ratio
number theory
least common multiple

Here's the problem:

Superman and Flash are running around the world. Flash can make it around the world in 1.5 hours and Superman can make it in 2.5 hours. Assuming that they start at the same time and same place, how many times will they meet if they keep running for 24 hours?

[hide="My first (wrong, and not at all water tight) Solution"]
If we can find the first time they meet, then we can find how many times they meet in a 24 hour period since when they meet its like starting all over again because they're at the same place at the same time.

If we image that flash and superman are going at the same speed, then its obvious that they will meet at half the circumference of the sphere.

Then, the ratio of the speed of one person the combined speed of the two is 1/2, and they meet at 1/2 the diameter of the circle.

So, if we do $ \frac{1.5}{2.5}$ then we get $ 0.6$ or $ \frac{3}{5}$.

Then, we multiply this by 1.5 to get when they will meet (I know this makes no sense whatsoever but keep reading please)

Now, we have 9/10 hours. We convert that to minutes to get 54.
and finally we convert 24 hours into minutes and divide it to get 54 and we get 26 2/3 (now please read the next solution).
[/hide]

[hide="The actual, real solution"]
Flash covers 1/1.5 of the world in an hour.
Superman covers 1/2.5 of the world in an hour.

If they meet after x hours, then,
Superman covers x/2.5 of the world
Flash covers x/1.5 of the world

Together both of them have covered the whole world so,

x/1.5+x/2.5=1

Solving that, we get,
x=25.6

But, we can't half a meeting so, we have 25!

[/hide]

The thing is, even though my bogus solution seems completely wrong its very close to the real answer. I have a couple of questions,

What things make my solution wrong?
Is there a reason why the answers are so closely related? [/hide]

I'm not convinced either solution is correct.  The first solution assumes that they can only meet at one point, which I would check (hint: it's false), and the second doesn't use $ 24$.  It also appears as if the problem is ambiguous as to whether Flash and Superman start in the same [b]direction[/b] or not.

They go in opposite directions (forgot to mention that, sorry :blush:)

anyway, this is from intro to algebra.

For the second one, it uses 24 hours but it converts it to minutes so that's probably why you didn't notice it.

Find the least common multiple?

2.5, 5, 7.5
1.5, 3, 4.5, 6, 7.5

LCM=7.5

$ 60\div7.5\equal{}8$

I am almost sure this is wrong, because it doesn't count when they meet somewhere other than their starting points.

Suppose Flash is $ k$ times faster than Superman for a positive integer $ k$.  I claim that Flash and the Superman have $ k \plus{} 1$ possible meeting points.  So this is indeed a possibility that needs to be checked.

[hide="My Solution"]Let the world be $ 1$ unit of distance. The Flash's rate is then $ \frac{2}{3}$, and Superman's rate $ \frac{2}{5}$. However, this problem is equivalent to Superman standing still and the Flash's rate being $ \frac{2}{3} \plus{} \frac{2}{5} \equal{} \frac{16}{15}$. Therefore, every hour, the distance between them increases by $ \frac{16}{15}$, so the distance between them is $ 24 \cdot \frac{16}{15} \equal{} 25.6$ after $ 24$ hours. Since, for every increase of $ 1$ in the distance between them, Superman and Flash meet each other exactly once, our answer is $ 25$. [/hide]

The actual solution is from intro to algebra and I fully understand it. I posted the thread to find out why my (wrong) solution is so close to the real thing.

[quote="Poincare"]

Then, the ratio of the speed of one person the combined speed of the two is 1/2, and they meet at 1/2 the diameter of the circle.

[/quote]

I don't really understand this.

Chances are that the answers were so close just by coincidence.

I also agree that the answers are close just due to coincidence. $ 24 \cdot \frac{60}{54} \approx 24 \cdot \frac{16}{15}$. There are a lot of numbers that are very close together in this problem, so it's easy to come up with a number that's close to $ 1$.


Two different integers are randomly chosen from the set $ \{ \minus{}5, \minus{}8, 7, 4, \minus{}2 \}$. What is the probability that their product is negative? Express your answer as a common fraction.

There are $ 3$ negative integers and $ 2$ positive integers. We need one negative integer and one positive integer to get a negative product, so the probability is $ \frac{3\cdot2}{\binom52}\equal{}\frac6{10}\equal{}\boxed{\frac35}$.

There are $3$ negative numbers in the list $ \{ \minus{}5, \minus{}8, 7, 4, \minus{}2 \}$. and $2$ positive numbers. $2+3$ is $5$, and as we said before there are $3$ numbers that are negative so the answer is $\boxed{\frac35}.
$

The way to do this is basically find the probability both and negative and then both are positive and then subtract that from 1.


[quote="thesunflower"]Consider the equation $ ax^3 \minus{} x^2 \plus{} bx \minus{} 1 \equal{} 0$
where $ a,b$ are real numbers, $ a \neq b, a \neq 0$ such that all of its roots are positive real numbers.
Find the smallest value of $ P \equal{} \frac {5a^2 \minus{} 3ab \plus{} 2}{a^2(b \minus{} a)}$[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now
Thanks!
[hide]http://www.mathlinks.ro/Forum/viewtopic.php?t=170552[/hide]


What is the probability that when dealt 5 cards from a standard 52 card deck, exactly two clubs are picked?


[quote="b-flat"]As you may or may not know, state MathCounts will be held on Friday.  We will be leaving after school tomorrow and staying overnight in the hotel.

Best of luck to everyone who is competing!

I will post the results some time after I get back.[/quote]

Florida owns more.  we miss two days of school, and get 2 nights in the hotel.  :D

good luck b-flat!

Well, here are the results:

[hide="before countdown"]1.  Grey Simon
2.  Jenny Chen
3.  Brendan Fletcher
4.  David Lucia
5.  Bryce Taylor
6.  Arjun Bhattacharya (please excuse the spelling error)
7.  ? Yang
8.  Calvin Deng
9.  Christine Hong
10.  Claire Zeng[/hide]

[hide="after countdown"]1.  Grey Simon
2.  Jenny Chen
3.  Brendan Fletcher
4.  Bryce Taylor
5.  David Lucia
6.  Calvin Deng
7.  Arjun Bhattacharya
8.  ? Yang
9.  Christine Hong
10.  Claire Zeng[/hide]

[hide="top teams"]1.  Arrendell Parrot
2.  Ligon
3.  Hanes
4.  RD & Euzelle P Smith
5.  West Cary
6.  Charlotte Home Educators
7.  Daniels
8.  Martin
9.  Metrolina Regional School
10.  Brawlers[/hide]

[quote="PenguinIntegral"]Good job making nats b-flat.[/quote]

Umm, I didn't  :(

I made it in 6th grade, but not last year or this year.

Yay mathcounts! I am now completely done with mathcounts forever... I managed to hold on til sophomore year.

Well, you managed to spell Arjun's name right, but it's Gray Symon, not Grey Simon.

Sorry you didn't make nats... :( But congratulations to those who did!

no, as b-flat who was 5th in written got 4th in countdown.

btw, ? Yang=Allen Yang.

or, as both damien and jb05 know, 2 years ago i got 6th written and 1st countdown.  i think that i'll stop regretting that now that jb05 does too...

wtf how do you not find the buzzer? don't you ahve it in your hand?


Prove that if f(x) is periodic function and continue on $R$ and f(x) is not constant then f(x) has least period.


this is a problem from our college's programming contest bytecode  ..(it's there on spoj)

A major cosmic battle was getting over. The InterGalactic SuperPower had been under attack, but it had defended itself quite well. It was about to launch its final retaliatory assault. But the number of enemy ships was quite large and they could scatter very easily. Their only hope, or so their Space Warfare expert said, was to bomb the enemies (who happened to be lined up in a long line!) using the strategy described below.

Because the number of ships will be a power of 2, to bomb all the ships (numbered 0 to 2N -1), the strategy to be used, which we will call BombStrat, goes like this:
1. Bomb it’s first half, [0 to 2N-1 -1], in the left to right direction.
2. Of the remaining half, bomb its latter half part in reverse direction, i.e., bomb ships 2N-1, 2N-2,...., 2N-1+2N-2 in that order.
3. Then use BombStrat on the remaining ships: [2N-1 to 2N-1 + 2N-2 -1 ]

For example, when N=3, i.e., with ships numbered from 0 to 23 -1, this is what happens:
Step 1: Ships 0,1,2,3 get bombed in that order.
Step 2: Ships 7, 6 go down next.
Step 3: Now, the remaining ships [4, 5] are destroyed using the same strategy.

So the bombing is done in the order 0 -> 1 -> 2 -> 3 -> 7 -> 6 -> 4 -> 5. To make the job easier for the InterGalactic SuperPower’s ships’ pilots, they want to find which ship should be bombed when. This is your task. Given N, and the description of a ship, return the 0-based serial number of the bomb will blast it.
Input

T – the number of test cases, T<=50.
For each test case:

One line containing a binary number, describing the number of the place. The length of this string will equal N (it will be padded with leading zeroes if necessary). N<=30000.
Output

For each test case, output the index of a bomb, represented in the same format, as binary digits, whose length is exactly N.

ideas anyone

my idea is something which goes like this
first initialize a counter value to point to first bit from the left..
if first digit is zero...return as it is
if it is 1 and next degit is 0,make it 11 and complement the rest of the string
if it is 1 and next digit is 1 ,make it 10
increment string counter by 2;
repeat

it seems to be giving wrong answer
so its wrong obviously :D

Right idea, you just have a couple of things in the wrong order.
If it starts with 0, return.
If it starts with 11, then it is in the third quarter, so replace that with 10 and complement the rest of the string, then return.
Otherwise if it starts with 10, replace that with 11 and move forward two positions.
(If you just have a 1 on the end, then it stays as 1.)

oh yes stupid me :D
thx anyways
i just cudnt see why i was going wrong
(the effects of waking up at three and tryin to do programming)

best regards
vasu


Okay there might be an easy trick to solve this, but I can't see it.

Solve (all solutions, ie use a k):  sin(4t)cos(t)=sin(t)cos(4t) for t.

Supposedly the correct answer is all reals, but I get t=k*pi, an by TI89 gives a bunch of approximate answers.  And by graphing it, I get 0, 60, 120, 180, 240, etc. so I assume the correct answer is: t=k*pi or t=pi/3+k*pi or t=2pi/3 + k*pi .....

Any insight would be appreciated =).  My work is attached as a gif.

First notice when cos t and cos (4t) are both zero.

Then assume they are not, and divide by both of them to get tan t = tan (4t). Then try solving that using the translational symmetry of tan.

Sorry I don't really understand your response.  Can you explain what you are saying.  Sorry I know I posted it in the intermediate forum but im not really at this level.

Sorry, my post was kind of brief.

We're trying to solve sin(4t)cos(t)=sin(t)cos(4t) for t. So lets divide both sides by cos(t)cos(4t).

This will give us:
sin(4t)/cos(4t) = sin(t)/cos(t)
tan(4t) = tan (t)

Lets look at when the tangent of two angles can be equal. Well tan(t :pm: k*:pi:) = tan(t), for any integer k. So this means that:

4t = t :pm: k*:pi:
3t = k*:pi:
t = k*:pi:/3, for any integer k.

But wait a minute! Remember when we divided by cos(t)cos(4t)? We have to make sure that one of those was not zero, because we can't divide by zero. See if you can find out where those happen.

If one of them (cos(t) or cos(4t)) is zero, then the other side of the equation also has to be zero for that same t. See if that happens for any of the zeros that you found above.

Yes, gauss's method was the first method I saw.  Another way is to use the trig identity sin(x - y) = sin(x) cos(y) - cos(x) sin(y).  Plug in x = 4t and y = t.

cos(4t)=0
let y=4t

cos(y)=0
y= pi/2 + k*pi

4t = pi/2 + k*pi
t = pi/8 + k*pi/4

cos t =0
t= pi/2+k*pi

So:

k*pi/3  :!=: pi/2 + k*pi  solve to k  :!=: -3/4 dosent matter b/c its not an integer

k*pi/3  :!=: pi/8 + k*pi/4  solve to k :!=: 3/2 dosent matter b/c its not an integer.

So t= k*pi/3

I feel like im doing something wrong.

BTW  :!=: means "not equal to" not "not congruent to"

Also Im a little confused about Ravi's method for using the difference identity of sine.  Why do we want sin(3t)?

We don't particularly want sin(3t).  It is the right-hand side where we get the most simplification.  Trust me, go back and plug in the values I suggested, and look closely at the right side.  You will get a pleasant surprise.  When you are done, you can compare your solution with mine below.

Spoiler: [hide]We will use the difference identity sin(x - y) = sin(x) cos(y) - cos(x) sin(y).   Plug in x = 4t and y = t.  We get sin(3t) = sin(4t) cos(t) - cos(4t) sin(t).  But we are assuming that sin(4t) cos(t) = cos(4t) sin(t).  Thus the right-hand side of the difference identity is 0 (!)  So we are left with the equation sin(3t) = 0.  The sine function has zeros at k:pi: (where k is an arbitrary integer).  Thus we get the equation 3t = k :pi:.  Dividing by 3 gives the solution t = k:pi:/3, where k is an arbitrary integer.

[/hide]

Thanks!  That's a really easy way to solve it, once you realize what to do...


limit
trigonometry
LaTeX
AMC
USA(J)MO
USAMO
number theory

$\lim_{x \rightarrow 0}{ \frac{\sin{(a+x)}\cdot \sin{(a+2x)}-\sin^2{a}}{x}}$

[color=red]Latex edited by mod.[/color]

[quote="amirhtlusa"]for $a\in \mathbb{R}$:
(click the smile)
 [hide=" ;)"]
$\frac{3\sin{(2a)}}{2}$[/hide][/quote]
motmaeni ke javabesh faghat addad namishe? :?:

$\\\lim_{x\rightarrow 0}\frac{\sin(a+x)\sin(a+2x)-\sin^2a}{x}\\ =\lim_{x\rightarrow 0} 2\sinh\cos a+\sin (a+2x)\cos (a+x)\\ =2\sin a\cos a+\sin a+\cos a\\ =\sin 2a+\frac{1}{2}\sin 2a\\ =\frac{3\sin (2a)}{2}$

Masoud Zargar

[quote="boxedexe"]Merci.  Rasti, ketabe olampiade riazi, physic, va chimi soragh dari?[/quote]
yani ye ketab ke ham dar morede olampiade riazi dashte bashe ham olampiade phisic va ham ...?
ya inke too har zamine manzooret ye ketabe jodaganast?

[quote="boxedexe"]Merci.  Rasti, ketabe olampiade riazi, physic, va chimi soragh dari?[/quote]

baraye riyazi ke kheyli ziyad hast:
(man midoonam shoma dar Canada hasti, pas baazi az in ketaaba ro mitooni az internet begiri)

math:
Mathematical Olympiad Challenges
by: Titu Andreescu (U.S. IMO coach), Razvan Gelca (USAMO writer)

Mathematical Olympiad Treasures
Titu Andreescu, Bogdan Enescu

102 Combinatorial Problems: From the Training of the USA IMO Team(Illustrated)
Titu Andreescu, Zuming Feng

Number Theory And Its Mathematical Structures: A Problem-solving Approach
Titu Andreescu, Dorin Andrica

Chemistry:

General Chemistry Vol I and II by Charles Mortimer

Physics:
Fundamentals of Physics
by David Halliday, Robert Resnick, Jearl Walker


algebra
polynomial
number theory unsolved
number theory

Let  $P(x)=a_0+a_1x+...+a_nx^n$ and  [b]positive integer [/b]$p$st:
i/$(a_1,p)=1$
ii/$a_2,..,a_n$ are divisible by all of prime divisors of$p$
Show that:for each k there exists a such that$P(a)$ is divisible by $p^k$

$Easyly \ find \ x_1 : p|P(x_1), \ because \ p \not |P^{'}(x_1), \ by \ Gensel's \ Lemm, \ we \ can \ find \ x_k: \ p^k|P(x_k).$


I'm sure he means this: given any five points in the plane, no three of which lie on a line, there is a convex quadrilateral with vertices among the five points.

Sorry, confused myself while typing, its edited now.

[quote="seamusoboyle"]Prove that given any 9 points in the plane, that there exists a convex pentagon with all its vertices on these 5 points.[/quote]

This is still unsolved? Anyway, see the older discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=21981 , and the related problems at

http://www.mathlinks.ro/Forum/viewtopic.php?t=26701
http://www.mathlinks.ro/Forum/viewtopic.php?t=16456
http://www.mathlinks.ro/Forum/viewtopic.php?t=47979
http://mathworld.wolfram.com/HappyEndProblem.html

  Darij


geometry
circumcircle
cyclic quadrilateral

Congratulations!
The students solving the following in 5 min.

4. Circles K and L intersect at points P and Q. Distinct points C and D (not at P and Q) are selected on K. The lines CP and DP meet L again at E and F respectively, and the lines CD and EF meet at O.
Prove that, as C and D vary, the circumcenters of triangles CEO all lie on one fixed circle.

First, assume that the point $C$ is fixed and the point $D$ moves, WLOG, on the arc $CQ$ of the circle $(K)$ opposite to the point $P$. From the triangles $\triangle COE, \triangle CDP, \triangle EFP$,

$\angle COE = 180^o - (\angle CDP + \angle CPD) - [180^o - (\angle EFP + \angle EPF)] =$

$= \angle EFP - \angle CDP$

Since the angles $\angle CDP, \angle EFP$ spanning fixed arcs $CP, EP$ of the circles $(K), (L)$ are fixed, the angle $\angle COE$ is also fixed. Since the side $CE$ of the triangle $\triangle COE$ is fixed as well, the circumcircle $(M)$ of this triangle does not change either, i.e., the circumcenter $M$ does not depend on the position of the point $D$.

Second, assume that the point $D$ is fixed and the point $C$ moves, WLOG on the arc $DP$ of the circle $(K)$ opposite to the point $Q$. By considering the triangle $\triangle DOF$ with the fixed side $DF$, we can show in exactly the same way as above that the angle

$\angle COE \equiv DOF = \angle DCP - \angle FEP$

is still fixed and has the same constant measure as before. Since the angle $\angle OEC \equiv FEP$ spanning the fixed arc $FP$ of the circle $(L)$ is also fixed, the triangle $\triangle COE$ now remains similar for all positions of the point $C$. Consider the quadrilateral $COEQ$. Since the angles $\angle PCQ, \angle PEQ$ spanning the fixed arcs $PQ$ of the circles $(K), (L)$ are also fixed, the quadrilateral $COEQ$ also remains similar for all positions of the point $C$. This quadrilateral is cyclic, because

$\angle COE + \angle CQE = (\angle EFP - \angle CDP) + (\angle CQP + \angle EQP) = 180^o$

Moreover, the vertex $Q$ of the cyclic quadrilateral $COEQ$ does not move at all.

[color=blue][b]Theorem. [/b][/color] Let the figure $\Phi$ move in such a way that it is always directly similar to a figure $\Phi_0$ and one common point $A$ of the similar figures $\Phi \sim \Phi_0$ does not move at all. Let the locus of a point $B$ of the figure $\Phi$ be a curve $\Gamma$. Then the locus of any other point $C$ of the figure $\Phi$ is a curve $\Gamma'$ similar to the curve $\Gamma$.

For a simple proof, see Yaglom, Geometric Transformations II (chapter 1, \$1, pp. 68-69, theorem without a number between theorems 2 and 3). Since the loci of vertices $C, E$ of the quadrilateral $COEQ$ are the circles $(K), (L)$, it follows that the locus of circumcenters $M$ of the triangles $\triangle COE$ is also a circle. Label it as the circle $(S)$. When the point $C$ coincides with the point $P$, the line $CE$ becomes a tangent to the circle $(K)$ at the point $P$, the point $O$ coincides with the point $F$ and the circle $(L)$ becomes the circumcircle of the triangle $\triangle COE$, because it is the circumcircle of the triangle $\triangle PQE \equiv \triangle CQE$ and the quadrilateral $COEQ$ is cyclic. Thus the point $L$ lies on the locus circle $(S)$. When the point $E$ coincides with the point $P$, the line $CE$ becomes a tangent to the circle $(L)$ at the point $P$, the point $O$ also coincides with the point $P$ and we can similarly deduce that the circle $(K)$ becomes the circumcircle of the triangle $\triangle COE$, i.e., the point $K$ also lies on the locus circle $(S)$. When the point $C$ coincides with the point $Q$, the triangle $\triangle COE$ and the quadrilateral $COEQ$ degenerate to the point $Q$, regardless of the position of the point $D$. Hence, the point $Q$ also lies on the locus circle $(S)$. As a result, the locus circle $(S)$ is the circumcircle of the triangle $\triangle KLQ$.


Prove the following by principle of induction :

$ 7\plus{}77\plus{}777\plus{}7777\plus{}.....................\plus{}777..........n$ $ digits7\equal{} \frac{7}{81} (10^{n\plus{}1}\minus{}9n\minus{}10).$

Please help. Thanks a lot.

[hide="Solution"]For $ n \equal{} 1$ we have $ \frac {7}{81}(10^{n \minus{} 1} \minus{} 9n \minus{} 10) \equal{} \frac {7}{81}(100 \minus{} 9 \minus{} 10) \equal{} \frac {7}{81}(81) \equal{} 7$ so base case done.

Now, assume the result proven for $ k$ and prove for $ k \plus{} 1$.

We are trying to show that $ 7 \plus{} 77 \plus{} 777 \plus{} 7777.... \plus{} 77777... k \plus{} 1$ digits $ \equal{} \frac {7}{81}(10^{k \plus{} 2} \minus{} 9(k \plus{} 1) \minus{} 10)$.

We have $ 7 \plus{} 77 \plus{} 777... \plus{} 777777... k \plus{} 1$ digits $ \equal{} \frac {7}{81}(10^{k \plus{} 1} \minus{} 9k \minus{} 10) \plus{} 7 \cdot 10^{k} \plus{} 7 \cdot 10^{k \minus{} 1}... \plus{} 7 \cdot 10 \plus{} 7$.

This is equivalent to $ \frac {7}{81}(10^{k \plus{} 1} \minus{} 9k \minus{} 10) \plus{} 7 \cdot 1111111... k \plus{} 1$ digits.

Now consider the number $ 11111... k \plus{} 1$ digits. This can be written as $ \frac {10^{k \plus{} 1} \minus{} 1}{9}$.

Now we have $ \frac {7}{81}(10^{k \plus{} 1} \minus{} 9k \minus{} 10) \plus{} \frac {7}{9}(10^{k \plus{} 1} \minus{} 1) \equal{} \frac {7}{81}(10^{k \plus{} 1} \minus{} 9k \minus{} 10) \plus{} \frac {7}{81}(9 \cdot 10^{k \plus{} 1} \minus{} 9)$.

Adding the two parts:

$ \frac {7}{81}(10^{k \plus{} 1} \plus{} 9 \cdot 10^{k \plus{} 1} \minus{} 9k \minus{} 9 \minus{} 10) \equal{} \frac {7}{81}(10^{k \plus{} 2} \minus{} 9(k \plus{} 1) \minus{} 10)$.

$ 7 \plus{} 77 \plus{} 777 \plus{} 7777.... \plus{} 77777... k \plus{} 1$ digits $ \equal{} \frac {7}{81}(10^{k \plus{} 2} \minus{} 9(k \plus{} 1) \minus{} 10)$.

Therefore, by the principle of mathematical induction, the result is true for all n.[/hide]

EDIT: Beaten to it.


linear algebra
matrix
linear algebra unsolved

consider this 2n dimensional matrix :

$\Gamma=\begin{bmatrix} 0& 1_n\\-1_n&0\end{bmatrix}$
its determinant is 1

Now suppose $J$ is invertible and $J^T \Gamma J=\Gamma$

Proof that the determinant of $J$ is 1

Is it also true when $\Gamma$ is an unspecified invertible asymmetric matrix?

So uhm, does anyone have :?  any ideas

I am starting to assume that my statement about skewsymmetric matrices will be wrong in such a general way?

I am also starting to feel I am missing something huge here

sorry for pushing, i still don't know anything more

If it makes it any more interesting, it is related to canonical transformation, in hamiltonian systems


First, sorry for not using LaTeX. :D

|x| - |x-7| = 7

x ≥ 0 for x ≥ 0
x-7 ≥ 0 for x ≥ 7

My breakpoints are 0 and 7.

...

I honestly do not quite understand what I have to do next... I am confused with the intervals and use of breakpoints. Anyone care to help me out (that is, show me a thorough and concise solution with explanations)? Thanks a lot!

[u]EDIT[/u]

I tried doing the following.

For x ≥ 0,

x - (-x+7) = 7
x = 7
Since 7 ≥ 0, x = 7 is a valid solution.

For x ≥ 7,
x - (x-7) = 7
7 = 7
(Now, I don't know what to do if I get 7 = 7 (similarly, 0 = 0))

Did I miss something? Did I do anything wrong? Please help me thanks!

You're on the right track.

[quote="I Hate Watermelons"]

|x| - |x-7| = 7

x ≥ 0 for x ≥ 0
x-7 ≥ 0 for x ≥ 7

My breakpoints are 0 and 7.

[/quote]

[hide="using intervals"]So you have two separate conditions $ x \ge 0$ and $ x \ge 7$. List out the possibilities: if none of them were true, if only the first one were true, if only the second were true, and if both were true. Obviously you can't have the second one but not the first one. So we have $ 3$ possibilities.

An easier way to think of it is: since you have $ 2$ breakpoints, you need at three intervals. (Think of a number line: if you cut it at two points, you get three sections.) The three intervals are the ones split by the breakpoints:
$ x \le 0$
$ 0 \le x \le 7$
$ 7 \le x$

Simply use logic to deduce what you should do.[/hide]

[hide="first interval"]In the first interval, $ x \le 0$, so the expression $ |x|$ would come out to be $ \minus{} x$. At the same time $ |x \minus{} 7| \equal{} \minus{} (x \minus{} 7) \equal{} \minus{} x \plus{} 7$. Thus
$ |x| \minus{} |x \minus{} 7| \equal{} 7$
$ \minus{} x \minus{} ( \minus{} x \plus{} 7) \equal{} 7$
$ \minus{} x \plus{} x \minus{} 7) \equal{} 7$
$ \minus{} 7 \equal{} 7$
The variables have canceled out. If you forget what to do when this happens, just think of it this way: we just deduced that for any $ x \le 0$, when you plug the $ x$ into $ |x| \minus{} |x \minus{} 7| \equal{} 7$ you get $ \minus{} 7 \equal{} 7$ (a false statement). So our $ x$ did NOT satisfy the equation. Thus there are no solutions for $ x \le 0$.[/hide]

[hide="second interval"]In the second interval, $ x \ge 0$ but $ x \le 7$. We have $ |x| \equal{} x$ and $ |x \minus{} 7| \equal{} \minus{} (x \minus{} 7) \equal{} \minus{} x \plus{} 7$. Doing the same thing:
$ |x| \minus{} |x \minus{} 7| \equal{} 7$
$ x \minus{} ( \minus{} x \plus{} 7) \equal{} 7$
$ x \plus{} x \minus{} 7) \equal{} 7$
$ 2x \equal{} 14$
$ x \equal{} 7$
So we have one solution, namely $ x \equal{} 7$, in the interval $ 0 \le x \le 7$. [/hide]
[hide="third interval"]
The third interval, which you already worked out correctly, leads to $ 7 \equal{} 7$, a true statement. Thus for any $ x$ that is greater than or equal to $ 7$, when you plug it in in checks. So all numbers greater than or equal to $ 7$ are a solution.[/hide]

(Our intervals overlapped, but that's ok.)

$ \boxed{x \ge 7}$ or $ \boxed{x \in [7,\infty)}$


Hi I was wondering how does one pronounce these two French phrases:

1) [b]il faut[/b] is this "eel foh", and this mean [u]it is necessary[/u], correct?
2) [b]il vaut[/b] [i]I don't know this one ...[/i], and this one means [u]it must[/u]?

[color=blue]Il vaut[/color] means “it is worth”; e.g. [color=blue]il vaut mieux aller[/color] (it is better to go). It’s pronounced like [color=blue]il faut[/color] but with /v/ instead /f/. :)


I think this problem is very hard . isn't it? :|

I think we need to show that among any 1999 diagonals of a 2000-gon there must exist three that intersect each other. I also guess that this is true, but find hard to prove. anyone help us :?:

[quote=DapperPeppermint][quote=M.A]No three diagonals of a convex 2000-gon meet at a point.the diagonals(but not the sides)are each colored with one of 999 colors.show that there is a triangle whose sides are on three diagonals of the same color. :)[/quote]

What? I think I am drunk. See that number of diagonal are $2020*2017$ and so $\exists$ a color such that at least $\left \lceil \dfrac{2020*2017}{999} \right \rceil = 4080$ diagonal are colored in that color. Now direction statement works well. Observe that at most $1000$ diagonals at once are parallel. Aren't wr done now?[/quote]

I reckon it must be 2020*2017/2.


linear algebra
matrix
IMO Shortlist
algebra
system of equations

In a certain city, age is reckoned in terms of real numbers rather than integers. Every two citizens $x$ and $x'$ either know each other or do not know each other. Moreover, if they do not, then there exists a chain of citizens $x = x_0, x_1, \ldots, x_n = x'$ for some integer $n \geq 2$ such that $ x_{i-1}$ and $x_i$ know each other. In a census, all male citizens declare their ages, and there is at least one male citizen. Each female citizen provides only the information that her age is the average of the ages of all the citizens she knows. Prove that this is enough to determine uniquely the ages of all the female citizens.

Let the graph $ G$ be defined as follows. Every person is a vertex. Two vertices are connected iff they are acquainted. The graph is a tree because it is connected and there are no loops. Let $ \parallel{}v\parallel{}$ denote the real number associated with the vertex $ v$. We shall use the property that every tree has at least two vertices of degree 1.

(i) If the vertex with degree 1 is a woman say $ w_1$:If the person acquainted to $ w_1$ is a man say $ m_1$, then we have $ \parallel{}w_1\parallel{}\equal{}\parallel{}m_1\parallel{}$. We can now eliminate $ w_1$ from $ G$ and look out for the vertex with degree 1 in $ G$.
If the person acquainted to $ w_1$ is a woman say $ w_2$ and if $ w_3,\dots,w_i,m_1,\dots,m_j$ are the acquainted to $ w_2$ ,then
\[ (i\plus{}j\plus{}1)\parallel{}w_2\parallel{}\equal{}w_1 \plus{}\sum_{k\equal{}3}^{k\equal{}i}w_k\plus{}\sum_{k\equal{}1}^{k\equal{}j}m_k\]
\[ or\quad (i\plus{}j)\parallel{}w_2\parallel{}\equal{}\sum_{k\equal{}3}^{k\equal{}i}w_k\plus{}\sum_{k\equal{}1}^{k\equal{}j}m_k\quad \dots \quad (1)\]
Since $ w_1$ has nothing to do with $ G$ we may remove it and then look for the vertex with with degree 1.

(ii) If the vertex with degree 1 is a man say $ m_1$: If the vertex acquainted to $ m_1$ is again a man, we may safely eliminate the vertex $ m_1$ as it is of no consequence and look out for the vertex with degree 1. If the vertex acquainted to $ m_1$ is a woman say $ w_1$ then $ \parallel{}w_1\parallel{}\equal{}\parallel{}m_1\parallel{}$. We may eliminate the vertex $ m_1$ and look out for the vertex with degree 1.

This process is ends after some time as the population is finite. In the last step we would  determine the real number associated with some woman say $ w_i$ by using the calculation analogous to (1). Going back one step previous, we can now get the real number associated with $ w_j$ and so on to get the real number associated with every woman.

[quote="srikanth"]The graph is a tree because it is connected and there are no loops.[/quote]

Sorry, but I don't understand how you deduced this.  Could you explain your method?

[quote="orl"]... Every two citizens $ x$ and $ x'$ either know each other or do not know each other. Moreover, if they do not, then there exists a chain of citizens $ x = x_0, x_1, \ldots, x_n = x'$ for some integer $ n \geq 2$ such that $ x_{i-1}$ and $ x_i$ know each other. [/quote]

@cosinator:
$ G$ has people as vertices and  there is an edge between any two vertices if they are acquainted.

Firstly, $ G$ is connected.

Secondly there is no loop in the graph. If there existed a loop in the graph then there exists an element $ x_1$ which is connected to some $ x_k\neq x_1$ through two different ways. Let $ x_0$ and $ x_n$ be the first elements one encounters when moving from $ x_1$ to $ x_k$ in the two ways respectively. $ x_1$ is acquainted to $ x_0$.We may reach from $ x_1$ to $ x_0$ by looping through the second way that passes through $ x_n$. But the existence of the second way is contradicted by the statement in quote which says that "Every two citizens $ x$ and $ x'$ either know each other or do not know each other. [b]Moreover, if they do not, then [/b]there exists a chain of citizens"

[quote="srikanth"]But the existence of the second way is contradicted by the statement in quote which says that "Every two citizens $ x$ and $ x'$ either know each other or do not know each other. [b]Moreover, if they do not, then [/b]there exists a chain of citizens"[/quote]  You are misunderstanding this sentence.  It does [i]not[/i] say that "$ x$ and $ x'$ are not acquainted [i]if and only if[/i] there exists such a chain," rather it says that if $ x$ and $ x'$ are not acquainted then there is such a chain (but there could also be such a chain for some pair of acquainted vertices).  This sentence gives you a connected graph; it does not give you a tree.

[quote="JBL"]but there could also be such a chain for some pair of acquainted vertices[/quote]

I would not agree with this. On the other hand, if the graph is not a tree then the proposotion does not hold. Consider the graph $ G'$(image titled 'nottree'). Then the system of equations does not have a solution.

[i]mathematica code:

In[87]:=
m={{2,-1,0,0,0},{-1,3,-1,0,-1},{0,-1,2,-1,0},{0,0,-1,2,-1},{0,-1,0,-1,2}}

Out[87]=
{{2,-1,0,0,0},{-1,3,-1,0,-1},{0,-1,2,-1,0},{0,0,-1,2,-1},{0,-1,0,-1,2}}

In[88]:=
m.{x,y,z,r,h}=={a,0,0,0,0}

Out[88]=
{2 x-y,-h-x+3 y-z,-r-y+2 z,-h+2 r-z,2 h-r-y}\[Equal]{a,0,0,0,0}

In[89]:=
Solve[%,{x,y,z,r}]

Out[89]=
{}[/i]

I hope orl would tell us what the question actually intends.

I don't know what you told Mathematica to do, but I do know that the problem is perfectly fine for any kind of connected graph, not just trees :).

It all boils down to showing that a certain linear system with the same number of equaqtions and unknowns (namely the number $ k$ of female citizens) has a unique solution. Write this sistem as $ Ax \equal{} b$, where $ A$ is a $ k\times k$ real matrix and $ x$ and $ b$ are column vectors of length $ k$. Now, if $ Ax \equal{} 0$ has the unique solution $ x \equal{} 0$, then you're done: it follows that $ A$ is invertible, so $ Ax \equal{} b$ does indeed have a unique solution. Now use the connectedness of the graph and the fact that there is at least one male citizen to prove that $ Ax \equal{} 0$ has the unique solution $ x \equal{} 0$. In male/female citizens languange, this means: show that in the hypotheses of the problem, if all the males declare their ages to be $ 0$, then the ages of all citizens are $ 0$. This should be easy enough: assuming the maximal age is $ M > 0$, consider a female with age $ M$ and see what happens (its neighbors must also have age $ M$, then the neighbors' neighbors also have age $ M$, etc., until you reach a male citizen, whose age is $ 0 < M$; you got the desired contradiction); similarly, you can show that the minimal age is $ 0$ as well.

Another way to state the problem:

The discrete [url=http://planetmath.org/encyclopedia/DirichletProblem.html]Dirichlet problem[/url] has a unique solution on a finite connected graph as long as the boundary is non-empty.

Here the graph is the one we've all been working on, and the "boundary" is the (non-empty, according to the hypothesis) set of male citizens.

This may be trivial, but I think there is one more step to be made: we need $ A$ to be invertible and we need for each positive vector $ b$ that the solution $ x$ to $ Ax \equal{} b$ also be a positive vector.

srikanth, I also don't understand what you were trying to get Mathematica to do, but whatever it was, it didn't work.  If there's only one male citizen then it's trivial that at least one solution exists, when all residents have the same age.

I did forget that we were working with ages and so the solution has to be positive, but technically, we don't need to prove that. The problem asks to show that we can determine all ages given the ages of the males. We already know the solution exists and it's positive (the people have positive ages - that much we know), so we only need to prove the uniqueness. Admittedly, my "reformulation" of the problem isn't really an exact reformulation, but it's in the ballpark :).

Here is a not-solution that is pretty amusing nonetheless:

After converting everything to a connected graph $G$, view the men as voltage sources whose voltages whose potential is equal to their age, and view the edges of the graph as $1 \ \Omega$ resistors.  We claim that the potential at each women is equal to the average of the potential of its neighbors.  Indeed, $V=IR=I$, so by Kirchoff's junction rule, \[ \sum_{\text{neighbors}} V_{\text{neighbor}} - V_{\text{her}} = 0 \] and the claim follows.

Now by the properties of the physical world, there is one equilibrium point, so we win.

[quote=vickymano]Let a1, a2, . . . , am be the ages of the male citizens (m ≥ 1). We claim that
the age of each female citizen can be expressed in the form c1a1+· · ·+cmam
for some constants ci ≥ 0, and we will prove this by induction on the
number n of female citizens.
The claim is clear if n = 1. Suppose it holds for n and consider the case
of n+1 female citizens. Choose any of them, say A of age x who knows k
citizens (at least one male). By the induction hypothesis, the age of each
of the other n females is expressible as c1a1 + · · · + cmam + c0x, where
ci ≥ 0 and c0 + c1 + · · · + cm = 1. Consequently, the sum of ages of the k
citizens who know A is kx = b1a1 + · · · + bmam + b0x for some constants
bi ≥ 0 with sum k. But A knows at least one male citizen (who does not
contribute to the coefficient of x), so b0 ≤ k−1. Hence x = b1a1+···+bmam/k-b0
and the claim follows.[/quote]

Why we need sum(ci)=1?

Let there be $n \geq 3$ female citizens (cases $n=1$ and $n=2$ can be checked directly). Let $f_i$ be the age of female citizen $i$. Let $a_{ij}=1$ if female citizens $i,j$ are connected and $a_{ij}=0$ otherwise. Then we have a system of equations of kind $p_if_i-a_{1i}f_1- \ldots -a_{ni}f_n=q_i$, where $p_i \geq a_{1i}+ \ldots +a_{ni}$. This system has a unique solution unless there is a linear combination of such equations with coefficients of all $f_i$ being $0$. Let such a linear combination exist where we multiply $i$-th equation by $c_i$, and not all of them $0$. Then take $j$ such that $|c_j|$ is largest. Then since $|c_jp_j| \geq |c_j(a_{1j}+ \ldots +a_{nj})| \geq |c_1a_{1j}+ \ldots +c_na_{nj}|$, so the coefficient of $f_j$ is not $0$ unless $|c_k|=|c_j|$ for all $k$ such that $a_{kj}=1$ and $j$th female isn't connected to any males. Then apply same logic to all her female friend, then to their friends and so on until we hit a female who is connected to a male (this will happen as there is at least one male and the graph is connected).

[hide=Sketch]
Let $L'$ be the Laplacian when we delete the rows and columns for males. The goal is to show that
\[L'v=x\]
has a unique solution where $x$ is some vector that you get from the males. If $L'v=x$ and $L'u=x$, then $L'w=0$ for some $w$. In this scenario, consider the oldest female. She is at most the average of her neighbors, so in fact all the entries of $w$ are equal, but since we deleted some males, this fails unless $w=0$. Thus, the solution is unique, so $L'$ is nonsingular, so there is a solution.
[/hide]


trigonometry AMC USA(J)MO USAMO geometry

Οι μεσοκάθετοι των πλευρών $ AB,AC$ ενός τριγώνου $ ABC$ τέμνουν τη διάμεσο $ AM$ στα σημεία $ D,E$ αντίστοιχα . Η

ευεθεία  $ CE$  τέμνει την  $ BD$ σημείο   $ K$ .Αν $ O$ είναι το περίκεντρο του $ ABC$ , να αποδειχθεί ότι $ AK \perp KO$.

 Μπάμπης

Ωραίο πρόβλημα!

Αφιερώνω την "ημιτελής" λύση μου στον Τάσο (Vaf) :wink: (ο ίδιος θα καταλάβει το γιατί...)

Λοιπόν κατ' αρχάς χωρίς βλάβη της γενικότητας έχω θεωρήσει $ AC>AB$

Για χάρη συντομίας θέτω $ \angle{DBA}\equal{}B_1,\angle{DBM}\equal{}B_2,\angle{ECA}\equal{}C1, \angle{ECA}\equal{}C_1, \angle{ECM}\equal{}C_2$, $ \angle{KAB}\equal{}A_1, \angle{KAC}\equal{}A_2,\angle{KOD}\equal{}O_1,\angle{KOE}\equal{}O_2$.

Εχουμε, χρησιμοποιώντας το νόμο ημιτόνων:

$ \frac{O_1}{O_2}\equal{}\frac{\sin{KDO}}{\sin{KEO}}\cdot\frac{DK}{KE}\equal{}$

$ \equal{}\frac{\cos{B_1}}{\cos{C_1}}*\frac{\sin{2C_1}}{\sin{2B_1}}\equal{}\frac{\sin{C_1}}{\sin{B_1}}$

ARA $ \frac{O_1}{O_2}\equal{}\frac{\sin{MAC}}{\sin{MAB}}$ κι επειδή $ \angle{BAC}\equal{}\angle{DOE}$ έπεται ότι
$ \angle{O_1}\equal{}MAC$ and $ \angle{O_2}\equal{}MAB$ (αυτό το τελευταίο συμπέρασμα το αφήνω για άσκηση)

Τώρα επειδή αρκεί να αποδείξουμε ότι $ O_1\equal{}A_1$ (από εγγράψιμα), αρκεί να αποδείξουμε ότι $ A_1\equal{}MAC$, δηλαδή ότι $ AK$ συμμετροδιάμεσος.

Εχω ιδιέταιρα σε κανένα μισάωρο και δεν προλαβαίνω να την ολοκληρώσω :( θα επιστρέψω μετά τις 8:30...

H askisi einai gnosti. einai palio thema tis usamo. gia tin akriveia usamo 2008 problem 2 ...see here http://www.mathlinks.ro/viewtopic.php?p=1116181#1116181

oso gia tin prosegisi tou mim...re mim den xreiazetai na blekseis imitona kai sinimitona gia na ftaseis oti arkei i ak na einai symmetrodiamesos. ftaneis sto idio me polu aples goniakes sxeseis.

[quote="bilot4"]Νίκο πιθανότατα έχεις δίκιο, αλλά ξέρεις πως είναι όταν σου κολλάει κάτι στο μυαλό :mad:  :oops: .

Anyway θα έπρεπε να το σκεφτώ να φέρω το ισογώνιο συζυγές του $ K$...αλλά έχει περιοριστεί πολύ ο διαθέσιμος χρόνος μου πλέον :([/quote]

e nai re...prepei na diavaseis kai gia ti molpi tis amfilikis...o ime!

[quote="Nick Rapanos"]

e nai re...prepei na diavaseis kai gia ti molpi tis amfilikis...o ime![/quote]

 :rotfl:  :rotfl: ...μαθαίνεις γρήγορα LOL!!!. BTW για λίγο ακόμα γέλιο σου παρουσιάζω ένα κομμάτι απ' τις σημειώσεις:P
"To ρομαντικό στοιχείο διαδέχεται το συμβολιστικό και επισφραγίζει το ποίημα με νοσταλγικούς τόνους καθώς πρόκειται για μια ουτοπική τάση φυγής χωρίς δικάιωση" !!! λολ

Sorry gia to spam alla ayti akrivws ti stigmi o dimitris kai o nikos exoyn ton idio arithmo posts... :rotfl:

[quote="silouan"]Λοιπόν για να μην φέρω εδώ το raστko και πει το αμίμητο  :wink:  βάζω μια άσκηση που ήρθε μετά από αυτή της USAMO .

Έστω τρίγωνο $ ABC$ και έστω $ C_1$ ο κύκλος που περνά από τα $ A,C$ και εφάπτεται της $ AB$ στο $ A$ και έστω $ C_2$ ο κύκλος που περνά από τα $ A,B$ και εφάπτεται της $ AC$ στο $ A$ . Έστω ότι αυτοί οι δύο κύκλοι τέμνονται στο $ K$ . Αν $ O$ το περίκεντρο του $ ABC$ νδο $ AK\bot KO$

Είναι πολύ εύκολη αλλά είναι ωραία  :)[/quote]

nice problem!

8a doso ena hint!

[hide]Αντιστροφή με πόλο $ A$  :wink: (οι εφαπτόμενοι κύκλοι πάνε σε παράλληλες ευθείες...) [/hide]

[quote="silouan"][quote="bilot4"]
8a doso ena hint!

[hide]Αντιστροφή με πόλο $ A$  :wink: (οι εφαπτόμενοι κύκλοι πάνε σε παράλληλες ευθείες...) [/hide][/quote]

Ωραία σκέψη ... :)   . Έχω όμως μια λύση που χρησιμοποιεί εντελώς elementary geometry . Try it ![/quote]
Νομίζω ότι πιο elementary, δεν γίνεται.  :wink:

[hide]Έστω $ (O_{1}),$ και $ (O_{2}),$ οι κύκλοι με χορδές $ AC,\ AB$ αντιστοίχως, που εφάπτονται στις πλευρές $ AB,\ AC,$ στο σημείο $ A.$

Αρκεί να αποδείξουμε ότι $ KO\parallel O_{1}O_{2},$ γιατί $ AK\perp O_{1}O_{2}$

Αποδεικνύεται εύκολα ότι το τετράπλευρο $ AO_{1}OO_{2}$ είναι παραλληλόγραμμο και προκύπτει έτσι ότι τα τρίγωνα $ \bigtriangleup OO_{1}O_{2}$ και $ KO_{2}O_{1}$ είναι ίσα, γιατί έχουν την  $ O_{1}O_{2}$ ως κοινή πλευρά και $ OO_{1} \equal{} AO_{2} \equal{} KO_{2}$ $ ,(1)$ και  $ OO_{2} \equal{} AO_{1} \equal{} KO_{1}$ $ ,(2)$

Από την ισότητα των τριγώνων αυτών, παίρνουμε $ \angle OO_{1}O_{2} \equal{} \angle KO_{2}O_{1}$ $ ,(3)$

Από $ (1)$ και $ (3),$ συμπεραίνουμε ότι $ KO\parallel O_{1}O_{2}$ και η πρόταση έχει αποδειχθεί.[/hide]
Κώστας Βήττας.

Πολύ ωραία , αυτό ακριβώς εννούσα . :)


Big Ben, a famous 12-hour clock in London, England, sounds four notes on the quarter-hour, eight notes on the half-hour, twelve notes on the three-quarter hour, and sixteen notes on the hour. Furthermore, it strikes one note for each hour on the hour. for example, it strikes five additional notes at 5 o'clock. How many notes will Big Ben strike in a twenty-four hour period?

Well as it is a 12 hour clock we should find the number of times in a 12 hour period and double it.

It will ring (1+2+3...+12)=78 bonus notes on the hour.

On quarter-hours it will ring (4x12)=48 times

Similar logic shows 96 on the half hour, 144 on the three quarter, and 192 on the hour.

48+96+144+192+78=48+240+270=558 per 12 hours.

558x2=1116


prove that the number 0.1011011101111... is irrational

D.

Did you mean the sequence
$ 10;110;1110,....;11..110$
Suppose it is rational then the sequence (not above) is period --->contrad.


If $a+b = 1$ and $a^2 + b^2 = 2$ then what is the value of $a^3 + b^3$?

Errr...I got different answer than amirhtlusa.  :?
[hide]
$(a^2+b^2)(a+b)=2\times1=a^3+a^2b+ab^2+b^3=2$
$a^3+b^3+ab(a+b)=2$
Since $a+b=1$, $a^3+b^3+ab=2$, so we need to find $ab$.

$(a+b)^2-(a^2+b^2)=1^2-2$
Simplifying gives $2ab=-1$, so $ab=-\frac{1}{2}$.
Plugging in gives $a^3+b^3=2+\frac{1}{2}=2\frac{1}{2}$[/hide]

Edit: wow 2 people beat me to it while I was typing this up  :lol:

akhhh...yes,
$\frac{5}{2}$ is correct :)

[quote="Valentin Vornicu"][quote="Iversonfan2005"]You are given that a+b=1 and a^2+b^2=2. So then you know that:

a^2+b^2=(a+b)^2-2ab, so that would mean that (1)^2-2ab=2, so then ab=-1/2

You know that a^3+b^3=(a+b)^3-3(ab)(a+b), so then just plug in the numbers, thus getting: 1^3-3(-1/2)(1) which is 1+3/2 which is equal to [b]5/2[/b].[/quote]How about using $\LaTeX$ ... I see you know the language ...  :roll:[/quote]

Sorry, I just am too lazy to install and use it. That's why I try to stay out of too many exponential problems.  :blush:

[quote="tarquin"]Using Latex on these forums does not require you to install anything at all.[/quote]Exactly. Follow the link in my signature. You already know $\LaTeX$, you just need to add the $\textdollar$ tags :)

[quote="Iversonfan2005"]You are given that a+b=1 and a^2+b^2=2. So then you know that:

a^2+b^2=(a+b)^2-2ab, so that would mean that (1)^2-2ab=2, so then ab=-1/2

You know that a^3+b^3=(a+b)^3-3(ab)(a+b), so then just plug in the numbers, thus getting: 1^3-3(-1/2)(1) which is 1+3/2 which is equal to [b]5/2[/b].[/quote]

Let me try this in latex version now:

$a^2+b^2=(a+b)^2-2ab$, so that would mean that $1^2-2ab=2$, so then $ab=-1/2$

You know that $a^3+b^3=(a+b)^3-3(ab)(a+b)$, so then just plug in the numbers, thus getting $1^3-3(-1/2)(1)$ which is $1+3/2$ which is equal to [b]$5/2$[/b]

Yay, my first use of Latex and I did great  :showoff:  :surf:  :gleam:  :clap2:  :10:  :thumbup:

Yeah, well writing $3/2$ is not really much $\LaTeX$. Instead write $\frac 32$ ( that is $\textdollar$\frac 32$\textdollar$):P


logarithms
calculus
calculus computations

yello  :D here is a question i asked for HW yesterday, have fun.  :)

given that

$ \beta(p) \equal{} \sum_{j\equal{}0}^\infty \frac{(\minus{}1)^j}{(2j\plus{}1)^p}$,

and $ \eta(p) \equal{} \sum_{j\equal{}1}^\infty\frac{(\minus{}1)^{j\minus{}1}}{j^p}$

show
$ \frac{\beta'(1)}{\beta(1)} \equal{} \gamma \plus{} 3\ln\pi \plus{} 2\,\eta(1)\minus{}4 \ln   \Gamma \left(\frac{1}{4}\right)$

hence,  find the  limit

$ \sum_{k\equal{}1}^\infty \; \ln \left[ \frac {(4k\minus{}1)^{\frac{1}{ 4k\minus{}1} }} {(4k\plus{}1)^{\frac{1}{4k\plus{}1}} } \right]$

I think your sum is something like $ \approx \beta'(1)$  :)  with simple correction (say $ \plus{}1$ or $ \minus{}1$ or.. )


combinatorial geometry
real analysis
real analysis unsolved

Can we find simple hypothesis on a space such that any point of the convex hull of a set is the barycenter of at most $n$ (and not $n+1$, as in the theorem) points of the set? Study the cases: connected, arc connected, etc

Ok, let's prove that "connected" works.

Take any point $P$ in the convex hull $\cal H(S)$ of the connected set $\cal S$ of points under consideration. By Caratheodory's Theorem, there are $n+1$ points $S_0,\ldots,S_n\in\cal S$ s.t. $P=\sum_{i=0}^n\alpha_iS_i$ with $\alpha_i\in[0,1]$ and $\sum_i\alpha_i=1$. If one of the $\alpha_i$'s is $0$, we're done. We may thus assume that all of them are positive.

Now let $\mathcal A\subset\mathbb R^n$ be the subset of the Euclidean $n$-space consisting of those points $X$ which, when written in the affine basis $P,S_1,\ldots,S_n$, have positive $P$-coefficient, while the other coefficients are negative. Also let $\mathcal A^c$ be the complement of $\cal A$ in $\mathbb R^n$.

Notice that $S_0\in \cal A$, while $S_i\in \mathcal A^c,\ \forall i\in\overline{1,n}$, so both $\mathcal A\cap\cal S$ and $\mathcal A^c\cap\cal S$ are non-empty. Since $\cal A$ is obviously open, $\mathcal A\cap\cal S$ is an open subset of $\cal S$. $\cal S$ is connected and neither $\mathcal A\cap\cal S$ nor $\mathcal A^c\cap\cal S$ are empty, so $\mathcal A\cap\cal S$ cannot be closed (in $\cal S$). This means that there is a point $X\in\cal S$ which belongs to the frontier $\delta\cal A$ of $\cal A$. This point $X$ can be written as $X=\sum_{i=1}^n\beta_iS_i+\beta P$, with $\sum_i\beta_i+\beta=1,\ \beta\ge 0,\ \beta_i\le 0,\ \forall i$, and at least one of the $\beta_i$ is $0$. This, in turn, means that $P$, when written as an affine combination of $X,S_1,\ldots,S_n$, has only non-negative coefficients, at least one of which is null (obe of the $S_i$-coefficients), so it’s a convex combination of at most $n$ points of $S$.


[quote="JRav"]Just a note, which I've noticed in not only your posts, but others: there is no word "slove."  The correct word is "solve."[/quote]
 Sorry for english of me so bad ...... :oops:

i) has no solutions.
 $ 9A^4_{ij}\plus{}15B^5_{ij}\equal{}11$ is impossible.
ii) $ 3A^2\plus{}5B^2\equal{}\begin{pmatrix}6&6\\6&6\end{pmatrix}$ has solutions ; for example:
$ A\equal{}\pm\begin{pmatrix}1&1\\1&1\end{pmatrix},B^2\equal{}0$


Hi,

I was wondering if it's possible to create a function that only does things if the argument is a specific thing?

For example, to create a function \p so that

\p{0} returns $ e^{\minus{}\lambda}$
\p{1} returns $ \lambda e^{\minus{}\lambda}$
\p{2} returns $ \frac{\lambda^2e^{\minus{}\lambda}}{2}$
\p{n} returns $ \frac{\lambda^ne^{\minus{}\lambda}}{n!}$ for n > 2

Thanks

It depends how specific you want to be. For example
[code]\newcommand{\poisson}[1]{\ensuremath{\frac{\lambda^{#1} e^{-\lambda}}{#1 !}}}[/code]produces for \poisson{0},\poisson{1},\poisson{2},\poisson{n}
$ \frac {\lambda^{0} e^{ - \lambda}}{0 !},\frac {\lambda^{1} e^{ - \lambda}}{1 !},\frac {\lambda^{2} e^{ - \lambda}}{2 !},\frac {\lambda^{n} e^{ - \lambda}}{n !}$

But if you want the more specific
${{ e^{ - \lambda}},\lambda e^{ - \lambda}},\frac {\lambda^{2} e^{ - \lambda}}{2},\frac {\lambda^{n} e^{ - \lambda}}{n !}$
then you'll need to use the ifthen package with \usepackage{ifthen} and the following code:
[code]\newcommand{\pzero}{\ensuremath{e^{-\lambda}}}
\newcommand{\pone}{\ensuremath{\lambda e^{-\lambda}}}
\newcommand{\ptwo}{\ensuremath{\frac{\lambda^{2} e^{-\lambda}}{2}}}
\newcommand{\pn}[1]{\ensuremath{\frac{\lambda^{#1} e^{-\lambda}}{#1 !}}}
\newcommand{\poisson}[1]{\ifthenelse{\equal{#1}{0}}{\pzero}{\poissona{#1}}}
\newcommand{\poissona}[1]{\ifthenelse{\equal{#1}{1}}{\pone}{\poissonb{#1}}}
\newcommand{\poissonb}[1]{\ifthenelse{\equal{#1}{2}}{\ptwo}{\pn{#1}}}[/code]Note that you can't have numbers in command names hence the use of \pzero, \poissona etc.

ifthen documentation can be found in your system at yourmiktexpath\doc\latex\base\ifthen.dvi.

[size=75]Edited to update commands with roadnottaken's changes.[/size]

That code doesn't look quite right.  The line that defines \pn should read:
[code]\newcommand{\pn}[1]{\ensuremath{\frac{\lambda^{#1} e^{-\lambda}}{#1 !}}}[/code]
and the line defining \poissonb should read:
[code]\newcommand{\poissonb}[1]{\ifthenelse{\equal{#1}{2}}{\ptwo}{\pn{#1}}}[/code]
Otherwise, any argument to \poisson other than 0, 1, or 2 would yield $ \frac{\lambda^ne^{\minus{}\lambda}}{n!}$, whereas the above adjustment replaces all instances of n with the argument to the command.

Yes, I only checked 0,1,2,n so have updated the code.

Yeah, that's basically what I wanted. Thanks guys!


Well, you can re-write the problem as

$ \minus{}1(ab\minus{}2ab)(\minus{}2)$

if it makes life easier.

First, order of operations, which means you perform the subtraction operation, leaving

$ \minus{}1(\minus{}ab)(\minus{}2)$

or

$ \minus{}2ab$

The beautiful thing is that you can switch things around however you please; you'll always get the same answer if you follow the rules.

\begin{align*} - (ab - 2ab)( - 2) & = - ( - 2)(ab - 2ab) \\
& = 2(ab - 2ab) \\
& = 2( - ab) \\
& = \boxed{ - 2ab} \\
\end{align*}

\begin{align*} - (ab - 2ab)( - 2) & = ( - ab + 2ab)( - 2) \\
& = ( - 2)( - ab) + ( - 2)(2ab) \\
& = 2ab + ( - 4ab) \\
& = \boxed{ - 2ab} \\
\end{align*}

\begin{align*} - (ab - 2ab)( - 2) & = - ( - 2)(ab - 2ab) \\
& = - \left(( - 2)(ab) + ( - 2)( - 2ab)\right) \\
& = - ( - 2ab + 4ab) \\
& = - (2ab) \\
& = \boxed{ - 2ab} \\
\end{align*}

[quote="ReceiverOfCanineRevenue"]-(ab-2ab)(-2)

Do you:
(sum with first negative applied)(-2)
-(-2)(sum)... becoming 2(sum)

What to do. I wrote a maths exam today and I'm not sure what happens and if I got it right...[/quote]

commutative property of multiplication :D

so either way you do it, you get -2ab


When bagged in groups of 4, 3 left over. When bagged in groups of 5, 2 left over. When bagged in groups of 7, 4 left over. Fast way to do this? With no or minimal guessing and checking.

From the last two conditions, $ x\equiv \minus{} 3\pmod5$ and $ x\equiv \minus{} 3\pmod7$, so $ x\equiv \minus{} 3\pmod{35}\implies x \equal{} 35a \minus{} 3$, and $ x \equal{} 4b \plus{} 3$ as well, so $ 35a \minus{} 3\equiv \minus{} a \minus{} 3\equiv4b \plus{} 3\equiv3\pmod4$
$ \implies a\equiv \minus{} 6\equiv2\pmod4\implies a \equal{} 4c \plus{} 2\implies x \equal{} 140a \plus{} \boxed{67}$.


Find all postive integer $m, n$ such that:

$m^{2}+n^{2}=1989$

[hide="hint"]
Use mod 3 to show that it's really equivalent to $a^{2}+b^{2}=221$. From here it's probably brute-forceable, although using mods of 13 and/or 17 might reduce the casework.
[/hide]


Let $n$ be a positive integer, $n\geq 4$. $n$ cards are arranged on a circle and the numbers $1$ or $-1$ are written on each of the cards. in a $question$ we may find out the product of the numbers on any $3$ cards.  What is the minimum numbers if questions needed to find out the product of all $n$ numbers?

I think that these are the answers:
$ n=3k \longrightarrow k$
$ n=3k+1\longrightarrow k+3$
$ n=3k+2\longrightarrow k+2$

[quote="lorincz"]I think that these are the answers:
$ n \equal{} 3k \longrightarrow k$
$ n \equal{} 3k \plus{} 1\longrightarrow k \plus{} 3$
$ n \equal{} 3k \plus{} 2\longrightarrow k \plus{} 2$[/quote]

I'm afraid you guessed wrongly. Say when $ n\equal{}7$, I'll multiply in the following way: $ \{1,2,3\},\{1,4,5\},\{1,6,7\}$.
[hide]
When $ n\equal{}3k$, it is clear that the answer is $ k$.
When $ n\equal{}3k\plus{}1$, where $ k\ge 2$, We can use $ \{1,2,3\},\{1,4,5\},\{1,6,7\}$, followed by group each three of the other numbers. So the answer is $ k\plus{}1$.
When $ n\equal{}4$, it can be easily check that the answer is $ 4$.
When $ n\equal{}3k\plus{}2$, the answer must be an integer $ m$ such that $ 3m>n$. If $ m\equal{}k\plus{}1$, then exactly one of the integer is multiplied twice, and thus it is excluded from the product. Hence $ m\ge k\plus{}2$. Finally, one possible way is $ \{1,2,3\},\{1,2,4\},\{1,2,5\}$ and group each three of the other numbers. So the answer is $ k\plus{}2$.
[/hide]


factorial geometry trigonometry function SFFT probability symmetry

To me, a week before AMC is not about solving any new problem. It's more of reviewing what you already know and learn others' tips on problems to get everybody's favorite score ---- 75*2! (note that factorial doesn't change my point  :rotfl: ). Anyway, let's talk about strategy both working on problems AND thinking about problems (that is, physical work and mental work). Try to be specific on what category.

I'll go first.

Algebra: Definitely think about small cases like $ x \equal{} 1$ to at about $ x \equal{} 3$ or 4. But if you find pattern, don't simply go with it unless you are fairly sure of it (especially with later ones, it could be a trick!). Making conjectures help too.

Geometry: Not my favorite thing but I've improved a lot in past two weeks or so than before. Drawing a big diagram, although it may sound obvious, definitely helps. It makes me easier to think. Drawing additional lines help but unless you're an experienced person, I don't recommend it right away. It often confuses you more than help you.

Trigonometry: Definitely use in geometry and for strict algebra ones, change to functions that you can mess around easily. That is, reduce powers by using Pythagorean Identity, and so on.

Counting: Pray to God. I have nothing to tell you.  :rotfl:

NT: Look for pattern and use some algebra but don't overuse it. NT problems often involve algebra but not whole bunch of it.

Overall, get good sleep and don't do too much math this week. Don't stress out either.

I'm interested in more geometry and counting tips.   :)  :)  :)  :)  :)  :)

NT: never hesitate to difference of squares, take a discriminant, or sfft something. Know fermat's little theorem and other associated things.

For word problems, write out all variables that are integers and look at simple divisibilities. You will most likely see a problem that is like given N sandwiches and M sodas that cost x and y amounts, where Nx+My=something and solve for N, M...

Trig: These are always contrived. Try doing some tricky manipulation like squaring and using sin^2+cos^2=1...etc. Use sum to difference.

Now that there is no calculator they will probably have a few tricky telescoping products or whatever.

Counting...

Yea I suck that these...the probability ones are usually pretty straightforward. Look for symmetry and don't overcount. Perhaps make a recursion.

Geometry:

Know all your theorems...adding a line can help simplify things a lot. Come up with a plan before brute-forcing something.

Algebra: These are all tricks. Know your division algorithm and other such polynomial rules.

Random Sequence questions: find a pattern by looking at small values.

Since Counting advice is lacking...[list][*]Definitely do small cases and find recursions. This solves a lot of them.
[*]This is probably too personal a method to be good advice, but... Try to visualize how each case is related to others. This may allow you to come up with a geometrical way to arrange all the cases, and then it should be a lot easier to figure out how many there are. For instance, when picking two distinct members of a set (order does not matter), you can arrange the cases in a triangle.
[*]After you solve a counting problem (or part of a problem), find a second way to do it. This is the only real reliable checking method I have found for harder counting problems.
[*]Probability is only two counting problems. How many possibilities are there total? (this is not always trivial; be careful) How many ways are there to satisfy the problem's condition? Quotient them.[/list]

[quote="MellowMelon"]Since Counting advice is lacking...[list]
[*]This is probably too personal a method to be good advice, but... Try to visualize how each case is related to others. This may allow you to come up with a geometrical way to arrange all the cases, and then it should be a lot easier to figure out how many there are. For instance, when picking two distinct members of a set (order does not matter), you can arrange the cases in a triangle.[/list][/quote]

can you elaborate on that? Perhaps an example?

On further reflection, the geometrical arrangements I spoke of are mostly just a way of visualizing the algebra/recursions (said visualization helps me avoid stupid mistakes). For instance when doing [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=784569#784569]this AIME problem[/url] from last year I had some geometrical way visualize a recurrence relation (there were lots of $ 5$s and $ 4$s), but most people can probably work with the algebra just fine without my need to visualize.

In any case, one of the main shapes that comes up is having to find the number of lattice points on and in an $ m$-dimensional tetrahedron with vertices on $ (0,0,0,\ldots,0), (n,0,0,\ldots,0), (0,n,0,\ldots,0), \ldots, (0,0,0,\ldots,n)$, which happens to be the same as the number of ways that $ x_1, x_2, \ldots x_m$ nonnegative integers can be picked such that $ x_1 \plus{} x_2 \plus{} \ldots x_m \leq n$ for a given nonnegative integer $ n$.

MellowMelon, your strategy helps! I'm talking about solving the counting problem one way and counting with second way. Can you (or anyone) take a look at this?

Problem
(Hong Kong Heat Group #10) 3 shoes are chosen randomly from 6 pairs of shoes with different models, find the probability that exactly two out of the three shoes are of the same model.

[hide="My Solution"]
Even for me  :P , it was clear that the denominator (or the sample space) is $ \binom{12}{3}$. Now, I named the twelve shoes as $ A_1,A_2,B_1,B_2,C_1,C_2,D_1,D_2,E_1,E_2,F_1,F_2$. Then, I don't know if this is allowed but I decided to pick $ A_1,A_2$ as my two shoes. There are 6 ways to do this since there are 6 pairs. Now, for rest of the shoes, I realized that I have 10 options for each so I found that the answer is $ \frac{6 \cdot 10}{\binom{12}{3}} \equal{} \frac{3}{11}$.

Now, using MellowMelon's checking idea, I decided to check my answer. Keeping the denominator same, I know that I can choose 1 pair from 6 pairs by $ \binom{6}{1}$. Then, I, for some reason, though that I still needed two more shoes (which I don't since I just got a pair!) so I multiplied it by $ \binom{10}{2}$ to get the final answer of $ \frac{15}{22}$.

Realizing the difference, I went back and saw where I did wrong to multiply by $ \binom{10}{1}$ and got the same answer.

Did I do this right? I don't know... Checking definitely helps but I get this weird feeling with counting problems that I'm wrong...  :huh: [/hide]

Anyone want to verify if I'm right on that problem or not?

Here seems to be another interesting problem (I haven't solved it yet)

(Canada 96)

(a) Given positive numbers $ a_1,a_2,a_3,...,a_n$ and the quadratic function

\[ \displaystyle f(x) \equal{} \sum^{n}_{i\equal{}1} (x \minus{} a_i)^2\]

show that $ f(x)$ attains its minimum value at $ \frac{1}{n}\sum^{n}_{i\equal{}1} a_i$ and prove that

$ \sum^{n}_{i\equal{}1} a_i^2 \geq \frac{1}{n} \left(\sum^{n}_{i\equal{}1} a_i\right)^2$

(for reply two posts before): Your answer of $ 3/11$ is correct. The way that had me most confident of this was complimentary counting. What's the probability of not getting a pair? You can pick any first shoe ($ 12/12$), but only ten of the remaining eleven shoes won't make a pair ($ 10/11$), and then eight of the remaining ten won't make a pair ($ 8/10$). So we get $ 1 \minus{} 12/12 \cdot 10/11 \cdot 8/10 \equal{} 3/11$. When using this checking method and getting different answers from the two ways, note that either method you used can be wrong, so make sure your suspicions aren't too focused on one of them.

EDIT: A big "wtf" goes to the two posts below, especially the observation that we are replying to a post from yesterday (9 hours -> 21 hours). Must be Mathcamp craziness...

[hide="Solution to problem in previous reply"]Quadratics $ ax^2 \plus{} bx \plus{} c$ obtain their minimum value at $ \frac { \minus{} b}{2a}$ (easy to see by examining the quadratic formula). Expanding $ f(x)$, we get that $ a \equal{} n$ and $ b \equal{} \minus{} 2(a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)$, so the conclusion follows. For the inequality, take AM-QM, square both sides, and multiply both sides by $ n$. This is probably cheating since I didn't use $ f$. :P [/hide]
EDIT: For the real reason I'm editing this post
[hide="Solution to the inequality using f"]Write the inequality as $ \frac {1}{n} \left(\sum^{n}_{i \equal{} 1} a_i\right)^2 \minus{} \frac {2}{n} \left(\sum^{n}_{i \equal{} 1} a_i\right)^2 \plus{} \sum^{n}_{i \equal{} 1} a_i^2 \geq 0$, equivalent to $ f(\frac {1}{n} \sum^{n}_{i \equal{} 1} a_i) \geq 0$, which is true since $ f$ is the sum of squares. Ha, no cheating this time.[/hide]

[quote="Silverfalcon"]MellowMelon, your strategy helps! I'm talking about solving the counting problem one way and counting with second way. Can you (or anyone) take a look at this?

Problem
(Hong Kong Heat Group #10) 3 shoes are chosen randomly from 6 pairs of shoes with different models, find the probability that exactly two out of the three shoes are of the same model.

[hide="My Solution"]
Even for me  :P , it was clear that the denominator (or the sample space) is $ \binom{12}{3}$. Now, I named the twelve shoes as $ A_1,A_2,B_1,B_2,C_1,C_2,D_1,D_2,E_1,E_2,F_1,F_2$. Then, I don't know if this is allowed but I decided to pick $ A_1,A_2$ as my two shoes. There are 6 ways to do this since there are 6 pairs. Now, for rest of the shoes, I realized that I have 10 options for each so I found that the answer is $ \frac {6 \cdot 10}{\binom{12}{3}} \equal{} \frac {3}{11}$.

Now, using MellowMelon's checking idea, I decided to check my answer. Keeping the denominator same, I know that I can choose 1 pair from 6 pairs by $ \binom{6}{1}$. Then, I, for some reason, though that I still needed two more shoes (which I don't since I just got a pair!) so I multiplied it by $ \binom{10}{2}$ to get the final answer of $ \frac {15}{22}$.

Realizing the difference, I went back and saw where I did wrong to multiply by $ \binom{10}{1}$ and got the same answer.

Did I do this right? I don't know... Checking definitely helps but I get this weird feeling with counting problems that I'm wrong...  :huh: [/hide][/quote]

Looks like Palmer beat me  :roll:

Your answer is correct. The method of checking you used though is pretty much the same as the approach you began with, except replacing $ 6$ with $ {6\choose 1}$, $ 10$ with $ {10\choose 1}$, so it might not have caught an error of some other sort.
[list]
[*]A good way to check counting in a different method is to use the complement principle. Often times complement is the best way to tackle the problem anyway.

[hide="For the problem, by complement"]
If we want to pick three shoes such that none of them are from the same pair, there are $ {6\choose 3} \equal{} 20$ ways to pick the three different brands. From each brand, there are two possible shoes to pick, so this is $ 20 \cdot 2^3$. Thus the probability is
\[ 1 \minus{} \frac {2^3 \cdot 20}{{12 \choose 3}} \equal{} \frac {3}{11}
\]
(yay!)
[/hide]
[*]Use smaller numbers to visualize the problem, or wind the problem into smaller parts of the same nature. The latter is sort of like casework and recursion. For example, rather than listing $ A_1, \cdots F_2$, smaller numbers can be used to better visualize how to tackle the problem.  [/list]

As for the other question, the second part is just Cauchy-Schwarz (I cheated also)
\[ (1^2 \plus{} 1^2 \plus{} \cdots \plus{} 1^2)(a_1^2 \plus{} \cdots \plus{} a_n^2) \ge (a_1 \cdot 1 \plus{} \cdots \plus{} a_n \cdot 1)^2
\]

Edit: We're beating each other over a post 9 hours ago .. thats pretty weird   :lol:

For the Hong Kong problem, you are right.  Here is yet another method, using complementary counting:
[hide]
Use complementary counting.  The number of ways to choose them from $ 3$ different pairs is $ \binom{6}{3}*2^3\equal{}160$.  So the probability is $ 1\minus{}\frac{160}{\binom{12}{3}}\equal{}\frac{3}{11}$.
[/hide]

EDIT: MellowMelon beat me to the complementary counting . . . but mine is slightly different.

EDIT2: azjps did too . . .  :rotfl:

Algebra: OK I hope I'll get that
Geometry: I can probably do the 10's geometry ones
Trig: I don't know any trig tables :-(
Counting: I hope there are a lot of these :P
NT: My worst subject

and
random guessing: hmm I think ill get like 1/5 of these right.

Generally for questions that you think need a lot of work: Look at choices in multiple choice, the one closest to what you think the range is just might be the answer :)
Geo: Generally messy algebra isn't that bad...(I actually once drew a very good sketch, and saved a ton of messy algebra work by counting up segments but this is obviously not going to work a lot of the time)

Oh and for those of you that make dumb mistakes a lot, several practice AMC12s should fix the problem ;)


geometry
circumcircle
geometric transformation
rotation
trigonometry
homothety
ratio

Given $ \triangle ABC$. Let $ D,E$ be two moving points on $ AB,AC$ such that the perpendicular projection of $ DE$ on $ BC$ is equal to $ \frac{BC}{2}$. Prove that the circumcircle of $ \triangle ADE$ passes through a fixed point.

$ DE$ cuts $ BC$ at $ F.$ By Clifford circle theorem, circumcircles $ (O) \equiv \odot(ABC), (P) \equiv \odot(ADE), (R) \equiv \odot(DBF), (S) \equiv \odot(ECF)$ concur at $ Z.$  $ AO$ cuts $ (O), (P)$ again at $ X, Y.$ $ \angle DYE \equal{} 180^\circ \minus{} \angle A \equal{} \angle BXC ,$ $ \angle YED \equal{} \angle YAD \equal{} \angle XAB \equal{} \angle XCB$ $ \Longrightarrow$ $ \triangle DYE \sim \triangle BXC$ are spirally similar with center $ Z,$ rotation angle $ \phi \equal{} \angle DFB \equal{} \angle PYA \equal{} \minus{} \angle PAY \equal{} \minus{} \angle PAO,$ and similarity coefficient $ k \equal{} \frac {DE}{BC} \equal{} \frac {PY}{OX} \equal{} \frac {PA}{OA}.$
 $ DE \cos \phi \equal{} \frac {_1}{^2} BC$ $ \Longleftrightarrow$  $ PA \cos \phi \equal{} \frac {_1}{^2} OA$ $ \Longleftrightarrow$ $ O \equiv Y \in (P).$

Why could you conclude $ O \equiv Y$ since $ \frac {1}{2}OA \equal{} PA\cos\phi$???

I think this is obvious, since the homothety-rotation centre $ Z$, ratio $ \frac {r_{(P)}}{R_{(O)}}$ and angle $ \phi$ turn $ (P)\mapsto (O)$ :maybe:

Because $ PA \cos \phi \equal{} \frac{_1}{^2}YA$ always ?

It is a very nice Problem (I think :D ). I solve it by inversion.
Here is my solution:

Let I be the center of the circumcircle of triangle ADE, H and K prespectively be the projection of D and E onto BC. It is easy to find out that (I) goes through O- the center of circumcenter of triangle ABC. My "mission" right now is to prove that A, D, E, O concyclic.
Invert around A with radius AO that maps the point O to itself, maps D to D*, E to E*.
To prove A, D, E, O concyclic, we will try to prove O, D*, E* collinear.
Now through the inversion of pole A, radius AO, it maps I To I*, maps D*E* to the (ADE), therefore, I* is the reflection of A through D*E*. Hence, to prove O belongs to D*E*, we need to prove that OA = OI* which means I* lies on (O). In the other words, let I* (different from A) be the intersection of AI with (O), we will prove that AI.AI*=[tex] AO^2\ [tex].
It is easier now  :lol: . Consider it as the lemma, or an excercise.  :D :lol:  :lol:

Happy Lunar New Year :P  :P

Let $ \{S\} \equal{} (ABC) \cap (ADE)$ and $ I$ is the reflection of $ B$ through the symmetry through line $ DH$, since $ HK \equal{} \frac {BC}{2}$, $ K$ is also the midpoint of $ IC$.

We have $ \triangle SDB \sim \triangle SEC$, i.e. $ \frac {SD}{DB} \equal{} \frac {SE}{EC}$, but $ DB \equal{} DI$ and $ EC \equal{} EI$ so $ \frac {SD}{ID} \equal{} \frac {SE}{IE}$. On the other hand, $ \widehat{DIE} \equal{} 180^o \minus{} \widehat{DIB} \minus{} \widehat{EIC} \equal{} 180^o \minus{} \widehat{ABC} \minus{} \widehat{ACB} \equal{} \widehat{BAC} \equal{} \widehat {DSE}$, so $ \triangle DIE \sim \triangle DSE$, but $ DE$ is their common side, i.e. $ \triangle DIE$ and $ \triangle DSE$ are congruent.

Therefore, $ ES \equal{} EC$, but $ OS \equal{} OC$ so $ \triangle SEC$ is isosceles at $ E$, so $ \widehat{SEA} \equal{} 2\widehat{SCE} \equal{} \widehat{SOA}$, i.e. $ S,A,O,E$ are concylic, or $ (ADE)$ passes through fixed point $ O$, which ends the proof.

Oh very nice solution, thank for your solution, Mashimaru :10:


I am... I'm so out of practice, though. We'll see what I can pull off.


[quote=Fermat -Euler]$p(k)$ is a polynomial of degree $n$ such that $p(k) = \frac{(n+1-k)!k!}{(n+1)!}$ for $k = 0, 1, ... , n$.
Find $p(n+1)$.[/quote]
It doesn't really seem to be a shortlist problem (at least not from [url=http://artofproblemsolving.com/community/c3935_1984_imo_shortlist]1984[/url]). But anyway, see [url=http://artofproblemsolving.com/community/c6h366958]here [/url]for a solution.


If there are 5 books on a bookshelf, how many ways can you arrange them?

[quote="anirudh"]Reread the problem  :wink:[/quote]
I don't get it.  :huh:  :ninja:

[quote="anirudh in the problem"]If one of the books can only be in the [size=200][b]beginning or the end[/b][/size][/quote]
 :wink: _

[hide]for #1, I did 5 X 4 X 3 X 2 X 1=5!=120 ways[/hide]

[hide]for extension, I did 4! X 2=24 X 2=48 ways[/hide]

[quote="i_like_pie"][quote="anirudh"]Both are correct :D

Extension:
If one of the books can only be in the beginning or the end, what are the remaining ways to arrange the books?[/quote]
[hide]$4!=24$[/hide][/quote]

[quote="anirudh"]Reread the problem  :wink: [/quote]

The trick in the problem was that the book could be placed either in the beginning or in the end, so the answer would be, as mickeythedunce and kheg_k said, $4!\cdot2=48\text{ ways.}$

[hide]5! times, or 120 different times[/hide]


geometric sequence
Pascal\u0027s Triangle

this result can be generalised:

$\sum_{n=m}^z \binom{n}{k}$ for some constant positive integer $k<n$

$\implies \sum_{n=m}^z (\binom{n+1}{k+1} - \binom{n}{k+1})$

This is written as the difference of $f(n) - f(n-1)$

$\sum_{n=m}^z \binom{n}{k} = \binom{z+1}{k+1} - \binom{m}{k+1}$

Oo! That's even nicer than I thought :)! In fact $\binom{10}{6}$ is the coefficient of $x^6$ in $(1+x)^{10}$ and so on. Notice that $S=(1+x)^{10}+...+(1+x)^{19}$ is a geometric sequence we find $S=\frac{(1+x)^{20}-(1+x)^{10}}{x}$ so $A$ must be the coefficent of $x^7$ from $S$. Hence $A=\binom{20}{7}-\binom{10}{7}$ and now using your first relation we get indeed $B$.

What sen states is also known as the Hockey Stick Identity. It states that

$\sum_{i=0}^k \binom{n+i}{n}=\binom{n+k+1}{n+1}$

Although this can be proved using the telescoping series from above, try looking at it as terms on Pascal's Triangle. This representation on Pascal's Triangle results in the name of the identity.

It should be noted that the combinatorial proof of the hockey stick identity is quite intriguing.  See if you can find it.


The sequence is a list of the ordered pairs $ (p, q) \in \mathbb{N}^2, q > 1, p < q$ by ordering on $ q$ and then on $ p$, which contains all such ordered pairs nearly by definition.  In fact, every rational number appears infinitely often (as written).


This propably has something to do with method of finding coefficient.Seen this here before

Multiply the first equation by $1$, the second by $-3$ and the third by $3$ and then add them all up, to obtain $16a+25b+36c+49d=334$

for the coefficients of $a$, we have $(n-1)^2$, $(n)^2$, $(n+1)^2$, and we want $(n+2)^2$, with $n=2$

it would be nice if we had a way of combining the first 3 by adding multiples that way we could use the operation for the coefficients for b,c,d also as they are those squares with $n=3,4,5$

say we have x multiples of the first, y of the second, z of the thrid...

$x(n-1)^2+y(n)^2+z(n+1)^2=(n+2)^2$

comparing coefficients, we have

$x+y+z=1$
$-2x+2z=4$
$x+z=4$

solving easily, we get $(x,y,z)=(1,-3,3)$

then the answer is $(1)(1)+(-3)(12)+(3)(123)=334$

That's what i'm talking about,nicely done altheman

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=method++undetermined+coefficients&t=75882]here is something exactly like it[/url]

[quote="José"][quote="Farenhajt"]Multiply the first equation by $1$, the second by $-3$ and the third by $3$ and then add them all up, to obtain $16a+25b+36c+49d=333$[/quote]

Nice! But I got $334$ ........[/quote]

Yup, a typo. Corrected. Sorry

[quote="shadysaysurspammed"]Farenhajit,remember you had solved one question like this using method of undetermined coefficients?[/quote]

That's right, Shady, but I couldn't be bothered with finding the topic :) Thanks for doing that though :)

[quote="Farenhajt"]Yup, a typo. Corrected. Sorry[/quote]

OK


1)Right Triangle ABC is inscribed in a circle, the radius of the circle is 5 centimeters and BC = 5 cm.  What is the number of cm in the length of arc BC?  Express your answer as a common fraction in terms of $ \pi$.

2)The floor of a square room is covered with congruent square tiles.  the diagonals of the room are drawn across the floor, and the two diagonals intersect a total of 9 tiles.  How many tiles are on the floor?

[hide="1"]
If the radius of the circle is 5, then the hypotenuse of the right triangle is 10.  Therefore, we have a 30-60-90 right triangle. Angle A must be 30 degrees, as sin(30)=.5. The arc must be 2*30=60 degrees, which is 1/6 of the circumference, so arc BC has length 10pi/6, which reduces to 5pi/3.
[/hide]

[hide="2"]
If the diagonals intersect 9 tiles, then each one intersects 5, and there is one overlap, for 5*2-1=9 tiels iintersected. Thus, the room is 5 by 5, and there are 25 tiles.
[/hide]

Edit: Thanks, 007math. I made a tyop.

[quote="Discrete_Math"] Thus, the room is 5 by 5, and there are [b]24[/b] tiles./quote]
 :wink:


aghayoon va khanoomhaye olampiad ba arze salam va khaste nabashid..
bebinid olampiad tashkil shode az $tarkybiat$,hendese,nazarye adad,jabr.
midoonam hame ino midoonid vali shomaham vaghan tarkybiato vel kardid dar sooraT ke bishtare bache ha too tarkybiat moshkel daran..ye nim negahi be marathone tarkybiat bendazid man ye mahe soal gozashtam vali yeki nazashet aghaln in sakhte asoone ya keC hal nakarde.jooone har ki doost darid ye zarre be tarkybiat ahammiat bedid .kheili vaze tarkybiate inja kharabe :!:  :!:

baraye Post kardane masale, Bakhshe Olympiad hastesh:

http://www.mathlinks.ro/Forum/index.php?f=217


Solve for x:

$ |3\minus{}\frac{x}{2}| \geq 6$

Oh wow, stupid moment of the day.

I was waaaaaaaaaaaaaay off. xD

hello, write your equation in the form
$ |6\minus{}x|\geq12 \Leftrightarrow |x\minus{}6|\geq12$ and consider the cases $ x\geq 6$ and $ x<6$.
Sonnhard.

hello, your solution is correct.
Sonnhard.

Hey!  That was my solution!  :lol:

That was not my stupid moment of the day, editing it out was.  :rotfl:


Hi Everyone,

Congrats to all who beat me on the COMC, and anybody has any idea when the olympiad invites would be sent out?

My math teacher doesn't particularly care, and I wouldn't want to miss anything because of him..
It'd be nice if someone would tell me when they get the invites..

Good luck, and hope to see you all at the Seminar!

Jenn

Hey Jenn! It's so nice to see you on the forum.  :D

Apparantly the certificates and medals are already sent out (though my teacher wants to hold on to the medals until the school awards night). However, I don't think that the CMO invitations are out yet. I'll keep you updated if I hear anything.

oh, okay..

I remember my teacher forgetting to give me CMO invitation last year..
I think I ended up sending it in way after the deadline.. So I'm becoming a little paranoid about it...

Anyways, I'll try to stay in touch with all of you...

Jenn

Are there any ways for me to check contest results online? I can't seem to find COMC results online x3

Jenny

You may check the general stats, winners, and top scores [url=http://cemc.uwaterloo.ca/english/contests/open.shtml]here[/url].

Though to view your exact score online, you will need the school's password  ;)

Hi Jenny, haven't talked to you for a while!

And is there any way to obtain the school ID/password when the math teacher refuses to tell it to you??  :D

hmm.. that question sounds.. wrong!!
But i'm really curious!!!

jenn


calculus
integration
trigonometry
logarithms
function
calculus computations

$\int{\sin{\ln{(x)}}}dx$

Probably a trig substitution, I just don't see what it is (I need to look over my identities/formulas).

Elaborating on what kunny said ...

The "natural" substitution is for the "inside" function in a composite function. Let's try that: $t=\ln x,\ dt=\frac{dx}{x}.$ No $x$ in the denominator, so the substitution doesn't go through simply - but we can always "force" any substitution if we really want to. Let's try that here - write it backwards:

$x=e^{t},$ so $dx=e^{t}\,dt.$ The integral becomes:

$\int e^{t}\sin t\,dt.$

This substitution didn't solve the problem by itself - it wasn't the last step - but it gave our problem a different appearance. Looking at it now, do you know the problem? Do you have an idea about how to approach it?

So now you integrate by parts and get (I think)
$\frac{1}{2}x(\sin{\ln{x}}-\cos{\ln{x}})+c$


Prove that $ (1\minus{}\cos^2 x)(1\plus{}\cos^2 x) \equal{} 2\sin^2 x \minus{} \sin^4 x$ using only fundamental identities.

[hide="Pretty Easy..."]
Using the identity $ \sin^2 x \plus{} \cos^2 x \equal{} 1$, we find that $ (1 \minus{} \cos^2 x)(1 \plus{} \cos^2 x) \equal{} (1 \minus{} 1 \plus{} \sin^2 x)(1 \plus{} 1 \minus{} \sin^2 x)$
$ \equal{} \sin^2 x(2 \minus{} \sin^2 x) \equal{} 2\sin^2 x \minus{} \sin^4 x$, as we desired.
[/hide]

You shouldn't really have to expand anything.
Just use the simplest identity as shown above...


[hide="Solution"]We have to use case work in this situation:
[hide="CASE 1"]
[b]Only 1 drawer is open.[/b]
We can open $ 1$ drawer in $ \binom{5}{1} \equal{} 5$ ways, with $ 0$ of them where a drawers contents cannot be reached.  So we have a total of $ \boxed{5}$ ways.[/hide]
[hide="CASE 2"]
[b]Only 2 drawers are open[/b]
Here, we take a more systematic approach.  Name the drawers, from top to bottom, $ 1, 2, 3, 4, 5$.  We must choose a pair $ (a, b)$ to open (where $ a$ and $ b$ are not consecutive numbers).  Since the order will not matter, simply assume $ a < b$.  If we have $ a \equal{} 1$,  then we have $ b \equal{} 3, 4, \text{ or } 5$.  If $ a \equal{} 2$, we have $ b \equal{} 4 \text{ or } 5$.  If we have $ a \equal{} 3$, then $ b \equal{} 5$.  So we have a total of $ 3 \plus{} 2 \plus{} 1 \equal{} \boxed{6}$ ways.[/hide]
[hide="CASE 3"]
[b]3 drawers are open[/b]
We can only do this when drawers $ 1, 3, \text{ and } 5$ are open, so only $ \boxed{1}$ way.[/hide]
[hide="Total"]Notice that we cannot reach all of the drawers' contents if 4 or 5 are open at a time.  So now we simply add up the number of ways in each case.  We have $ 5 \plus{} 6 \plus{} 1 \equal{} \boxed{\boxed{12}}$[/hide]
[/hide]

What exactly do you mean by order? If you mean the order they are opened in, that does not matter because we are only looking at the end result.

[hide=eee]so i created a table (left side bold numbers indicate which drawer is kept open for the case) and the combos that work (also avoiding repeats)\begin{tabular}{l|l|l|l|l|l}
\textbf{1} & 1 & 1 3 & 1 4 & 1 5 & 1 3 5 \\ \hline
\textbf{2} & 2 & 2 4 & 2 5 &     &       \\ \hline
\textbf{3} & 3 & 3 5 &     &     &       \\ \hline
\textbf{4} & 4 &     &     &     &       \\ \hline
\textbf{5} & 5 &     &     &     &
\end{tabular}

there's 12 combos in total[/hide]

[quote=GameBot]A dresser has five drawers stacked vertically. To be able to reach the contents in an open drawer, no drawer that is adjacent to the open drawer may be open at the same time. In how many ways can one or more drawers be open so that the contents in each of the open drawers can be reached?

[asy]draw((0,0)--(10,0)--(10,26)--(0,26)--(0,0)--cycle,linewidth(1));

draw((1,1)--(9,1)--(9,5)--(1,5)--(1,1)--cycle,linewidth(1));
draw((1,6)--(9,6)--(9,10)--(1,10)--(1,6)--cycle,linewidth(1));
draw((1,11)--(9,11)--(9,15)--(1,15)--(1,11)--cycle,linewidth(1));
draw((1,16)--(9,16)--(9,20)--(1,20)--(1,16)--cycle,linewidth(1));
draw((1,21)--(9,21)--(9,25)--(1,25)--(1,21)--cycle,linewidth(1));

filldraw(circle((5,3),.25));
filldraw(circle((5,8),.25));
filldraw(circle((5,13),.25));
filldraw(circle((5,18),.25));
filldraw(circle((5,23),.25));

filldraw(circle((2,-.5),.5));
filldraw(circle((8,-.5),.5));[/asy][/quote]

[hide=Casework Sol]We can use casework.

There are $5$ ways to just open one drawer.

There are $6$ ways to open two drawers.

Finally, there is only one way to open $3$ drawers.

Unless I messed up somewhere, the answer should be $12$.[/hide]

For Case I, there are $\binom{5}{1}=5$ ways to choose one drawer to open out of five.

For Case II, we can calculate $\binom{5}{2}=10$. However, this counts the situations in which two adjacent drawers are open. There are four sub-cases (CCOOO, OCCOO, OOCCO, and OOOCC) in which two adjacent drawers are open. Therefore, there are a total of $10-4=6$ ways to open two drawers.

For Case III, there is one sub-case (OCOCO) by inspection.

The answer is $5+6+1=\boxed{12}$.


function
geometric series
complex analysis

i need help finding all the laurent expansions of z/(z^2 - 3z + 2) about z_0 and give the region of convergence of each.

i do not understand any of this and need major help. if anyone can, can someone explain to me how to do this step by step please. i've been looking everywhere online and i'm still confused on this. i dont even really understand how to get the geometric series or taylor series of functions. i know the formulas for them: the sum from n=0 to infinity of z^n = 1/1-z and the sum of f^(k) (a)/k! (z-a)^k. i know the formulas but i dont know how to apply them to the functions. is there anyone out there that can help me please, i need major help.

Let's look at a simpler example: the expansions of $ \frac1{z\minus{}c}$ around zero, where $ c$ is not zero. For $ |z|<c$, we have the geometric series $ \minus{}\frac1c\minus{}\frac{z}{c^2}\minus{}\cdots\minus{}\frac{z^n}{c^{n\plus{}1}}\plus{}\cdots$. For $ |z|>c$, we have the geometric series $ \frac1z\plus{}\frac{c}{z^2}\plus{}\cdots\plus{}\frac{c^{n\minus{}1}}{z^n}\plus{}\cdots$

How does this relate to the function we care about? We can split into two copies of the above by partial fractions. Each part has the inside/outside choice made separately, so there's an inner region in which all exponents of $ z$ are nonnegative, an outer region in which all are nonpositive, and an annulus between the two roots in which the series is infinite in both directions (since we made one of each choice). It matters which root is closer, so writing down the general answer is an annoying exercise in cases.


One thing that has been bothering me lately is that if I send a PM to somebody...[with say a solution partly in Latex], I am unable to recover the code for the message. [In particular, I am unable to recover the messages that I have sent when they are in either the sentbox or savebox, they can be recovered in the outbox, but that is useless once the person has gotten their message ]. The quote button does not seem to be there. Am I missing something? Is there some reason for this? Or could this be fixed?

Thanks,
--Alex

When I select and copy text (including LaTeXed images), when I paste, the LaTeX code is still there (although I have to put in new dollar signs).  This is using FireFox.

And, I don't think there's a quote button anywhere there isn't a "reply" or "new post" button, since quote executes that command, as well.


What is the sum of all integers $x$ that satisfy $-5 \le \frac{x}{\pi} \le 10$?

[quote="jimli"]10 pi is 31.415...
31/pi is less than 10
-5 pi is 15.7
15/pi is less
-15+-14+-13...29+30+31
496-120=376[/quote]

  you should really take the time to lean $%Error. "Latex" is a bad command.
$

[quote="jimli"]10 pi is 31.415...
31/pi is less than 10
-5 pi is 15.7
15/pi is less
-15+-14+-13...29+30+31
496-120=376[/quote]

  you should really take the time to lean $\LaTeX$

[quote] you should really take the time to lean [Unparseable or [u]potentially dangerous latex formula[/u]].
Error 5 : 1020x1320][/quote]

What about you ?   :rotfl: lol

[quote="gr8sk8r"]yea, that's what i got with math92's help.[/quote]
yea, gr8sk8r and i got that answer

Math92, please try to avoid double posting. Thanks.

I agree with G-Unit. $%Error. "LATEX" is a bad command.
$


arithmetic sequence
number theory
relatively prime
number theory unsolved

Do there exist positive integers $ a,b,c$ such that:

\[ a(a\plus{}b)(a\plus{}2b)(a\plus{}3b)(a\plus{}4b)\equal{}c^{2008}\]

(Do there exist 5 distinct consecutive members of an arithmetic progression wihch product is a power of 2008?)

Let $ a \equal{} 2^{401} 3^{803} 5^{803}$ and $ b \equal{} a$. Then the left side is $ 2^3 3^1 5^1 a^5 \equal{} 2^{2008} 3^{4016} 5^{4016} \equal{} (2^1 3^2 5^2)^{2008}$.

[hide="Funny story"]A problem very similar to this took this year's US IMO team and the rest of the candidates a [b]very[/b] long time to solve, because we all thought the answer was no. Why else would they have given us the problem?[/hide]

Thx for the answer :) I also thought the answer was no, thats why i spent more than 3 hours on the competition itself to solve it :P I have to say that im kinda disappointed with the answer ;) meh

Here is the generalization of the problem and its proof:

Prove or disprove that there exists an arithmetic sequence $ a_1, a_2, \cdots, a_n$ so that $ a_1a_2\cdots a_n$ is a perfect $ k$th power given that $ k$ and $ n$ are relatively prime.
[hide="Solution"]
The answer is that there is such a sequence.  We let $ a_1$ be the common difference as well as the first term.  It then transforms into proving that $ n!(a_1^n)$ may be an $ n$th power for some $ a_1$.  We let $ n! \equal{} \prod_{i \equal{} 1}^x {p_i^{e_i}}$.  Then, we let $ p_i$ be the prime factors of $ a_1$ when $ 1\le i\le x$.  Let $ p_i^{j_i}\parallel{}a_1$, so $ p_i^{nj_i}\parallel{}a_1^n$.  Now, we want for $ nj_i \plus{} e_i\equiv 0\bmod k$ so that $ n!(a_1^n)$ is a perfect $ k$th power, which means that $ j_i\equiv \frac {\minus{}e_i}{n}\bmod k$, which is possible since $ k$ and $ n$ are relatively prime, meaning that $ n$ is invertible $ \bmod k$.  We have found a suitable $ a_1$, so the sequence exists. [/hide]

How about we make things a little trickier?

When does there exist an arithmetic sequence $ a_1, a_2, \ldots a_n$ so that the product of the terms is an $ n$th power?

Hmmm, the method The QuattoMaster 6000 used doesnt work here. Also, I know for $ n\equal{}4$ there doesnt exist(it was on ML contest) such a sequence. Im checking it for $ n\equal{}3$ and I guess there doesnt exist such $ n$...

[hide="A simpler solution"]Take $ a \equal{} b \equal{} 120^{803}$, then $ c \equal{} 14400$. So the answer is yes[/hide]

i dont think these problems should be discussed
some cities in canada havent written the contest yet


As some of you may know (especially those doing Higher Level Maths in the IB curriculum, where this is the topic of an investigation), there is an interesting property about the roots of a cubic function. If the three roots of a cubic function are $ r_{1}$, $ r_{2}$, and $ r_{3}$, the line tangent to the cubic function at the average of the two roots, $ (\frac{{r_{1}}+{r_{2}}}{2})$ intersects with the line again at the third root $ r_{3}$.

Shown below is an example (the function is $ {2x^{3}}+{6x^{2}}-4.5x-13.5$ and its roots are -3, -1.5 and 1.5

[img]http://mcraefamily.com/MathHelp/FactoringCubic3RealRoots.gif[/img]

This is not difficult to prove and is therefore not my question :P

The question is, what are the x-values (in terms of its three other roots) at which a line tangent to a [u]quartic[/u] intersects with the quartic's fourth root.

This is clearly a question of calculus and as such belongs in the calculus forum.


Adam and kris play a little banana-based game. Initially, Adam has 201 banans in his pile, while Kris has 156 bananasin his pile. Everyday, each of them eats one banna from his pile if possible, and then simultaneously gives the other half of what they have remaining in their own pile, rounded down. After all bananas are eaten, how many did Kris eat?

if i understand your question correctly, i am thinking it's (201+156) / 2 = 178.5 and rounding down would be 178.

It's actually $ 179$. I don't have a very good proof; because my proof is induction-based, and it is based on the fact that the number of bananas is cyclic and, modulo $ 4$, has a period of $ 4$.

One of them must eat one less than the other, and that person is the one with the greatest amount, since they give the most first.

So, we have $ \frac {201 + 156}{2} = 178.5 \text$ which is approximately $ 179$.

Nope, 178

[quote="orangefronted"]Adam and kris play a little banana-based game. Initially, Adam has 201 banans in his pile, while Kris has 156 bananasin his pile. Everyday, each of them eats one banna from his pile if possible, and then simultaneously gives the other half of what they have remaining in their own pile, [b]rounded down.[/b] After all bananas are eaten, how many did Kris eat?[/quote]

Regardless, I wrote a Java program to confirm my result; the answer is $ 179$. One person eats $ 178$, the other $ 179$.


The Taylor series of a function $ f(x)$ is $ \sum_{n\equal{}2}^{\infty}\dfrac{(\minus{}1)^nx^{2n}}{(n\minus{}1)!}$. Find an explicit formula for $ f(x)$.

I think it is $ x^2(1\minus{}e^{\minus{}x^2})$.

Yep, that's the right answer.
$ f(x)\equal{}\sum_{n\equal{}2}^{\infty}\dfrac{(\minus{}1)^nx^{2n}}{(n\minus{}1)!}$.

Now, $ \exp(\minus{}x^2)\equal{}\sum_{n\equal{}0}^\infty \frac{(\minus{}x^2)^n}{n!}\equal{}\sum_{n\equal{}0}^\infty \frac{(\minus{}1)^n x^{2n}}{n!}\equal{}\sum_{n\equal{}1}^\infty \frac{(\minus{}1)^{n\minus{}1} x^{2n\minus{}2}}{(n\minus{}1)!}$
so we have to multiply by $ \minus{}x^2$ to get the same form as $ f(x)$:

$ \minus{}x^2\exp(\minus{}x^2)\equal{}\sum_{n\equal{}1}^\infty \frac{(\minus{}1)^{n} x^{2n}}{(n\minus{}1)!}$. Now substract the n=1 term (which is -x^2), to get

$ f(x)\equal{}x^2\minus{}x^2\exp(\minus{}x^2)\equal{}x^2(1\minus{}\exp(\minus{}x^2))$

thanks  :D  i thought it was something like $ e^x$ but didn't manage to think it was $ e^{\minus{}x^2}$  :P


There is a (2n+1)*(2N+1) square.we cut a 1*1 square from the corner of (2n+1)*(2N+1) square.
we cover the square with dominos(1*2 rectangular).for which n the number of vertical dominos is equal to horizental dominos.

n - even number, but how to prove it???  :maybe:


abstract algebra
superior algebra
superior algebra unsolved

[quote="1234567a"]let $ R$ be a ring with unit and $ I$ be the unique [b]right ideal[/b] of $ R$
prove that $ I$ is two-sided ideal of $ R$?

Any idea? thanks[/quote]

Did you leave out some words?  Both 0 and R are right ideals of R, so if I is the only one, then I=0=R is clearly a two-sided ideal of a very silly ring.

Depending on what words you left out, you might try considering the right ideal rI for arbitrary r in R.  If it is other words you left out, you might consider the annihilator of the right R-module R/I as the intersection of the annihilators of the elements of R/I, all of which are R or maximal right ideals.


probability
calculus
integration
number theory open
number theory

A number is called super-even if it is a power of 2 and all of its digits are even.  Find all super-even numbers.

For example, $2^1=2, 2^2=4, 2^3=8, 2^6=64, 2^{11}=2048$ are all super-even.  Note that it may seem easier to prove that there are no more super-even numbers after 2048...but I am looking for a counter-example, which is hard, because you would need a super computer.

But $1024$ is not super-even, since it has the digit $1$, and $2048>1024$ is, so the assertion that there are no more after $1024$ is wrong :? Am I missing something here?

Well...
There aren't any that have fewer than 1000 digits - I used a calc program to check all the way to 2^3322, when it ran out of memory.
A probability integral tells us there should be 6 1/7, and we've only found five. Interesting.

Definitely this is still an open question!
Someone named Bertram Felgenhauer (IMO participant years ago) check it for $2^n$, $n \leq 2^{69}$ or more. He is still searching. No other values were found and it is supposed there is no other number.
On the other hand there is a claim of proof using some probabilistic reasons. You may see more other facts about at mathforum.org.


$ABCD$ is a square side $1$. Points $A', B', C', D'$ are taken on the sides $AB, BC, CD, DA$ respectively so that $\frac{AA'}{AB}=\frac{BB'}{BC}=\frac{CC'}{CD}=\frac{DD'}{DA}=\frac{1}{n}$. The strip bounded by the lines $AC'$ and $A'C$ meets the strip bounded by the lines $BD'$ and $B'D$ in a square area $1/1985$. Find $n$

[hide]
$n=\sqrt{\frac{1985}{2}}$
Use the triangle sides relations: $1:1:\sqrt{2}$.[/hide]

[quote="Arne"]Shobber, do you see any angles of 45 degrees?[/quote]
 :wallbash_red:  :wallbash:
Let's say the side of the little square is $d$. Hence:
\[\frac{d}{BB'}=\frac{AB}{BD'}\]
\[\frac{d}{\frac{1}{n}AB}=\frac{AB}{\sqrt{1+(1-\frac{1}{n})^2\cdot AB}}\]
\[\frac{d^2}{AB^2}=\frac{1}{n^2+(n-1)^2}=\frac{1}{1985}\]

Hence $n=32$.

[quote="gotztahbeazn"]oo i remember seeing this problem on the chicago ARML hw  :lol:[/quote]
Oh really? I was on chicago ARML team, too :) But I don't think I saw this problem on hw...

By the way, just so you know, this is originally from 1985 AIME.

frt, it seems like you know all AIME paper problems,how come?
Shobber, you are right :first:

[quote="riddler"]frt, it seems like you know all AIME paper problems,how come?[/quote]
Because I've been practicing for AIME with old AIME problems from kalva :)

I know Kalva too :D  did you try the Dudeney problems? they are fun!

I solved two geometry problems here last night, one is this thread and the other one is the area ratio thing. Both are from AIME 1985.
They appeared in a Chinese book about solving geometry problems using area methods. Both of them are very easy problems in the book.


The sum of three diffe4rent positive unit fractions is 6/7. What is the least number that could be the sum of the denominators of these fractions?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=111745558&t=242776]here[/url]. Many problems' solutions can be found by using the search function.  :wink:


algebra
polynomial
number theory
greatest common divisor
induction
algorithm
Euclidean algorithm

Let $ P(x)\equal{}b_px^p\plus{}b_{p\minus{}1}x^{p\minus{}1}...\plus{}b_1x\plus{}b_0$  be a Polynomial such that  $ b_0,b_1...b_p$ are a positive integers.We definite the sequence $ a_n$ :

$ a_0\equal{}0$  and  $ a_{i\plus{}1}\equal{}P(a_i)$.

Prove that: $ \text{gcd}(a_m,a_n)\equal{}a_{\text{gcd(m,n)}}$

[hide="Solution"]
Lemma: if $ n \equal{} mq \plus{} r$, where $ q,r$ are nonnegative integers and $ 0\leq r < m\rightarrow a_n\equiv a_r (mod a_m)$.
Proof: induction on $ q$.
So, notice that obviously $ a_i \plus{} 1 > a_i$, hence $ a_r$ is the remainder obtained when $ a_n$ is divided by $ a_m$. Reasoning in this way, if we make the Euclidean Algorithm with $ a_n$ and $ a_n$, it'll stop when the Euclidean Algorithm with $ m$ and $ n$ stop, and it will happen in $ gcd(m,n)$. So, it's proved.
[/hide]


If the polynomial $ P\left(x\right)$ satisfies $ 2P\left(x\right) \equal{} P\left(x \plus{} 3\right) \plus{} P\left(x \minus{} 3\right)$ for every real number $ x$, degree of $ P\left(x\right)$ will be at most

$\textbf{(A)}\ 0   \qquad\textbf{(B)}\ 1   \qquad\textbf{(C)}\ 2   \qquad\textbf{(D)}\ 3   \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]Construct $ \Lambda (x) \equal{} P(0) \plus{} \left( \frac {P(3) \minus{} P(0)}{3} \right) x$.

It is easy to show inductively that $ \Lambda (x) \equal{} P(x)$ for all $ x \equal{} 3n$

Thus, the polynomial $ (P \minus{} \Lambda)(x)$ has infinitely many roots, and must be identically 0.

$ \therefore \; P(x) \equal{} \Lambda (x) \implies \boxed{\text{deg} \; P \leq 1}$.[/hide]

[hide="Another approach"]$ P(x\plus{}3)\minus{}P(x)\equal{}P(x)\minus{}P(x\minus{}3)$, hence $ P$ maps an arithmetic progression into an arithmetic progression, so it must be either constant or linear.[/hide]

$$\begin{array}{rcl}
P(x) &=& a_nx^n + a_{n-1}x^{n-1}+\dots + a_1x + a_0 \\
P(x+3) &=& a_n(x+3)^n + a_{n-1}(x+3)^{n-1}+\dots + a_1(x+3) + a_0 \\
P(x-3) &=& a_n(x-3)^n + a_{n-1}(x-3)^{n-1}+\dots + a_1(x-3) + a_0
\end{array}$$
Let's calculate the coefficient of the term $x^{n-2}$.
$$\begin{array}{rcl}
P(x+3) &=& \dots + (a_n \cdot {n \choose 2} \cdot 3^2 +a_{n-1} \cdot {{n-1} \choose 1}\cdot 3 + a_{n-2})x^{n-2} + \dots \\
P(x-3) &=& \dots + (a_n \cdot {n \choose 2} \cdot 3^2 -a_{n-1} \cdot {{n-1} \choose 1}\cdot 3 + a_{n-2})x^{n-2} + \dots \\
P(x+3)+P(x-3) &=& \dots + (2a_n \cdot {n\choose 2} \cdot 3^2\cdot x^{n-2} + 2a_{n-2}\cdot x^{n-2}) + \dots \\
2P(x) &=& \dots + 2a_{n-2}\cdot x^{n-2}+ \dots
\end{array}$$
So we get $$2a_n \cdot {n\choose 2} \cdot 3^2\cdot x^{n-2} = 0$$ This is possible if the $(n-2)$th term is undefined.
So $n-2<0 \Rightarrow n \leq 1$.
For $n=1$, it is clear that $P(x) = x$ satisfies the equation.