inequalities
number theory proposed
number theory

Let $ a > b > 1, b$ is an odd number, let $ n$ be a positive integer. If $ b^n|a^n\minus{}1,$ then $ a^b > \frac {3^n}{n}.$

We only need to consider b as an odd prime number . Let $ d \equal{} {ord}_a mod p$ then we have $ d|n$ .By the lifting exponent lemma ,we have : $ v_p(a^n \minus{} 1) \equal{} v_p(a^d \minus{} 1) \plus{} v_p(\frac {n}{d})$ . Because $ p^n|a^n \minus{} 1$ so $ v_p(a^d \minus{} 1) \plus{} v_p(n)\geq n,\rightarrow a^d \minus{} 1\geq \frac {p^n}{n}\geq \frac {3^n}{n}$.
Other hand $ d\leq \varphi(p) \equal{} p \minus{} 1$ so $ a^p > a^d \minus{} 1 > \frac {3^n}{n}$ (QED)

This solution seems too easy. Am I being crazy or is the problem actually this simple?

Since $ b^n | a^n\minus{}1$, write $ a^n \equal{} kb^n \plus{} 1$. Then $ a^b \equal{} (1\plus{}kb^n)^{b/n} \geq 1 \plus{} \frac{kb^{n\plus{}1}}{n} > \frac{3^n}{n}$. The first inequality is Bernoulli's Inequality and the second is because $ b \geq 3$.

[quote="Hamster1800"]This solution seems too easy. Am I being crazy or is the problem actually this simple?

Since $ b^n | a^n \minus{} 1$, write $ a^n \equal{} kb^n \plus{} 1$. Then $ a^b \equal{} (1 \plus{} kb^n)^{b/n} \geq 1 \plus{} \frac {kb^{n \plus{} 1}}{n} > \frac {3^n}{n}$. The first inequality is Bernoulli's Inequality and the second is because $ b \geq 3$.[/quote]
sorry but we have (1+x)^n> 1+ nx if n>1 and x>-1,so b>n?,Mr.Hamster1800 :?:

This  problem  is  very  easy,and  let's  prove  that   for a^b>3^n(1) ,also  we  can  show  that   for  every  constant  c>0,  we  can  find  infinitely  many  integers  n,s.t  a^b>3^(cn).(Also  for  infinite  set  we  can  replace  3  by  every  integer).
It  is  enough  to  prove  this  for  b =p>2  ,  prime.
(1)  is  true ,when  n<=p-1,Let  assume ,that  n>p-1.Let  r=ordp(a)<=p-1<n.If vp(a^r-1)=t,then  vp(a^n-1)=t+vp(n/r)>=n=>t>=n-vp(n/r)=A,then  a^p=(a^r)^(p/r)>(p^t)^(p/r)>p^(pA/r)>=p^n.
Such  that  B=pA/r>=p/r*(n-n/rc)=n*(p/r*(1-1/rc)=nC,where  C  can  be  as  large  as  we  want ,for  diffirent  n's  and  p's.
If p-1>r ,then  p>2r=>C>2(1-1/rc)>=2(1-1/1*2)=1,Such  that  vp(k)<=k/c ,and  c=2  holds  for  every  k  and  p>2.
If  p-1=r,then  C=(1+1/r)(1-1/rc)=1+1/r(1-1/c-1/cr)>=1.

[quote="[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2197621]fedja[/url] on problem [b][i]25[/i][/b]"]Let $P$ be the set of all prime divisors of $b$. For $p\in P$, let $s_p$ be the least integer such that $p|a^{s_p}-1$. We have $s_p|n$, $s_p|p-1$ and $n\le v_p(a^{s_p}-1)+v_p(n/s_p)$. Now note that
$b\ge\prod_{p\in P} p>\prod_{p\in P} s_p=S$ and that $a^{s_p}-1|a^S-1$ for all $p\in P$. Thus,
\[a^b>a^S-1\ge \prod_{p\in P}p^n \prod_{p\in P}p^{-v_p(n)}\ge 3^n/n.\][/quote]

Good and easy to observe part is we only need to prove for prime case,since $a^d$ is increasing, rest will follow immediately now $p=b\implies v_p(a^n-1)\geq p^n$. Suppose $d$ be order of $p$ and $v_p(a^d-1)=m\implies dp^{n-m}|n$ and so $a^b>a^d>p^m>\frac{p^n}{n}$ and so done

[b]China TST 2009 Quiz 6 P1.[/b] Let $n$ be a positive integer and let $a>b>1$ be integers such that $b$ is odd and $b^n\mid a^n-1$. Prove that $a^b>\frac{3^n}{n}$.

[b]Solution.[/b] Let $p$ be a prime divisor of $b$, then it follows that $p$ is odd.
Let $k=ord_p a$, then it follows that $k\mid p-1$ and $k\mid n$.
Now by LTE
\[a^b>a^b-1\ge a^p-1>a^{p-1}-1\ge a^k-1=p^{log_p(a^k-1)}\ge p^{\nu_p(a^k-1)}=p^{\nu_p(a^k-1)+\nu_p(\frac{n}{k})-\nu_p(n)}=p^{\nu_p((a^k)^{\frac{n}{k}}-1)-\nu_p(n)}=p^{\nu_p(a^n-1)-\nu_p(n)}\ge p^{\nu_p(b^n)-\nu_p(n)}\ge p^{n-\nu_p(n)}\ge\frac{p^n}{n}\ge \frac{3^n}{n}.\]

Solved with [b]Dimath27[/b]  :) .
 WLOG $b$ is a prime so $v_p(a^n-1) \ge n$ then let $d$ the order of $a$ in mod $p$ so by LTE: $v_p(a^n-1) = v_p(a^d-1)+ v_p(\frac{n}{d}) $ but $d \mid p-1$ so we have: $$n \leq v_p(a^n-1) = v_p(a^d-1)+ v_p(n) \implies p^{n-v_p(n)} \leq a^d-1 < a^p $$
$$a^p > \frac{p^n}{p^{v_p(n)}} \ge  \frac{p^n}{n} \ge \frac{3^n}{n} $$
as desired. $\blacksquare$

[quote=Sprites]Let $p$ be a prime with $p|b$
By LTE,$n \ge \nu_p(b^n) \ge \nu_p(a^n-1) \ge \nu_p(a^{n(p-1)}-1^n) = \nu_p(a^{p-1}-1)+\nu_p(n)$ hence $a^b>a^{p-1} \ge p^{\nu_p(a^{p-1}-1)} \ge \frac{p^n}{n} \ge \frac{3^n}{n}$[/quote]

I think it should be $n \le \nu_p(b^n) \le \nu_p(a^n-1) \le \nu_p(a^{n(p-1)}-1^n) = \nu_p(a^{p-1}-1)+\nu_p(n)$.


Let $ S$ be the set of points $ (a,b)$ with $ 0\le a,b\le1$ such that the equation \[x^4 \plus{} ax^3 \minus{} bx^2 \plus{} ax \plus{} 1 \equal{} 0\] has at least one real root. Determine the area of the graph of $ S$.

[quote="worthawholebean"]([b]7[/b]) Let $ S$ be the set of points $ (a,b)$ with $ 0\le a,b\le1$ such that the equation

$ x^4 \plus{} ax^3 \minus{} bx^2 \plus{} ax \plus{} 1 \equal{} 0$

has at least one real root. Determine the area of the graph of $ S$.[/quote]

anyone know the formula for the global minimum of a quartic? (if it's less than or equal to 0 and a is positive, or it's greater than or equal to 0 and a is negative, there is a root)

[quote="azjps"][hide="Hint"]
It looks symmetric, so divide through by $ x^2$ and then substitute $ y \equal{} x \plus{} \frac {1}{x}$. Don't know if that gets anywhere.
[/hide][/quote]

[hide="it does, kinda"]
it leads to $ y^2\plus{}ay\minus{}(b\plus{}2)\equal{}0$
using discriminant or whatever it's called
$ a^2\plus{}4b\plus{}8\ge 0$
or $ b\ge \minus{}2\minus{}\frac{a^2}{4}$
treat this as a function, with b=y and a=x

it's a downward facing parabola

now i just have to add in the restriction that $ y\equal{}x\plus{}\frac{1}{x}$ does not work for all values (or does it)

[/hide]

[hide=Solution]
The equation's symmetric, so we apply the standard method here. Dividing both sides by $x^2$, we have that \[\left(x^2+\frac{1}{x^2}\right)+a\left(x+\frac{1}{x}\right)-b=0.\] Let $y=x+\frac{1}{x}$. Notice that $y\leq -2$ and $y\geq 2$. We have that \[y^2+ay-b-2=0\implies y=\frac{-a\pm \sqrt{a^2+4b+8}}{2}.\] We need the maximum absolute value of this to be at least $2$, so \begin{align*}\frac{a+\sqrt{a^2+4b+8}}{2}\geq 2 \\ \implies a^2+4b+8\geq 16-8a+a^2 \\ \implies b\geq -2a+2.\end{align*} Combining with $0 \le a, b\le 1$, we have a triangle with vertices $\left(\frac{1}{2}, 1\right), (1, 1), (1, 0)$, so the area is $\boxed{\frac{1}{4}}$.
[/hide]


The problem is available [url=http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10A_Problems/Problem_12]here[/url].

I was confused about whether the triangles were included or not - it doesn't say anything specific. With the triangles, the answer is $ \frac{2}{3} \pi \plus{}1$, but $ \frac{2}{3} \pi$ if they aren't included. At first I thought the triangles were included, but after looking at the solution

Am I missing something? Did this become a problem on the actual test?

The triangles are included if I read it correctly, but I'm pretty sure the solution is still $ \frac{2}{3}pi$

Oh... my mistake about the answers, but still: the triangles could be excluded depending on the interpretation.

It says "The [b]figure shown[/b] is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length ?"

figure shown means the whole figure, and besides, if it didn't include the triangles, the answer would have to be $ (something) \pi \minus{} \sqrt {3}$ (Sorry, too lazy to find the "something")


ratio
geometry
incenter
national olympiad

[b]TURKEY NMO 2006[/b]
December 16-17
Each day 4½ hours to work
Day 1

[b]1)[/b]$ABCD$ is convex quadrilateral. On $[BA]$ we have points $P$ and $Q$ such that $AP=BQ$. $(APD) \cap(BQD)=K$ and $(APC)\cap(BQC)=L$ (different from $D$ and $C$ also). Show that $D, C, K, L$ are concyclic.($(ABC)$ shows the circle that pass through $A,B,C$)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=722406#722406[/url]
[b]2)[/b]There are $2006$ students and $14$ teachers. There are some relationships among the students and the teachers such that all of the students know at least one teacher. If for all the correlations between students and teachers there is one student, teacher pair ,each knows other, such that the ratio of the number of students that the teacher knows; to the number of teachers that the student knows is at least $t$. Find the maximum of $t$.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=722416#722416[/url]
[b]3)[/b]Find all positive integer n such that, all of the coefficients of $P(x)$are divisible by $7$
$P(x)=(x^{2}+x+1)^{n}-(x^{2}+1)^{n}-(x+1)^{n}-(x^{2}+x)^{n}+x^{2n}+x^{n}+1$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=722413#722413[/url]
Day 2
[b]4)[/b]$x_{1},...,x_{n}$ are positive reals such that their sum and their squares' sum are equal to $t$. Prove that $\sum_{i\neq{j}}\frac{x_{i}}{x_{j}}\ge\frac{(n-1)^{2}\cdot{t}}{t-1}$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=722414#722414[/url]
[b]5)[/b]$ABC$ be a triangle. Its incircle touches the sides $CB, AC, AB$ respectively at $N_{A},N_{B},N_{C}$. The orthic triangle of $ABC$ is $H_{A}H_{B}H_{C}$ with $H_{A}, H_{B}, H_{C}$ are respectively on $BC, AC, AB$. The incenter of $AH_{C}H_{B}$ is $I_{A}$; $I_{B}$ and $I_{C}$ were defined similarly.
Prove that the hexagon $I_{A}N_{B}I_{C}N_{A}I_{B}N_{C}$ has all sides equal.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=722411#722411[/url]
[b]6)[/b]Find all the triangles such that its side lenghts, area and its angles' measures (in degrees) are rational.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=722412#722412[/url][/b]

according to results of the competition
4+ solutions gold (2 people)
3+ solutions silver (4 people)
2+ solutions bronze (8 people)
also what can you say for the problems.


[quote="nononana"]a,b,c>0, a+b+c=2

prove that

$ \frac {ab}{1 \plus{} c^2} \plus{} \frac {bc} {1 \plus{} a^2} \plus{} \frac {ca}{1 \plus{} b^2} \leq 1$[/quote]

I have a proof with AM-GM and Schur.It is nice. :)


there would be two cases, one where the length of the hypotenuse = 21, and one where one of the other sides = 21..

Correct?

the original question says the length of one LEG is 21.  :lol:

[quote="iamagenius"]how many distinct right triangles exist if the lengths of all sides are integers and the length of one leg is 21?[/quote]

[hide]I think there's that thing where it's like $a=m^{2}-n^{2}$, $b=2mn$, $c=m^{2}+n^{2}$ so $21=m^{2}-n^{2}$ so $a=21$, $b=20$, $c=29$ and $21=2mn$ can't work...but 21 isn't prime so it could be the leg of a non-primitive right triangle...so we find the ones with 3 and 7...$3=m^{2}-n^{2}$, the sides are $3,4,5$ giving $21,24,27$, and then $7=m^{2}-n^{2}$, $m=4$, $n=3$ giving sides $7,24,25$ or $21,72,75$ so there are $\boxed{3}$?[/hide]

i didn't read you solution, but my answer sheet says it is 4.

[hide]
$21$ is not prime and could be the leg of a non primitive

$[7-24-25] * 3$
$[3-4-5] * 7$

21 could be the smallest leg in the odd side pythagorean triple formula yielding
$21-220-221$

and finally, the obscure pythagorean triple
$20-21-29$

proving that there are no other solutions... I don't know how to do that, but I'll work on it

[/hide]

[hide]
$21^{2}= x^{2}-y^{2}= (x-y)(x+y)$

we must find all factors of 441 which are
$21\cdot 21$, $9\cdot 49$, $7\cdot 63$, $3\cdot 147$ and $1\cdot441$
So there are 5 such sets of factors, therefore $y-x$ equals the smaller factor and $y+x$ equals the larger factor.

21 and 21 doesnt work becuase no two positive integers add to and subtract to 21, which leaves 4 sets of factors and thus 4 distinct triangles with sides:
$21, 20, 29$
$21, 28, 35$
$21, 72, 75$
$21, 220, 221$
[/hide]


Prove by combinatorial reasoning that

$ \sum_{k\equal{}1}^{n}k\binom{n\plus{}1}{k\plus{}1}\equal{}(n\minus{}1)2^{n}\plus{}1$.

Umm... not completely combinatorial, but

[hide]
Notice the LHS = $ \sum_{k \equal{} 1}^{n \plus{} 1}{(k \plus{} 1)\binom{n \plus{} 1}{k \plus{} 1}} \minus{} \sum_{k \equal{} 1}^{n \plus{} 1}{\binom{n \plus{} 1}{k \plus{} 1}}$. We can apply combinatorial arguments to these :/

For the first sum, let's add $ \binom{n \plus{} 1}{1}$ to make it prettier. Then, that sum is equivalent to the number of ways of picking $ k \plus{} 1$ people out of a group of $ n \plus{} 1$ and designating one a leader. However, we could count this like "pick a leader out of the n+1 people and then pick any number of people," which is $ (n \plus{} 1)(2^{n})$. The second sum is $ (1 \plus{} 1)^{n\plus{}1} \minus{} (n \plus{} 1) \minus{} 1 \equal{} 2^{n \plus{} 1} \minus{} (n \plus{} 2)$. So, the LHS is
\[ (n \plus{} 1)(2^{n}) \minus{} (n \plus{} 1) \minus{} 2^{n \plus{} 1} \plus{} (n \plus{} 2) \equal{} (n \minus{} 1)2^n \plus{} 1
\]
[/hide]


Prove that the number $\underbrace{111\ldots 11}_{1997}\underbrace{22\ldots 22}_{1998}5$ (which has 1997 of 1-s and 1998 of 2-s) is a perfect square.

Looool... http://www.mathlinks.ro/Forum/viewtopic.php?t=5797 . Although, thinking about it once again, probably knowing the initial problem was not that much of an advantage in solving the ISL problem.

  Darij

[quote="darij grinberg"]Looool... http://www.mathlinks.ro/Forum/viewtopic.php?t=5797
  Darij[/quote]You got to be kidding me  :rotfl: :)

[quote="tµtµ"][hide]A solution is to prove by induction that :

$\underbrace{111\ldots 11}_{n}\underbrace{22\ldots 22}_{n+1}5 = (\underbrace{333\ldots 33}_{n}5)^2$[/hide][/quote]

Your solution is the most easier to picture out and I appreciate your innovative approach.Do you know some more kinds of ways in which induction works

@above its a square that ends in 5 it should be 33...3335


pigeonhole principle
combinatorics proposed
combinatorics

Suppose otherwise.

Let $F$ be a vertex that is never visited by a frog. Let $S_i$ denote the set of vertices that have a distance of $i$ edges from $F$. Since the graph is connected, it is partitioned into $V, S_1, S_2, \ldots S_n$ for some $n$.

A vertex $V$ is "filled" if it contains $\deg V$ frogs.

Since the process goes on indefinitely, let $k$ be the least integer such that there are infinity many times for which some vertex in $S_k$ filled. Clearly, $k>1$ as no vertex in $S_1$ can ever be filled (as $F$ is never occupied).

The choice of $k$ implies that after some time $T$, no vertex in $S_1, \ldots, S_{k-1}$ will ever be filled.

But then, each time some vertex in $S_k$ is filled, some frog jump into $S_{k-1}$. So the number of frogs in $S_{k-1}$ increases indefinitely without filling, an impossibility.

(Yay! 400th post  :D )

pretty pretty extremal proof for the 400th post  :lol:

btw, you only need 2m - (no. of vertices) + 1 frogs.

[quote="vess"]Let $G$ be a connected graph with $m$ edges. Consider $2m+1$ frogs, each staying on some vertex of $G$. At each second, if $v \in G$ contains at least $d(v)$ frogs, then $d(v)$ of the frogs on $v$ jump, one on each of the $d(v)$ adjecent vertices. Show that every vertex will be visited by a frog at some point.[/quote]
Indeed, that is what I did for a 'chinese' problem :)

[quote="vess"]Let $G$ be a connected graph with $m$ edges. Consider $2m+1$ frogs, each staying on some vertex of $G$. At each second, if $v \in G$ contains at least $d(v)$ frogs, then $d(v)$ of the frogs on $v$ jump, one on each of the $d(v)$ adjecent vertices. Show that every vertex will be visited by a frog at some point.

--Vesselin[/quote]
nice problem

but I think "2m+1"should  be replaced by "∑d(vi) -the point's number+1" :)

[quote="Myth"]but $2m+1=1+\sum d(v_i)$  ;)[/quote]

Actually you just need $1+\sum (d(v_i)-1)$ frogs to use the pigeonhole principle to show the process never ends? It seems the proof runs through perfectly all the same.


Eleven distinct points $ P_1,P_2,...,P_{11}$ are given on a line so that $ P_i P_j \le 1$ for every $ i,j$. Prove that the sum of all distances $ P_i P_j, 1 \le i <j \le 11$, is smaller than $ 30$.

$ n \equal{} 2m \plus{} 1 > 1$ is odd, $ m > 0.$ WLOG, $ P_1, P_2, ..., P_n$ follow on the line in this order. First, suppose that these points are not necessarily all distinct. Then $ \sum_{j \neq i \equal{} 1}^{n} |P_iP_j| \le m(m \plus{} 1) |P_1P_n|.$ For $ m \equal{} 1,$ $ |P_1P_2| \plus{} |P_2P_3| \plus{} |P_1P_3| \equal{} 2 |P_1P_3|,$ the proposition is true. Suppose the proposition holds for sufficiently large $ m \minus{} 1$ $ \Longrightarrow$

$ \sum_{j \neq i \equal{} 1}^{n} |P_iP_j| \equal{} |P_1P_n| \plus{} \sum_{k \equal{} 2}^{n \minus{} 1} \left(|P_1P_k| \plus{} |P_kP_n|\right) \plus{} \sum_{j \neq i \equal{} 2}^{n \minus{} 1} |P_iP_j| \le$

$ \le 2m |P_1P_n| \plus{} (m \minus{} 1)m |P_2P_{n \minus{} 1}| \le m(m \plus{} 1)|P_1P_n|.$

Suppose now that $ P_1, P_2, ..., P_n$ are all distinct and $ n \equal{} 2m \plus{} 1 > 3, m > 1.$ Then

$ \sum_{j \neq i \equal{} 1}^{n} |P_iP_j| \le 2m|P_1P_n| \plus{} (m \minus{} 1)m |P_2P_{n \minus{} 1}| < m(m \plus{} 1) |P_1P_n| \le m(m \plus{} 1).$

Plugging in $ n \equal{} 11, m \equal{} 5$ yields $ \sum_{j \neq i \equal{} 1}^{11} |P_iP_j| < 30.$
____________________________________________________________________________

I regret that the problem is not for 13 points, because then the answer would be ... [color=red][size=200][b]42 !!![/b][/size] [/color]
(Most definitely, this is the Answer.)


[quote="Done"]A point is moving randomly inside an equilateral triangle whose side length is 3 cm. What is the probability that its distance to any corner is greater than 1 ?[/quote]
[hide]
The circumcenter of the triangle is the point which is, in this case, is rt3 from all the corners. Therefore the set of points where the distance to any corner is greater than 1 is the circle with radius (rt3-1) and the same center as the circumcircle. The area of this area is pi*(rt3-1)^2= (4-2rt3)pi. The area of the triangle is 9rt3/4. Therefore the probability is (16-8rt3)pi/9rt3.
[/hide]

I probably messed up somewhere... given that I'm doing this at 8:30 in the morning right before class starts  :( .

I tried it like this.

[hide] For each vertex of the triangle, draw the region around that point that contains the set of points in the triangle <= 1 cm from the vertex.    Each region is one sixth (from the 60 degree angle) of a circle with radius 1.  The area of each region is pi/6, and since there are 3, the total area is pi/2.  That means the area of the region in the triangle containing points>1 cm is just (9/4)rt3 - pi/2=(9rt3 - 2pi)/4.  The probability wanted is just this area over the total area.

{(9rt3 - 2pi)/4}/{(9/4)rt3} = (9rt3 - 2pi)/9rt3 = {27-2pi(rt3)}/27
[/hide]


1-Let n be a positive integer and x_1, x_2,...,x_n be integer numbers such that
(x_1) 2  + (x_2) 2  +...+ (x_n) 2 +n^3 ≤ (2n-1)(x_1 + x_2 +...+ x_n) + n 2
Show that
a) x_i for i=1,2,...,n are nonnegative integers
b) the number x_1 + x_2 +...+ x_n + n + 1 is not a perfect square

I think I may have a solution for a) (not very experienced with this stuff, so I make things up as I go)

[hide]

inequality can be rearranged as

 :Sigma: n :^2: + :Sigma: x_i :^2: - :Sigma: 2nx_i :le:  :Sigma: n- :Sigma: x_i

 :Sigma: (n-x_i) :^2:  :le:  :Sigma: (n-x_i)

 (n-x_i) - (n-x_i)^2 :le: 1/4

so let a_i=(n-x_i) - (n-x_i)^2, a_i :le: 1/4

0 :le: :Sigma: a_i :le: 1/4n

now lets consider a x_k which is negative, for this value

a_k=(n-x_k) - (n-x_k)^2 less than n-n^2

and the sum of the others values of a is :le: 1/4(n-1)

thus the sum is less than 5/4n-n^2-1/4 = -1/4(4n :^2: -5n+1)=-1/4(4n-1)(n-1) less than 0 for n=1,2...

contradiction, thus all x_i are non-negative

[/hide]

edit: did I do something wrong? because for integer x_i

a_i=(n-x_i) - (n-x_i)^2, a_i :le: 0 since n is also an integer, and that would mean all x_i=n or n-1, so i don't see why all x_i have to be integers...

edit: now I am going to try b) but x_1,x_2... do not have to be integers

[hide]

let x_i=(n-b_i) so now we have to prove n^2+n+1- :Sigma: b_i isn't a perfect square

0 :le: a_i=(b_i) - (b_i)^2 :le: 1/4

0 <=  :Sigma: b_i(1-b)  <= 1/4n

0 <= (:Sigma: (b_i(1-b)) )/n <= ( :Sigma: (b_i)/n(1- :Sigma: (b_i)/n) )

thus  :Sigma: (b_i)/n is between 0 and 1

and n^2+1 <= n^2+n+1- :Sigma: b_i  <= (n+1)^2

[/hide]

so do x_1, x_2....x_i have to be integers? because I think my proof shows that x_1, x_2....x_i are nonegative and that their sum+n+1 can't be a perfect square even if they are not integers? or is my proof just horribly wrong?

I understood only the first two lines of your solution. :maybe:
Help!

The problem is I don't know if my solution is right, perhaps someone more experienced can tell me that ... if the problem is that it is unreadable b/c i didn't TeX it, then I can do that now (learning to use LateX was really easy, 10 minutes and having the guide up and ur set!)


Two sides of a right triangle have the lengths 4 and 5. What is the product of the possible lengths of the third side? Express the product as a decimal rounded to the nearest tenth.

There are two possible lengths: Where it's the shortest and where it's the longest side. So the longest side would be $ 16\plus{}25\equal{}41$ because of the Pythag, but you have to take the radical of $ 41$. $ \sqrt{41}\equal{}6.403124237..$. And then the other side would be $ 25\minus{}16\equal{}9$ because of again: The pythag theorem. So then you take the radical of $ 9$ and then  you get $ 3$.

$ \frac{3\plus{}6.40312437..}{2}\equal{}4.7$ Roughly. Cause it asks for the nearest tenth.

uh I think this will have to be clarified a little...

If the side length of $ 5$ is the longest, the other leg must be $ \sqrt{5^2\minus{}4^2}\equal{}\sqrt{25\minus{}16}\equal{}\sqrt{9}\equal{}3$.
If $ 5$ and $ 4$ are the legs, the hypotenuse will be $ \sqrt{4^2\plus{}5^2}\equal{}\sqrt{16\plus{}25}\equal{}\sqrt{41}\approx6.4$.

Taking the product gives $ 3\times6.4\equal{}\boxed{19.2}$

Lol. I'm sorry. I didn't read the question thoroughly.  :blush:  :blush:


geometry
incenter
circumcircle
trigonometry
geometry proposed

This is my first post here and for those of you who don't know me, I'm Lee and currently am on the UK IMO Squad (not team...yet).

Every year we have a training session down in Trinity College Cambridge and this question appeared on the sheet of background problems courtesy of the veteran problem setter David Monk:

[color=red]$\bigstar \ \bigstar \ \bigstar$[/color]

The triangle $ABC$ has circumcentre $O$ and incentre $I$. The altitude from $A$ meets the circumcircle again at $R$. The lines $BI$ and $CO$ meet at $P$; $CI$ and $BO$ meet at $Q$. Given that the line $PQ$ passes through $A$ and $OR$ is perpendicular to $AI$, find the angles of the triangle.

Let K, L, M, be the feet of the AI, BI, CI (the angle bisectors), X, Y, Z be the feet of AO, BO, CO and S the foot of APQ. The line AO is the isogonal conjugate of the A-altitude, AI bisects also the angle $\angle OAR$. Since $RO \perp AI$, the triangle $\triangle ARO$ is isosceles with AR = AO, but it is also isosceles with OA = OR (the circumradius), hence it is equilateral, $\angle AOR = 60^\circ,$ $\angle ACR = 30^\circ,$ and $\angle ABR = 150^\circ.$ It follows that

$150^\circ = \angle ABR = \angle ABC + \angle CBR = \angle B + \angle CAR = \angle B + 90^\circ - \angle C$

$\angle B - \angle C = 60^\circ,\ \ \angle A = 180^\circ - (\angle B + \angle C) = 240^\circ - 2 \angle B$

Using the sine theorem for the triangles $\triangle BCY, \triangle ABY$,

$\frac{CY}{BC} = \frac{\sin(90^\circ - A)}{\sin (90^\circ + A - C)} = \frac{\cos A}{\cos (A - C)}$

$\frac{AY}{AB} = \frac{\sin(B + A - 90^\circ)}{\sin (90^\circ + C - A)} = -\frac{\cos(A + B)}{\cos (A - C)} = \frac{\cos C}{\cos (A - C)}$

$\frac{CY}{AY} = \frac{BC}{AB} \cdot \frac{\cos A}{\cos C}$

and by cyclic exchange,

$\frac{AZ}{BZ} = \frac{CA}{BC} \cdot \frac{\cos B}{\cos A}$

By Ceva's theorem,

$\frac{AM}{BM} \cdot \frac{BS}{CS} \cdot \frac{CY}{AY} = \frac{AZ}{BZ} \cdot \frac{BS}{CS} \cdot \frac{CL}{AL} = 1$

$\frac{CA}{BC} \cdot \frac{CY}{AY} = \frac{BC}{AB} \cdot \frac{AZ}{BZ}$

$\frac{CA}{BC} \cdot \frac{BC}{AB} \cdot \frac{\cos A}{\cos C} = \frac{BC}{AB} \cdot \frac{CA}{BC} \cdot \frac{\cos B}{\cos A}$

$\frac{\cos A}{\cos C} = \frac{\cos B}{\cos A}$

$\cos^2 A = \cos B \cos C = \frac{\cos(B + C) + \cos(B - C)}{2} = \frac{\cos(2B - 60^\circ) + \cos 60^\circ}{2} =$

$= \frac{\cos(180^\circ - A) + \cos 60^\circ}{2} = \frac 1 4 - \frac 1 2 \cos A$

$\cos A = \frac{-1 \pm \sqrt 5}{4} = -\frac G 2$ or $\frac{1}{2G},\ \ \ \angle A = 144^\circ$ or $72^\circ$.

The 1st root leads to $\angle B = \frac{240^\circ - \angle A}{2} = 48^\circ,$ $\angle C = \angle B - 60^\circ = -12^\circ$, hence this root is not acceptable. The 2nd root leads to $\angle B = 84^\circ,$ $\angle C = 24^\circ.$


Let $ a,b,c>0$.

Prove that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\geq \frac{6}{\sqrt{1\plus{}abc}}$

I think, for positive $ a,$ $ b$ and $ c$ the following inequality holds:
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {8}{a \plus{} b \plus{} c} \ge \frac {17}{3}\sqrt {\frac 3{1 \plus{} 2abc}}
\]

but $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {9}{a \plus{} b \plus{} c} \ge 6\sqrt {\frac 3{1 \plus{} 2abc}}$ is wrong already.

[quote="Vasc"]It suffices to show that
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {9}{2(a \plus{} b \plus{} c)} > \frac {6}{\sqrt {1 \plus{} abc}}$.

[/quote]

How to show this, can you hint?

Also, what about this similar-looking one?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173131[/url]

For $ a,b,c$ positive real numbers prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {9}{2(a \plus{} b \plus{} c)} \ge \frac 9{2}\sqrt {\frac 3{1 \plus{} 2abc}}$

Have you a nice proof , Vasc?

Thank you very much. :)

Can anyone please post solution to

$ a, b, c \geq 0$

Prove that

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\geq \frac{6}{\sqrt{1\plus{}abc}}$

I am very curious about this inequality.. Your help is much appreciated  :oops:

[quote="Vasc"]It suffices to show that
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {9}{2(a \plus{} b \plus{} c)} > \frac {6}{\sqrt {1 \plus{} abc}}$.

The following similar inequality holds with equality for $ a \equal{} b \equal{} c \equal{} 1$:

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac 1{c} \plus{} \frac {9}{2(a \plus{} b \plus{} c)} \ge \frac 9{2}\sqrt {\frac 3{1 \plus{} 2abc}}$.[/quote]
We can prove it by
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {9}{2(a \plus{} b \plus{} c)} \ge \frac{9}{2\sqrt [3]{abc}}
\]
:)


Starting at the origin, a beam of light hits a mirror (in the form of a line) at $ A (4, 8)$ and is reflected to the point $ B (8, 12)$. Compute the exact slope of the mirror.

[hide="Solution"]
Extend $ AB$ to some point $ B'$ on $ \vec{AB}$ such that $ OA \equal{} AB'$, forming an isosceles triangle. The slope of $ OB'$ is the slope of the mirror.

$ \vec{AB} \equal{} (4, \; 4)$ and $ \frac{\|OA\|}{\|AB\|} \equal{} \frac{\sqrt{10}}{2}$

Hence, $ \vec{AB'} \equal{} \frac{\sqrt{10}}{2} \vec{AB} \equal{} (2 \sqrt{10}, \; 2 \sqrt{10})$

Then, $ OB' \equal{} (2 \sqrt{10}\plus{}4, \; 2 \sqrt{10}\plus{}8)$, and it's slope is $ \frac{2 \sqrt{10}\plus{}8}{2 \sqrt{10}\plus{}4} \equal{} \frac{ \sqrt{10}\plus{}4}{ \sqrt{10}\plus{}2}$

We can go ahead and rationalize this to get $ \boxed{\frac{1}{3}(1 \plus{} \sqrt{10})}$
[/hide]

magnitude of the acute angle between the beam and the mirror and the image and the mirror is same.
$ \left| \frac {2\minus{}x}{1\plus{}2x} \right|\equal{}\left| \frac {1\minus{}x}{1\plus{}x} \right|$
You get 2 cases of which one yields nonsense and the other $ \frac 13(1\pm \sqrt {10})$ simple observation tells us that $ \frac 13(1\plus{} \sqrt {10})$ is the slope of the mirror while $ \frac 13(1\minus{} \sqrt {10})$ is the slope of the normal to the mirror.


Each lattice point of the plane is labeled by a positive real. Each of these numbers is the arithmetic mean of its four neighbors. Show that all the labels are equal, and also give a counter example of this problem in the case positive real is replaced by real.

Posted before and rediscussed in the old thread recently.

http://www.mathlinks.ro/Forum/viewtopic.php?t=5327
(The first post in that thread contains only a particular case, but later on the general version is solved.)


Construct a triangle ABC whose lengths of heights $ h_a$ and $ h_b$
(from A and B, respectively) and length of median $ m_a$ (from A) are given.

IMO 1960/4

http://www.mathlinks.ro/viewtopic.php?t=54830


x,y,z is positive,and \sqrt { \frac {1-x}{yz}}+\sqrt { \frac {1-y}{zx}}+\sqrt { \frac {1-z}{xy}}=2
find the max of xyz.[/quote]

[quote="duj12358"]$ x,y,z$ is positive,and $ \sqrt {\frac {1 \minus{} x}{yz}} \plus{} \sqrt {\frac {1 \minus{} y}{zx}} \plus{} \sqrt {\frac {1 \minus{} z}{xy}} \equal{} 2$
find the max of $ xyz.$[/quote]
$ 2\sqrt {xyz} \equal{} \sum_{cyc}\sqrt {x \minus{} x^2}\leq\sum_{cyc}\sqrt {\frac {3}{8} \minus{} \frac {x^2}{3}}\leq\sqrt {\frac {27}{8} \minus{} (x^2 \plus{} y^2 \plus{} z^2)}\leq$
$ \leq\sqrt {\frac {27}{8} \minus{} 3\sqrt [3]{x^2y^2z^2}}.$ It gives $ xyz\leq\frac {27}{64}.$
But for $ x \equal{} y \equal{} z \equal{} \frac {3}{4}$ we obtain $ xyz \equal{} \frac {27}{64}.$
Id est, $ \max xyz \equal{} \frac {27}{64}.$


A function $f$ satisfy the following conditions:

i) For each rational number $x$, $f(x)$ is a real number

ii) $f(1988) \neq f(1987)$

iii) $f(x+y)=f(x)f(y)-f(xy)+1$ holds for all rational numbers $x$ and $y$

Show that $f(\frac{-1987}{1988})=\frac{1}{1988}$.

Well you divided by $1-f(0)$ in one of those steps, so by contradicting the second condition you determine that $f(0) = 1$ but not $f(x) = 1$ for all $x \in \mathbb{Q}$.

[hide="Solution"]If in (iii) we let $y = 0$, then
$f(x) = f(x)f(0)-f(0)+1$
$(f(x)-1)(f(0)-1) = 0$
So either $f(x) = 1$ or $f(0) = 1$. The former contradicts (ii), so $f(0) = 1$.

If in (iii) we let $y =-x$, then
$f(x)f(-x) = f(-x^{2})$
Suppose $x = \pm 1$. Then
$f(1)f(-1) = f(-1)$
Then we have either $f(-1) = 0$ or $f(1) = 1$.

But if $f(1) = 1$, then plugging in $y = 1$ to (iii) shows
$f(x+1) = f(x)-f(x)+1 = 1$.
A quick induction shows $f(n) = 1$ for all integers $n$, again contradicting (ii). Therefore $f(-1) = 0$.

Now by plugging $y = 1$ into (iii),
$f(x+1) = f(x)f(1)-f(x)+1$.
Then $x =-2$ gives $0 = f(-2)f(1)-f(-2)+1$.
But plugging $x = y =-1$ into (iii) also gives
$f(-2) = 1-f(1)$.

Substituting that equation into the previous one, we get $0 =-f(1)f(1)+f(1)-1+f(1)+1$, from which we eventually get $f(1)(f(1)-2) = 0$. Then either $f(1) = 2$ or $f(1) = 0$.

Suppose $f(1) = 0$. Then $f(x+1) =-f(x)+1$. So $f(x+2) =-f(x+1)+1 =-(-f(x)+1)+1) = f(x)$. Since $f(0) = 1$, we get $f(m) = 1$ for all even integers $m$.
Now let $x = n$ and $y = \frac{1}{n}$ for some even integer $n$. Plug that into (iii) to get $f(n+\frac{1}{n}) = f(\frac{1}{n})-f(1)+1$. Because $f(n+\frac{1}{n}) = f(\frac{1}{n})$, we get $0 = 1$. Contradiction. So $f(1) \neq 0$.

This leaves $f(1) = 2$. Then plugging $y = 1$ into (iii) gets
$f(x+1) = f(x)+1$.
Induction on this starting with $f(0)$ going in both positive and negative directions gets both $f(n) = n+1$ and $f(x+n) = f(x)+n$ for all integers $n$.

Now plug $x = \frac{a}{b}$ and $y = b$ for integers $a,b$ into (iii) to get

$f(\frac{a}{b}+b) = f(\frac{a}{b})f(b)-f(a)+1$

$f(\frac{a}{b})+b = (b+1) \cdot f(\frac{a}{b})-f(a)+1$

$f(\frac{a}{b}) \cdot b = a+b$

$f(\frac{a}{b}) = \frac{a}{b}+1$

As $\frac{a}{b}$ encompassed the entire domain, we have found the solution for $f$. This definition quickly solves the problem.[/hide]

Very nice solution!  Two minor comments, which have nothing to do with the validity of the solution:

[quote="MellowMelon"]If in (iii) we let $y = 0$, then
$f(x) = f(x)f(0)-f(0)+1$
$(f(x)-1)(f(0)-1) = 0$
So either $f(x) = 1$ or $f(0) = 1$. The former contradicts (ii), so $f(0) = 1$.[/quote] Why not just put $x = y = 0$?  One fewer sentence to write.

[quote="MellowMelon"]But if $f(1) = 1$, then plugging in $y = 1$ to (iii) shows
$f(x+1) = f(x)-f(x)+1 = 1$.
A quick induction shows $f(n) = 1$ for all integers $n$, again contradicting (ii). Therefore $f(-1) = 0$.[/quote] Induction?  Since every rational is one less than some other rational, this gives directly that $f(x) = 1$ for all rationals $x$, no induction needed.

Haha, you can tell I'm not the most experienced at this. I always have inefficiencies like this.  :blush: Thank you for your comments.

$f(x+1) = 1$ requires induction... not going to forgive myself for that one for awhile.  :rotfl:


limit
real analysis
real analysis unsolved

Sequence $ (x_{n})$ is determined: for all $ k, n\in N$    $ x_{n}\geqslant 0, \ \ \ \ \ x_{n\plus{}k}\leqslant x_{n}\plus{}x_{k}$.
Show that $ \lim_{n\to\infty}\frac{x_{n}}{n^{2}}\equal{}0$.

fix a random natural number r,

$ \lim_{n\to \infty} \frac {x_n}{n^2} \equal{} \lim_{rn\to \infty} \frac {x_{rn}}{(rn)^2}$

$ \le \lim_{rn\to \infty} \frac {rx_{n}}{r^2n^2} \equal{} \frac {1}{r} \lim_{n\to \infty}\frac {x_n}{n^2}$

implies the limit is zero.. or infinity.. doh

I do not understand your 'solution' (and probably not only me)... Any other ideas?

$ x_{n\plus{}1}\leq x_n\plus{}x_1\Rightarrow x_{n}\leq nx_1\Rightarrow \lim_{n\to \infty}\frac{x_n}{n^2}\leq \lim_{n\to \infty}\frac{x_1}{n}\equal{}0$.

[quote="Ununuquantium"]I do not understand your 'solution' (and probably not only me)... [/quote]
No need to be so negative.  Repeated application of the first condition gives $ x_{rn} \le \underbrace{x_n \plus{} x_n \plus{} ... \plus{} x_n}_{\text{r times}}$.  So if the limit $ L$ exists, then $ L \equal{} \frac{L}{r}$ for any positive integer $ r$, which is true if and only if $ L \equal{} 0$.  But JoeBlow's solution also allows you to show that the limit exists.


conics
parabola
geometry
rectangle
power of a point
radical axis
geometry proposed

Let $\Gamma_{1}$ be a circle. $AB$ is a diameter, $\ell$ is the tangent at $B$, and $M$ is a point on $\Gamma_{1}$ other than $A$. $\Gamma_{2}$ is a circle tangent to $\ell$, and also to $\Gamma_{1}$ at $M$.

a) Determine the point of tangency $P$ of $\ell$ and $\Gamma_{2}$ and find the locus of the center of $\Gamma_{2}$ as $M$ varies.

b) Show that there exists a circle that is always orthogonal to $\Gamma_{2}$, regardless of the position of $M$.

Let $O_{1},\ O_{2}$ be centers and $r_{1},\ r_{2}$ radii of  $\Gamma_{1},\ \Gamma_{2}.$ Tangent r of $\Gamma_{1}$ at M is the radical axis of $\Gamma_{1},\ \Gamma_{2},$ it cuts BP in half at $N \in l$, BP = 2BN. NB = NP = NM, hence $BM \perp PM,$ but AB is a diameter of $\Gamma_{1}$, $BM \perp AM$ and A, M, P are collinear, AM cuts $l$ at P. Let $d \parallel l$ cut AB at D, different from $O_{1}$, such that $BD = BO_{1}$ and let $O_{2}P \perp l$ cut d at Q. $O_{1}O_{2}= r_{1}+r_{2}= QO_{2},$ $O_{2}$ lies on a parabola $\mathcal P$ with focus $O_{1},$ vertex B and directrix d. Inversion in a circle (A) with radius AB carries $\Gamma_{1}$ to $l,$ $l$ to $\Gamma_{1}$ and $\Gamma_{2}$ to a circle tangent to $\Gamma_{1},\ l$ at M, P, i.e., into itself, hence $\Gamma_{2}\perp (A)$ for any M.

Two circles are orthogonal iff the triangle formed by their centers $ O_{3}$, $ O_{2}$, and either of their points of intersection $ T$ is rectangle at $ T$, or iff $ O_{3}O_{2}^{2}\equal{} r_{3}^{2}\plus{}r_{2}^{2}$, where $ O_{3}$, $ O_{2}$ are the centers of the circles and $ r_{3}$, $ r_{2}$ their respective radii.  We just need to show that taking $ O_{3}\equal{} A$, $ r_{3}\equal{} 2r_{1}$, then $ AO_{2}^{2}\equal{} r_{2}^{2}\plus{}4r_{1}^{2}$, or since $ AO_{2}^{2}\equal{} BP^{2}\plus{}(r_{2}\minus{}2r_{1})^{2}$, we just need to prove that $ BP^{2}\equal{} 4r_{1}r_{2}$.  But this is easy, since $ (r_{1}\plus{}r_{2})^{2}\equal{} O_{1}O_{2}^{2}\equal{} BP^{2}\plus{}(r_{2}\minus{}r_{1})^{2}$, and we are done.


$\displaystyle (\frac{1}{2}x+1)^{23}=0
\Leftrightarrow (\frac{1}{2}x+1)=0
\Leftrightarrow x=-2$

[hide]
Not sure how to use logarithms, maybe if it equaled another number?

anyway:
$\displaystyle (\frac{x}{2}+1)^{23})=0\\\frac{x}{2}+1=0\\frac{x}{2}=-1\\x=-2$

[/hide]

[quote="guile"]sorry, i know this is a dumb question but i guess i dont have the time to review my logarithm.

problem:

 (1/2 x + 1)^23 = 0

please kindly explain, im not  a math major.[/quote]

Logarithms aren't necessary, because it $=0$.

[hide]
Take the 23rd root of both sides:
$\sqrt[23]{(\frac{x}{2}+1)^{23}}=\sqrt[23]{0}$
Simplify:
$\frac{x}{2}+1=0$
Now, as a linear equation, the rest of the solution is trivial.
$\frac{x}{2}=-1 \iff x=-2$
[/hide]

Yeah, but how would you use logs? I just kinda did it in my head because I knew that since 0 is the only number that is raised to a power would be equal to 0. So -2 makes the number being raised to the 23rd power 0, so I just kinda did it like that. No real technique though  :huh:  :surrender:

You can't apply logarithms to this question.  Your reasoning, Iversonfan, is exactly correct: there is no number other than 0 which when raised to the 23rd power equals 0, so we must have the base equal to 0 and from there we can find x.

Here's a question where you can apply logarithms.

$8^{2x-5}=5^{x-5}$

[quote="guile"]sorry, i know this is a dumb question but i guess i dont have the time to review my logarithm.

problem:

 (1/2 x + 1)^23 = 0

please kindly explain, im not  a math major.[/quote]
Take the 23rd root of each side

$\sqrt[23]{(\frac{1}{2}x+1)^23}=\sqrt[23]{0}$

that simplifies to:

$\frac{1}{2}x+1=0$

$\frac{1}{2}x=-1$

$x=-2$

i'm not sure how you would do this with logarithms

That's what I get. You have to use a calculator from there.

Here's one that is much simpler.

$11^x = 554$
Calculators are neccesary.

Yes, I believe you can also take natural logrithms to solve this too, right?
the answer still remains the same.


Find all pair of numbers (x, y) that satifies equation:
(3+2cos(x-y))=squart(3+2x-x*x).(1+cos(x-y))+(1-cos(x-y))

[quote="hien"]Find all pair of numbers (x, y) that satifies equation:
$(3+2cos(x-y))=squart(3+2x-x*x).(1+cos(x-y))+(1-sin(x-y))$


Let $ a,$ $ b$ and $ c$ are positive numbers, such that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}3.$ Prove that:
\[ \sqrt[3]{\frac{a^3\plus{}b^3}{2}}\plus{}\sqrt[3]{\frac{b^3\plus{}c^3}{2}}\plus{}\sqrt[3]{\frac{c^3\plus{}a^3}{2}}\leq2(a\plus{}b\plus{}c)\minus{}3\]

does anybody know how to prove this. i got it down to an inequality, to which i'm pretty sure can be solved easily, but it wasn't pretty to get there and i don't know latex (well i do, but i forgot all the commands)

The problem is obvious. :wink:
Here is my solution:(I'm sure not the best)
At first i'll prove that:
$ \sqrt[3]{\frac{a^3\plus{}b^3}{2}}\leq\frac{a^2\plus{}b^2}{a\plus{}b}$.
Which is equivalent to:
$ (a^3\plus{}b^3)(a\plus{}b)^3\leq 2(a^2\plus{}b^2)^3$.
But the last one is equivalent to:
$ (a\minus{}b)^4(a^2\plus{}ab\plus{}b^2)\geq 0$.
So we have to prove that:
$ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{a^2\plus{}c^2}{a\plus{}c}\leq 2(a\plus{}b\plus{}c)\minus{}3$.Or
$ \frac{2ab}{a\plus{}b}\plus{}\frac{2bc}{b\plus{}c}\plus{}\frac{2ac}{a\plus{}c}\geq 3$.Or
$ \frac{2}{\frac{1}{a}\plus{}\frac{1}{b}}\plus{}\frac{2}{\frac{1}{c}\plus{}\frac{1}{b}}\plus{}\frac{2}{\frac{1}{a}\plus{}\frac{1}{c}}\geq 3$,
now applying CBS inequality:
$ \frac{2}{\frac{1}{a}\plus{}\frac{1}{b}}\plus{}\frac{2}{\frac{1}{c}\plus{}\frac{1}{b}}\plus{}\frac{2}{\frac{1}{a}\plus{}\frac{1}{c}}\geq \frac{(3\sqrt{2})^2}{\frac{2}{a}\plus{}\frac{2}{b}\plus{}\frac{2}{c}}\equal{}3$.
Thus we are done.

\sqrt[3]{\frac{a^3+b^3}{2}}\leq\frac{a^2+b^2}{a+b}
that was what i was missing i was trying to find something suitable but never thought of that. thanks. and that cyclic inequality i reduced it to, it came out with murihead, so no hassle my was worked as well  :D thanks tho, urs is sleeker  :ninja:

I guess,that you would like to use LaTeX,to use it,you must just write dollar signs and only between them write,something like \frac ,\sqrt[3], e.t.c
Thank you

ok i did latex for a week , but after i didn't use it for 8 months i totally forgot evrything :blush:


darkquantum wrote:If x^2+y^2=1, and a^2+b^2=1, find the maximum value of ax+by

[hide] let x=sin p, y=cos p, a=sin q, b=cos q (you stole all the good variables already!). Well now, we want to maximize cos p cos q+sin p sin q=cos(p-q). The maximum is 1.[/hide]

Wow, MysticTerminator, I'm impressed with your use of trig substitutions. That's nice.

Anyway, a different method:

[hide]By Cauchy-Schwarz

(ax+by)^2 :le: (a^2+b^2)(x^2+y^2).

Thus, ax+by :le: 1. [/hide]

His solution is elegant as hell but you could also do 2 = x^2 + y^2 + a^2 + b^2 (addition) >= 2ax + 2by (straight AM-GM), hence 1 >= ax + by (magical technique of the ancient gods). You also need to prove that ax + by can achieve the value 1, but this is trivial because you may just let a = x = 1, b = y = 0.

ya....coincidentally enough, i though of it because of the last problem. Interesting....

[quote="teratomato"]I forgot to mention that this isn't number theory by any stretch of the imagination.[/quote]
Indeed.


inequalities
combinatorics unsolved
combinatorics

In an exam with k questions, n students are taking part. A student fails the exam
if he answers correctly less than half of all questions. Call a question easy if more than half of all students answer it correctly. For which pairs (k, n) of positive integers is it possible that
(a) all students fail the exam although all questions are easy;
(b) no student fails the exam although no question is easy?

There are a total of $ nk$ student-question pairs, and call such a pair "good" if the student solves the question.  If every question is easy, then (counting over questions) more than $ \frac{n}{2} \cdot k$ student-question pairs are good, while if every student fails then (counting over students) fewer than $ n \cdot \frac{k}{2}$ student-question pairs are good.  So this situation is impossible.  When we replace strict inequalities with weak inequalities in part (b), a similar analysis shows that we must have that every question is solved by exactly half the students and every student solves exactly half the questions.  This is possible whenever $ k, n$ are both even: e.g., have the first half of the students solve the first half of the questions and the second half of the students solve the second half of the questions.

Question related to part (b): how many (labeled, or up to isomorphism) bipartite graphs are there with partition sizes $ k, n$ even integers such that every vertex on each side is connected to exactly half the vertices on the other?


Where can I find old physics olympiad (USAPhO, IPhO, etc.)? Thanks!

You can see here from 1967.
[url]http://www.jyu.fi/kastdk/olympiads/[/url]


abstract algebra
geometry
Ring Theory
superior algebra
superior algebra unsolved

I have a problem understanding the following  statement:

"Let[b] f:A -> B[/b] be a flat ring homomorphism. [b]q[/b] is a prime ideal of B, and [b]p = q^c[/b] (contraction of [b]q[/b] in A). ....[b]Bp[/b] is a flat [b]Ap[/b] module (this is clear)...
Since [b]f(A-p)[/b] is mult. closed and does not meet [b]q [/b]we see that [b]Bq[/b] is a localization of [b]Bp[/b]".

The last statement is not clear....localization of Bp with respect to what prime ideal of Bp??

$ B_q$ is the localization of $ B_p$ with respect to the multiplicative set $ f(A\minus{}p)$.

thanks. ...I thought that localization always means localization of some ring [b]A [/b]with respect to a prime ideal [b]p[/b].

Anyhow, ho do we consider a ring of fractions of [b]Bp [/b]with respect to [b]f(A-p).[/b]

As I understand, [b]Bp [/b]is an [b]Ap [/b]module. It also has a ring structure since B is a ring. Now, [b]f(A-p)[/b] is a multiplicative subset of [b]B[/b].

How do we construct a localization of [b]Bp [/b]with respect to [b]f(A-p) [/b]and why is it the same as [b]Bq[/b]?

sorry, I need to make sure I understand this before moving any further  :oops:

use the yoneda-lemma and the universal property of localizations to show that $ f(A\minus{}p)^{\minus{}1} R_p \cong B_q$ are isomorphic as rings. note that $ B_q$ is the localization of the ring $ B$ at the prime ideal $ q$, whereas $ B_p$ is the localization of the $ A$-module $ B$ at the prime ideal $ p$.

thanks. I think I understand it better now.

Can I ask you what book you would recommend alongside with Atiyah - MacDonald?
I am self-studying this book, but find it a bit dry. even though the set of exercises is great.

I think Sharp's "Steps in commutative algebra" is easier to read but A-M contains more information. Eisenbud's "Commutative algebra with a view towards algebraic geometry" might also by useful.

thanks. I've just bought Eisenbud's book.


$ a,b,c,d$ be the real numbers such that

$ (1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)(1\plus{}d^2) \equal{} 16$

Prove that

$ \minus{}3 \leq ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd \minus{} abcd \leq 5$

[quote="arqady"][quote="colorfuldreams"]$ a,b,c,d$ be the real numbers such that

$ (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2)(1 \plus{} d^2) \equal{} 16$

Prove that

$ \minus{} 3 \leq ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \minus{} abcd \leq 5$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78246[/quote]

thanks :lol:


In acute $ \triangle{ABC}$ denote by $ M$ and $ N$ the midpoints of the altitudes $ BB_{1}$ and $ CC_{1}$ respectively,$ P \equal{} AM \cap CC_{1}$ and $ Q \equal{} AN \cap BB_{1}$.If the points $ B,C,P$ and $ Q$ are concyclic then prove that $ \triangle{ABC}$ is isosceles

What is isoceles? Could you mean please?

Sorry :blush: ,It must be [b]isosceles[/b]


[quote="jnApp66"]When I checked the anwers for the AMC 12b online, my score was 6 points higher than what my actual score that my teacher told me was, and I'm 99.99% sure I bubbled everything right because I was extremely careful.  This happened to someone else at my school too and I've seen a couple other complains about it on this forum.  Is there a chance that there could have been a grading mistake? :mad:  :mad:  :mad:  :mad:  :mad:[/quote]
You can request a hand rescoring if you are really that sure.
It'll cost you like 10 bucks I think.

[quote="Leao"]Maybe its like they said, they curved the AMC 12 scores...Btw, where else did you see the complaints on this forum?[/quote]Check the "Did you make the AIME thread".

Nobody curved the AMC scores. To make a convincing case that the test was scored other than how you expected, show, problem by problem, what answer you recall marking for each problem; how that would be scored as a correct answer, wrong answer, or omission; and how that totals up under this year's scoring rules for a total score.

But remember that the AMC computer is the authoritative "reader" of your answer sheet, unless you request hand scoring. Maybe what you remember doing and what is actually marked on your answer sheet is different. Experienced coaches have seen that happen before.

The most likely cause is one of the following:
On your answer sheet
(1) you did not darken the circle, i.e. writing too lightly with a too firm pencil
(2) you did not erase completely, leaving what the scanner believed to be two marks in a single row.
(3) you did not darken the circle completely, leaving a small hole in the middle
(4) you erased too vigorously leaving a rough spot on the paper which is mis-read by the scanner.

You may certainly request a rescoring, the request form is in the Teachers' Manual, about page 12.  It does involve a fee, since it means one of the staff has to locate your individual paper out of approximately 250,000, then hand score it, and make the appropriate changes if necessary in the database.  It takes the combined effort of about 3 of our staff, and usually takes about an hour.  We do not charge enough to recover the costs of the re-scoring.

I usually am called upon to additionally look at the paper and verify the judgment of the staff person who hand-scores these special requests.  In all the cases I have seen, the scanner has never mis-read the answer sheet if it was marked and/or  erased properly.  In many cases, it takes some head-scratching and holding the paper up to a bright light to properly determine what was intended.

I am not saying that the scanner did not mis-read your paper.  It is possible, but other causes are more likely.  If you still have dounbts, submit the re-scoring request.

Steve Dunbar
Director, American Mathematics Competitions

yeah, i probably did not erase cleanly enough, so i got 132 instead of 138 on the 10A.


Hey,
A singular 1-simplex $ \sigma$ on a topological space $ X$ is called a loop if $ \sigma(e_0)\equal{}\sigma(e_1)$.
Two loops $ \sigma_0$ and $ \sigma_1$ are called freely homotopic if there is a continuous map $ F: [0,1]*[0,1] \rightarrow X$ s.t
$ F(0,t)\equal{}\sigma_{0}((1\minus{}t)e_0\plus{}te_1)$ and $ F(1,t)\equal{}\sigma_{1}((1\minus{}t)e_0\plus{}te_1)$. Now the question is to prove that two freely homotopic lops are homologous, i.e they differ by an element in the image of $ d_{2}$ which goes from $ C_2(X)$ to $ C_1(X)$.

Does any one have a short/elegant proof for this?

The elegant way is perhaps to use the homotopy-axiom:

Interpret the homotopy $ F$ as map $ h: S^1\times I \to X$. Then let $ \sigma: \Delta_1\to S^1, (t,1\minus{}t)\mapsto e^{2\pi i t}$. $ \sigma$ is a 1-cycle and $ \sigma_t\equal{}f_t\circ\sigma$, i.e. $ [\sigma_t]_{H_1(X)}\equal{}(h_t)_\ast([\sigma]_{H_1(S^1)})$. The homotopy-axiom implies $ (h_0)_\ast\equal{}(h_1)_\ast \implies [\sigma_0]_{H^1(X)}\equal{}[\sigma_1]_{H_1(X)}$.

It is also possible to write down explicitely a 2-simplex $ \Sigma$ with $ d\Sigma\equal{}\sigma_0\minus{}\sigma_1$. I just "draw" a picture:
[code]
 -----<-----
|         / |
|  +    /   |
v     /     v
|   /   -   |
| /         |
 -----<-----
[/code]
That's an (affine) 2-simplex on the unit-square. Define a 2-simplex on $ X$ with $ F$. Check that the boundary is the right one. (Maybe I messed up the signs so that $ \sigma_1\minus{}\sigma_0$ is the boundary)

Ok thanks, I have never heard of the axiom. I will have a look at it.

by the way,  suppose we have a topological pair $ (X,Y)$ and let $ Z$ be topological space. Then $ H_k ((X,Y)xZ)\equal{}H_k ((XxZ,(YxZ))$ so there is some distribution between the comma and 'x' I would like to know why this is so.


How to solve this problem:

Prove that there is no $ n$ from $ N$ such that: $ 2n^2 \plus{} 1$ ,$ 3n^2 \plus{}1$, $ 6n^2 \plus{} 1$ ale all perfect squares simultaneously.

See http://www.mathlinks.ro/viewtopic.php?p=150539#150539 .
For a more general version, see http://www.mathlinks.ro/viewtopic.php?t=156645 .


In a certain company 20% of the men and 40% of women attended the annual company picnic,If 35% of all the employess are men,what % of al the employee went to the picnic?
cheers,
Rosy. :)
---------------------------------
[url]http://www.esumz.com/[/url]

Let's say that here $100$ people in the company. That means there are $35$ men and $65$ women. So $35 \cdot \frac{1}{5}=7$ men went to the picnic and $65 \cdot \frac{2}{5}=26$ So the percentage is: $\frac{26+7}{100}=\frac{33}{100}$ This becomes 33%

The other way is:
\begin{eqnarray*}\left(\frac{7}{20}\cdot\frac{20}{100}\right)+\left(\frac{26}{40}\cdot\frac{40}{100}\right)&=&\\ \frac{7}{100}+\frac{26}{100}&=&\\ &=&\frac{33}{100}\end{eqnarray*}
This is the same as 33%

[hide=Post #4 by zandrabhernandez][url=aops.com/community/user/779998][b]zandrabhernandez[/b][/url] · 15 minutes ago [url=aops.com/community/p21511569](view)[/url][color=transparent]helo[/color]
I magine men getting paid for changing light bulbs, putting furniture together, carrying heavy grossery bags...now that would be a salary :

-----------
[color=#5b7083][aops]x[/aops] 4[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@zandrabhernandez][img]https://avatar.artofproblemsolving.com/avatar_779998.png[/img]
AoPS User[/tip] Please don't bump 13 year old threads. :)


Find the area of the region bounded by the graph of the equation $|x-4|+|y+3|=3$.

[i]No Calculators
Solutions Necessary[/i]

[hide]
If you just draw the equation, you will see that the region is a 45-45-90 right triangle.
The three vertices are $(4,0)$, $(1,-3)$, and $(7,-3)$.
The legs are equal to $3\sqrt{2}$.
$\frac{(3\sqrt{2})^{2}}{2}=\boxed{9}$[/hide]

Consider the cases when x+4 and y-3 or whatever it is are positive and negative. I don't feel like actually doing it.


Find positive integers $ n$ with exactly 12 divisors and $ d_i  < d_{i\plus{}1}$ where $ i \equal{} 1,2,3..............11$
such that $ d_{d_4 \minus{} 1} \equal{} (d_1 \plus{} d_2 \plus{} d_4 )d_8$

[hide="Hint"]Prove that $ d_8 (d_4 \plus{} d_2 \plus{} d_1 ) \equal{} d_{12}$[/hide]


calculus
integration
function
derivative
calculus computations

Hello Everybody!

I'm studying Taylor Expansions (Lagrange/Cauchy form of the remainder etc.) and I have this exercise which I can't seem to make a progress on. I'll be grateful for a hint, if anybody can give me one (just a hint, since I'd like to struggle with it a bit more before completely giving up)  :) .

Let $ f$ be a differentiable function on $ [0,1]$ with a bounded derivative $ |f'(x)| \leq 8$. Prove that if $ \int_{0}^{1}f(x)dx \equal{} 0$ then $ \left|\int_{0}^{t}f(x)dx \right| \leq 1$ for all $ t \in [0,1]$.

So far I've only managed to prove that there's a $ c \in (0,1)$ such that $ f(c)\equal{}0$ and then $ |F(t) \minus{} F(c)| \leq 1$   (where $ F(t) \equal{} \int_{0}^{t}f(x)dx$). I'm not sure whether it helps at all, however  :maybe:.

Thanks for reading!
Perrin

The problem is better restated as follows. [i]Given that $ F(0)\equal{}F(1)\equal{}0$, $ |F''|\le 8$, prove that $ |F|\le 1$.[/i]. Now just take the point $ a$ where $ F$ attains its  maximum (or minimum) and evaluate $ F(0)$ if $ a\le\frac 12$ and $ F(1)$ if $ a>\frac 12$ using the Taylor formula at $ a$ with remainder in the Lagrange form.

One can also avoid the use of Taylor expansion (possibly defeating the purpose of the exercise). Indeed, let $ G_{\epsilon}(x)\equal{}F(x)\minus{}(4\plus{}\epsilon)x(1\minus{}x)$ where $ \epsilon>0$. This function cannot have a point of maximum in $ (0,1)$ because $ G_{\epsilon}''>0$. Since its values at the endpoints are $ 0$, we have $ G_{\epsilon}(x)\le 0$ for all $ x\in [0,1]$. And so on.

This feels like an old Putnam... I remember doing this one at some point, I think.

Maybe I'm completely off base. I'll go look through the database.


In triangle $ ABC$, $ \angle ABC \equal{} 120^{\circ}$. Let the angle bisectors of angles $ A, B$ and $ C$ meet the opposite sides in $ D, E$ and $ F$ respectively.

Prove that the circle on diameter $ EF$ passes through $ D$.

It is an old problem ,see  http://www.mathlinks.ro/viewtopic.php?search_id=427092691&t=114998


integration
calculus
calculus computations

for $ \alpha>0$ show that

$ e^{\alpha} \int_{2\alpha}^\infty u^{\alpha\minus{}1}e^{\minus{}u} du \le 2^{\alpha} \Gamma(\alpha)$

just so you dont have to download.


AMC
AIME
geometry
quadratics
analytic geometry
geometric transformation
angle bisector

Here are the answers:

1.051
2.958
3.458
4.XXX -- anything goes. free point.
5.120
6.401
7.531
8.513
9.821
10.778
11.000
12.314
13.169
14.500
15.764

Go ahead and vote for your scores (so everyone can see the distribution). If anything is wrong, please just post it in this thread. The solutions are attached. Feel free to discuss the problems here as well.

EDIT -- reworded #4 to make the lines concurrent. sorry about that. :blush:

[quote]By Ptolemy's, which you don't need at all[/quote]

[hide="My number 13"]
By $Area=\frac{1}{2}absin(C)$ we have $A(ABC)=\frac{1}{2}(28\sqrt{3})(42)(\frac{1}{2})=294\sqrt{3}$
$ADE$ is a 30-60-90 right triangle, so $AE=14\sqrt{3}$ and $AD=28$ so $DB=42-328=14$, so triangle $BED$ is isosceles, with obtuse angle $EDB=120^o$. The angle bisector of $EDB$ will divide The angle into two $60^o$ angles, and perpendicularly bisecting $EB$ at a point $F$, making angle $BED=30^o$. Then $EF=BF=7\sqrt{3}$. For consistency, use $A=\frac{1}{2}absin(C)$ for triangle $BED=\frac{1}{2}(14)(14\sqrt{3})(\frac{1}{2})=49\sqrt{3}$

Seeing that $BEC$ is the area of the whole triangle $ABC$ minus the areas of triangle $BED$ and $ADE$

To find the area of triangle $ADE$ (again, for consistency), use $A=\frac{1}{2}absin(C)$
$A(ADE)=\frac{1}{2}(14)(14\sqrt{3})(\frac{1}{2})=98\sqrt{3}$
So $A(BEC)=294\sqrt{3}-98\sqrt{3}-49\sqrt{3}=147\sqrt{3}$

Thus $A(BEC)-A(BED)=147\sqrt{3}-49\sqrt{3}=98\sqrt{3}$
=$98(1.73)$
=$\bold{169}$
[/hide]

[hide="My number 8"]

$4^{1/3}x+\frac{2^{1/3}}{x}=3$
$4^{1/3}x^2+2^{1/3}=3x$
$4^{1/3}x^2-3x+2^{1/3}$
By quadratic formula:
$x=2^{-2/3}$ or $x=2^{1/3}$

[hide="Checking"]
$4^{1/3}x+\frac{2^{1/3}}{x}=3$
$4^{1/3}(2^{1/3})+\frac{2^{1/3}}{2^{1/3}}=3$
$2^{2/3}(2^{1/3})+1=3$
$2+1=3$ (And it works the other way too)
[/hide]

So $64x^9+\frac{8}{x^9}$
$=64(2^{1/3})^9+\frac{8}{(2^{1/3})^9}$
$=64(2^3)+\frac{8}{2^3}$
$=64(8)+\frac{8}{8}$
$=512+1$
$=\bold{513}$
[/hide]

[quote="origamimasterjared"][quote="beta"]The solution for #4 is incorrect, it didn't say in the problem that the lines are concurrent[/quote]

What's not concurrent? Didn't the problem say extend AC to G?, thus AC, CG, and AG are all on a line, right?[/quote]

The problem didn't say AD, BE, CF all intersect at one point. To use Ceva's Theorem, AD, BE, CF have to intersect at one point

[quote="Rep123max"]And I am curious, on #3, do we have to take into account the color of the points for finding the total number of polygons formed?[/quote]
no, because the total number of polygons will account for every possible outcome, so wlog you can assume any distribution pattern of the colors.

I got an 11 (spent 2.5 hours...then had to do homework), so I can safely say that this was a relatively easy mock AIME. :P I solved 12 but overlooked some things on #10. I tried #4, #8, and #15 for a while and gave up. (but I guess I could have easily solved #4 had i been told that the lines were concurrent.)

Thank you very much for the AIME, paladin8, and good luck to all taking AIME this Tuesday.  :)

For 4, Ceva's theorem is for concurrent cevians. A cevian is a segment joining a vertex to the opposite side. If the cevians aren't concurrent, then there are (I think) infinitely many solutions. I just got lazy and assumed concurrent cevians, even though it's definitely unjustified.
My score is 9, but I made several arithmetic errors. I spent about 2 hours and didn't check my work. I got the basic concept for 12-13 of them. Now if I can only learn to add, subtract, multiply and divide.

Thanks for the warm up, Paladin.

Whoops. I guess I forgot to mention that they were concurrent >.< Whatever, everyone can give themselves that point  :P.

And also, for the #11 solution -- it should read tan(x) = -tan(180-x).

[quote="filletwho"]Is it assumed that you have to know root three?[/quote]

Pretty much. Or you have to approximate it well. That's also a skill, you know =P. And besides, most people do know it to at least two decimal places which is all you need.

[quote="no_one_2000"]I have a quick quesiton about #7.  Now, I'm probably wrong because I average 4-5 on these AIMEs, but...

The equation for the circle is x^2+y^2-14x+4y-523=0.  So I completed the square and got (x-7)^2+(y+2)^2=24^2.  The answer says it's (y-2)^2 instead of (y+2)^2.  Because 4y is positive, shouldn't the 2 in the squared binomial be positive also?  Sorry if this question is stupid, but I'm just a little confused... (btw, my resultant answer was 480pi+144sq(3), so I was close)[/quote]

Thank you. As you can see, I was writing this in a hurry. However, the answer remains the same, as the y-coordinate does not have anything to do with the translation to the origin. The revised version is now posted.

[quote="xxreddevilzxx"]nice test jeffrey. i got a 10.... (hey! i only spend an hour taking the test, which is really weird)

the test had a lot of long questions, but they were too easy to find ways to condense it into my own understanding... which helped me solve the problem.[/quote]

Yeah, I wrote this test so that every question had a relatively quick solution. There weren't any grueling counting questions like on the actual AIME -- even #15 isn't that bad when you look at the solution (after I edited it, at least).

this post is really late (i was getting ready for alternate AIME's), but in #4, there are infinitely many points G and H that fit the problem


Of the 60 students in the drama club, 36 take mathematics, 27 take physics and 20 students take both mathematics and physics. How many drama club students take neither mathematics nor physics?

By applying PIE, it is $ 60\minus{}(36\plus{}27\minus{}20)\equal{}60\minus{}43\equal{}\boxed{17}$

[hide=Using a Venn Diagram][img]https://i.ibb.co/6nxJdY0/Venn1.png[/img][/hide]


Let function $ f$ be defined over the nonnegative integers such that:

If $ x$ is odd then $ f(x) \equal{} \lfloor{\dfrac{x}{2}\rfloor}$.

If $ x$ is even then $ f(x) \equal{} \dfrac{x}{2} \plus{} 1$

What is the value of $ \dfrac{f^{2n \plus{} 1}(2^{2n \plus{} 1}\cdot n \plus{} 1)}{n}$?

Where $ f^n(x) \equal{} f^{n \minus{} 1}(f(x))$

[hide="Solution"]Note that if $ x$ is odd, $ f(x) \equal{} \frac{x \minus{} 1}{2}$. We can inductively show that for any integer $ k$, $ f^{2k \plus{} 1}(2^{2k \plus{} 1}n \plus{} 1) \equal{} n$. To do this, for the base case, note that $ f(2n \plus{} 1) \equal{} n$. For the inductive step, note that $ f^{2k \plus{} 3}(2^{2k \plus{} 3}n \plus{} 1) \equal{} f^{2k \plus{} 2}(2^{2k \plus{} 2}n) \equal{} f^{2k \plus{} 1}(2^{2k \plus{} 1}n \plus{} 1) \equal{} n$, by the inductive hypothesis. As a consequence, the numerator is simply $ n$, so our answer is just $ \boxed{1}$. [/hide]


Let (x,p)  be a compact metric space and f:X -> X is an isometry, which means p(f(x), f(y)) = p(x,y) for every x,y in X. Prove that f is bijective.

Often times, an isometry is defined to be a bijective continuous function with a continuous inverse.  To see injectivity, suppose that $ \rho(f(x),f(y))\equal{}\rho(f(w),f(z))$.  Then by definition, $ d(x,y)\equal{}d(w,z)$.  But by definition of metrics, these must be the same points.

[quote="JRav"]Often times, an isometry is defined to be a bijective continuous function with a continuous inverse. [/quote]
I would call that "homeomorphism" not "isometry".

So, injective has been done.  That leaves surjective.  Here, the compactness must be used.  For example $ f(x) \equal{} x\plus{}1$ maps $ [0,\infty)$ isometrically into itself, but is not surjective.

this problems was discussed here : [url]http://www.mathlinks.ro/viewtopic.php?t=300234[/url]


Let $F_{n}$ be the $n-$ th Fibonacci number (where $F_{1}= F_{2}= 1$). Consider the functions
$f_{n}(x)=\parallel . . . \parallel |x|-F_{n}|-F_{n-1}|-...-F_{2}|-F_{1}|, g_{n}(x)=| . . . \parallel x-1|-1|-...-1|$ ($F_{1}+...+F_{n}$ one’s).
Show that $f_{n}(x) = g_{n}(x)$ for every real number $x.$

[quote="Rust"]Obviosly $ f_{3}(x) \equal{}\parallel\parallel x|\minus{}2|\minus{}1|\minus{}1|\not \equal{}\parallel\parallel x\minus{}1|\minus{}1|\minus{}1|\minus{}1| \equal{} g_{3}(x)$.[/quote]
But why?  :maybe:

For example, consider $ f_{3}(\minus{}4)\equal{}0\not \equal{}2\equal{}g_{3}(\minus{}4)$.


modular arithmetic
invariant
AMC
USA(J)MO
USAMO
number theory
algebra

Well, as the deadline's past, I don't see why not.

Hm, I re-checked the website...Test B is still up. I guess people can submit it late for first come first serve decisions...Darn.

Nah, test A was up long after the application deadline was over. In fact, if I recall correctly, it was up until test B came up.

EDIT:

So I'll discuss the test now. In my opinion, this test was much, [i]much[/i] easier than admission test A. I answered every question on this test in less than an hour.

I'll only give my comments on the nontrivial problems.

[hide="4"]There are a number of ways to do this. I didn't like it, as it was way too easily bashable by a computer. On that note, we could just do a straightforward Chinese remainder theorem on it. [/hide]

[hide="5"]In my opinion, this problem was the hardest and the most interesting. (Unfortunately, the answer was on Wikipedia. XD) Note that in the magic square, the sum is always the same: $ 1 \plus{} 2 \plus{} \ldots \plus{} n^2 \equal{} \frac {n^2 (n^2 \plus{} 1)}{2}$. Let $ d$ be the sum of the terms in both diagonals.

(For example, if the square were
2 7 6
9 5 1
4 3 8
d would be 6 + 5 + 4 + 2 + 5 + 8 = 30.)

Let $ k$ be the sum of one row/diagonal/column. If we were to add up the sums of each term in each row and column and diagonal, each entry would be counted twice, except for the diagonals. Therefore, the sum obtained in this case would be $ n^2(n^2 \plus{} 1) \plus{} d$. However, this number is also $ (2n \plus{} 2)k$, since the sum in each row/column/diagonal is the same, and there are $ n$ rows, $ n$ columns, and 2 diagonals. Note also that $ d \equal{} 2k$ (since the sum of the numbers in the diagonals has to be twice the magic square constant.) Thus, $ n^2(n^2 \plus{} 1) \plus{} 2k \equal{} 2nk \plus{} 2k$, so $ k \equal{} \frac {n^3 \plus{} n}{2}$.
[/hide]

[hide="6"]Consider the factoring $ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1) \equal{} abc \minus{} ab \minus{} ac \minus{} bc \plus{} a \plus{} b \plus{} c \minus{} 1$. Use this fact and just manipulate the given equations. [/hide]

[hide="7"]We can angle chase to show that $ \angle ABX \equal{} \angle XCY \equal{} \angle YDA$; by SAS, they're all congruent, and since their bases are the sides of the triangle, we're done. [/hide]

[hide="8"]This was also a nice problem. Color the chessboard so that one column is entirely red, the next entirely blue, the next entirely red, etc. Suppose WLOG the camel starts on a red square. Note that after an even number of jumps, he lands on a red square, and after an odd number, he lands on a blue square. Since the square he starts no has to be red, he needs an even number of steps.
[/hide]

[hide="9"]At first glance, this problem looks quite hideous. Of course, if it really were hideous, it wouldn't be on this test. So I just went ahead and applied the general technique; I rationalized the demoniator and saw that I had to add the square roots of a bunch of terms. From here, we can kind of deduce that this has to telescope somehow. Writing it in the form $ \sqrt {a} \minus{} \sqrt {b}$ and solving for a and b, we see this does indeed telescope and arrive at the answer easily. [/hide]

[hide="10"]Note that after n moves, h coins have been discarded, and there are n coins remaining. Suppose it is possible for there to be n piles of 3 coins; then the number of coins discarded is also n. Thus, the total number of coins in the original pile must be 3n + n = 4n = 2009, a contradiction. [/hide]

Hope I didn't do anything stupid. :/

Actually, for number 5, can't you just do it like:

[hide]

The total sum is $ 1 \plus{} 2 \plus{} \cdots \plus{} n^2 \equal{} \frac {n^2(n^2 \plus{} 1)}{2}$. Since there are $ n$ rows (or columns) with the same sum, we divide by n to get the constant, or $ \frac {n(n^2 \plus{} 1)}{2}$

[/hide]\

Edit: darn, turns out I somehow misinterpreted number 10

I don't see how that's true. The total number of rows/columns/diagonals is 2n + 2, not n.

If that really does work, then d'oh. >_<

Sorry about not being clear, but we're only looking at rows. (or you can pick columns) Summing all the rows is equivalent to summing the whole thing.

Ah crap.  :wallbash_red:

Props to that solution though.

the magic square one seemed really easy for me. 1+2....n^2=n^2*(n^2+1)/2. n rows, the constant is n^2*(n^2+1)/2n= n^3+n/2. On that note, #2 didnt seem trivial to me. I did some wierd substitution and got 16. is that the only solution? also, i screwed up on #4, did a really weird and really slow way that used casework, analysis in mods, and multiplying three digit numbers. i still got it right, but,yeah..... don't ask.

[hide="2"]Note that if $ a$ and $ b$ are perfect squares, then $ \sqrt {a} \plus{} \sqrt {b}$ is not an integer. Therefore, we first need that $ n \equal{} k^2$ and $ n \plus{} 2009 \equal{} j^2$ for some integers $ j$ and $ k$; we now apply difference of squares and eliminate the extraneous solutions. [/hide]

My bad if I called that trivial. I've been working hard on my number theory recently. Heheh.

As for #5, yes, on hindsight, it is very easy. I simply did not recognize that solution that you and lifeisacircle posted.

By the way, I didn't actually get the answers to most of the problems (particularly the computation ones; I actually skipped #1. Incidentally, #1 is probably the one I'd most likely get wrong for some stupid reason.) I just reasoned it out and asserted that my solutions were correct. Hopefully they are.  :P

lol i spent like 30 minutes for number 1. I first tried to find a general formula to find the angle, but then gave up because i realized it would be really hard to explain. Then i split it up into 3 smaller angles and solved it.

[quote="AIME15"]The answer to number 5 was on Wikipedia?[/quote]

All I see though is an answer, without a solution

to tonypr:
nice solutions! i only have a few (minor) objections:
4) I did the same way as you. However, you have to explain how you got from 53, 03, 47, 97 with manipulations in mods. I actually made mod equations to explain them; otherwise, it might seem that you brute-forced with a calc.
5) If you looked at the discussion above, there is a smart and simple way to do it, because honestly, it seems like your solution is a brute-force look-for-a-pattern.
8)Ok solution, but you didn't really explain how a 0 has to go to 1 or 1 has to go to 0. I like my solution, it transformed the chessboard to a cartesian plane, with the camel starting at 0,0 (or (even, even)) and using a parity argument since the camel always moves +or-1 or 3 for x and y axis. It looks kinda messy written here, but its much better on paper. :)
10) Your explanation is a bit iffy, especially the "this works" part. i used a contradiction-mod analysis. basically, assume there are  n piles of 3, look at mod 4, when adding, the bigger pile is still 3 (mod 4). Since 2009 is not congruent to 3 (mod 4) it is not able to be split up into piles of 3.
Other than that, your solutions are really good. You solved more than me, so  :D

Thanks for your objections, I wish I could change my solutions, but I already sent it.

Would you think it's enough to get in?

probably. ive seen ppl solve 6 and make it. I solved 8, pretty good solutions, with 2 brute-forcy ones, but i havent gotten the admission results yet. pretty nervous :maybe:

I considered #5 as the hardest (but still easy...) and most interesting problem (but too well known...) among the 10 problems; this is what I did:

We divide the problem by 2 cases: 1) n=odd, 2) n=even.

1) For n=odd, there are three kinds of squares: One in the center, numbers in the diagonals besides center, and all the others.

The one in the center is used 3 times, numbers in the diagonals besides center are used 2 times, and others are used only 1 time.

Set up an equation, and solve.

2) For n=even, everything is same, except since there is no 'center square', it will be a little easier.

EDIT: I just saw lifeisacircle's solution...
Wow.

P.S. I solved every problems, and pretty sure I got them all right except #6... I thought the answer to #6 was $ \minus{} 1500$?

[quote="lukezli"]ermmm.... fantasylover, i dont know if youve been reading this thread, because a solution thay i and someone else posted is much more simpler that yours. honestly, this was (to me) much easier than many problems. I thought #9 was hard.[/quote]
Yes, I just read lifeisacircle's solution...

I thought all the problems were easy.

For problem ten, I worked
[hide]
backwards. Basically, combine piles of three. Combine two piles, add one chip. Combine three piles of three takes two steps, so two extra chips. It then follows that if you combine n piles of three, you add n-1 chips, resulting in a stack of 3n+(n-1), or 4n-1 chips. Test: is 2009 a number of form 4n-1?

4n-1=2009
4n=2010
n=502.5. This is not an integer, so a stack of 2009 cannot be built. It then follows that a stack of 2009 cannot be broken down.
[/hide]

I had the same solution as lifeisacircle.  Apparently it's the same as the Wikipedia solution.

http://en.wikipedia.org/wiki/Magic_constant

Ah snap, it looks like for #10, I undercounted by 1. :/ ak090909 said what I meant to say. And the mod argument still works. =D

[quote]All I see though is an answer, without a solution[/quote]
I meant the answer was on Wikipedia.

so yeah, these problems were really easy, i solved 9 of them (all but #6). i spent a total of around 4-5 hours on this.

Err, what was the answer to #6?

I got $ \minus{}1500$...

At first I got what you had, but what you have to do is, instead of factorizing $ 2008$ into $ 2\cdot 2\cdot 502$ you change it into $ (\minus{}2)\cdot 2\cdot (\minus{}502)$ and this gives different odd values for $ a,b,c$ which would be $ (\minus{}1,3,\minus{}501)$ instead of what you probably got $ (3,3,503)$ (If remember correctly, I might be wrong).

[hide="Number 6"]

Let $ a \equal{} 2a_0 \plus{} 1$, $ b \equal{} 2b_0 \plus{} 1$, and $ c \equal{} 2c_0 \plus{} 1$, with $ a_0, b_0, c_0$ being integers. Hence,

$ n \plus{} 2009 \equal{} 2(a_0 \plus{} b_0 \plus{} c_0) \plus{} 3$

and

$ n \plus{} (2a_0 \plus{} 1)(2b_0 \plus{} 1)(2c_0 \plus{} 1) \equal{} (2a_0 \plus{} 1)(2b_0 \plus{} 1) \plus{} (2b_0 \plus{} 1)(2c_0 \plus{} 1) \plus{} (2a_0 \plus{} 1)(2c_0 \plus{} 1)$

Expanding, simplifying, and factoring these two equations, we have:
\[ n \equal{} 2(a_0 \plus{} b_0 \plus{} c_0 \minus{} 1003)
\]
and
\[ n \equal{} 2(a_0 \plus{} b_0 \plus{} c_0 \plus{} 1 \minus{} 4a_0b_0c_0)
\]
Setting equal, $ 1 \minus{} 4a_0b_0c_0 \equal{} \minus{} 1003$, or $ a_0b_0c_0 \equal{} 251$, which is prime. Hence, $ a_0,b_0, c_0 \equal{} 1,1,251$, and $ a,b,c$ is some permuatation of $ 3,3,503$. Plugging this into $ n \plus{} 2009 \equal{} a \plus{} b \plus{} c$, we have $ \boxed{n \equal{} \minus{} 1500}$.

[/hide]

Did I do something wrong? I'm fairly certain I got the expanding correct, and I even checked the numbers.

Yeah, I realize my solution sort of sucks

The problem states that $ a,b$ and $ c$ have to be different.. so you can't have $ (3,3,503)$ as possible values for $ a,b$ and $ c$.

:wallbash_red: I didn't see it say distinct.

Also, what I did for number 8:

[hide]

Consider the camel going only up and down. The camel must go 3 or 1 steps up or down in one move. For the camel to return to its original position, it must have equal steps up and down. Let $ w$ be the number of 3 step jumps upward, $ x$ the number of 1 step jumps upward, $ y$ the number of 3 step jumps downard, and $ z$ t5he number of $ 1$ step jump downwards. We have $ 3w\plus{}x\equal{}3y\plus{}z$. Taking modulo 2, $ w\plus{}x\equiv y\plus{}z \pmod 2$. If $ w\plus{}x$ and $ y\plus{}z$ have the same parity, their sum must be even, and we are done.

[/hide]

My solution is different than what other people would have probably did, but I think it works.

dude #10 was trivial (unless i failed it  :maybe: ) the number of stones plus piles is invariant, so that invariant must be 3n + n = 4n. but we know it is 2009 + 1 = 2010. and no 4n = 2010 so it is impossible. legit.

Curses, I need to work on my chessboard style proofs. Why must lifeisacircle always somehow use mods and number theory stuff in his solutions? GAH.

Hmm? What heresy are you speaking of? Anyways, I prefer the chessboard technique more than mine, although mine does show an alternative way to do it.

I managed to botch 10. Two cheers for me!

lol i shouldve used mods in that chessboard question. Then i would of used it three times during the test!  :P But for my chessboard answer, i did sorta the same thing as lifeisacircle, but i usede a parity argument instead, which is i guess is the same as looking at it mod 2, because 0 (mod 2) is even and 1 (mod 2) is odd.

I really epic-phailed on this test.. I actually solved 7 and 1 partial proof. The ones I left blank, and you people who got into USAMO at age 12 or whatever are going to say, "Wow, this guy is stupid, why is he coming to AMSP?".. but... they are 7 and 9.. Solved rest with 9 as a kind of partial proofish with a suckish answer.. Hm..  :wallbash_red: I just feel stupid now -.-

dont feel too bad i only got 7 and 2 partial proofs

[quote="tonypr"]Well I don't know if I got in or not yet, so I'll just post the pdf I sent, here and hopefully you guys can tell me if I have enough to get in or not.[/quote]
I didn't read all of your solutions but one comment:
2. In your solution, how do you know that $ x,y$ are integers?

Hmm..

Well if they weren't then $ z$ would be the square root of the sum two positive irrational numbers, which would make $ z$ also an irrational number, and not an integer.


Let $ x_1, x_2, ..., x_5$ be a non-negative real numbers such that $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_5 \equal{} 100.$ Let $ M$ be a maximum of the numbers $ x_1 \plus{} x_2, x_2 \plus{} x_3, x_3 \plus{} x_4,$ and $ x_4 \plus{} x_5$. The least possible value of $ M$ lies in the interval

A. [0,32)
B. [32, 34)
C. [34, 36)
D. [36, 38)
E. [38, 40]

[hide]Let $ a_1 \equal{} x_1 \plus{} x_2, a_2 \equal{} x_2 \plus{} x_3, \ldots, a_5 \equal{} x_5 \plus{} x_1$. Then
\[ a_1 \plus{} \ldots \plus{} a_5 \equal{} 200 \implies \max(a_1,a_2,a_3,a_4)\ge \frac {200 \minus{} x_5 \minus{} x_1}{4}
\]
We want to minimize $ M \equal{} \max(a_1,a_2,a_3,a_4)$, that is, we want to maximize $ a_5 \equal{} x_1 \plus{} x_5$. However, we also know that $ M\ge x_1,x_2,x_3,x_4,x_5$. We can use a smoothing argument to show that it is optimal when $ x_2 \equal{} x_4 \equal{} 0, x_1 \equal{} x_5 \equal{} 50 \minus{} \dfrac{x_5 \plus{} x_1}{4} \implies x_1 \equal{} x_5 \equal{} \dfrac{100}{3}$. Only $ x_3$ is left, and since the sum of the $ x_i$ is 100, $ x_3\equal{}\dfrac{100}{3}$ as well.  (Note that this could be guessed fairly easily...) Therefore, the minimum lies in B.[/hide]

[hide="Official Solution by Creators"]

$ 3M \geq (x_1 \plus{} x_2) \plus{} (x_3 \plus{} x_4) \plus{} (x_4 \plus{} x_5) \equal{} 100 \plus{} x_4 \geq 100$

So, $ M \geq \frac{100}{3}$. From here, it can be easily shown that $ x_1 \equal{} x_3 \equal{} x_5 \equal{} \frac{100}{3}$ and $ x_2 \equal{} x_4 \equal{} 0$ to show that $ \frac{100}{3}$ is indeed a possible value. So, the answer is B [/hide]


Well, I searched the forum and couldn't find a thread on multivariable calculus textbooks. So does anyone have any recommendations? I have heard that Stewart's is ok for beginners, but are there any other good textbooks out there?

Thanks.

What is your level of experience with calculus so far, and what do you want this book for? That might help people give you recommendations.

There are thousands of Multivariable Calculus books - from the "doesn't really prove things rigorously - but here are the facts and how to use them" kind of textbooks (like high school calculus textbooks are). To the "here are the guts and details of the theory and this book is about proving things" kind of textbooks (as advanced college multivariable calculus courses use). And then there are lots of books in between - including books geared towards certain applications, like business or engineering.

On the former end (the high school style textbooks) I would say that Stewart or Larson are good choices. The first halves of the books are regular single variable calculus books, with the second halves being devoted to multivariable calculus.

On the latter end of the spectrum (the advanced college multivariable calculus style textbooks), I would say that Apostol's Calculus Vol. II is one of the best. The first half of the book (which is essential to read before you get into the second half) is all about Linear Algebra; and then the second half is the Multivariable Calculus. And when I say multivariable... I mean [i]really[/i] multivariable, not just into 2 and 3 dimensions as Stewart and Larson will be.

And then in between, there is Vector Calculus by Marsden and Tromba. This is the book that I first learned Multivariable Calculus from (after taking BC Calculus) and I would recommend this one. It gives you more explaination than the Stewart and Larson book, but not as much abstraction and technical detail as the Apostol book. In my opinion it is the best place to start for someone who hasn't had exposure to a hard core "Lemma, Theorem, Proof" book with no pictures what-so-ever, but who has a strong grasp (i.e. a 5) on BC Calculus material.

But again, it depends on where you are in your studies and what you want the book for.

Thanks for the suggestions gauss202. Well, I took BC Calculus last year and was introduced to some multivariable calculus (with applications to physics) this year. I will be studying engineering in college next year, so I'd like a textbook either for pure mathematics or a bit geared towards engineers. I will take a look at some of your suggestions.

Thanks again, and I see that you are still very much active on the forum (I haven't frequented the forum for nearly a year, so I mostly see a lot of new usernames now).

[quote="Moubinool"]What about Spivak ?[/quote]

Spivak's [u]Calculus[/u] book covers only single variable calculus. He has a book called [u]Calculus on Manifolds[/u] - which is [i]sort of[/i] a multivariable calculus book, but it is very advanced and it should only be used after you already have at least an intuitive understanding of linear algebra and multivariable calculus.

Don't quote me, but you could try the one that EPGY uses (Anton - Calculus: A New Horizon)... I've just started MV Calc, but I found the book quite interesting (which = more interesting than others)... someone should put their views on though


I am not able to download GSView from any of the two sites mentioned in the Asymptote Download Wiki Page. Could anyone help?

The sites seem to be down at the moment so try downloaiding gsv49w32.exe from http://tug.ctan.org/tex-archive/support/ghostscript/ghostgum/


You mean the surface area of the cube, right?

Any ideas?  What have you tried?

The space diagonal of a cube is $ x\sqrt{3}$ where $ x$ is the length of the side of the cube.

[quote="thomasklau"]Hi,
    I was wondering how to find the area of a cube when you are given the space diagonal.

Thanks.[/quote]

We will have a right triangle with legs $ s$ and $ s\sqrt{2}$.

The hypotenuse, the space diagonal, wil then be $ s\sqrt{3}$.

So, If you are given the space diagonal,

the area, letting the space diagonal be $ x$, is

$ \frac{x^3}{3\sqrt{3}}$.


[hide]Just notice that the stuff in that radical has to come out 0, so the roots would be 1, 2, and 3, which makes the final answer $\fbox 6$[/hide]

[hide="answer"]Notice that if you subtract 6 from each side, you get 6* :sqrt: (x-1)(x-2)(x-3)=0. The radical has to be zero. Using the zero product property, x-1 or x-2 or x-3 must be 0. so x can be 1 or 2 or 3. 1+2+3=[u][b]6[/b][/u].[/hide]

[hide]That's pretty easy. I can't believe that came from a nationals sprint round. You know that 6 * sqrt((x-1) (x-2) (x-3)) has to equal 0, and you can do that by making any one of (x-1), (x-2), or (x-3) = to 0. So you can have 1, 2, or 3 respectively. 1 + 2 + 3 = 6. [/hide]  :)  :)

The first page of a national sprint is usually really, really easy.

not always... unless your math is at a really really high level

[quote="09husbri"]Yeah, but this was the last problem of a round. I say the first page of a nationals sprint is like the last page of a chapters sprint. Pretty easy.[/quote]
Yeah the first 15 or so are easy. (I get all of them) but then they get harder.

Interesting though how this simple problem was a #30, isn't it?

yea
i think that older tests are easier too
last year's test for some reason was extremely hard compared to the 2003, 2002, etc. tests

[quote="tarquin"]I think older nationals sprint tests were somewhat easier than the newer ones. This is something I've noticed because I've taken every 30-problem nationals sprint test and some 40-problem ones, that is, prior to 1987.[/quote]Tarquin, where'd you get them? Can you scan them and send them please. I really NEED some for prep and my coach only has 2003 set. :?  :?  :?  :?  :?  :?

[quote="tarquin"]1. I don't have a scanner.
2. They're at Mr. Hay's and all the copies I have are written on with answers all over them.[/quote]
Thats ok. Maybe Ill steal them some how :ninja: :ninja: :ninja:

[quote="09husbri"]Here's a nice easy problem for you guys from an old, old nationals sprint round. Good luck:

30. Find the sum of the value(s) of x that satisfy the given equation.
6 * sqrt( (x - 1) (x - 2) (x - 3) ) + 6 = 6

[/quote]

[hide="Solution"]
6sqrt( (x - 1)(x - 2)(x - 3) = 0
sqrt( (x - 1)(x - 2)(x - 3) = 0
(x - 1)(x - 2)(x - 3) = 0

x = {1, 2, 3}

1 + 2 + 3 = [b]6[/b]
[/hide]

The only new sprint I've actually taken in whole is 2004, and it hung me out to dry. :lol: Without any clothes on. :o Just kidding. :rotfl:

But I stunk it up anyway. I did take some older ones and did really well on 'em.

[hide]6 or 1+2+3 which are the roots of the equation.[/hide]
On the practice nationals problems I did, I usually get the first and last page right.

[quote="09husbri"]Here's a nice easy problem for you guys from an old, old nationals sprint round. Good luck:

30. Find the sum of the value(s) of x that satisfy the given equation.
6 * sqrt( (x - 1) (x - 2) (x - 3) ) + 6 = 6

[hide="Solution"]Simplifying the equation easily gets sqrt( (x - 1) (x - 2) (x - 3) = 0, so (x - 1) (x - 2) (x - 3) = 0 after taking the square of each side. Now, since these three binomials when multiplied together equal zero, (at least) one of them must equal zero. If x-1=0, x=1. If x-2=0, x=2. If x-3=0, x=3. These are all values of x. Thus the sum of all possible values is 1+2+3=[b]6[/b].[/hide][/quote]

[hide]I think this is how you do it:

6 * sqrt( (x - [b]1[/b]) (x - [b]2[/b]) (x - [b]3[/b]) ) + 6 = 6

Add those numbers in bold together and you get your answer. (Forget the negatives)[/hide]


Sorry ! Is there are typo ?  :maybe:
In this problem $c,d$ unnecessary and
$P \leq a^{4}+b^{4}$
Let $a \to 0$ and $b\to \infty$ so $P \to \infty$
$\Rightarrow$ problem's un solve  :(
I think this problem's
  Let $1\leq a\leq b \text{ : }\frac{1}{a}+\frac{2}{b}\geq 2$
  Find the maximum of $P=a^{4}+b^{4}$
 What do you think about it ??? :oops:

[quote="Do Tien Dang"]  Let $1\leq a\leq b \text{ : }\frac{1}{a}+\frac{2}{b}\geq 2$
  Find the maximum of $P=a^{4}+b^{4}$
 What do you think about it ??? :oops:[/quote]
   There are no solution, isn't there

[quote="Flove"]Just in a second, I see that it's exactly Karamata inequaity.  :lol:  There's nothing here.[/quote]
I don't understand what did you say ?. can you solve it?

Okie, I had edited this problem. Can you prove it?

[u]Firt solution[/u]
We have $1\leq a\leq \frac{3}{2}$
Let $x= \frac{1}{a}\Rightarrow \frac{2}{3}\leq x\leq 1$ and $b\leq \frac{2}{2-x}$
 $\Rightarrow a^{4}+b^{4}= \frac{1}{x^{4}}+\frac{2}{(2-x)^{4}}=f(x)$
$f'(x)=\frac{8}{(2-x)^{5}}-\frac{4}{x^{5}}\geq  0 \Rightarrow$ max$(a^{4}+b^{4})=17$  :)
[u]Second solution[/u]
$a^{4}+b^{4}-1^{4}-2^{4}=a^{4}(1-\frac{1}{a^{4}})+b^{4}(1+\frac{1}{b^{4}}=(a^{4})(1-\frac{1}{a^{4}})+b^{4}(2-( \frac{1}{a^{4}}+\frac{1}{b^{4}})) \leq 0 \Rightarrow$
 max$(a^{4}+b^{4})=17$ :)
I think second solution is very nice but I don't think using Karamata inequality is good because it very strong

Yes, but it really nice. Can you post your solution?

[quote="dt20022006"]I think it will be solved by AM-GM inequalities[/quote]
  I don't think so ! I think we have to use the Abel's expression and it is a nice solution  :) .Can you solve it by AM_GM inequality ?

No, there is a solution just using Am-Gm ineq. Let's thinking... :idea:


inequalities
calculus
calculus computations

[quote="t0rajir0u"]Certainly; the inequality is obvious if the signs don't match, and otherwise it's a straightforward application of Muirhead or Rearrangement or weighted AM-GM.[/quote]

Hi T0rajir0u,

Thank you for your comment. But I could not still prove it. Please specify something more! :)

The difference equals $ (a^{p\minus{}1}\minus{}b^{p\minus{}1})(a\minus{}b)$ for $ a,b>0$. The expressions in the parentheses always have the same sign.

[quote="fedja"]The difference equals $ (a^{p \minus{} 1} \minus{} b^{p \minus{} 1})(a \minus{} b)$ for $ a,b > 0$. The expressions in the parentheses always have the same sign.[/quote]

Hi Fedja,

Excellent! Thank you :lol:


geometry
3D geometry
symmetry
algebra
factorization
sum of cubes

How do you factor completely:

$x^{15} + y^{15}$ - My math teacher claims, and so does my book that there are 4 factors, but I can only derive a maximum of 3 at any time.

I've tried sum of cubes, and sum of fifths.

And even $(x+y)(x^{14} - x^{13}y + \ldots + x^2y^{12} - xy^{13} + y^{14})$, but I don't know how to approach it from there.
Any help would be appreciated.

And also: Edit: No radicals in the factorization.

I said it in a misleading manner.

I basically meant: $(x^3)^5 + (y^3)^5$.

So, that becomes $(x^3+y^3)(x^{12} - x^9y^3 + x^6y^6 - x^3y^9 + y^{12})$, But it didn't work.

Ok...

You can factor a little like this:

\[(x^5+y^5)(x^{10}-x^5y^5+y^{10})=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)(x^{10}-x^5y^5+y^{10}).\]

Or you could do this:

\[(x^3+y^3)(x^{12}-x^9y^3+x^6y^6-x^3y^9+y^{12})=(x+y)(x^2-xy+y^2)(x^{12}-x^9y^3+x^6y^6-x^3y^9+y^{12}).\]

Ok, now this is where my way gets foggy but let's go ahead anyways.  Knowing that it has 4 factors helps.  We know one is going to be $x+y$ and looking at those factorizations it seems likely that $x^2-xy+y^2$ and $x^4-x^3y+x^2y^2-xy^3+y^4$ are going to be factors (the others are of a high degree and have more of a chance to factor.  So:

\[x^{15}+y^{15}=(x+y)(x^2-xy+y^2)(x^4-x^3y+x^2y^2-xy^3+y^4)P(x).\]

Expanding we get:

\[x^{15}+y^{15}=P(x)(x^7-x^6y+x^5y^2+x^2y^5-xy^6+y^7).\]

So, now we use some deductive skills.  There is obviously a $+x^8$ and $+y^8$.  Now when we multiply out the $x^8$ we will get a $-x^{14}y$ so to offset this we need an $x^7y$ and by symmetry a $y^7x$.  The $x^{13}y^2$ terms will cancel if we multiply out terms, thus the coefficient of $x^6y^2$ is $0$.   Looking at what the coefficient of $x^{12}y^3$ will be in the expansion we see that we need to offset it by adding in a $-x^5y^3$ term (and thus a $-x^3y^5$ also).   Using the same procduere we find that we need a $-x^4y^4$ terms also.  Everything works out, yay!!!  So our factorization looks like:

\[(x+y)(x^2-xy+y^2)(x^4-x^3y+x^2y^2-xy^3+y^4)(x^8+x^7y-x^5y^3-x^4y^4-x^3y^5+xy^7+y^8).\]


Let $n$ be a natural number. Find the smallest vallue of $n$, where $\cos(frac{\Pi}{n})$ cannot be represented in the form $p+sqrt{q}+sqrt[3]{r}$, where $p,q$ and $r$ are rational.

You proved nothing, Rust. He asks for the smallest. Can you prove that $7$ is not ok? Or $21671282$?

[quote="Rust"][quote="rumtreiber"]Let $n$ be a natural number. Find the smallest vallue of $n$, where $\cos(\frac{\pi}{n})$ cannot be represented in the form $p+\sqrt{q}+\sqrt[3]{r}$, where $p,q$ and $r$ are rational.[/quote]
Why?
$cos\frac{\pi }{2}=0 \ (p=q=r=0),$
$cos\frac{\pi }{3}=\frac{1}{2}\ (p=\frac{1}{2}, q=0,r=0)$,
$cos\frac{\pi}{4}=\frac{1}{\sqrt 2 }, \ p=0, q=\frac{1}{2}, r=0$,
$cos\frac{\pi}{6}=\frac{\sqrt 3 }{2}, \ p=0,q=\frac{3}{4}, r=0$.[/quote]
you prove that $n\ge 7$.

I read end: $cos\frac{\pi}{n}$ can not be represent in form $p+\sqrt q+\sqrt[3] r . For n=5$x=cos\frac{\pi }{5}=\frac{1+\sqrt 5 }{4}.
For n=8 $cos\frac{\pi}{8}=\sqrt{\frac{\sqrt 2-1}{2\sqrt 2}}\not =p+\sqrt q+\sqrt[3] r .$
For n=7 if $x=cos\frac{\pi }{7}$, then $0=cos\frac{4\pi}{7}+cos\frac{3\pi}{7}=x^{4}-6x^{2}(1-x^{2})+(1-x^{2})^{2}+4x^{3}-3x$ or
$f(x)=8x^{4}+4x^{3}-8x^{2}-3x+1=0$. If it can be represent in these form, then r=0 and f(x) is reducible in Q.
If y=2x we get $g(y)=y^{4}+y^{3}-4y^{2}-3y+2=0$, but g(y) is irreducible.


topology
real analysis
real analysis unsolved

Prove that every compact metric space is a continuous image of the Cantor set

That's very nice ! Do you have a proof ?

Here's a sketch:

Denote the Cantor set by $C$, and the compact metric space by $(X,d)$. Cover $X$ with finitely many closed balls of diameter $\le 1,\ X_{1},X_{2},\ldots,X_{t},\ t\ge 1$. Clearly, we can partition $C$ into $t$ uncountable closed subsets of diameter $\le 1$, say $C_{1},\ldots,C_{t}$. Then, for each $i\in\overline{1,t}$, cover $X_{i}$ with finitely many balls $X_{ij}$ of diameter $\le\frac{1}{2}$. Again, partition each $C_{i}$ into finitely many uncountable closed subsets $C_{ij}$ of diameter $\le\frac{1}{2}$, and continue the process.

Each point $x\in X$ is determined by some (maybe more than one) sequence $n_{1},n_{2},\ldots$ of positive integers in the sense that it is the intersection of the closed sets $X_{n_{1}},X_{n_{1}n_{2}},\ldots$. Map the intersection of the sets $C_{n_{1}},C_{n_{1}n_{2}},\ldots$ onto $x$. It shouldn't be hard now to show that this map from $C$ onto $X$ is indeed continuous, as desired.

I don't see why one can partition the compact set into $t$ disjoints closed subsets ?

You only partition the Cantor set. This is certainly possible, becaue it's totally disconnected. Just look at some pictures of it :). You don't partition the other compact space ($X$) or its subsets; you merely [i]cover[/i] those with sets of small diameter.


Can someone point out some book that would tell me something about two-player strategy games.
There are many tasks about this games but i never saw any book about them.

what do you mean?
do you mean like 2 people pick up certain numbers of objects and the person who picks up the last object wins/loses?

I don't know how in-depth of a book you want. The only book I have on the subject is [url=http://www.amazon.com/Two-Person-Game-Theory-Anatol-Rapoport/dp/0486406865]Two Person Game Theory[/url] by Anatol Rapoport. It's okay, but I wouldn't recommend it as a "textbook".

Here's a bibliography of books on Game Theory, but I can't recommend any one of them over another: http://www.ericweisstein.com/encyclopedias/books/GameTheory.html


$m\ ,\ n\ \in N^*\ ;\ x\ ,y\ ,z\ >\ 0\ ;\ xyz=1\Longrightarrow \sum \frac{x^n}{1+x^{m+2n}}\le \frac 12\cdot \sum x^m\ .$

I may be wrong but dont we have
$\sum \frac{x^n}{1+x^{m+2n}}\le \frac{1}{2}\sum \frac{x^n}{x^{n+\frac{m}{2}}}\le \frac{1}{2}\sum x^m$
 :?:
 :huh:


We all know the pledge of allegiance.
But in truth our country is divided very severly: Democrats and Republicans.
There is almost a 1:1 ratio of polls leading to the canidates.

Why are we evading from our pledge that many of us make every day?

I don't think the pledge is advocating a single political party. It's a lot more general than that.

[quote="ProtestanT"]The pledge is more to a general purpose of America (and its government) than to specific parties.

That means that whether John McCain or Barack Obama is elected, most Americans will support and respect their president once inaugurated. [b]It's similar to how many Americans don't like Bush, but still respect him and stand behind him.[/b][/quote]

NO.  JUST NO.

I know many people who don't like Bush, don't respect him, and don't stand behind him.

I don't understand how having different views on how to best approach the nation's problems has anything to do with the pledge of allegiance.

Also, pledging allegiance to the country does not mean supporting the decisions of the president. The country is not equivalent to the government, and furthermore, the president represents only the executive branch of government.

The phrase "United we stand, divided we fall" is not in the Pledge of Allegiance.  It was said by Abraham Lincoln concerning the deep divisions in the United States during the time of the Civil War.  (I don't know whether it was first said before or after the start of the war, but that is not relevant here.)  The only way the Pledge of Allegiance is relevant to the topic of solidarity is when we pledge to an "indivisible" nation.

I don't believe the US is as severely divided by party as the OP makes it appear, and certainly not nearly to the point of starting another civil war.  There are a good number of Americans who do not see themselves as a strong supporter of either party, but stand somewhere in the middle and share some of the values and beliefs of each party (myself being one of them).

[quote="cf249"]It was said by Abraham Lincoln concerning the deep divisions in the United States during the time of the Civil War.[/quote]  Closer, but no.  See [url=http://en.wikipedia.org/wiki/United_we_stand,_divided_we_fall]the Wikipedia article[/url] for the history of the phrase.  The line you're thinking of is "A house divided cannot stand," from the 1858 Lincoln-Douglas debates.  See [url=http://en.wikipedia.org/wiki/Lincoln's_House_Divided_Speech]another wikipedia article[/url]

Arguably, we are united in hating Bush. 11% approval rate? SInce when does 89% of America agree on anything?

The last [url=http://www.gallup.com/poll/1723/Presidential-Job-Approval-Depth.aspx]Gallup[/url] has him at 25%, the lowest it's been.  If you look at the second chart, I think it suggests quite starkly that the country is indeed very much divided, even on the question of Bush's performance (though I think the effect ProtestanT mentioned may be skewing the numbers).

Just out of curiosity, where did you read that it's at 11%?  I wonder if out there somewhere is some equally absurd Republican-sponsored poll claiming he's at an even 50%.

Sorry for :spam: ing, but that gallup poll is super funny.

I always thought the statement meant that we're united when we stand, but divided when we fall. Not that we stand when united, but fall when divided.

We need our country to be divided between democrats and republicans.

The 'united' you are referring to would be a single party system...

That's one step away from dictatorship. Thing is, if things in government get done too fast and too easily, the government has too much power. This often leads to freedoms being taken away, etc.

[quote="ProtestanT"]The pledge is more to a general purpose of America (and its government) than to specific parties.

That means that whether John McCain or Barack Obama is elected, most Americans will support and respect their president once inaugurated. It's similar to how many Americans don't like Bush, but still respect him and stand behind him.[/quote]I think it's more like: No matter what happens to our country, we still stand behind it and hope it succeeds.

[quote="bbgun34"]some people are so religiously intolerant that they take their children out of schools that say the pledge daily. yikes.[/quote]

Why are you reviving a six month-old thread with an only vaguely coherent statement to add to it?  (This is a rhetorical question and doesn't require an answer.  My hope is that this thread will sink back down where it belongs.)