function
parameterization
geometry
logarithms
3D geometry
prism
analytic geometry

Hi EveryOne,
Thank you very much for your watching and solving the VIETNAMESE ENTRANCE COLLEGE EXAM, School year 2005 – 2006, Subject Mathematics - Block A. Here are one another. I hope you will enjoy it.

[b]VIETNAMESE ENTRANCE COLLEGE EXAM
School year 2005 – 2006
Subject Mathematics - Block B
Time Allowed 180 minutes [/b]

[b]Problem I[/b]: (2 points)
Let $(C_m )$ be a graph of a function: $ y=\frac {x^2+(m+1)x+m+1} {x+1} $ (*) ($m$: parameters)
1.	Discuss the geometry of the graph of the function (*) and construct its graph when $m = 1$.

2.	Prove that for arbitrary $m$, the graph $(C_m)$ always has the maximum, the minimum and the distance between them equals $ \displaystyle \sqrt {20}$.

[b]Problem II[/b]: (2 points)
1.	Solve the equation system: $\displaystyle \{\begin{array}{c} \sqrt{x-1}+\sqrt{2-y}=1\\3\log_{9} (9x^2)-\log_{3}y^3=3\end {array}$.

2.	Solve the equation: $\displaystyle 1+ \ sinx +\ cosx+ \sin2x +\cos2x=0$

[b]Problem III[/b]: (3 points)
1.	Given in a plane Oxy are two points: $A(2;0)$ and $B(6;4)$. Write the equation of a circle $(C)$ tangent to the axis Ox at the point $ A $ and the distance between the centre of the circle $ (C) $ and the point $ B $ equals 5.

2.	Given in a space Oxyz is a right prism $ABC.A_1B_1C_1 $ with $A(0;-3;0), B(4;0;0), C(0;3;0), B_1(4;0;4)$.

a.	Determine the coordinates of the vertices $ A_1, C_1 $. Write the equation of a spherical surface with the centre at $ A $ and tangent to the plane $(BCC_1B_1)$.

b.	$ M $ is the midpoint of the segment $A_1B_1$. Write the equation of a plane $ (P)$ passing through two points $ A, M $ and parallel to $BC_1$. $ N $  is the intersection of the plane $(P)$ and the straight line $A_1C_1$. Find the length of the segment $ MN $.

[b]Problem IV[/b]: (2 points)
1.	Compute: ${\displaystyle I=\int_ {0} ^ {\frac{\pi}{2}} \frac{sin2x.cosx}{1+cosx}}dx$

2.	There are 15 people consisting of 12 boys and 3 girls in a group of youth volunteers. How many ways are there to assign that group the task of coming to help 3 mountain provinces, such that each province has 4 boys and 1 girl?

[b]Problem V[/b]: (1 point)
	Prove that for all $ x \in R $,  $\displaystyle (\frac{12}{5})^x+(\frac{15}{4})^x+(\frac{20}{3})^x \geq 3^x+4^x+5^x$.

When does the equality occur?

[b]------------- The end -------------[/b]

Nice and very challenging!

I tried problem V as follows:
use $a^2+b^2+c^2\geq(ab+ac+bc)$
for $a=\frac{12}{5})^x$
     $b=\frac{15}{4})^x$
     $c=\frac{20}{3})^x$
and we are done! :D

Problem III part 1
We know that the center of the circle has x coordinate 2.  Therefore, there are two point with x coordinate 2 with distance 5 from (6, 4).  They would be $(2, 4+\sqrt{34})$ and $(2, 4-\sqrt{34})$.  Thus, the possible equations of the circle are

$(x-2)^2+(y-4+\sqrt{34})^2=50-8\sqrt{34}$
and
$(x-2)^2+(y-4-\sqrt{34})^2=50+8\sqrt{34}$

part2
a
The coordinates of $A_1$ and $C_1$ are $(0, -3, 4)$ and $(0, 3, 4)$.
The distance between the plane $BCC_1B_1$ and the point $A$ is the same as the distance between $A$ and the line $BC$.  That distance is $\frac{24}{5}$.  Thus, the equation of th sphere in question with center at $A$ and radius $\frac{24}{5}$ is
$x^2+(y+3)^2+z^2=\frac{576}{25}$
b
We know that two intersecting lines define a plane.  A line is parallel to a plane is the plane contains a line which is parallel to the line.  Thus, let us draw a line through M which is parallel to $BC_1$.  And I don't know how to graph in space, so I'll leave III2b up to someone else...

choose province A: $C\limits_{12}^4.C\limits_3^1$
choose province B: $C\limits_8^4.C\limits_2^1$
choose province C: $C\limits_4^4.C\limits_1^1$
result: 207900


function
calculus
integration
limit
inequalities
absolute value
calculus computations

My second year calculus class is starting halfway through Wade's "Introduction to Analysis" where, when trying to motivate the notation of the supremum norm, he makes reference to the fact that the supremum of a continuous function over a closed interval $ [a,b]$ can be computed by taking the limit of $ (\displaystyle\int^b_a |f(x)|^p\,dx)^\frac{1}{p}$ as $ p$ approaches infinity. Seeing as we started halfway through the textbook, we seemed to have missed this in our first year courses. Can anyone help me out in understanding how to prove this?

Let $ M \equal{} \sup_{x\in[a,b]}|f(x)|.$ (For the statement you made, you do need that absolute value, or you need $ f$ to be nonnegative.)

Let $ \|f\|_p \equal{} \left(\int_a^b|f(x)|^p\,dx\right)^{\frac1p}.$ You want to show that $ \lim_{p\to\infty} \equal{} M.$

Of course, an analyst usually proves an equality by proving two inequalities. Often, one of the two is pretty easy, but the other requires more work.

The "pretty easy" inequality is that $ \|f\|_p\le (b\minus{}a)^{1/p}M$ for all $ p.$ I won't say any more about that.

That leaves the other side. Here's a start to the argument: Given $ \epsilon > 0,$ there exists $ \delta > 0$ and an interval $ I$ of length $ \delta$ such that for $ x\in I,$ $ |f(x)| > M \minus{} \epsilon.$
(Note that this will use the given continuity of $ f.$)

I'll let you figure out what to do with that.


function
trigonometry
limit
topology
real analysis
real analysis theorems

Does there exist a continuous and one to one function from $ \mathbb{R}^2$ to $ \mathbb{R}$?

Let $ f: (0,1)\rightarrow\mathbb{R}$ be a bijection.  Define $ F: (0,1)\times (0,1)\rightarrow\mathbb{R}^{2}$ by $ F(x,y)\equal{}(f(x),f(y))$, which is also a bijection.  Now there exists a bijection $ G$ between $ (0,1)\times (0,1)$ and $ (0,1)$, for example, $ G(0.a_{1}a_{2}a_{3}\ldots,0.b_{1}b_{2}b_{3}\ldots)\equal{}0.a_{1}b_{1}a_{2}b_{2}\ldots$.  Then the function $ f\circ G\circ F^{\minus{}1}: \mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is a composition of bijections and thus is a bijection.

First note that if a function from $ \mathbb{R}$ to $ \mathbb{R}$ is continuous and one to one then it is strictly monotone. If a function is continuous and strictly monotone, then its range is necessarily an open interval.

Now suppose that a continuous and one to one function like $ f: \mathbb{R}^2\rightarrow\mathbb{R}$ exists. Take an arbitrary point like $ (x_0,y_0)$ in the plane. Define the functions $ \phi(t)\equal{}f(x_0,t)$ and $ \psi(t)\equal{}f(t,y_0)$ from $ \mathbb{R}$ to $ \mathbb{R}$. Since $ f$ is continuous and one to one, so are $ \phi$ and $ \psi$. Therefore, the range of $ \phi$ and $ \psi$ are necessarily open intervals. Because $ f(x_0,y_0)$ is in the range of both of these functions one can assume the intersection of their ranges is also an open interval.

Now pick a value differing from $ f(x_0,y_0)$ like $ C$ in that interval. Since $ \phi$ and $ \psi$ are one to one there exists a unique $ t_1$ and unique $ t_2$ such that $ \phi(t_1)\equal{}\psi(t_2)\equal{}C$. Also, $ t_1\neq y_0$ and $ t_2\neq x_0$ because $ C$ and$ f(x_0,y_0)$ are different. From this we deduce that $ f$ is not one to one, because $ f(x_0,t_1)\equal{}f(t_2,y_0)\equal{}C$, even though $ (x_0,t_1) \neq (t_2,y_0)$. Therefore we have reached a contradiction, and such a function cannot exist.

[quote="t0rajir0u"]What can the image of the unit circle be?[/quote]
To elaborate:  if $ f(x, y)$ is such a function, $ f(t) \equal{} f(\cos t, \sin t)$ is also continuous and injective on $ [0, 2\pi)$, hence monotonic, which means $ f(0)$ can't equal $ f(2\pi)$; contradiction.

[quote="t0rajir0u"]To elaborate:  if $ f(x, y)$ is such a function, $ f(t) \equal{} f(\cos t, \sin t)$ is also continuous and injective on $ [0, 2\pi)$, hence monotonic, which means $ f(0)$ can't equal $ f(2\pi)$; contradiction.[/quote]
I don't think you are right; since $ f(t)$ should only be one to one on $ [0, 2\pi)$; there is no reason for $ f(t)$ to be one to one on $ [0, 2\pi]$.

Yes, but if $ f$ is continuous and monotonic then $ \sup_{x \in [0, 2\pi)} f(x) > f(0)$ and since $ \lim_{t \to 2 \pi^{\minus{}}} f(t) \equal{} f(2\pi) \equal{} f(0)$ it follows that there exists small $ \delta$ such that $ \sup_{x \in [0, 2\pi)} f(x) > f(2\pi \minus{} \delta) > f(0)$, which contradicts monotonicity.  I'm sure this can be phrased in fancier topological language; what I want to say is something like $ \mathbb{R}^2$ can be disconnected by removing the unit circle (which is path-connected) into two parts, one of which is bounded, but any corresponding division of $ \mathbb{R}$ can't be done with a path-connected region.

Expanding on t0rajir0u, using some topology.

Fact:
If $ f: X \to Y$ is a continuous bijection, where $ X$ is compact and $ Y$ is Hausdorff, then $ f$ is a homeomorphism (in this case, inverse is continuous too).

So take the closed unit circle $ B^2$ in $ \mathbb{R}^2$. This is a compact set (closed and bounded using euclidean metric). $ \mathbb{R}$ is Hausdorff, being metrizable.

So now if we would have such a function $ f$, it would be a continuous bijection when restricted to $ B^2$ and its image. $ B^2$ is connected, so $ f(B^2)$ is too. We use the above fact to also conclude that $ f$ must be a homeomorphism (when restricted). The only connected subspaces of $ \mathbb{R}$ are the intervals. So we have that $ f(B^2) \equal{} [a,b]$ for some $ a, b$ ($ B^2$ is compact, so the interval must be as well). Now remove the point $ c\equal{} \frac{a\plus{}b}2$ from $ f(B^2)$. There is some $ p \in B^2$ with $ f(p) \equal{}  c$, so if we also remove that point from $ B^2$, we still have a homeomorphism between both residual spaces, using $ f$. But $ f(B^2) \setminus \{c\}$ is not an interval, thus not connected. However, $ B^2 \setminus \{p\}$ is still connected (easily seen by using path connectedness). Thus they can not be homeomorphic, contradiction. There can not exist such an $ f$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

It can be shown that for all $n \ne 6$, every automorphism of $S_{n}$ is inner. (This is on my homework so if you want to post a proof of it please wait at least until tomorrow.)

Exhibit an automorphism of $S_{6}$ that is not inner.

The reason I'm interested in this is that the step of my proof for which we have to exclude the $n=6$ case doesn't really illuminate that well for me why it screws up, and I wonder if seeing the actual guy who screws things up might help me understand what's going on.

 :ninja:

There is a subgroup of $S_{5}$ which has 20 elements (generated by a 5-cycle and 4-cycle). It has 6 conjugates, so the operation by conjugation yields a homomorphism $S_{5}\to S_{6}$ which must be injective (the kernel is obviously not $S_{5}$ and can't be $A_{5}$).

Therefore $S_{6}$ contains transitive subgroups isomorphic to $S_{5}$ which can't be the stabilzers of one point. There are again 6 conjugates of such a transitive subgroup and hence the operation by conjugation yields a homomorphism $S_{6}\to S_{6}$ which is again injective and hence an automorphism. In fact it is an outer one.

Here's an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=16615]old thread[/url] with an explicit construction.


My niece is very interested in math/physics (she is currently in high school). One option is to pursue an undergrad degree in engineering, while another is to apply for a Bsc in Math/Physics.  Besides being aware of the IITs, REcs and Bits Pilani I am quite ignorant about what other good quality college/university options she could pursue in India.

I wondered if i could take a poll of the india group on teh following questions:

1. Would anyone like to take a stab at the 5-10 best math/physics undergrad programs in south india?
2. Which are the next best engineering college options one could consider in south india other than the big 3 i just mentioned above?

Thanks for your help and thoughts.

Chitra

I too have posted a similar poll in the Round Table. I feel that opting for a pure science course in one of the following top-notch institutes would be great; CMI(Chennai Mathematical Institute), ISI(Indian Statistical Institute).
Any other views?

Not in India, but you might be interested in it anyway:

x posted:
FWIW: A news item and some perspective  that Ivy leagues     may be less expensive than many think.  To attract  the best and brightest, they tell that    you need not be very rich to go there: A story from Times of India:
[url=http://timesofindia.indiatimes.com/articleshow/1490065.cms]Enroll at Yale for free[/url][quote] Are you smart, intelligent and 18? Then stop worrying about quotas and think Ivy League — that too with full financial aid. That was Yale professor and director of its world fellows programme Daniel Esty making his pitch for India’s best and brightest.

[b]Yale[/b], which recently expanded its under-graduate financial aid programme along with [b]Harvard, Stanford and University of Pennsylvania, is now emphasising its affordability as well as accessibility[/b]. Parents of undergrads whose annual income is less than $\$$ 45,000   (or about Rs 20 lakh) no longer have to pay any portion of the cost of education. Full financial aid includes tuition, room and board. What’s more, Esty says there’s no cap on the number of foreign students. Currently, Yale has 125 Indian students out of 11,000 students and would be happy to hike the number. ‘‘We want to attract the best talent and lack of funds should not come in the way of a student applying,’’ he said. Yale offers aid — full or part —[b] to 60% of undergrads and its financial aid budget for 2006-07 is approximately $\$$59 million,[/b] an 8% increase over last year. In addition, all international students will get tickets for one return trip home every year.

‘‘We noticed that many foreign students were not going home during the vacation because they could not afford the journey. The tickets will help make their Yale experience as positive as possible,’’ said George Joseph, assistant secretary of international affairs at Yale.[/quote]

Interesting that Harvard President was pitching similar theme  recently.

Number of good "science/math" colleges in US are quite high, undergraduate admissons are a little tough than graduate schools  (besides some  Indian parents think a lot before they  send their babies that far away )anyway  something to think about.


floor function
inequalities
combinatorics unsolved
combinatorics

There are $ 30$ pupils in a class, and each of them has the same number of friends among the classmates. What is the greatest possible number of pupils, each of which studies more than a majority of his/her friends? (For any two pupils it can be determined who studies more).

There is a solution (which I can't find ..) somewhere in Engel, which bounded the number of favorable pairs [for the students in the majorities] against the number of total pairs (which is $ 15n$, where $ n$ is the number of friends per student). The best student beats his $ n$ friends, and the other $ x\minus{}1$ students who do better than the majority of their friends beat at least $ \frac {n\plus{}1}2$ of their friends.

[hide]
The answer is $ 25$.
[/hide]

[quote="azjps"]There is a solution (which I can't find ..) somewhere in Engel[/quote]

chapter 3 problem 37.

[quote="Desmo90"]
chapter 3 problem 37.[/quote]

Can you post it, please?  :)

All right, I've got it. Thanks.

Here goes the proof:

We call a student [b]good[/b] if he learns better than the majority of his friends. Let $ x$ be the number of good students and $ k$ the number of friends of each student. The [b]best[/b] student in class is the best of $ k$ pairs, and any other good student of at least $ \lfloor{\frac{k}{2}}\rfloor \plus{}1\geq \frac{k\plus{}1}{2}$ pairs. Hence, the good students are the best in at least $ k\plus{}\frac{(x\minus{}1)(k\plus{}1)}{2}$ pairs. This number cannot exceed the number of all pairs of friends in the class, which is $ 15k$. Hence $ k\plus{}\frac{(x\minus{}1)(k\plus{}1)}{2}\leq 15k$, or $ x\leq \frac{28k}{k\plus{}1}\plus{}1$. We observe that $ \frac{k\plus{}1}{2}\leq 30\minus{}x$ or $ k\leq 59\minus{}2x$, since the number of students who are better than the [b]worst[/b] among the good ones, does not exceed $ 30\minus{}x$, that is, $ x\leq \frac{28(59\minus{}2x)}{60\minus{}2x}\plus{}1$, or $ x^2\minus{}59x\plus{}856\geq 0$. The greatest integer $ x\leq 30$ satisfying the last inequality is $ x\equal{}25$. Find an example showing that $ 25$ can be attained.


im trying to find a general pattern between certain graphs

f(x) = 2x+5   f(x) = (2x+5) +3  f(x) = (2x+5) - 4
g(x) = x^2    g(x) = x^2 + 3     g(x) = x^2 -4
h(x) = x^3    h(x) = x^3 + 3     h(x) = x^3 - 4
i(x) = lnx      i(x) = lnx +3         i(x) = lnx - 4

now between these theres a distinct pattern i just cant really find any, i have noticed that the number infront of the x chnages its position along the graph for the first one its along the x axis and the next three its along the y axis. any help would be really great thank you!

Say you start out with a graph $ G_1: y\equal{}f(x)$. Now, I add a real number $ c$ so that my graph is $ G_2: y\equal{}f(x)\plus{}c$. Did I change anything to the $ x$ values? Nope. What about the $ y$'s? Yes, I added a real number $ c$. So the graph of $ G_2$ is a shift along the y-axis by $ c$ units from $ G_1$. Remember, $ c$ is a real number, in other words, it can be positive, negative, or zero. Try it yourself. Try $ c\equal{}1$ (shifting 1 unit), $ c\equal{}0$ no shift?, and $ c\equal{}\minus{}1$ (shifting 1 unit also but which direction???). Test it out yourself and you'll see. Use a function like $ f(x)\equal{}x^3$, you'll see better the patterns than a line. The keyword for the distinct pattern you are looking for is 'shifting', or more formally, translational transformation.


I am, somehow, having trouble with solving this question. Can someone please help me out?
It will be appreciated.
The question is

Let ABC be a regular triangle (angle A=angle B= angle C). Prove that for every point P inside this triangle the sume of distances from P to sides AB, BC and CA is the same.

I guess it has something to do with CEVA's theorem but I do not know how to start off.

Thanx for your help.

How did this ended up in algebra? Mods, please move it (preolympiad, I guess)

The answer: The key is in the areas. Let P be the point inside the triangle. Then area of the triagles PAB, PBC and PAB is $ \frac {a*p}{2} \plus{} \frac {a*q}{2} \plus{} \frac {a*r}{2} \equal{} \frac{a^2 \sqrt {3}}{4}$ where p, q and r are the distances to the sides. Divide the equality with a/2, and the result follows.

[quote="dluvbell"]I do not know how to move this post to pre-olympiad;;;;'

Anyways Thank you very much[/quote]
You are not a moderator so you can't do it.
hsiljak asked modersators do to it,not you. :rotfl:


[hide]$f(x)=0$ :D

We just got done covering this chapter in our math class.[/hide]

For those who wonder: the terms "odd" and "even" were taken after the power functions, since $(-x)^{n}=x^{n}$ if $n$ is even and $(-x)^{n}=-(x^{n})$ if $n$ is odd.

[quote="lingomaniac88"]An odd function $f$ is defined as follows: $f(-x)=-f(x)\text{ for all real }x$
An even function $f$ is defined as follows: $f(x)=f(-x)\text{ for all real }x$

There exists a function that is both odd and even.  What is it?[/quote]
Just a few manipulations:
[hide]We have $f(-x)=-f(x)$ and $f(-x)=f(x)$. Substituting, we get $f(x)=-f(x)$, so $2f(x)=0$, which means that
$\fbox{f(x)=0}$
is the only answer.

EDIT: When I use the code
[code]\fbox{f(x)=0}[/code], it produces the expression shown, which isn't what I'd like. However, when I try to use
[code]\fbox{$\$$f(x)=0$\$$}[/code], I get something weird like \[\fbox{$: d20f9695cdf(x)=0$: d20f9695cd}.\] Anyone know how to fix this?[/hide]

[quote="nebula42"][quote="lingomaniac88"]An odd function $f$ is defined as follows: $f(-x)=-f(x)\text{ for all real }x$
An even function $f$ is defined as follows: $f(x)=f(-x)\text{ for all real }x$

There exists a function that is both odd and even.  What is it?[/quote]
Just a few manipulations:
[hide]We have $f(-x)=-f(x)$ and $f(-x)=f(x)$. Substituting, we get $f(x)=-f(x)$, so $2f(x)=0$, which means that
$\fbox{f(x)=0}$
is the only answer.

EDIT: When I use the code
[code]\fbox{f(x)=0}[/code], it produces the expression shown, which isn't what I'd like. However, when I try to use
[code]\fbox{$\$$f(x)=0$\$$}[/code], I get something weird like \[\fbox{$: 9689f7b5c1: d20f9695cdf(x)=0$: 9689f7b5c1: d20f9695cd}.\] Anyone know how to fix this?[/hide][/quote]
I don't think it works in TeXnicCenter but \boxed command works on the AoPS forums, if I think I am understanding what you want.
\boxed{f(x)=0} turns out to be $\boxed{f(x)=0}$

f(x)=0 is half correct

For all odd and even functions, their domain must be symetrical  to (0,0)

eg" f(x)=0,x is in interval:(-2,4]  "is not a correct answer!


Hey, Guys!
Look at this problem:

Prove that there aren't x and y positives integers so that x³+y³ = 2^2009.

I wanna see someone do this problem without using Parity.

After I'm gonna post my solution without using parity.

Bye!

left side 0,1,-1,2,-2 mod 7  rhs: 4 mod 7

This technique is on Mathematical Circles - Russian experience, do you have?
It's interesting!


Hey guys,
does anyone know any good expected value problems?

Sure, here are a few:

1. Jake and Milo play a game. Jake rolls a die. If the number is a perfect square, he then flips an unfair coin, for which the probability of getting a head and tail is in the ratio 7:9. If he gets heads, he wins 8 dollars. If he gets tails, he loses 12 dollars. If he rolled a number other than a perfect square, he wins 3 dollars.

Milo rolls a pair of dice labeled with powers of 2: 1, 2, 4, 8, 16, and 32. If the sum of his numbers is odd, he gets 7 dollars. If the sum is even, he flips two fair coins. If the results on each coin are the same(Head-Head or Tail-Tail), he gets 15 dollars. If not, he loses 9 dollars.

To play this game, Jake and Milo both pay the expected value of one turn of their game. (One turn is the series of events listed above. For example, Jake's turn might consist of rolling a die, getting a perfect square, then flipping a coin, getting tails, and winning 8 dollars.) Who pays more and by how much?

2. ou first flip a coin. If it is heads, you roll a die. If you get a perfect number, you will receive 10 dollars. If you get an odd number, you will get 2 dollars. If you get an even number that is not a perfect number, you will get 3 dollars.

If you got a perfect number, you will not only get 10 dollars, you will roll the die once more. The same conditions stated above apply for the second time you roll the die, except you will not get 10 dollars for another perfect number, but 20, and you will roll again.

Again, the same conditions apply, except if you roll a perfect number for a third time, you will get the jackpot, 50 dollars!

Each time, if you don't get a perfect number, the game will end.

If when you flipped the coin, you got tails, you must pay 8 dollars. The game will end.

What is a fair price to play this game?

3. Blind spider is sitting on a cube's vertice, dead fly is on opposite cube's vertice. Spider can walk only on edges, he turns randomly then he reaches another vertice. What's the expected number of edges spider walks through to reach a fly?

1. Jake: The probability that Jake rolls a perfect square is 1/3 ({1,4} out of {1,2,3,4,5,6}).  The probability that he then gets a head is 7/16. The probability that he then gets a tail is 9/16.  The probability that he DOESN"T roll a perfect square 2/3.  Thus, the expected value is:
(1/3)(7/16)(8)+(1/3)(9/16)(-12)+(2/3)(3)=
(7/6)-(9/4)+2=[u]11/12[/u]
Milo: The probability that Milo rolls an odd sum is: (2*1*5)/(6*6)=10/36=5/18.  The probability that Milo rolls an EVEN sum is thus 13/18.  The probability that he flips two of the same results is 1/2.  Hence, the expected value of his return is:
(5/18)(7)+(13/18)(1/2)(15)+(13/18)(1/2)(-9)=
(35/18)+(13/36)(15-9)=
(35/18)+(39/18)=[u]37/9[/u]
[b]Thus, MILO pays more by 115/36[/b]

EDIT: And for #2, by perfect number you mean perfect square?

I think generally, 1 is not considered a perfect number.

[quote="007math"][quote="isabella2296"]A perfect number is a number such that the sum of its proper factors is the number itself, e.g. 6 or 28.[/quote]

So is 1 a perfect number?  (Just asking)[/quote]

A perfect number is number = $ \frac{1}{2}\times$(sum of it's positive divisor).
Like:
$ 6 \equal{} \frac{1}{2}(1\plus{}2\plus{}3\plus{}6) , 28 \equal{} \frac{1}{2}(1\plus{}2\plus{}4\plus{}7\plus{}14\plus{}28)$.

$ \Rightarrow$ 1 is not a perfect number.

[quote="mathson"][quote="007math"][quote="isabella2296"]A perfect number is a number such that the sum of its proper factors is the number itself, e.g. 6 or 28.[/quote]

So is 1 a perfect number?  (Just asking)[/quote]

A perfect number is number = $ \frac {1}{2}\times$(sum of it's positive divisor).
Like:
$ 6 \equal{} \frac {1}{2}(1 \plus{} 2 \plus{} 3 \plus{} 6) , 28 \equal{} \frac {1}{2}(1 \plus{} 2 \plus{} 4 \plus{} 7 \plus{} 14 \plus{} 28)$.

$ \Rightarrow$ 1 is not a perfect number.[/quote]
But $ 1 \equal{} \frac12(1 \plus{} 1)$

the definition of proper factors is:

-all factors of a number [b][u]except for the number itself[/u][/b]

thus, 1 is not a perfect number.


AMC
AIME
geometry
geometric transformation
reflection
USA(J)MO
USAMO

That title is more than a bit misleading -- I don't even think IMO people have the gift to express a full solution to an AIME in three digits.  (Although that sort of conciseness would be helpful for saving time on olympiads.)

Although I suppose a fairly elementary "solution" to the AIME would be to grab the nearest IMO person and make him do the problems for you :-)

P.S.  Sorry, I'm being hard on a newbie.  It's probably well enough that there is now a list of correct answers in its own thread instead of mixed up with other stuff.

[quote="mickeymouse"]It should be 817.  Concur?[/quote]

I agree with 817 for #10.

yeah ....

and #13 should be 482, not 582 ?

Oh, thank god I got 10 right.  I spent so much time on that one during the test.  I think I spent 45 minutes on it, because I couldn't qualify for USAMO because my AMC was too low, so I figured I'd try to work out some of the harder ones.


I'm writing up the solution packet for COP 4th Championship Round and I couldn't remember how I solved this problem.

10. Find the remainder when $(\tan^2 1^\circ + \tan^2 3^\circ + \tan^2 5^\circ + \cdots + \tan^2 89^\circ)$ is divided by 24.

Can anyone post non-calculator solution? I'll really appreciate that.

Thanks! :)

WOW ive spent wayy too much time on this problem -_______-

and i still havent solved it ugh

but here's where i got to:

[hide]

looking at

$(\tan 1 + \tan 89)^2 - 2\tan1\tan89 = \tan^2 1 + \tan^2 89$

$(\displaystyle\frac{\sin1}{\cos1} + \frac{\sin{89}}{\cos{89}})^2 - 2$

$(\displaystyle\frac{\sin {90}}{\cos1\cos89})^2 - 2$

$(\displaystyle\frac{4}{\cos^2{88}}) - 2$

which is the simplified version

adding all of these pairs up you have:

$ 4(\sec^2{88} + \sec^2{84} + ... + \sec^2{4}) - 43$

and from there i tried to simplify it some more but it just got messy and stuff

i also tried to use mods directly from there but all that did was tell me that this is equal to $1,5,9,13,17,$ or $21$ mod 24, which doesn't really help me

[hide]

using calculator answer is 9

 :P

[/hide]

so yea if anyone can find a clever solution or help me finish it

hope that helps a lil though silverfalcon... maybe it'll... jog ur memory or something

[/hide]

[quote="236factorial"]Is $\tan{\theta}=\frac{1}{\tan({90-\theta})}$[/quote]
Yup. $\tan\theta=\frac{1}{\tan (90^\circ-\theta)}=\cot (90^\circ -\theta)$

Hmm..

I remember my solution was something based on the expansion of:

$(\cos \theta + i \sin \theta)^{90}$ where $i = \sqrt{-1}$. I remember I took the real part of it but I can't remember anything after that (or how I took the real part).

Anyone has any idea?

you can verify this for your self,
$\tan^{2}(x) + \tan^2(90-x) = \frac{4}{\sin^{2}(2x)} -2$

then pairing up terms, like 1 and 89, 3 and 87, etc, but $\tan(45)^{2}$ does not have a match so we leave that alone

$\sum^{21}_{k=0}( 4\csc^{2}(2(2k+1)) - 2) + \tan(45)^{2}$

$\sum^{21}_{k=0}( 4\csc^{2}(2(2k+1)) - \sum^{21}_{k=0}(2) + \tan(45)^{2}$

$4\sum^{21}_{k=0}( \csc^{2}(2(2k+1)) -43$

$4\sum^{21}_{k=0}( \cot^{2}(2(2k+1)) +1) -43$

$4\sum^{21}_{k=0}( \cot^{2}(2(2k+1))) + 4\sum^{21}_{k=0}(1) -43$

$4\sum^{21}_{k=0}( \cot^{2}(2(2k+1))) + 45$

after that i cannot simplify it, but dividing 45 by 24 yields the remainder of 21 (which is correct)
i believe 21 is the correct answer, since i used my calc to find the sum
but somehow, you need to show that the remaining sum is divisible by 24, since there is the factored out 4, you just need to get the 6 factored out as well


inequalities
LaTeX
inequalities proposed

If $a,b,c\geq0$ such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$

demonstrate that $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq\frac{3}{2}$

Have fun!!
 :surf:

IRAN 2005
[url=http://www.mathlinks.ro/Forum/topic-49899.html]www.mathlinks.ro/Forum/topic-49899.html[/url]
 :roll:

hi Mr socrate.it's a very good example .i tryed a lot of things to solve it until i arrived to this:
let X=max(a,b,c) and let Y=min(a,b,c).
so we have:
2=1/(1+a)+1/(1+b)+1/(1+c) superior or equal to 3/(1+X) so X is superior or equal to 1/2        (1)
2=1/(1+a)+1/(1+b)+1/(1+c) inferieor or equal to 3/(1+Y) so Y is inferior or equal to 1/2          (2)
X=max(a,b,c) and Y=min(a,b,c) so (a+b+c) is in the interval (3Y;3X).and from (1) and (2) we find that
a+b+c=3/2         (3)

and we know that (a+b)/2 is superior or equal to \sqrt{ab}          (4)
                           (b+c)/2 is superior or equal to \sqrt{bc}          (5)
                            (c+a)/2 is superior or equal to \sqrt{ac}          (6)
so from 3,4,5,6 the inequality is proven

and excuse me because i don't use LATEX
 friendly
nawram88

aren't you convinced by mine Mr silouan ??

[quote="silouan"]Socrates can you post your solution. Thanks[/quote]

It is similar to the one given in [url]http://www.mathlinks.ro/Forum/post-316315.html#316315[/url]

I did the substitution
$a=\frac{x}{y+z}$
$b=\frac{y}{x+z}$
$c=\frac{z}{x+y}$....

[quote="cezar lupu"]Yes, actually this problem has been discussed in the romanian forum. ;)[/quote]

I didn't know that.

By the way, I think this subst. is one of your favorite....
(mine too) :D  :)


Let $ ABC$ be a triangle so that $ 2\sin A\plus{}\sin B\sin C\equal{}\frac{1\plus{}\sqrt{17}}{2}$ . Show that the triangle is [b]isosceles[/b] .

Keep up the good work, Constantin!

[b][u]Solution[/u][/b]

The equation is equivalent to
\[ 4sinA + cosA + cos(B - C) = 1 + \sqrt {17}\]
However, we have
\[ \{\begin{array}{cc} cos(B - C) \le 1 \\
 \\
4sinA + cosA \le \sqrt {17(sin^2A + cos^2A)} = \sqrt {17} \end{array}\]
Hence, the conclusion follows.


I think it's wrong.Just try $a=b=c=\frac{1}{4}$.You might have forgotten a condition

Sorry ,my problem is:
Prove that:
$\frac{a+bc}{b+c}+\frac{b+ca}{c+a}+\frac{c+ab}{a+b}\geq 2$
where $a;b;c$ are positive real numbers satis fying $a+b+c=1$

$\sum \frac{a+bc}{b+c}=\sum \frac{a^{2}+ab+bc+ca}{b+c}=\sum \frac{(a+b)(a+c)}{b+c}\geq 2(a+b+c)=(a+b)+(b+c)+(c+a)$
If we denote $x=a+b$,$y=b+c$ and $z=c+a$ we have \[\sum\frac{xy}{z}\geq x+y+z\] $\leftrightarrow$ $\sum (xy)^{2}\geq \sum xy\cdot yz$

we have a+bc=1-b-c+bc=(1-b)(1-c) ,the inequality is equivalent to
(1-b)(1-c)/(1-a)+(1-c)(1-a)/(1-b)+(1-b)(1-a)/(1-c)>=2
in the other hand 1/2(1-b)[(1-a)/(1-c)+(1-c)/(1-a)]>=1-b and simirly..
adding this tree inequality, we are done


The usual coloring pattern on an 8 x 8 checkerboard is changed so that 20 unit square are now colored red, and the rest are colored white.  When the board is foled in half along a line parallel to one edge of the board, exactly seven [i]pairs[/i] of red unit squares coincide.  Compute the number of [i]pairs[/i] of white unit squares that coincide.

[hide="Solution"]
Each pair of squares are either both red, white, or different. $ 7$ pairs are both red, $ 20 \minus{} 7 \cdot 2 \equal{} 6$ pairs have one red, and so $ 32 \minus{} 7 \minus{} 6 \equal{} \boxed{19}$ pairs are coinciding white squares.
[/hide]


inequalities
rearrangement inequality
inequalities unsolved

[color=darkblue]Let $ a,b,c>0$. Prove that: $ \frac{a^8\plus{}b^8\plus{}c^8}{a^3b^3c^3}\ge \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$[/color]

But this is equivalent to $ a^8\plus{}b^8\plus{}c^8\ge a^3b^3c^2\plus{}b^3c^3a^2\plus{}c^3a^3b^2$ which is quite obvious from rearrangement inequality..

OK.

First solution: Verify that the sequences $ (a^2,b^2,c^2)$ and $ (a^6,b^6,c^6)$ are increasing in the same direction. Hence by Rearrangement inequality we have
\[ a^2\cdot a^6 + b^2\cdot b^6 + c^2\cdot c^6\ge a^6b^2 + b^6c^2 + c^6a^2
\]
and
\[ a^2\cdot a^6 + b^2\cdot b^6 + c^2\cdot c^6\ge b^2c^6 + c^2a^6 + a^2b^6
\]
Then adding the two inequalities and applying AM-GM we get
\[ 2\sum a^8\ge \sum (a^6b^2 + b^2c^6)\ge 2\sum a^3b^2c^3
\]
Second solution: From Chebyshev's inequality
\begin{align*}3(a^8 + b^8 + c^8) & \ge (a^2 + b^2 + c^2)(a^6 + b^6 + c^6) \\
& \ge 3a^2b^2c^2(a^2 + b^2 + c^2) \\
& \ge 3a^2b^2c^2(ab + bc + ca)\end{align*}
where the last two inequalities follow from AM-GM.

Third solution: Using Holder's inequality we get
\[ (a^3b^3c^2 + b^3c^3a^2 + c^3a^3b^2)^8\le (a^8 + b^8 + c^8)^3(b^8 + c^8 + a^8)^3(c^8 + a^8 + b^8)^2 \\
\implies a^8 + b^8 + c^8\ge a^3b^3c^2 + b^3c^3a^2 + c^3a^3b^2
\]
Fourth solution: From AM-GM,
\[ \sum\left(3\frac {a^5}{b^3c^3} + 3\frac {b^5}{c^3a^3} + 2\frac {c^5}{a^3b^3}\right)\ge \sum\sqrt [8]{\left(\frac {a^5}{b^3c^3}\right)^3\left(\frac {b^5}{c^3a^3}\right)^3\left(\frac {c^5}{a^3b^3}\right)^2} = 8\sum\frac {1}{c}
\]

[quote="thanhnam2902"][color=darkblue]Let $ a,b,c > 0$. Prove that: $ \frac {a^8 \plus{} b^8 \plus{} c^8}{a^3b^3c^3}\ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$[/color][/quote]

Let us assume that $ a \leq b \leq c$. Then $ X\geq0$ and $ Y\geq0$, where
\[ X \equal{} {{c}^{3} \left( {c}^{2} \minus{} {a}^{2} \right) \plus{} {c}^{2}\left( {c}^{3} \minus{} {b}^{3} \right) \plus{} ({b}^{5} \minus{} {a}^{5})} \\
Y \equal{} {{a}^{3} \left( {c}^{2} \minus{} {a}^{2} \right) \plus{} {c}^{2}\left( {c}^{3} \minus{} {b}^{3} \right) \plus{} {a}^{3} \left( {c}^{2} \minus{} {b}^{2} \right)}
\]
But it's easy to check for all $ a,b,c$ the following identity is true:
\[ {a^8 \plus{} b^8 \plus{} c^8 \minus{} (a^3b^3c^2 \plus{} a^3b^2c^3 \plus{} a^2b^3c^3) \equal{} (b^3 \minus{} a^3)\,X \plus{} (c^3 \minus{} b^3)\,Y}
\]
This finishes the proof that for all positive $ a,b,c$
\[ a^8 \plus{} b^8 \plus{} c^8 \geq a^3b^3c^2 \plus{} a^3b^2c^3 \plus{} a^2b^3c^3
\]

Thank you for the nice inequaltity!

Sergey Markelov
markelov@mccme.ru

It follows by rearrangement on the sequences

$ (a^3,b^3,c^3)$, $ (a^3,b^3,c^3)$, $ (a^2,b^2,c^2)$

[color=darkblue]THANK YOU VERY MUCH.[/color]

[hide=" how does the world make the best ?"]How can we extend this problem?
I suggest one: $ \frac {a^n \plus{} b^n \plus{} c^n}{3} \geq (\frac {a \plus{} b \plus{} c}{3})^n$[/hide]

[quote="great math"]How can we extend this problem?
I suggest one: $ \frac {a^n \plus{} b^n \plus{} c^n}{3} \geq (\frac {a \plus{} b \plus{} c}{3})^n$[/quote]
$ \sqrt[n]{\frac{a^n\plus{}b^n\plus{}c^n}{3}} \geq \frac{a\plus{}b\plus{}c}{3}$ for $ n\geq 1$.

2 posts up: Lol, that's just the Power Mean inequality, hehe.

It's jensen with $ f(x)\equal{}x^n$  :rotfl:


I wonder when the Answer Form is returned to Lincoln and school administrator keeps the student contest booklet and solution pamphlet after the test.

When you got score in late December, will you know which questions you did wrong?

Sorry to ask these questions as I took the 2009 AMC 8 first time this year.

Do you have your booklet or did your teacher keep it? If you got it back, you can check your answers on the website [url=http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2009-8a/2009-amc8answers.shtml]here[/url]. If you don't have your booklet or didn't mark your answers in it, I don't think you can figure out which questions you got wrong, unless you remember them.

You should get your answer booklet back pretty soon (when school starts again). Your teacher (administrator) probably has a copy of the solutions (and problems), so you could try asking him/her.

I don't get my answer booklet back in our school.I think.


Can you please explain number 1? I am not really good at probability.

1.) An allergist claims that 50% of the patients she tests are allergic to some type of weed.  Waht is the probability that

a.) three out of her netx 4 patients are allergic to weeds
b.) none of her four patients are allergic to weeds.

2.) If the probability that Tom will b alive in 20 years in 0.7 and the probability that Nancy will be alive in 20 years is 0.9, what is the probability that neither will be alive in 20 years.

[hide="Answers"]
1.) a.) 1/4   b.) 1/16
2.) 0.03[/hide]

[hide="1a"]Ther are $ \binom{4}{3}$ ways we can choose the 3 that aren't allergic. That probability that each one of those 3 is allergic and the other isn't is $ (\frac{1}{2})^{3}\cdot\frac{1}{2}=\frac{1}{16}$. Multiplying, the final answer is $ \frac{4}{16}=\frac{1}{4}$[/hide]

[hide="1b"] There is a $ \frac{1}{2}$ probability that each one isn't allergic, so it's $ (\frac{1}{2})^{4}=\frac{1}{16}$[/hide]

[hide="2"]There is a $ 1-0.7=0.3$ chance Tom is not alive, and a $ 1-0.9=0.1$ that Nancy is not alive. Multiplying, the answer is $ 0.03$[/hide]


if $a,b,c,d$ are reals and $a^2+b^2+c^2+d^2\leq 1$ prove that

$ab+bc+cd+da+ac+bd\leq 4abcd+\frac{5}{4}$

Did this one appeared also in Crux? I sent it at least two years ago... I thought it vanished. Anyway... it's weak.

Harazi,
You have 4 problems in Crux 5 and 3 problems in Crux 6.
This inequality is published in Crux 5. :lol:

Does anyone has a nice solution to this nive problem ?My teacher told me it is difficult


Prove that the below equation:

$(x+1)^{y}-x^{z}=1$ $x,y,z>1$ has only one set of positive integers solution. What is the solution?

[quote="Franz Joseph"]Prove that the below equation:

$(x+1)^{y}-x^{z}=1$ $x,y,z>1$ has only one set of solution. What is the solution?[/quote]
For example: $x=2,$ $y=3$ and $z=4.700439718...$ :lol:

Well, the solution is $(2, 2, 3)$ by Mihailescu's Theorem.   :rotfl:


Prove that $ \sin{\varphi}+\sin{(\varphi+\alpha)}+\sin{(\varphi+2\alpha)}+\cdots+\sin{(\varphi+n\alpha)}=\frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)}\cdot \sin{(\varphi+\frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}$
Credit goes to wikipedia for that formula

[hide="hint"]
1. Multiply the sum by $ 2\sin\frac{\alpha}{2}$ and then transform the products of sines in differences of cosines.
[/hide]

[hide="Method 3"] Complex numbers ;) [/hide]

[quote="sylow_theory"][hide="Hint"]
Sorry, t0rajir0u Im trying to steal answers from you.  :wink:

Here is a hint.

See what happens when you sum,
$ e^{ix}+e^{2ix}+...+e^{nix}$ and $ x\not = 2\pi n$ for $ n\in \mathbb{Z}$.
By using the geometric series formula.
[/hide][/quote]
thanks!


geometry
incenter
ratio
geometry proposed

Let $ F_a,F_b,F_c$ be three Feuerbach points of triangle $ ABC$ and $ F_a',F_b',F_c'$ be three Feuerbach points of triangle $ F_aF_bF_c$. Prove that $ F_aF_a',F_bF_b',F_cF_c'$ are concurrent.

It's a special case of this problem:
Given triangle $ ABC. (O_1)$ is tangent to ray $ AB, AC$, similar with $ (O_2), (O_3); (O)$ is tangent to $ (O_1),(O_2),(O_3)$ externally (internally) at $ A_1, B_1,C_1$, respectively then $ AA_1, BB_1, CC_1$ are concurrent.
Note that $ AF_a, BF_b, CF_c$ concur at $ P$, $ P$ lies on $ IE$ ($ I$ is the incenter, $ E$ is the center of Nine-point circle) and divide $ IE$ a ratio $ \frac{2r}{R}$


What is this row's limit:
$ x_n \equal{} \sqrt{x^4\plus{}2}\ \minus{} \sqrt[3]{n}$

1)  Do you mean $ x_n \equal{} \sqrt{x_{n\minus{}1}^4 \plus{} 2} \minus{} \sqrt[3]{n}$?
2)  The English term is "sequence," not row.
3)  This problem is a little too difficult for this forum.

1) yes, that's what i mean
2) i knew that the term is row :blush:
3) I don't think the problem is too difficult. it's intermediate  :wink:


oh well.. I saw $ 10405071317$ and I recognized it was the sum and I know it's prime. I'm not certain if there is a rule for the remainders..

[hide="EDIT: Wrong"]
Let's try the small values first:
When divided by two we take the modulo residues:
$ 1 \plus{} 0 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 0 \equal{} 5\equiv1$
Take them for 3:
$ 1 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 1 \plus{} 0 \plus{} 1 \equal{} 7\equiv1$
Now 4:
$ 1 \plus{} 0 \plus{} 3 \plus{} 0 \plus{} 1 \plus{} 0 \plus{} 3 \plus{} 0 \plus{} 1 \plus{} 0 \equal{} 9\equiv1$

I'm sensing a pattern, now prove it.
[/hide]
EDIT: Misread problem.
@Quatto: I'm sure there's some formula for it, brute forcing each one seems.... wrong.

[quote="Christine"]What is the remainder if 1 + 2^2 + 3^3 + ... + 10^10 is divided by 2,3,4,5,6,7,8 and 9?[/quote]
[hide="2"] So, we have that we wish to find $ (1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10})\bmod 2$.  Note that every $ n^n$ where n is odd is odd and where n is even is even.  There are thus 5 odds and 5 evens, so the number is odd and thus congruent to $ \boxed{1\bmod 2}$. [/hide]

[hide="3"] Okay, so
\[ 1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10}\equiv 1 \plus{} 2^2 \plus{} 4^4 \plus{} 5^5 \plus{} 7^7 \plus{} 8^8 \plus{} 10^{10}\equiv 1 \plus{} ( \minus{} 1)^2 \plus{} (1)^4 \plus{} ( \minus{} 1)^5 \plus{} (1)^7 \plus{} ( \minus{} 1)^8 \plus{} (1)^{10}\equiv 1 \plus{} 1 \plus{} 1 \minus{} 1 \plus{} 1 \plus{} 1 \plus{} 1\equiv 5\equiv \boxed{2\bmod 3}
\]
[/hide]

[hide="4"]
Noting that when n is even, $ n^n$ is divisible by 4, we have that
\[ 1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10}\equiv 1 \plus{} 3^3 \plus{} 5^5 \plus{} 7^7 \plus{} 9^9\equiv 1 \plus{} ( \minus{} 1)^3 \plus{} (1)^5 \plus{} ( \minus{} 1)^7 \plus{} (1)^9\equiv \boxed{1\bmod 4}
\]
[/hide]

[hide="5"] Okay, so we have that $ 1\equiv 1\bmod 5$, $ 2^2\equiv 4\bmod 5$, $ 3^3\equiv 2\bmod 5$, $ 4^4\equiv 1\bmod 5$, $ 5^5\equiv 0\bmod 5$, $ 6^6\equiv (1)^6\bmod 5\equiv 1\bmod 5$, $ 7^7\equiv 2^7\bmod 5 \equiv 3\bmod 5$, $ 8^8\equiv 3^8\bmod 5\equiv 9^4\bmod 5\equiv ( \minus{} 1)^4\bmod 5\equiv 1\bmod 5$, $ 9^9\equiv ( \minus{} 1)^9\bmod 5\equiv 4\bmod 5$, $ 10^{10}\equiv 0\bmod 5$.  Adding everything up, we get that our number is $ 1 \plus{} 4 \plus{} 2 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 3 \plus{} 1 \plus{} 4 \plus{} 0\equiv \boxed{2\bmod 5}$ [/hide]

[hide="6"] YAY!!  We don't need to do anything too ugly.   So, we have that our number leaves a remainder of 2 when divided by 3 and one when divided by 2.  So, if you add one to our number, it would be divisible by 2 and 3 and therefore, 6.  So, adding one number makes our number divisible by 6, so our number is congruent to $ \minus{} 1\bmod 6\equiv \boxed{5\bmod 6}$ [/hide]

[hide="7"] Back with the messiness:
\[ (1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10})\equiv 1 \plus{} 4 \plus{} 6 \plus{} 4 \plus{} ( \minus{} 2)^5 \plus{} ( \minus{} 1)^6 \plus{} 0 \plus{} (1)^8 \plus{} (2)^9 \plus{} (3)^{10}\equiv 15 \minus{} 32 \plus{} 1 \plus{} 1 \plus{} (8^3) \plus{} (9^5)\equiv \minus{} 15 \plus{} 1^3 \plus{} 2^5\equiv 18\equiv \boxed{4\bmod 7}
\]
[/hide]

[hide="8"] Okay, so we have that all $ n^n$ such that n is even other than 2 is divisible by 8.  So, we have that
\[ (1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10})\equiv (1 \plus{} 4 \plus{} 3 \plus{} ( \minus{} 3)^5 \plus{} ( \minus{} 1)^7 \plus{} (1)^9)\equiv (8 \minus{} 3 \minus{} 1 \plus{} 1)\bmod 8\equiv \boxed{5 \bmod 8}
\]
[/hide]

[hide="9"] Notice that $ n^n$ is divisible by 9 when 3 is a factor of n.
\[ (1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10})\equiv 1 \plus{} 4 \plus{} 0 \plus{} 256 \plus{} ( \minus{} 4)^5 \plus{} 0 \plus{} ( \minus{} 2)^7 \plus{} ( \minus{} 1)^8 \plus{} 0 \plus{} 1^{10}\equiv 1 \plus{} 4 \plus{} 2 \plus{} 5 \plus{} 6 \minus{} 1 \minus{} 0 \minus{} 2 \minus{} 4 \minus{} 1 \minus{} 2 \minus{} 8 \plus{} 1 \plus{} 1\equiv \boxed{2\bmod 9}
\]
[/hide]
EDIT:
At CircleSquared: Well, there are tricks for a few of them I suppose... like 6 and 2... but I can't see a pattern...
You are probably right though... there must be a pattern... this solution took me a really long time to write up...
EDIT AGAIN: Yeah, sorry about that, Temperal.

[quote="The QuattoMaster 6000"]
[hide="9"] Notice that $ n^n$ is divisible by 9 when 3 is a factor of n.
\[ (1 \plus{} 2^2 \plus{} 3^3 \plus{} 4^4 \plus{} 5^5 \plus{} 6^6 \plus{} 7^7 \plus{} 8^8 \plus{} 9^9 \plus{} 10^{10})\equiv 1 \plus{} 4 \plus{} 0 \plus{} 256 \plus{} ( \minus{} 4)^5 \plus{} 0 \plus{} ( \minus{} 2)^7 \plus{} ( \minus{} 1)^8 \plus{} 0 \plus{} 1^{10}\equiv 1 \plus{} 4 \plus{} 2 \plus{} 5 \plus{} 6 \minus{} 1 \minus{} 0 \minus{} 2 \minus{} 4 \minus{} 1 \minus{} 2 \minus{} 8 \plus{} 1 \plus{} 1\equiv \boxed{2\bmod 10}
\]
[/hide][/quote]

Typo?

And CircleSquared, I suppose a more elegant solution could be derived by factoring.

No factoring is possible.  The solution by modular arithmetic is the intended one.

:idea:
Maybe it means divided by 2,3,4,5,6,7,8,9 at the same time?


we have $ 2R \cdot  \sin A \equal{} a$, $ p \cdot \tan \frac{A}{2} \equal{} r_a$ and $ h_a \equal{} \frac{2r_a (p\minus{}a)}{a}$. The sideline BC is one of the 2 internal tangent line between the A-excircle and the circle with center A and radius $ h_a$.


conics
ellipse
geometry
geometric transformation
reflection
parabola
calculus

From the definition of an ellipse, prove that if a ball is shot from one focus to the elliptical wall, it will reflect off and travel straight to the other focus.

From the definition of a parabola, prove that if a beam of light travels perpendicular to the directrix, it will reflect off the parabola and travel straight to the focus.

I'm looking for non-calculus proofs, preferably using symmetry/reflection.

Of course, the question is: which definition?

Ellipse: the locus of points the sum of whose distance to two fixed points is constant.

Parabola: the locus of points such that the distance to a fixed point is equal to the distance to a fixed line.

consider an ellipse with fixed foci and the tangent line to the ellipse, the ellipse is tangent at the point where teh sum of the distance from the foci to the line back to the other focus is at a minimum (otherwise, the locus would go beyond the line), and now that it is the minimum, the other way to find the minimum point is to reflect one foci over the line, and connect the reflection with the other foci, easy congruent angles finish the proof

take any point A on the parabola (but not the vertex) extend a perpendicular down to the directrix, and a line segment from A to the focus, then put a line  perpendicular to that segment through the focus.  let the intersection of that line with the directrix be P.  From here you can show that PA is the tangent line at A, and some simple properties of reflective surfaces will complete the proof, except for the vertex case but thats trivial.


vector
trigonometry
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parallelogram
analytic geometry

I have problems understanding the concept of a scalarproduct, and for that reason i can't find a way of proving that

$\det (\vec{a},\vec{b})=(\vec a \cdot \vec b)\cdot \tan v$

I know

$\vec a \cdot \vec b=|\vec a|\cdot |\vec b| \cdot \cos v=a_{1}b_{1}+a_{2}b_{2}$

But the problem is i don't know what the scalarproduct actually is. If $\vec a \cdot \vec b=0.9$, what is it that is $0.9$?

So can anyone help me proving this?

[quote="Sporally"]But the problem is i don't know what the scalarproduct actually is.[/quote]

For example, the scalar product of a force vector and a distance vector gives a quantity of work done.

The scalar product doesn't have an immediate graphical description in the same way as the cross product.  Simply put, it's a measure of the "closeness" of two vectors (this is easy to see from the $\cos$ definition).  It is maximal if the two vectors point in the same direction, and minimal (in fact, equal to $0$) if they are perpendicular.

OK, i think i understand your example. However, i don't really like i can't get a gfx overview of the scalar product, but i will try and live with that, kinda have to :wink:

Could you help me with proving the above stated?

$(| \vec{a}| | \vec{b}| \cos v) \tan v = | \vec{a}| | \vec{b}| \sin v$

This is the (magnitude of) the [b]cross[/b] product, which is simply the area of the parallelogram formed by the vectors.  Then you'll need to do some coordinate geometry to calculate this in another way.  (Or, better yet:  what does $det( \vec{a}, \vec{b})$ actually [b]mean[/b]? )

i would assume that that means that a and b are the column vectors and you are taking the determinant of the whole thing


$ p$ is odd,
then $ p| C_p^k , k \equal{} 1,\cdots,p \minus{} 1$ ?

thanks

Perhaps you mean $ p|^pC_k$ or $ p | {p \choose k}$?

Please do not double post. Your original post was moved to another part of this site because it's not calculus.


linear algebra
matrix
algebra
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linear algebra unsolved

Determine the number of possible values for the determinant of $A$, given that $A$ is a $n\times n$ matrix with real entries such that $A^3 - A^2 - 3A + 2I = 0$, where $I$ is the identity and $0$ is the all-zero matrix.

Three easily computable "possible" eigenvalues, which allow you to compute the possible characteristic polynomials in $C[X]$, and from there you get the possible determinants. I guess you have to be careful with the roots because the product of two of them is (-1), and we don't want to take into account the same determinant twice.

Using the theorem that tells you that there is a matrix $ M$ such that $ A\equal{}MTM^{\minus{}1}$ where $ T$ is a triangular matrix the problem becomes pretty simple.

I think the following works:

We can start with factoring this equation as
\[(A-2I)(A^2+A-I)=0\]

So one matrix is $A=2I$ which has $\det(A)=2^n$.

The other matrices must satisfy $A(A+I)=I$.

It is known that any matrix in $\mathbb{C}^{n\times n}$ is triangulizable. Thus let $A=MTM^{-1}$ as Alfonso Band suggested.

Also let's assume $T$ is upper triangular. We get
\[MTM^{-1}(MTM^{-1}+MIM^{-1})=I\implies MTM^{-1}M(T+1)M^{-1}=I\]
\[\implies MT(T+I)M^{-1}=I\implies M^{-1}MT(T+I)M^{-1}M=M^{-1}IM\implies T(T+I)=I\]

Now the eigenvalues of $T+I$ are those of $T$ increased by 1 and so

\[\prod_{i=1}^n (\lambda_i+1)=\prod_{i=1}^n \frac{1}{\lambda_i}\]

Also, $T+I$ and $T^{-1}$ agree in trace. So we have

\[\sum_{i=1}^n (\lambda_i+1)=\sum_{i=1}^n \frac{1}{\lambda_i}\]

From this we can derive for any $k\in\mathbb{N}$ we wish, that

\[\sum_{i=1}^n (\lambda_i+1)^k=\sum_{i=1}^n \frac{1}{\lambda_i^k}\]

Thus $\lambda_1=\cdots=\lambda_n=\lambda$.

This gives $(\lambda+1) = 1/\lambda$ and so $\lambda=\frac{-1\pm\sqrt{5}}{2}$ and so

\[\det{A}=\det(T)=\bigg(\frac{-1\pm\sqrt{5}}{2}\bigg)^n\]

Thus there are three possible determinants for any $n$ and they are

\[2^n, \bigg(\frac{-1+\sqrt{5}}{2}\bigg)^n , \bigg(\frac{-1-\sqrt{5}}{2}\bigg)^n\]

Nope. Counterexample to your claim:
$A=\begin{bmatrix}0&0&-2\\1&0&3\\0&1&1\end{bmatrix}$ is $3\times 3$ and has determinant $-2$. We can have more than one different root appear.

You are right. I guess I was too tired when I was coming up with an answer. It was past midnight.

Loup Blanc is right for the diagonalizable case. But I think all the notation can be dropped since all that is needed is this:
\begin{align*}
A^3-A^2-3A+2I=0&\implies SD^3S^{-1}-SD^2S^{-1}-3SDS^{-1}+2SIS^{-1}=S0S^{-1}\\
&\implies S(D^3-D^2-3D+2I)S^{-1}=0\\
&\implies D^3-D^2-3D+2I=S^{-1}0S=0\\
\end{align*}

Since powers of a diagonal matrix is as easy as just raising the diagonal terms to the powers. All eigenvalues of $D$, and hence of $A$ satisfy $\lambda_i^3-\lambda_i^2-3\lambda_i+1=0$ independently. So for the diagonal matrix we choose all the eigenvalues from $\{2,\frac{-1\pm\sqrt{5}}{2}\}$ $n$ times without replacement. Order doesn't matter, only how many of each get chosen, so in the diagonalizable case, it is in fact $\binom{n+2}{n}$, which is Loup Blanc's result.

We must show that $A$ is diagonalizable or that the non-diagonalizable cases lead to the same determinants as in the diagonalizable case.

$A$ is diagonalizable because its minimal polynomial divides $x^3-x^2-3x+2=(x-2)(x+1/2+\sqrt{5}/2)(x+1/2-\sqrt{5}/2)$, hence it's separable. This is well known. However if we only care about the determinant then we don't need to know that.

Yo are right.

In fact for any equation

\[a_dA^d+\cdots+a_1A+a_0I=0\]

with $n>0$, all solutions are diagonalizable.

Proof:

Let $V $ be the span of all of $A$'s eigenvectors (the closure of the union of it's eigenspaces).

Then since $A^k $ has the same eigenvectors, it is $V $ for every term except possibly the $
a_0I $, and so of the sum of those terms. Since the equation implies that this sum is $-a_0I $, whose single eigenspace is $\mathbb {R}^n$, this implies $V =\mathbb {R}^n $, which is equivalent to $A $ being diagonalizable.

Corollary:

For an $n\times n $ matrix $A $ satisfying
\[a_dA^d+\cdots+a_1A+a_0I=0 ,\]
if there are exactly $k $ distinct roots (ignoring multiplicity) to
\[a_dx^d+\cdots+a_1x+a_0=0,\]
then there are $\binom {n+k-1}{k} $ possible values for $\det (A)$.

You're wrong in both.

The matrix $A^k$ can have more eigenvectors than $A$, take for example $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $k=2$. Also the part about the sum is wrong, think $\begin{pmatrix}0&1\\0&0\end{pmatrix}+\begin{pmatrix}0&-1\\0&0\end{pmatrix}$.

In general, [i]every[/i] square matrix with elements from some field (in fact any commutative ring but let's stick to fields for simplicity's sake) satisfies some polynomial equation. This is easy to see since the dimension of the space of $n\times n$ matrices over the given field is $n^2$ so the matrices $I,A,\ldots,A^{n^2}$ are linearly dependent. However it turns out that even more is true: every such matrix satisfies some polynomial equation of degree $\le n$; this follows from the Cayley-Hamilton theorem. As for diagonalizability, it turns out that $A$ is diagonalizable iff it satisties some [i]separable[/i] polynomial equation (which happens iff its minimal polynomial is separable).

Your corollary isn't correct either. First of all, the formula in the generic case (as in the problem above) is $\binom {n+k-1}{n}=\binom {n+k-1}{k-1}$, not $\binom {n+k-1}{k}$. But this fails in some special cases, for example for the equation $A^2-I=0$ the formula predicts $n+1$ different values of determinant, but there are only two: $\pm1$. However as I've said above, this has nothing to do with diagonalizability, the determinant is always equal to the product of eigenvalues (counted with algebraic multiplicities) no matter whether the matrix is diagonalizable or not.

... and that was with the simplification of implicitly assuming an algebraically complete field. If a $10\times 10$ real matrix $A$ satisfies $(A^2-4A+5I)(A^2-8A+17I)=A^4-12A^3+54A^2-108A+85I=0$, then its determinant can be $5^5$, $5^4\cdot 17$, $5^3\cdot 17^2$, $5^2\cdot 17^3$, $5\cdot 17^4$, or $17^5$, a total of six choices. If you try to apply that formula - well, it's a $10\times 10$ matrix and a polynomial with no (real) roots, so it claims $\binom{9}{-1}$, a far cry from $\binom{6}{1}$.

In the general case, we have to worry about the degrees of the irreducible factors of that polynomial, and we get a count of certain kinds of partitions which cannot easily be reduced to a formula, even without coincidences of equal determinants in different ways.

I think I see your point. Thank's for the explanation.

[quote=loup blanc]Let $ a\equal{}\frac{\minus{}1\minus{}\sqrt{5}}{2},b\equal{}\frac{\minus{}1\plus{}\sqrt{5}}{2}$; $ A$ is diagonalizable, then $ det(A)\equal{}2^{\alpha}a^{\beta}b^{\gamma}$ with $ \alpha\plus{}\beta\plus{}\gamma\equal{}n$ and $ \alpha,\beta,\gamma\in\mathbb{N}$.
The function $ (\beta,\gamma)\in\mathbb{N}^{2}\rightarrow{2^{n\minus{}\beta\minus{}\gamma}a^{\beta}b^{\gamma}}$ is one to one.
Proof: assume that $ a^{\beta}b^{\gamma}\equal{}2^{\beta\plus{}\gamma}$ with $ \beta,\gamma\in\mathbb{Z}$. $ ab\equal{}\minus{}1$ then $ |a^{\beta}b^{\gamma}|\equal{}b^{\gamma\minus{}\beta}$; this last number is rational only if $ \beta\equal{}\gamma$. Therefore $ (\minus{}1)^{\beta}\equal{}2^{2\beta}$ and $ \beta\equal{}\gamma\equal{}0$. QED.
Conclusion: there exist $ (n\plus{}1)\plus{}n\plus{}\cdots\plus{}1\equal{}(n\plus{}2)(n\plus{}1)/2$ possible values for $ det(A)$.[/quote]

Nice


algebra
polynomial
linear algebra
linear algebra unsolved

$ A,B\in M_n(R)$ ,($ n\ge2$) and $ \lambda_0 , \lambda_1,..,\lambda_n$are distincts real numbers such that :
1) $ A$ is diagonalizable.
2)$ \forall i\in${$ 0,1..,n$} we have : $ rank(A\plus{}\lambda_i B) \le k$
Proof that  $ rank(A)\le k$

no one ?!!!  :blush: i thought it will be very easy for you mathlinkers

$ A$ diagonalizable is useless.
We assume $ rank(A)>k$. Let $ t\in\mathbb{R}$; we fix a submatrix $ S_t$ of $ A\plus{}tB$ of dimension $ k\plus{}1$.
Let $ f: t\in\mathbb{R}\rightarrow{det(S_t)}$. $ f$ is a polynomial of degree at most $ n$ that have $ n\plus{}1$ roots: $ \{\lambda_0,\cdots,\lambda_n\}$; then, for all $ t$, $ det(S_t)\equal{}0$. In particular $ det(S_0)\equal{}0$ where $ S_0$ is a submatrix of $ A$ of dimension $ k\plus{}1$.
Conclusion: $ rank(A)\leq{k}$.


The number of real solutions of the equation

\[ |x-2| + |x-3| = 1 \]

is

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{more than 3}$

[hide]
You can easily deduce from the equations that $x=2$ and $x=3$ work.  If $2<x<3$, canceling out the absolute values yields $x-2-x+3=1$, which is true.  So there are more than 3 solutions.  $\boxed{E}$
[/hide]

[quote="tz"]Do the answers have to be integers?[/quote]
No, it asks for the number of real solutions. ;)


The table shows the number of days per week households eat their dinner together.  What percent of households eat their dinner together at least once a week?

$ \begin{tabular}{|c|c|}
\multicolumn{2}{c}{}\\\hline
7 days&40\%\\\hline
5\minus{}6 days&30\%\\\hline
1\minus{}4 days&20\%\\\hline
0 days&10\%\\\hline
\end{tabular}$

[quote="GameBot"]The table shows the number of days per week households eat their dinner together.  What percent of households eat their dinner together at least once a week?

$ \begin{tabular}{|c|c|} \multicolumn{2}{c}{} \\
\hline 7 days & 40\% \\
\hline 5 \minus{} 6 days & 30\% \\
\hline 1 \minus{} 4 days & 20\% \\
\hline 0 days & 10\% \\
\hline \end{tabular}$[/quote]

Our answer is simply $ 20 \plus{} 30 \plus{} 40 \equal{} 90$ percent.


calculus
integration
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floor function

Let's say we have a1+a2+a3...+am=n.  a is an addend, m is the last addend (or also the total amount of addends), and n is any positive integral sum of the addends.  Assume that all of the addends are nonnegative integers.

I was told the useful formula that the amount of ordered pairs, triples, or however many addends there are that add up to a certain sum is as follows, where each addend is a positive integer (so the addends can't be zero):
(n-1)C(m-1).
This formula is true.

However, now I ask for the formula that gives the same information as the previous formula, except that the addends CAN be zero along with positive integers.  Sorry that I can't use LaTeX yet; I'll learn soon.

Does anyone know this formula?

no... I'm just explaining what the answer would be if people could score 0 as well....

[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=634]T0rajir0u's Blog[/url]
Yes. He is a genius.

I know that this is a type of balls and urns problem, but sticking back to my original question, can someone tell me the formula for the case that I described?

I understand your thinking, ac-king.  However, if the problem was a bit more complex, involving four or five numbers, it would be harder to go case-by-case and easier to go with a single formula.  But thanks anyways.

This is balls and urns.
Say you have x objects, and you want to have them given to y people. The formula is:
$ \binom{x\plus{}y}{y\minus{}1}$
I think.

If you simply know the concept, you should get it anyways. And this is a generalized balls and urns, it's not a type.

You want your case: $ \binom{n \plus{} m \minus{} 1}{n} \equal{} \binom{n \plus{} m \minus{} 1}{m \minus{} 1}$.

[quote="zolojetto"][quote="unimpossible"]this is called balls and urns.[/quote]

I know i sound stupid but i have never understood the balls and urns, could someone explain???[/quote]

Basically, balls and urns is a way to divide $ n$ things into $ k$ groups. This is done by having the $ n$ objects and $ k\minus{}1$ dividers(for example, if there are 2 groups, you only need 1 divider). To calculate how many arrangements of these objects, take

$ \dbinom{n\plus{}k\minus{}1}{k\minus{}1}$ to choose how many places the dividers can go.

[quote="ac-king"]idk the formula.. but...

[hide]"During football season, 25 teams are ranked by three reporters (Alice, Bob and Cecil). Each reporter assigned all 25 integers (1 through 25) when ranking the twenty-five teams. A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting one point for a 25th place ranking. The Hedgehogs earned 27 total points from the three reporters. How many different ways could the three reporters have assigned their rankings for the Hedgehogs? One such way to be included is Alice - 14th place, Bob - 17th place and Cecil - 20th place."

This is an example that was taken from a previous post.... Now let's assume that you can score 0 too!!!!
According to your formula... if it was only 1-25 that can be scored... then  it would be....
26 choose 2 or 325... yay!!! It's right....
Now... you can just find all the possibilities where there is one zero (there can't be two zeros or else it won't be able to add up to 27...)
So, it's all the possibilites where 2 people can score 27....
That would be.... 26 choose 1 or 26!!!! Now, you have to also include the ones that aren't possible: such as 1,26.... that takes away 2 possibilities...
so the total amount is 325+24=349!!!!
Hope this helps (it should be right :P)[/hide][/quote]

Using the balls and urns concept, $ \dbinom{25\plus{}3\minus{}1}{3\minus{}1} \equal{}$ $ 27\choose2$ $ \equal{} 351$, not 349. Am I misunderstanding something here?

Where are you getting the 27?
Edit: Ah, thanks, I didn't fully read the problem...

The team scored 27, so you need the numbers to sum to that.

more important is that you understand the formula.  some problems require a clever use of the idea.

[quote="the future"]Yea try this

How many ways, using pennies, nickels, and dimes are there to get 50 cents?[/quote]

don't see how balls and urns can be applied to this...

[quote="alanchou"]more important is that you understand the formula.  [b]some problems require a clever use of the idea.[/b][/quote]
THat problem does

Well If I saw that problem I would just immediately start listing.

[quote="the future"]Yea try this

How many ways, using pennies, nickels, and dimes are there to get 50 cents?[/quote]

There's a way to do this with generating functions that my friend told me, but even with that method, the problem would be extremely long.  I don't see any way to use balls and urns apart from just listing systematically, which doesn't take too long.  The future, if you know a method with balls and urns, can you post it?

Hmm...  Are you sure this method works, because if it does, that would be a lot of help for me.  Can anyone help confirm this?  Thanx! :D

[quote="math154"]i listed the first few out and i kind of came with a pattern--i think it works, but i'm not sure.
1-4 cents:1 way
5-9 cents:2 ways
10+ cents:3+[n/5]([n/5]-1)/2 ways where [n/5] is the greatest integer less than n/5[/quote]
[hide="Lol, so it's been a while"]I was bored last year so I decided to do this stuff. I didn't post this last year since I gave it to a MCer who is now in high school.

Anyway, we can find a formula by starting out with the problem of just pennies, nickels, and dimes. This is equivalent to finding the number of nonnegative solutions of $a+5b+10c=n$. Let the number of solutions be $f(n)$. Also let (by the division algorithm) $n=5\lfloor{n/5}\rfloor+k$ where $0\le k<5$. Then considering modulo five, $a=5a'+k$, so $a'+b=\lfloor{n/5}\rfloor-2c$, which has $\lfloor{n/5}\rfloor-2c+1$ solutions in nonnegative integers as long as $\lfloor{n/5}\rfloor\ge2c$ and of course $c\ge0$. But
\[\sum_{c=0}^{j}\left(\left\lfloor\frac{n}{5}\right\rfloor-2c+1\right)=(j+1)\left(\left\lfloor\frac{n}{5}\right\rfloor-j+1\right).\]If $\lfloor{n/5}\rfloor=2k$ is even (i.e. $n=10k,10k+1,10k+2,10k+3,10k+4$), then $j=k$ so the number of solutions is $(k+1)^2$. If $\lfloor{n/5}\rfloor=2k+1$ is odd (i.e. $n=10k+5,10k+6,10k+7,10k+8,10k+9$), then $ j=k$ so the number of solutions is $(k+1)(k+2)$. With floor functions it's not difficult to see that the number of solutions to this problem in general is $\lfloor(\lfloor{n/5}\rfloor/2+1)^2\rfloor$.

With quarters as well, we are finding the number $g(n)$ of solutions to $a+5b+10c+25d=n$, or $a+5b+10c=n-25d$. Then we see that
\[g(n)=\sum_{d=0}^{\lfloor{n/25}\rfloor}f(n-25d).\]With some casework, we find that (note $g(5j)=g(5j+1)=g(5j+2)=g(5j+3)=g(5j+4)$)
\begin{align*}
g(50k)&=\frac{1}{6}(4k+3)(5k+1)(5k+2)&=&\binom{k+3}{3}+45\binom{k+2}{3}+52\binom{k+1}{3}+2\binom{k}{3}\\
g(50k+5)&=\frac{1}{6}(4k+1)(5k+3)(5k+4)&=&2\binom{k+3}{3}+52\binom{k+2}{3}+45\binom{k+1}{3}+\binom{k}{3}\\
g(50k+10)&=\frac{1}{6}(k+1)(100k^2+95k+24)&=&4\binom{k+3}{3}+57\binom{k+2}{3}+39\binom{k+1}{3}\\
g(50k+15)&=\frac{1}{6}(k+1)(5k+4)(20k+9)&=&6\binom{k+3}{3}+63\binom{k+2}{3}+31\binom{k+1}{3}\\
g(50k+20)&=\frac{1}{6}(k+1)(100k^2+155k+54)&=&9\binom{k+3}{3}+67\binom{k+2}{3}+24\binom{k+1}{3}\\
g(50k+25)&=\frac{1}{6}(k+1)(5k+6)(20k+13)&=&13\binom{k+3}{3}+69\binom{k+2}{3}+18\binom{k+1}{3}\\
g(50k+30)&=\frac{1}{6}(k+1)(5k+4)(20k+27)&=&18\binom{k+3}{3}+69\binom{k+2}{3}+13\binom{k+1}{3}\\
g(50k+35)&=\frac{1}{6}(k+1)(100k^2+245 k+144)&=&24\binom{k+3}{3}+67\binom{k+2}{3}+9\binom{k+1}{3}\\
g(50k+40)&=\frac{1}{6}(k+1)(5k+6)(20k+31)&=&31\binom{k+3}{3}+63\binom{k+2}{3}+6\binom{k+1}{3}\\
g(50k+45)&=\frac{1}{6}(k+1)(100k^2+305k+234)&=&39\binom{k+3}{3}+57\binom{k+2}{3}+4\binom{k+1}{3}.
\end{align*}Note that the coefficients of the binomial coefficients appear in the generating function
\[(1+x+\cdots+x^{48}+x^{49})(1+x^5+\cdots+x^{40}+x^{45})(1+x^{10}+x^{20}+x^{30}+x^{40})(1+x^{25}).\]With a similar method we would be able to do this with half dollars as well, but it's a little more tedious, obviously. An explanation of the binomial coefficient representations is found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=866422384&t=161837]here[/url].

In fact, we can get a closed form which is not of much use except for programming and computation given that you have the formula. Indeed, let $ k=\lfloor{n/50}\rfloor$ and $ j=\lfloor{n/5-10k}\rfloor$. Define $c=2$ if $j=0$, $c=4$ if $j=1$, and $c=j^2-j+6$ otherwise. Then
\[g(n)=\frac{1}{6}(100k^3+15(2j+9)k^2+(3j^2+27j+53)k+3c).\]
[b]Edit:[/b] A cleaner way to define $c$ is
\[j^2-j+6-2\left\lfloor\frac{2}{j+1}\right\rfloor.\][/hide]

[quote="resurrection"]Well actually, assuming that their scores can be 0, it should be $ \binom{27\plus{}3\minus{}1}{2}\equal{}\frac{29}{2}\equal{}29*14\equal{}406$

Otherwise, if the scores can't be 0, we add one to $ a,b,c$ to obtain $ a\plus{}b\plus{}c\equal{}24$. So $ \binom{26}{2}\equal{}325$[/quote]

Okay ... let's do this one correctly first:
"During football season, 25 teams are ranked by three reporters (Alice, Bob and Cecil). Each reporter assigned all 25 integers (1 through 25) when ranking the twenty-five teams. A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting one point for a 25th place ranking. The Hedgehogs earned 27 total points from the three reporters. How many different ways could the three reporters have assigned their rankings for the Hedgehogs? One such way to be included is Alice - 14th place, Bob - 17th place and Cecil - 20th place."

1st:  Assuming that the Hedgehogs can be unranked (receiving 0 points ... which is not what the problem is asking) we consider the 27 points as "balls" and the three reporters as "urns".  We need two dividers to separate the balls into three urns ... for example, oo|ooooo|oooooooooooooooooooo represents 2 points, 5 points, and 20 points awarded.  It first seems like there are $\binom{27}2=406$ ways this can be done, but some of those ways are impossible ... 27-0-0 is not possible (nor is 0-27-0 or 0-0-27) because you cannot get 27 points from a single judge ... so we eliminate these cases.  Additionally, there are 6 ways to arrange the impossible 26-1-0.  Eliminate these nine cases and we have $\boxed{397}$ ways to get 27 total points from the three judges, where a judge may assign a score of 0 through 25.

2nd:  The intended problem is much easier ... consider again the "balls" as points, separated by two dividers |.  Each divider must be placed in one of the 26 spaces between two points ... as shown below, we need to put 2 dividers | into the gaps represented by the 26 blanks below.
o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o

So there are $\binom{26}2= \boxed{325}$ ways that this can be done.

I stink with the formulas ... if you understand the idea you will have no difficulty solving these sorts of problems without a formula.  I'll look at the coins one a bit later and see if there is a nice way to approach it ... but I am pretty certain there is not a "balls-and-urns" approach to that one.

[quote="the future"]Yea try this

How many ways, using pennies, nickels, and dimes are there to get 50 cents?[/quote]

Okay ... clearly not a balls-and-urns problem ...

Consider the much easier problem:  How many ways, using nickels, and dimes are there to get 50 cents?
We can list them easily (I am assuming that you don't need at least one dime or nickel):
DDDDD
NNDDDD
NNNNDDD
NNNNNNDD
NNNNNNNND
NNNNNNNNNN
=6 ways.

Now, throw pennies into the mix and it gets annoying.  Any nickel can be replaced with 5 pennies, and we can only have a number of pennies that is divisible by 5.  We don't worry about the dimes ... the dimes fix our 6 cases to 5, 4, 3, 2, 1, and 0 dimes.

This gives us:
DDDDD (1 way ... 0 pennies)
NNDDDD (3 ways ... 0, 5, or 10 pennies)
NNNNDDD (5 ways)
NNNNNNDD (7 ways)
NNNNNNNND (9 ways)
NNNNNNNNNN (11 ways)

$1+3+5+7+9+11 = 6^2=\boxed{36}$ ways.

Now ... copeland spent some time with me explaining the way to use a generating function to solve this ... I will not try to reconstruct what he just explained as I will butcher it, but ... the conclusion was:

# of ways to make 0-4 cents:  1
# of ways to make 5-9 cents:  2
# of ways to make 10-14 cents:  4
# of ways to make 15-19 cents:  6
# of ways to make 20-24 cents:  9
...
the pattern 1, 2, 4, 6, 9, 12, 16, 20, is
$1^2, 1\cdot2, 2^2, 2\cdot3, 3^2,$ etc. more [url=http://www.research.att.com/~njas/sequences/index.html?q=1%2C+2%2C+4%2C+6%2C+9%2C+12%2C+16&language=english&go=Search]here[/url].

Sorry to be reviving, but does this pattern happen when you can only use pennies, nickels, dimes, and quarters?


How would I make four circles tangent (not look-like, really tangent) using GeoGebra if I know the radius length of each circle?

Thanks!!


Is there any site that I can find British Math Olympiad solutions?

I know their site has a lot of problems but no answers or solutions (although I like solution way better.)

Thanks.

[quote="orl"]Patience is a virtue as Valentin prefers to say. You will find them on ML soon.  :)[/quote]

Yay.. :lol:

I searched the forum and found like two problems that had answers for British Math Olympiad but I wasn't sure if I did the search right.  :roll:

Thanks for the response.


limit
inequalities
real analysis
real analysis unsolved

maybe $x\to+oo$ ???
answer is $2e$ by taylor
use substitution: $x=\frac{1}{y}$
then you will get
$e\frac{e^{3y}-e^{y}}{y}\to 2e$

Extremal , you are wrong

Using $\lim_{n\to\infty}(1+\frac{1}{x})^{x}=e$ , we get that answer is $e$ too .

yes I agree with you .  :blush:
when I use Taylor then I forgot something
limit is really equal to $e$
because our limit is equivalent to the next:
$x(e^{(x+2)(2/x-2/x^{2})-(x+1)(1/x-1/2x^{2})}-e^{(x+1)(1/x-1/2x^{2})})\to e$

An analogous formula

${f(n)=n^{n}}$
${\lim_{n\to \infty }\, \left(\frac{f(n+3)}{f(n+2)}-\frac{f(n+2)}{f(n+1)}\right)}$

For larges n's (by stirling approximation)

${f(n)=n^{n}\approx \frac{e^{n}n!}{\sqrt{n}\sqrt{2 \pi }}=g(n)}$

${\frac{g(n+3)}{g(n+2)}-\frac{g(n+2)}{g(n+1)}=\frac{e \sqrt{n+2}(n+3)!}{\sqrt{n+3}(n+2)!}-\frac{e \sqrt{n+1}(n+2)!}{\sqrt{n+2}(n+1)!}=e \sqrt{n+2}\left(\sqrt{n+3}-\sqrt{n+1}\right)}$

${\lim_{n\to \infty }\, \sqrt{n+2}\left(\sqrt{n+3}-\sqrt{n+1}\right)=1}$

So the limit is $e$

is it fine to do that, no?

this just shows that if limit exists, it's $e$.
you need to show now that limit actually exists, and then you're done :)

for what limit? the last i posted? or the first?

By Squeeze theorem

${\left(\frac{g(n+3)}{g(n+2)}-\frac{g(n+2)}{g(n+1)}\right)^{\frac{n-1}{n}}\leq \left(\frac{f(n+3)}{f(n+2)}-\frac{f(n+2)}{f(n+1)}\right)\leq \left(\frac{g(n+3)}{g(n+2)}-\frac{g(n+2)}{g(n+1)}\right)}$

but the first limit is a little bit worst than the third :)

but is solvable in the same way

[quote="Love U"]Calculate this

$ \lim_{n\to+\infty}\left(\frac{(x+2)^{x+2}}{(x+1)^{x+1}}-\frac{(x+1)^{x+1}}{x^{x}}\right)$[/quote]
[hide="solution :P"]
Let $ x\in\mathbb{N}$, then let $ e_{x}=\left(1+\frac{1}{x}\right)^{x}$, ofc $ \lim_{x}e_{x}=e$, then we have:
$ \lim_{x\to\infty}\bigg[(x+2)e_{x+1}-(x+1)e_{x}\bigg]=\lim_{x\to\infty}\bigg[ x(e_{x+1}-e_{x})+2e_{x+1}-e_{x}\bigg]=$
And now:
$ 0<e_{x+1}-e_{x}<\frac{2}{x^{2}}$ (it isn't so hard to prove the right inequality), so we have:
$ =0+2e-e=e$

It's can be done for $ x\in\mathbb{R}$, but for $ x\in\mathbb{N}$ it's easier to prove everyting, for $ x\in\mathbb{R}$ we'll have bit more to calculate.
Q.E.D. [/hide]


Sa se rezolve in nr.  reale sistemul:
x^2+y^2=4
2^(x^4+y^2)+2^(y^4+x^2)=128 :(

Din a doua relatie se obtine cu inegalitatea mediilor ca $x^{4}+y^{4}\leq 8$. Pe de alta parte, $x^{4}+y^{4}\geq \frac12(x^{2}+y^{2})^{2}=8$. Sistemul de rezolvat devine deci \[\left\{\begin{array}{l}x^{2}+y^{2}=4\\ x^{4}+y^{4}=8 \end{array}\right.\,,\] care are multimea solutiilor $S=\{(\pm\sqrt{2},\pm\sqrt{2})\}$.


A Fibonacci-like sequence is defined as a sequence where the first term
is $ a$, the second term is $ b$, and all subsequent terms are equal to the sum of the two previous terms. The sum of any ten consecutive terms in a Fibonacci-like sequence equals eleven times the seventh term of the ten. In a Fibonacci-like sequence with first term $ a$ and second term $ b$, the sum of the first ten terms is $ 11n$. Express $ n$ in terms of $ a$ and $ b$.

The terms are: a, b, a+b, a+2b, 2a+3b, 3a+5b, 5a+8b, .....etc
We only need the 7th term to use the given fact of 10 consec. terms sum = 11(7th term).
11(5a+8b) is the sum of the series, or 11n. So, $ n \equal{} \boxed{5a\plus{}8b}$


A number is increased by 32. The new number is one-third of
$ 60\%$ of the original number. What was the original number?

$ x \plus{} 32 \equal{} \frac {1}{3}\times{\frac {3}{5}}\times{x} \implies x \plus{} 32 \equal{} x(\frac {1}{5}) \implies ( \minus{} \frac {4}{5})(x) \equal{} 32 \implies x \equal{} \boxed{ \minus{} 40}$


What is the least four-digit positive integer, with all different digits, that is divisible by each of its digits?

But how did we arrive at this number?

Clearly the first digit is 1.

To minimize we make the 2nd digit 2 and the 3rd digit 3.

Thus we have 123x

For 123x to be divisible by 1,2, and 3, it must be divisible by 6. Thus x=0,6

1230 is not divisible by 0, but 1236 is divisble by 6, so it works.


Is anyone participating in USAMTS this year?
I might, but not completely sure...

Yes, I am doing it.

Last year I started in round 3 two days before they were due and got 12 points, but in round 4 I got 25 points.

I don't think I can even get an honorable mention this year :( But it never hurts to try.

I'm doing it, even though I'm not in Ohio right now...

I'm doing it, but it's taking a while, considering that I started learning LaTeX yesterday, but...yeah.


search
number theory proposed
number theory

Solve this equation in natural numbers

$ x^2 \plus{} y^2 \equal{} 1997(x\minus{}y)$

http://www.mathlinks.ro/viewtopic.php?search_id=1326770934&t=239405
http://www.mathlinks.ro/viewtopic.php?search_id=472821406&t=317045
 :)
search: bulgaria 1998


quadratics
algebra
quadratic formula
absolute value

If m and n are roots of 2x^2 - 5x + 1 = 0, what is the value of
|m - n|?

A. 17/4
B. sqrt(17)/2 or -sqrt(17)/2
C. -sqrt(17)/2
D. sqrt(17)/2

[hide]

quadratic formula gives the roots as

( 5 + :sqrt: (17) ) / 4 and ( 5 - :sqrt: (17) ) / 4

subtract the two gives ( 2 :sqrt: (17) ) / 4 and take absolute value, which is  :sqrt: (17) / 2

D

[/hide]

That is correct.

Now, can you find another way to do this?

Good Job!

Personally, I like the sums and products, but the main thing is to use whatever method suits you best.

Really, you know the discriminant will be the only thing left (other stuff cancels), and the sign doesn't matter, so all you need is $ \frac {\sqrt{b^2 \minus{} 4ac}}{2a}$, which happens to be $ \boxed{\frac {\sqrt {17}}{4}}$.


how come math jams dont have the math symbols to choose from on the side like the forums. it would hurry things up. i may be out of the loop or somat on why it may not work, cause i dont know how to program. hmmmmmmm. i really should read that arml math jam. but ive been up for more than 48 hours. but i must post. aaaaaargh

Part of the reason is that it would be pretty difficult to port that over to the transcripts (which are just html) - if we had a bunch of little math smilies all over the chat room, we'd have to convert them in HTML, which would be a nightmare.  If we didn't convert them, the transcripts would be nearly unreadable.

ahhhh, but how about the mathematical symbols. the smilies "we no need". at least i dont

that would result in the same thing, I think.

-interesting_move

Notice that the math symbols fall in the same category as smilies - they're all emoticons...


Can someone please explain to me the concept of mod and how we use it in math problems? i don't have a specific math problem though...unless you can show me one that needs mod. :D

Think of it as "clock arithmetic".  When you write the expression $ 15\equiv 6 (\mathrm{mod} 9)$, this is read as "15 is equivalent to 6, modulo 9".  What does that mean?  It means that both 15 and 6 have the same remainder when divided by 9.  It also means that 9 divides 15-6.  Frequently, mods are used in number theory problems, and problems that involve finding remainders, primes, and last digits.

Modulos (mods for short) in general is just the mathematical way to express remainders. For example, $ 5\equiv{2}\pmod{3}$, because it has a remainder of 2 when divided by 3 (@JRav: I thought it was always read as ...congruent to... not equivalent). So for example, 49 and 14 are the same in modulo 7 because they have the same remainder when divided by 7. You can add, subtract, and multiply the congruencies as long as you do it to both sides. Dividing, on the other hand, is a different story. It can be done, but not necessarily (because you can't be sure if the result of the division is an integer). Lots of problems use modulos; for example, in order to prove the divisibility rule for 3 and 9, you observe that 10 is congruent to 1 in modulo 3 and 9. Another example is finding the smallest positive integer that has a remainder of 2 when divided by 3, a remainder of 1 when divided by 5, and a remainder of 3 when divided by 7.

First of all, no one yet has even given you the definition of mod.  Mod is indeed like remainder, but they are not the same (note the difference in negative numbers).

$ a\equiv b\pmod{n}\Leftrightarrow\frac{a\minus{}b}n\in\mathbb{Z}$

In other words, $ \frac{a\minus{}b}n$ is an integer.

That is basically all that mods are; however, there are very useful to use in solving number theory problems.

For example, $ a^p\equiv a\pmod{p}$ and $ (p\minus{}1)!\equiv \minus{}1\pmod{p}$ for any prime $ p$.

Nerd_of_the_Ages, it is "congruent".  My mistake.  The $ \text{\LaTeX}$ syntax was still in my head.

[quote="igiul"]Even better than all of those links that aren't user friendly  :P  I have a practical application for you. Sometimes when my teacher gives assignments in math he says 'every other odd' but doesn't know if the last number is in the sequence. (Don't be too critical though, he just working on the fly)

Well? If we begin at a and end on b then $ a\equiv b\pmod{4}$[/quote]

I'm not quite sure how that explains mods, but whatever...

Also, if you want some sample problems, try this one (if you can solve it, then you probably understand mods):

Find all positive integers, $ n$ such that $ n^6\plus{}2n^4\plus{}3n^2\minus{}2n\plus{}2$ is a multiple of $ n\plus{}1$.

Or this one:

Prove that $ 100,003$ cannot be written as the sum of 2 perfect squares.

[quote="b-flat"][quote="igiul"]Even better than all of those links that aren't user friendly  :P  I have a practical application for you. Sometimes when my teacher gives assignments in math he says 'every other odd' but doesn't know if the last number is in the sequence. (Don't be too critical though, he just working on the fly)

Well? If we begin at a and end on b then $ a\equiv b\pmod{4}$[/quote]

I'm not quite sure how that explains mods, but whatever...

Also, if you want some sample problems, try this one (if you can solve it, then you probably understand mods):

Find all positive integers, $ n$ such that $ n^6 \plus{} 2n^4 \plus{} 3n^2 \minus{} 2n \plus{} 2$ is a multiple of $ n \plus{} 1$.

Or this one:

Prove that $ 100,003$ cannot be written as the sum of 2 perfect squares.[/quote]

For a person thats only known about mods since yesterday, this looks pretty hard. Try easier examples. Like try to find and prove the divisibility rule for 11.

[quote="Nerd_of_the_Ages"]For a person thats only known about mods since yesterday, this looks pretty hard. Try easier examples. Like try to find and prove the divisibility rule for 11.[/quote]

Actually, these are fairly basic problems; easier than proving the divisibility rule for 11, imo.


Let $p(x)$ be a polynomial with integer coefficients.
If $p(0)$ and $p(1)$ are odd
then show that $p(x)$ does not have any integer root.!

 :)

p(0) = last coefficient odd
p(1) = sum of coefficients odd
so the other coefficients should be even in sum, so any integer has p(n) odd, and therefor <> 0?

The sum of the coefficents as well as the constant coefficent is odd.

Suppose $f(x) = \sum_{i=0}^n a_ix^i$ and that x is an integer root.

If x is even, then $f(x) = \sum_{i=0}^n a_ix^i = (\sum_{i=1}^n a_ix^i) + a_0$ which is odd.  (every term in the brackets has a factor x which is even.)

If x is odd, then $f(x) =  (\sum_{i=1}^n a_ix^i) + a_0$ is odd, (because, if $a_i$ is odd, then $a_ix^i$ is odd, and if $a_i$ is even, then $a_i x^i$ is even.  So the parity of the stuff in the brackets is the same as the parity of $\sum_{i=1}^n a_i$, which is even.  So an even + $a_0$ is odd as required.)

Contradiction.

$f(ax+b) \equiv f(b) \pmod a$ (basic properties of congruences)
So
$f(2x) \equiv f(0) \equiv 1 \pmod 2$
$f(2x+1) \equiv f(1) \equiv 1 \pmod 2$
So f(x) is odd for all integer x.  But 0 is even.  Contradiction.  So there are no integer solutions.


[quote="Ngông"]$a,b,c,d,e \geq 0$ and $a+b+c+d+e=1$.
Find max of $f(a,b,c,d,e) = abc+bcd+cde+dea+eab$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112980

I didn't see any solution there. I don't know Russian.

sorry for my stupidness, but can you write it more clearly?

you can see it at phan thanh viet' e_book (diendantoanhoc)
it is " phuong phap don bien "


abstract algebra
superior algebra
superior algebra unsolved

Assume that a group $ G$ acts transitively on each of two sets $ \Omega$ and $ \Lambda$.
Prove that the natural induced action of $ G$ on the cartesian product $ \Omega \times \Lambda$ is transitive if and only if $ G_{\alpha}G_{\beta} \equal{} G$ for some choice of $ \alpha \in \Omega$ and $ \beta \in \Lambda$.

As always, $ G_{\alpha} \equal{} \left\{g \in G \mid \alpha^g \equal{} \alpha\right\}$.

I think this is a lot easier to see if instead of letting G act transitively on arbitrary sets, you just take the isomorphic action on cosets.  Let G_a = H, G_b = K, and let G act on the left on the cosets of H and K.

The action on G/H is g -> ( xH -> gxH ) and on G/H x G/K is g -> ( (xH,yK) -> (gxH,gyK) )

For the most part now you just write down the definitions and solve some easy equations.  Sweeping such things under the rug you get:

If the action on G/H x G/K is transitive, then in particular there is some g taking (1H,1K) to (1H,xK) for every x in G.  This means that gH = H and gK - xK, so that g in H and x = gk for some k in K.  That is, x is in HK.  That is G=HK.

In the other direction if G=HK and x,y in G, then G=KH too so write y^-1*x = k*h, and let g = xh^-1 = yk. Then g*(1H,1K) = (gH,gK) = (xh^-1H,ykK) = (xH,yK).  That is, G acts transitively on G/H x G/K.

This is interesting Jack Schmidt, your solution is different from mine for both directions. This is exactly why I post these problems  :)

Just a tiny comment: I think you mean the right cosets, since $ xH \mapsto gxH$ is not an action unless $ G$ is abelian, is it? In this case, it is clear that the action you introduce is indeed equivalent. Nice to know!


Where can I find the shopping cart button? Right now the only way i can get to it is to enroll a class then click the link following.

You only have the "View Cart" button on the sidebar when there is something in your cart.

That's the way it should work; please let us know if that's not happening on your computer.

Sorry, but I have a small problem with the shopping cart and my enrollment.
When I add the class to my shopping cart, I can only view the cart until I log out. When I log back in, I have to re-add it to my cart. So it's like logging out "resets" my shopping cart. I sent an email to classes@artofproblemsolving.com outlining this problem, can someone please help?  :lol:

Thank you


topology
real analysis
real analysis unsolved

How do I show that any open set in R cannot be homeomorphic to a open set in $R^2$

Let $A\subset \mathbb R$ and $B\subset \mathbb R^2$ be the two sets. Any open subinterval of $A$ is mapped homeomorphically onto an open connected subset of $\mathbb R^2$, but this is not possible, since taking a point out of an open interval disconnects it, while taking a point out of an open connected subset of $\mathbb R^2$ does not disconnect it.

Using this argument we can show that any open set in $R^n$ cannot be homeomorphic to a open set in $R^m$ whenever n is not equal to m

I seriously doubt that. You need more involved arguments, relying on algebraic topology, for larger $m$ and $n$, since you can't disconnect sets in these higher-dimensional spaces by removing one point.

Aww is it really very difficult.can you give me an ouline of the proof.


What is the greatest possible value of $ x$ for the equation \[ \left(\frac{4x\minus{}16}{3x\minus{}4}\right)^2\plus{}\left(\frac{4x\minus{}16}{3x\minus{}4}\right)\equal{}12?\]

Let $ y\equal{}\frac{4x\minus{}16}{3x\minus{}4}$. Applying the substitution, we get $ y^2\plus{}y\minus{}12\equal{}0\implies(y\plus{}4)(y\minus{}3)$. By the problem's restraint, $ \frac{4x\minus{}16}{3x\minus{}4}\equal{}\minus{}4$. This gives that $ 4x\minus{}16\equal{}\minus{}12x\plus{}16\implies\boxed{x\equal{}2}$


Find all triplets $(x,y,z) \mathbb \in{N}$ such that

$x+2y+2z=xyz$


i have to write a personal statement because i'm going to aply by ucas to cambridge but i'm not english, i'm italian and i've never written nothing like this. da you know where could i see some examples?

Just visit :
http://www.top-law-schools.com/forums/viewtopic.php?f=18&t=4353


Hello,

Sorry for bad english.

I do getting ready for olympiad mathematics, I have done many so far. I have question from teacher friend and need help. The only questions I have left and cannot possibly solve. Your help thanked

1.
If a,b and c are positive real numbers so that a + b + c = 3... prove;

1/sqrt(a) + 1/sqrt(b) + 1/sqrt(z) is greater or equal to 3

2.
Find all real numbers 'y' so that these equations have common root;
x(x^4 - y)^2    -    x^3(1 - y)^2    +    x^2      -     4    = 0
AND
x^3  +   x^2   +   x   +   y   = 0

3.
If x,y,z form arithmetic sequence, in which x,y,z are real numbers so that 2y^2 + xz = 1,

prove;

1/(x^2  +  1)  ,   1/(y^2   +   1) ,   1/(z^2   +   1)

form arithmetic sequence.

Your help greatly thanked. Programming answer aseptable, process of ansering also appreciated.

Thank You many times.

[hide="1."]
Using the convexity of $\frac{1}{\sqrt{x}}$ and Jensen's, we have

$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}} \ge 3\left(\frac{1}{\sqrt{\frac{a+b+c}{3}}}\right) = 3$.[/hide]

for the 2nd question use the 2nd equation to solve for $y$, plug that mother into the first equation.  then you'll see that the terms $x(x^4-y)^2-x^3(1-y)^2$ actually cancel, so you're left with $x^2-4=0$, so use that and go back to solve for $y$ using 2nd equation.  $(x,y)=(2,-14), (-2,6)$ or something like that.

Thank You all very much, that was very helpful. Very quick reply. Thanks You.

[quote="cats..."]#3 was done here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41436&postorder=asc[/quote]
Also for 1, Cauchy+AM-HM still works here.


inequalities
function
geometry
calculus
integration
logarithms
complex analysis

Let $ f$ be analytic and injective on $ \mathbb{D}$ with $ f'(0) \equal{} 1$ and $ f(0) \equal{} 0$, then
$ |f(z)| \leq \frac{2}{(1\minus{}|z|)^2}$

Here is a somewhat failed attempt, advised to try looking at the $ h$ below by my teacher. Maybe it can be fixed (the constant C does tend to 3 at the boundary, but I don't know how to patch it up).

Note: I tried not making the estimate (*) and plugging in the resulting line integrals with $ 1 \minus{} |a|^2$ instead of $ 1 \minus{} |a|$ in Maple, and you get some gross expression, but I did get that $ |f(z)| \leq \frac {D}{(1 \minus{} |z|)^2}$ for a constant $ D$ independent of $ f$.

Let $ \phi_a(z) \equal{} \frac {a \minus{} z}{1 \minus{} \bar{a}z}$ and let $ h(z) \equal{} \frac {f \circ \phi_a \minus{} f(a)}{f'(a)(|a|^2 \minus{} 1)}$

$ h$ is univalent and satisfies $ h(0) \equal{} 0$ and $ h'(0) \equal{} 1$, and $ h''(0) \equal{} \frac {f''(a)}{f'(a)}(|a|^2 \minus{} 1) \plus{} 2\bar{a}$.  By a version of the area theorem $ |h''(0)| \leq 4$ so that for all $ a \in \mathbb{D}$

$ \left| \frac {f''(a)}{f'(a)} \minus{} \frac {2\bar{a}}{1 \minus{} |a|^2} \right| \leq \frac {4}{1 \minus{} |a|^2}$

Through a poor estimate
$ \left| \frac {f''(a)}{f'(a)} \right| \leq \frac {6}{1 \minus{} |a|^2} \leq (*) \frac {C}{1 \minus{} |a|}$ with the bad $ C \equal{} 6$.

Now then,
$ \ln|f'(z)| \leq \int_0^z \left| \frac {f''(w)}{f'(w)} \right| \, |dw| \leq C \int_0^1 \frac {|z|}{1 \minus{} t|z|} \equal{} C\ln \frac {1}{1 \minus{} |z|}$

And

$ |f(z)| \leq \int_0^z \left|f'(w)\right| \,|dw| \leq \int_0^1 \frac {|z|}{(1 \minus{} t|z|)^C} \, dt \equal{} \frac {1}{(1 \minus{} |z|)^{C \minus{} 1}(C \minus{} 1)} \minus{} \frac {1}{C \minus{} 1}$

When you do it without the $ (*)$ estimate, you get

$ |f(z)| \leq \frac { \minus{} \ln(1 \minus{} |z|)(6|z|^2 \minus{} 12|z| \plus{} 6) \minus{} |z|^3 \minus{} 5|z| \plus{} 10|z|^2}{(1 \minus{} |z|)^2}$
and while I haven't proved this, it seems like the numerator is $ \leq 4|z|$ which is pretty cool. I guess that is pretty enough, especially considering how crudely I did this.


algebra
polynomial
calculus
factoring polynomials

I'm having trouble understand factoring polynomials with rational roots - or rather, how to test the factors using division. For example, factor $ P(x)=2x^{3}+x^{2}-13x+6$. I get the possible rational zeros of $ \pm{1}$, $ \pm{2}$, $ \pm{3}$, $ \pm{\frac{1}{2}}$, $ \pm{\frac{6}{1}}$, and $ \pm{\frac{3}{2}}$. Now, I am testing each by dividing $ P(x)$ by the factors.

This is from Precalculus: Mathematics for Calculus 5th ed. Stewart et al and their posted solution uses $ (x-2)$ as a factor. However, I tested the other ones too and found that ${ (x-\frac{1}{2}})$ is also a factor since the remainder was 0. I verifed this by multiplying it out and getting the original polynomial. So ${ P(x)=2(x-\frac{1}{2}})(x+3)(x-2)$. I'm not sure if I made any mistakes but it doesn't seem like I did. Also, I'm just learning this so I'm not sure if I'm allowed to have a factor with $ c$ being a fraction or not...

So, my question is this: Is there more than one solution? And did the textbook only post one solution? Thanks.

Your polynomial is correct. :D There are three solutions, namely $ \frac{1}{2}$, $ \minus{}3$, and $ 2$. Your book has posted the wrong solution for some strange reason.
Fractions in factors are okay, but using your polynomial as an example: $ 2(x\minus{}\frac{1}{2})(x\plus{}3)(x\minus{}2)\equal{} (2x\minus{}1)(x\plus{}3)(x\minus{}2)$ is also a valid factorization.


calculus
derivative
function
analytic geometry
calculus computations

Hey all,

I have a few questions in the suggested exercise for my course that I am stuck on:

a) Find the extreme values of $ f(x,y,z)\equal{}xyz$ subject to the constraint $ x^2\plus{}2y^2\plus{}3z^2\equal{}6$

So, taking the gradients and setting them equal with the multiplier in there yields the system:

$ yz\equal{}2\lambda x$
$ xz\equal{}4\lambda y$
$ xy\equal{}6\lambda z$

And I am stuck on how to go about solving the system.

b)Find the extreme values of $ f(x,y,z)\equal{}x^2\plus{}y^2\plus{}z^2$ subject to the constraint $ x^4\plus{}y^4\plus{}z^4\equal{}1$

You get the system:

$ x^2\equal{}y^2\equal{}z^2\equal{}\frac{1}{2\lambda}$

And there I thought I was able to solve just by plugging that into the constrating and solving for lambda, but I ended up with the wrong answer, so maybe someone else could help me.

Thanks

by Lagrange multipliers do you mean :
defining the $ L$ such that $ L \equal{} f \minus{} \lambda(x^2 \plus{} 2y^2 \plus{} 3z^2 \minus{} 6)$ and computing the derivative of $ L$ wrt variables $ x,y,z$?
if so then
a)obviously none of $ x,y,z$ are zero
find $ xyz$ as a function of $ \lambda$ which is $ xyz \equal{} 48 \lambda^3$
and then we get $ 24 \lambda^2 \equal{} x^2$ and so on which implies :
$ \lambda \equal{} \sqrt {\frac {1}{12}}$
and finally we'll get the extreme values of $ f(x,y,z) \equal{} xyz$ which are $ \pm 48.(\sqrt {\frac {1}{12}})^3$
b)u didn't consider whether $ x,y,z$ can be zero or not before dividing both sides of the equation $ 2x \minus{} 4 \lambda x^3 \equal{} 0$ by $ x^2$.for example take $ (x,y,z) \equal{} (0,0,1)$
the rest seems trivial

One thing to remember in solving these is that you really don't care what the value of $ \lambda$ is. Hence, it often makes sense to start by eliminating $ \lambda$ between equations.

I'm looking at your first problem. If $ \lambda\equal{}0,$ then either $ y\equal{}0$ or $ z\equal{}0,$ and if either $ y$ or $ z$ is nonzero, then $ x\equal{}0.$ But all of the points for which two out of three of $ x,y,z$ are zero have $ f(x,y,z)\equal{}0,$ and it seems unlikely that that is the max or min.

If $ \lambda\ne 0,$ and at least one of $ x,y,z$ is zero, then it turns out that two of the three are zero anyway, and we don't learn anything new. So now assume that none of $ x,y,z$ are zero.

Rewrite the equations as:

$ 2\lambda \equal{} \frac{yz}{x}$

$ 2\lambda \equal{}\frac{xz}{2y}$

$ 2\lambda \equal{} \frac{xy}{3z}$

Put the first two together: $ \frac{yz}{x}\equal{}\frac{xz}{2y},$ or $ 2y^2z\equal{}x^2z.$ Since we're assuming $ z\ne 0,$ we get that $ 2y^2\equal{}x^2$ or $ y\equal{}\pm\frac{x}{\sqrt{2}}.$

A very similar computation with the first and third equations gets us $ z\equal{}\pm\frac{x}{\sqrt{3}}.$

Now put that in the constraint equation: $ x^2\plus{}\frac{2x^2}{2}\plus{}\frac{3x^2}{3}\equal{}6,$ or $ x^2\equal{}2,$ or $ x\equal{}\pm\sqrt{2}.$

This gets us 8 critical points, one in each octrant: $ \left(\pm \sqrt{2},\pm 1,\pm\sqrt{\frac23}\right).$

The value of $ f(x,y,z)$ at those points is $ \pm\frac{2}{\sqrt{3}}.$ That's positive in the first octant and in the three octants in which exactly one variable is positive, and negative in the other four octants. We have found four local and global maxima and four local and global minima.

One note: the set $ \{(x,y,z)\,|\,x^2\plus{}2y^2\plus{}3z^3\equal{}6\}$ is a compact set (an ellipsoid) and $ f$ is a continuous function. By the Extreme Value Theorem, $ f$ must have a global maximum and global minimum.

(Note: this does agree with shoki's result. I just wanted to offer a different organization.)

Now, (b): As you said, we have $ x^2\equal{}y^2\equal{}z^2\equal{}\frac1{2\lambda}$. Ignore $ \lambda$, and that gives $ x^4\equal{}y^4\equal{}z^4\equal{}\frac13$ and $ x^2\plus{}y^2\plus{}z^2\equal{}3\sqrt{\frac13}\equal{}\sqrt{3}$.

Uh, that's only one value. We should have both a maximum and a minimum- what's going on?

Let's look back at the original equation $ (2x,2y,2z)\equal{}\lambda(4x^3,4y^3,4z^3)$. You divided by $ x$ in the first coordinate- but $ x$ can be zero. The same is true for the other coordinates, which leaves us with more possibilities: $ x^2\equal{}y^2$ and $ z\equal{}0$, or $ y^2\equal{}z^2\equal{}0$ (and the same around each axis). What values do these give?
For the first, $ x^4\equal{}y^4\equal{}\frac12$ and $ x^2\plus{}y^2\plus{}z^2\equal{}2\sqrt{\frac12}\equal{}\sqrt{2}$. For the second, $ x^4\equal{}1$ and $ x^2\plus{}y^2\plus{}z^2\equal{}1$.

The minimum is $ 1$, achieved at the six points where one of $ x,y,z$ is $ \pm 1$ and the others are zero. The maximum is $ \sqrt{3}$, achieved at the eight points where $ x,y,$ and $ z$ are all in $ \pm 3^{\minus{}\frac14}$.