}$, $ S_k \to \frac {1}{2}$ faster as $ \mu \to \frac {1}{2}$.[/hide]

For #4, did anyone else get a solution that proved $ w\plus{}x$ could not be positive by showing it was less than $ \frac{y\plus{}z}{a_n}$ where $ a_n$ is a sequence that approaches $ \infty$?

I found such a neat solution the day of the deadline (sounds like a title to a comedy), then failed to get a similarly neat (or otherwise) solution to #5 even later the day of.

Yeah that was also my solution to #4. I got the sequence you speak of to be $ a_{n\plus{}1}\equal{}4a_n\minus{}1$

Can someone give me a proof that the the sum required in number 3 converges?

heheh I attached my solutions (my solution to 5 is scary)


If $x,y>0$ then prove $ 2( \frac{x^2}{y^2}+ \frac{y^2}{x^2} ) \geq \frac{x}{y} + \frac{y}{x}+2 $.

cheers! :D :D :cool:

You can multiply x^2y^2 and apply Muirhead's right away, but you can also do this:

Let x/y + y/x = a.

Suffices to prove 2(a^2 - 2) \geq a + 2
<=> (2a + 3)(a - 2) \geq 0. True because a > 0 implies 2a + 3 > 0 and a = x/y + y/x \geq 2 (AM-GM) implies a - 2 \geq 0.


geometry
circumcircle
angle bisector
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle,0 the circumcircle of ABC, O the center of 0, A' the midpoint of the arc BAC, A" the antipole of A' wrt 0, M the midpoint of BC and A* the intersection of the tangents to 0 at B and C.
Prove or disprove that   A"M/A"A* = OM/OA'.
Sincerely
Jean-Louis

$ BA'CA'', BOCA^*$ are cyclic kites $ \Longrightarrow$ $ (\overline{A'O} \plus{} \overline{OM}) \cdot \overline{MA''} \equal{} \minus{}\overline{MB} \cdot \overline{MC} \equal{} \overline{OM} \cdot (\overline{MA''} \plus{} \overline{A''A^*})$ $ \Longrightarrow$ $ \overline{MA''} \cdot \overline{A'O} \equal{} \overline{OM} \cdot \overline{A''A^*}$ $ \Longrightarrow$ $ \frac{\overline{A''M}}{\overline{A''A^*}} \equal{} \frac{\overline{OM}}{\overline{OA'}}.$

Dear Mathlinkers,
a synthetic proof of this result has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis

$BA"$ is angle bisector of $\angle MBA^*$, hence $\frac{A"M}{A"A^{*}}=\frac{BM}{BA^{*}}$ $(1)$, but $\triangle OBA^*$ is right-angled, so $\triangle MBA^* \sim \triangle MOB$, with $\frac{MB}{BA^*}=\frac{MO}{OB}$ $(2)$, but $OA'=OB$, done.

Best regards,
sunken rock


Solve the system of equations: $\begin{cases}3x^{3}+5y^{3}=6+2xy\\2x^{3}+3y^{3}=8-3xy\end{cases}$

[hide="Outline"]Solve for $x^{3},y^{3}$ in terms of $xy$, then multiply the obtained equations and substitute $t=xy$ to get a cubic in $t$. One nice solution is $t=1$, giving $x=y=1$, and there are two much uglier, but nevertheless real solutions for $t$:

\begin{eqnarray*}&& t_{2,3}=-137\pm\sqrt{19033}\\ &\implies& (x_{2},y_{2})=(\sqrt[3]{2899-21\sqrt{19033}},\sqrt[3]{-1793+13\sqrt{19033}})\\ && (x_{3},y_{3})=(\sqrt[3]{2899+21\sqrt{19033}},\sqrt[3]{-1793-13\sqrt{19033}})\end{eqnarray*}[/hide]

EDIT: A calculation error was corrected.


This is my first attempt with geogebra. So forgive me if I have gone wrong anywhere.

[geogebra]5c21840ab9459ec6d237c354d7648e7029dbf1ec[/geogebra]

$ In  \triangle ABC , \angle C \equal{} 3 \angle A$

$ i.e. \alpha \equal{} 3 \cdot \beta$

and $ BC \equal{} 27$ units   while $ AB \equal{} 48$ units

Find $ AC$

In triangle, if $ \alpha\equal{}3\beta$, you can prove $ (a\minus{}b)^2(a\plus{}b)\equal{}bc^2$ by using Law Of Sine on small triangles if you put $ AD$ to be a line for which $ \angle BAD\equal{}\beta,\angle CAD\equal{}2\beta,D\in BC$

hello, we have $ \frac{\sin(3\alpha)}{\sin(\alpha)}\equal{}\frac{48}{27}$, applying the formula for $ \sin(3\alpha)$ and dividing by $ \sin(\alpha)$ we get $ \cos(\alpha)\equal{}\frac{5}{6}$ (the solution $ \minus{}\frac{5}{6}$ is meaningless),
from here we have $ b^2\equal{}48^2\plus{}27^2\minus{}2\cdot48\cdot27\cdot\frac{5}{6}$ or $ b\equal{}\sqrt{873}$.
Sonnhard.

[quote="harekrishna"]I am not keen on using Geogebra again.

Nor do I think I get the answer Dr Sonnhard Graubner got.

[hide=" Let me give you a hint"]

First construct a line segment from A to a point D on the opposite side such that $ \angle ACD$ measures $ \angle CAD$

And so you get two isosceles triangles.

[/hide][/quote]

Did you see my solution? I also get 2 isosceles triangles, and get everything perfect (I am sure) just I used $ \angle A\equal{}3\angle B$. Substitute $ B$ for $ C$ and corresponding sides and it's your problem.


2^16^x = 16^2^x
I don't at all understand the explanation from the book. :( Thank you for any help.

[hide="Solution"]Given: $ 2^{16^x}\equal{}16^{2^x}$

We know $ 16\equal{}2^4$, so the RHS can be rewritten as $ 2^{4\cdot2^x}$, or $ 2^{2^{x\plus{}2}}$.

Now, take the $ \log_2$ of both sides of $ 2^{16^x}\equal{}2^{2^{x\plus{}2}}$ to get $ 16^x\equal{}2^{x\plus{}2}$.

Again using $ 16\equal{}2^4$, we can rewrite the LHS as $ 2^{4x}$.

Take the $ \log_2$ of both sides again to get $ 4x\equal{}x\plus{}2$, so $ x\equal{}\tfrac{2}{3}$.[/hide]

Thank you very much! It all makes sense again.


Don is penniless.
Cao feels sorry for Don.  Cao gives Don $ \$ 363.60$.  Cao now has $ 4/5$ of the money he started with.
Ben is even more generous.  Ben gives Don $ 1/4$ of his money.
Not to be outdone, Abe donates to Don $ 1/3$ of his money.
However, each gave Don the same exact amount of money.
What is the ratio of Don's new money to the total remaining money of Don's benefactors?

[hide="Solution"]Don acquired $ 3\cdot\$363.60$.

His benefactors have $ (2\plus{}3\plus{}4)(\$363.30)$.

$ \frac{3}{2\plus{}3\plus{}4}\equal{}\boxed{\frac{1}{3}}$[/hide]

[hide="semi solution because of poor vocabulary"]x/3=y/4=z/5=363.6
im stuck. what does benefactors mean[/hide]

[hide="solution"]
A: benefactors are the ppl who give u stuff if yr like poor or needy.
B: he has 3*363.6 dollars(why cant they use round #s?) his benefactors had (3+4+5)*363.6. they now have (2+3+4)*363.6

in total, Don now has 3/(2+3+4)=1/3 of his benefactor's money.[/hide]

[quote="mz94"]

B: he has 3*363.6 dollars(why cant they use round #s?) [/quote]
To discourage people from multiplying everything out.


What is the perfect square trinomial for
16x^4 -56^2 *y + __  = ( ___ - ___ )^2


[quote="arqady"][quote="bbbbbbbbbb"]x, y, z >0.Prove this inequality
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}z+y^{3}x+z^{3}y \]
It's not like vasc-inequality.
His inequality is
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}y+y^{3}z+z^{3}x \]
:).

It's easier :).[/quote]
I think, it's a same inequality. :wink:[/quote]
Can you prove first ineq by cauchy :-?

[quote="bbbbbbbbbb"]
Can you prove first ineq by cauchy :-?[/quote]
No, because I cannot prove the second inequality by  Cauchy's inequality. If you are able to do it, show as. Thank you.

:rotfl: Of course it is [b]much[/b] easier...
How did you solve it with Cauchy?

As a sidenote, in Vasc's inequality x, y, z are reals (not necessarily positive). I have no idea why people always forget this.

  darij

[quote="darij grinberg"]:rotfl: Of course it is [b]much[/b] easier...
How did you solve it with Cauchy?
[/quote]
I agree with you, darij grinberg, it's much easier.
But what your proof by Cauchy's inequality? Thank you.

[quote="arqady"][quote="bbbbbbbbbb"]x, y, z >0.Prove this inequality
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}z+y^{3}x+z^{3}y \]
It's not like vasc-inequality.
His inequality is
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}y+y^{3}z+z^{3}x \]
:).

It's easier :).[/quote]
I think, it's a same inequality. :wink:[/quote]
No argady, it's very diff inequality because vasc inequality has equality when a=b=c and (a,b,c)=k.sin^2 ... (do u remember).
Above inequality has equality when a=b=c only :).

And I have a stronger  Inequality
\[ \& \frac43\left({a^{2}+b^{2}+c^{2}}\right)^{2}\ge{{\left({a^{2}+b^{2}+c^{2}}\right)^{2}}}+{}\left({a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}\right) \ge 4\left({a^{3}c+b^{3}a+c^{3}b}\right) \cr \& \cr}\]
Proof is very nice.
PS: This inequality is true when a, b, c be real numbers.

[quote="MithitxaoA-One"][quote="arqady"][quote="bbbbbbbbbb"]x, y, z >0.Prove this inequality
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}z+y^{3}x+z^{3}y \]
It's not like vasc-inequality.
His inequality is
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}y+y^{3}z+z^{3}x \]
:).

It's easier :).[/quote]
I think, it's a same inequality. :wink:[/quote]
No argady, it's very diff inequality because vasc inequality has equality when a=b=c and (a,b,c)=k.sin^2 ... (do u remember).
Above inequality has equality when a=b=c only :).[/quote]

Wrong. "Above inequality" is just Vasc's inequality, applied to the numbers x, z, y instead of x, y, z.

[quote="MithitxaoA-One"]And I have a stronger  Inequality
\[ \& \left({a^{2}+b^{2}+c^{2}}\right)^{2}\ge{{\left({a^{2}+b^{2}+c^{2}}\right)^{2}}\over 2}+{3 \over 2}\left({a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}\right) \ge 3\left({a^{3}c+b^{3}a+c^{3}b}\right) \cr \& \cr}\]
Proof is very nice.
PS: This inequality is true when a, b, c be real numbers.[/quote]

Wrong for a = 2, b = 2, c = 3.

  Darij

[quote="darij grinberg"][/quote][quote="MithitxaoA-One"][quote="arqady"][quote="bbbbbbbbbb"]x, y, z >0.Prove this inequality
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}z+y^{3}x+z^{3}y \]
It's not like vasc-inequality.
His inequality is
\[ \frac{{\left({x^{2}+y^{2}+z^{2}}\right)^{2}}}{3}\ge x^{3}y+y^{3}z+z^{3}x \]
:).

It's easier :).[/quote]
I think, it's a same inequality. :wink:[/quote]
No argady, it's very diff inequality because vasc inequality has equality when a=b=c and (a,b,c)=k.sin^2 ... (do u remember).
Above inequality has equality when a=b=c only :).[/quote]

Wrong. "Above inequality" is just Vasc's inequality, applied to the numbers x, z, y instead of x, y, z.

[quote="MithitxaoA-One"]And I have a stronger  Inequality
\[ \& \left({a^{2}+b^{2}+c^{2}}\right)^{2}\ge{{\left({a^{2}+b^{2}+c^{2}}\right)^{2}}\over 2}+{3 \over 2}\left({a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}\right) \ge 3\left({a^{3}c+b^{3}a+c^{3}b}\right) \cr \& \cr}\]
Proof is very nice.
PS: This inequality is true when a, b, c be real numbers.[/quote][quote="darij grinberg"]

Wrong for a = 2, b = 2, c = 3.

  Darij[/quote]
"Above ineq" mean topmost inequality, it's not  vasc's inequality :)
[quote]Wrong for a = 2, b = 2, c = 3.[/quote]
Sorry, I'm drowsing so my caculation be mistaked :D. I edited

[quote="MithitxaoA-One"]\[ \& \frac43\left({a^{2}+b^{2}+c^{2}}\right)^{2}\ge{{\left({a^{2}+b^{2}+c^{2}}\right)^{2}}}+{}\left({a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}\right) \ge 4\left({a^{3}c+b^{3}a+c^{3}b}\right) \cr \& \cr}\][/quote]

Still wrong, try a = 10, b = 11, c = 12.

And I did speak about the topmost inequality.

  Darij


Let ABC be a triangle for which there is the point D which lies inside (ABC)
such that m(<DAB)=m(<DCA) and m(<DBA)=m(<DAC). We note the points E, F which belong to the rays (AB and (CA respectively such that AB=BE and CA=AF. Prove that the points A, E, D, F are placed on a circle.

[color=red][Moderator edit: Slight correction.][/color]

The figures $DABE$ and $DCAF$ are similar, so $\angle DEA=\angle DFA$, which means that $DAFE$ is cyclic, as desired.


calculus
integration
algebra
function
domain
superior algebra
superior algebra solved

Let $ F$ be a field and let $ D$ be an integral domain which is a finite dimensional $ F$-algebra. Prove that $ D$ is a field.

$ d\mapsto ad$ is a F-linear map. It is injective if $ a\neq 0$ because D is an integral domain. It is surjective because the dimension is finite. Hence there is an inverse for a.

I found the proof confusing for a moment because it almost seems like you are saying that since multiplication by $ a$ is a bijection, an inverse for the liniar transformation gives an inverse for $ a$ when in fact you were saying that since the map is surjective, there exists $ b$ such that $ ab\equal{}a(b)\equal{}1$.

I only post this so that nobody else makes the same mistake and I can move this to solved with a clear conscience


integration
function
calculus
logarithms
real analysis
real analysis unsolved

[color=darkred]Ascertain without the decomposition of the denominator in irreducible factors (over $ \mathcal R$) :
$ 1\blacktriangleright A_k \equal{} \int\frac {x^k}{x^6 \plus{} 1}\ \mathrm {dx}$ where $ k\in \overline {0,5}\ .$

$ 2\blacktriangleright B_k \equal{} \int\frac {x^k}{x^8 \plus{} 1}\ \mathrm {dx}$, where $ k\in \overline {0,7}\ .$

[b][u]Problem.[/u][/b] Let $ f: \mathcal R\rightarrow\mathcal R$ be the function with the correspondence $ f(x) \equal{} \frac {x^4 \plus{} 1}{x^6 \plus{} 1}$.
Prove that on the interval $ [2,\infty )$ the function $ f$ admits a primitive as the form
$ F(x) \equal{} \frac 13\cdot \arctan\frac {P(x)}{Q(x)}$, where $ \{P,Q\}\subset \mathcal Z[X]$. But on the interval $ (\minus{}\infty , 2]\ ?$[/color]

$ A_2\equal{}\dfrac{1}{3}\cdot\int\frac{\mathrm{dx^3}}{1\plus{}(x^3)^2}
\equal{}\frac{1}{3}\cdot\arctan x^3\plus{}C$. :P

I don't understand something,
Virgil Nicula we have already answer for  this integral $ \int\frac {x^{m}}{1 \plus{} x^{n}}dx$

[quote="sos440"]Let $ m, n$ be integers with $ 0 \leq m < n$, and $ \theta_{k} \equal{} \frac {2k \minus{} 1}{n}\pi$. Then

$ \int \frac {x^{m}}{1 \plus{} x^{n}}\, dx \equal{} \frac {1}{n}\sum_{k \equal{} 1}^{n}\left( \sin \left((m \plus{} 1) \theta_{k}\right) \tan^{ \minus{} 1}\left(\frac {x \minus{} \cos \theta_{k}}{\sin \theta_{k}}\right) \minus{} \frac {1}{2}\cos \left((m \plus{} 1) \theta_{k}\right) \log \left(x^{2} \minus{} 2 \cos \theta_{k}x \plus{} 1\right)\right)$

For odd $ n$, $ \sin \theta_{k}$ becomes zero for $ k \equal{} (n \plus{} 1)/2$. In this case, we consider the term

$ \sin \left((m \plus{} 1) \theta_{k}\right) \tan^{ \minus{} 1}\left(\frac {x \minus{} \cos \theta_{k}}{\sin \theta_{k}}\right)$

to be also zero.[/quote]

[color=darkblue][b]Very well, [u]Extremal[/u] ! Sorry, and I have what you have.[/b]  You are a bit exclusively.
Exist many high school - students who are interested by same problems.
With other words, these are some exercises for the students of high school.

Exists a section on this site for calculus in high school ?!
If [b]yes[/b], please you move there the all my proposed problems.[/color]

[color=darkred]Thank you for the your housing, more exactely, hospitality ![/color]

$ \blacktriangleright A_2 \equal{} \int\frac {x^2}{x^6 \plus{} 1}\ \mathrm {dx} \equal{} \frac 13\cdot\int\frac {\left(x^3\right)'}{\left(x^3\right)^2 \plus{} 1}\ \mathrm {dx}\implies \boxed {\ A_2 \equal{} \frac 13\cdot\arctan\left(x^3\right) \plus{} \mathcal C\ }\ .$

$ \blacktriangleright A_5 \equal{} \int\frac {x^5}{x^6 \plus{} 1}\ \mathrm {dx} \equal{} \frac 16\cdot\int\frac {\left(x^6 \plus{} 1\right)'}{x^6 \plus{} 1}\ \mathrm {dx}\implies \boxed {\ A_5 \equal{} \frac 16\cdot\ln\left(x^6 \plus{} 1\right) \plus{} \mathcal C\ }\ .$

$ \blacktriangleright A_4 \minus{} A_0 \equal{} \int\frac {x^4 \minus{} 1}{x^6 \plus{} 1}\ \mathrm {dx} \equal{}$ $ \int\frac {x^2 \minus{} 1}{x^4 \minus{} x^2 \plus{} 1}\ \mathrm {dx} \equal{}$ $ \int\frac {1 \minus{} \frac {1}{x^2}}{x^2 \plus{} \frac {1}{x^2} \minus{} 1}\ \mathrm {dx} \equal{}$

$ \int\frac {\left(x \plus{} \frac 1x\right)'}{\left(x \plus{} \frac 1x\right)^2 \minus{} 3}\ \mathrm {dx}\implies$ $ \boxed {\ A_4 \minus{} A_0 \equal{} \frac {\sqrt 3}{6}\cdot\ln\left|\frac {x^2 \minus{} x\sqrt 3 \plus{} 1}{x^2 \plus{} x\sqrt 3 \plus{} 1}\right| \plus{} \mathcal C\ }\ .$

$ \blacktriangleright A_4 \minus{} A_2 \plus{} A_0 \equal{} \int\frac {x^4 \minus{} x^2 \plus{} 1}{x^6 \plus{} 1}\ \mathrm {dx} \equal{} \int\frac {1}{x^2 \plus{} 1}\ \mathrm {dx}\implies \boxed {\ A_4 \plus{} A_0 \equal{} A_2 \plus{} \arctan x\ }\ .$[hide="Otherwise."]$ A_4 \plus{} 2\cdot A_2 \plus{} A_0 \equal{}$ $ \int\frac {(x^2 \plus{} 1)^2}{x^6 \plus{} 1}\ \mathrm {dx} \equal{}$ $ \int\frac {x^2 \plus{} 1}{x^4 \minus{} x^2 \plus{} 1}\ \mathrm {dx} \equal{}$ $ \int\frac {1 \plus{} \frac {1}{x^2}}{x^2 \plus{} \frac {1}{x^2} \minus{} 1}\ \mathrm {dx} \equal{}$ $ \int\frac {\left(x \minus{} \frac 1x\right)'}{\left(x \minus{} \frac 1x\right)^2 \plus{} 1}\ \mathrm {dx} \equal{}$ $ \arctan\frac {x^2 \minus{} 1}{x} \plus{} \mathcal C$ $ \implies$ $ \boxed {\ A_4 \plus{} A_0 \equal{} \minus{} 2\cdot A_2 \plus{} \arctan\frac {x^2 \minus{} 1}{x}\ }\ .$[/hide][color=darkblue]From the last two relations obtain $ A_0$ and $ A_4\ .$

$ \blacktriangleright A_3 \plus{} A_1 \equal{} \int\frac {x(x^2 \plus{} 1)}{x^6 \plus{} 1}\ \mathrm {dx} \equal{} \int\frac {x}{x^4 \minus{} x^2 \plus{} 1}\ \mathrm {dx}$ $ \implies$

$ \boxed {\ A_3 \plus{} A_1 \equal{} \frac {3}{\sqrt 3}\cdot\arctan\frac {2x^2 \minus{} 1}{\sqrt 3} \plus{} \mathcal C\ }\ .$

$ \blacktriangleright$ $ A_5\minus{}A_3 \plus{} A_1 \equal{} \int\frac {x(x^4\minus{}x^2\plus{}1)}{x^6 \plus{} 1}\ \mathrm {dx} \equal{}$ $ \int\frac {x}{x^2 \plus{} 1}\ \mathrm {dx}$ $ \implies$

$ \boxed {\ A_3\minus{}A_1\equal{}A_5\minus{}\frac 12\cdot\ln (x^2\plus{}1)\ }\ .$ From the last two relations obtain $ A_1$ and $ A_3\ .$

[/color]


It is now 12:00:00 midnight, as read on a 12-hour digital clock.
In 122 hours, 39 minutes and 44 seconds the time will be $ A: B: C$.
What is the value of $ A \plus{} B \plus{} C$?

[quote="kabi"]Hello
The time is $ 02: 39: 44$.
Answer:$ 85$
Thank u.[/quote]

yes that solution is soooo obvious....

proper solution:

taking 122 mod 24 (24 hrs in a day) we get 2, so going ahead 122 hrs is equivalent to going ahead only 2 hrs, or 2AM. the rest is trivial. 00+39=39 min and 00+44=44 sec

so 2+39+44=85


Yesterday and today, when I have attempted to login via IE5, the login screen accepts my Username and Password, and then IE goes to the "Cannot find server" page.  [Screen shot attached.]

When I click "Refresh", IE gives a little more detail in the Tooltip Text for the IE Button on the Taskbar.  [Screen shot attached.]

Since a similar problem of a few days ago, I have had my DNS server logging requests.  The apparently bad DNS "www.artofproblemsolving" was not submitted to my DNS server for resolution.  So I think it's more likely that IE was supplied a syntactically invalid redirect address, which it couldn't submit for resolution.

In case you guys have logs, the time shown in the lower right corner is correct and shows Mountain Daylight Time (MDT).  [One hour ahead of Pacific Daylight Time.]

BTW, the workaround is trivial -- I just type the URL by hand or I click back twice and then "Refresh".  I find that I have been logged in.  Cheers!

tried http://www.mathlinks.ro ?  If one is down, the other usually works :)


what is a finite difference formula where can it be used? i googled it but i still dont get it .can someone explain in simple terms please.


[quote="April"]Let $a,\,b,\,c,\,d$ are positive numbers. Prove that:
\[\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\]
[/quote]
For $a_{1},$ $a_{2},...,$ $a_{n}$ are positive numbers and  $n\geq2$  the following inequality holds:
$\sum_{cyc}\frac{a_{1}}{a_{2}+2a_{3}+...+(n-1)a_{n}}\geq\frac{2}{n-1}.$ :wink:

[quote="arqady"]For $a_{1},$ $a_{2},...,$ $a_{n}$ are positive numbers and  $n\geq2$  the following inequality holds:
$\sum_{cyc}\frac{a_{1}}{a_{2}+2a_{3}+...+(n-1)a_{n}}\geq\frac{2}{n-1}.$ :wink:[/quote]
hmm... nice generalization, but can you explain or provide a proof :?:

oh nice.. thanks!
being new to inequalities, it took me a while to understand your first line:[quote="SplashD"]By Cauchy in Engel form,
$\sum_{cyc}\frac{a_{1}}{a_{2}+2a_{3}+...+(n-1)a_{n}}\geq \frac{(\sum_{cyc}a_{1})^{2}}{n \sum_{sym}a_{1}a_{2}}$
[/quote]i guess you used cauchy on $ \sum_{cyc}\frac{a_{1}^{2}}{a_{1}a_{2}+2a_{1}a_{3}+...+(n-1)a_{1}a_{n}}$

[quote=sqing]Let $a,\,b,\,c,\,d$ are positive numbers. Prove that:
\[\frac{a}{b+c+4d}+\frac{b}{c+d+4a}+\frac{c}{d+a+4b}+\frac{d}{a+b+4c}\ge\frac{2}{3}.\][/quote]
By C-S $\sum_{cyc}\frac{a}{b+c+4d}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ac+4da)}=\frac{(a+b+c+d)^2}{5\sum\limits_{cyc}ab+2(ac+bd)}$.
Hence, it remains to prove that $3(a^2+b^2+c^2+d^2)-4(ab+bc+cd+da)+2(ac+bd)\geq0$, which is
$(a-b+c-d)^2+\sum_{cyc}(a-b)^2\geq0$. Done!

[quote=sqing]Let $a,\,b,\,c,\,d$ are positive numbers. Prove that:
\[\frac{a}{b+c+4d}+\frac{b}{c+d+4a}+\frac{c}{d+a+4b}+\frac{d}{a+b+4c}\ge\frac{2}{3}.\][/quote]

We use Rearrangement inequality, let $a\ge b\ge c\ge d$ then
$\frac{a}{b+c+4d}+\frac{b}{c+d+4a}+\frac{c}{d+a+4b}+\frac{d}{a+b+4c}\ge\frac{b}{b+c+4d}+\frac{c}{c+d+4a}+\frac{d}{d+a+4b}+\frac{a}{a+b+4c}$
$\frac{a}{b+c+4d}+\frac{b}{c+d+4a}+\frac{c}{d+a+4b}+\frac{d}{a+b+4c}\ge\frac{c}{b+c+4d}+\frac{d}{c+d+4a}+\frac{a}{d+a+4b}+\frac{b}{a+b+4c}$
$\frac{4a}{b+c+4d}+\frac{4b}{c+d+4a}+\frac{4c}{d+a+4b}+\frac{4d}{a+b+4c}\ge\frac{4d}{b+c+4d}+\frac{4a}{c+d+4a}+\frac{4b}{d+a+4b}+\frac{4c}{a+b+4c}$

By adding these 3 inequalities we get the inequality to prove.

[Let $a,\,b,\,c,\,d$ are positive numbers. Prove that:
\[\frac{a}{a+b+c+3d}+\frac{b}{b+c+d+3a}+\frac{c}{c+d+a+3b}+\frac{d}{d+a+b+3c}\ge\frac{2}{3}.\]
[url=http://www.artofproblemsolving.com/community/c6h1136323p5308336]here[/url]


logarithms
inequalities
limit
algebra unsolved
algebra

of my friend :
$lim_{n\to  \infty }1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}-lnn$

[quote="tranthanhnam"]of my friend : $\lim_{n\to \infty }\left[1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}\right]-\ln n$[/quote]

[color=darkblue]Denote $A_{n}=\sum_{k=1}^{n}\frac{1}{k}$ and $E_{n}=A_{n}-\ln n\ .$ Suppose well-known the double inequality for $k\in N^{*}\ ,$ $\left(1+\frac{1}{k}\right)^{k}<e<\left(1+\frac{1}{k}\right)^{k+1}$ $\Longleftrightarrow$ $\boxed{\ \frac{1}{k+1}<\ln\frac{k+1}{k}<\frac{1}{k}\ }\ \ (*)\ .$ Therefore, $\sum_{k=1}^{n}\frac{1}{k+1}<\ln (n+1)<\sum_{k=1}^{n}\frac{1}{k}$ $\implies$ $A_{n}+\frac{1}{n+1}-1<\ln (n+1)<A_{n}$ $\implies$ $\ln (n+1)<A_{n}<\ln (n+1)+\frac{n}{n+1}\ \ (1)\ .$ From this inequality obtain $\boxed{\ \lim_{n\to\infty}A_{n}=\infty\ }\ .$ From the inequality $(*)$ obtain $E_{n+1}-E_{n}=\frac{1}{n+1}-\ln \frac{n+1}{n}<0\ ,$ what means that the sequence $E_{n}$ is decreasing $(\searrow)$  with positive terms, therefore is convergent. From the relation $(1)$ obtain : $\ln (n+1)<E_{n}+\ln n<\ln (n+1)+\frac{n}{n+1}$ $\implies$ $\ln \frac{n+1}{n}<E_{n}<\ln\frac{n+1}{n}+\frac{n}{n+1}\ .$ For $n\to\infty$ in this double inequality obtain $E_{n}\searrow E\in (0,1)\ .$ The number $E$ is named the [b][u]Euler's number[/u][/b] and $E=0,577215\ldots\ .$[/color]


If $ a$ is a constant such that $ 9x^2 \plus{} 24x \plus{} a$ is the square of a binomial, then what is $ a$?

Remember that a trinomial that represents the square of a binomial always comes in the form $ a^2\plus{}2ab\plus{}b^2$, with the variables taken from the original expression $ (a\plus{}b)^2$. Looking at the first term, $ a^2\equal{}9$, so $ a\equal{}3$. Substituting that into the second term, $ b(2(3))\equal{}24$, or $ 6b\equal{}24$. Dividing by $ 6$ on both sides of that, $ b\equal{}4$. [Hilariously,] $ b$ in our square-of-a-binomial formula represents the $ a$ in the question, so the value of $ a$ would be $ \fbox{4}$.

As we have the term $ 9x^2$, we know the binomial is of the form $ (3x\plus{}n)$. Squaring this gives us $ 9x^2\plus{}6xn\plus{}n^2\equal{}9x^2\plus{}24x\plus{}a$. We observe that $ 6xn\equal{}24x$ and this holds true only when $ n\equal{}4$. Now, since the last term, $ a$, is $ n^2$, $ a\equal{}n^2\equal{}16$.


The set of points $ (x, y)$ that satisfy
$ x^3\minus{}3xy^2 \ge 3x^2y\minus{}y^3$ and $ x \plus{} y \equal{} \minus{}1$
is a line segment. Find its length.

$ x^3\minus{}3xy^2\geq3x^2y\minus{}y^3\implies x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)$

$ \equal{}(x\plus{}y)((x\plus{}y)^2\minus{}3xy)\equal{}3xy\minus{}1\geq3xy(x\plus{}y)\equal{}\minus{}3xy\implies xy\geq\frac16$.

Graphing this, we see that the segment that satisfies this has endpoints satisfying $ xy\equal{}\frac16$ and $ x\plus{}y\equal{}\minus{}1\implies x\equal{}\minus{}1\minus{}y$.

Plugging this in, we get $ \minus{}(1\plus{}y)y\equal{}\frac16\implies y^2\plus{}y\plus{}\frac16\equal{}0$, which has solutions

$ y\equal{}\minus{}\frac12\pm\frac{\sqrt3}6$, which, by symmetry, correspond to $ x\equal{}\minus{}\frac12\mp\frac{\sqrt3}6$.

By the distance formula, the length of the segment is $ \sqrt{\left(\frac{\sqrt3}3\right)^2\plus{}\left(\frac{\sqrt3}3\right)^2}\equal{}\frac{\sqrt3}3\cdot\sqrt2\equal{}\boxed{\frac{\sqrt6}3}$.


[hide="solution"]The harmonic mean of two numbers $a$ and $b$ is given by $\frac{2ab}{a+b}$, so we have $\frac{2\cdot8\cdot4}{8+4}={\frac{16}{3}}$. [/hide]

[quote="bpms"]What is the harmonic mean of 4 and 8?[/quote]

[hide]formula for harmonic mean is $\frac \frac{2ab}/{a+b}$,
so, it would be $\frac \frac{2*4*8}/{4+8}$ so, that is equal to $\frac\frac{16}/{3}$[/hide]

[quote="funcia"]just wondering? how do harmonic means ever help us? in comp.? :lol:[/quote]

I don't know. Personally, I never seen a problem needing to know what harmonic means are.

me neither...strange people even know about it in this forum i'm gonna go wiki it

[quote="funcia"]just wondering? how do harmonic means ever help us? in comp.? :lol:[/quote]
just because something won't help you in a competition doesn't mean its not important  :wink:
also, I think (don't really know) they become much more important later  :D

haha but i want to filter my brain! i don't want to memorize things that won't help me win things....ok that''s not true
http://en.wikipedia.org/wiki/Harmonic_Mean

[quote="funcia"]just wondering? how do harmonic means ever help us? in comp.? :lol:[/quote]
I used harmonic means on this years AMC 8 on the book reading problems.  Their good in general for rate problems and stuff.

Gary Niger runs to school at the speed of 8 miles per hour because he is late. However, on the way back from school, he does not need to rush so he walks at the speed of 4 miles per hour. What was his average speed for the entire trip, in miles per hour?

[hide]$H = \frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}}.$
now just plug in the numbers[/hide]

[hide="solution"]The harmonic mean of n numbers is the number of numbers all over the sum of the reciprocals of the numbers.
SO...

$\frac{2}{\frac{1}{4}+\frac{1}{8}}=\frac{2}{\frac{3}{8}}=\frac{16}{3}$[/hide]

Ok, remember that $GM\geq HM$

Proof for two terms:$\sqrt{ab}\geq \frac{2\sqrt{a^{2}b^{2}}}{a+b}\\(a+b)\sqrt{ab}\geq 2\sqrt{a^{2}b^{2}}\\a+b\geq 2\sqrt{ab}$
which is AM-GM

ok. Since he brought up the Geometric mean, I'll make a problem on it.

Find the geometric mean of a and b.

a is the arithmetic mean of 1 through 10.

b is the harmonic mean of 1,2, and 3.


function
algebra
domain
algebra unsolved

Are there infinitely many functions $ f$ with the domain the set of all natural numbers (and the range whatever) and

$ f(nf(k) \plus{} kf(n)) \equal{} f(n^2 \plus{} k^2)f(n \plus{} k \minus{} 1)$

for any $ k,n\in \mathbb{N}$?

Yes!

$ f(2k\minus{}1)\equal{}1,f(2k)\equal{}e$ is a solution, where $ e$ can be any positive even number.
The equation always holds, because
1) $ n^2\plus{}k^2$ and $ n\plus{}k\minus{}1$ have different parity, $ RHS\equal{}e$
2) $ nf(k)\plus{}kf(n)$ is always even, $ LHS\equal{}e$.


I joined online class "Intro to Counting and prob" on March 5.
I cannot access the message board for the class.

This is what I do:

Community-> Online School ->  ?????

I am supposed to see it here. But I dont. Am I missing something?

Even the link provided on the class page does not work for me.

You should see it now; please let me know if you don't.

will learning from a Ph.D/ former 1991 USAMOer be better than taking this class and practice for the AIME? :ninja:

Most instructors on AoPS have qualified for USAMO and MOSP multiple times...

As for 1-on-1 training, I don't know. However, some very bright people are not very good at teaching.  AoPS instructors are very good at teaching.


Erm could someone describe the compound properly?  :D

i actually read that long ago so don't remeber the exact rule

[quote="Cyber-Shreyas"][quote="pardesi"]i remember reading this a long long long long time ago.....
it should be something like this [1,3',3'',3''',3'''',3''''']hexakis cyclohexane[/quote]

Right track da.. one or 2 small changes.

Six Cyclohexane groups linearly bonded to each other.[/quote]

So is the compd like Decalin? See
http://en.wikipedia.org/wiki/Decahydronaphthalene

if linearly then the bonding should be in para postion with respet to each bond

Pardesi, yes I am sorry, I meant para. Typed it in a hurry. Anyway here is the name

1,1',4',1'',4'',1''',4''',1'''',4'''',1'''''-sexicyclohexane

However not sure if this is surely right. Cause different books suggest different things. Is carcul a member of IUPAC?

don't know about the molecule but the name looks s**y :P

just ask Carcul...should i move this to the chemistry forum

Well, its a trivial issue. Dunno if this is really worth his time. :)

About the name, you echoed my thoughts man. :D


Let $S$ be a set of $2n+1$ points in the plane so that no three are collinear and no four are concyclic. A circle is called "good" if it has $3$ points on its circumference, $n-1$ points in the interior and $n-1$ points on the exterior. Prove that the number of good circles has the same parity as $n$.

Sorry :blush: , the error has been fixed.

[quote="A cute angle"]The total number of points you have is 2n + 5 ** versus the
2n + 1 points in the plane mentioned at the beginning.  At least one of these is not right.

** (n + 1) + (n + 1) + 3 = 2n + 5

Also, once you determine what quantities should be in the places of "n + 1," if any changes are needed there, then type in "points" immediately after them as well because they are missing at present.[/quote]

Pleaser... ehm, A cute angle, you could have computed the "missing number" on your own, either assuming that the n + 1 is correct, or assuming that the 2n + 1 is correct. Anyway I am going to lock this topic now since there is an older one for this problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=79789 .

  Darij


The only thing I know (away from simple trying) is to use "quadratic residues" (but this is not really a intermediate topic I think)

[quote="devious_"]How do I find perfect squares in a certain modulo? For example, is 7 a perfect square modulo 321?[/quote]
I think, you should use reciprocal law and Legendre symbol (I forgot how to write it in right way).


I checked the website and couldn't find the answer to this question, How important is the team competition?

The team competition basically earns your team a ranking. If your team ranking is good enough, you earn an automatic spot to the next level of the competition, except for state --> nats, because that's only the top 4 individuals. So if you're having trouble from chapter-->state, having a good team helps.

I am guessing that you are asking about the scoring.

The scoring is as follows:
Individuals get 1 point for correct answers on the Sprint (30 max) and 2 points per Target Round question (16 max).
The team score is the average of the four individual scores plus twice the team round.  This makes each correct answer in the team round equivalent to 8 correct sprint round answers or four correct target round questions (for the team as a whole).

[quote="Maybach"]Of course, if your team always does well every year, or your team stinks, then the team round isn't important.  And if states- nats is hard(I'm sure it is) then you need to be focused on sprint and target(sprint especially)[/quote]

Well, if your team is very good, then you get a free pass to states.  :)

What if I didn't have a team and did it on my own. Would the team round count for me?

Nope. If your not on a team, you take the team round with a couple other individuals only for fun.

The team round also is multiplied by 2 and added to your individual score so it is very important


calculus
algebra
polynomial
function
inequalities
LaTeX
calculus computations

1) Let F(x) be a polynomial function so that for all real (x)
f(x^2+2)=X^4+10X^2+4
Evaluate f(X^2-2)
Show all the steps, step by step

2) a function f(X) is defined for all real numbers and satisfies the condition:
f(X+2)=f(X) for all real X
If X=1 is a 0 of X, how many zeros does f(X) in the interval [0,100].
Show all your reasoning step by step!

3)Consider any 5 points all of which are in, or on , a square of side 1 . Prove that there must be at least one pair of them who will be no further than [b]Square root of 2[/b]/2 apart.
SHOW ALL YOUR REASONING, STEP BY STEP.

4)Prove that the inequality 3a^4-4a^3b+b>=0 holds for all real numbers a and b.
Show all your work step by step.

5)The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,0) and 2, respectively. Circumference Q is internally tangent to circumference M and externally tangent to circumference P.Of course that there are infinitely many circumferences Q which have this property. Determine and graph the equation of the curve that contains the centers of all the circumferences of the Q kind.
SHOW ALL YOUR WORK, STEP BY STEP.

6) Consider a collection of piles of bananas consisting of one pile of 9 bananas, one pile of 6, and one pile of 2. Such a collection, denoted as C1 , could be expressed as (9,6,2). Obtain a new collection C2 by harvesting C1, where harvesting is defined to mean remove one banana from each pile to form a new pile. Thus we have C2 = (8,5,3,1) , and if is harvested we C2 we obtain (7,4,4,2).
A)Let C1=(8,5,2) and determine C100
B)Let C1=(7,6,5) and determine C1995
Show all your work, step by step.

Note: you can use LaTex type setting on this website by enclosing your equations with $ \$\ldots\$$.

[quote="rfacusse"]1) Let F(x) be a polynomial function so that for all real (x)
f(x^2+2)=x^4+10x^2+4
Evaluate f(x^2-2)
Show all the steps, step by step[/quote]

We know that $ f(x^2 \plus{} 2) \equal{} x^4 \plus{} 10x^2 \plus{} 4$, and we want to know what $ f(x^2 \minus{} 2)$ is, so let's determine what to put in for $ x$ (we'll call this value $ y$) that changes the argument of $ f$ from $ x^2 \plus{} 2$ into $ x^2 \minus{} 2$. We have
\[ y^2 \plus{} 2 \equal{} x^2 \minus{} 2
\]

\[ y^2 \equal{} x^2 \minus{} 4
\]

\[ y \equal{} \pm\sqrt { x^2 \minus{} 4}
\]
Now we put this value of $ x$ into $ f(x^2 \plus{} 2) \equal{} x^4 \plus{} 10x^2 \plus{} 4$ (that is, replace each $ x$ in this equation with $ \pm\sqrt { x^2 \minus{} 4}$) to get
\[ f\left( \left( \pm\sqrt { x^2 \minus{} 4}\right)^2 \plus{} 2\right) \equal{} \left( \pm\sqrt { x^2 \minus{} 4}\right)^4 \plus{} 10\left( \pm\sqrt { x^2 \minus{} 4}\right)^2 \plus{} 4
\]

\[ f\left( (x^2 \minus{} 4) \plus{} 2\right) \equal{} (x^2 \minus{} 4)^2 \plus{} 10( x^2 \minus{} 4) \plus{} 4
\]

\[ f\left( x^2 \minus{} 2\right) \equal{} (x^4 \minus{} 8x^2 \plus{} 16) \plus{} 10x^2 \minus{} 40 \plus{} 4
\]

\[ \boxed{f\left( x^2 \minus{} 2\right) \equal{} x^4 \plus{} 2x^2 \minus{} 20}
\]
[color=red]Another way to do this problem is:[/color]

Since $ f(x^2 \plus{} 2) \equal{} x^4 \plus{} 10x^2 \plus{} 4$, $ f(x)$ is a second degree polynomial. (Why?) Let's determine this polynomial:

$ f(x^2 \plus{} 2) \equal{} x^4 \plus{} 10x^2 \plus{} 4$
$ .\qquad\qquad \equal{} (x^4 \plus{} 4x^2 \plus{} 4) \plus{} 6x^2$
$ .\qquad\qquad \equal{} (x^2 \plus{} 2)^2 \plus{} 6x^2 \plus{} 12 \minus{} 12$
$ .\qquad\qquad \equal{} (x^2 \plus{} 2)^2 \plus{} 6(x^2 \plus{} 2) \minus{} 12$

so $ f(x) \equal{} x^2 \plus{} 6x \minus{} 12$, and hence
\[ f\left( x^2 \minus{} 2\right) \equal{} \left( x^2 \minus{} 2\right)^2 \plus{} 6\left( x^2 \minus{} 2\right) \minus{} 12
\]

\[ f\left( x^2 \minus{} 2\right) \equal{} (x^4 \minus{} 4x^2 \plus{} 4) \plus{} (6x^2 \minus{} 12) \minus{} 12
\]

\[ \boxed{f\left( x^2 \minus{} 2\right) \equal{} x^4 \plus{} 2x^2 \minus{} 20}
\]
[quote="rfacusse"]2) a function f(x) is defined for all real numbers and satisfies the condition:
f(x+2)=f(x) for all real x
If x=1 is a 0 of x, how many zeros does f(x) in the interval [0,100].
Show all your reasoning step by step![/quote]

$ f(x)$ is a periodic function with a period of $ 2$. Hence $ f(1)\equal{}0$ implies that $ f(1\plus{}2) \equal{} f(3)\equal{}0$ as well. In fact, we have that $ f(1)\equal{}f(3)\equal{}f(5)\equal{}f(7)\equal{} \cdots \equal{} f(97)\equal{}f(99) \equal{}0$ by the same reasoning. Thus there are [i]atleast[/i] ($ f$ might have some other zeros elsewhere) $ 50$ zeros of $ f$ on $ \left[ 0, 100\right]$.

[quote="rfacusse"]3)Consider any 5 points all of which are in, or on , a square of side 1 . Prove that there must be at least one pair of them who will be no further than [b]Square root of 2[/b]/2 apart.
SHOW ALL YOUR REASONING, STEP BY STEP.[/quote]

[quote="rfacusse"]4)Prove that the inequality 3a^4-4a^3b+b>=0 holds for all real numbers a and b.
Show all your work step by step.[/quote]

[quote="rfacusse"]5)The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,0) and 2, respectively. Circumference Q is internally tangent to circumference M and externally tangent to circumference P.Of course that there are infinitely many circumferences Q which have this property. Determine and graph the equation of the curve that contains the centers of all the circumferences of the Q kind.
SHOW ALL YOUR WORK, STEP BY STEP.[/quote]

[quote="rfacusse"]6) Consider a collection of piles of bananas consisting of one pile of 9 bananas, one pile of 6, and one pile of 2. Such a collection, denoted as C1 , could be expressed as (9,6,2). Obtain a new collection C2 by harvesting C1, where harvesting is defined to mean remove one banana from each pile to form a new pile. Thus we have C2 = (8,5,3,1) , and if is harvested we C2 we obtain (7,4,4,2).
A)Let C1=(8,5,2) and determine C100
B)Let C1=(7,6,5) and determine C1995
Show all your work, step by step.[/quote]

3) You can use Pegion Hole theorem.

4) Just factor the left side or you can use  a tangent line.

By the way, are these problems your homework?

they are some problems which we have to do for pre calculus class. as a project. i cant get them and everyone in my class is seeking for help with engineers, college professors, and some math teachers in school even though these teachers cant do the problems.

For 3, seeing as this is pre calc, the following should be sufficient.

Consider the first 4 points on a square of side 1. The furthest they can be from each other is at the vertices's of the square. Now the furthest the 5th point can be from the 4 vertices is in the centre of the square, because being further from one corner makes it closer to another. So the maximum distance between points is given by placing the points in the vertices and the center, which by pythagoras easily gives a distance of sqrt2 /2.

That's not a solution to #3- there's no convincing reason why a corner has to be filled.
The right way: Cut the square into four pieces, each with diameter* $ \frac{\sqrt{2}}{2}$.

*The diameter of an arbitrary set is the supremum (maximum) of the distance between pairs of points in the set.


Find the maximum value of $ T\equal{}(\frac{a\plus{}b}{c\plus{}d})(\frac{a\plus{}c}{a\plus{}d}\plus{}\frac{b\plus{}d}{b\plus{}c})$ ,where $ \frac{1}{2}$≤$ a,b,c,d$≤$ \frac{2}{3}$

hello, i have computed $ T\le \frac{8}{3}$ and the equal sign holds for $ a\equal{}b\equal{}\frac{2}{3}$ and $ c\equal{}d\equal{}\frac{1}{2}$.
Sonnhard.


function
abstract algebra
integration
inequalities
limit
logarithms

In some article, I'm reading:
[color=red]
We denote by $ \mathcal{L}_{\Omega}^{p}(G)$ the set of all functions $ f \in L^{1}(G)\cap L^{p}(G)$ such that $ \text{supp} \mathcal{F}f$
is a compact subset of $ \Omega$. Let $ G$ be a compact Abelian group, $ \Gamma$ the dual group. Let $ \Omega$ be a subset of $ \Gamma$. We denote by
$ r_{G}(\Omega)$ the infimum of the set of all $ p \in (0,1]$ such that $ \int_{G}(|1\plus{}f|^{p}\minus{}1)dm \geq 0$  [b] for all[/b] $ f \in \mathcal{L}_{\Omega}^{1}(G)$
if $ 0 \in \Omega$,  then there is no such $ p$, and we put $ r_{G}(\Omega) : \equal{}1$.

Now let $ 0$ is not belong to $ \Omega$, then the infimum is attained  provided $ r_{G}(\Omega)>0$
[/color]
If I understand well if $ 0$ is not belong to $ \Omega$ then  $ r_{G}(\Omega)>0$.
I think that in this case $ r_{G}(\Omega) \equal{}1$
because we always can find  function $ f$ (which is not identically equal to $ 0$) such that $ |f|<1$ and $ \mathcal{F}f(0)\equal{}0$ therefore by using inequality
$ (1\plus{}x)^{p} \leq 1\plus{}px$ we obtain :
\[ \int_{G}(|1\plus{}f|^{p}\minus{}1)dm \leq p\int_{G} f dm \equal{}0\]
where equality holds iff when $ p\equal{}1$ thus infimum of such $ p$ is equal to $ 1$

There is no Bernoulli inequality for complex numbers! :roll: As an example, take $ G\equal{}\mathbb T$, $ \Omega\equal{}\{1\}$. Then you are talking about $ \int_{\mathbb T}(|1\minus{}az|^p\minus{}1)\,dm(z)$, which is non-negative for all $ p>0$. A good exercise is to show that in this case $ r_G(\Omega)\equal{}1$ for finite sets $ \Omega$ if and only if $ \Omega\cap(\minus{}\Omega)\ne \varnothing$.

Oh thank you  :) .
if $ \Omega \in \mathbb{Z}$ and $ \Omega \cap \minus{} \Omega \neq \emptyset$
Then there exist $ m \neq 0$ and $ m \in \mathbb{Z}$  such as $ m, \minus{} m \in \Omega$

then note that $ f(z) \equal{} \frac {z^{m} \plus{} z^{ \minus{} m}}{2} \in \mathcal{L}_{G}^{1}(\mathbb{T})$  and $ f < 1$  and
$ \int_{\mathbb{T}} f(z) \equal{} 0$ Hence  by Holder's inequality
$ \parallel{}1 \plus{} f(z)\parallel{}_{L_{p}} < \int_{\mathbb{T}}(1 \plus{} f(z)) dm(z) \equal{} 1$ where $ m(\mathbb{T}) \equal{} 1$
 for all $ p < 1$
Thus we obtain :
$ \int_{\mathbb{T}} (|1 \plus{} f(z)|^{p} \minus{} 1) d m(z) < 0$ for all $ p < 1$ done.
I note that I have not used the fact that  $ \Omega$ is finite set.

[quote="Extremal"]
I note that I have not used the fact that  $ \Omega$ is finite set.[/quote]
And obtained the result in only one (easy) direction ;).

Oh sure  :) .  Then really  we need the fact : $ \Omega$ is finite.
We need to prove : if $ r_{G}(\Omega) \equal{} 1$ then $ \Omega \cap \minus{} \Omega \neq \emptyset$.
from contrary: suppose $ \Omega \cap \minus{} \Omega \equal{} \emptyset$

We can conctruct monotone increaing positive sequance
$ \{p_{n}\}$ with $ 0 < p_{0} < 1$ and $ \{f_{n} \} \in \mathcal{L}_{\Omega}^{1}(\mathbb{T})$ such that:
 $ \lim_{n \to \infty} p_{n} \equal{} 1$ and $ \parallel{}1 \plus{} f_{n}\parallel{}_{L^{p_{n}}} < 1$ for all $ n$, since $ r_{\mathbb{T}}(\Omega) \equal{} 1$.
Note that :
$ \parallel{}f_{n}\parallel{}_{L^{p_{0}}}^{p_{n}}\leq \parallel{}f_{n}\parallel{}_{L^{p_{n}}}^{p_{n}}\leq 1 \plus{} \parallel{}1 \plus{} f_{n}\parallel{}_{L^{p_{n}}} < 2$
thus, the sequence $ \{ f_{n} \}$ is bounded in the finite-dimensional (since $ \Omega$ is finite) space $ \mathcal{L}_{\Omega}^{1}(\mathbb{T})$ hence we can find converging subsequence $ \{ f_{n_{k}}\}$therefore we have :
put: $ f \equal{} \lim_{k \to \infty} f_{n_{k}}$ then :
$ \int_{\mathbb{T}} |1 \plus{} f| dm(z) \leq 1$, therefore $ \Im f \equal{} 0$ hence $ f \equal{} 0$ because $ \Omega \cap \minus{} \Omega \equal{} \emptyset$.
But on the other hand:
Let $ f \in \mathcal{L}_{\Omega}^{1}(\mathbb{T})$
then
1) $ \int_{\mathbb{T}} f \equal{} \int_{\mathbb{T}} \bar f\bar \equal{} 0$
since $ 0$ is not belong to $ \Omega$.
2) $ \int_{\mathbb{T}} f^{2} \equal{} \int_{\mathbb{T}} \bar f^{2}\bar \equal{} 0$
by using Plencherels theorem since $ \Omega \cap \minus{} \Omega \equal{} \emptyset$ .

 There exist increasing functions $ C > C_{1}(p),C_{2}(p),C_{3}(p),C_{4}(p) > 0$ for all $ p \in (0,1]$ such as :
\[ |1 \plus{} z|^{p} \geq 1 \plus{} C_{1}(p)(z \plus{} \bar z \bar) \minus{} C_{2}(p)(z^{2} \plus{} \bar z^{2} \bar) \plus{} C_{3}(p)|z^{2}| \minus{} C_{4}(p)|z^{3}|
\]
therefore
\[ 1 > \int_{\mathbb{T}}|1 \plus{} f_{n}|^{p_{n}} \geq 1 \plus{} C_{3}(p_{n})\int_{\mathbb{T}}|f_{n}|^{2} \minus{} C_{4}(p_{n}) \int_{\mathbb{T}}|f_{n}|^{3}
\]

\[ \geq 1 \plus{} (C_{3}(p_{n}) \minus{} C_{4}(p_{n})\parallel{}f_{n}\parallel{}_{L^{\infty}})\parallel{}f_{n}\parallel{}^{2}_{L^{2}}
\]
thus we obtain :
\[ \parallel{}f_{n}\parallel{}_{L^{\infty}} \geq \frac {C_{3}(p_{n})}{C_{4}(p)} \geq \frac {C_{3}(p_{0})}{C}
\]
hence :
\[ \parallel{}f\parallel{}_{L^{\infty}} \geq \frac {C_{3}(p_{0})}{C} > 0
\]
contradiction.

[quote="fedja"] As an example, take $ G = \mathbb T$, $ \Omega = \{1\}$. Then you are talking about $ \int_{\mathbb T}(|1 - az|^p - 1)\,dm(z)$, which is non-negative for all $ p > 0$.[/quote]
is it true that if $ \Omega \in \mathbb{N}$ or ($ \in - \mathbb{N}$) then $ r_{\mathbb{T}}(\Omega) = 0$ ? :maybe:
I believe that it's same as  to show $ \int_{\mathbb{T}} \ln |1 + f(z)| dm(z) \geq 0$ for all $ f$ such that  $ \text{supp} \mathcal{F}f \in \mathbb{N}$ but why if
$ \text{supp} \mathcal{F}f \in \mathbb{N}$ then $ \int_{\mathbb{T}} \ln |1 + f(z)| dm(z) \geq 0$? :huh:

The logarithm of the absolute value of an analytic function is subharmonic ;).


What is usually the score in chapter to make top 10

is chapter harder than school

because it seems like it over the years

thanks!!!

Difficulties of MATHCOUNTS Rounds (easiest to hardest):

School - Chapter - States - Nationals

meh, because school is sometimes used to determine who will be attending chapter, some say that chapter is actually easier. personally, they are both easy, states is relatively medium, and nats gets medium-hard.

in my chapter, score to make countdown (top 10): like, 42+. i got a 44 last year and was ranked 6th. then again, my chapter is pwnage =D.

Oh dang, same.  Only I failed Chapter last year (39, 11th place).

In my opinion, School is just a load of computation.

meh. almost said something about this years school.

anyways, in order to make nats, your scores should be:

school:44+
chapters:44+
states:40+

i may be speaking solely on behalf of michigan here, though. 40 might not be it, 35 could get ya there.

I don't think you need to score that high but it depends on which state. I heard in California, you need to get like 45 in states just to make countdown... Last year, the cutoff at states in my state was 31 and at chapter it was 42.

Score to make top 10 depends on the test and who's competing in the chapter

As for which is easier (school or chapter), it depends on what kinds of problems appear on the tests.  Personally I find chapter tests to be the same level, if not easier, than school tests.

No offense, but the highest score in California was a 44 in 2008...

In my chapter, you needed 41 for CD in 2008 but lower in 2007.

[quote="athunder"]hmm well my score for two diffferent school tests were 45 and 44 AND i didnt/wont make nats....care to explan?!?!??!!?[/quote]

first of all, explain is spelled with an i. second, different only has 2 f's.

third, its not my fault your stupid teacher decided that your stupid amc 8 score should be counted as a stupid zero, for no stupid reason. the whole thing is....stupid. and no, i dont care to explain, because i have absolutely nothing to do with your stupid teacher and stupid school.

 :P just kidding athunder =D.

I think the total you can get is 30 (sprint) + 16 (target) = 46. Last year in my chapter, I got 28 sprint and 16 target (8 x 2) and came in ONLY 7TH PLACE OVERALL. So it just depends where you live (my place is Texas).

Yes, I think school round can sometimes be tougher than chapter. How well people do is just based upon your state. In my state last year, I only got 36 chapter (3rd), 33 state (2nd), 20 nationals (140-149th).

what place do u have to get in chapter to make states

Even though it will be better to start a new topic on this, I will still answer it:

To get to state you either have to come in top 4 as a school team or top 4 if you are competing as an individual.

I think...

It matters on how many people there are at your chapter. For example, my chapter lets the top 2 teams and top 2 individuals, but it differs.

its not necessarily individuals, its the members of the top 4 (or however many) teams, and then the next 4 contestants whose teams DID NOT MAKE STATES. for example, my team failed, nbut i made states.

In here, the top 4 teams move on to state, then the top6 individuals move on to State.  Usually only 16 people go each year because the people on the team are generally the top 6 too.

Well, I'm just saying, when the chapter sends $ x$ teams and $ n$ individuals, its $ x$ teams, then the top-scoring $ n$ individuals [b]not part of any of the $ x$ teams[/b].

school is easy
chapter team is easy, in you want to make state by individual maybe 45+
45+ in state to make to cd, then depending on whether cd is official, win cd (if cd isn't official then too bad; hope for a perfect)

[quote="AIME15"]Um, after 4 teams go on, the NEXT 6 individuals move on...[/quote]

then your saying my chapter rigged me and i could have gone to state? (i got 13th)

EDIT: we had around 200 people at are chapter

GUYS FOR THE LAST TIME, IT DIFFERS FROM CHAPTER TO CHAPTER.

How about you all stop worrying about this and go practice? It won't help you to know that you missed state by one point.

smaller chapters send less people.
bigger chapters send more people. obviously if your chapter consists of 21 people it cant send the top four teams and the next 6 individuals
and in my opinion it is MUCH easier to make state via team.
in my sixth grade year everyone on our team did average but we got to state
EDIT: xpmath summed it up pretty nicely. except for the part about practicing.

[quote="Wickedestjr"]I made it to states with a score of 33, and I had ranked 9th, but at state I got a score of 21 I think and got 7th at state. I think it all depends on where you live and the difficulty of the school/chapter/state/national tests.[/quote]

i wish i lived in your state...

[quote="AIME15"]Uh, wicked, what state is that?  21 is a pretty low score to even place at states...[/quote]

Well, his location says Antarctica...was there even anyone else there?  :rotfl:  :rotfl:  :rotfl:

Yeah, but there were only six other people. That's why I got seventh...[hide="..."]Nah, I'm just kidding. That was in Nevada. [/hide]


Do you mean this one

$\sum_{i=1}^{n}\sum_{j=i}^{n-1}ji$ ?


What is the number of posts you must have to acheive levels of post numbers (I don't know what else to call it) that appears below one's avatar. Also, where did the names come from?

Check the 15th post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266888&search_id=993592065]here[/url]. They are the seven millennium problems.

Thanks. Do you know where the names came from?

.....

The 7 millenium problems? Ring a bell?

r15 put that in his last post. I bet you didn't even read the link!

Or were you meaning that where did they come up with the names?
Well who knows? They made it up!

[quote="PowerOfPi"]Actually, I read all the posts in the linked forum. And I was wondering that last question. Thanks.[/quote]
i think they made the names up. or they're all some kinds of theories in math. not sure...

Your thought has already been answered above.


Let {$ x_n$}:
 i)$ x_1\equal{}c>0$,$ x_2\equal{}d>0$
 ii)$ x_{n\plus{}1}\equal{}x_n\plus{}\frac{1}{x_n^k}$  , $ k$ is a constant >0.
Prove that exist only one pair($ a,b$) is real such that $ lim(\frac{x_n^a}{nb})\equal{}1$

wy you say that $ d>0$
we have $ d\equal{}c\plus{}\frac{1}{c^k}>0$

but i think that you want say $ x_{n\plus{}2}\equal{}f(x_{n\plus{}1},x_n)$ ??

[quote="nguoigo"]Let {$ x_n$}:
 i)$ x_1 \equal{} c > 0$,$ x_2 \equal{} d > 0$
 ii)$ x_{n \plus{} 1} \equal{} x_n \plus{} \frac {1}{x_n^k}$  , $ k$ is a constant >0.
Prove that exist only one pair($ a,b$) is real such that $ lim(\frac {x_n^a}{nb}) \equal{} 1$[/quote]
Posted before .

Sorry we don't need $ d$.
@ TTsphn:Can you give me the link to proof?


Let $ P$ be the product of all non-zero digits of the positive integer $ n$. For example, $ P(4) \equal{} 4$, $ P(50) \equal{} 5$, $ P(123) \equal{} 6$, $ P(2009) \equal{} 18$.
Find the value of the sum: P(1) + P(2) + ... + P(2008) + P(2009).

Notice that the sum is essentially the same as

2(1+1+2+3+4+5+6+7+8+9)^3+2(1+1+2+3+4+5+6+7+8+9)-1=194763.

The first one can be treated as a place holder for 0.