percent
ratio
geometry
analytic geometry
probability
graphing lines
slope

(I am compiling this for the Resources section)

3. If x, y, and z are real numbers such that

$(x-3)^2+(y-4)^2+(z-5)^2=0$

then $x+y+z=$

A) -12  B) 0  C) 8  D) 12  E) 50

Solution:
[hide]
D) 12

The only solution $(x,y,z)$ is $(3,4,5)$. Since the square of any rational number is positive, any other values would result in a sum greater than zero; thus no other solution exists. $3+4+5=12$
[/hide]

4. If $a$ is 50% larger than $c$, and $b$ is 25% larger than $c$, then $a$ is what percent larger than $b$?

A) 20%  B) 25%  C) 50%  D) 100%  E) 200%

Solution:
[hide]
A) 20%

Based on the given information,
$a=\frac{3}{2}c$
$b=\frac{5}{4}c$

Thus the ratio $\frac{a}{b}=\frac{\frac{3}{2}}{\frac{5}{4}}=\frac{6}{5}=1.2=$120%

Thus $a$ is 120% of $b$, or 20% larger than $b$.
[/hide]

6. Consider the sequence
$1, -2, 3, -4, 5, -6,...,$
whose $n^{th}$ term is $(-1)^{n+1}n$. What is the average of the first 200 terms of the sequence?

A) -1  B) -0.5  C) 0  D) 0.5  E) 1
Solution:
[hide]
B) -0.5

Average=$\frac{\text{sum of the first 200 terms}}{200}$

To get the sum, note that starting with the first term (1), every pair of consecutive terms sums to -1. Since there are 200 numbers, there are 100 such pairs, and the sum of the first 200 terms is thus -1(100)=-100.

Average=$\frac{\text{sum of the first 200 terms}}{200}= \frac{-100}{200}$
$= -0.5$
[/hide]

7. The sum of seven integers is -1. What is the maximum number of the seven integers that can be larger than 13?

A) 1 B) 4 C) 5 D) 6 E) 7

Solution:
[hide]
D) 6

Let the seven integers be {a, b, c, d, e, f, g} and let all but $g$ be greater than $13$. Then the sum of those 6 integers can be expressed as $(14)(6)+x$ for some $x$ that is greater than or equal to 0.  Then $g=-((14)(6)+x)-1$ and the sum is still -1; thus, 6 integers are possible.

All Seven integers cannot be greater than 13; if $g>13$, then $(14)(6)+x+g>-1$.
[/hide]

8. Mientka Publishing Company prices its best-seller [i]Where's Walter[/i] as follows:
$C(n)= 12n$ if 1[u]<[/u]n[u]<[/u]24; $11n$ if 25[u]<[/u]n[u]<[/u]48; $10n$ if 49[u]<[/u]n

where $n$ is the number of books ordered and $C(n)$ is the cost of $n$ books. Notice that 25 books cost less than 24 books. For how many $n$ is it cheaper to buy more than $n$ books than to buy exactly $n$ books?

A) 3 B) 4 C) 5 D) 6 E) 8

Solution:
[hide]
D) 6

Clearly the conditions may only be met for values of $n$ close to changes in price. This is because for some price k, where k>0, C(n)=nk, C(n+1)=(n+1)k, and nk<(n+1)k. The price, k, must change.

Following that reasoning, we see that the following cases work:

12(24)>11(25)
12(23)>11(25)
11(48)>10(49)
11(47)>10(49)
11(46)>10(49)
11(45)>10(49)

That is, buying 25 books is cheaper than buying $n={23, 24}$ books, and buying 49 books is cheaper than buying $n={45, 46, 47, 48}$ books.

Thus $6$ values of n meet the conditions.
[/hide]

9. (Figure provided but not necessary.) ABCD is a 2x2 square, E is the midpoint of AD, and F is on BE. If CF is perpendicular to BE, then the area of quadrilateral CDEF=

A) 2  B) $3-\frac{\sqrt{3}}{2}$  C) $\frac{11}{5}$  D) $\sqrt{5}$  E) $\frac{9}{4}$

Solution:

[hide]
C) $\frac{11}{5}$

Impose the figure in the coordinate plane with B on the origin, BA on the positive y axis, and BC on the positive x axis. Then,
$B=(0,0)$
$C=(2,0)$
$D=(2,2)$
$A=(0,2)$

We may solve for point F by noting that F is the intersection of BE, which lies on the line $y=2x$, and CF, which lies on $y=-\frac{1}{2}x+1$. Solving for the intersection gives
$F=(\frac{2}{5},\frac{4}{5})$.

We may find the area of CDEF by taking the area of square ABCD and subtracting right triangles AEB and BFC.

Area of AEB= $\frac{1}{2}(1)(2)=1$
Area of BFC= $\frac{1}{2}(2)(\frac{4}{5})=\frac{4}{5}$

Area of quadrilateral CDEF= $(2)(2)-(1+\frac{4}{5})=\frac{11}{5}$.
[/hide]

I think this should be in the AMC Forum.

10. Two six-sided dice are fair in the sense that each fave is equally likely to turn up. However, one of the dice has the 4 replaced by 3 and the othetr die has the 3 replaced by 4. When these dice are rolled, what is the probability that the sum is an odd number?

A) $\frac{1}{3}$  B) $\frac{4}{9}$  C)$\frac{1}{2}$  D) $\frac{5}{9}$  E) $\frac{11}{18}$

Solution:
[hide]
D) $\frac{5}{9}$

The dice have faces {1,2,3,3,5,6} (let this be #1) and {1,2,4,4,5,6} (let this be #2), respectively.

Since each die has six faces, there are $(6)^2=36$ combinations of faces total.

The sum of the two rolls will be odd if and only if the sum includes one even number and one odd number.

There are 4 odd numbers on die #1 and 4 even numbers on die #2, making for $(4)(4)=16$ combinations that will sum to an odd number.

There are 2 even numbers on die #1 and 2 even numbers on die#2, making for $(2)(2)=4$ combinations that will sum to an even number.

The probability of an odd sum is thus $\frac{16+4}{36}$=$\frac{5}{9}$.
[/hide]

11. In the sixth, seventh, eigth, and ninth basketball games of the season, a player socred 23, 14, 11, and 20 points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than 18, what is the least number of points she could have scored in the tenth game?

A) 26  B) 27  C) 28  D) 29  E) 30

Solution:
[hide]
D) 29

Her points-per-game average for games 6,7,8, and 9 was $\frac{23+14+11+20}{4}=17$ points. For this to have increased her average from the first five games, her average from the first five games must have been less than 17. Since we are looking for the least number of points she could have scored in game 10, we want to assume the highest possible number of points before hand; thus, she could have scored $17(5)-1$ points in the first 5 games.

To have a ten-game average above 18, she must have scored at least $18(10)+1$ points total. Her lowest possible tenth game score is thus:

$181-(17(5)-1+23+14+11+20)=(181-152)=29$.
[/hide]

12. If $m$ and $b$ are real numbers and $mb>0$, then the line whose equation is $y=mx+b$ [u]cannot[/u] contain the point:

A) (0, 1997)  B) (0, -1997)   C) (19, 97)  D) (19,-97)  E) (1997,0)

Solution:
[hide]
E) (1997,0)

If $mb>0$, then $m$ and $b$ are either both positive or both negative. Thus the slope and y-intercept are either both positive or both negative, and this is impossible if a line is to contain any point on the positive x-axis. Any such line with a positive slope would have a negative y-intercept, and a negative slope would yield a positive y-intercept.

A is possible for any line through (0, 1997) with a positive slope.
B is possible for any line through (0, -1997) with a negative slope.
C is possible for any line through (19, 97) with $0<m<\frac{97}{19}$; the restraint on m ensures a positive y-intercept to match the positive slope. Any line with a negative slope would have a positive y-intercept and thus would not work.
D is possible for any line through (19, -97) with $\frac{-97}{19}<m<0$; the restraint on m ensures a negative y-intercept to match the negative slope. Any line with a positive slope would have a negative y-intercept and thus would not work.

E is thus the only answer.
[/hide]

13. How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of $N$ is a perfect square?

A) 4   B) 5   C) 6   D) 7   E) 8

Solution:
[hide]
E) 8

Let any two-digit number $N$ be expressed as $N=10a+b$, where $a$ is the tens digit and $b$ is the ones digit. The reverse of $N$ is thus $10b+a$, and the sum of the two numbers is $11a+11b=11(a+b)$.

Since $a$ and $b$ are integers, $a+b$ is an integer, so $11(a+b)$ is divisible by 11 and a perfect square; since 11 is prime, it follows that $(a+b)=11k^2$ for some positive integer $k$, which would make the perfect square $(11k)^2$. Since 0[u]<[/u]a,b<10, and $a$ and $b$ cannot both be $0$, it follows that $k=1$ is the only possible value for $k$ and so $a+b=11$ in all cases.

The two-digit positive integers that fit these constraints are:
{92, 83, 74, 65, 56, 47, 38, 29}

And thus $8$ numbers fit the constraints.
[/hide]

[quote="bthings_2000"]you're probably right, I was just stashing it here while I worked it up... it's not like the GA forum is super active anyway.[/quote]

Actually the GA forum is the 5th most popular state forum, judging by number of posts in it.

14. The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$. If the populations in the years 1994, 1995, and 1997 were 39, 60, and 123, respectively, then the population in 1996 was

A) 81  B) 84  C) 87  D) 90  E) 102

Solution:
[hide]
B) 84

Let $x$ be the population of geese in 1996. Based on the given proportional relationship, we know that

$\frac{x-39}{60}=\frac{123-60}{x}=k$.

Solving for $x$ yields $x=84$.
[/hide]

15. (diagram given but not necessary.) Medians BD and CE of triangle ABC are perpendicular, BD=8, and CE=12. (Medians BD and CE intersect at G inside triangle ABC.) The area of triangle ABC is

A) 24  B) 32  C) 48  D) 64  E) 96

Solution:
[hide]
D) 64

Since BD is a median, it divides triangle ABC into two triangles with equal bases and heights; thus the areas of triangles BDC and BAC are equal. Triangle BDC has base BD and height GC. BD=8 (given), and GC=$\frac{2}{3}EC=8$.

Thus the area of triangle BDC is $\frac{1}{2}(8)(8)=32$, and the area of triangle ABC is $32(2)=64$.
[/hide]

16. The three row sums and the three column sums of the array

4 9 2
8 1 6
3 5 7

are the same. What is the least number of entries taht must be altered to make all six sums different from one another?

A) 1  B) 2  C) 3  D) 4  E) 5

Solution:
[hide]
D) 4

Changing only three numbers must leave at least a pair of equal sums, because at least one row and column will either be unchanged or have a common change and no other change; this common change keeps their sums equal. Making four changes in any number of ways can make all the sums unequal.
[/hide]

17. A line $x=k$ intersects the graph of $y=\log_{5}x$ and the graph of $y=\log_{5}{x+4}$. The distance between the points of intersection is $0.5$. Given that $k=a+\sqrt{b}$, where $a$ and $b$ are integers, what is $a+b$?

A) 6  B) 7  C) 8  D) 9  E) 10

Solution:
[hide]
A) 6

Since the line $x=k$ is vertical, the distance between it's intersections with any $f(x)$ and $g(x)$ can be expressed simply $f(x)-g(x)$. Also, both of these log functions increase as $x$ increases, and since $\log_{5}{x+4}$ is a 4-unit down translation of $\log_{5}{x}$, we know that $\log_{5}{x}<\log_{5}{x+4}$. Thus, we have

$0.5=\log_{5}{x+4}-\log_{5}{x}$

Then using log identities,

 $\log_{5}{\sqrt{5}}=\log_{5}{\frac{x+4}{x}}$
$\sqrt{5}=\frac{x+4}{x}$
$x=\frac{4}{\sqrt{5}-1}(\frac{\sqrt{5}+1}{\sqrt{5}+1})=1+\sqrt{5}$.

$a=1$, $b=5$,
$a+b=6$.
[/hide]

18. A list of integers has mode 32 and mean 22. The smallest number in the list is 10. The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10$, the mean and median of the new list would be $24$ and $m+10$, respectively. If $m$ were instead replaced by $m-8$, the median of the new list would be $m-4$. What is $m$?

A) 16  B) 17  C) 18  D) 19  E) 20

Solution:
[hide]
E) 20

let $S$ be the sum of the members of the list and $n$ be the number of members in the list. We are given that that $\frac{S}{n}=22$ and $\frac{S+10}{n}=\frac{S}{n}+\frac{10}{n}=24$; thus, $n=5$.

So in our 5-member list we have a 10, an $m-4$ (which must be a member of the list even though it is not initially the median), an $m$, and two 32s (mode). Since the sum of the list is $(22)(5)=110$, We find $m=20$.

Thus the final list is {10, 16, 20, 32, 32}. This fits all of the criteria.
[/hide]


Let ma,mb,mc be the length of the center lines of triangleABC which is not an obtuse triangle.Prove that a^2/ma+b^2/mb+c^2/mc>=8R-4r

Dear yixinshen
If I am right, I think $ m_{a},m_{b},m_{c }$ should be the lengths of the medians of triangle $ ABC$.
right?
prove that  $ \frac {a^2}{m_{a}} \plus{} \frac {b^2}{m_{b}} \plus{} \frac {c^2}{m_{c}}\geq 8R \minus{} 4r$


look in the archives because I have already solved this one! :D

here is a sol :

write 1/a+1/b+1/c=2 as:

sum (a-1)/a =1

then apply Cauchy to:

sqrt(a+b+c)*sqrt(sum (a-1)/a)>=sum(sqrt(a)*sqrt((a-1)/a))=sum(sqrt(a-1)).

cheers! :D

Sorry about that Lagrangia! I guess my memory is not that good huh... :)


[quote="kohjhsd"][quote="CheeseIsGood"][quote="mwpl11"][hide]
$x^{6}$ has to be less than $1000$, and
the largest $6th$ power less than
$1000$ is $3^{6}= 729$, so the
answer is $\boxed{3}$.[\hide][/hide][/quote]

I'm not sure why you did that.  :huh:[/quote]

[hide]Number that is both a perfect square and a perfect cube should be a $2 \cdot 3=6$th power as well.

When factorized, same factor should be there in multiples of 2 as well as in multiples of 3.[/hide][/quote]
Oh. Thanks. :)

[quote="vamathletes"]How many numbers from 1-1000 are both perfect cubes and perfect squares?[/quote]
[hide=" myanswer"] you know that the perfect cube from 100~1000 has to be 5~10 to the 3rd power, so the only one that is also a square root is 729.
Therefore there is one  100~1000.
THen, among the remaining 1~4, 64(4*4*4) and 1(1*1*1) work so from 1~100 there are 2.
2+1=3, so there are 3.[/hide]

I found a way that proves that there are only 3

[hide]Look at the numbers that are cubed in the problem:

$1^{3}, 4^{3}, 9^{3}$

Now look at the numbers that can be squared to get the same answer as the ones above.

$1^{2}, 8^{2}, 27^{2}$

If we look at them closer, we see that all of the numbers in the first row are perfect squares themselves, and all of the numbers in the second row are perfect cubes themselves.  Thus we can say that for every number $x$, $((x^{2})^{3})=((x^{3})^{2})$[/hide]

How does that prove that there are only three?  That proves that there [i]are[/i] three, but how do you know there aren't more?

Sorry, I forgot to add this:

[hide]By computing the propblem above in  my last post, you see that for a number to be both a perfect square and a perfect cube, it has to be equal to $x^{6}$.  The only numbers that can be plugged in are:

$1$
$-1$
$2$
$-2$
$3$
$-3$

All of these numbers translate into these answers:

$1$
$64$
$729$

Meaning that no other number can fit to make the problem true.[/hide]

Right, and those numbers are all perfect squares.... Good job!  :lol:


[quote="Muhammadaminov"]$ABC$ is inscibed triangle of circle $S$. Let $A_0$ be the midpoint of arc $BC$ not including $A$ and $C_0$ be the midpoint of arc $AB$ not including $C$. Circle $S_1$ with center $A_0$ touchs $BC$, circle $S_2$ with center $C_0$ touchs $AB$. prove that point $I$ which is the center of inscribed circle of $ABC$ lie on one of the tangents of $S_1$ and $S_2$.

Draw it too please!!!!!!!! Jimbo[/quote]

Let $N$ be the midpoint of $BC$. $U$ the point where the tangent from $I$ touches $S_1$, $V$ the point where the tangent from $I$ touches $S_2$(not depicted). $A_0B = A_0I$ this is angle chasing. $A_0N = A_0U$. Then triangles $A_0BN$ and $A_0UI$ are equal.
$\angle A_0IU = \angle A_0BC = \frac{\angle A}{2}$.  Now you can prove that $\angle A_0IC = \frac{\angle A + \angle C}{2}$ so $\angle UIC = \frac{\angle C}{2}$. Do the same with the other circle and you'll find that points $U$, $I$ and $V$ are actually collinear.

Daniel


geometry
parallelogram
analytic geometry
geometry proposed

Parallelogram ABCD.  Within it, EG parallel to AD, E on AB, G on CD.  Also within it, FH parallel to AB, F on BC, H on AD.  EG and FH meet in J.  CH and AG meet in K.  Show that B,J,K are collinear.

I have a coordinate solution, but not Euclid, ...if someone might point the way.

This problem is the particular case of another well known [b][size=100]Brianchon's theorem[/size][/b], where the segment lines $EG,$ $FH,$
pass through the intersection points of the opposite sidelines of an arbitrary quadrilateral $ABCD.$

I have been used this particular case as a Lemma, for the solution of another problem. For an elementary proof,
you can the [b][size=100]Lemma 2[/size][/b] at: http://www.mathlinks.ro/Forum/viewtopic.php?t=107997

Kostas Vittas.

Thank you for the response, but as I said I already did a proof using coordinates.  It is simple if we accept EG and FH as coordinate axes, and J as the origin. I was looking to see if there was anything strictly Euclidean.

Is it not Euclidean the proof, of the [b][size=100]Lemma 2[/size][/b], I mentioned?

Kostas Vittas.

[quote="Davidd"][color=darkred]Parallelogram $ABCD$.  Within it, $\{\begin{array}{c}EG\parallel AD\mathrm{,\ where\ }E\in AB\ ,\ G\in CD\\\\ FH\parallel AB\mathrm{,\ where\ }F\in BC\ ,\ H\in AD\end{array}$.  Denote $\{\begin{array}{c}J\in EG\cap FH\\\ K\in CH\cap AG\end{array}$. Show that $B\in JK$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Apply the direct Menelaus' theorem to the transversal $\overline{AKG}$ in $\triangle CDH$ : $\frac{AH}{AD}\cdot\frac{GD}{GC}\cdot\frac{KC}{KH}=1$. But $\{\begin{array}{c}AH=BF\ ,\ AD=BC\\\ GD=JH\ ,\ GC=JF\end{array}$.
Thus, $\frac{BF}{BC}\cdot \frac{KC}{KH}\cdot\frac{JH}{JF}=1$, where $B\in FC$, $K\in CH$, $J\in HF$. From the reciprocal Menelaus' theorem applied in $\triangle CFH$ obtain $B\in JK$.

[b]Remark.[/b] Vittasko, I think that the name of the extension of this problem is [u][b]Pappus' theorem [/b][/u]: $\{\begin{array}{c}Z\in XY\ ;\ C\in AB\\\ U\in YC\cap ZB\\\ V\in XC\cap ZA\\\ W\in XB\cap YA\end{array}\|$ $\implies$ $W\in UV$.[/color]

[quote="Virgil Nicula"][color=darkblue][b]Remark.[/b] Vittasko, I think that the name of the extension of this problem is [u][b]Pappus' theorem [/b][/u]: $\left\{\begin{array}{c}Z\in XY\ ;\ C\in AB\\\ U\in YC\cap ZB\\\ V\in XC\cap ZA\\\ W\in XB\cap YA\end{array}\right\|$ $\implies$ $W\in UV$.[/color][/quote]

I have been saw ( in some Hellenic textbooks ) the extension of [b][size=100]Davidd's[/size][/b] problem as [b][size=100]Brianchon's theorem[/size][/b], but you are right its proof is based on [b][size=100]Pappus's theorem[/size][/b] (at least the one I have in mind).

Kostas Vittas.

[color=darkblue]How said Vittasko, here is the [u][b]Brianchon's theorem [/b][/u]:[/color]

[color=darkred][b][u]The Brianchon's theorem.[/u][/b] Let $ABCD$ be a convex quadrilateral for which denote$\{\begin{array}{c}M\in AD\cap BC\\\ N\in AB\cap CD\end{array}$.
Let $\{\begin{array}{c}E\in AB\ ,\ F\in BC\\\ G\in CD\ ,\ H\in DA\end{array}$ be four points so that $\{\begin{array}{c}M\in EG\\\ N\in FH\end{array}$. Define $\{\begin{array}{c}J\in EG\cap FH\\\ K\in AG\cap CH\end{array}$. Then $B\in JK$.[/color]
[color=darkblue][b][u]Proof.[/u][/b] Apply the Pappus' theorem to the lines $\overline{MAH}$ and $\overline{NCG}$ : $\{\begin{array}{c}B\in MC\cap AN\\\ J\in MG\cap HN\\\ K\in AG\cap HC\end{array}$ $\implies$ $B\in JK$.[/color]

[quote="Davidd"][color=darkred]Parallelogram $ABCD$.  Within it, $\{\begin{array}{c}EG\parallel AD\mathrm{,\ where\ }E\in AB\ ,\ G\in CD\\\\ FH\parallel AB\mathrm{,\ where\ }F\in BC\ ,\ H\in AD\end{array}$.  Denote $\{\begin{array}{c}J\in EG\cap FH\\\ K\in CH\cap AG\end{array}$. Show that $B\in JK$.[/color][/quote]
[color=indigo][b]Proof. (Only use Thales Theorem!)[/b]  :wink:
Let $M=AG\cap CD,\,N=BJ\cap CD$
We have: $\frac{NG}{CG}=\frac{NG}{BE}=\frac{JG}{JE}=\frac{JG}{HA}=\frac{JM}{HM}$
On the otherhand, $HM\parallel CG$, thus $K\in JN$ i.e. $B\in JK$[/color]


It is known that $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots = \frac{\pi^2}{6}$. Given this fact, determine the exact value of: \[ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots  \]

-mandelbrot 1994 meet 1 #5


algebra
function
domain
algebra proposed

[quote="esmalls"]1.Find $ h^{ \minus{} 1}( \minus{} 1)$ without finding $ h^{ \minus{} 1}$ given that $ h(x) \equal{} \frac { \minus{} x \plus{} 2} {3x \plus{} 3^{h(x)^{2}}}$ where $ x\neq1$.Then, find $ h^{ \minus{} 1}(x)$ .What can you conclude about $ h$ ?
[/quote]

Hello esmalls!

You again have the same problem : not enough precision about domains and codomains.

Now, there exist solutions to your functional equation.

But some solutions have codomains which dont contain $ \minus{} 1$, so  $ h^{ \minus{} 1}( \minus{} 1)$ is not defined for any solution ...


geometry
inequalities proposed
inequalities

$1/$ Given two triangles $ABC,A'B'C'$ with any $k>1$ prove that $a''=\sqrt[k]{a^{k}+a'^{k}},b''=\sqrt[k]{b^{k}+b'^{k}},c''=\sqrt[k]{c^{k}+c'^{k}}$ are three side of a triangle and it has area $S''$ prove that
$\sqrt{S''^{k}}\ge \sqrt{S^{k}}+\sqrt{S'^{k}}$
$2/$ Prove that similar problem for quadrilateral!

I know following problem: Assume $A_{1}B_{1}C_{1},A_{2}B_{2}C_{2}$ are two triangles have sides $a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}$ and areas $S_{1},S_{2}$. Let $a=\sqrt{\frac{a_{1}^{2}+a_{2}^{2}}{2}},...$. Prove that $a,b,c$ are sides of a triangle and if note $S$ is its area then $S\geq \frac{S_{1}+S_{2}}{2}$.

And your problem maybe solve same as. :D

Hey I think you are so selfconfident, why don't you try to post your solution by that way?


algebra
polynomial
Ring Theory
algebra theorems

Hello,

if $ p(x)\equal{}x^3\plus{}2x\plus{}1$ is a polynomial in $ \mathbb{Z}_3[x]$, is $ \mathbb{Z}_3[x]/p(x)$ a field?

Thanks in advance,

Los

Do you mean $ \mathbb{Z}_3[x]/\langle p(x)\rangle$? This is a field as $ \langle x^3 \plus{} 2x \plus{} 1\rangle$ is a maximal ideal of $ \mathbb{Z}_3[x]$. This follows from the fact that $ p(x)$ is irreducible in $ \mathbb{Z}_3[x]$.


Putnam
number theory
least common multiple
invariant
induction
linear algebra
matrix

Start with a finite sequence $ a_1,a_2,\dots,a_n$ of positive integers. If possible, choose two indices $ j < k$ such that $ a_j$ does not divide $ a_k$ and replace $ a_j$ and $ a_k$ by $ \gcd(a_j,a_k)$ and $ \text{lcm}\,(a_j,a_k),$ respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made. (Note: $ \gcd$ means greatest common divisor and lcm means least common multiple.)

The ending sequence is where all the exponents in the prime factorizations of $ a_1, a_2, \ldots$ for a given prime are in nondecreasing order. You can show that the set of these exponents across the sequence is invariant under the operation pretty easily, so the final nondecreasing sequence is unique. Proving the process terminates is basically proving bubble sort works, which is annoying.

[quote]Proving the process terminates is basically proving bubble sort works, which is annoying.[/quote]
Yeah, I had this one pegged as "I understand what is going on, but it would be annoying to write up."

Here's a simpler way of showing the process terminates: the sum of the $ n$ numbers strictly increases, while the product $ p$ stays constant. That means the sum is bounded by $ p\plus{}n\minus{}1$, and must eventually stop increasing (after no more than $ p$ steps).
Now that we know the process terminates, the ending state must be sorted.

Hmm; when I was writing this up, I found proving termination a lot easier than uniqueness (probably because I tried to be too tricky).  Induction on $ n$ gives the horrible bound that the process terminates after at most $ (n\minus{}1)!$ moves, but the proof is almost automatic:  the moves that don't involve $ a_n$ terminate, and $ a_n$ can be involved at most $ n\minus{}1$ times.

To show that it terminates, you can ascribe some value to the sequence for the number of divisions you can write out. Performing the process with a, b will increase this number, because a'|b' now and for any other c:
if c|a but not b, c|b' at least.
if c doesn't divide either, then no loss.
if c|a, c|b, then c|a' and c|b' (easy to prove).
thus the number of divisions will increase, but it is bounded by (n-1)*(n-1). (equality when all values are equal).

[quote="jmerry"]Here's a simpler way of showing the process terminates: the sum of the $ n$ numbers strictly increases, while the product $ p$ stays constant. That means the sum is bounded by $ p \plus{} n \minus{} 1$, and must eventually stop increasing (after no more than $ p$ steps).
Now that we know the process terminates, the ending state must be sorted.[/quote]
Actually I don't think that's true. If $ a_k$ divides $ a_j$ then the sum will remain constant. So in the worst case we'll have to bubble sort the list before the sum increases. This approach still works, but we can only conclude that the sum increases every, say, $ \binom{n}{2} \plus{} 1$ steps.

Here's another interesting fact $ \prod a_i$ is constant but $ \prod a_i^i$ increases. :)

Termination:

The number, $ N$, of pairs of indices $ (i,j)$ s.t. $ a_i|a_j$ is non-decreasing (and bounded by $ n^2$):
If $ a_i, a_j|a_k$, then $ gcd(a_i,a_j), lcm(a_i,a_j)|a_k$
Else if  $ a_i|a_k$, then only $ gcd(a_i,a_j)|a_k$
Else if neither divides $ a_k$, the neither gcd nor lcm divides $ a_k$
The reverse of the above 3 hold as well (when $ a_k$ divides/does not divide $ a_i,a_j$)

(excuse my avoidance of the "does not divide" symbol)

Thus, $ N$ in fact increases since $ gcd(a_i,a_j)|lcm(a_i,a_j)$ unless $ a_j|a_i$. However, if $ a_j|a_i$, then the number of pairs of "out of order" elements under the divisibility ordering strictly decreases and is bounded below by 0, and so we can swap such elements only finitely many times. Thus we can preform finitely many swaps between each increase of $ N$, which there are finitely many of, so the entire process is finite.

Note on my solution: I didnt actually write down the three If/else statements in my proof--do you think I'm going to lose points for that? I didn't supply my uniqueness proof because it was identical to MellowMelon's.

For what it's worth, I was determined to complete the proof of uniqueness without any explicit reference to prime factorization, so again I proceeded by induction:  the operations involving $ a_n$ commute with the operations not involving $ a_n$ with respect to the first $ n\minus{}1$ elements of the sequence, so perform them last.  The moves on $ a_1, a_2,... a_{n\minus{}1}$ give a unique result by the inductive hypothesis.  After the moves involving $ a_n$ have been completed in whatever order, $ a_n \equal{} [a_1, a_2, ... a_n]$ and the first $ n\minus{}1$ elements have all been replaced by their gcd with $ a_n$.

[quote="jmerry"]I think that this one, more than any other question this year, is the "How well can you write?" question. I predict high rates of partial credit, both near ten and near zero.[/quote]

English test on the Putnam, whodathought?

[quote="obscurans"]I went and decomposed them into common prime factors $ a_i \equal{} p_1^{q_{i1}}p_2^{q_{i2}}...p_m^{q_{im}}$, created the matrix $ B \equal{} [b_{ij} \equal{} q_{ij}]$, and imposed the metric "number of adjacent-element swaps (in the same row) to convert one matrix to another". Then it's a matter of showing the gcd/lcm sorts the two considered columns entry-wise and is contractive, if $ a_k \not| a_j$.

And invoked Banach's fixed point theorem. It's complete because it's a discrete metric. Likewise a contractive operation will only take a finite number of steps before hitting the actual fixed point, which is the sorted matrix. The special case where the process could never start was if $ a_i | a_{i \plus{} 1}\,\forall i$ which interestingly was a reverse-sorted matrix.

PS: how to correctly write the symbol for "does not divide" without that horrendous thing above?[/quote]

hm, does this work? looks like it. pretty cute! (especially as, as everyone else is saying, writing this up in the obvious way is a huge headache)

I couldn't figure out how to prove uniqueness, but I think I proved that it terminates.

I basically said that the prime factors could only "move" right, and they couldn't keep going right forever, because there are only finitely many numbers in the list.
(What I turned in was proofier than that, though.  "Proofier" should be a word.)

Well if you break it into $ k$ sequences, where $ k$ is the number of distinct primes dividing the product of the terms of the sequence, and where the $ i$th sequence consists of the powers of the $ i$th prime dividing each term, then we want to show that eventually this set of sequences, under the given operation, is invariant.  The given operation will take any two terms of the sequence and will, for the corresponding two terms of each of these sequences, put them in nondecreasing order.  If the two were already in nondecreasing order for all $ k$ sequences, then we aren't allowed to perform the operation.  I thought it was pretty clear to prove that the process terminates from here, and would be a waste of the graders' time to go ahead and prove it (you could use induction on $ k$ here), so I just stated that.  But does that need to be proven here?

My proof was the type of thing you'd find in a very boring textbook :D

Basically, I defined the set $ F(a_1, ..., a_n)$ to be the set of ordered triples $ (j, k, p)$ with $ j<k$, $ p$ prime, such that there exists $ i$ with $ p^i | a_j$ and $ p^i\nmid a_k$. Then I found an injection from the new $ F$ into the original $ F$, along with an element which had no inverse.

I blame my lack of sleep the night before for this solution...

My entire answer with all the details:

Decompose $ \{a_i\}$ into a common prime basis $ a_i = p_1^{q_{i1}}p_2^{q_{i2}}...p_m^{q_{im}}$ and define the $ n\times m$ matrix $ B = [b_{ij} = q_{ij}]$. The condition that $ \exists j < k$ such that $ a_j\not| a_k$ is equivalent to the column-vector $ \vec{B_j}\leq\vec{B_k}$ entrywise with one strict.

Consider the case that no such $ j,k$ exist. Then the process cannot start at all, terminates and the result is independent of the (0) choices.

Suppose not, then define the metric space $ (RS(B),d)$ where $ RS(B)$ (I named it for "row-wise symmetries") is the set of all matrices generated by action of $ S_n^m$ on $ B$ with each copy of $ S_n$ acting on a different row. Define the function $ h_k: RS(M)\times RS(M)\rightarrow\mathbb{N}$ as the number of swaps of adjacent elements of the $ k$th row to take one matrix to another. Then define $ d(A,B) = \sum_{i = 1}^nh_i(A,B)$.

To see that this is a metric, identify $ h_k$ as the distance between $ \vec{A_k}$ and $ \vec{B_k}$ on the Cayley graph $ \mathcal{C}(S_n,\{(12),(23),(34),...,(n - 1\;n)\})$, that graph distance is a metric, and that the sum of metrics is a metric. Call the sorted matrix $ S\in RS(B)$ with $ S_{ij}\leq S_{i,j + 1}$

Now consider the action of picking $ j,k$ for the process, and call the new matrix $ B'$. $ \gcd(\vec{B_j},\vec{B_k})$ has entries $ \min(B_{ij},B_{ik})$ while $ \text{lcm}(\vec{B_j},\vec{B_k})$ has entries $ \max(B_{ij},B_{ik})$. Thus it sorts the lower entries into $ j$ and higher entries into $ k$. I purpote that if the operation is nontrivial $ d(B',S) < d(B,S)$. I first claim the $ h_i$th subcomponent of the distance is nonincreasing. To see that consider the $ ij$th and $ ik$th entries. Since the sorted matrix obeys $ S_{ik}\geq S_{ij}$ but ours need not before but necessarily after, that implies the distance of $ B_{ij}$ from its corresponding entry of $ S$ does not increase nor does $ B_{ik}$, -1 if they moved or 0. Since every element is at most the same distance away the inversion distance cannot increase. To see that the inequality is strict, if nothing moves the operation was trivial.

Now, we have that the sequence of distances is strictly decreasing unless an operation was trivial, and hence will eventually hit $ 0$, whence further operations cannot do anything. The number of nontrivial steps is bounded by $ d(B,S)$, the sum of the number of inversions in each row considered separately.

PS: I guess I can't actually invoke Banach's fixed point theorem, just the well-ordering principle. I didn't prove it for arbitrary target $ A$, just the sorted $ S$

[quote="jmerry"]Now that we know the process terminates, the ending state must be sorted.[/quote]
I also proved that the process terminates first, and then that there is only one possible ending state, though I did it a different way. I said that each operation always moves at least one prime factor to the right in the sequence (e.g. if you pick 5 and 7, you move the 5 right over to the box with 7 in it and get 1 and 35). Since the sequence is finite and each number has finitely many prime factors, you can only move something to the right finitely many times...

I still don't see why it's annoying to write up -- if you just think of prime factors as objects and the $ a_i$s as boxes, it's quite obvious what's going on.

Copied over (I'd prefer not to start new topics):

[quote="mli"]Did anyone else use directed graphs to prove the uniqueness part? After showing that it ends in finite steps, you can represent the moves with a directed graph whose size is finite, since, from each node, there are finitely many choices to make (finite degree?), and there is no infinite path, from the first part. (this follow from cond'l axiom of choice) Thus, you can do induction on the size of the graph to show uniqueness.

For example, if you start out with (2,3,4), there are 3 moves to be made from the first node, etc.[/quote]

[quote="obscurans"]I went and decomposed them into common prime factors $ a_i \equal{} p_1^{q_{i1}}p_2^{q_{i2}}...p_m^{q_{im}}$, created the matrix $ B \equal{} [b_{ij} \equal{} q_{ij}]$, and imposed the metric "number of adjacent-element swaps (in the same row) to convert one matrix to another". Then it's a matter of showing the gcd/lcm sorts the two considered columns entry-wise and is contractive, if $ a_k \not| a_j$.[/quote]
Rather than identify a metric, there's a monovariant available:  the sum $ \sum_{i} \prod_j b_{ij}$ (or the other way around; it's annoying to keep track) is strictly increasing and has a unique maximum when all of the columns are similarly sorted by Rearrangement.  So either the columns are all sorted in increasing order or in decreasing order at the end.  But it's obvious which equality case is possible.

Now if only I'd thought of this on the test  :|

Sounds like [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=199668]this problem[/url] posted on my blog ages ago, which Ravi B answered in the affirmative, has the answer. Any finite poset can be embedded in the poset $ \mathbb{N}$ under divisibility, and I think you can associate any finite set of points on a lattice with a finite poset.

EDIT: Oh, I didn't realize there was a "for all" quantifier in front of the finite sequence.

The embedding in a finite-dimensional poset shouldn't work generically, since I'm asking about any finite sequence in the lattice.  I think finite-dimensionality is the correct constraint, but I'm wondering under what conditions it can be weakened.

Anyway, an interesting interpretation of this problem is in terms of the question of finding a particular canonical form for the group $ G \equal{} \mathbb{Z}_{a_1} \times \mathbb{Z}_{a_2} \times ... \times \mathbb{Z}_{a_n}$.  The uniqueness part then follows by the [url=http://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Invariant_factor_decomposition]structure theorem[/url] - it is in particular a consequence of the computation of the [url=http://en.wikipedia.org/wiki/Smith_normal_form]Smith normal form[/url] - and one will need to prove the lemma

$ \mathbb{Z}_p \times \mathbb{Z}_q \simeq \mathbb{Z}_{\gcd(p, q)} \times \mathbb{Z}_{\text{lcm}(p, q)}$.

The final set of numbers is known as the set of invariant factors of $ G$.

[quote="t0rajir0u"]So the natural question from here is to ask under what conditions a finite sequence of points in a [url=http://en.wikipedia.org/wiki/Lattice_(order)]lattice[/url] has the same properties.  In this particular problem, the lattice we are looking at is the poset $ \mathbb{N}$ ordered by divisibility, with the join operation being $ \text{lcm}$ and the meet operation being $ \gcd$.  I say natural, but really I'm trying to figure out how many of the proofs given in this thread are specific to this situation and don't generalize :)[/quote]

I had asked the same question, and we ended up showing that it works precisely for distributive lattices. Check the putnam archive solutions [url=http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml](link)[/url] for the proof.

[quote="MellowMelon"]Any finite poset can be embedded in the poset $ \mathbb{N}$ under divisibility, and I think you can associate any finite set of points on a lattice with a finite poset. [/quote]

Not by a lattice morphism. Since otherwise you could conclude that any lattice is distributive.

Let $N$ be the number of pairs $a_j, a_k$, such that $a_i$ does not divide $a_k$. We prove $N$ is strictly decreasing under the operation in the problem.

Suppose we replace $a_k$ and $a_j$ by $a =\gcd(a_k,a_j)$ and $b =\text{lcm}(a_j,a_k)$, then we have $a | a_k,a_j | b$. So far $N$ decreased by $1$. We have 3 cases:

If $a_j | a_i$ and $a_k |a_i$ for some $i$ then we also have $a,b | a_i$. No net change in $N$.
If $a_j | a_i$ and $a_k \nmid a_i$ then we at least have $a | a_i$. $N$ is decreased.
If $a_j \nmid a_i$ and $a_k \nmid a_i$ then $N$ is decreased.

From this, $N$ is a strictly decreasing sequence, since the starting $N$ is always finite, eventually, it will be $0$.

You effectively end with an array, where in sorted order, the prime factorization ascend.


number theory
relatively prime
number theory unsolved

Let $ a,b,c$ are  positive integers such that
$ c(ac \plus{} 1)^2 \equal{} (5c \plus{} 2b)(2c \plus{} b)$
Prove that $ c$ is an odd square.

[hide="Solution"]
Assume that $ c$ is even, say $ 2k$.  Then, we have that $ 2k(2ak\plus{}1)^2 \equal{} 2(5k \plus{} b)(4k \plus{} b)$, so $ k(2ak \plus{} 1)^2 \equal{} (5k\plus{}b)(4k\plus{}b)$.  Now, notice that $ \gcd (5k \plus{} b, 4k \plus{} b) \equal{} \gcd (k, 4k \plus{} b) \equal{} \gcd (k, b)$.  Now, we have that $ k$ and $ 2ak \plus{} 1$ are coprime, so any prime dividing $ k$ may not divide $ 2ak\plus{}1$.  Consider a prime, $ p$, so that $ p^l \parallel{} k$.  Then, let $ p^x \parallel{} (5k \plus{} b)$.  If $ x > l$, then $ (5k \plus{} b)(4k\plus{}b)$ will have more factors of $ p$ than $ k(2ak\plus{}1)^2$, which is a contradiction, so $ x\le k$.  If $ x \equal{} k$, then we would have that $ p|b$, so $ p|(4k \plus{}b)$, so $ (4k \plus{} b)(5k\plus{}b)$ has more factors of $ p$ than $ k(2ak\plus{}1)^2$, which is a contradction.  Therefore, $ x < k$.  Then, $ p^x \parallel{} b$, so $ p^x\parallel{}(4k\plus{}b)$.  It follows that $ p^{2x}\parallel{}(5k\plus{}b)(4k\plus{}b)$, so $ p^{2x}\parallel{}k$, so $ 2x \equal{} l$, so any prime dividing $ p$ divides it twice, and hence, it is a perfect square.  Let $ k \equal{} n^2$.  Then, from the above reasoning, it is seen that $ n|b$, so say that $ mn \equal{} b$.  This means that $ n^2(2an^2\plus{}1)^2\equal{}(5n^2 \plus{}nm)(4n^2 \plus{} nm)$, so $ (2an^2 \plus{}1)^2 \equal{} (5n \plus{} m)(4n \plus{} m)$.  Notice $ n$ and $ m$ are relatively prime, because if they share a prime factor, $ p$, we would have that $ p|(5n \plus{} m)(4n \plus{} m)$, so $ p|(2an^2 \plus{}1)^2$, so $ p|\gcd (n, 2an^2 \plus{} 1) \equal{} 1$, which is a contradiction.  Then, $ \gcd (5n \plus{} m, 4n \plus{} m) \equal{} 1$, so $ 5n \plus{} m$ and $ 4n \plus{} m$ are perfect squares.  Let them be $ x^2$ and $ y^2$ respectively.  We have that $ x^2 \minus{} y^2 \equal{} n$ and $ 5y^2 \minus{} 4x^2 \equal{} m$ after solving.  Now, we get that $ 2a(x\plus{}y)^2(x\minus{}y)^2 \plus{} 1 \equal{} xy$.  This condition is equivalent to $ 2(x\plus{}y)^2(x\minus{}y)^2|(xy\minus{}1)$.  Without loss of generality, say that $ x\ge y$ (because of the squares, it doesn't make a difference).  Then, let $ x \equal{} y \plus{} z$, so $ 2(2yz \plus{} z^2)^2 | (y^2 \plus{} yz \minus{} 1)^2$.  However, $ (2yz \plus{} z^2)^2 > (y^2 \plus{} yz \minus{} 1)$, which is a contradiction and we are done.

Now, $ c$ is odd.  We practically may do the same thing to obtain that $ c$ is a perfect square.  Let $ p$ be a prime dividing $ c$ and let $ p^l \parallel{} c$.  Consider $ x$ so that $ p^x \parallel{} (2c \plus{} b)$.  If $ x \ge l$, then $ p^l |b$, so $ p^l | (2c \plus{}b)$, so $ p^{2l} | (5c \plus{} 2b)(2c \plus{} b)$, which is a contradiction since $ p^l \parallel{}(c)(ac\plus{}1)^2$.  This implies that $ x < l$, so $ p^x \parallel{} b$.  Since $ c$ is odd, $ p^x \parallel{} (5c \plus{} 2b)$, so $ p^{2x} \parallel{} (5c \plus{} 2b)(2c \plus{} b)$, so $ l \equal{} 2x$, so $ c$ is a perfect square. [/hide]


[quote="GlenthemanN"]I was just thinking and posed this question to myself..

How many numbers exist such that the sum of the three greatest factors of the number other than the number itself is infact the number. Example: 18, factors are 18, 9, 6, 3, 2 and 1; The greatest three other than 18 being 9, 6 and 3. $9+6+3 = 18$
How many other such number exist?[/quote]
this seems somewhat related to perfect numbers.
We know that the number has to be an abundant number.

[hide="hint"]
I noticed that numebers like 6, 66, 666, etc all work so Im guessing theoretically there are an infinite amount?
[/hide]

infinite amount.
the complete solution set is:

$n=12k+6$ for any positive integer k

That's not correct. For $k=2$, the sum is $15+10+6=31$, where the # is $30$.

my bad
correction:

any number of one of the following forms for any positive integer k:

$60k+6$
$60k+18$
$60k+42$
$60k+54$

this removes the divisibility by 5 problem.

alternately u could just use my previous solution and stipulate that k is not congruent to 2 modulo 5.


This is a problem from a summer packet for AP calc. It's the only problem I have left. Here it is..

A curve is traced by a point P(x,y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve.

I'm pretty sure I got one point on the curve as (1.25,-0.5).

this should be posted in the calculus section

I was going to say that, but the solution actually requires only algebra -- no calculus at all.

This doesn't require calculus.
I think it's fine in this forum.

[hide="Hint"] Distance Formula [/hide]

he said summer packet for AP calc, so it's like Alg II/pre calc.

I got a summer packet too, but i didn't even start it

I used the distance formula and came up with

8y² + 20y = -8x² + 38x - 43


symmetry
combinatorics proposed
combinatorics

This proof is wrong so I delete everything.

What you achieved is to show, for a particular sitting (of high symmetry), how to partition into groups.

I think the problem wants you to do it, whatever the sitting were!


what is the best way in maple to solve problems analytically? I know maple has built in functions for this (plane, line, point, etc.) but our university forces us to write them down as equations and matrices, to be solved with linear algebra.

Is this just a strange idea of them or is it commonly used? I'd like to learn the best way right from the start...

Hi Peter VDD, Everyone,

Have you ever used the linalg package in Maple?  I am not quite sure what you are asking for in your post, but you can do a lot of linear algebra with the procedures that get loaded when you do the following in Maple:

with(linalg);

By using the semi-colon, you will get to see all the different procedures you can use.  Then go to the help files to find out about the syntax for each procedure in the linalg package.

By the way, I am a big fan of Maple.  I guess some of this has to do with the fact that MapleSoft is a Canadian company formed by people associated with a good school --- the University of Waterloo.  The University of Waterloo has a [i]Faculty of Mathematics[/i]  --- neat!  I would not say that I know how to use Maple very well, but I have a little experience.  I will probably be using Maple more in the future when I start researching mathematical questions on my own, devoid of any interference from school.

Yours truly,

mathfreak


floor function
modular arithmetic
number theory unsolved
number theory

Let $n$ be a positive integer, and let $p$ be a prime, such that $n>p$.
Prove that :
\[ \displaystyle \binom np \equiv \left\lfloor\frac{n}{p}\right\rfloor \ \pmod p. \]

Immediate with use of Lucas' Theorem. If $n=a_0+a_1p+...$ is a representation of $n$ in a base of $p$. Then $\binom{n}{p} \equiv a_1 \mod p$ by Lucas' Theorem, $[\frac{n}{p}] \equiv a_1 \mod p$ is obvious. QED

Let $n = \left \lfloor \frac{n}{p} \right \rfloor \cdot p + r$ with $0 \leq r < p$. [b]1.[/b] Consider, the product $( n - p + 1 ) \cdot \ldots \cdot n$ first.
\begin{eqnarray} \prod_{j = 0}^{p - 1} \left( n - j \right) & = & ( n - r ) \prod_{j \neq r} \left( n - j \right) \nonumber\\ & = & \left( \left \lfloor \frac{n}{p} \right \rfloor p + r - r \right) \prod_{j \neq r} \left( n - j \right) \nonumber\\ \prod_{j = 0}^{p - 1} \left( n - j \right) & = & \left \lfloor \frac{n}{p} \right \rfloor p \prod_{j \neq r} \left( n - j \right) \end{eqnarray}
[b]2.[/b] We have $p - 1$ terms in $\prod_{j \neq r} \left( n - j \right)$, each of them is different and $\not \equiv 0 ( \mbox{ mod } p )$, therefore
\begin{eqnarray*} \prod_{j \neq r} \left( n - j \right) & \equiv & \left( p - 1 ) ! ( \mbox{ mod } p ) \right. \end{eqnarray*}
Now, it's just routine,
\begin{eqnarray*} \left(\begin{array}{c} n\\ p \end{array}\right) & = & \frac{\prod_{j = 0}^{p - 1} \left( n - j \right)}{p!}\\ \left(\begin{array}{c} n\\ p \end{array}\right) & = & \frac{\left \lfloor \frac{n}{p} \right \rfloor \prod_{j \neq r} ( n - j )}{( p - 1 ) !}\\ \left(\begin{array}{c} n\\ p \end{array}\right) & \equiv & \left \lfloor \frac{n}{p} \right \rfloor \cdot 1 ( \mbox{ mod } p ) \end{eqnarray*}
Note that since $\gcd ( ( p - 1 ) !, p ) = 1$ we are in right to assert that $\frac{( p - 1 ) !}{( p - 1 ) !} \equiv 1 ( \mbox{ mod } p )$.

We  could  use  lucas theorem  and poof!! the problem  goes away!! :P


I need CMO paper ASAP :help: . I couldn't find mine. Can anyone scan the paper and send it to me immediately(or you can post the paper here). I don't need the problems, but I need the actual scanned paper.

My email address is lightrhee@yahoo.co.kr
--------------------------------------------------
EDITED: I really need this as soon as possible. Please hurry!

[quote="TheDreamer"]Ok I sent it to you :)

Just out of curiosity, if you don't mind telling, what is this for?  Why do you need a scanned copy?[/quote]
Dr. Andy Liu(he was a Canadian IMO team leader for few times) asked me to give him a copy, so that he can give it to his friends.

Thank you very much!


[quote="junggi"][hide="not sure if this is right"]WLOG let c be the longest side of the triangle. From the triangle inequality, we have
$a+b>c$
substituting a+b=1-c in this
$1/2>c\rightarrow-c>-1/2$
substitute this in,
$a^{2}+b^{2}+1/4-2ab=(a-b)^{2}+1/4<1/2\rightarrow(a-b)^{2}<1/4$
Since c is the longest side, the difference between a and b is always less than 1/2, so $(a-b)^{2}<1/4$ and the ineq. is true...[/hide][/quote]
You have $+4abc$ and not $-4abc$ :!:

[quote="junggi"][hide="not sure if this is right"]WLOG let c be the longest side of the triangle. From the triangle inequality, we have
$a+b>c$
substituting a+b=1-c in this
$1/2>c\rightarrow-c>-1/2$
substitute this in,
$a^{2}+b^{2}+1/4-2ab=(a-b)^{2}+1/4<1/2\rightarrow(a-b)^{2}<1/4$
Since c is the longest side, the difference between a and b is always less than 1/2, so $(a-b)^{2}<1/4$ and the ineq. is true...[/hide][/quote]
You mean WOLOG, WLOG means with loss of generality. :lol:
I'm not sure if you proved it right, how does $(a-b)^{2}<1/4$ prove that the ineq. is true?[hide=" :( "]I'm confused[/hide]

[quote="junggi"]I'm probably wrong :lol:
I'm bad at substituting stuff in inequalities(directions confuse me)[/quote]
I'm confused too. :(
(with directions too)

[quote="Inspired By Nature"]You mean WOLOG, WLOG means with loss of generality. :lol: [/quote]

No, "WLOG" means "without loss of generality."

[quote="arkhammedos"]If  $a$,$b$ and $c$ are lengths of a triangle  such that $a+b+c=1$
Prove that  $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$. :roll:
 :rotfl:[/quote]

Let homogenize it;
$(a+b+c)^{3}-2((a+b+c)(a^{2}+b^{2}+c^{2})+4abc) \geq 0$
<--> $(b+a-c)(a+c-b)(b+c-a) >0$ which is obviously true.


Let $F$ be a figue in the plane with [b]AREA[/b] =$A$ and [b]PREMETER[/b]=$P$.

Prove that $\frac{A}{P^2} \ge \frac1{4\pi}$ ,and equality just occurs for circle.

[quote="lomos_lupin"]Let $F$ be a figue in the plane with [b]AREA[/b] =$A$ and [b]PREMETER[/b]=$P$.

Prove that $\dfrac{A}{P^2} \ge \frac1{4\pi}$ ,and equality just occurs for circle.[/quote]

That is precisely the isoperimetric theorem, not an inequality used in proving it!    The inequality is reversed, it should be $\leq$ because the area can be close to $0$ for any given $P$.


Prove that for every integer $ n>1, 1 \cdot 3 \cdot 5 \cdot ... \cdot (2n\minus{}1)<n^n.$

[hide="Solution"]$ n$ even, notice that the average of the $ n$ numbers is $ n$, and in pairs $ (n\minus{}k)(n\plus{}k)<n^2\Leftrightarrow \minus{}k^2<0$ (1) which is true. Multiplying over $ k\equal{}1,3,\dots ,n\minus{}1$ yields the desired inequality.

$ n$ odd, the average of the $ n$ numbers is still $ n$, but this is an element of $ \{1, 3, \dots, 2n\minus{}1\}$. But we can multiply (1) over $ k\equal{}2, 4, \dots n\minus{}1$ to obtain $ 1\times 3\times \cdots \times (n\minus{}2) \times (n\plus{}2) \times \cdots \times (2n\minus{}1) < n^{n\minus{}1}$. Multiplying by $ n$ yields the desired inequality.[/hide]

[b]Case 1:[/b] If $n$ is even, we can pair "opposite" terms of the LHS together. Therefore, we have
$$(n-a)(n+a)(n-b)(n+b)...<n^2\cdot n^2...$$
For each of these pairs of terms, we have $(n-a)(n+a)<n^2$, since $n^2-a^2<n^2$ for all real $a$ other than $0$, which will never happen since no two numbers on the LHS are equal. Therefore, this case is proven.

[b]Case 2:[/b] If $n$ is odd, we have all of the same terms as in case 1, except for the middle term $n$, which is equal to one of the $n$s on the RHS. Therefore, this does not affect the end result, proving that this case is also true.

Both cases have been proved, so we are finished. $\square$

By AM-GM
$$1\cdot3\cdot5\cdot\dots\cdot(2n-1)\leq\left(\frac{1+3+5+\dots+(2n-1)}{n}\right)^n=\left(\frac{n^2}{n}\right)^n=n^n$$
with equality iff $1=3=5=\dots=2n-1$, which is impossible provided that $n>1$. $\blacksquare$

The induction work fine too, the induction step it is ${{n}^{n}}(2n+1)\le {{(n+1)}^{n+1}}\Leftrightarrow {{\left( 1+\frac{1}{n} \right)}^{n}}\ge 2-\frac{1}{n+1}$ true


Let $E$ be an euclidean space and $F\subset E$ such that: $\forall (x,y)\in F^{2},\quad x\neq y\Rightarrow (x|y)<0$.
Find: $\max (\text{card }F)$.

already posted. Try to find it on the forum ..


For me, I do not know, because I have never taken such a test. Is this a test that must be done in person?

The Stanford-Binet L-M's norms are severely outdated and don't give a correct score for anyone in the early twenty-first century. There is also a HUGE problem of disparity of scoring from one test-giver to another, so there is double reason not to regard L-M scores seriously. And there is triple reason not to regard those scores seriously, as the item content of the L-M is no longer secure.

Check the online bibliography "[url=http://learninfreedom.org/iqbooks.html]Books on IQ and Human Intelligence[/url]" for more on this subject.

I was referring to up-to-date tests (such as Mensa, High IQ Society, http://www.intelligencetest.com) which give scores according to the original Stanford-Binet scale.

IQ tests in general can be considered subjective...

Most of the supposed high-IQ tests have never had a proper validation study to back them up. At least one such test is very suspect, because there is a member of one high-IQ society who is a well-known [url=http://www.crank.net/harris.html]net.crank[/url] who makes a fool of himself on the sci.math newsgroup.

It takes intelligent behavior in the real world to validate an IQ test result.

[quote="tokenadult"]It takes intelligent behavior in the real world to validate an IQ test result.[/quote]

I agree completely.

But there are some select ones which are free and online that are normed.

Or you can pay a bit of money to take any of Paul Cooijmans' tests.

[quote="sponge008"]Well, I got a 192 on an online test I took, but I doubt that's the truth... :?[/quote]

probably not true unless you are a super-genius.  the online-tests are pretty bogus. i've scored 160 and 90 on online tests.  my IQ is in the 130-140 range (i think).

There are two common scales that measure IQ. The Cattel scale gives a score much higher than the Stanford-Binet scale. For example, my 148 on the Stanford-Binet is a 171 on the Cattel.

An accurate tests can be found at http://www.highiqsociety.com and http://www.intelligencetest.com

For http://www.highiqsociety.com, take the Ultimate IQ test, it's the most accurate one, unless you feel that your IQ is above 160, in which case you should take the Test for Exceptional Intelligence.

[quote="banblink182"]No, I don't believe one's IQ determines their "fate" or limits their abilities; I'm just curious as to the IQs of most of you.

Please don't lie, no one will know what you put down. I want this to be as accurate as possible.

[i]I recognize that such a poll has been done before, but I want to repeat such a poll to test for accuracy.[/i]

Personal IQ - 148[/quote]

You've started too many threads on IQ already.  Let it die.

Yeah what's up with the obsession about IQ? My IQ's 65, so what?

never have a test...
I wonder what it is used for?

[quote="probability1.01"]It's what Americans do in their spare time to feel better about themselves. :)[/quote]

I personally feel that the satisfaction of doing well on an IQ test is less than the satisfaction of doing well on a math competition.

[quote="bubala"]"People who brag about their IQ are losers"--Stephen Hawking.  I agree 100%.  Doing well on a math competition is much more rewarding than doing well on an IQ test.  I have never taking an IQ test.  I don't see the point.  If the IQ test results told me that I was bad at math, would that be true? I hope not.[/quote]

Stephen Hawking is hilarious. Anyone who gets the chance should read some of his books.

An IQ test would not say that you're bad at math, it would say that you scored lower than x% of the people who took this test in the logical/mathematical section. It would not mean any more or less than if you scored less on the USAMTS than someone.

Who's bragging?  :)


[hide]look at the middle column, in order for b+b to equal _b, there has to be a 1 carried from b+a. Also, the tens digit carried from b+b has to be a 1.

So, b+a=10+c
a+b+1=10c+a

We see that 1 works for c, so b=9 and a=2 and the product is 18.[/hide]

[quote="davidli"]The addition problem below has a unique solution. Each of the letters A, B and C represents a different nonzero digit. What is the product of A, B and C?

[code] abb
+bba
-----
cabc[/code][/quote]
[hide]We immediately see $c=1$ because $1$ is the most you can carry over. The digits are nonzero, so $a+b=11.$ Looking at the second column, we see $b+b+1=b.$ Again, since $b$ can't be zero or negative, the equation must be $b+b+1=b+10,$ and we find $b=9.$ We now substitute $b$ into the first equation to find $a=2.$ The product is $\boxed{18}.$[/hide]

[hide="solution"]Since C is 1, we want to find the product of A and B. We have this equation:

[code]

 abb
 bba
-----
1ab1
[/code]

20b+101(a+b)=1001+100a+10b

a+111b=1001

Since a is a digit and b is a digit, b has to equal 9. That gives us

[code]

 a99
 99a
-----
1a91[/code]

We try 0 for a. It doesn't work. We try 1 for a:

199+991=1090

That doesn't work either.

We try 2:

299+992=1291

2 works.  We want to find abc. That equals 1*2*9=18[/hide]

[hide]
We know right away that c = 1.
So we have:
             abb
           +bba
           -------
           1ab1
we then know that b must be greater than 5, we find that the more it is over 5, the closer it sum is to being its own, so choosing 9 yields this:
             a99
           +99a
           ------
           1a91
Put in a as 2 so that 9+2 = 11...
           299
         +992
         ------
          1291

This is true, so a=2,b=9,and c=1.
Their product is then $\boxed{18}$

[/hide]


Three positive integers have an arithmetic mean of 26 and median of 27. If the median is 5 less than the largest number, what is the smallest number?

The median is $ 27$, so the largest number is $ 32$, and $ 26\equal{}\frac{x\plus{}27\plus{}32}3\implies78\equal{}x\plus{}59\implies x\equal{}\boxed{19}$.


I don't spend more than an hour each day  :D

only 23 hours a day  :oops:

fine more like 2, then more if there is lesson

Like 2-3 hrs in the summer, but only around 30 minutes during school days.

[quote="Poincare"]I used to be on for almost 3 hours! But, nowadays, its more like 30-50 minutes.[/quote]

 :( Why the inactivity?

Since it's summer, I'm usually on for like 3-4 hours a day, and the rest of the day, doing math offline. I like the extra math time in the summer :D

about 30 minutes. I really should spend more time on.

About the same amount of time im on facebook. Which is too much time

Except most of the time im on here is when my computer screen/browser is still on and im far far away doing completely useless stuff. or on facebook.

How long my attention span is...which is quite short  :blush:

Read my signature. Impossible to stop...

It depends:
No-class day: 30mins-1hr
Class: About 2 hours (not including class time)

[quote="pingala"]uhhh  like 2 HR max?   my dad hates when im on the computer especially when im on the artofproblemsolving.com[/quote]
Why would your dad hate you doing math? It doesn't make sense.

Not all parents consider all features of the website math-like (like some threads in this forum).

If you do 2-3 hours a day, you actually have an extreme ability to self control.  :)

used to be 2 hrs about, now, like 15-30 min per week. so like 2 mins a day =]

I used to spend like 3-4 hours a day.
Now, since its summer, i spend less time.....

not that much time actually...
well, at least compared 2 some other people on the website
like an hour??? not really sure (at most)


calculus
AMC
AIME
summer program
PROMYS
Ross Mathematics Program
USA(J)MO

Hello everyone.  I'm pretty new to this forum and really love it.  In truth I wish I could have found it sooner.  I'm a ninth grader and scored a 104 on the AMC 10 this year, answering 17 questions and getting 3 wrong.  I love math, and I'm taking AP Calculus BC this year (some may call this the calculus trap, but I don't know).  I really want to get into the AIME next year as well as do better on other competitions like the ASMA and MANDELBROT on my school's math team.  I've enrolled in the introduction to number theory course here this spring semester, and I've purchased the art of problem solving books as well as some others including Mathematical Circles and the The Art and Craft of Problem Solving.  I'm applying to PROMYS and the Ross Program this summer, but I have a sort of dilemma.  I am also in science research and am thinking about doing research this summer.  Another goal of mine is to get into ISEF (and be an STS finalist but I have a while for that).  I'm posting here hoping that someone could lend me some advice with whether to pursue research this summer or to focus on math.  Another thing is if I do decide to go for the math, which program would be better, PROMYS or Ross?  And would there be some sort of mathematical research project I could form from the number theory work I'd be doing?  Thanks for any help anyone can offer.

Well it's hard to judge that.  This is my first year taking science research at school since it's for ninth graders and above.  In our class we have to keep a journal everyday, called a process journal, with at least 6 lines a day, as well as 2 long reflective entries per week.  Each student presents two powerpoint presentations to the class over the course of the year, a five minute and a ten minute.  I presented my five minute earlier this year and got the first 100 in the class.  My ten minute is in April.  The teacher is really really good and pushes us as well as grades us harshly.  On the first draft of my paper [u]How Exercise Stimulates the Mind, Affecting Mathematical Accuracy[/u] I got an 88 which is above the average grade.  I am going to enter this project into Manhattanville later this year.  For the Long Island Math Fair, I wrote a paper [u]Projectile Motion Modeled Under Varying Air Resistance[/u] which I have to present about to judges in mid-March.  What do i do? =(

Most of all, welcome to the forum.

For your specific question, I suggest you that if you want to get better on AIME or Mandelbrot, you should spend fair amount on learning problem solving. No offense but it is true that many people who take calculus are not the ones who are making USAMO (although knowing calculus certainly wouldn't hurt) or even scoring high on AMC's. This is because math competition is heavily on the problem solving.

Now, how do you develop your problem solving? Well, that's pretty tough question. You have to read a lot of books on techniques (AoPS books are excellent materials. So as Mathematical Circles (Russian Experience)) and you have to do a lot of problems. By word "a lot" I mean A LOT.

Lucky that you're not directly heading for proof now. If you don't have a basic problem solving, it is NOT worth to try the proofs because proofs require the basic problem solving and creativity. I'm still working on it too except for proof, you can ask other people in this forum, who is much better on it than me. :)

For your research, I have similar situation here. I'm hoping to get high score on USAMTS and AIME so I can make USAMO next year (with floor score :D ) but I'm hoping to attend a science contest too. Not sure if I'm interested as much as math competitions but why not? Experience is always good thing. :)  So, choose wisely. If your goal tends to lean against the AIME, work more on math problem solvings. If you're leaning to other way, set aside math for sometimes and do a research. Research does take a lot of time and trust me, it is not like you can do both of these easily. Each one of them is very time-consuming thing to do.

If you have any more concerns, you may post here to hear other people's response. You can pm me as well.

Hope this helps!

Regards,

Silverfalcon

Thanks a lot Silverfalcon for your information and advice.  But more specifically, I would like to get some feedback on what I should do for the summer.  I understand that problem solving requires A LOT of time, and I'm going to try to put as much time as I can into it.  For the summer, my two options are a math camp or science research.  If when you were talking about proofs you meant that PROMYS and Ross are highly proof involved, then I'm really unsure of what to do.  I know one person who went to PROMYS and one person who went to Ross, and the one who went to PROMYS qualified for the USAMO while the one who went to Ross got first place in the USAMTS (this was quite a while ago).  As I said before, I am really hoping to at least reach the AIME next year, and I assume that PROMYS or Ross would definitely aid me in doing so (by the way, should i stick with the AMC 10 next year or go for the AMC 12?).  At the same time, if I do summer research I can enter the Long Island Science and Engineering Fair and take a shot at making ISEF.  I am planning to take more courses through AoPS so maybe that will be enough if I spend time working through the books as well, without having to go to PROMYS or Ross so that I could do summer research.  Does this sound too ambitious, or could it really be killing two birds with one stone...or something like that?

I think it's a solid plan for the summer. Math is a lot more self-study-able than research (although studying math with smart kids, like in PROMYS or Ross, would help), so if you really want to to do both math and research, I'd suggest research for the summer, and math here at AoPS when you come back.

I'm not going any mathcamps and never been one so I'm not sure what kind of thing you do on PROMYS or Ross.

It seems to me that you want to get pass AIME but you're also leaning a lot to science. Because of this reason, I agree with sky9073. Go to your research first and when you come back, see how much math you can do.

For the taking classes, it is very good idea. I took one this year but it is also fairly time-consuming thing too. It is great class but trust me, just going to school or classes doesn't mean you learn those stuffs completely. ;) You have to use it on real-life situation. In this case, doing on the problem. The classes have message boards but generally, doing lots of problems is best thing.

So, my final conclusion is this:

Give more time on the research and if you get time, I suggest you doing some math. Now, if your goal was like USAMO, then you might want to spend more time on math also but since you said:

[quote="You"]I am really hoping to at least reach the AIME next year[/quote]

And that you have basic math curriculum done plus caculus now so if you just do some problems, you'll be fine.

For qualification, if you really want to get to AIME, take AMC-12. That's easier to qualify.

Not this means easier to go from AIME but according to cutoff, AMC-12 is easier and since you know calculus and stuffs, that should give you a lot of benefits.

Regards,

Silverfalcon

Thanks a lot for all the feedback!  Wow do i love this forum =).  In that case, I think I will try and find research for this summer, but still work independently with problem solving and advancing myself in math.  I am hoping that I can find research somewhere close enough that I can commute.  This will also allow me to take a course with AoPS in the summer semester.  Now I just have to decide what course to take!  Maybe an independent study would be very worthwile, but this does take a lot of time.  Thanks again, and if anyone has anything else to add I'd be glad to hear it.

ffdbzathf, since you are taking BC Calc as a freshman, what do you plan on taking for the rest of your highschool years?  I ended up running out of high level math courses and have to take Statistics as my only math for my senior year.  If there are any on-line classes that will give credit, let me know.

Well to start, I plan to take AP Statistics next year.  There are 2 high schools in my town, and together they do a shared program where one night a week advanced students from both high schools who've finished the math in their school can go to take a college level course.  They rotate what that course is between Abstract Algebra, Combinatorics, and Multivariable Calculus.  I have heard very mixed reviews about how good a class this is, and it also gives no college credit, therefore I am not sure if I will decide to do this.  But, Stanford has an online program called the Education Program for Gifted Youth where you can take university level courses in math and physics among other things.  Direct credit is given for future Stanford students and most colleges will accept their courses as credits in that you do not have to repeat the course in that such university.  The link for this is http://www-epgy.stanford.edu/ and I certainly recommend it (I took Introductory Physics online through them last year).  I hope to have time to enroll in some of their university level math courses throughout the rest of my high school career, that is if I have enough time considering all the problem solving I hope to be doing =).

If you want, try to go to the Siemens Westinghouse Competition and Scholarship Program math jam on April 16.  THat might help you a little with research.

welcome to the forum, ffdbzathf. i would like to say that i like your attitude and enthusiasm towards AIME. I dont really know alot about researches you were talking about(although I want to go to RSI or Westinghouse ---------> make sure u check those out and see if they fulfill your interest), but i do know alot about taking AIME. I was just like you last year - freshman, enthusiastic, qualifying for AIME with 140.5 AMC10 - but i was not able to do well on AIME becuase my enthusiasm died. make sure if you want to be better, do not lost interest. THat is the worst thing that could happen. If you are stuck on something, dont get frustrated and quit. That is what happened to me last year, solving in the AoPS book. Go back, look at the solution, read over, do it yourself, consult(a lot of experts in this forum, i would like to consider myself somewhat experienced ;)) etc.

Well, i like that you are very interested. keep up the good work. see you in MOP in the future.

reddevil

[quote="Jimmy"]If there are any on-line classes that will give credit, let me know.[/quote]

http://epgy.stanford.edu/courses/math/


function
limit
calculus
derivative
complex analysis

Suppose that f is entire and that $ \lim _{z\rightarrow \infty }{\frac {f \left( z \right) }{z}}\equal{}0$. Prove that f is constant.

[hide="Taylor"]
Because $ f$ is entire, then we can write $ f$ in the form : $ f(z) \equal{} a_0 \plus{} a_1z\plus{}a_2z^2\plus{}\cdots$ then $ \frac {f(z)}{z}\equal{} \frac {a_0}{z} \plus{} a_1 \plus{} a_2z \plus{} \cdots$, then we obtain that $ a_i\equal{}0$ for $ i \geq 1$, and this compelete my proof.
[/hide]

[hide="Thoughts"]Since $ f$ is entire, it can be represented as

$ f(z) \equal{} \sum_{n \equal{} 0}^{\infty} a_{n}z^{n}$;

dividing by $ z$, we get

$ z^{ \minus{} 1}f(z) \equal{} \sum _{n \equal{} 0}^{\infty} a_{n}z^{n \minus{} 1} \equal{} a_{0}z^{ \minus{} 1} \plus{} a_{1} \plus{} a_{2}z^{1} \plus{} ...$ ,

as [b]Qruni[/b] mentioned. Now, suppose there exist one or more non-zero coefficients $ a_{k}$ (where $ k > 0$); this would imply that $ z^{ \minus{} 1}f(z)$ either diverges as $ z \rightarrow \infty$ (which is true when $ k > 1$), or converges to a nonzero number (which is true when $ k \equal{} 1$). Thus, the only way $ z^{ \minus{} 1}f(z)$ can converge to zero as $ z \rightarrow \infty$, is if $ a_{k} \equal{} 0$, for all $ k > 0$; in other words,  $ f$ must be constant.

Letting $ f(z) \equal{} a_{0}$,

$ \lim_{z \rightarrow \infty } z^{ \minus{} 1}f(z) \equal{} \lim_{z \rightarrow \infty } a_{0}z^{ \minus{} 1} \equal{} a_{0} \left[\lim_{z \rightarrow \infty } \frac {1}{z}\right]$ ,

which is indeed zero.[/hide]

probably the more standard argument would be to bound $ f'(a)$ by $ \oint_C \frac{f(z)}{(z \minus{} a)^2}dz$ for large circles $ C$ surrounding $ a$ that go to $ \infty$ to show the derivative identically vanishes


symmetry
algebra
polynomial
binomial theorem

Find the roots of the following equation:
\[ f(x,y,z)=3(xy^2-x^2y-xz^2+x^2z-y^2z+yz^2) \]
Masoud Zargar

looking at the symmetry, i thought of saying that $x=y$...this makes it 0...

suppose this is a polynomial about $x$, then $(x-y)$ is a factor

by symmetry we have that it factors into:

$C*(x-y)(y-z)(z-x)$ for some constant C

(note that it is a constant because the degree of the initial expression was 3 for each term)

easily verifying, C=1

so the general solution is $(t,t,u)$, $(t,u,t)$ and $(u,t,t)$ where $t$ and $u$ are arbitrary numbers

[quote="boxedexe"]You mean $C=3$.

Here's my solution.

Add $x^3$, $y^3$, and $z^3$ and then subtract them.  By the binomial theorem, we have
$f(x,y,z)=(x-y)^3+(y-z)^3+(z-x)^3$. Now since $(x-y)+(y-z)+(z-x)=0$, it is clear that $f(x,y,z)=3(x-y)(y-z)(z-x)$.
The conclusion then follows.

Masoud Zargar[/quote]

oh right...i was wondering what that 3 was for...


The angle formed by the hands of a clock at $ 2: 15$ is:

$ \textbf{(A)}\ 30^\circ \qquad
\textbf{(B)}\ 27\frac{1}{2}^\circ\qquad
\textbf{(C)}\ 157\frac{1}{2}^\circ\qquad
\textbf{(D)}\ 172\frac{1}{2}^\circ\qquad
\textbf{(E)}\ \text{none of these}$

[hide="My Solution"]
The initial position of the hour hand is $ 60^\circ$ (from 12, clockwise) and it moves $ 30\cdot\frac{m}{60} \equal{} \frac{m}{2}$ additional degrees based on the number of minutes $ m$ that have passed. The position of the minute hand is simply $ 6m$ because each tick represents a traversal of $ 6^\circ$. Therefore, the angle formed is

$ 6(15) \minus{} \left(60 \plus{} \frac{15}{2}\right) \equal{} 90 \minus{} 60 \minus{} \frac{15}{2} \equal{} 30 \minus{} \frac{15}{2} \equal{} \frac{45}{2} \equal{} 22\frac{1}{2}^\circ.$ Therefore, the answer is $ \fbox{(E)}$.
[/hide]

You just answered your own question.

[hide="My Solution"]
Angle of minutes hand=$ \frac{15}{60}\equal{}90^o$
Angle of hours hand=$ \frac{2\frac{1}{4}}{12}\equal{}\frac{67.5}{360}\equal{}67.5^o$
Difference=$ 90\minus{}67.5\equal{}22.5$

So the answer is (E)
[/hide]

The hour hand is at 2*30 degrees so 60 degrees while the minute hand is on 15 so 90 degrees but hour hand also moves at 1/2 a degree per minute so 7.5 degrees. 90-67.5= 22.5 degrees which is E.


[quote="Davron"]Well, my mathematical friends ... today there is a competition which is not really well known but by the way known :) Olympiad [Zhayutykovs International Math Olympiad]. Let's discuss it here ...
Who is there rais your hands and tell us about it ??

[i hope that we will see the problems as soon as possible]

Thanks!
davron latipov[/quote]

Hey Davron, I think I have never heard about this olympiad. Can you give us more detales about it such as where it is beeing organized, which counteries are taking part and so forth?

Hello Davron
I am here in olympiad.Today was the second day of exam.Questions were not so hard.But I haven't time to write questions here.I hope someone will write. :lol:

The problems have been posted here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128806

Can anyone please tell what is this olympiad and who participates in it? :maybe:

Hi, everybody!

It is an Olympiad which is organised by the Republic of Kazakhstan. Every year in January. This Olympiad still is not so famous, but a lots of countries participate in it. The countries like: Russia, Turkey, Azerbaijan, Romani, Uzbekistan, Tajikistan... In this MO, every country can be represented by their teams, but the number of teams can be more than one. Every team can have four stedents [for MO] and three [for PO]. Thus every team there will be, will consist of seven students.

p.s.: Kazakhstan always has a lots of teams in this MO

davron latipov

[quote="Tiks"]Oh, now I see that I have actually heard about this olympiad-although have never participated.

So, I am wondering, is Armenian team presented there or no? I know that our team was there for the last two olympiads but I am not sure for this time.[/quote]

This time too  :wink:

That's good Vardan :lol: , let's hope our mathematicans will succeed there;though, I am a bit dubious about it :maybe: .

Actually, I was planning to participate in this olympiad either. However, my application to the US universities didn't let me to do so, anyhow.

Guys I am wondering when the the results will be available? I saw the problems and I think that they are rather easy comparing to the problems of the last olympiad. Please, informe us as soon as you can( the people who are there now) about the results and especially about the cut-off for gold medal.

Regards
Tigran :wink:

Is there a page with the results ? :) Who participated for Romania?

Hmm Cefer was first of first with 42 point!!! and this is second fof of his at ZIMO,i hope he'll also get gold medal in Vietnam!!

P.S Romania was the first and Azerbaijan 3rd in math,i exactly know only this information.. :)

Jafarbey afferim! I congratulate you from the bottom of my heart! Inshallah, you will get a perfect score in the IMO, too.

davron

Congrats Jafar.Wish you success at IMO 2007.

Here are some results (only medals)

$1.Turea Lucian Gabriel-Romania-Buxaresti-7-7-7-7-7-7-GOLD$
$2.Jafarov Jafar-Azerbaijan-Baku-7-7-7-7-7-7$
$3.Nikolai Beluxov-Bulgaria-Sofia-7-7-7-2-7-7$
$4.Bajev Alen-Kazakstan-Almata-7-7-7-2-3-7$
$5.EnxbaatarTemuqe-Monqolia-UlanBator-7-3-7-0-5-7$
$6.Tixonov Ulij-Russia-Moscow-7-0-0-7-7-7$
$7.Artamanov Stepan-Tatarstan-Kazan-7-7-0-7-7-0$
$8.ChristopherRaymond-Indonesia-Jakarta-7-7-3-7-0-2SILVER$
$9.Qalinovskiy Evqeni-Belarussia-Minsk-7-7-3-2-7-0$
$10.Smirnov Evqeni-Russia-Moscow-2-7-0-7-7-2$
$11.Han Beng Koe-Indonesia-Jakarta-7-7-0-7-2-2$
$12.Boltajev Utkir-Uzbekistan-Karaqulski-7-5-7-3-2-0$
$13.Klochkov Eqor-Kazakstan-Povlador-7-7-0-1-7-2$
$14.Arabidze Qioqri-Georgia-Tbilisi-7-7-0-1-5-4$
$15.Aleksandr Daskalov-Bulgaria-Sofia-7-1-3-3-2-7$
$16.Zine Juman-Kazakstan-Aktobe-7-7-0-2-0-7$
$17.Anachkiy Anatolij-Resp.Saxa-Jakutsk-7-1-0-7-7-0$
$18.Missarova Alsu-Tatarstan-Kazan-7-7-0-0-7-1$
$19.Maxambetov Bauirjan-Kazakstan-Almata-7-7-2-2-2-1$
$20.Akopjan Tigran-Armenia-Erevan-7-7-0-2-2-2$
$21.Shulchevskiy Dmitrij-Russia-Moscow-7-0-3-7-2-0-BRONZE$
$22.Qafarov Rustam-Tatarstan-Kazan-7-0-0-3-7-2$
$23.Balijev Timur-Tatarstan-Kazan-7-0-0-5-7-0$
$24.Feldman Qriqorij-Russia-Novosibirsk-7-7-0-2-2-0$
$25.Kalin Landjev-Bulgaria-Sofia-7-2-0-1-6-2$
$26.Stasiy Iqor-Kazakstan-UstKameneqorsk-7-7-0-2-0-2$
$27.Chuluun Altan-Ozlij-Monqolia-UlanBator-7-7-0-1-0-3$
$28.Tusubekov Erken-Kazakstan-Almata-2-7-0-2-0-7$
$29.Daliev Aset-Kazakstan-Almata-7-1-0-3-5-1$
$30.Stoican Stefan Alexandru-Romania-Buxaresti-7-3-0-7-0-0$
$31.NurazxanovChingiz-Kazakstan-Astana-7-2-3-1-2-2$
$32.Ilias Dastan-Kazakstan-Aktobe-7-2-0-1-0-7$
$33.Baxalmetov Dias-Kazakstan-Karaqanda-7-0-7-0-3-0$
$34.Ljamkin Vitalij-Kazakstan-UstKameneqorsk-7-7-0-1-2-0$
$35.Musaqalijev Aset-Kazakstan-Almata-7-6-0-2-2-0$
$36.Viktorov Qleb-Russia-Novosibirsk-7-0-0-0-7-2$
$37.Turmuxametova Ajzana-Kazakstan-Almata-7-0-0-7-2-0$
$38.Kemeshov Erbol-Kazakstan-Almata-2-0-0-7-5-2$
$39.Kaniskin Serqej-Russia-Moscow-7-0-0-1-7-0$
$40.Kesenov Eskender-Kazakstan-Astana-7-0-0-0-2-6$
$41.Dijarov Arman-Kazakstan-Aktobe-6-7-0-2-0-0$
$42.Kim Vitalij-Kazakstan-Almata-7-4-0-0-3-1$

1. Romania,Tudor Vianu High school of computer science   $GRAND-PRIX$
2.Kazakhstan,РСФМСШИ_1________________$1st$ $PRIZE$
3.Russia,СУНЦ МГУ______________________$2nd$ $PRIZE$
4.Belarussia, lyceum БГУ__________________$2nd$ $PRIZE$
5.Khazakshtan,РСФМСШИ_2
6.Ukrain,Ukranien math-phis school KНУ_____$3rd$ $PRIZE$
7.Khazakhstan team
8.Russia,МOУ gimnazium-Novosibirsk________$3rd$ $PRIZE$
9.Azerbaijan, Baku Tukish lyceum___________$3rd$ $PRIZE$
10.Khazakhstan,Aktobe Turkish lyceum______$Honourable$ $mention$
11.Bulgaria,Sofia math gimnazium___________$Special$ $prize$
12.Republic Sakha,republic lyceum__________$Honourable$ $mentinon$
13.Khazakhstan,school No-20______________$Honourable$ $mention$
14.Indonesia,Smak 1 BPK Penabur__________$Honourable$ $mention$
15.Uzbekistan,Special school No-1,Bukhar____$Honourable$ $mention$
16.Tatarstan,University of Kazan's lyceum
17.Armenia,phys-math school (EpГУ)
18.Russia,СУНЦ НГУ
19.Khazakhstan,team of Almaty
20.Khazakhstan, team of Pavlodor


number theory
greatest common divisor
Diophantine Equations

Solve in integers the following equation \[n^{2002}=m(m+n)(m+2n)\cdots(m+2001n).\]

For $ n\equal{}0$ we have the solutions $ m\equal{}n\equal{}0$.
For $ n\not\equal{}0$, we have no solution: since the factors in the product differ from eachother by a multiple of $ n$, they either all divide $ n$, or none of them does. In the first case we have $ \left|\prod (m\plus{}kn)\right > n^{2002}$, in the latter case we get a contradiction by looking at the equation modulo $ n$.

We may assume the gcd of m, n is 1 else we can divide it out. the equation mod n shows $m^{2002} = 0$ mod n $\rightarrow m = 0$ mod n. m = kn. ${n^{2002} = \prod_{j = 0}^2001}((k+j)n) \rightarrow \prod_{j = 0}^{2001}(k+j) = 1$ which is absurd.


Solve the equation $ \frac{\left[{x\over 2}\right]}{\{x\}}\equal{}x\minus{}1$

[hide="Straightforward"]Write $ x \equal{} i \plus{} f$, where $ i$ is an integer and $ 0\le f < 1$. Now suppose $ i$ is even; we then have that $ \frac {i}{2f} \equal{} i \plus{} f \minus{} 1$ so that $ f^2 \plus{} (i \minus{} 1)f \minus{} \frac {i}{2} \equal{} 0$. Solving this as a quadratic in $ f$, we find that $ f \equal{} \frac {1 \minus{} i\pm \sqrt {i^2 \plus{} 1}}{2}$. Since $ 0\le f < 1$, we need the numerator to be smaller than the denominator and the numerator positive, so the $ \minus{}$ doesn't work for positive $ i$; let us first consider positive $ i$. Using the smaller than $ 1$ condition, we get that our condition is $ \sqrt {i^2 \plus{} 1} < 1 \plus{} i$. We quickly see that this is true for all positive $ i$ but no other $ i$, so if $ i$ is a positive even, then we get an answer of $ \boxed{\frac {1 \plus{} \sqrt {i^2 \plus{} 1}}{2}}$ for all positive even integers $ i$. We can go through a similar chain of inequalities for negative even $ i$ to discover that the only negative $ i$ which work are positive negative $ i$, which of course don't exist.

For odd $ i$, the equation is $ f^2 \plus{} (i \minus{} 1)f \minus{} \frac {i \minus{} 1}{2} \equal{} 0$. Solving this as a quadratic in $ f$ again, we find that $ f \equal{} \frac {1 \minus{} i\pm \sqrt {i^2 \plus{} 4i \minus{} 1}}{2}$. We see immediately that with the plus sign, the expression is always greater than or equal to $ 1$ for integers (since when $ 1$ is plugged in for $ i$ the expression is $ 1$ and it's increasing from there and noncomplex negative values of the expression are all trivially greater than $ 1$), so there are no solutions with the plus sign. If it's a minus sign, we see that $ i$ needs to be positive for the expression to be less than $ 1$ but negative for it to be greater than $ 0$, so when $ i$ is odd, there are no solutions. Our final answer is hence $ \boxed{\frac{1\plus{}\sqrt{i^2\plus{}1}}{2}}$ for all positive even integers $ i$.[/hide]

There's an error in the solution formula (check it), and there are other solutions as well.

Oh, is it $ \frac {1 \plus{} i \plus{} \sqrt {i^2 \plus{} 1}}{2}$? I don't see how there could be other solutions, though, unless I messed up one of the positive/negative cases.

Don't ask ME what is it :) check and double check and give the final answer. Also, check for any potential messing up along the way.

Is anyone interested in correcting and finishing this? :)

[hide="OK then"]For $ i$ even, the solution is $ x\equal{}{1\plus{}i\plus{}\sqrt{i^2\plus{}1}\over 2}$

For $ i$ odd, the discriminant in the above solution is wrong, and it should be $ f\equal{}{1\minus{}i\plus{}\sqrt{i^2\minus{}1}\over 2}\implies x\equal{}{1\plus{}i\plus{}\sqrt{i^2\minus{}1}\over 2}$

Hence we can combine the two formulas thus: $ x\equal{}{1\plus{}i\plus{}\sqrt{i^2\plus{}(\minus{}1)^i}\over 2},\mathbb{N}\owns i\geqslant 2$[/hide]


geometry
trigonometry
inradius
Pythagorean Theorem
trig identities
Law of Sines

[quote="erik-the-red"]What is the formula for the radius of a circle inscribed in a right triangle?[/quote]
$\dfrac{1}{2} \cdot r \cdot (a+b+c) = \dfrac{1}{2}ab \Longleftrightarrow r=\dfrac{ab}{a+b+c}$

using congruent tangents, you find that it is $\dfrac{c-a-b}{2}$ where c is hyp, a,b legs
and this does also agree with shobber's equation, just it is specific to right triangles, whereas his is for any triangle

My formula is also for right triangles. Because only the area of right triangle can be expressed as $\dfrac12 ab$.

correct...my mistake, also i think i remember this problem from the aops books...

Isn't he simply asking for the radius if sides of right triangle are already known?

Thus you can use $rs=A$, $r$ is the radius you're trying to find, $s$ is the semiperimeter of the triangle, and $A$ is the area of the triangle.

Yeah, and $rs=A$ will give $r(a+b+c)=ab\Longleftrightarrow r=\dfrac{ab}{a+b+c}$, which is what shobber wrote. Also, by using Pythagorean Theorem, we can actually obtain the other one ($r=\dfrac{c-a-b}{2}$) that Altheman gave us.

So, adapting off shobber's formula, the formula for a general triangle would be
$\dfrac{1}{2}ab\sin{C}=r\dfrac{a+b+c}{2} \implies r=\dfrac{ab\sin{C}}{a+b+c}$...
Which indirectly proves the Law of Sines  :D

It seems that the formula for the inradius is popular here, so I'll post another one:
\[ abc \cdot r = \dfrac{1}{2}(a+b+c) \cdot AI \cdot BI \cdot CI \]

Using Heron's, for a general triangle, we can obtain $\displaystyle r=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}$.  However, if we let $x$, $y$, and $z$ be the tangents that are cut off by the incircle, and note that $a=x+y$, $b=x+z$, etc., we can find that it simplifies to \begin{eqnarray*} r=\sqrt{\dfrac{xyz}{x+y+z}}\end{eqnarray*},

which is my personal favorite.


function
real analysis
real analysis unsolved

Expand this function: $ f(x) \equal{} \prod_{i\equal{}1}^{\infty} \frac{1}{1\minus{}x^i}$

Ah, sorry, I forgot. Expand it as $ f(x) \equal{} \sum_{i\equal{}0}^{\infty} a_i x^i$

[hide="answer"]For $ |x|<1$ we have $ f(x)\equal{}\prod_{n\equal{}1}^\infty \sum_{k\equal{}0}^\infty x^{nk}$ so \[ f(x) \equal{} \begin{cases} \sum_{k \equal{} 0}^\infty a_kx^k & |x| < 1 \\
0 & |x| > 1 \end{cases}\] where $ a_k$ is the $ k$th [url=http://www.research.att.com/~njas/sequences/A000041]partition number[/url].[/hide]


Find the 100th term of the arithmetic sequence $ \minus{}3,4,11,18,...$.

You don't need to use formulas.  We know that each term is $ 7$ larger than the previous; thus, the $ 100$th term is $ 99 \times 7\equal{}693$ larger than the first.  Thus, the $ 100$th term is $ 693\plus{}(\minus{}3)\equal{}\boxed{690}$.  Note that I basically used the formula, except I explained why it works and how it works.


now we have shared the TST problems from China,Vietnam,Romania,France,etc,just except the USA,I wonder why??? :(

I find a little bit strange that absolutely nobody wants to post these problems. Are they secret of state or what?  :D

[quote="harazi"]Excuse me, does anyone have the problems from the TST in USA 2005? Thanks a lot. :)[/quote]

I asked that last and this year twice, [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=286407#p286407]1[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=256850#p256850]2[/url], but nobody bothered to reply. I just saw two MOP practice tests in [url=http://web.mit.edu/rwbarton/Public/mop/]Reid Barton's MOP folder[/url] from 2004. BTW there are some [url=http://www.mathlinks.ro/Forum/index.php?f=177]ISL problems[/url] in his tutorial where he did not indicate the source. So what can we conclude about the profileration of the USA TSTs? Maybe the USA just want to benefit from other countries via ML but is not very keen on contributing, or they don't have original [url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]USA TST problems[/url], ...

Also the Russian TST. Russian geometry are very popular, I am waiting for them.

Is it possible that the USA TST problems are copyrighted and the MAA does not allow them to be posted?

look, even I don't have the TST problems (I was at MOP, but in blue group)

so I very much doubt they'd get posted here  :(  sorry.

[quote="MysticTerminator"]look, even I don't have the TST problems (I was at MOP, but in blue group)
[/quote]
What is MOP?

Also, mildorf has the problems right?

MOP = Maths Olympiad Program  :D  :D

Bomb

I have some materials, and in it solutions of all problems on A, C, N of USATST 00-03 could be found!

[quote="shobber"]How do I open a .rar file?  :?[/quote]

You can also go to http://www.zxxk.com to download it

I   have   a    copy    of    USA TST 03, but  it's  in   chinese   ,if  you  need  ,please  e-mail  to  me.

w2h_94@163.com

i think you may mail to Zuming Feng,and the resourse of qsy is also from the thinkpad of him.
he give the coach materials in Zhuhai&Nanhai,Guangdong.

Does any member in this forum have participated in it,could you ask them for the problems?Can we find it in some USA Mathematical magazines?Does anyone know some participants in this competition?

Uhhh... the AMC guys like to keep the most recent TST questions a secret for some reason, so that's why no one has given out the 2006 questions. I don't know about 2005, but maybe they'll wait a while before releasing that too?

In fact,as Zuming Feng was from China,we got the 2006 USA TST when we(our national team) were training in Beijing.It's a little difficult,but I think the order of the problems isn't very good,not from the easy one to the hard one in both days.
But maybe it's not good for me to post it here for it's still secret.
So maybe you guys should wait for some days.

I don't think there's really a "secrecy" kind of thing ... it's just that usually only around 12 kids see the USA TST plus 4 or 5 graders, and unfortunately nobody bothers to post them anywhere. I took USA TST 2005 and I don't even know where I put it anymore, otherwise I'd post it... (don't see why not, there was nothing special about the problems either)

In this case, could you post the problems jin?

[quote="harazi"]Excuse me, does anyone have the problems from the TST in USA 2005? Thanks a lot. :)[/quote]

you  can  mail to w2h_94@163.com
I have a hard copy.

Well, now they're here : http://www.mathlinks.ro/Forum/viewtopic.php?t=107443


Given non-negative integers $ n,N$ with $ n<<N$. need to solve equation:
$ \sum_1^n a_k\equal{}N$, where all $ a_k$ are non-negative integers.
Prove: the number of solutions is less than $ n^N$.

Number of solutions is $ C_{N\plus{}n\minus{}1}^{n\minus{}1}\le n^N.$


Compute $ (1\cdot 5\cdot 25\cdot 125)\cdot (16\cdot 8\cdot 4\cdot 2\cdot 1)$.


For a positive integer n,let r(n) denote the sum of the remainders when n is divided by 1,2,....,n. respectively.Prove that r(k)=r(k-1) for infinitely many positive integers k.

Let $ k\equal{}2^m$, for natural $ m$.

$ r(k)\minus{}r(k\minus{}1)\\\equal{}\sum_{i\not|k}1\minus{}\sum_{i|k,i<k}(i\minus{}1)\\\equal{}k\minus{}1\minus{}\sum_{i|k,i<k}i\\\equal{}2^m\minus{}1\minus{}\sum_{j\equal{}0}^{m\minus{}1}2^j\\\equal{}0$.

[quote="SUPERMAN2"][quote="xxp2000"]Let $ k \equal{} 2^m$, for natural $ m$.

$ r(k) \minus{} r(k \minus{} 1) \\
\equal{} \sum_{i\not|k}1 \minus{} \sum_{i|k,i < k}(i \minus{} 1) \\
\equal{} k \minus{} 1 \minus{} \sum_{i|k,i < k}i \\
\equal{} 2^m \minus{} 1 \minus{} \sum_{j \equal{} 0}^{m \minus{} 1}2^j \\
\equal{} 0$.[/quote]
Thanks.But can you explain more please,I really don't understand.[/quote]

xxp2000 means:
Let $ R_i(k)$ be the remainder of $ k$ when divided by $ i$, then $ R_i(k)\equal{}R_i(k\minus{}1)\plus{}1$ unless $ i|k$ ,
So the second line, i.e. $ \equal{}\sum_{i\not|k}1 \minus{} \sum_{i|k,i < k}(i \minus{} 1)$ comes from dividing $ R_i(k)\minus{}R_i(k\minus{}1)$ into two cases :
[b](i)[/b]$ i\nmid k$ or [b](ii)[/b]$ i|k, i < k$


AMC
USA(J)MO
USAMO
calculus
AoPS Books
AMC 10

Will the aops books + intermediate independent study A and B + part of art and craft + 1-3 subject classes be good enough preparation for amc 12/aime. I am going directly to intermediate and I am worried I will miss a concept from one of the intro independent study classes. Will I learn everything?

[quote="xxreddevilzxx"]I am sure there has been a lot of discussions on this in this forum. However, I will say that AoPS books 1 and 2 will be enough for AIME.

at least thats how i made USAMO (I didnt take any AoPS classes, although I regret that i would have scored higher)[/quote]

Hope you're right reddevilz :P
I also have the exact same question as Calculus. I'm definitely taking a look at some of these classes, though I am waiting to take the AMC-12 class instead of the AMC-10, as I feel it would be a better experience to take the class for the test I actually am taking (though I understand the two are similar).

Calculus, even though I'm not so sure as to what's enough prep, that much should be enough to make it.


symmetry
geometry
geometric transformation
rotation

In how many ways can 4 distinct beads be placed on a bracelet that has a knot in it?  (Beads cannot slide past the knot.)

Never mind, just delete this someone. The problem just seemed ambiguous. But it's not.

[quote="mewto55555"]Its just 4!/2. We divide by 2 becuz we can flip so we get 12.[/quote]

Unfortunately, this is not the case.

$ \dfrac{4!}{4(2)}\equal{}3$

There are 4! ways to line up the beads. This is then divided by 2 for the flip [b]and 4 for the rotations.[/b]

[quote="Dojo"][quote="mewto55555"]Its just 4!/2. We divide by 2 becuz we can flip so we get 12.[/quote]

Unfortunately, this is not the case.

$ \dfrac{4!}{4(2)} \equal{} 3$

There are 4! ways to line up the beads. This is then divided by 2 for the flip [b]and 4 for the rotations.[/b][/quote]

Unfortunately, this is not the case because the balls cannot slide past the knot so [b]there are no rotations.[/b]

Just treat the knot as another bead. There are ways to place the beads and the knot on the bracelet, but we must divide by 5 for rotational symmetry (5 rotations for each arrangement), and by 2 for reflectional symmetry (we can flip the bracelet to get the same arrangement). The answer is $ \frac{5!}{5 \cdot 2}\equal{}\boxed{12}$, like mew said.

I think I got the right answer the wrong way.  I figured, if you can't do rotations since the beads can't slide past the knot, it's 4!, but if it's flipped it's the same, so you have to divide by 2.  4! (24) divided by 2 is 12.

Can I just treat the bracelet with a knot as a line? Like _ _ _ _, as in row of beads.

Quick thing I made-up for these problems.

$b=\text{number of beads}$.\[\frac{(b-1)!}{2}\]

[quote=Mewto55555]Its just 4!/2. We divide by 2 becuz we can flip so we get 12.[/quote]

lol how 12 ways, show me. :P

[quote=saphiresky]How did GameBot post this question in 2008? It says GameBot joined in 2009.[/quote]

gamebot last visited 2015, but he posts problems in 2016

[quote=chu42][quote=saphiresky]How did GameBot post this question in 2008? It says GameBot joined in 2009.[/quote]

It's a conspiracy, they just got careless.[/quote]

because maybe it is another gamebot

not likely because when I searched the gamebot users, I only found one. :P

[hide=my sol]We can image this diagram: -_|_-. The first bead can go on two positions (middle and bottom), the second can go on all the rest(they're all unique), the third can go on two, and the last can go on one. We get $2\cdot3\cdot2\cdot1=\boxed{12}$[/hide][hide=why my sol is good]It's simple and logical. Please upvote if you like it.[/hide]

Hello math friends. I am having trouble accepting/understanding the official solution. I know I am probably wrong, but I cannot see the rotations in this case. The knot cannot be treated as another bead in my eyes. Please check my logic and poke holes in it.

Lets simplify the problem to one with 3 distinct beads, and lets start with a case without the knot and then move on to a case with a knot.

[u]Case 1[/u]: Bracelet has no knots.
Since there is no knot, it follows that we have 3! = 6 distinct arrangements for the beads. These are: ABC, ACB, BAC, BCA, CAB, CBA. In such a case, ABC, BCA and CAB are all the same, since in a cyclical arrangement they follow the sequence A->B->C (or B->C->A or C->A->B). Similarly, ACB, BAC and CBA are identical arrangements. Thus, we divide 3! by 3 to account for the rotational symmetry. This yields 2 unique arrangements: ABC and ACB. Furthermore, we must account for the reflectional (flip) symmetry, hence we divide by 2, yielding only one possible arrangement. Its hard for me to show that ABC and ACB are reflections of each other in a text format, but if you can draw them in a circular form you can convince yourselves of this.

[u]Case 2[/u]: Bracelet has a knot. Beads cannot go past the knot.
Let X denote then knot. Then the possible arrangements are: XABC, XACB, XBAC, XBCA, XCAB, XCBA. Note that the X never moves. This "brakes" the rotational symmetry that was present in the case without the knot. For example, XABC is not the same as XBCA since we go from A->B->C->X in the former arrangement, while we go from A->X->B->C in the later. Therefore, we still have 3! = 6 distinct arrangements but only reflectional symmetry in this case. Thus, we must divide 3! by 2 yielding 3 distinct arrangements. Again, you can visualize the reflections on a piece of paper and see that XABC and XCBA are reflections of each other.

In conclusion, this is the reasoning I followed to solve the problem (in my second attempt). However, I was surprised that the official solution suggests to use the knot as a extra bead. It then proposes that the computation is 5! / 5 * 2 (5 for rotational symmetry and 2 for reflectional symmetry).

After writing the above, and thinking through the problem thoroughly, I believe I have understood the reasoning of the official solution (correct me if I am wrong). The knot can be treated as a bead if we consider that we can add beads from both ends of the the bracelet. Indeed, in such a case the X will be "moving" and therefore, XABC and ABCX will be identical.

I will still post this comment in the hopes that it will help other math enthusiasts.

Thank you!


geometry
analytic geometry
geometry solved

consider a point $A(3,2,1)$  and a plane $\pi:2x-y+z=1$
what is the equation of the plane that passes through $A$ and is paralel with plane $\pi$ ?

$\pi:\ \ \frac{x}{p} + \frac{y}{q} + \frac{z}{r} = \frac{x}{\frac 1 2} + \frac{y}{-1} + \frac{z}{1} = 1$

where $p, q, r$ are the segments $OP, OQ, OR$ the cut by the plane $\pi$ on the x-, y-, z-axes from the origin $O$.
Since $\pi' \parallel \pi$,

$\pi':\ \ \frac{x}{ap} + \frac{y}{aq} + \frac{z}{ar} = \frac{x}{\frac a 2} + \frac{y}{-a} + \frac{z}{a} = 1$

Since $A \in \pi'$, substituting the coordinates of the point $A = (3, 2, 1)$ into the equation of the plane $\pi'$,

$\frac{3}{\frac a 2} + \frac{2}{-a} + \frac{1}{a} = 1$

$6 - 2 + 1 = a,\ \ a = 5$

$\pi':\ \ \frac{x}{\frac 5 2} + \frac{y}{-5} + \frac{z}{5} = 1,\ \ 2x - y + z = 5$