Find all functions $f: \mathbb{R}\to{R}$ such that for any $a<b$, $f([a,b])$ is an interval of length $b-a$

first,it is easy to prove $f$ is continuous.
(just notice $|f(a)-f(b)|\le |a-b|$)
Then fixed $f(0)$,we will get the $f(1)$.
We know that $\{f(x),x \in [0,1]\}$ is closed interval.
and the length of the interval is 1.
We assume:$m=f(a)=\min \{f(x),x \in [0,1]\},M=f(b)=\max \{f(x),x \in [0,1]\}$,And $M-m=1$
So we must have:$|b-a|=1$(because $f([a,b])$ is length 1)
so $b=1,a=0$,or $a=1,b=0$
then we have:$f(1)=f(0)+1$,or,$f(1)=f(0)-1$
Case (1):
$f(1)=f(0)+1$,Now consider $f(x)$,use the similar method we have $f(x)=f(0)+x$  or $f(0)-x$,but also consider $f(1)$ is fixed,then we have:$f(x)=f(1)+x-1=f(0)+x$ or $f(x)=f(1)-x+1=f(0)+2-x$
Thus:$f(x)=f(0)+x$
same discussion with $f(1)=f(0)-1$.
So we proved the only function is:
$f(x)=x+k$ or $f(x)=-x+k$,where $k$ is the fixed real number.

An alternative way to write this down:

Shift the function so that $f(0)=0$. Considering the interval $[0,x]$ for positive $x$, we see that $|f(x)| \le x$, but there must be an $y$ in $[0,x]$ with $|f(y)|=x$. This already gives $f(x)=\pm x$  for each $x$.
If the function takes the positive sign for one $x$, it must also do so for all smaller positive ones. The same argument for the other signs already gives all solutions after excluding the functions $f(x)=\pm |x|$.

another way to think about this problem.

Consider three x-values $a<b<c$. Note that $f([a,c])=f([a,b]) \cup f([b,c])$. Since they are intervals, $|f([a,b]) \cup f([b,c])| \le |f([a,b])|+|f([b,c])|$ with equality iff $f([a,b])$ and $f([b,c])$ intersect in exactly one point or not at all. But $f([a,c]) \le f([a,b])|+|f([b,c])|=(b-a)+(c-b)=c-a$, hence there is equality. Thus  $f([a,b])$ and $f([b,c])$ are disjoint. These intervals share the common point $f(b)$, thus their intersection is a single point.

Suppose that $f(a) \le f(b)$. If $f(c)<f(b)$ then $f([a,b])$ and $f([b,c])$ both contain the interval $[\max\{f(a),f(c)\},f(b)]$, which is clearly not a single point. Hence $f(c) \ge f(b)$.  Thus $f$ is non-decreasing on $[b, +\infty]$ and by the same logic $f$ is non-decreasing on all of $\mathbb{R}$. Hence for all $0<y$, $|f([0,y])|=f(y)-f(0)=y$ thus $f(y)=f(0)+y$. Similarly this holds for all $y \in \mathbb{R}$.

If $f(a) \ge f(b)$ we get that $f$ is non-increasing and by the same logic, $f(y)=-y+f(0)$.


a) Two positive integers are chosen. The sum is revealed to logician Α, and the sum of squares is revealed to logician Β . Both  and  are given this information and the information contained in this sentence. The conversation between  and  goes as follows:  starts

B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` Now I can tell what they are.'

What are the two numbers?

b) When  first says that he cannot tell what the two numbers are,  receives a large amount of information. But when  first says that he cannot tell what the two numbers are,  already knows that  cannot tell what the two numbers are. What good does it do  to listen to ?

Επειδή η διατύπωση στα αγγλικά δν είναι πολύ καλή θα μπορούσες ν τν γράψεις στα ελληνικά σε παρακαλώ........

α) Επιλέγονται δυο θετικοί ακέραιοι αριθμοί. Το άθροισμά τους γνωστοποιείται στον λογικολόγο Α, ενώ το άθροισμα των τετραγώνων τους στον λογικολόγο Β. Και στους δυο δίνεται αυτή η πληροφορία και η πληροφορία που περιέχεται σε αυτήν την πρόταση. Ακολουθεί η συζήτησή τους:

Β: "Δεν είναι δυνατό να γνωρίζω ποιοί είναι."
Α: "Δεν είναι δυνατό να γνωρίζω ποιοί είναι."
Β: "Δεν είναι δυνατό να γνωρίζω ποιοί είναι."
Α: "Δεν είναι δυνατό να γνωρίζω ποιοί είναι."
Β: "Δεν είναι δυνατό να γνωρίζω ποιοί είναι."
Α: "Δεν είναι δυνατό να γνωρίζω ποιοί είναι."
B: "Τώρα είναι δυνατό να γνωρίζω."

Ποιοί είναι οι δυο αριθμοί;

The reasoning must be as follows: Let the numbers be a and b. Then A receives a number c=a+b and B receives a number d=a^2 +b^2. Note that there are only certain numbers that B can receive. Listing them, beginning with the smallest (I assume that a=b is allowed): 2, 5,8, 13,18, etc. If he receives any of the above he will immediately know the solution because they only have one (for example 13 has the solution 2 and 3, and no others). So when B first says that he doesn't know what they are, that means that the equation d=a^2+b^2 must have multiple solutions (at least 2). A then makes a list of all d values that have multiple solutions for a and b and the corresponding sums a+b. He notes if any correspond to the value c=a+b that he has. If he finds only one possibility, he will know the answer, but he must have found more than one because he says he doesn't know. B then makes a list of all numbers that can be expressed as a^2+b^2 in more than one way, and that give the same value for the sum a+b. Again he finds more than one possibility, so he says again that he doesn't know the solution. A repeats what B did, but from the list counts only those that give the same values of a+b. And so on, until only one possibility is left. However I still cannot find which numbers specificly they are.....

Για το ερώτημα β νομιζω οτι πλησιάσα λιγο............Ο Β γνωρίζει οτι ο Α δε μπορεί ν διακρίνει τους δύο αριθμούς χωρίς στοιχεία π ν αφορούν τους αριθμούς του Β. Ωστόσο οταν ο Α το πρωτολεει αυτό, μπορούν ολα τ ζευγάρια των αριθμών α,β των οποίων το αθροισμα ειναι η πληροφορία του Α να είναι ένα μοναδικό άθροισμα τέτοιο ώστε το (α^2)+(β^2) να γράφεται με δύο διαφορετικούς τρόπους . Στην περίπτωση δλδ που ο Α γνωρίζει οτι ο Β δν μπορεί ν διακρίνει ποιοι αριθμοί είναι (και οτι (α^2)+(β^2) μπορεί ν γραφεί ως άθροισμα δύο τετραγώνων με δύο τρόπους) ειναι σ θέση ν μειώσει τους δυνατούς αριθμούς. Το γεγονός οτι ακόμα ο Α δν μπορεί να βρεί τους αριθμούς, ενημερώνεται ο Β οτι δν ειναι αυτή η περίπτωση. Ετσι ο Β αποκομίζει καποιες πληροφορίες.

Kritikos, μάλλον χαζομάρες γράφω αλλα νομίζω ότι στο πρώτο ερώτημα πρεπει ν προθέσεις κάτι (μπορεί ν κάνω κ λάθος :maybe: )......

Wallace δεν χρειάζεται να προσθέσω κάτι έτσι είναι η εκφώνηση.


probability
inequalities
inequalities proposed

Given an unfair 6 sided dice. When the dice is rolled up two times, let $ P$ be the probability such that same number comes out and $ Q$ be the one such that the number of odd in the first and that of even in the second.

(1) Prove that $ P\geq \frac{1}{6}$,  then find the necessary and sufficient condition for which the equality holds.

(2) Prove that $ \frac{1}{4}\geq Q\geq \frac{1}{2}\minus{}\frac{3}{2}P$.

By the condition, there exist 6 positive reals $ a,b,c,d,e,f \in [0,1]$ such that $ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\equal{}1$ and $ P\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}e^{2}\plus{}f^{2}$, $ Q\equal{}(a\plus{}b\plus{}c)(d\plus{}e\plus{}f)$.
(1) =>Cauchy-Schwarz
(2) left =>AM-GM
(3) right=>Apply Cauchy to $ (a\plus{}d)^{2}\plus{}(a\plus{}e)^{2}\plus{}(a\plus{}f)^{2}\plus{}(b\plus{}d)^{2}\plus{}(b\plus{}e)^{2}\plus{}(b\plus{}f)^{2}\plus{}(c\plus{}d)^{2}\plus{}(c\plus{}e)^{2}\plus{}(c\plus{}f)^{2}$


Given $ a, b, c \geq\ 0$. Prove that:
$ \sqrt{\frac{b^2\minus{}bc\plus{}c^2}{a^2\plus{}bc}}\plus{}\sqrt{\frac{a^2\minus{}ab\plus{}b^2}{c^2\plus{}ab}}\plus{}\sqrt{\frac{c^2\minus{}ca\plus{}a^2}{b^2\plus{}ac}}\plus{} \frac{2(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2} \ge 4$
If it's trues, it'll be a strong ineq  :)
@babyloverain: [hide]Trc het anh muon test thu? xem no co dung' hay ko da~, luoi` test qua' :P[/hide]

This inequality is valid, nguoivn and it follows from applying AM-GM as follow:
$ \sqrt{\frac{b^2\minus{}bc\plus{}c^2}{a^2\plus{}bc}} \ge \frac{2(b^2\minus{}bc\plus{}c^2)}{(a^2\plus{}bc)\plus{}(b^2\minus{}bc\plus{}c^2)} \equal{}\frac{2(b^2\minus{}bc\plus{}c^2)}{a^2\plus{}b^2\plus{}c^2},$
;) :)

@[hide]Bài này em chế từ lời giải của em sau khi làm MIC round 1,bài 1 của anh Lộc[/hide]


geometry
geometric transformation
reflection
geometry unsolved

angle of $ \triangle ABC$ is 20,80,80. Let $ X$ be a point on $ CA$ such $ \angle ABX\equal{}20$ and $ Y\equal{}AB\cap CO$. It is well known that $ \angle BXY\equal{}30$. Let $ D$ be reflection of $ X$ wrt $ AB$. Then $ D$ lies on $ AO$ because $ \angle OAB\equal{}\angle CAB$. Since $ \angle BYD\equal{}\angle BYX\equal{}130$,$ \angle BYC\equal{}50$, $ D,Y,C$ are collinear. Hence $ D\equal{}O$. Thereore,$ \angle BOC\equal{}30$.


There is probably a much more elegant way to do this, but this is my way. We compare the terms on the LHS and RHS based on the powers of r. My conjecture is that each of these will hold by AM-GM.

For any i, we will gather all the terms of the form r^i. On the RHS, that is easy. Binomial theorem gives that it is C(n,i)*r^i*g^(n-i). On the LHS, it will be r^i*(n-i of the a terms multiplied together). We cancel the r^i term from each side. Now, on the LHS, there are C(n,i) terms. Within these terms, each a(j) occurs exactly C(n-1,i-1). I believe this is just AM-GM.
Correct?

That's it !
This inequality is not particularly elegant but often comes in handy.

For example, you can prove with this inequality that

(a + 1)(b + 1)(c + 1) :ge: (ab + 1)(bc + 1)(ca + 1)

if a, b, c are positive reals. Try it !

For the last one you posted: we add ineqs of the type (1+a^3)(1+a^3)(1+b^3)>=(1+a^2b)^3 to get what we want. To prove these we use, of course, the ineq you posted first on this topic.

Notice that also
(r + a(1))(r + a(2))...(r + a(n)) :le: (r + a)^n
where a is the arithmetic mean of a(1), a(2), ..., a(n).
This is even easier to prove :)


parameterization
conics
ellipse
trigonometry
calculus
calculus computations

here are the two curves

16x^2 + 25y^2 = 32000 ('z' unrestricted)

and

z = 5.4x - 3.2y - 16

how do i find the equation of the curve i get by the intersection of the above two curves

What do you mean by "the" equation for the curve? You can't express a curve in 3-space with a single equation, the way you can in 2-space. You either need two equations in $x,y,z$ (which you already have) or you need a parameterization. Is that it - do  you want a parameterization?

(This is the intersection of an elliptic cylinder with a tilted plane. The curve will be an ellipse - lying in that tilted plane, of course.)

how do i find the equation of this ellipse? should i just set them equal to each other or something?

Again, "the" equation is the wrong wording. I think you've got a fundamental misunderstanding about curves in 3-space.

About the only way I could think of to indicate this set with a single equation would be something like this:

$\left(16x^{2}+25y^{2}-32000\right)^{2}+(5.4x-3.2y-z-16)^{2}=0$

- and that would be ridiculous.

I think [b]james21[/b] may be looking for something like $\vec r(t)=\langle a\cos t, b\sin t, 5.4a\cos t-3.2b\sin t-16\rangle$, $0\le t\le 2\pi$, where $a=\sqrt{2000}$ and $b=\sqrt{1280}$.


Yeah, a frustum is basically a region of a 3-d object (most commonly cones and pyramids) cut by 2 parallel planes. The question is basically asking for the volume of the pot.

EDIT: Thanks for correcting me, ZhangPeiJin

......
Cylinders?
You do know that cutting a cylinder wouldn't give you a frustum, it would just give you another cylinder?
Unless you meant the other way, which would just give you a section of the cylinder, not a frustum.
Anyways, Bases of 12 and 18.
That means the bases' radii are $ \frac {6}{\pi}$ and $ \frac {9}{\pi}$
And the height is 12 inches.
Now we need to find the height of the original cylinder.
Seeing as the radii went from $ \frac {9}{\pi}$ to$ \frac {6}{\pi}$.
$ \frac {h}{\frac {9}{\pi}} \equal{} \frac {h \minus{} 12}{\frac {6}{\pi}}$

$ \frac {9h \minus{} 108}{\pi} \equal{} \frac {6h}{\pi}$

$ 9h \minus{} 108 \equal{} 6h$
$ 3h \equal{} 108$

$ h \equal{} 36$
Therefore, the original cylinder had radius $ \frac {9}{\pi}$ and height 36, while the cylinder that was taken off had radius $ \frac {6}{\pi}$ and had height 24
Therefore, the area of the frustum is:
$ (\frac {1}{3}(\frac {9}{\pi})^{2}\pi\cdot 36) \minus{} (\frac {1}{3}(\frac {6}{\pi})^{2}\pi\cdot 24)$

$ 12(\frac {81}{\pi}) \minus{} 8(\frac {36}{\pi})$

$ \frac {972 \minus{} 288}{\pi}$

$ \frac {684}{\pi}$
And I still think I probably did something wrong, but I can't check my work because I have an English project due tomorrow...
That I didn't start yet.

Zhang Peijin is correct with the answer, I just checked it.

[quote="ZhangPeijin"]$ (\frac {1}{3}(\frac {9}{\pi})^{2}\pi\cdot 36) \minus{} (\frac {1}{3}(\frac {6}{\pi})^{2}\pi\cdot 24) 12(\frac {81}{\pi}) \minus{} 8(\frac {36}{\pi})$[/quote]

You can make your parentheses look better by doing "\left(blablabla\right)"

[quote="ZhangPeijin"]
And the height is 12 inches.
Now we need to find the height of the original cylinder.
Seeing as the radii went from $ \frac {9}{\pi}$ to$ \frac {6}{\pi}$.
$ \frac {h}{\frac {9}{\pi}} \equal{} \frac {h \minus{} 12}{\frac {6}{\pi}}$

$ \frac {9h \minus{} 108}{\pi} \equal{} \frac {6h}{\pi}$

$ 9h \minus{} 108 \equal{} 6h$
$ 3h \equal{} 108$

$ h \equal{} 36$
[/quote]
[hide="Simpler way for this part"]
Every 12 inches the height increases, the circumference decreases by 6 inches (this is for the cone, not the flowerpot).  For the circumference to equal 0 (since a point has no circumference), the height is $ \frac{18}{6}\times 12 \equal{} 36$ inches.  [/hide]

yeah. The pies get really messy if you work with them too much.

why are you guys talking about cylinders if it is shaped like the frustum of a.
"right cylinder cone?"

I got a completely different answer :(
(702/pi)

[quote="dingzhou"]dingzhou decided to buy a flower pot to hold his worms, the flower pot has the shape of a frustum of a right circular cone. the circumference of the bases are 12 inches and 18 inches, the height is 12 inches. How many cubic inches of dirt should dingzhou buy to fill up the pot? dingzhou want you to express your answer in terms of $ \pi$.[/quote]

yeah dude, it says right circular cone...

and I got... $ \frac{648}{\pi}$

[quote="Ihatepie"]why are you guys talking about cylinders if it is shaped like the frustum of a.
"right cylinder cone?"[/quote]

The reason is because I was defining a frustum (badly) and accidentally put cylinder. Then ZhangPeijin corrected me right before he wrote the solution. So that's why he was putting cylinder instead of cone...but the answer should have been right??

yep, the answer is right, good job! :thumbup:  :thumbup:


Show that for reals $ x,y,z$ which are not all positive,

$ \frac{16}{9}\left(x^{2}\minus{}x\plus{}1\right)\left(y^{2}\minus{}y\plus{}1\right)\left(z^{2}\minus{}z\plus{}1\right)\geq(xyz)^{2}\minus{}xyz\plus{}1$

 :)

usa tst 2006
http://www.mathlinks.ro/viewtopic.php?p=841248#841248


Area of shaded region = $ \frac{2 \cdot 2}{2} \plus{} \frac{1 \cdot 2}{2} \equal{} 3$

Total area = $ 2 \cdot 4 \equal{} 8$

Ratio = $ \boxed{\frac{3}{8}}$


[quote="eXistenZ"]1) Figure out how much the chocolate cost. (Please explain your way.)
2) Figure out the next letter in this series.[/quote]

Answer to the second question by all chances should be letter "B".
I suppose ,that is  famous Fibbonacci sequence indicated by  the graphic design :

1,1,2,3,5,8,13,?

Logically the next one is 8+13=21

B=3x7

How did you get that sequence from the path? :huh:

I guess I just know how to count.  ;)

Add up number of sticks in touch with considered letter,and follow the path indicated by the sequence.
Adding up occurs only if the contact letter-stick is a bump-free.The sums obtained in this way are multipled by factor corresponding to how many times the letter is encircled.Following this principle,one points out:

J=1x1= $1$
I=1x1= $1$
H=2x1= $2$
G=3x1= $3$
K=5x1= $5$
F=4x2= $8$
A=13x1=$13$
$B$=7x3=$21$
 [url=http://mathworld.wolfram.com/FibonacciNumber.html]Fibonacci Sequence[/url]

"B" is the only term that continues presented sequence and has my strong vote to be choosen as an adequate solution.

B is the correct answer, due to the Fibonacci sequence:
J has 1 circle around it and touches 1 stick that it does not touch at a bump
1x1=1
I has 1 circle and 1 stick touching it that it does not touch at a bump
1x1=1
H has 2 circles, 1 stick
2x1=2
G has 1 circle, 3 sticks
1x3=3
K has 1 circle, 5 sticks
1x5=5
F has 2 circles, 4 sticks
2x4=8
A has 1 circle, 13 sticks
1x13=13
we now need 21
B has 3 circles, 7 sticks
3x7=21
B is therefore correct

[quote="JesusFreak197"][hide="1"]Vowels cost five cents, consonants cost three cents. I can't explain how to look at it in this way from the beginning, but once you try it, you can easily prove that it works. Therefore, [b]chocolate[/b] costs 35 cents.[/hide]

I've taken this test before, and I've never been able to figure out the second problem. :huh:[/quote]
This doesn't work. $\text{GUM}=13\neq3+5+3$


algorithm
number theory
relatively prime
combinatorics proposed
combinatorics

Let p and q be relatively prime positive integers and let n \geq p + q - 1.
Do there exist n real numbers such that
-the sum of any p real numbers, chosen from the set of n real numbers, is positive, and
-the sum of any q real numbers, chosen from the set of n real numbers, is negative?

The same solution as on the old IMO problem?
Let a[n] be the required sequence.
Take b[i]=sum(a[j]) [1<=j<=i], another sequence.
Then b[k] - b[k+p] < 0 & b[k]  - b[k+q]>0 <=>
b[k] < b[k + p] ,  b[k] > b[k + q].
If the sequence has more terms, by the Euclidian algorythm, when starting to write b[1]  < b[ p + 1] bla bla bla, we will get a contradiction, sth like b[1] < b[1].
I think so :)


Prove the distance to a line theorem. From line Ax+By+C=0 and point (x0,y0)

d=|Ax0+By0+c|/(A^2+B^2)^1/2

You could prove it by calculating the perpendicular line to $0 = ax+by+c$ trough the point $(x_{0},y_{0})$, but that's very computational.

This one's better:

The distance from $A(x_{0},y_{0})$ to an arbitrary point $B(x_{1},y_{1})$ on $ax+by+c = 0$ is equal to $\sqrt{(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}$. The definition of "distance to a line" says that we're looking for the minimal value of this expression. But according to Cauchy-Schwarz we know that \[\begin{aligned}\left((x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}\right)\left(a^{2}+b^{2}\right) & \geq \left(a(x_{0}-x_{1})+b(y_{0}-y_{1})\right)^{2}= \left((ax_{0}+by_{0})-(ax_{1}+by_{1})\right)^{2}\\ & = \left((ax_{0}+by_{0})-(0-c)\right)^{2}= \left(ax_{0}+by_{0}+c\right)^{2}\end{aligned}\] Therefore, \[d(A,B) = \sqrt{(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}\geq \frac{|ax_{0}+by_{0}+c|}{\sqrt{a^{2}+b^{2}}}.\] Therefore, the minimal value of $d(A,B)$ is $\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^{2}+b^{2}}}$, and that's what we wanted to prove.

cool but I don't know where you got the other set of numbers as in where does the $(x_{1},y_{1})$ come from?

[quote="moogra"]where does the $(x_{1},y_{1})$ come from?[/quote]

[quote="Kurt Gödel"]an arbitrary point $B(x_{1},y_{1})$[/quote]

;)

In fact, the proof by dot product also uses an arbitrary point.  The specific coordinates are irrelevant.


I took the MATHCOUNTS Problem Series class during the Summer. Is the class being offered on Nov. 18th the same class?

Yes, it is.

If you are interested in more material at a similar level, it's introductory level classes you should look into.

If you are feeling more ambitious this time, you might give the AMC 10 Problem Series a shot.

I signed up for the fall Mathcounts Problem Series course. This is the first time that I've used an online course, and I'd like all the information that I can get about how to take this course. I'd also like to know when the class begins and ends. :?:

In order to take the course, log in with your username before class time.  A few minutes before class begins, click on the Classroom button on the left sidebar.  The virtual classroom will then open.  You might consider testing the virtual classroom by attending tomorrow night's Math Jam (a free class that will be taught by 2002 MATHCOUNTS National Champion Albert Ni).

Information about when the course begins and ends can be found on the enrollment page.


A triangle with sides 3cm, 4cm, and 5cm is rotated around one of its legs to form a cone. What is the number of cubic centimeters in the volume of the larger of the two possible cones? Express your answer in terms of $ \pi$.

$ 3^2\pi\times4\equal{}36\pi$.
$ 4^2\pi\times3\equal{}48\pi$.
Thus, $ \boxed{48\pi \ \text{cm}^3}$

Koiuchi isn't right here - you have to divide by 3 because of the cone volume formula: bh/3.
48pi/3=[u]16pi^3[/u].

We know the cone with radius $4$ will produce the greatest volume because the radius is a squared factor in the volume which impacts it more than anything else.  Therefore we must have $4$ as the radius.


How can I find angle Y with only the given angles and arc measures?

A circle is cut into 3 pieces by 2 secant lines meeting at a point outside the circle. the arc measures of the arcs outside the lines are 80 and 130. Then there are 2 lines drawn between the intersections of the lines and the edge of the circle, forming an x. how do I find the angle of the secant lines that intersect outside the circle(y)? I tried finding some angles (1,2,3,4,5,6,y)but all i did was make a bunch of equations that didn't help me. I'm thinking that there is a way to find the position of the interection of the two segments that make 105,75,105 and 75 angles. probably by using triangles but i cant figure it out.

Here is a diagram[url=http://imageshack.us][img]http://img102.imageshack.us/img102/4717/problemhl3.png[/img][/url]

The sum of the arc lengths is 360 degrees, so:

$80+130+B+A = 360$

$B+A = 170$

But we know that $B=150-A$, so substitute that into that to get:

$(150-A)+A= 170$

$150=170$

???

I don't get this problem.

Hmmm... Unless I read incorrectly, there is not enough information with which to solve the problem:

$\frac{\alpha-\beta}{2}$ would be the measure of the angle in question where $\alpha$ is the longer unknown arc and $\beta$ is the shorter unknown arc. All we know is that $\alpha+\beta = 150$. I don't see how this is enough to come to one definite solution.

I get a similar result as vishalarul.

Using the diagram above, the top middle arc is $A$ and the bottom middle arc is $B$.  Then $A+B=170$ using the arcs.  And using the triangles (angle 1 must equal $\frac{B}{2}$), $A+B=150$.

This either means that some of the information is wrong... I believe...

well, math6genius and vishalarul would have a point but 80+130+a+b=360               and 360-130-80 is 150, not 170.
so a+b is 150

I think there is a way to solve it by using the angle 75 and the angles incribed in the circle that make a triangle with that a angle. Then using those angles to find the location of the intersection(in proportion with the center of the circle) with the angle 75. I know its possible just not something commonly known(I guess). its a challenging problem. It makes me think.

I'm gonna work on it but i'd like some big help.

this problem is obviously not uniquely defined; consider chords AB, and CD, which subtend arcs of 80 and 130, respectively, clearly, we can move these chords around (i.e. rotate one of them), and we still retain all properties of the diagram; clearly the angle between chords varies

Oooooohhh I just figured I wouldnt be able to rotate them. I should have thought of it that way first. I think its impossible now.


Say that, hypothically :wink: , there was a contest, say, at your local cd store. In the contest, pretend that you were given a giant picture that looked like the cover from Sargent Pepper's Lonely Hearts Club Band, pretty much with a ton of random people on it, and you had to name all of the people to win 500 dollars worth of music. Strictly hypothically, would it be ethical to post such a picture on a forum and ask people to help you? Keeping in mind that last year, the winners were three guys who got them all right by working together, and you got about half right, even though you asked everyone that you knew, and you REALLY REALLY LIKE MUSIC. Strictly hypothically, of course :wink: .

completely ethical. society defines ethics essentially, but feel free to create your own guidelines. The contest is meant to test your skill at gathering information, and you show that precisely. Go for it!

Well the three guys are probably each getting part of the prize money. I don't know about the forumers.

And I have a doubt or two about the word 'hypothically'.

do it :P nah we would have to do it somewhere else... :P

How about this. If you are interested, and you don't have any problems ethically, pm me and I'll give you a magical e-mail address. If you feel that it is ethically wrong to help me, just don't do anything and let me rot in hell alone.

I know nothing about bands, but I am ready to help you if I can. You count on me for that.


[color=blue]i have this problem that i need help with.

the number you have is 45678.
you arrive at the number by subtracting one number from another.
the two numbers must contain 0,1,2,3,4,5,6,7,8, and 9
and you can only use each digit once

okay.
i got
48615-02937=45678
59460-13782=45678
57984-12306=45678

but the problem says there are 24 solutions
but i only have three.
ugh. :?:

help [b]please?[/b] :][/color]

Um...Potentially, you could start with all the possible combinations for the units place.

I found the 24 solutions but I used brute force. Anyway, If you want, I'll write them  :)

Let $ A\equal{}(a_4a_3a_2a_1a_0)$ and $ B\equal{}(b_4b_3b_2b_1b_0)$ be the two numbers

Let $ X\equal{}(x_4x_3x_2x_1x_0)\equal{}45678$. Then $ A\minus{}B\equal{}X$

We can use the recursive formulas:

$ b_k \equal{} (a_k\minus{}x_k\plus{}c_{k\minus{}1}) \mod 10$

$ c_k \equal{} \left[\frac{a_k\minus{}b_k}{10}\right]$ with $ c_{\minus{}1}\equal{}0$

$ c_k$ is the carry for each digit. It is $ a_4>b_4$, so it should be $ c_4\equal{}0$

So we choose arbitrarily the digits $ a_k$, but we sould take care that all $ a_k$ and $ b_k$ are different

[quote="pontios"]I found the 24 solutions but I used brute force. Anyway, If you want, I'll write them  :)

Let $ A \equal{} (a_4a_3a_2a_1a_0)$ and $ B \equal{} (b_4b_3b_2b_1b_0)$ be the two numbers

Let $ X \equal{} (x_4x_3x_2x_1x_0) \equal{} 45678$. Then $ A \minus{} B \equal{} X$

We can use the recursive formulas:

$ b_k \equal{} (a_k \minus{} x_k \plus{} c_{k \minus{} 1}) \mod 10$

$ c_k \equal{} \left[\frac {a_k \minus{} b_k}{10}\right]$ with $ c_{ \minus{} 1} \equal{} 0$

$ c_k$ is the carry for each digit. It is $ a_4 > b_4$, so it should be $ c_4 \equal{} 0$

So we choose arbitrarily the digits $ a_k$, but we sould take care that all $ a_k$ and $ b_k$ are different[/quote]

um, i feel stupid but can you elaborate please? :]
thanks(:

Ok.. Let's see the example 48615-02937=45678

It is $ A \equal{} (a_4a_3a_2a_1a_0) \equal{} 48615$ and  $ B \equal{} (b_4b_3b_2b_1b_0) \equal{} 02937$
$ X \equal{} (x_4x_3x_2x_1x_0) \equal{} 45678$

Suppose that we don't know the value of $ B$ (in fact, we'll do the subtraction $ A \minus{} X$). Also suppose that we chose $ A \equal{} 48615$ randomly

We'll calculate $ B:$

$ b_k \equal{} (a_k \minus{} x_k \plus{} c_{k \minus{} 1}) \mod 10 \Rightarrow$

$ b_0 \equal{} (a_0 \minus{} x_0 \plus{} c_{ \minus{} 1}) \mod 10 \Rightarrow$
$ b_0 \equal{} (5 \minus{} 8 \plus{} 0) \mod 10 \Rightarrow$
$ b_0 \equal{} ( \minus{} 3) \mod 10 \Rightarrow \boxed{b_0 \equal{} 7}$ (correct!)

$ c_k \equal{} \left[\frac {a_k \minus{} b_k}{10}\right] \Rightarrow$

$ c_0 \equal{} \left[\frac {a_0 \minus{} b_0}{10}\right] \equal{} \left[\frac {5 \minus{} 7}{10}\right] \equal{} \minus{} 1$

Now go to the next step
------------------------------------------------------------------------------------------------
It is $ a_1 \equal{} 1$

$ b_1 \equal{} (a_1 \minus{} x_1 \plus{} c_{0}) \mod 10 \Rightarrow$
$ b_1 \equal{} (1 \minus{} 7 \minus{} 1) \mod 10 \Rightarrow$
$ b_1 \equal{} ( \minus{} 7 ) \mod 10 \Rightarrow \boxed{b_1 \equal{} 3}$ (correct!)

$ c_1 \equal{} \left[\frac {a_1 \minus{} b_1}{10}\right] \equal{} \left[\frac {1 \minus{} 3}{10}\right] \equal{} \minus{} 1$

If we continue this procedure, we'll find consecutively the digits $ 9,2,0$, so the number $ B \equal{} 02937$ is formed

------------------------------------------------------------------------------------------------

But if we had started with $ A \equal{} 28614$, we would not find a solution. Indeed, we would get
$ b_0 \equal{} (a_0 \minus{} x_0 \plus{} c_{ \minus{} 1}) \mod 10 \Rightarrow$
$ b_0 \equal{} (4 \minus{} 8 \plus{} 0) \mod 10 \Rightarrow$
$ b_0 \equal{} ( \minus{} 4) \mod 10 \Rightarrow \boxed{b_0 \equal{} 6}$

But the digit $ 6$ is already used in $ 28614$

------------------------------------------------------------------------------------------------

Watch another example in the attachment (yellow:a, green:b, rose:c, blue:result), 5 steps

We create the number $ A$ step-by-step, starting from the last digit
When we add $ 0$, the digit $ 2$ is created and we don't have any problem
Then we add $ 3,8,9$ and the digits $ 5,1,4$ are created and we don't have problems
But when we add $ 7$, the digit $ b_4$ becames $ 3$, which is already used! This is wrong.

The result is the desired of course.
But this configuration with $ A \equal{} 79830$ yields no solution


algebra
polynomial
induction
complex numbers

Find all real polynomials p(x) such that p(p(x)) = p(x)^n for some positive integer n.

I didn't understand the last part of the kalva solution for this one either.

Edited for spoiler:

Edited again because I know that there is a flaw in the logic near the end, but I am way too tired and sleepy right now to think about it. I'll try to fix it in the morning .

[hide]

Let p(x) = a0x^n + a1x^(n-1) + ... + an

Then p(p(x)) = a0p(x)^n + a1p(x)^(n-1) + ... + an

Thus, the highest degree term of p(p(x)) is a0p(x)^n or a0x^(n :^2:)

The highest degree term of p(x)^n is a0^n * x^(n :^2:)

It is clear that for these two polynomials to be equal to each other, n=1 OR a0 = 1. a0 = 1 is the more general choice (p(x) = x is a trivial solution).

This means that:

[highest term of p(p(x))] + [other terms of p(p(x))] = [highest term of p(x)^n] + [other terms of p(x)^n].

other terms of p(p(x)) = other terms of p(x)^n

This is true if all other terms are 0. To prove that this is the only case, see that for the second term of p(p(x)) is NOT equal to the second term of p(x)^n. The exponents will be different.

Thus, the only polynomial p(x) that satisfies the equation is x^n.[/hide]

[quote]To prove that this is the only case, see that for the second term of p(p(x)) is NOT equal to the second term of p(x)^n. The exponents will be different.[/quote]

Hmm...could you please clarify this? Thanks.

Why can't we just explain this in a very simple way.[hide]Let's say p(x)=M, where M is not constant.sub M in for p(x), we have p(M)=M^n.[/hide] See if this makes it any clearer.

this is by no means rigorous, and you'll still have to see if there are other polynomials satisfying this, but you could get an idea what the general form of P(x) is by (temporarily) ignoring the fact that some polynomials do not have proper inverses, and just stick in
P^(-1)(x), which you then get as P(x)=x^n

Ah, nevermind, I get it now. Thanks everyone.

What about this :

p =  :Sigma: ak x^k  K[X], deg(p)= l, p != 0 (p=0 obvious solution)

Looking near +oo :
p(p(x)) ~ al^(l^2+1) x^(l)
p(x)^n ~ al^n x^(l*n)

So n=l, al=1

Let x0 a root of p (in the algebraic closure of K if needed  :wink:) then p(p(x)) = p(0) = a0 = 0.

p(x)  = x * q(x), q != 0,

p(p(x)) = x*q(x) q(xq(x)) = x^l q(x)^l or q(xq(x)) = q(x)^(l-1) x^(l-1)

So q is solution of q(xq(x)) = x^(l-1) q(x)^(l-1), d(q)=l-1.
Again q(0) = 0, q(x) = x*r(x), r(xr(x)) = x^(2l-4) r(x)^(l-2), r(0) = 0, .... By induction p(x) = x^l, which works.

Is it just me, or did anyone else not understand a single word of belenos' solution at all :).

here's how I did it:

[hide]if a is any root of the polynomial P, then P(P(a))=P(a)^n implies that P(0) = 0, so 0 is a root, so we may write P(x)=x Q(x) for some polynomial Q. Then, P(P(a))=P(a)^n implies that Q(x Q(x)) = x^(n-1) Q(x)^n. Putting 0 into this shows that 0 is a root of Q, so Q(x) = x R(x). Continuing in this mannar, we see that 0 is a root again and again and again... Since the leading coefficient must be 1 (as shown in previous posts), the only possibilities are P(x)=x^n.[/hide]

zabelman -- I don't see how 0 as a root follows trivially from the given.  What if P(0) = 1, P(1) = 1?  Or for n = 1, P = constant?

rlee -- in your original proof, you said that the highest degree term of P(P(x)) was a0x^(n2) -- I  think it's actually a0n + 1x^(n2), although you get the same result from that.  I don't think the last bit of your proof is correct, though, or at least I don't think it's well justified.

beta, I have no idea what you mean.

I also did not understand most of what belenos wrote -- either characters are messed up or they need much more explanation.  I think the C with two cross-portions is supposed to be "is an element of," and K[X] is the set of polynomials in X.

I'm very sorry for providing a messy (to be polite) solution  :(

The main ideas were :

. from behaviour near +oo, deduce the degree of p and highest degree  coefficient.

p(x) ~ q(x) means p(x)/q(x) -> 1
If p(x) ~ q(x) (near +oo) and p and q are both polynoms then degree(p)=degree(q) and p and q have same highest degree coefficient.

. prove that 0 is a root of p, then deduce a (very similar) equation for p(x)/x = q(x). Then proceed until getting a constant polynom.

I'm not sure this makes things clearer  :?  I did my best !

I hope this clearifies the first part of my solution:

[hide]Let a be any root of the polynomial P, which is guaranteed to exist by the fundamental theorem of algebra. So P(a)=0. Since P(P(x))=P(x)^n holds for all x, it holds for a. On the left side of this equation, we get P(P(a))=P(0). On the right side, we get P(a)^n = 0^n = 0. So, left side equals right side, i.e. P(0)=0.[/hide]

How about this solution, based on zabelman's idea, but simpler?  [hide]The case when p is a constant is trivial, so let's assume p is not a constant.  Let b be an arbitrary number.  By the Fundamental Theorem of Algebra, there is a (possibly complex) number a such that p(a) = b.  Thus p(b) = p(p(a)) = p(a)n = bn.  Because this is true for all b, the polynomial p equals the polynomial xn.[/hide]

It's not clear from the problem statement whether p(p(x)) = xn is assumed to be true for all complex x or for all real x.  But it's easy to prove that if a polynomial identity is true for all real numbers (or just an infinite set of numbers), then it must be true for all complex numbers too.

By the way, Fiery, which competition is this question from?

[quote]By the way, Fiery, which competition is this question from?[/quote]

Canada 1975

I'm trying to get through all of these Canadian problems in the first few days of my spring break. I'm not doing too badly--ive got 9 under my belt and i still have 6 more hours in this second day. :D

After that I'll go off and do all of the AIME's.  :mrgreen:


Calculate the sum
\[6+66+666+\cdots+666\ldots6\]
where the last term has $n\,\,6$'s.


Let S equal the number of five-pointed stars on the front cover of this test.  This number S is a factor of  , where N is a whole number.  What is the smallest possible value of N?

A.6	B.8	C.10	D.12	E. 16

Hi sri340, is there something missing here?

ya, we need some more information.Posted twice  :mad:


Through the courtesy of a participant on the College Confidential Web site, I have learned of a Web [url=http://www.sonicpixel.net/ahs04/]page linking to scatterplots[/url] of college admission rates for differing combinations of SAT scores and high school grade averages. Visit the page and click on the name of your favorite college. Then see [url=http://www.collegeconfidential.com/discus/messages/5/77777.html]the discussion[/url] on the College Confidential site. And there are [url=http://www.collegeconfidential.com/discus/messages/5/77924.html]links to other scattergrams[/url] in another post there.

GPA combined with SAT is not enough.

Wow.  On University of Chicago, there is someone with <1000 SAT and <3 GPA.

Noble Peace Prize winner?

Yeah, after all U Chicago is rather obsessed w/their number of Nobel Laureates...
Probably has their last name on a building.


limit
integration
function
calculus
real analysis
real analysis unsolved

I am trying to prove a thing, and got into a situation as follows, that I wonder about.

Suppose $ f \in L^1(\mathbb{R})$. Suppose $ \theta_n$ is a sequence tending to $ \theta$. Then it's well known that
$ \lim_{n\to\infty}\int_{\mathbb{R}} |f(t \plus{} \theta_n) \minus{} f(t \plus{} \theta)| \, dt \equal{} 0$

Now, I also have functions $ g_n$ that have a uniform integrable bound, and such that $ g_n$ tends to $ g$ pointwise. I wish to conclude that
$ \lim_{n\to\infty}\int_{\mathbb{R}} |g_n(t)f(t \plus{} \theta_n) \minus{} g(t) f(t \plus{} \theta)| \, dt \equal{} 0$

In the problem I was trying to do, I figured out that it was fairly easy by subtracting and adding $ g_n(t)f(t \plus{} \theta)$ since the $ g_n$ and $ g$ were compactly supported and uniformly bounded. Is it true in the general?

YES, I prefer you to the excellent book entitled "MEASURE AND INTEGRAL : An Introduction to Real Analysis" due to R.L. Wheeden and A. Zygmund.

In this book, problem 11 in page 144 states that: For $ 1 \leqq p < \plus{} \infty$, if $ f_k \to f$ in $ L^p$ and $ g_k \to g$ pointwise with $ \|g_k\|_\infty \leqq M$ for all $ k$, then $ f_k g_k \to fg$ in $ L^p$.

 :lol:

[quote="Kalle"]Suppose $ f \in L^1(\mathbb{R})$. Suppose $ \theta_n$ is a sequence tending to $ \theta$. Then it's well known that
$ \lim_{n\to\infty}\int_{\mathbb{R}} |f(t \plus{} \theta_n) \minus{} f(t \plus{} \theta)| \, dt \equal{} 0$[/quote]

This comes from the continuity in $ L^p$: if $ 1 \leqq p < \plus{} \infty$ and $ f \in L^p(\mathbb R^N)$ then
\[ \mathop {\lim }\limits_{\left| \textbf{h} \right| \to 0} \left\| {f\left( {\textbf{x \plus{} h}} \right) \minus{} f\left( \textbf{x} \right)} \right\|_{L^p \left( {\mathbb{R}^N } \right)} \mathop \equal{} \limits^\Delta \mathop {\lim }\limits_{\left| \textbf{h} \right| \to 0} \left( {\int_{\mathbb{R}^N } {\left| {f\left( \textbf{x \plus{} h} \right) \minus{} f\left( \textbf{x} \right)} \right|^p } }d\textbf{x} \right)^{\frac {1} {p}} \equal{} 0.
\]

[quote="Kalle"]Well, as I said in the last paragraph, I did figure out how to do it under those conditions. Is the book written in such a way that this suggests the conditions cannot be weakened?[/quote]

That's a problem, no hint, no solution, no no no thing. Why dont you try to solve the problem I posted above and then deduce...

As I said twice now, I did solve the problem you gave, and came to those sufficient conditions on my own. The proof method doesn't say anything about more general things, and relies heavily on the uniform bound in $ L^\infty$. However, it seems to me that this thing could be true under more general settings. If $ g_n$ simply has an integrable majorant, then it is true for continuous functions, or simple functions, or other classes. Then we would need a soft approximation argument for any $ f$, but this does not seem to quite work ...

[quote="Kalle"]it seems to me that this thing could be true under more general settings. If $ g_n$ simply has an integrable majorant, then it is true for continuous functions, or simple functions, or other classes. Then we would need a soft approximation argument for any $ f$, but this does not seem to quite work ...[/quote]

So that you need to weaken the hypothesis on $ g_k$, for e.g., only integrable instead of bounded uniformly?


calculus
integration
trigonometry
logarithms
derivative
calculus computations

Evaluate the integral: \[\int_{\pi/4}^{\pi/3}{x \over \sin^{2}x}dx.\]

[quote="Lovasz"]Explain this one, people! Thanks!
[quote="coffeym"]$\int{\frac{x}{\sin^{2}x}}dx=\ln{|\sin{x}|}-x\cot{x}]$[/quote][/quote]

integration by parts of course :)

Ok, I'll show the working  :lol:

So $\int{\frac{x}{\sin^{2}{x}}}\,dx$

=$\int{x\,\csc^{2}x}\,dx$

Which by parts
$=-x\cot{x}+\int{\cot{x}}\,dx$

as the integral of $\csc^{2}x$ w.r.t. $x$ is $-\cot{x}$

Now $\int{\cot{x}}\,dx=\int{\frac{\cos{x}}{\sin{x}}}\,dx$

The numerator is the derivative of the denominator, so $\int{\cot{x}}\,dx=\ln{|\sin{x}|}+const$

So, overall, the integral $=\ln{|\sin{x}|}-x\cot{x}+const$

Plug the limits in to get the answer. Hope that's O.K  :lol:


Find the value of $ k$ for which $ kx^2 \minus{}5x\minus{}12 \equal{} 0$ has solutions $ x\equal{}3$ and $ x \equal{} \minus{}\frac{4}{3}$.

So you know you are going to have $ (x\minus{}3)(3x\plus{}4)$. So then you expand to $ 3x^2\minus{}5x\minus{}12$. So $ k\equal{}3$.

use the $ 3$. $ k(3^2)\minus{}5(3)\minus{}12\equal{}0$

$ 9k\minus{}15\minus{}12\equal{}0$

$ 9k\equal{}27$

$ \boxed{k\equal{}3}$

we can verify that this is true for the $ \frac34$ also.

you can also use jonathan's method, although i feel mine could be done in one's head, using the $ 3$.


If constructing a box with a square base with a volume of 1000 inches cubed:
The material for the top and bottom is 3 dollars per 100 inches squared and the
material for the sides costs 1.25 dollars per 100 sq. in.
1) If x is the length of the side of the base, what is the cost in x?
2)If the side of the base must be at least six inches long, for what
value of x will the cost be $20?

If there is a rectangular garden with an area of 250 sq. meters next to
a building and fenced on three sides, with the building acting as the
fourth fence...
1) If the side of the garden parallel to the building has length x
meters, express amount of fencing required as a function of x
2) for what values of x will less that 60 meters of fencing be needed?
-------------------------
I started working out the first part for the second situation and got (although this is probably specious) y=(250-x)/x by setting y equal to the length of one of the sides, and because you don't need fencing for the building side of the garden, and 250=xy...

Thanks for your time.

im not sure i quite understood the question...  the [i]base[/i] is 1000 in :^3: .  by base do you mean the bottom of the box?

or i guess you mean the total volume of the box is 1000 in :^3: , and the base is square?

I mean the whole box is 1000 inches cubed.  And, that the bottom of the box is the shape of a square.

1) y=3(x 2 /100)

2)should be  x=:pm: 26?

i may have done this wrong.  i may have not have comprehended it correctly.  i often don't.

Can you tell me how you derived that?  Any thoughts on the second one?  And can you explain your derivation for that one if you do....

Thanks

Note: I might have made a mistake. The results I got were a bit complex... I'll read over what I've written

Well for #1 Part 1, a hint that I would give you would be to write the height in terms of $x$, which you can do since you know the volume. Basically, solve the volume formula for $h$. Then, look for the surface area.

Don't look at this until you've tried every avenue:
[hide]Let's call the unknown height of the box $h$ and the side of the square base $x$. Then, we know that $V=lwh \implies 1000=x^2h$. Solving for $h$, we get $h=\frac{1000}{x^2}$.

Now, we know that the material for the base is 3 dollars/100 square inches. The surface area for both bases is obviously $2x^2\text{ square inches}$. To find how much it costs, multiply it by the conversion factor and we have:

$2x^2\text{ square inches} \cdot \frac{3\text{ dollars}}{100\text{ square inches}}=\frac{3x^2}{50}\text{ dollars}$.

Now, we turn our attentions to the sides of the box. It is clear that there are 4 lateral faces. It turns out because our base is a square, the surface area of those 4 faces are all equivalent to $xh$.

Therefore, the total surface area for the lateral faces is $4xh\text{ square inches}$. Since we know that $h=\frac{1000}{x^2}$, we can substitute and get the area for the lateral faces to be $\frac{4000}{x}\text{ square inches}$.

Now, we can multiply this by the conversion factor of 1.25 dollars/100 square inches to get:

$\frac{4000\text{ square inches}}{x} \cdot \frac{1.25\text{ dollars}}{100\text{ square inches}}=\frac{50}{x}\text{ dollars}$.

Finally, since we know how much this costs, we can put it together and get the total cost in terms of $x$ to be $\frac{3x^2}{50}+\frac{50}{x}\text{ dollars}$.
[/hide]

PART 2 IS POSSIBLE! I MADE A CALCULATION ERROR BEFORE!
[hide]We know that the formula for finding the cost is now:

$C=\frac{3x^2}{50}+\frac{50}{x}$.

We want to find $x$ when $C=20$. So setting $C$ to 20, our equation is now:

$20=\frac{3x^2}{50}+\frac{50}{x}$. Multiplying by the LCD ($50x$) to get rid of denominators, we have

$1000x=3x^3+2500 \implies 3x^3-1000x+2500=0$.

This means that the three real roots (after putting in cubic calculator) are approximately:

$16.84853231017, -19.398260531710193, 2.5497282215401924$. The one we want is $16.84853231017$ and that is our answer.
[/hide]

--------------------------------------------------------------------------------------

As for the second question part 1, I will again give you the same hint as above. Basically, find $y$, the second side, in terms of $x$ by using the area formula.

As is:
[hide]We want to find the amount of fencing, this means we need to find $x+2y$. We know that $xy=250$ as you mentioned in your post. This means that $y=\frac{250}{x}$ and the amount of fencing $F$ is:

$F=x+\frac{500}{x}$. (I think you were getting there in your post)
[/hide]

Now for part 2, once you know the formula for finding the amount of fencing based on $x$, we can solve this easily (try it yourself first);
[hide]
$F=x+\frac{500}{x}$

We are looking for when $F<60$. Therefore, we are solving the inequality:

$F<60 \implies x+\frac{500}{x}<60$. Clearing fractions and rearranging, we get:

$x^2-60x+500<0 \implies (x-10)(x-50)<0$

Now, we observe that the only time that this will be less than 0 is when $(x-10)$ and $(x-50)$ have opposite signs. No matter what $x$, $(x-10)>(x-50)$. This means that $(x-10)$ must be positive and $(x-50)$ must be negative.

We get the two inequalities:
$x-10>0 \implies x>10$
$x-50<0 \implies x<50$

This means that x can be between 10 and 50. Our range for x is:

$10<x<50$ and that is our answer.
[/hide]

Zyloch, your post was temendously helpful.  I did get a solution for the second part of part one: 2.55.  I used my graphing calculator.  Try the substitution, and it seems to check out.

Yes, 2.55 does seem to come extremely close. Let me reread my post.

Edit: Yep! I made a stupid calculation error (20*50!=100 lol). Now, the problem [i]is[/i] possible, with an answer around 16.


geometry
perpendicular bisector
geometry unsolved

In triangle $ABC$, $\angle B = 2\angle C$. Let $P$ and $Q$ be points on the perpendicular bisector of segment $BC$ such that rays $AP$ and $AQ$ trisect $\angle A$. Prove that $PQ < AB$ if and only if $\angle B$ is obtuse.

Assume $\angle QAB=\frac{\angle CAB}3$, let $D$ be the point where the perpendicular bisector of $BC$ cuts $AC$, and $E$ the projection of $A$ on this perpendicular bisector. Notice that $\frac{AE}{\frac{BC}2}=\frac{AD}{DC}=\frac{AB}{BC}$, so $AB=2AE\ (*)$. We also know that $\angle APE=30^\circ\ (\#)$ (a simple angle chase), so $AP=2AE$. Together with $(*)$, this shows that $AP=AB$. Now, since $Q$ lies on the bisector of $\angle BAP$, which is also the perpendicular bisector of $BP$, we have $QP=QB=QC$. This means that $PQ<AB\iff\angle AQB>\angle QAB$, and, since $\angle AQB+\angle QAB=30^\circ$ (a consequence of $(\#)$), we get $PQ<AB\iff\frac{\angle A}3<15^\circ$, and the conclusion follows.

:lol: ,dr.trig avoids construction operation,does it with usual,but patiennt ease.


geometry
trapezoid
parallelogram
similar triangles
geometry unsolved

Let ABCD be a trapezium $ (AB//CD)$. Suppose that M is a point on the side AD and N is interior to the trapezium such that $ \widehat{NBC} \equal{} \widehat{MBA}, \widehat{NCB} \equal{} \widehat{MCD}$. Let P be the fourth vertex of the parallelogram MANP. Prove that P is on the side CD

Construct a line through $ M$ and parallel to $ AB$, it cuts $ (MBC)$ at $ H, BC$ at $ I$.
We have $ \angle CBH\equal{}\angle CMH\equal{}\angle MCD\equal{}\angle NCB \Rightarrow NC//BH$.
Similarly, $ NB//CH \Rightarrow CNBH$ is a parallelogram.
Construct a line through $ N$ and parallel to $ BC$, it cuts $ CD$ at $ P$.
$ \Rightarrow \Delta PNC\equal{}\Delta IBH \Rightarrow NP\equal{}BI\equal{}AM \Rightarrow AMPN$ is a parallelogram. QED

Let $ BM$ meet $ CD$ at $ E$ and $ BN$ meet $ CD$ at $ F$. From the given conditions, observe that triangles $ CME$ and $ CNB$ are similar and triangles $ CFB$ and $ CBE$ aresimilr. Let the parallel line to $ BF$ through $ M$ meet $ CD$ at $ G$.
From the similar triangles in the figure, $ \frac{GM}{ME}\equal{}\frac{FB}{BE}\equal{}\frac{CB}{CE}\equal{}\frac{NB}{ME}$ . Hence, $ MG\equal{}BN$ and this implies that a halfturn about the midpoint of $ MN$ carries ray $ CD$ into ray $ AB$ and the result follows.

[img]http://img42.imageshack.us/img42/4482/imageop.png[/img]


inequalities
logarithms
function
algebra
polynomial
real analysis

1. Show that $log_{e}x-x+1 \leq 0$

2. Prove the AM-GM Inequality using the result from 1.

1. $y=\ln x\ (x>0)$ is an increasing function and concave up. Then $y=x-1$ is the tangent line at the point $(1,\ 0)$ on the curve $y=\ln x\ (x>0).$

[quote="N.T.TUAN"][hide]1) Use real Analysis
2) By 1. $\ln{b_{i}}\leq b_{i}-1(i=1,2,...,n)$ where $b_{i}=\frac{na_{i}}{a_{1}+a_{2}+...+a_{n}}(i=1,2,...,n)$[/hide][/quote]

Do you mean looking at taylor polynomial?

[quote="me@home"]Do you mean looking at taylor polynomial?[/quote]
If $f(x)=\ln{x}-x+1$ then $f'(x)=\frac{1}{x}-1=\frac{1-x}{x}$, therefore $\max f(x)=0$ when $x=1$.

Solving [b]2[/b] requires a clever (sort of) bit of substitution.

[quote="t0rajir0u"]Well, [b]1[/b] is equivalent to Bernoulli's inequality, isn't it?[/quote]
I am not understand your think  :maybe:

[quote="t0rajir0u"]I mean $\ln x-x+1 \le 0 \Leftrightarrow e^{u}\ge 1+u$ (where we perform the substitution $u = x-1$).  And this is well-known.[/quote]
Oh, I know it. I understand mistake your think, 1. is NOT $1$ :D

Would anyone like to post a full solution to the original problem?


Find the value of k for which the two equations

   (x - 2)^4 - (x - 2) = 0    and    x^2 - kx + k = 0

have two roots in common.

Source:  MATC Mathematics Club

[hide]Replacting x-2 with y in the first equation, and factoring, we are left with:

y(y-1)(y^2+y+1)=0

Plugging x-2 back in:

(x-2)(x-3)(x^3-3x+3)=0

This last factor fits x^2-kx+k, making [b]3[/b] the answer.[/hide]


function
floor function
algebra
functional equation
number theory

Find a function $ f: \mathbb{N}\to \mathbb{N}$ such that $f(1)=2$ and for every positive integer $n$,
\[f(f(n))=f(n)+n \quad \text{and} \quad f(n)<f(n+1).\]

[quote="Rust"]Obviosly $ f(n)=[\phi n+1]$, were $ \phi =\frac{1+\sqrt 5 }{2}$ is solution.
I think it posted before.[/quote]

Obviosly $ f(n)=[\phi n+1]$, were $ \phi =\frac{1+\sqrt 5 }{2}$ [u][b]is not [/b][/u]a solution : $ f(1)=2$, $ f(2)=4$ and $ f(f(1))\neq f(1)+1$

In fact, an answer could be $ f(n)=[\phi n+\frac{1}{2}]$, were $ \phi =\frac{1+\sqrt 5 }{2}$

[quote="pco"]In fact, an answer could be $ f(n)=[\phi n+\frac{1}{2}]$, were $ \phi =\frac{1+\sqrt 5 }{2}$[/quote]
If $ f(n)=[\phi n+c]$, then $ -(\phi-1)-1+n+\phi c<f(f(n))-f(n)=[(\phi-1)[\phi n+c]+c]<n+\phi c$,
 therefore work c, suth that $ c\phi <1$ and $ (\phi-1)(c-1)+c\ge 0$. It give $ 2-\phi\le c<\phi-1$. (c=1/2 work).
But these functional equation had infinetely many solutions. For example $ f(4)$ may be 6 or 7.

[quote="cckek"]Find a function $ f: \mathbb{N}\to \mathbb{N}$, $ f(1)=2,f(f(n))=f(n)+n,\quad f(n)<f(n+1).$[/quote]

In fact  $ f(n)=\lfloor\phi n+a\rfloor$, with $ \phi =\frac{1+\sqrt 5 }{2}$ and $ a\in[2-\phi,\phi-1)$ is a possible solution :

1) $ f(1)=2$ :
$ f(1)=\lfloor\phi+a\rfloor=2$ since $ 2-\phi\leq a<\phi-1<3-\phi$ and so $ 2\leq \phi+a<3$
Q.E.D.

2) $ f(n)<f(n+1)$
We have $ f(n)=\lfloor\phi n+a\rfloor$ $ < \lfloor\phi n+1+a\rfloor$ $ \leq \lfloor\phi n+\phi+a\rfloor = f(n+1)$
Q.E.D.

3) $ f(f(n))=f(n)+n$
Since $ f(n)=\lfloor\phi n+a\rfloor$, we have $ \phi n+a=f(n)+\{f(n)\}$ and so $ f(n)=\phi n+a-\{f(n)\}$
So $ f(f(n))=\lfloor\phi f(n)+a\rfloor$ $ =\lfloor\phi^{2}n+\phi a-\phi\{f(n)\}+a\rfloor$
So $ f(f(n))=\lfloor\phi n+n+\phi a-\phi\{f(n)\}+a\rfloor$ (remember $ \phi^{2}=\phi+1$)
So $ f(f(n))=n+\lfloor(\phi n+a)+\phi a-\phi\{f(n)\}\rfloor$
So $ f(f(n))=n+\lfloor f(n)+\{f(n)\}+\phi a-\phi\{f(n)\}\rfloor$
So $ f(f(n))=f(n)+n+\lfloor \phi a-(\phi-1)\{f(n)\}\rfloor$

It remains to show that $ \lfloor \phi a-(\phi-1)\{f(n)\}\rfloor=0$, which is easy :
$ \{f(n)\}\geq 0$ $ \implies$ $ \phi a-(\phi-1)\{f(n)\}\leq \phi a <\phi(\phi-1)=1$
$ \{f(n)\}< 1$ $ \implies$ $ \phi a-(\phi-1)\{f(n)\}>\phi a-(\phi-1)\geq \phi(2-\phi)-(\phi-1)=0$

So $ 0<\phi a-(\phi-1)\{f(n)\}<1$ and so $ \lfloor \phi a-(\phi-1)\{f(n)\}\rfloor=0$

And so $ f(f(n))=f(n)+n$
Q.E.D.


Is it true that if $\forall X \in M_n \left( \mathbb{C} \right)$ we have $AX=XA$ then $A = t I_n$ ?

Yes, it is true.  You can prove it using matrices with a single nonzero entry.

and this shows that it works on arbitrary fields ..

Let $A,B \in \mathcal M _k ( \mathbb C )$ be matrices such that $A^m B^n = B^n A^m$, where $m,n$ are fixed numbers $\geq 1$.

What conditions should we impose on $A$ and $B$ such that the conclusion will be $AB=BA$?


Vieta
geometric sequence
algebra proposed
algebra

hi all :
sorry if it in the wrong place
 :blush:

hello, with
$ x_1\equal{}x_1$
$ x_2\equal{}qx_1$
$ x_3\equal{}q^2x_1$
i have got the following system
$ x_1^3\plus{}7x_1^2\plus{}cx_1\minus{}8\equal{}0$
$ q^3x_1^3\plus{}7q^2x_1^2\plus{}qcx_1\minus{}8\equal{}0$
$ q^6x_1^3\plus{}7q^4x_1^2\plus{}q^2cx_1\minus{}8\equal{}0$
this system have you to solve.
Sonnhard.

Using Viete's relations we obtain:

$ x_1\plus{}x_2\plus{}x_3\equal{}7$

$ x_1x_2\plus{}x_2x_3\plus{}x_3x_1\equal{}c$

$ x_1x_2x_3\equal{}8$

Because $ x_1,\ x_2,\ x_3$ are in geometric progression $ \implies x_2^2\equal{}x_1x_3$, so $ x_2\equal{}2$.

$ \implies x_1x_3\equal{}4$ and $ x_1\plus{}x_3\equal{}5\implies c\equal{}14$ .

Solving the sistem $ x_1\plus{}x_3\equal{}5$ and $ x_1x_3\equal{}4$ obtain easily that $ x_1\equal{}1,\ x_3\equal{}4$ or $ x_1\equal{}4,\ x_3\equal{}1$ .

[quote="Dr Sonnhard Graubner"]hello, but i have found more solutions as you.
Sonnhard.[/quote]

Which are those soluions?

Because I get $ c\equal{}14\implies x\in\{1,2,4\}$ .

Let the roots are $ \frac{a}{r},a,ar$
By vieta's theorem
$ \frac{a}{r}\plus{}a\plus{}ar\equal{}7$
$ \frac{a^2}{r}\plus{}a^2\plus{}a^2r\equal{}c$
$ a^3\equal{}8$, so $ a\equal{}2$

From the first equation,
$ \frac{1}{r}\plus{}1\plus{}r\equal{}\frac{7}{2}$
From the second equation, $ c\equal{}4(\frac{1}{r}\plus{}1\plus{}r)\equal{}14$
 :D

hello, i have also found
$ c\equal{}\minus{}14$
$ q\equal{}\minus{}9/4\plus{}1/4\sqrt{65}$
$ x_1\equal{}\minus{}9/2\minus{}1/2\sqrt{65}$
$ x_2\equal{}\minus{}9/2\plus{}1/2\sqrt{65}$
$ x_3\equal{}2$, which fulfills the conditions.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have also found
$ c \equal{} \minus{} 14$
$ q \equal{} \minus{} 9/4 \plus{} 1/4\sqrt {65}$
$ x_1 \equal{} \minus{} 9/2 \minus{} 1/2\sqrt {65}$
$ x_2 \equal{} \minus{} 9/2 \plus{} 1/2\sqrt {65}$
$ x_3 \equal{} 2$, which fulfills the conditions.
Sonnhard.[/quote]
But then $ x_1\plus{}x_2\plus{}x_3 \equal{} \minus{}7$.. :huh:

[quote="Dr Sonnhard Graubner"]hello, i have also found
$ c \equal{} \minus{} 14$
$ q \equal{} \minus{} 9/4 \plus{} 1/4\sqrt {65}$
$ x_1 \equal{} \minus{} 9/2 \minus{} 1/2\sqrt {65}$
$ x_2 \equal{} \minus{} 9/2 \plus{} 1/2\sqrt {65}$
$ x_3 \equal{} 2$, which fulfills the conditions.
Sonnhard.[/quote]

Dear Sonnhard

Those solutions are for the equation $ x^3\plus{}7x^2\plus{}cx\minus{}8\equal{}0$. Our equation is actually $ x^3\color{red}{\minus{}}$ $ 7x^2\plus{}cx\minus{}8\equal{}0$ .

hello, thank you for your hint, that was the reason, the minus sign, ok.
Sonnhard.

Let $ x_1,x_2,x_3$ be the roots an assume wlog $ x_2^2 \equal{} x_1x_3 \iff x_2^3 \equal{} x_1x_2x_3\equal{}8$. So $ x_2 \in \{2,2\omega,2\omega^2\}, \omega\equal{}e^{i\frac{2\pi}{3}}$. In any case: $ x_2^3 \equal{} 8$, so $ x_2^3\minus{}7x_2^2\plus{}cx_2\minus{}8\equal{}0 \iff c \equal{} 7x_2$. So the only solutions are $ c \in \{14,14\omega,14\omega^2\}$, and it's easy to check that they are indeed solutions.

thanks for all who particibate in thes solutions
Cndy : u have asmal mistak my dear a^3 k^3 = 8 not -8 !!
thaanks to :
Dr sonnhard
mateescu constantin
IW@IT
mathias_DK
my solution is like IW@IT
wailt the next
almonkez
 :10:


A triangle is formed by connecting the points $ (5,0)$, $ (0,6)$, and $ (5,0)$. The resulting triangle is then rotated about the $ x$-axis to form a solid. The original triangle is then rotated about the $ y$-axis to form a different solid. What is the positive difference between the number of cubic units in the volumes of the two resulting solid? Express your answer in terms of $ \pi$.

Assuming that first point is $ (0,0)$...

One of the cones has radius $ 5$ and height $ 6$, so it has volume \[ \frac{5^2\pi \cdot 6}{3}\equal{}50\pi\] The other cone has radius $ 6$ and height $ 5$, so it has volume \[ \frac{6^2\pi \cdot 5}{3}\equal{}60\pi\]  The positive difference is $ 60\pi \minus{}50\pi \equal{}\boxed{10\pi }$

Actually, the second solid is not a cone because it's rotated about the x-axis. It's actually two cones sharing a common base. The volume is 2*60pi =120pi.

The positive difference is 120pi-50pi=70pi

[quote="neutrus"]Actually, the second solid is not a cone because it's rotated about the x-axis. It's actually two cones sharing a common base.[/quote]
I don't understand why the second solid is actually two cones sharing a common base.    Can you elaborate please?  Thank you.

ummmmm, (5,0) and (5,0) are the same point.  We have a line rotated about the x and y axes, not a triangle.


Find a closed form for $\sum_{k=1}^{n}kx^{k}$

[hide="here is what i got"]
$\frac{(n+1)x^{n+1}(x-1)-x^{n+2}+x}{(x-1)^{2}}$

ill post my proof i just want to see if anyone else gets the same answer and if so if they got it in a different way.
[/hide]

Yes, that is correct. Note that the infinite sum (what you get by taking the limit as $n\to\infty$ for $|x|<1$) gives a cleaner and easier to remember answer.

$\frac{1}{1-x}=\sum_{k=0}^\infty x^{k}$
$\frac{x}{(1-x)^{2}}=\sum_{k=1}^\infty kx^{k}$
$\sum_{k=n+1}^\infty kx^{k}=x\cdot\frac{d}{dx}\frac{x^{n+1}}{(1-x)}$
$\sum_{k=1}^{n}kx^{k}=\sum_{k=1}^\infty kx^{k}-\sum_{k=n+1}^\infty kx^{k}$
Subtract the corresponding closed forms in the last line.


Dear Math Enthusiast,

Our second competition of the year, the 2005 Team Scramble, took place on Thursday, November 3rd. Because teams are asked to solve 100 problems in just 30 minutes, teamwork and management skills are just as essential as mathematical knowledge, requiring teams to quickly assign and double-check problems. The recognized schools were:

1   Vestavia Hills High School
2   Case High School
3   Saratoga High School
4   Eastlake High School
5   Grissom High School
6   Hoover High School
7   Honourable Vincent Massey Secondary School
8   Parkview High School
9   Palo Alto High School
10  Texas Academy of Math and Science
11  Thomas Jefferson High School
12  Moses Lake High School
13  Parkland Sr. High School
14  Singapore American School
15  Klein High School
16  Science Academy of South Texas
17  Parkway South High School
18  Milpitas High School
19  Ladue Horton Watkins High School
20  Homestead High School
21  Menlo School
22  Garfield High School
23  White Station High School
24  Memphis University School
25  Fort Myers High School
26  Summerville High School

The recognized scores were: 42-79

     Scores tended to change a bit during our double-check, partly due to an error on our part.  The diagram for problem 76 was mislabeled, causing the answer to be 0 instead of our stated answer of 350.  We apologize for this error, but take some solace in the fact that "trick" problems are part of this test and many schools caught this one.

Thank you,

Tom Clymer
Director of Academic Competitions
National Assessment & Testing
http://www.natassessment.com

P.S. Our third contest, the 2005 Ciphering Time Trials, consists of ten rounds in which individuals are challenged to solve three problems in three minutes. Problems in each round cover a range of topics and difficulties, offering confidence and challenges to students of all abilities. The 2005 Ciphering Time Trials will occur on Thursday, December 8th, 2005, and has a registration fee of fifty dollars.

[quote="chess64"]Surprising, TJHSST only got 11th :D[/quote]

Surprisingly, you are referring to the wrong school.

[quote="chess64"]Surprising, TJHSST only got 11th :D[/quote]

yeah, the school that is 11th is the one in WA, which is a really strong school.
but on this one they didn't have their strongest person (xxreddevilzxx) on the team so that might be the reason they didnt do as well.

[quote="eryaman"]when are you guys updating the website?[/quote]

looks like it has been done now... :)


[quote="aznmonkey1992"]joe sleeps for 5/13 of the day. How many days of sleep does he get after 377 days[/quote]
[hide]
5/13 x 377= 1885/13 = 145

[/hide]

[hide][quote="aznmonkey1992"]joe sleeps for 5/13 of the day. How many days of sleep does he get after 377 days[/quote]

$\frac{5}{13} \times 377 = 145 \text{ days of sleep.}$[/hide]

[quote="aznmonkey1992"]joe sleeps for 5/13 of the day. How many days of sleep does he get after 377 days[/quote]

[hide="math92s spamming told me to do this"]
$\frac{5}{13}\cdot377=145$
[/hide]

[hide]
All you have to do is $\frac{5}{13} * 377 = \framebox{145}$[/hide]

[quote="math92"]use the hide thing ;)[/quote]

:spam: potentially useless.


Andrew can paint a room in $ 3$ hours. Bob can paint the same room in $ 6$ hours. How many hours will it take Andrew and Bob to paint the room together?

We can write $ x(\frac{1}{3}\plus{}\frac{1}{6})\equal{}1$, and $ x\equal{}\frac{1}{\frac{1}{2}}$, and $ x\equal{}\boxed{2}$.

How did you arrive at your first equation?

Andrew can paint $1/3$ of a room in $1$ hour, and Bob can paint $1/6$ of a room in $1$ hour, so together, they paint $1/3 + 1/6 =1/2$ of a room in one hour. Thus, it will take $2$ hours.

[hide=My take]
This means Andrew can paint $\dfrac{1}{3}$ of the room per hour, and Bob can do $\dfrac{1}{6}$ per hour. Adding these, they can do $\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{3}{6}$ or $\dfrac{1}{2}$ of the room per hour together. Thus, it takes two hours to paint the room with Andrew and Bob.[/hide]


I cannot see anyone's, or my geogebra. The applet loads, and nothing happens. It's just white and says Applet Loaded.

It's your computer at fault. Reinstall Java on your computer.


superior algebra
superior algebra solved

Is it true that:

For any (finite) group G, and any element x in G, then the size of the centralizer of x times the size of the conjugacy class of x will equal the size of the group?

So, |Z_G(x)|*|C_G(x)|=|G|

thank you

Let $H$ be the conjugacy class of $g$ in $G$, where $G$ is not necessarily finite.  Then the fundamental counting principle says that $|H|=|G: C_G(g)|$.  By Lagrange, if $G$ is finite, then

$|H||C_G(g)|=|G|$

I think I have a detalied solution. Correct me if  I'm wrong.

$Z_G(g)=\{xgx^{-1}|x \in G \}={g_1,g_2...,g_m}$. Now I count the number of x such that $xgx^{-1}=g_i$.
$xgx^{-1}=ygy^{-1} \Leftrightarrow xy^{-1} \in C_G(g)\Leftrightarrow x \in C_G(g)y$ so this number is $|C_G|$. Since $xgx^{-1}$ takes the value $g_i$ for $C_G(g)$ times when x takes all values in G we have that $m\cdot |C_G(g)|=|G|$ q.e.d.

Another solution is by using the action of G on G by conjugation and writing that
\[ O(g)=G/\text{Stab}_g(x) \]
we can easily get that  O(g) is the conjugacy class of g,and $\text{Stab}_g(x)$  is the centralizer of element g.


Show that if $ f(x,y)$ is symmetric and $ x \minus{} y \mid f(x,y)$ then $ (x \minus{} y)^2 \mid f(x,y)$

Write $ f(x,y)\equal{}(x\minus{}y)g(x,y)$. Since $ f$ is symmetric, we have $ (x\minus{}y)g(x,y)\equal{}(y\minus{}x)g(y,x)$, so $ g(y,x)\equal{}\minus{}g(x,y)$ for all $ x\neq y$, and thus (by continuity) for $ x\equal{}y$ as well. setting $ x\equal{}y$, we get $ g(x,x)\equal{}\minus{}g(x,x)$, implying $ g(x,x)\equal{}0$ for all $ x$. but this means $ (x\minus{}y)|g(x,y)$, and therefore $ (x\minus{}y)^2|f(x,y)$.


I need help for this mathcounts problem

Square ABCD has side lengths of 13 units.  Point E lies in the interior of the square such that AE = 5 units and BE =12 units. What is the distance from E to side A?  Express your answer as a mixed number.

I presume by "side $ A$", you mean "side $ AB$".

 Drop a perpendicular from $ E$ to  $ AB$, meeting $ AB$ at $ F$.

Let $ AF\equal{}x$ so that $ FB \equal{} 13\minus{}x$ and let $ EF \equal{} y$.

Then by Pythagoras in triangle $ AEF$, $ x^{2} \plus{} y^ {2} \equal{} 25$ so that $ y \equal{} \sqrt{25\minus{}x^{2}}$ and in triangle $ EFB$, $ y^{2} \plus{} (13\minus{}x)^{2} \equal{} 144$.

Therefore  $ 25\minus{}x^{2} \plus{}169\minus{}26x\plus{}x^{2} \equal{}144$

$ 26x\equal{}50$

$ x\equal{}\frac{50}{26} \equal{} \frac{25}{13}$.

Therefore $ y\equal{} \sqrt{25\minus{}(\frac{25}{13})^{2}}$

$ \equal{}\frac{\sqrt{25\times13^{2} \minus{}25^{2}}}{13}$

$ \equal{} \frac{\sqrt{25(169\minus{}25)}}{13}$

$ \equal{}\frac{5 \times 12}{13}$

$ \equal{}\frac{60}{13}$

$ \equal{}4\frac{8}{13}$.

I hope this is helpful, guilou25.


find all function $ f: R\rightarrow R$ that: $ f(2x)\plus{}f(1\minus{}x)\equal{}5x\plus{}9$

[quote="ZVIL"]find all function $ f: R\rightarrow R$ that: $ f(2x) \plus{} f(1 \minus{} x) \equal{} 5x \plus{} 9$[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=317255


There is a pile of $ 2009$ chips on a table. You are allowed to perform repeatedly the following operation: choose a pile containing more than two chips, throw away a chip from the pile, the divide into two smaller (not necessarily equal) piles. Is it possible that eventually all the piles on the table consist of exactly three chips?

[quote="serialk11r"]2009 is divisible by 7, and if you take 7 chips, throw one out, and split into groups of 3... :)[/quote]
That's what I initially thought too, but note that the first pile you must choose has to be the pile of 2009 chips. You're not allowed to split up the pile of 2009 before performing the operation. The "if you take 7 chips" part of your solution is flawed.  :ninja:

In that case:
[hide]
Since we are splitting the piles in 2 every time, we add one pile per operation. At the same time we reduce the number of chips by 1.
We want to find a solution to the equation in integers:
2009-n=3(n+1)
2009-n=3n+3
2006=4n
However 2006=2 mod 4, so this is not possible.[/hide]

Here's a nice proof by contradiction written by me:  :D
assume 2009 can be split up into n groups of 3. examining in mod 4, we see all piles are congruent to 3 (mod 4) since they are all 3. working backwards, instead of subtracting one and dividing into two smaller piles, we take two smaller piles (namely two piles of three) and add one. 3+3+1=7; notice 7=3(mod 4). Now we have n-1 piles all congruent to 3 (mod 4). So, we at last get the original pile, 2009, which by our assumption must be congruent to 3 (mod 4). However, it isn't, so 2009 can not be split up into piles of 3.


modular arithmetic
algebra
polynomial
calculus
derivative
number theory open
number theory

Let $a_{k}$ be the smallest natural solution to the equation $x^{3}+2x+1\equiv 0\pmod{2^{k}}$

Let $f(k)$ be the sum of the digits of the binary representation of $a_{k}$

Computer analysis shows that:

$f(n)\approx \frac{n}{2}$

For instance:

$f(10)=6$ and the mistake is $\frac{1000}{10000}$

$f(100)=49$ and the mistake is $\frac{100}{10000}$

$f(1000)=507$ and the mistake is $\frac{70}{10000}$

$f(10000)=5025$ and the mistake is $\frac{25}{10000}$

Is there a particular reason for this? Is this true also for larger n?

Can this be generalized to other polynomial?

A possibility:

Given $a_{k}$ we know that either $a_{k+1}= a_{k}$ or $a_{k+1}= a_{k}+2^{k}$.  On average, we would expect these two events to occur equally often, so on average we increase the digital sum of $a_{k}$ by $\frac{1}{2}$.  We would expect that for "most" polynomials this result would hold.

[quote="t0rajir0u"]We would expect that for "most" polynomials this result would hold.[/quote]

I did not have time previously to clarify what I meant; here are some necessary conditions on the polynomial $P(x)$.

- $P(x)$ cannot have any integer roots.  Otherwise, these integer roots would determine the roots $\bmod 2^{k}$ for sufficiently large $k$.

- $P(x)$ must have solutions $\bmod 2^{k}$ for all $k$.  For example, the polynomial $x^{2}-5$ has no solutions for $k \ge 3$.  I think the general condition is that the formal derivative $P'(x)$ cannot be identically zero $\bmod 2$.


This seemed kind of hard to me for GS.  Then again, I'm probably solving it in too difficult a fashion.

[hide="Solution"]
$3abc + a^2 (b + c) + b^2 (c + a) + c^2 (a + b) = a^3 + b^3 + c^3 - (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + a^2 (b + c) + b^2 (c + a) + c^2 (a + b) = (a + b + c)(ab + bc + ca)$
[/hide]

[quote="shadysaysurspammed"]Factorise

$(a+b)(b+c)(c+a)+abc$

 :)[/quote]

consider a polynomial $f(a)=(a+b)(b+c)(c+a)+abc$
$f(-b-c)=0$, so $(a+b+c)$ is a factor, it is pretty easy to factor that way, and you find the other factor to be $(ab+bc+ca)$

Exactly Altheman :) ,though some might argue that you did that after seeing the answer but it is right; nevertheless,I have another solution as well....

$(b+c)(c+a)=c^2+ab+bc+ca$

So,

$(a+b)(b+c)(c+a)=(a+b)(ab+bc+ca)+(a+b)c^2$

$(a+b)(b+c)(c+a)+abc=(a+b)(ab+bc+ca)+(a+b)c^2+abc$

                                    $=(a+b)(ab+bc+ca)+c(ac+bc+ab)$

                                     $=(a+b+c)(ab+bc+ca)$

your solution is so large
we can substitude a+b+c = s
then equation becomes (s-a)(s-b)(s-c) + abc
which can be written as s^3 - s^2(a+b+c) + s(ab + bc + ac)
which is s(ab+bc+ac)

hence it is  (a+b+c)(ab+bc+ac)

[quote="shadysaysurspammed"]I was hoping for a one-liner :D

Yeah,your method is a little complicated to what I've done.The way I solved it it seemed like a G.S level problem. :)[/quote]

All factorizations have a one-liner consisting of the factorization itself, but I wanted to show at least a few of the nonintuitive intermediate steps :D


quadratics
number theory unsolved
number theory

(a) If $m, n$ are positive integers such that $2^n-1$ divides $m^2 + 9$, prove
that $n$ is a power of $2$;
(b) If $n$ is a power of $2$, prove that there exists a positive integer $m$ such
that $2^n-1$ divides $m^2 + 9$.

This one is from ISL 98 :
http://www.kalva.demon.co.uk/short/soln/sh98n5.html

Pierre.

Well, this is a beautiful problem. I will use Quadratic Reciprociti to solve it.

Let $p$ be a prime factor of $2^n-1$. Then $p|2^n-1|m^2+9$ and so $\left(\frac{9}{p}\right) = 1$.

Typo: it should be $-9$ instead of $9$.
But how do you continue then¿ Similar as in the link¿

Well, this is a beautiful problem. I will use Quadratic Reciprocity to solve it.

[b]LEMMA[/b]: if $p$ is a prime which divides $2^n-1$(in the assertive of this problem), then $p=3$ or $4|(p-1)$ .

Let $p \not = 3$ be a prime factor of $2^n-1$. Then $p|2^n-1|m^2+9$ and so $\left(\frac{-9}{p}\right) = 1$.
Then $1=\left(\frac{-9}{p}\right) = \left(\frac{-3}{p}\right)\cdot\left(\frac{3}{p}\right)$ and so $\left(\frac{-3}{p}\right)=\left(\frac{3}{p}\right)$. But by Quadratic Reciprocity, this is equivalent to
$(-1)^{\frac{-3-1}{2}\frac{p-1}{2}}=(-1)^{\frac{3-1}{2}\frac{p-1}{2}}$ (because $\left(\frac{p}{3}\right)=\left(\frac{p}{-3}\right)$). Then $\frac{p-1}{2}$ is even, or $p \equiv 1 (mod\ 4)$.

Then, with this lemma is easy to prove that $n$ is a 2-power.
In fact, in $k$ is an odd divisor of $n$, then $2^k-1 | 2^n-1$ and $2^k-1 \equiv 3 (mod\ 4)$. But the divisors of $2^n-1$ leave remainder 1 modulo 4, and the product of numbers of this form is always of this same form. Then $2^k-1 \equiv 1 (mod 4)$, a liar! An absurd!

Well, to prove that any 2-power is sufficient, try to factor $2^{2^k}-1$. This is the most funny part!The factors of it is exactly the Fermat numbers $F_i=2^{2^i}-1$


A sequence of positive integers is defined by $ a_0\equal{}1$ and $ a_{n\plus{}1}\equal{}a_n^2\plus{}1$ for each $ n\geq 0$.  Find $ \gcd(a_{999},a_{2004})$.

hmm... unless i dont understand your problem it seems like the numbers in the sequence increase by 1?

Sorry never mind.  I see what i did.  Sorry again

[quote="math_explorer"][hide="Not sure"]
Note that $ a_{n \plus{} 1} \equiv 1 \mod a_{n}$.
and $ a_{n \plus{} 2} \equal{} a_{n \plus{} 1}^{2} \plus{} 1 \equiv 1^{2} \plus{} 1 \mod a_{n}$
$ a_{n \plus{} k} \equiv a_{k \minus{} 1} \mod a_{n}$
$ \gcd(a_{999}, a_{2004}) \equal{} \gcd(a_{999}, a_{1004}) \equal{} \gcd(a_{999}, a_{4}) \equal{} \gcd(a_{994}, a_{4}) \equal{} \cdots \equal{} \gcd(a_{4}, a_{4}) \equal{} a_{4} \equal{} 677$
[/hide]
Is that right?[/quote]

Yes, that's right.  Nice solution.


algebra
polynomial
greatest common divisor
quadratics
quadratic formula
number theory
relatively prime

This new teacher in my school calls this method the "rainbow method". But what is it REALLY called? and whos method is this?

It is used to factor any polynomial in the form ax^2 + bx + c where a, b, and c are not equal to 0. Here is an example:

Factor 24y^2 - 2y - 1

1. Multiply the "a term" with the constant "c term". Remove the "a term" from the polynomial.

ie. y^2 - 2y - 24

2. Factor the new polynomial.

ie. (y-6)(y+4)

3. Place the "a term" you removed earlier back into each factor, that is, before the variable.

ie. (24y - 6)(24y + 4)

4. Find the GCF for each factor. Divide each factor by that GCF.

ie. (4y - 1)(6y + 1)

5. That is 24y^2 - 2y - 1 in factored form.

Has anyone heard of or seen this? Using this method, it is possible to factor polynomials with insanely large coefficients such as 84x^2 - 62x + 48. Please let me know what this method is really called and who came up with it. Thanks.

i don't really see why you would do this
here is a better way to factor...that might be related or it might not
given $ax^2 + bx + c$, it can be written in the form $a(x-r_1)(x-r_2)$ where $r_1$ and $r_2$ are the roots of the quadratic (use the quadratic formula to do this), the way that we know that it can be written that way is from the factor theorem
this really just takes no thinking what so ever, but it works, and it can be used for something with 2 variables too

i will look at this method more, but i don't really understand the first step
is it take $ax^2 + bx + c$ and make it into $x^2 + bx + ac$ ?
(if you could show it with variables, that would be nice, since with the -1, there could be some confusion)

Yes. Make ax^2 + bx + c into x^2 + bx + ac. I guess this method is useful when you're simplying complex fractions where you have to cancel terms of the top and bottom.

you could just show this with the quadratic formula, that being,
$ax^2 + bx + c$

$x^2 + bx + ac$, factoring...
$\left(x - \dfrac{-b+\sqrt{b^2 - 4ac}}{2}\right)\left(x - \dfrac{-b-\sqrt{b^2 - 4ac}}{2}\right)$

then placing the a term in...

$\left(ax - \dfrac{-b+\sqrt{b^2 - 4ac}}{2}\right)\left(ax - \dfrac{-b-\sqrt{b^2 - 4ac}}{2}\right)$

hmm, this does not look like it will always simplify to the correct answer, are you sure that this works all the time?

I remember learning about a method like this in my Algebra II class last year, it went something like this:

Ex. say you have the equation $5x^2+38x+21$.  Multiply $5\cdot 21=105$.  Then find two numbers whose product is $105$ that sum to $38$.  We find $35$ and $3$.  Then split up the $38x$ into $35x+3x$.  So then we have $5x^2+35x+3x+21$.  Factoring now we have $5x(x+7)+3(x+7)$, and factoring out the $(x+7)$ finally gives us $(5x+7)(x+7)$, which is correct.

I think airelemental's method is basically the same as this.

continuing from my last post, i actually see that this works, here is why, in a quadratic where the x coefficient is 1, then the product of the roots is the constant term, and that would be ac in this case, so if we were reducing this, we know that with the expansion of
$\left(ax - \dfrac{-b+\sqrt{b^2 - 4ac}}{2}\right)\left(ax - \dfrac{-b-\sqrt{b^2 - 4ac}}{2}\right)$
there will be an extra a on the x^2 term, but since we know that among both the roots, there is a factor of a (from the product ac), then taking the gcf of the roots and a, will always have that extra a cancel, so it does work, except
if a and c are not relatively prime, then this creates a problem, this is because when you take out the gcf, you will also be taking out some of the c, and changing the expression

comment: i think that this is a reasonably clever way to do things, but at the same time, there are much better ways to factor than this, and also this only works when a and c are relatively prime, so it is quite limited in its usefulness


function
real analysis
real analysis solved

Determine f:R-->R with the properties:
i) there exists the limit : lim(x-->oo) (f(x)-x)
ii) f(f(x))=x+4; for any x real.

exists and is finite lim{x->oo} (f(x)-x) ???
or this limit can be oo too?
the function is
f(x)=x+2???

it only says that it exists, it doesn't say that it is finite!

It is easy to see that f(x+4)=f(x)+4, thus f(x)-x is periodic and has limit , thus is constant.

When it says it exists, it has to be finite, infinity is not a value.

I came back to edit just to say this: not a hard problem, cmon people you can do it. Olympiads are even harder than this.


Να βρεθούν όλες οι λύσεις της $x^2+2\cos \phi \cdot x+1=0$

αν $x \in \mathbb{R}$ και $\phi \in [0,2\pi)$

Προφανώς  , εννοείς να βρεθούν όλα τα ζεύγη  $( x , \phi)$. Αφήνω τη λύση της απλής αλλά όμορφης αυτής άσκhσης για τους μαθητές μας. Αν χρειαστεί - το αποκλείω βέβαια - θα επανέλθω .

 [b]Μπάμπης[/b]

ναι, εννοώ όλα τα ζεύγη  :)

Πολύ μου άρεσε ως άσκηση.
Να η λύση που βρήκα

$x^2+2cos \phi\cdot x+1=0$   <=>  $x^2+2cos \phi\cdot x+(sin\phi)^2 +(cos\phi)^2=0$

<=>$(x+cos\phi)^2+(sin\phi)^2=0$

άρα και τα δύο είναι μηδέν και οι λύσεις είναι $\phi =0$ kai $x=-1$
Πώς σας φάνηκε ?Λουκά έχεις άλλη λύση?

nice!  :)

Ναι! Έχω κι άλλη λύση  :D

$\phi = \pi$ και $x=1$

η άσκηση έτσι μου ... ήρθε προχθές και σκέφτηκα ότι ήταν καλή ιδέα να τη βάλω στο φόρουμ.

Δείτε επίσης το http://www.mathlinks.ro/Forum/viewtopic.php?t=78173 (έχει λυθεί)


geometry
parallelogram
circumcircle
floor function
ratio
trigonometry

In parallelogram $ABCD,$ let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}.$  Angles $CAB$ and $DBC$ are each twice as large as angle $DBA,$ and angle $ACB$ is $r$ times as large as angle $AOB.$  Find the greatest integer that does not exceed $1000r.$

[hide="I think I did something wrong"]Let angle $DBA=x$ and angles $CBD=CAB=2x$. Angle $AOB=180-3x$ by degrees in a triangle. This means angle $BCA$ is $r(180-3x)$. Because adjacent angles of a parallelogram have to be complementary, angle $CAD$ is $180-5x$, and because alternate interior angles are the same, $180-5x=r(180-3x)$ so $r=\frac{180-5x}{180-3x}$ which, decreases as $x$ increases, so the biggest $r$ can be is when $x \to 0 = 1$ The most $1000r$ can be then is $1000$ as $x \to 0$ so the largest integer that does not exceed $1000r$ is $999$?[/hide]

Is it possible that $\triangle AOB$ is a 4-5-6 triangle?

[hide="solution"]Note that i am going to focus on $\triangle ABC$, let $\angle ABO=x$, then $\angle BAO=\angle OBC=2x$. Draw perpendicular from $C$ intersecting $AB$ at $H$. Let $AO=CO=1$. Then $HO=1$ since $O$ is the circumcenter of $\triangle AHC$. Then $\angle OHA=2x$.

By the exterior angle theorem, $\angle COB=3x$ and $\angle COH=4x$. That implies that $\angle HOB=x$. That makes $HO=HB=1$. Then since by AA ($\angle HBC=\angle HOC=3x$ and reflexive on $\angle OCB$), $\triangle OCB \sim \triangle BCA$.
$\frac{CO}{BC}=\frac{BC}{AC} \implies 2=BC^2 =\implies BC=\sqrt{2}$.

Then by the pythagorean theorem, $1^2+HC^2=\sqrt{2}^2\implies HC=1$. That makes $\triangle HOC$ equilateral. Then $\angle HOC=4x=60 \implies x=15$. Then $\angle AOB=180-3*15=135$ and $\angle ACB=180-5*15=105$

Then $\frac{\angle ACB}{\angle AOB}=r=\frac{7}{9}$. Then it follows that$\left\lfloor \frac{7000}{9}\right\rfloor=\boxed{777}$[/hide]

Uh...I think there's a typo...

$HOC = 4x$ in your solution, but then later you say $HOC = 3x$.

[hide="A Different Method"]
Let $ \angle ABD\equal{}x$. We obtain $ \angle ADB\equal{}\angle BAO\equal{}2x$, and see that triangles $ ADB$ and $ OAB$ are similar.  Also, we get $ \angle ACD\equal{}2x$ and $ \angle ODC\equal{}x$.

From the pair of similar triangles, we obtain

$ \frac{AD}{AO}\equal{}\frac{AB}{BO}\equal{}\frac{DB}{AB}$.

Note that $ DB\equal{}2BO$.  So $ 2AB^2\equal{}DB^2$, and the similarity ratio is $ \sqrt{2}$.

Thus $ AD\equal{}AO\sqrt{2}$, or$ AD\sqrt{2}\equal{}AC$.  Now, by the Law of Sines,

$ \frac{AC}{AD}\equal{}\sqrt{2}\equal{}\frac{\sin 3x}{\sin 2x}$.

Expanding this out, we get

$ \sqrt{2}\equal{}\frac{\sin 2x\cos x\plus{}\sin x\cos 2x}{2\sin x\cos x}$.

Expanding even more, this equals

$ \frac{2\sin x\cos^2 x\plus{}2\sin x\cos^2 x \minus{} \sin x}{2\sin x\cos x}\equal{}2\cos x\minus{}\frac{1}{2\cos x}$.

Let $ 2\cos x\equal{}a$, we solve $ a\minus{}\frac{1}{a}\equal{}\sqrt{2}$ to find

$ a\equal{}\frac{\sqrt{6}\plus{}\sqrt{2}}{2}$.

Thus $ \cos x\equal{}\frac{\sqrt{6}\plus{}\sqrt{2}}{4}$, which is easily recognizable as $ \cos 15$.  So $ x\equal{}15$.

Let $ \angle AOB\equal{}y$, then $ \angle CAD\equal{}ry$.  In triangles $ ADB$ and $ AOB$ we get

$ y\plus{}3x\equal{}180\Rightarrow y\equal{}135$.

$ ry\plus{}5x\equal{}180\Rightarrow ry\equal{}105$.

So $ r\equal{}\frac{105}{135}\equal{}\frac{7}{9}$ and the answer is $ 777$.  [/hide]

[geogebra]6199aea640f7f0dffedb7c26032724d699f4b3f8[/geogebra]

Let $ \angle DBA\equal{}\angle BDC\equal{} x$ and $ \angle CAB\equal{}\angle DBC\equal{} \angle ADB \equal{} \angle ACD \equal{}2x$. Let $ BC\equal{}AD\equal{}Z$.

Pick point $ P$ on $ DB$, so that $ CP\equal{}BC\equal{}Z$. Now we notice $ \triangle{BPC}$ is isosceles, so $ \angle CBP \cong \angle CPB$ or $ \angle CPB\equal{}2x$.

$ \angle PDC \plus{} \angle PCD \equal{} \angle CPB$, from the exterior angle theorem, so $ \angle PCD\equal{}x$. $ \angle PCD \cong \angle PDC$ so $ \triangle PDC$ is isosceles with $ PD\equal{}Z$.

Now we notice $ \triangle ADP$ is isosceles because $ DP\equal{}DA\equal{}Z$, so $ \angle DAP \cong \angle DPA$ and they both equal to $ 90\minus{}x$ because angles in a triangle sum to $ 180$.

From before, we had $ \angle BPC\equal{}2x$ and $ \angle BAC\equal{}2x$ so $ \angle BPC \cong \angle BAC$ and therefore quadrilateral $ BAPC$ is cyclic and so $ \angle CAP \cong \angle CBP\equal{}2x$

From the parallelogram, $ \angle DAC \cong \angle ACB$. $ \angle ACB\equal{} 180\minus{}2x\minus{}3x\equal{}180\minus{}5x$ and $ \angle DAC \equal{} 90\minus{}x\plus{}2x\equal{}90\plus{}x$

So $ 180\minus{}5x\equal{}90\plus{}x$ and $ x\equal{}15$.

$ \angle AOB\equal{}180\minus{}3x\equal{}135$ and $ \angle ACB\equal{}180\minus{}5x\equal{}105$

$ r\equal{} \frac{\angle ACB}{\angle AOB}\equal{}\frac{105}{135}\equal{}\frac{7}{9}$

And so $ \left\lfloor1000r\right\rfloor\equal{} \left\lfloor\frac{7000}{9}\right\rfloor\equal{}\boxed{777}$

Sorry for the massive revive, but I feel like I'm not stating anything new, but using a method from [b]vincenthuang75025[/b]'s excellent article, which can be found here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=721&t=609795
This idea is likely inspired from the following problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2969813&sid=a45d0778dd9897a2ca10999554fb74f7#p2969813

So, as in the above solutions, note that we must solve $\sin{5\theta}\cdot\sin{\theta} = \sin{2\theta}^2$

We can do this with messy trig identities, but consider the fact that if $e^{ix}=z$ is on the unit circle in the complex plane, that $\frac{z-\frac{1}{z}}{2i} = \sin{x}$. Applying this, we have,

$\frac{z^5-\frac{1}{z^5}}{2i}*\frac{z-\frac{1}{z}}{2i} = (\frac{z^2-\frac{1}{z^2}}{2i})^2$

$(z^{10}-1)*(z^2-1) = (z^5-z)^2$

This comes out to $z^{10}-z^8-z^6+z^4+z^2-1=0$.
Factoring out a $z^2-1$ gives
$z^8-z^4+1=0$.

which implies $z^{12}=-1$, or $z^{24} = 1$. One solution is $z=cis(15)$, which suffices, and we are done, $\theta = 15$

sorry for the revive, but at the solution above, how does va2010 go from $(z^{10}-1)*(z^2-1) = (z^5-z)^2$ to $z^{10}-z^8-z^6+z^4+z^2-1=0$.? I get a different equation.

Let $\angle OBA = x$, so $\angle CAB = \angle DBC = 2x$. Let $E$ on $\overline{BO}$ be such that $\angle EAO = \frac{\angle BAO}{2} = x$. Since $\angle CAB = \angle DBC$, the circumcircle of $\triangle OAB$ is tangent to $\overline{BC}$. By power of a point, $\frac{1}{2}CA^2 = CA\cdot CO = CB^2$. We also get $\triangle CBO \sim \triangle CAB$, so $BO = AB\cdot \frac{CB}{CA} = \frac{AB}{\sqrt{2}}$. Since $\angle EAO = x = \angle DBA$, the circumcircle of $\triangle BAE$ is tangent to $\overline{AC}$. By power of a point and the angle bisector theorem, we see that
\[OA^2 = OE\cdot OB = \frac{OB\cdot OA}{OA + AB}\cdot OB.\]
Equivalently, we have $\frac{AB^2}{2} = OB^2 = OA^2 + OA\cdot AB$. Dividing by $OA^2$ and letting $y = \frac{AB}{OA}$ yields $\frac{1}{2}y^2 = 1 + y$. Rearranging and multiplying through by 2, we get $y^2 - 2y + 1 = 3$. Since $y$ is positive, it equals $\sqrt{3}+1$.
Thus, up to scaling, the ratio $AB:BC:CA$ is $\sqrt{3}+1:\sqrt{2}:2$. Consequently, $\triangle ABC$ can be made by putting a $1:1:\sqrt{2}$ and $1:\sqrt{3}:2$ triangle and putting them together along the 1, so $\angle BCA = 105^\circ$. Since this angle is $180^\circ - 3x - 2x = 180^\circ - 5x$, $x = 15^\circ$. The value of $\angle AOB$ is then $180^\circ - 2x - x = 180^\circ - 45^\circ = 135^\circ$, so $r = \frac{105}{135} = \frac{7}{9}$.


Did anyone see today's Foxtrot comics? LOL.

EDIT: Here's a link: [url]http://www.gocomics.com/foxtrot/2009/04/19[/url]

There's another comic with the same property...

It's posted on [url=http://mathworld.wolfram.com/FoxTrotSeries.html]MathWorld[/url]. Check it out.

What's crazy is that the problem actually has an answer.

Haha, I remember getting that comic Sunday and working it a bit to see the message.  It was kind of tedious but also sort of cool.  Did you see the other one FoxTrot ran, the one about Merry Christmas?


Plz tell me the complete logic to each question.

[hide="Question 2"]A, C, and D give syn-elimination.

A - if the reaction conditions are heating to about 300ºC, then this is the pyrolysis of an esther.

C - this is the Chugaev Reaction: heating a methyl xanthate to about 200ºC brings about a syn elimination to give an alkene.

D - this is the Cope reaction.[/hide]

Oh so we have to learn the stereoselectivity/specificity of such reactions?

Can you also tell me the other two ques.

In first I think since E1 and E2 both have almost the same preference for the degree of C , so I think all 4 can go E2 if base is strong and/or under E2 like conditions.

In 3rd I think more sterically hindered base/substrate will cause hoffman elimination. But this way I could convince myself that only A) would give hoffman.

[hide="Question 1"]
I think it can be both (a) and (b).
For elimination to take place the leaving group should be in the axial.
If it is equatorial then it will flip so that elimination takes place. but here the t-butyl group freezes the chair form so that no flipping takes place. So answer can be only (a) or (b)
[/hide]

[hide="Question 3"]
The answer is (a),(c),(d).
For (a)- The base is bulky and thus prefers the less hindered proton.
Both (c),(d)-The leaving group is bulky and thus the base prefers the less hindered proton for attack
[/hide]


algebra
polynomial
inequalities
inequalities solved

Prove or disprove for $a,b,c\ge0$ and every positive integer $n$:

$(a^{n}+b^{n}+c^{n})+(a^{n-2}bc+ab^{n-2}c+abc^{n-2})\ge(a^{n-1}b+ab^{n-1}+b^{n-1}c+bc^{n-1}+c^{n-1}a+ca^{n-1})$

[hide="some notes"]note that this is equivalent with $(n,0,0)+(n-2,1,1)\ge(n-1,1,0)+(n-1,1,0)$
Second note that this is exactly Schur[/hide]

This is true, it is general Schur.

EDIT: oops, didn't see your note. There you go. :lol:


Does anybody know any good tutorials for algebra?

[quote="Silverfalcon"]Here is a REALLY good site for algebra.

http://www.purplemath.com/

I think it teaches you from very basic to pretty advanced level though. I never used it myself but I heard it's good. Hope this helps.[/quote]

I've heard of it too.  :lol:


geometry
3D geometry
geometric transformation
rotation
octahedron

[quote="bthings_2000"]6. What is the number of ways to color the six faces of a cube six different colors? (Two ways are considered to be the same if one can be obtained from another by rotating the cube).

Here is the solution, and what I don't understand:
[hide]
There are $6!$ ways to color the cube and $8\times{3}$ rotations, so the answer is $\frac{6!}{8\times{3}}=30$ ways. I do not understand where $8\times{3}$ comes from.
[/hide][/quote]

Wait, do you have the link so we can better see the problem.

I can see that it must be multiplicative.

That is the exact problem, including their parenthesis, and there is no picture.

What it means is that for any cube with 6 different colored faces, you can rotate it in 24 different ways. A nice way to see this is to realize that there can only be one face on top, and for each face on top, it can be rotated 4 different ways, because there are 4 sides. There are 6 faces, the there are $6*4$ ways to rotate the cube.

I've seen this problem before, and remember how I solved it:

Pick two colors. The cube can always be rotated so that the first color is on top.

Now the second color can either be opposite the first color (on the bottom), or on one of the sides.

If it's on one of the sides, there are $4!=24$ ways to distribute the colors on the remaining 4 sides.

If it's opposite, there are again $4!$ ways to distribute the colors, but $4$ rotations of it, so there are $3!=6$ distinct ways to do this.

Answer: $24+6=30$

To see that there are $8\cdot 3$ rotations, you can visualize the cube as standing on a corner rather than a face (like the octahedron is usually shown as). There is one vertex on the top, one vertex on the bottom, and two equilateral triangles in the middle. So when rotating, we can:
first send any vertex to the top (8 such rotations because 8 vertices of the cube) and then:

perform a second rotation that does not move the top vertex. But such a rotation also fixes the center, so it must be a rotation around the vertical line connecting the top and bottom vertices. So this just becomes a rotation of the equilateral triangles, and there are clearly 3 ways to do this.

More generally, if a regular polyhedron has $v$ vertices and $d$ edges meet at each vertex, then there are $v\cdot d$ rotations.

That mostly makes sense, except I'm not quite envisioning the equilateral triangles you speak of... it's a cube.

They're not really equilateral, and they're not really triangles either.

What he wants to do to count rotations is hold the cube by two opposite corners and spin it around the axis containing those two corners.

There are 8 ways to choose the top corner, and once you pick the top corner, if you spin it around, there are 3 positions because 3 faces come together at each vertex.

Fix one face of the cube and give it color 1. The face opposite it has 5 choices. Then, the faces in between have $3!$ distinct choices because of rotational symmetry. So there are $3! \cdot 5=30$ permutations of colors total.

i understood agolsme solution, i hav no idea the others are saying, can someone explain?


calculus
derivative
group theory
abstract algebra
superior algebra
superior algebra solved

$G$ a group, $S=<\{x^2| x \in G \}>$ the subgroup generated by the squares.
Show that if $G$ is generated by the involutions (elements such that $g^2=Id$), $S=G'$ (the derived group).

First of all, we notice that $S$ always contains $G'$: it is clearly normal, and all the elements of the quotient group $G/S$ have order $\le 2$, so $G/S$ is abelian, i.e. $G'\le S$.

What we want now is to prove that whenever $G$ is generated by involutions, it must be that $S\le G'$, which is equivalent to $x^2\in G',\ \forall x\in G$. Now, $\tilde G=G/G'$ is an abelian group generated by involutions. If we show that in such a group all elements have order $\le 2$, we're done. This is trivial: take some $x\in \tilde G$ and let $x=a_1\cdot\ldots\cdot a_n$, where $a_i$ are involutions. Since all the $a_i$ commute to one another, $x^2=1$, as desired.

yep  :)  I find it is a nice way to find the derivated group of $S_n$, $A_n$, $O_n$, $SO_n$

grobber, i'm sorry to trouble you with such trifles-it must be because of stupidness:
could you explain why $S$ is normal ?

The generating set is closed under conjugation: $a^{-1}x^2a=(a^{-1}xa)^2$.

truly..., thanks-it was blindly not to see it.


AMC
AIME
probability
geometry
3D geometry
MATHCOUNTS

So, who got lazy and followed the pattern on #13 (thus getting the answer as 81)?

I drew a 5x5 lattice and got 27, ugh. I used the method described in the Math Jam.

I got 27 too! But I don't know how they did it in math jam...
I just did a 4 wide/4 high triangular lattice and figured the # of ways to get to each of the rightmost four points.  Then multiplied by two, and added one.

Except I did the first step one... more specifically, I forgot that once you reach the side you can still go up =P

So yeah, it sucked.  And I got 11 for #2.  I'm still kicking myself for that one. And 76 for #9 - I accidentally used the probability of the edges cubes have a NON-painted side O_o. I got a five... heh. But I'm a freshman, so it's all good.

w00t. i followed the pattern because i ran out of time. grr..

I got #13 right, but not the most efficiant way

I also drew a 5x5 and got 27. I can't believe I made such a stupid mistake. Getting it right would have brought me up from a 6 to a 7.

the pattern used get u a wrong answer if u don't look at the direction of the previous step for no right turns can be made...

so do wut i did and label each one as up, across, or diagonal...and solve it in 10min

I hate spring break...made me miss such a easy test...AIME2 better be just as easy...errr

I used a counting method: x for move right, y for move up, and z for diagonal. We know that x and y can't be adjacent and there must be five x and five y(or replace an x and a y with z). I calculated permutations, but the answer i got was 106. anyone kno y this was wrong?

As an 8th grader doing Mathcounts, patterns are what you LOOK FOR, so I really fell into the trap of the pattern on 13. O well I still have the next four years to qualify...

[quote="Mildorf"]Tempting though that may be, you should probably think to yourself that they would try to make the answer unguessable on an AIME #13.[/quote]

well considering how easy number two was (yes it was number two, but comparatively) some people very well could have fallen for the trap.... because it's irritating to do the case by case thing...

I drew the lattice and used a recursion type counting system to count the number of ways to reach each point, disregarding those paths that included right angles.

I got a little scared when I noticed that the number of ways to reach (1,1), (2,2), (3,3), and (4,4) followed a pattern, but I went with 83.

Haha for some reason I thought you could never cross the diagonal from (0, 0) to (5, 5). It bothered me a lot that the diagonal was a boundary, and made my life a lot harder on the problem.

Then I finished the test and realized that I had been hallucinating.

Not the first time it's happened either ... maybe the AIME needs a warning label.

i don't have enough time..but i figure it out by pattern at home after 20minutes(....)

a way to do it without patterns is to represent the possible movements with characters
a- move right
b-move left
c-diagnol
so basically ccccc would be a straight line to (5,5)
then you replace one c with ab to get abcccc. Find all possible permutations of this where a and b are not next to each other (if a and b are next to each other it breaks the rule). then do it for ababccc, abababcc, ababababc

I drew a grid, drew lines connecting the dots going up, right, and diagonal, and just wrote my numbers on the lines.  It wasn't too hard to count recursively until I got to (5,5), but that breakdown of the pattern made me kinda nervous.

I actually wrote out permutations for the case of two up, two right, and three diagonals, i guess i missed a few cases because I got 67 total. (Where in the world did the other 16 come from?  :lol: )