[quote="jli"]halo 3 beta is coming out on may 16

here's the countdown:

[hide="really long so i hid it"]http://www.7is7.com/otto/countdown.html?year=2007&month=5&date=16&hrs=00&min=0&sec=0&tz=-420&title=Countdown%20To%20Halo%203%20Beta&lang=en&show=dhms&mode=r&cdir=down&bgcolor=%23CCFFFF&fgcolor=%23000000[/hide]

btw, you have to get crackdown to get the invite :( but im pretty sure it will be worth it

the real things coming out in October or something :o[/quote]

uh is this only for xbox360 or for both the regular and taht.

[quote="jli"]halo 3 beta is coming out on may 16

here's the countdown:

[hide="really long so i hid it"]http://www.7is7.com/otto/countdown.html?year=2007&month=5&date=16&hrs=00&min=0&sec=0&tz=-420&title=Countdown%20To%20Halo%203%20Beta&lang=en&show=dhms&mode=r&cdir=down&bgcolor=%23CCFFFF&fgcolor=%23000000[/hide]

btw, you have to get crackdown to get the invite :( but im pretty sure it will be worth it

the real things coming out in October or something :o[/quote]

oh yeah...glad to see im not the only one on AoPS crazy about this  :D .

my friend has XBL i can just go over to his house to play Beta but im looking more forward to the campagin....i wanna see how it ends!

[quote="jli"]halo 3 beta is coming out on may 16

here's the countdown:

[hide="really long so i hid it"]http://www.7is7.com/otto/countdown.html?year=2007&month=5&date=16&hrs=00&min=0&sec=0&tz=-420&title=Countdown%20To%20Halo%203%20Beta&lang=en&show=dhms&mode=r&cdir=down&bgcolor=%23CCFFFF&fgcolor=%23000000[/hide]

btw, you have to get crackdown to get the invite :( but im pretty sure it will be worth it

the real things coming out in October or something :o[/quote]

oh did u buy Crackdown? if u are registered on for Beta I'll mite play you once it comes out!


Euler
symmetry
geometry
incenter
geometry proposed

We start my own problem with follow problem in MYM(Vietnam)
 Let four circle $(O) (O_{1}) (O_{2}) (O_{3})$such that $: (O_{i})$ intouch $(O)$ at $A,B,C$ respectively and $(O_{i})$extouch $(O_{i+1})$ at $M,N,P (i=1,2,3)$ respectively.Prove that $AM,BN,CP$ are concurent at poin $(X_{1})$

[quote="nsato"]They concur at the [url=http://mathworld.wolfram.com/FirstEppsteinPoint.html]First Eppstein Point[/url] of triangle $O_{1}O_{2}O_{3}$.[/quote]
sorry! can you post the sol for this pro because i can't find solution in above link

That link was just meant to be a reference.  If you want to see a proof, see this link:
[url]http://www.cut-the-knot.org/Curriculum/Geometry/Eppstein.shtml[/url]

Thanks
Now we continue
let circle $(O_{4})$ intouch $(O)$ , ectouch $(O_{2}) ,(O_{3})$ ,similarly with circle$(O_{5})$ and $(O_{6})$.Prove that $O_{1}O_{4},O_{2}O_{5}$ and ${O_{3}}O_{6}$  concurent at $X_{1}$ (in the first problem

Beautiful problem!

The tangents to $(O)$ at $A$, $B$, and $C$ determine a triangle, say $XYZ$.  Lines $XO_{1}$, $YO_{2}$, and $ZO_{3}$ concur at the Eppstein point.  ("The Euler-Gergonne-Soddy Triangle of a Triangle," AMM 103, 319-329, 1996.)  Hence, by symmetry, it suffices to show that $O_{4}$ lies on $XO_{1}$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/2397a9ba70760d6aeda84d7c339fdf47.png[/img]

Let $\omega_{1}$ be the circle with centre $X$ and radius $XB = XC = XM$.  Upon inversion through circle $\omega_{1}$, circles $(O)$, $(O_{2})$, and $(O_{3})$ get mapped to themselves, and circle $(O_{1})$ gets mapped to circle $(O_{4})$.  Therefore, points $X$, $O_{1}$, and $O_{4}$ are collinear.

On saturday i will post full problem of my own
I hope will have nice soln  :lol:

sorry i forgot this topic
here is all my problem
Let circle $(O)$and circles$(O_{i})(i=1,2,3,4,5,6)$such:
 $i) (O_{i})$intouch $(O)$at$A,B,C,D,E,F$respectively
  $ii) (O_{1}) (O_{2})$ extouch  at $M,(O_{2})$extouch $(O_{3})$ at N, $(O_{3})$ extouch $(O_{1})$at $P$
 $iii)(O_{2}),(O_{3}),(O_{4})$extouch other
$iiii) (O_{2}),(O_{1}),(O_{6})$extouch other
  $iiiii) (O_{1}),(O_{5}),(O_{3})$ extouch other
prove that
$1) AM,BN,CP,O_{1}O_{4},O_{2}O_{5},O_{3}O_{6}$are concurent at$J$
 $2) O{1}M,O_{2}N,O_{3}P$ are concurent at $G$
 $3) O_{4}M,O_{5}N,O_{6}P$ are concurent at $H$
$4) AD,BE,CF$ are concurent at $T$
$5) AD$ met$BC$at $Z$ .prove that$O,M,Z$are conllinear
$6) I$be incenter of$MNP$ .Prove that$O,I,J,G,H,T$are conllinear
 $7) AM$met$(O)$at$A_{1}$ .prove that $OA_{1}\perp O_{2}O_{3}$
 i hope a nice sol for this pro :lol:


I have never qualified for the AIME by means of the USAMTS so I have a few questions if someone would answer them.  First of all, how do I find out if my score was high enough to qualify?  I have a 73 after 3 rounds which I imagine is pretty good.  I have searched the AMC and USAMTS websites for information but did not find anything helpful.  Does it usually take a while for the info to get posted or will I automatically be notified?  And lastly does anyone know how the scores are converted to the index used for USAMO qualifiers?  Any information would be appreciated. Thanks.

1. You will qualify.  You may have already qualified with a 73.

2. You will get 100 points towards your USAMO index, maximum.  Probably if you score anywhere in the 80-90 range they'll give you the 100 points.

Also, could you delete the duplicate of this thread that you double-posted?

hmm, i don't think the 4th round matters at all with 73/75 you get 100/150 on AMC12

Assuming things haven't changed since 2 years ago:
You can qualify for AIME with at least 90% on the first three rounds. Your USAMTS score will be counted as a 97% on AMC-12 would be since you received 97% of the possible points.


A customer at a 7-11 store selected 4 items to buy, and was told that the cost was $7.11, He was curious that the cost was the same as the store name, so he enquired as to how mathe figure was derived. The shop asiis said that he had simply multiplyed the prices of the four individual items. The customer protested and said that the four prices should have been added, not multiplied. But when he did the result was exactly the same: $7.11 What were the 4 invidual prices?

a moderator should lock this topic, since it is being discussed at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48818

hi; uh sorry for the double post; i was trying to edit this one and it didnt edit but rather made a new topic for some reason  :?  :?.
this topic should be deleted if possible.

yea, i posted this problem a while ago too, the solution is basically, you have a,b,c,d
and if you multiply them by 100, you get integers, then you factor the product and for the sum you know that either 1 or 3 of the numbers are odd, then you go through case analysis and eventually get something that works


calculus derivative algebra polynomial function logarithms conics

I'm helping the Algebra II kids at my school study for the state contest and these are a few problems that I'm unsure how to solve using their methods.  Nothing above Algebra II, I can solve most of them with the higher methods  :lol: .  No 89's either.

$10.$  If $log_{8}3=p$ and $log_{3}5=q$, then find $log_{10}5$ in terms of $p$ and $q$.

Answer:
[hide]$\frac{3pq}{1+3pq}$

I know that you can just store values and see which of the answers it works for (multiple choice), but how do you solve that algebraically?[/hide]

$11.$ If $f^{-1}(\frac{2}{x})=x+2$, then $f(2x)=$

Answer:
[hide]$\frac{1}{x-1}$

I know how to do these types of problems but if someone could show me a way to solve these types of problems in general (particularly how to find $f^{-1}(x)$ from what is given.  I have to teach this to others.)[/hide]

$17.$ The maximum value of $f(x)=-x^{2}+2bx+c$ is

Answer:
[hide]$c+b^{2}$

I know how to do this using derivatives and calculus but how should I explain this to some Algebra II students?[/hide]

$21.$ Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $(x-19)$, the remainder is $99$; and when $P(x)$ is divided by $(x-99)$, the remainder is $19$.  What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$?

Answer:
[hide]$-x+118$

I know that you can do $P(19)=99$ and $P(99)=19$, but how do you solve this?[/hide]

$23.$ The first four terms in an arithmetic sequence are $x+y$, $x-y$, $xy$, $\frac{x}{y}$ in that order.  What is the fifth term?

Answer:
[hide]$\frac{123}{40}$[/hide]

$25.$ For each real number $x$, let $f(x)$ be the minimum of the numbers $3x+1, x+1,-2x+5$.  Find the maximum value of $f(x)$.

Answer:
[hide]$2+\frac{1}{3}$[/hide]

$34.$ Find $i+2i^{2}+3i^{3}+...+2006i^{2006}$ where $i = \sqrt{-1}$

Answer:
[hide]$-1004+1003i$[/hide]

$40.$ Let the function $A(x,y)$ be defined by the following rule:

$A(0,n)=n+1$
$A(m,0)=A(m-1,1)$
$A(m,n)=A(m-1,A(m,n-1))$

Find the numerical value of $A(2,3)$.

Answer:
[hide]$9$[/hide]

Well, I scrolled down, and saw 34. This is a nice one if you look at it right.
[hide="Solution"]
Well, basically, we have coefficients of $i$, which should IMMEDIATELY be SCREAMING at you look at the powers mod 4.
Just so you're clear on where I'm coming from, we have
$i^{1}= i$
$i^{2}=-1$ (by definition)
$i^{3}=-i$
$i^{4}= (i^{2})^{2}= (i^{2})(i^{2}) = (-1)(-1) = 1$
Then after that, we're just repeating this cycle again.
So powers of $i$ are periodic mod 4.
This means, that writing out the first few terms, we see
$(i^{1}+2i^{2}+3i^{3}+4i^{4})+(5i^{5}+6i^{6}+7i^{7}+8i^{8})+\ldots (2001i^{2001}+2002i^{2002}+2003i^{2003}+2004i^{2004})+2005i^{2005}+2006i^{2006}$

Notice that this is grouped by terms of four, except the last two expressions, so we'll look at those last.
These expressions all follow the same sign pattern, because we're dealing with repeating patterns of i.

$(i-2-3i+4)+(5i-6-7i+8)+\ldots (2001i-2002-2003i+2004)+2005i-2006$
This is $(2-2i)+(2-2i)+(2-2i)$ added 501 times total, with the additional $2005i-2006$
So we have $501(2-2i)+2005i-2006$
which evaluates easily to $-1004+1003i$
[/hide]

[quote="skimnc"]
$11.$ If $f^{-1}(\frac{2}{x})=x+2$, then $f(2x)=$[/quote]

Use $\boxed{f^{-1}(a)=b\Longleftrightarrow a=f(b)}$, we have $f(x+2)=\frac{2}{x}$. Thus plugging $x-2$ into this equation gives $f(x)=\frac{2}{x-2}$.

$\therefore f(2x)=\frac{2}{2x-2}=\boxed{\frac{1}{x-1}}$.

Thanks thusfar.  Yeah, I know all the basic formulas and stuff like $f^{-1}(x) = y$ then $f(y)=x$, change of base, and stuff like that.  I just need help explaining it in their terms.  And thanks for that one with the imaginaries.. For some reason I did just that but kept getting one.. probably added wrong.  So whats there is good.

Those left: $17, 21, 23, 25, 40$ to those interested.

number 17, its a parabola opening downward, so all you need to do is find the vertex (-b/2a) and plug that back in to get that answer

number 21, set up two things of f(x)=p(x)q(x)+r(x) and find the roots of p(x) and r(x) must be of degree 1 (so ax+b) for the (x-19)(x-99) and you will have two system of equations to solve for a and b because p(x)q(x) will be 0 because you can plug the roots of p(x) back in..kind of confusing ill do it if you dont understand

number 23, three equations, 3 variables, r,x, and y, you should  be able to solve.  You should get x=-9/8, y=-3/5 and r=6/5....

number 25, find the intersection points, plot the points and then bold in where the minimum is (should be 3 piecewise functions) and you'll find the maximum of the minimum is the second largest y value out of the three intersection points.

Only 40 left..hmm you work backwords.....

heres a solution for 21(my hints kinda hard to understand)..
[hide]
We'll go from the step where you stopped, you know
P(19)=99
P(99)=19

So P(x)=(x-19)(x-99)+ax+b because the remainder is of degree n-1 (which is 2-1=1)

Plug in x=19, so we have P(19)=19a+b
and P(99)=99a+b when we plug in 99

so 99=19a+b
19=99a+b,

subtrating the equations we have 80=-80a, so a=-1, plugging in 99=-19+b, b=118, so remainder is ax+b, $-x+118$[/hide]

[quote="kunny"][quote="skimnc"]

$40.$ Let the function $A(x,y)$ be defined by the following rule:

$A(0,n)=n+1$
$A(m,0)=A(m-1,1)$
$A(m,n)=A(m-1,A(m,n-1))$

Find the numerical value of $A(2,3)$.[/quote]

$A(1,\ n)=A(0,\ A(1,\ n-1))=A(1,\ n-1)+1$, so the sequence $\{A(1,\ n)\}$ is the arithmetic progression with common difference 1, we have $A(1,\ n)=A(1,\ 0)+1\cdot n=A(0,\ 1)=n+2$. Thus $A(2,\ n)=A(1,\ A(2,\ n-1))=A(2,\ n-1)+2$, so the sequence $\{A(2,\ n)\}$ is the arithmetic progression with common difference 2, we have $A(2,\ n)=A(2,\ 0)+2\cdot n=A(1,\ 1)+2n=2n+3.$ $\therefore A(2,\ 3)=\boxed{9}.$

I don't know  Algebra II level in your country, maybe this solution seems difficult a little.[/quote]

well it is 40, so it should be a little hard, anyways nice solution..this reminds me of an old aime problem that used bigger numbers.

[quote="DiscreetFourierTransform"][url]http://mathworld.wolfram.com/AckermannFunction.html[/url][/quote]

Lol ok...


Hi there, i was wondering how would you guys create a network matrix eg.[0 1 0 1] for this diagram

[img]http://i9.photobucket.com/albums/a61/luft/Cityairplaneticketmatrix.jpg[/img]

The question asks, "Construct a network matrix A for these routes."

See that chart as a graph and construct the costs matrix:
To each city assign a number from $1$ to $5$, then construct a matrix $A=(a_{ij})_{i,j=\overline{1,5}}$ with $a_{ij}=a_{ji}$ being the cost of transport by direct route from city $i$ to city $j$. I guess that where you have no direct route, you assign $\infty$.


function
real analysis
real analysis unsolved

[quote="johnjohn"]What does it mean that something converges in the sense of $ L^2$?[/quote]

$ u_n  \to u$ means that $ \left\| {u_n } \right\|_{L^2 }  \to \left\| u \right\|_{L^2 }$.

No. It means that the $ L^2$ norm of the difference converges to zero. For a sequence of functions $ f_n$, that would be $ \|f_n\minus{}f\|\to 0$


inequalities
induction
inequalities unsolved

let $ x_1,x_2\cdots x_n \in \mathbb{R}$

prove that:

$ \frac{ |\sum{x_i}| }{1\plus{}|\sum{x_i}|}\leq \sum_{i\equal{}1}^{n}{\frac{|x_i|}{1\plus{}|x_i|}}$

First of all we prove that $ \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|}\leq \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}$ for all real $ a$ and $ b$.Because of  $ |a| \plus{} |b|\geq|a \plus{} b|$,               $ \frac {|a| \plus{} |b|}{1 \plus{} |a| \plus{} |b|}\geq \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|}$                . And             $ \frac {|a|}{1 \plus{} |a| \plus{} |b|}\leq \frac {|a|}{1 \plus{} |a|}$ ,  $ \frac {|b|}{1 \plus{} |a| \plus{} |b|}\leq \frac {|b|}{1 \plus{} |b|}$ the proof is ended. Now by the induction $ a \equal{} x_1 \plus{} x_2 \plus{} ... \plus{} x_n$ and $ b \equal{} x_{n \plus{} 1}$ inequality equivalents to $ \frac {|x_1 \plus{} x_2 \plus{} ... \plus{} x_{n \plus{} 1}|}{1 \plus{} |x_1 \plus{} x_2 \plus{} ... \plus{} x_{n \plus{} 1}|}\leq \frac {|x_1 \plus{} x_2 \plus{} ... \plus{} x_n|}{1 \plus{} |x_1 \plus{} x_2 \plus{} ... \plus{} x_n|} \plus{} \frac {|x_{n \plus{} 1}|}{1 \plus{} |x_{n \plus{} 1}|}\leq \frac {|x_1|}{1 \plus{} |x_1|} \plus{} \frac {|x_2|}{1 \plus{} |x_2|} \plus{} ... \plus{} \frac {|x_n|}{1 \plus{} |x_n|} \plus{} \frac {|x_{n \plus{} 1}|}{1 \plus{} |x_{n \plus{} 1}|}$ and the main proof is ended.


$ \boxed{1 - 1}$  [b]領域における最大値[/b]
$ x \ge 2$, $ y \ge 2$, $ 8 \le xy \le 16$ のとき, $ z = \log_2\sqrt {x} + \log_2 y$ の最大値を求めたい.
$ s = \log_2 x$, $ t = \log_2 y$ とおく.
(1) $ s$, $ t$ の取り得る値の範囲を求めよ.
(2) $ s + t$ の取り得る値の範囲を求めよ.
(3) $ z$ を $ s$, $ t$ を用いて表せ.
(4) $ z$ の最大値と, そのときの $ s$, $ t$ を求めよ.
(5) $ z$ が最大値のとき, $ x$, $ y$ を求めよ.

$ \boxed{1 - 2}$  [b]三角方程式, 角度の大小関係[/b]
$ 0 \le \theta < 2 \pi$ の範囲で
\[ 5 \sin \theta - 3 \cos 2\theta = 3 \ \cdots \cdots (*)
\]
を満たしている.
(1) 方程式 $ (*)$ を $ \sin \theta$ を用いて表せ.
(2) $ \sin \theta$ の値を求めよ.
(3) $ 0 \le \theta < 2 \pi$ の範囲でこの等式を満たす $ \theta$ のうち, 小さい方を $ \theta_1$, 大きい方を $ \theta_2$ とする.
(3-1) $ \cos \theta_1$ の値を求めよ.
(3-2) $ \cos \theta_2$ の値を求めよ.
(4) $ \theta_1$ についての不等式で, 正しいものを以下の中から選べ.
ただし, 必要なら, 次の値を用いてもよい.
$ \cos \frac {\pi}{5} = \frac {1 + \sqrt {5}}{4}$, $ \cos \frac {\pi}{12} = \frac {\sqrt {6} + \sqrt {2}}{4}$
$ A$ : $ 0 < \theta_1 < \frac {\pi}{12}$
$ B$ : $ \frac {\pi}{12} < \theta_1 < \frac {\pi}{6}$
$ C$ : $ \frac {\pi}{6} < \theta_1 < \frac {\pi}{5}$
$ D$ : $ \frac {\pi}{5} < \theta_1 < \frac {\pi}{4}$
$ E$ : $ \frac {\pi}{4} < \theta_1 < \frac {\pi}{3}$
$ F$ : $ \frac {\pi}{3} < \theta_1 < \frac {\pi}{2}$
(5) 不等式 $ n\theta_1 > \theta_2$ を満たす自然数 $ n$ のうち最小のものを求めよ.

$ \boxed{2}$  [b]軌跡, 接線, 面積の最大, 図形の面積[/b]
放物線 $ y = 2x^2$ を $ C$, 点$ (1, \ - 2)$ を $ A$ とする.
(1) 点 $ Q(u, \ v)$ に関して, 点 $ A$ と対称な点を $ P(x, \ y)$ とするとき, $ x$, $ y$ を用いて点 $ Q$ を求めよ.
(2) $ Q$ が $ C$ 上を動くときの点 $ P$ の軌跡 $ D$ の方程式を求めよ. ($ D$ は放物線)
(3) 二つの放物線 $ C$ と $ D$ の交点を $ R$ と $ S$ とする. ただし, $ x$ 座標の小さい方を $ R$ とする.
点 $ R$, $ S$ の $ x$ 座標をそれぞれ求めよ.
(4) 点 $ R$, $ S$ における放物線 $ D$ の接線の方程式をそれぞれ求めよ.
(5) $ P$ を放物線 $ D$ 上の点とし, $ P$ の $ x$ 座標を $ a$ とおく.
$ P$ から $ x$ 軸に引いた垂線と放物線 $ C$ との交点を $ H$ とする.
$ a$ が 点 $ R$, $ S$ の $ x$ 座標の値の間にあるとき, 三角形 $ PHR$ の面積 $ S(a)$ を $ a$ を用いて表せ.
(6) S(a) が最大値となる $ a$ を求めよ.
(7) $ a$ が (6) で求めた値のとき, 直線 $ HR$ と放物線 $ D$ の交点のうち, $ R$ と異なる点の $ x$ 座標を求めよ.
(8) $ x$ が (6) と (7) の間の範囲のとき, 放物線 $ D$ と直線 $ PH$ および直線 $ HR$ で囲まれた面積を求めよ.

$ \boxed{3}$  [b]等比数列, いろいろな数列の和を求める工夫[/b]
$ \{a_n\}$ を初項 $ a_1$ が $ 1$ で公比が $ \frac {1}{3}$ の等比数列とする.
数列 $ \{a_n\}$ の偶数番目の項を取り出して, 数列 $ \{b_n\}$ を $ b_n = a_{2n} \ (n = 1, 2, 3, \cdots)$ で定める.
$ T_n = \sum_{k = 1}^n b_k$ とおく.
(1) $ \{b_n\}$ も等比数列である.
初項と公比を求めよ.
(2) $ T_n$ を求めよ.
(3) 積 $ b_1 b_2 \cdots b_n$ を求めよ.
(4) 数列 $ \{c_n\}$ を $ c_n = 2n \cdot b_n \ (n = 1, 2, 3, \cdots)$ で定め, $ U_n = \sum_{k = 1}^n c_n$ とおく.
$ c_n$, $ c_{n + 1}$ の関係を $ b_n$ を用いて表せ.
ただし, $ p$, $ q$, $ r$ の最大公約数が $ 1$ であるような $ p$, $ q$, $ r$ を 用いて, $ p c_{n + 1} + q c_n = r b_n$ の形で書くこと.
(5) (4) の答えの $ c_{n + 1}$, $ c_n$ の係数に対応する値を $ p$, $ q$ とする.
このとき $ \sum_{k = 1}^n (p c_{n + 1} + q c_n)$ を $ U_n$, $ c_{n + 1}$, $ c_n$ を用いて表せ.
(6) $ U_n$ を求めよ.

$ \boxed{4}$  [b]四角錐とその $ 1$ つの面に平行な平面との共通部分[/b]
$ O$ を原点とする座標空間における $ 5$ 点を $ A(0, \ 0, \ 1)$, $ B(1, \ 0, \ 0)$, $ C(0, \ 2, \ 0)$, $ D( - 1, \ 0, \ 0)$, $ E(0, \ - 2, \ 0)$ とする.
ひし形 $ BCDE$ を底面とする四角錐 $ A - BCDE$ と, 平面 $ ABC$ に平行な平面との共通部分ついて考える.
[asy]size(300);

import graph3;
import math;

currentprojection=orthographic(1,-0.5,0.5);

triple o=(0, 0, 0);
triple a=(0, 0, 1);
triple b=(1, 0, 0);
triple c=(0, 2, 0);
triple d=(-1, 0, 0);
triple e=(0, -2, 0);

draw (a--e--b--c--a--b);
draw (d--a^^d--e^^d--c, dashed);

draw ((-1.25, 0, 0)--(2, 0, 0), Arrow);
draw ((0, -2.25, 0)--(0, 2.5, 0), Arrow);
draw ((0, 0, 0)--(0, 0, 1.5), Arrow);

label("$x$",(2,0,0),SE);
label("$y$",(0,2.5,0),NE);
label("$z$",(0,0,1.5),N);

label("$O$",o,SSW);
label("$A$",a,NE);
label("$B$",b,S);
label("$C$",c,S);
label("$D$",d,S);
label("$E$",e,S);[/asy]
(1) $ \overrightarrow{BC} \cdot \overrightarrow{BA}$ の値を求めよ.
(2) 三角形 $ ABC$ の面積を求めよ.
(3) $ \overrightarrow{u} = \overrightarrow{BA}$, $ \overrightarrow{v} = \overrightarrow{BE}$ とおく.
$ 0 < a < 1$ とし, 点 $ B_1$ を線分 $ BE$ を $ a: (1 - a)$ に内分する点とするとき, $ \overrightarrow{BB_1}$ を $ \overrightarrow{v}$ を用いて表せ.
(4) 点 $ A_1$ を $ \overrightarrow{OA_1} = \overrightarrow{OA} + \overrightarrow{BB_1}$ と定める.
$ A_1B_1$ 上の点 $ P$ は, $ 0 \le b \le 1$ を満たす $ b$ を用いて, $ \overrightarrow{OP} = \overrightarrow{OB} + b \overrightarrow{u} + p \overrightarrow{v}$ と表される.
ただし, $ \overrightarrow{BB_1} = p \overrightarrow{v}$.
また, $ AE$ 上の点 $ Q$ は, $ 0 \le c \le 1$ を満たす $ c$ を用いてどのように表されるか.
$ \overrightarrow{OQ} = \overrightarrow{OB} + q \overrightarrow{u} + (r - c) \overrightarrow{v}$ の形で書くこと.
(5) $ P$ と $ Q$ は どのようなときに一致するか.
$ a$ と $ b$ と $ c$ の関係式で表せ.
(6) 線分 $ A_1B_1$ と $ AE$ は, $ AE$ をどのように内分する点で交わるか.
(7) (6) の点を $ E_1$ とする.
点 $ C_1$ を $ \overrightarrow{OC_1} = \overrightarrow{OC} + \overrightarrow{BB_1}$ で定めると, 線分 $ A_1C_1$ と線分 $ AD$ は, $ AD$ をどのように内分する点で交わるか.
(8) (7) の点を $ D_1$ とすると $ \overrightarrow{D_1E_1}$ を $ \overrightarrow{DE}$ を用いて表せ.
(9) 四角形 $ B_1C_1D_1E_1$ の面積を求めよ.
(10) $ |\overrightarrow{B_1D_1}|$ を求めよ.

$ \boxed{5}$  [b]資料の分析, 相関係数, 相関図[/b]
下の表は, $ 10$ 名からなるある少人数クラスを $ I$ 班と $ II$ 班に分けて, $ 100$ 点満点で $ 2$ 回ずつ実施した数学と英語のテストの得点をまとめたものである.
ただし, 表中の平均値はそれぞれ $ 1$ 回目と $ 2$ 回目の数学と英語のクラス全体の平均値を表している.
また, $ A$, $ B$, $ C$, $ D$ の値はすべて整数とする.
\[ \begin{array}{|c|c|cc|cc|}\hline \multicolumn{2}{|c|}{} & \multicolumn{2}{c|}{1\text{st}} & \multicolumn{2}{c|}{2\text{nd}} \\
\hline \text{Group} & \text{No}. & \text{Math} & \text{English} & \text{Math} & \text{English} \\
\hline & 1 & 40 & 43 & 60 & 54 \\
& 1 & 63 & 55 & 61 & 67 \\
I & 1 & 59 & B & 56 & 60 \\
& 1 & 35 & 64 & 60 & 71 \\
& 1 & 43 & 36 & C & 80 \\
\hline & 1 & A & 48 & D & 50 \\
& 1 & 51 & 46 & 54 & 57 \\
II & 1 & 57 & 71 & 59 & 40 \\
& 1 & 32 & 65 & 49 & 42 \\
& 1 & 34 & 50 & 57 & 69 \\
\hline \multicolumn{2}{|c|}{\text{Average}} & 45.0 & E & 58.9 & 59.0 \\
\hline \end{array}
\]
ただし, $ 1$st は $ 1$ 回目, $ 2$nd は $ 2$ 回目, Group は 班, No. は番号, Math は数学, English は英語, Average は平均値を意味する.
以下, 小数の形で解答する場合は, 指定された桁数の一つ下の桁を四捨五入し, 解答せよ.
(1) $ 1$ 回目の数学の得点について, $ I$ 班の平均値を求めよ.
(2) $ II$ 班の $ 1$ 番目の生徒の $ 1$ 回目の数学の得点 $ A$ を求めよ.
(3) $ II$ 班の $ 1$ 回目の数学と英語の得点について, 数学と英語の分散はともに $ 101.2$ である.
相関係数を求めよ.
(4) $ 1$ 回目の英語の得点について, $ I$ 班の $ 3$ 番目の生徒の得点 $ B$ の値が分からないとき, クラス全体の得点の中央値 $ M$ の値として何通りの値があり得るか.
(5) $ 1$ 回目の英語の得点のクラス全体の平均値 $ E$ が $ 54.0$ 点であったとき, $ B$ の得点を求めよ.
(6) (5) のとき, 中央値 $ M$ を求めよ.
(7) $ 2$ 回目の数学の得点について, $ I$ 班の平均値は $ II$ 班の平均値より $ 4.6$ 点大きかったとき,
$ I$ 班の $ 5$ 番目の生徒の得点 $ C$ から $ II$ 班の $ 1$ 番目の生徒の得点 $ D$ を引いた値は何点か.
(8) 以下の図はクラス全体の数学と英語の得点の相関図 (散布図) である.
[asy]size(250);

label("A", (50, 105), N);

pair a=(40, 43);
pair b=(63, 55);
pair c=(59, 62);
pair d=(35, 64);
pair e=(43, 36);
pair f=(36, 48);
pair g=(51, 46);
pair h=(57, 71);
pair i=(32, 65);
pair j=(34, 50);

for (int i = 0; i <= 100; i+=10)
{draw((i,0)--(i,100)^^(0,i)--(100,i));
label(format("\small$%i$",i), (i,0), S);
label(format("\small$%i$",i), (0,i), W);}

dot (a^^b^^c^^d^^e^^f^^g^^h^^i^^j);

label("Math", (50, -10), S);
label("English", (-15, 50), W);[/asy]
[asy]size(250);

label("B", (50, 105), N);

pair a=(60, 54);
pair b=(61, 67);
pair c=(56, 60);
pair d=(60, 71);
pair e=(69, 80);
pair f=(64, 50);
pair g=(54, 57);
pair h=(59, 40);
pair i=(49, 42);
pair j=(57, 69);

for (int i = 0; i <= 100; i+=10)
{draw((i,0)--(i,100)^^(0,i)--(100,i));
label(format("\small$%i$",i), (i,0), S);
label(format("\small$%i$",i), (0,i), W);}

dot (a^^b^^c^^d^^e^^f^^g^^h^^i^^j);

label("Math", (50, -10), S);
label("English", (-15, 50), W);[/asy]
[asy]size(250);

label("C", (50, 105), N);

pair a=(32, 42);
pair b=(34, 69);
pair c=(35, 71);
pair d=(36, 50);
pair e=(40, 55);
pair f=(42, 80);
pair g=(51, 58);
pair h=(57, 40);
pair i=(59, 60);
pair j=(63, 68);

for (int i = 0; i <= 100; i+=10)
{draw((i,0)--(i,100)^^(0,i)--(100,i));
label(format("\small$%i$",i), (i,0), S);
label(format("\small$%i$",i), (0,i), W);}

dot (a^^b^^c^^d^^e^^f^^g^^h^^i^^j);

label("Math", (50, -10), S);
label("English", (-15, 50), W);[/asy]
[asy]size(250);

label("D", (50, 105), N);

pair a=(48, 47);
pair b=(53, 57);
pair c=(55, 60);
pair d=(56, 69);
pair e=(58, 45);
pair f=(59, 67);
pair g=(60, 54);
pair h=(62, 68);
pair i=(64, 50);
pair j=(68, 76);

for (int i = 0; i <= 100; i+=10)
{draw((i,0)--(i,100)^^(0,i)--(100,i));
label(format("\small$%i$",i), (i,0), S);
label(format("\small$%i$",i), (0,i), W);}

dot (a^^b^^c^^d^^e^^f^^g^^h^^i^^j);

label("Math", (50, -10), S);
label("English", (-15, 50), W);[/asy]
(8-1) 1 回目の相関図を上図の中から選べ.
(8-2) 2 回目の相関図を上図の中から選べ.
(9) $ 1$ 回目のクラスの全体の数学と英語の得点の相関係数を $ r_1$, $ 2$ 回目のクラスの全体の数学と英語の得点の相関係数を $ r_2$ とするとき, 値の組 $ (r_1, \ r_2)$ として正しいものを以下の中から選べ.
$ A$ : $ (0.54, \ 0.20)$
$ B$ : $ ( - 0.54, \ 0.20)$
$ C$ : $ (0.20, \ 0.54)$
$ D$ : $ (0.20, \ - 0.54)$
(10) $ 2$ 回目のクラス全体 $ 10$ 名の英語の得点について, 採点基準を変更したところ, 得点の高い方から $ 2$ 名の得点が $ 2$ 点ずつ下がり, 得点の低い方から $ 2$ 名の得点が $ 2$ 点ずつ上がったが, その他の $ 6$ 名の得点に変更は生じなかった.
このとき, 変更後の平均値は変更前と比べてどのように変化したか.
(11) (10) のとき, 変更後の分散は変更前と比べてどのように変化したか.

$ \boxed{6}$  [b]異なる $ 2$ つの自然数 $ P$, $ Q$を用いて表される数[/b]
$ p$, $ q$ を異なる自然数とする.
このとき, 与えられた自然数 $ d$ について, $ d$ 以下の自然数 $ k$ のうちで
\[ k = mp + nq \ \cdots\cdots (*)
\]
(ただし, $ m$, $ n$ は $ 0$ 以上の整数)
のように表すことができるものを小さい順にすべて列挙し, 最後にその個数を表示したい.
そのために次のような [プログラム] を作った.
ここで, $ \text{INT(X)}$ は $ \text{X}$ を超えない最大の整数を表す関数である.

[code]100 INPUT PROMPT "p=": P
110 INPUT PROMPT "Q=": Q
120 INPUT PROMPT "D=": D
130 LET U=0
140 FOR K=1 TO D
150      IF K-INT(K/P)*P=0 THEN [甲]
160      FOR M=0 TO INT(K/P)
170           LET R=K-M*P
180           IF [乙] THEN [甲]
190      NEXT M
200      [丙]
210      PRINT K
220      [丁]
230 NEXT K
240 PRINT "総数="; U
250 END[/code]
(1) [プログラム] の [甲], [丙], [丁] に当てはまるものを, それぞれ以下の中から選べ.
$ A$ : $ \text{GOTO} \ 150$
$ B$ : $ \text{GOTO} \ 170$
$ C$ : $ \text{GOTO} \ 180$
$ D$ : $ \text{GOTO} \ 200$
$ E$ : $ \text{GOTO} \ 210$
$ F$ : $ \text{GOTO} \ 230$
$ G$ : $ \text{PRINT R}$
$ H$ : $ \text{PRINT U}$
$ I$ : $ \text{PRINT M}$
$ J$ : $ \text{LET R} = \text{R} + 1$
$ K$ : $ \text{LET U} = \text{U} + 1$
$ L$ : $ \text{LET K} = \text{K} + 1$
(2) [乙] に当てはまるものを, 以下の中から選べ.
$ A$ : $ \text{R} - \text{INT(R/M)}*\text{M} < > 0$
$ B$ : $ \text{R} - \text{INT(R/M)}*\text{M} = 0$
$ C$ : $ \text{R} - \text{INT(R/P)}*\text{P} < > 0$
$ D$ : $ \text{R} - \text{INT(R/P)}*\text{P} = 0$
$ E$ : $ \text{R} - \text{INT(R/Q)}*\text{Q} < > 0$
$ F$ : $ \text{R} - \text{INT(R/Q)}*\text{Q} = 0$
(3) [プログラム] を実行し, 変数 $ \text{P}$, $ \text{Q}$, $ \text{D}$ にそれぞれ $ 3$, $ 7$, $ 15$ を入力したとき, 整数の列 $ 3 \ \Box \ 7 \ 9 \ \Box \ 12 \ 13 \ 14 \ 15$ が出力される.
$ \Box$ に当てはまる数字を答えよ.
(4) 変数 $ \text{P}$, $ \text{Q}$, $ \text{D}$ にそれぞれ $ 3$, $ 7$, $ 100$ を入力したとき, 総数はいくつか.
(5) [プログラム] を部分的に変更して, 次のような $ 2$ 種類のプログラムを作る.
式 $ (*)$ のように表すことができないような $ d$ 以下の自然数 $ k$ を小さい順にすべて列挙し, 最後にその個数を表示したい.
そのためには, [プログラム] の $ 150$ 行および $ 180$ 行にあたる [甲] を何に置き換えて, $ 200$ 行を削除すればよいか.
当てはまるものを, 以下の中から選べ.
$ A$ : $ \text{GOTO} \ 190$
$ B$ : $ \text{GOTO} \ 200$
$ C$ : $ \text{GOTO} \ 210$
$ D$ : $ \text{GOTO} \ 220$
$ E$ : $ \text{GOTO} \ 230$
$ F$ : $ \text{GOTO} \ 240$
(6) 自然数 $ k$ に対して, 式 $ (*)$ を満たす組 $ (m, \ n)$ の個数を $ v_k$ とする.
$ d$ 以下の各自然数 $ k$ について $ v_k$ を出力し, 最後に総数として和 $ v_1 + \cdots + v_d$ の値を表示したい.
[プログラム] を変更するにあたって, 次の [ ] に当てはまるものを, 以下の中から選べ.
(6-1) $ 150$ 行を
[code]150      [戊][/code]
に変更する.
(6-2) $ 180$ 行の [甲] を [己] に置き換えて, $ 200$ 行を削除する.
(6-3) $ 210$ 行および $ 220$ 行を
[code]210      PRINT "k=" ;K; "のとき, " ;V; "個"
220      [庚][/code]
に変更する.
$ A$ : $ \text{GOTO} \ 210$
$ B$ : $ \text{GOTO} \ 220$
$ C$ : $ \text{GOTO} \ 230$
$ D$ : $ \text{LET V} = 0$
$ E$ : $ \text{LET V} = \text{U}$
$ F$ : $ \text{LET U} = \text{U + V}$
$ G$ : $ \text{LET V} = \text{V + U}$
$ H$ : $ \text{LET U} = \text{U} + 1$
$ I$ : $ \text{LET V} = \text{V} + 1$

$ \boxed{4}$
(1)
\begin{eqnarray*} \overrightarrow{BC} \cdot \overrightarrow{BA} & = & (\overrightarrow{OC} - \overrightarrow{OB}) \cdot (\overrightarrow{OA} - \overrightarrow{OB}) \\
& = & ((0,2,0) - (1,0,0)) \cdot ((0,0,1) - (1,0,0)) \\
& = & ( - 1,2,0) \cdot ( - 1,0,1) \\
& = & ( - 1) \cdot ( - 1) + 2 \cdot 0 + 0 \cdot 1 \\
& = & 1 + 0 + 0 \\
& = & \boxed{1}. \end{eqnarray*}
(2)
\begin{eqnarray*} S(\triangle ABC) & = & \frac {1}{2}\sqrt {\left|\overrightarrow{BC}\right|^2\left|\overrightarrow{BA}\right|^2 - \left(\overrightarrow{BC} \cdot \overrightarrow{BA}\right)^2} \\
& = & \frac {1}{2}\sqrt {\left|( - 1,2,0)\right|^2 \left|( - 1,0,1)\right|^2 - 1^2} \\
& = & \frac {1}{2}\sqrt {(( - 1)^2 + 2^2 + 0^2) \cdot (( - 1)^2 + 0^2 + 1^2) - 1^2} \\
& = & \frac {1}{2}\sqrt {5 \cdot 2 - 1} \\
& = & \frac {1}{2}\sqrt {9} \\
& = & \boxed{\frac {3}{2}}. \end{eqnarray*}
(3)
\begin{eqnarray*} \overrightarrow{BB_1} & = & a\overrightarrow{BE} \\
& = & \boxed{a\overrightarrow{v}}. \end{eqnarray*}
(4)
\begin{eqnarray*} \overrightarrow{OQ} & = & \overrightarrow{OB} + \overrightarrow{BQ} \\
& = & \overrightarrow{OB} + c\overrightarrow{BA} + (1 - c)\overrightarrow{BE} \\
& = & \boxed{\overrightarrow{OB} + c\overrightarrow{u} + (1 - c)\overrightarrow{v}}. \end{eqnarray*}
(5)
$ \overrightarrow{OP} = \overrightarrow{OQ} \iff \overrightarrow{OB} + b\overrightarrow{u} + a\overrightarrow{v} = \overrightarrow{OB} + c\overrightarrow{u} + (1 - c)\overrightarrow{v}$ より,
\[ \left\{\begin{array}{l} b = c \\
a = (1 - c) \end{array}\right. \iff \boxed{b = c = - a + 1}.
\]
(6)
(5) より, $ \overrightarrow{BQ} = c\overrightarrow{BA} + (1 - c)\overrightarrow{BE}$ であるので,
線分 $ A_1B_1$ と $ AE$ は, $ AE$ を $ \boxed{a: 1 - a}$ に内分する点で交わる.
(7)
同様に,
$ \overrightarrow{w} = \overrightarrow{CA}$ とおく.
$ \overrightarrow{v} = \overrightarrow{BE} = \overrightarrow{CD}$ に注意すると,
$ A_1C_1$ 上の点 $ R$ は, $ 0 \le d \le 1$ を満たす $ d$ を用いて, $ \overrightarrow{OR} = \overrightarrow{OC} + d \overrightarrow{w} + a \overrightarrow{v}$ と表される.
また, $ AD$ 上の点 $ S$ は, $ 0 \le e \le 1$ を満たす $ e$ を用いて, $ \overrightarrow{OS} = \overrightarrow{OC} + e \overrightarrow{w} + (1 - e) \overrightarrow{v}$ と表される.
$ \overrightarrow{OR} = \overrightarrow{OS} \iff \overrightarrow{OC} + d\overrightarrow{w} + a\overrightarrow{v} = \overrightarrow{OC} + e\overrightarrow{w} + (1 - e)\overrightarrow{v}$ より,
\[ \left\{\begin{array}{l} d = e \\
a = (1 - e) \end{array}\right. \iff \boxed{d = e = - a + 1}.
\]
$ \overrightarrow{CS} = e\overrightarrow{CA} + (1 - e)\overrightarrow{CD}$ であるので,
線分 $ A_1C_1$ と $ AD$ は, $ AD$ を $ \boxed{a: 1 - a}$ に内分する点で交わる.
(8)
\begin{eqnarray*} \overrightarrow{D_1E_1} & = & \overrightarrow{AE_1} - \overrightarrow{AD_1} \\
& = & a \overrightarrow{AE} - a \overrightarrow{AD} \\
& = & a (\overrightarrow{AE} - \overrightarrow{AD}) \\
& = & \boxed{a \overrightarrow{DE}}. \end{eqnarray*}
(9)
\[ \left\{\begin{array}{l} \overrightarrow{OA_1} = \overrightarrow{OA} + \overrightarrow{BB_1} \\
\overrightarrow{OB_1} = \overrightarrow{OB} + \overrightarrow{BB_1} \\
\overrightarrow{OC_1} = \overrightarrow{OC} + \overrightarrow{BB_1} \end{array}\right.
\]
より, $ \triangle A_1B_1C_1$ は $ \triangle ABC$ と平行であるから,
\begin{eqnarray*} S(B_1C_1D_1E_1) & = & S(\triangle A_1B_1C_1) - S(\triangle A_1D_1D_1) \\
& = & S(\triangle ABC) - a^2 S(\triangle ABC) \\
& = & (1 - a^2)S(\triangle ABC) \\
& = & \boxed{\frac {3}{2}(1 - a^2)}. \end{eqnarray*}
(10)
\begin{eqnarray*}
\left|\overrightarrow{B_1D_1}\right|^2
& = & \left|a\overrightarrow{B_1C_1} + (1 - a)\overrightarrow{B_1A_1}\right|^2 \\
& = & \left|a\overrightarrow{BC} + (1 - a)\overrightarrow{BA}\right|^2 \\
& = & \left|a^2\left|\overrightarrow{BC}\right|^2 + (1 - a)^2\left|\overrightarrow{BA}\right|^2 + 2a(1 - a)\overrightarrow{BC} \cdot \overrightarrow{BA}\right| \\
& = & \left|a^2\left|( - 1,2,0)\right|^2 + (1 - a)^2\left|( - 1,0,1)\right|^2 + 2a(1 - a) \cdot 1\right| \\
& = & \left|a^2(( - 1)^2 + 2^2 + 0^2) + (1 - a)^2(( - 1)^2 + 0^2 + 1^2) + 2a(1 - a) \cdot 1\right| \\
& = & \left|5a^2 + 2(1 - a)^2 + 2a(1 - a)\right| \\
& = & \left|5a^2 + 2a^2 - 4a + 2 - 2a^2 + 2a\right| \\
& = & \left|5a^2 - 2a + 2\right| \end{eqnarray*}
したがって, $ \left|\overrightarrow{B_1D_1}\right| = \boxed{\sqrt {5a^2 - 2a + 2}}$.

$ \boxed{6}$
(1)
$ 150$ 行は $ k$ が $ p$ で割り切れるかを判定しているので, 割り切れた場合は $ k$ を表示してして次の $ k$ の値の判定へ進むので, [甲] は $ \text{GOTO} \ 210$ であるので $ \boxed{E}$.
$ 200$ 行は $ k$ が条件を満たすか判定をした結果, 条件を満たさなかったためにこの行へ来たので, $ k$ の値を表示しないようにして次の $ k$ の値の判定へ進むので, [丙] は $ \text{GOTO} \ 230$ であるので $ \boxed{F}$.
$ 220$ 行は $ k$ を表示した次の行であり, 条件を満たす個数をカウントするために [丁] で $ \text{LET} \ \text{U}\equal{}\text{U}\plus{}1$ を実行しなければならないので $ \boxed{K}$.
(2)
$ 180$ 行は $ k$ が $ p$ で割り切れなかったため, $ k$ から $ p$ の倍数を引いた値が $ q$ で割り切れるか判定するための行である.
よって, $ k$ から $ p$ の倍数を引いた値を $ r$ としているので, $ r$ が $ q$ で割れるか判定する命令を作ればよい.
したがって, [乙] は $ \text{R}\minus{}\text{INT}(\text{R}/\text{Q})*\text{Q}\equal{}0$ を実行すればよいので $ \boxed{F}$.
(3)
$ 1 \sim 15$ のうち $ 3$ の倍数は $ 3$, $ 6$, $ 9$, $ 12$, $ 15$.
$ 1 \sim 15$ のうち $ 7$ の倍数は $ 7$, $ 14$.
$ 1 \sim 15$ のうち $ 3$ の倍数と $ 7$ の倍数の和で表せる物は $ 3\plus{}7\equal{}10$, $ 6\plus{}7\equal{}13$.
以上より条件を満たす整数の列は
\[ 3, \ \boxed{6}, \ 7, \ 9, \ \boxed{10}, \ 12, \ 13, \ 14, \ 15\]
(4)
$ 12 \sim 14$ までの数字は
$ 12\equal{}4\times3$
$ 13\equal{}2\times3\plus{}1\times7$
$ 14\equal{}2\times7$
と表すことができ,
$ 14$ より大きいすべての数字は $ 12 \sim 14$ のいずれかに $ 3$ の倍数を付け足せば必ず作ることが出来る.
よって, $ 12$ 未満のうち条件を満たす物は (3) より $ 5$ 個.
$ 12 \sim 100$ は $ 100\minus{}12\plus{}1\equal{}89$ 個の数字があるので, 合わせて $ 5\plus{}89 \equal{} \boxed{94}$.
(5)
$ 150$ 行は $ k$ が $ p$ で割り切れるかを判定している. ここで, 割り切れた場合は $ k$ を表示しないように次の $ k$ の値の判定へ進みたいので, [甲] を $ \text{GOTO} \ 210$ から $ \text{GOTO} \ 230$ に置き換える必要があるので $ \boxed{E}$.
(6-1)
$ v_k$ の値をカウントするために, まず始めに初期値を設定しなければならない.
よって, [戊] は $ \text{LET} \ \text{V}\equal{}0$ であるので $ \boxed{D}$.
(6-2)
条件を満たす個数をカウントするために [己] で $ \text{LET} \ \text{V}\equal{}\text{V}\plus{}1$ を実行しなければならないので $ \boxed{I}$.
(6-3)
総数として和 $ u_d\equal{}\sum_{k\equal{}1}^d{v_k}$ を計算するためには, $ u_{n\plus{}1}\equal{}u_n\plus{}v_{n\plus{}1}$ という計算をしていくことと同じなので, $ \text{LET} \ \text{U}\equal{}\text{U}\plus{}\text{V}$ を実行すればよいので $ \boxed{F}$.


geometry
circumcircle
geometric transformation
reflection
parallelogram
incenter
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, 0 the circumcircle of ABC, 2, 3 the B, C-mixtilinear incircles of ABC, B*, C* the point of contact of 2, 3 with 0,
B' the second point of intersection of  BB* with 2,
Pb the tangent to 2 at B' which is parallel to the tangent to 0 at B,
C' the second point of intersection of  CC* with 3,
Pc the tangent to 3 at C' which is parallel to the tangent to 0 at C,
and A" the meet point of 2 and 3.
Prove that AA" is the A-symmedian of ABC.
Sincerely
Jean-Louis

Dear Mathlinkers and Fang,
yes there is a typos in problem :

"A" is the meet point of Pb and Pc"

Sorry...
Sincerely
Jean-Louis

Tangents $ p_b, p_c$ of the mixtilinear incircles $ (2), (3)$ at $ B', C'$ cut $ AB, AC$ at $ P, Q.$ Lines $ BB'B^*, CC'C^*$ intersect at the external similarity center $ X_{56}$ of the circumcircle $ (O)$ and incircle $ (I).$ This is isogonal conjugate of the Nagel point $ Na.$ Thus, reflections $ t_b, t_c$ of the tangents $ p_b \equal{} B'P, p_c \equal{} C'Q$ of $ (2), (3)$ in the internal angle bisectors $ BI, CI$ are tangents of $ (2), (3)$ parallel to $ AC, AB,$ which allows to calculate $ BP, CQ.$ Let $ t_b$ cut $ BA, BC$ at $ A_2, C_2$ and let $ t_c$ cut $ CA, CB$ at $ A_3, B_3.$ Circles $ (2), (3)$ are the B-, C-excircles of similar $ \triangle A_2BC_2 \sim \triangle A_3B_3C.$ Let $ (2)$ touch $ BA, BC$ at $ F_2, D_2$ and let $ (3)$ touch $ CB, CA$ at $ D_3, E_3.$  $ D_2E_3F_2D_3$ is a parallelogram, its diagonals $ F_2D_2, D_3E_3$ cutting each other in half at the incenter $ I$, $ F_2E_3 \parallel D_3D_2 \equiv BC.$

$ \frac {BP}{CQ} \equal{} \frac {BC_2}{CB_3} \equal{} \frac {BD_2}{CD_3} \equal{} \frac {BF_2}{CE_3} \equal{} \frac {BA}{CA}$

This means that $ PQ \parallel BC.$ Let tangents of the circumcircle $ (O)$ at $ B, C$ intersect at $ A_0.$ $ \triangle A_0BC \sim \triangle A''PQ$ are centrally similar with similarity center $ A,$ having parallel sides, hence $ A, A'', A_0$ are collinear and the line $ AA''A_0$ is the A-symmedian.

Let $p_b \equiv A''B',p_c \equiv A''C'$ cut $AB,AC$ at $U,V,$ resp. B-mixtilinear incircle $(O_b)$ touches $BC,BA$ at $B_a,B_c$ and C-mixtilinear incircle $(O_c)$ touches $CB,CA$ at $C_a,C_b.$ As $B_aB_c,$ $C_aC_b$ cut each other in half at the incenter $I,$ then $B_aC_aB_cC_b$ is a parallelogram $\Longrightarrow$ $B_cC_b \parallel BC.$ If the B-excircle $(I_b)$ touches $BC$ at $X,$ then we have $\triangle ABC \cup (I_b) \sim \triangle (BC,BA,p_b) \cup (O_b)$ $\Longrightarrow$ $\tfrac{BU}{BB_c}=\tfrac{BC}{BX}=\tfrac{a}{s}$ and similarly we have $\tfrac{CV}{CC_b}=\tfrac{a}{s}$ $\Longrightarrow$ $UV \parallel B_cC_b \parallel BC.$ Thus if the tangents of $(O)$ at $B,C$ meet at $S,$ then $\triangle UVA''$ and $\triangle BCS$ are homothetic with center $A$ $\Longrightarrow$ $A'' \in AS,$ i.e. $AA''$ is the A-symmedian.


Let ABC be a triangle and M is an interior point. What do we do to draw a line (xMy) which divides the triangle ABC into two equal parts? :P  :roll:

That doesn't work: a line through the centroid parallel to one of the sides divides a triangle in the ratio $ 4: 5$ rather than into two equal halves.

[quote="bertram"]Connect $ M$ with the centroid ;)[/quote]
oh, it's not correct :?:


Find all $m\in N$ such that $(x+y)(y+z)(z+x)$ divides $x^m+y^m+z^m-(x+y+z)^m$

Not much to prove but if $(x+y)(y+z)(z+x)$ divides $f(x,y,z)=x^m+y^m+z^m-(x+y+z)^m$ then one of $(x+y), (y+z), (z+x)$ must be a zero of $f(x,y,z)$.  WLOG assume $(x+y)$ is a zero.  That is $f(x,-x,z)=0$.  So $x^m+(-x)^m=0$ which is true only if $m$ is odd.


algebra
polynomial
induction
algebra solved

With numbers u and v as input, a machine generates the number uv + v as output. In the first operation, the only number that can be used for input are 1,-1 and a fixed number c. In latter operations, numbers generated as output in preceding operations may also be used. Prove that for any polynomial f(x) = a_0x^n + a_1x^(n-1) + ... + a_n with integer coefficients, the machine can generate f(c) as output after a finite number of operations.

Let S be the set of the numbers which can be obtained after a finite number of steps.

Observe that if a is in S then so is a+1, by choosing u = a and v = 1. An easy induction shows that if a is in S then so is a+n for each integer n  \geq  0. (1)
Moreover, by choosing u = a+n and v = -1, we deduce that -a-n+1 is in S for each integer n  \geq  0.
In particular, it follows that Z c S.

By choosing u = n-1 and v = c, we deduce that for each integer n, we have nc is in S.
From above, it follows that nc + p is in S for all integers n and p, with p  \geq  0.
Thus, -nc-p+1 is in S too. That leads to nc+p is in S for all integers n,p.

Now, suppose that for a given k, all the numbers f(c) are in S where f is any polynomial with integers coefficients and degree k.
Let P(x) = a_(k+1)*x k+1  + ... a_1*x + a_0 be a polynomial with integer coefficients.
- Suppose that a_0 \geq 0 :
From the induction hypothesis, we know that
f(c) = a_(k+1)*c k + ... + a_2*c + (a_1 - 1) is in S.
Choosing u = f(c) and v = c, we deduce that P(c) - a_0 is in S.
From (1), since a_0 \geq 0, we deduce that P(c) is in S
- Suppose that a_0 < 0 :
From the induction hypothesis, we know that :
g(c) = -a_(k+1)*c k - ... - a_2*c - a_1 - 1 is in S.
Choosing u = g(c) and v = c, we deduce that
-a_(k+1)*c k+1 - ... - a_2*c 2  - a_1*c is in S.
Since -a_0 - 1 is a nonnegative integer, it follows from (1) that
h(c) = -a_(k+1)*c k+1 - ... - a_2*c 2  - a_1*c -a_0 - 1 is in S.
Thus, choosing u = h(c) and v = -1, we deduce that -(h(c)+1) = P(c) is in S.

Thus, in each case P(c) is in S. Since P is an arbitrary polynomial with integer coefficients and degree k+1, it follows that for each polynomial with integer coefficients and degree k+1, the number P(c) is in S, and we are done.

Pierre.


Thanks for the statistics. But you should also add some analysis. Should also scores with 5 and 6 points considered that the participant has solved the problem? Is there anything comparable to the [url=http://en.wikipedia.org/wiki/Flynn_effect]Flynn effect?[/url] For example I think there would be many more contestants these days solving 1988/6.  What about the increasing number of countries who participate in the IMO? Was about the improved access to olympiad materials online?

Source: http://www.imo-official.org

Here are the difficult problems under another measure of hardness.
On the following problems the students scored less than half a point on average.

[u]problem[/u]   [b][u]average number of points[/u][/b]
1996-5 ...  [b]0.491[/b]
2002-6 ...  [b]0.409[/b]
2003-3 ...  [b]0.405[/b]
2006-6 ...  [b]0.187[/b]
2007-3 ...  [b]0.304[/b]
2007-6 ...  [b]0.152[/b]
2008-6 ...  [b]0.260[/b]

[quote="ridgers"]I think IMO 2009 6 was really hard![/quote]

IMO 2009 #6 is not very hard when you give it to people while telling them it is Irish MO 2005 #2  :P

What does "Irish MO 2005 #2" really means?
A geometry problem?


calculus
integration
algebra
polynomial
calculus computations

Let $ y\equal{}(x\minus{}a)^{n}e^{bx}\sqrt{1\plus{}x^{2}}$, where $ n$ and $ a$ are constants and $ b$ is a non-zero constant. Show that

$ \frac{dy}{dx} \equal{} \frac{(x\minus{}a)^{n\minus{}1}e^{bx}q(x)}{\sqrt{1\plus{}x^{2}}}$,

where $ q(x)$ is a cubic polynomial.

Using this result, determine:

(i) $ \int \frac{(x\minus{}4)^{14}e^{4x}(4x^{3}\minus{}1)}{\sqrt{1\plus{}x^{2}}}\,dx$;

(ii) $ \int\frac{(x\minus{}1)^{21}e^{12x}(12x^{4}\minus{}x^{2}\minus{}11)}{\sqrt{1\plus{}x^{2}}}\,dx$;

(iii) $ \int \frac{(x\minus{}2)^{6}e^{4x}(4x^{4} \plus{} x^{3} \minus{} 2)}{\sqrt{1\plus{}x^{2}}}\,dx$.


A friend and I have taken a word game I used to play with some friends on paper and put it online!  The competition is pretty intense, and the WordTower site has over 40 registered users currently.  Feel free to make an account and [url=http://www.mogos.net/]join in[/url] once you read the rules.  We also have an IRC chat for the game at irc.webchat.org, channel #wordtower.

did you guys actually go and buy a domain name, and write that entire site yourself??

Wow. :) GJ.

Hmm... I signed up as kaoticity. This is pretty fun!  :)

Chong (my partner-in-crime) had the domain name already, but we did write the back-end for the site.  We had so much stuff that we had to go to a content management system called Drupal to organize it all.  I'm glad you guys are enjoying it!  Tell your friends!

A few of my friends like it too.
This would be the perfect game for a long bus ride!

I played it for several hours once on a long bus ride to a competition.  I had to use my laptop and a spreadsheet so we could keep track of the strings.

The site was down for a long time, but now it's back up! :D

I played just now!
But it must be made more popular!


calculus
integration
function
geometry
trigonometry
calculus computations

A function is defined by $f(x)=\int_{0}^{x}\frac{1}{1+t^{2}}dt.$

(1) Find the equation of normal line at $x=1$ of $y=f(x).$

(2) Find the area of the figure surrounded by the normal line found in (1),
the $x$ axis and the graph of $y=f(x).$

Note that you may not use the formula $\int \frac{1}{1+x^{2}}dx=\tan^{-1}x+Const.$

[quote="kunny"]
Note that you may not use the formula $\int \frac{1}{1+x^{2}}dx=\tan^{-1}x+Const.$[/quote] Then, how are we going to find out $f(1),$ even though we don't need the formula to find out $f'(x)$ ?


are you asking for help, or are you looking for a faster method, or are you giving us a challenge problem?

thats annoying....

$ 43\equal{}36\plus{}7\equal{}47_9$
$ 47\equal{}40\plus{}7\equal{}57_8$
$ 57\equal{}49\plus{}7\plus{}1\equal{}111_7$ i am obviously not going in the right direction. i am not sure i understand the question

it is a challenge, mathking123

abacadaea, remember that the 47 is still $ 47_9$

Wait--the first number is in base 10, right? Then we continue to convert the new number we get each time?

ok the number 43 when written in each base, is

$ 43_{10} \minus{}\minus{} \minus{} 47_9 \minus{} \minus{} \minus{} 53_8 \minus{}  \minus{} 61_7 \minus{} \minus{} 111_6 \minus{}  \minus{} 133_5 \minus{} \minus{} 223_4 \minus{} \minus{} 1112_3 \minus{}  \minus{} 101011_2$
so....starting from base 10, i added them and got 59

i probably added wrong lolz....too many numbers....

edit1:yep, thats the answer you posted...
edit2:i cant format this to look how i want it.....

[quote="asterday1"]take the number 43.
record the digits.
convert it to base 9.
record the digits.
convert to base 8.
record the digits.
repeat until you have recorded the binary (base 2) digits.
add all of the digits together.
what did you get?

[hide="answer"]59[/hide][/quote]

$ 43_{10}\equal{}47_9\equal{}53_8\equal{}61_7\equal{}111_6\equal{}133_5\equal{}223_4\equal{}1121_3\equal{}101011_2$

Now we add all these numbers:
$ 4\plus{}3\plus{}4\plus{}7\plus{}5\plus{}3\plus{}6\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}3\plus{}3\plus{}2\plus{}2\plus{}3\plus{}1\plus{}1\plus{}2\plus{}1\plus{}1\plus{}0\plus{}1\plus{}0\plus{}1\plus{}1\equal{}59$

@mathking123, i think one of your numbers was wrong (1112) but since u were using the same digits, the answer did not change

I converted all these numbers into their bases using their powers.


$\left(\forall\right) n\in N^*\ \ \wedge\ \ \left(\forall\right)\ x_k\in (0,1],\ k\in \overline {1,n} \Longrightarrow\boxed {\ \frac{1}{x_1x_2\ldots x_n}-x_1x_2\ldots x_n\ge \sum \limits_{k=1}^n \left(\frac{1}{x_k}-x_k\right)\ }$

Could you verify my interpretation of the formula?  I'm not familiar with all of those symbols.

It looks like "For all positive integers 'n', all 'x_k' in the range of (0,1], and all integers k in the range of 1 to n, **inequality**"

I see you understand very well !!! The symbol $\wedge$ means "and" . The symbol $\vee$ means "or".

I see you had a lot of fun not using any words :P. This problem looks extremely Jenseny.

[hide="Mathematical Induction (hint):"]

we have:
$\frac{1}{{ab}} - ab \ge \left( {\frac{1}{a} - a} \right) + \left( {\frac{1}{b} - b} \right) \Leftrightarrow \left( {1 - ab} \right)\left( {1 - a} \right)\left( {1 - b} \right) \ge 0$

and
it's easy to prove that :

$\frac{1}{{x_1 \cdots x_n }} - x_1 \cdots x_n \ge \left( {\frac{1}{{x_1 }} - x_1 } \right) + \cdots + \left( {\frac{1}{{x_n }} - x_n } \right)\mathop \Rightarrow \limits^?$

$\mathop \Rightarrow \limits^? \frac{1}{{x_1 \cdots x_n x_{n + 1} }} - x_1 \cdots x_n x_{n + 1} \ge \left( {\frac{1}{{x_1 }} - x_1 } \right) + \cdots + \left( {\frac{1}{{x_n }} - x_n } \right) + \left( {\frac{1}{{x_{n + 1} }} - x_{n + 1} } \right)$

[/hide]

It is O.K. ! In my opinion, this is the only solution. This problem was at the district stage of the National Olympiad from Romania.


I have started my preparations for mathematical contests nowadays,
and there is one problem that I never noticed before (maybe because
my ambitions are now higher than they were in past years, so I am
more serious in practising). Solutions?

When I am trying to solve problems from this forum, for which I
don't have a solution until somebody solve it (including myself)
there is nothing bad, but when I start solving problems from
textbooks with solutions, I don't know just how long should I stick
to one problem when I can't solve it.

If I look to a solution I lose possibility for enjoyment of solving
the problem myself, but then again, I maybe would never solve that
problem and so losing the valuable time that I can use for solving
some other problem.

Now the question is: What is your way for solving this kind of
problem. What time do you usually spend for solving one problem when
you have a solution for it on just 50 pages away. I know that this
depends a lot from the number of ideas you have, but what do you do
when you don't have any ideas left, do you look at the solution, go
to the next problem or just keep trying to solve the first problem.
In general, what is your way of preparing for contests from
textbooks?

Any kind of help would really mean a lot to me...

[quote="SpongeBob"]
If I look to a solution I lose possibility for enjoyment of solving
the problem myself, but then again, I maybe would never solve that
problem and so losing the valuable time that I can use for solving
some other problem.

Now the question is: What is your way for solving this kind of
problem. What time do you usually spend for solving one problem when
you have a solution for it on just 50 pages away. I know that this
depends a lot from the number of ideas you have, but what do you do
when you don't have any ideas left, do you look at the solution, go
to the next problem or just keep trying to solve the first problem.
In general, what is your way of preparing for contests from
textbooks?[/quote]

You should think about one problem exactly as long as you can enjoy solving it - not shorter, nor any longer. Your main target is not to win mathematical contests but to enjoy studying math - if you are bored working on some problem and you don't have any ideas how to approach it, look to a solution. But even if you think about some problem very long, and still it is very enjoyable for you - don't look!

I will think about it, thanks, and for now my main target is to go
to the IMO, so on the first place studding competition-like
mathematics (aka, if I don't need something for competitions, I
wouldn't lose my time in studding it) and problem-solving, i guess
later I will come to that level what you tried to tell me about:
main target is not to win mathematical contests but to enjoy
studying math. Don't get me wrong, I do enjoy it, very much :D

Looking for more opinions

What I do is try to solve a problem, then, if iI have no ideas left, I pass to another. After doing this 3 or 4 times, I return to the original one. Maybe I see sth different or I have another idea, but, if I don't have any idea, I look to the solution.

The problem is your enemy, you have 7 points to share with it and it is stupid to give him 4p for nothing like I did last year. Concentrate, analyze the problem and try to get as much overall knowledge as possible, then accordinhg to that, decide what to do, how to, get stipulated, and ATTACK! DESTROY!, that's that, but such thing takes much time, through you can be sure to solve everything, I learnt this at IMO slovenia, and after that I have met no problem to share my points with it, the biggest time is three days, thinking while sleeping too, it makes great sense... really... Unfortunately, I can't prove the effectiveness, because I have just one poor 18 at imo, but you'll get sure this way works next year!:) best regards!


Could you please either post it or pm to me? Thank you!!

Actually, if you look back on the Forum and go to 2008 National Mathcounts Countdown Round? thread, CuBeX posted it there. He also posted the rest of the 2008 National Rounds answer key w/ the National Cd Round. You can just download it there.


Espero que me ayuden con este problema.
Hallar todos los numeros primos tal que $p^{2}-p+1$ sea un cubo perfecto.
les agradeceria la solucion.

hallemos las congruencias modulo 8
por razones obvias, todo primo es congruente con 1, 3,5 ó 7 módulo 8 (excepto 2, el único par primo, y para 2, p^2-p+1 es 3)
las congruencias de los cubos perfectos modulo 8 son:
si p es congruente con 1 mod 8, p^3 es congruente con 1 mod 8
si p es congruente con 3 mod 8, p^3 es congruente con 3 mod 8
si p es congruente con 5 mod 8, p^3 es congruente con 5 mod 8
si p es congruente con 7 mod 8, p^3 es congruente con 7 mod 8
y de los cuadrados:
si p es congruente con 1 mod 8, p^2 es congruente con 1 mod 8
si p es congruente con 3 mod 8, p^2 es congruente con 1 mod 8
si p es congruente con 5 mod 8, p^2 es congruente con 1 mod 8
si p es congruente con 7 mod 8, p^2 es congruente con 1 mod 8

Entonces,
p^2-p+1 es congruente con p mod 8
p^2+1 es congruente con 2p mod 8

entonces, como p^2 es congruente con 1 mod 8,
2 es congruente con 2p mod 8
1 es congruente con 1 mod 2

entonces, tan sólo podrian ser los de la forma 8k+1

(8k+1)^2 -8k-1+1=64k^2+16k+1-8k=64k^2+8k+1=(8k)^3/(k+1)

Entonces
-o k+1 es uno, y k es 0 y p es 1
-o k+1 es cubo perfecto y divide a (8k)^3

pero (k,k+1) es 1, así que, como k+1 no divide a k^3, divide a 8^3=512

k+1 divide a 512, asíque k+1 = 1, 8, 64, 512.

k = 0, 1, 7, 63, 511
8k+1 = p = 1, 9, 57, 505, 4089

1 no es primo, 9 tampoco, 57 tampoco, 505 menos, y 4089 es múltiplo de 3

Entonces no hay!
perdon por lo largo :blush:  :blush:

Que te parece si pruebas con p=19???
Creo que has de tener algun error en tu prueba!

Sea $ p^2\minus{}p\plus{}1 \equal{} a^3$.
Tenemos que $ p(p\minus{}1)\equal{}(a\minus{}1)(a^2\plus{}a\plus{}1)$.
Entonces $ p|a\minus{}1$ o $ p|a^2\plus{}a\plus{}1$.

La primera opción nos da $ p\leq a\minus{}1$ y luego que $ p^2\minus{}p\plus{}1< a^2 < a^3$.

En la segunda, tenemos $ p|a^2\plus{}a\plus{}1$  y $ b\equal{}a\minus{}1|p\minus{}1$.
Si $ p.r \equal{} a^2\plus{}a\plus{}1\cong 3 mod (b)$ entonces $ r\cong 3 mod(b)$.
Si $ r\geq b\plus{}3 \equal{} a\plus{}2$ entonces $ p.r \geq (2a\minus{}1)(a\plus{}2) > a^2\plus{}a\plus{}1$.
Entonces debemos tener $ r\equal{}3$ y basta resolver el sistema
$ 3p\equal{}a^2\plus{}a\plus{}1$
$ p\minus{}1 \equal{} 3(a\minus{}1)$


I don't know it. In my book, connect 4 is like a super elaborate tic-tac-toe. Examining every case could bring you to the ultimate strategy, but it would take alot longer than tic tac toe. Because this "ultimate strategy" exists, I believe that no matter what, if both players play perfectly, the game ends in a draw. However, it may be that player 1 does have a way to win no matter what, but I stand by my idea of it being a dead game from the start (just like tic-tac-toe)

I don't know the strategy either (I've wondered about this before), but what are the dimensions of the board?  I believe a typical Connect-4 board is 7x6, but I've seen variations with others, like 8x8.

It doesn't seem like there would be [i]that[/i] many ways you could play out a full game of Connect-4, so it could probably be analyzed with a computer.  Using stacks, you could write a program in Java to play out every game, but I don't know how you could use that data to see if there's a perfect winning strategy for either player.

while playing out every game, it should analyze it and group it similarly to how we group organisms, by every move. If only the last moves differ, then the games should be stored in a set. Similarly, the set of sets of games that have the same moves up to the last two should be a group and so on up to the first move. You could then analyze the probability of winning for each strategy, and the one with the greatest probability would be "perfect"

oh, and considering computers can analyze things like texas hold-em, connect-4 shouldn't be that hard.

[quote="lightrhee"][quote="math92"]not me. tell us[/quote]
What I meant to say was that I read :read:  something that said there is a winning strategy. I don't know the strategy :(  and I couldn't find it there :mad: . So I was asking you to tell me the strategy.[/quote]
oh, i don't know hte stategy

Or even better, a guy who wrote his master thesis on the winning strategy.

I think there was an artifcle on this in the wolfram site, it has been proved.

[quote="lightrhee"]I found out that a winning strategy for first player exists. Does anyone know the strategy? :?[/quote]

Search for "James Allen".  Not only was the game solved but the strategy can be distilled into a small number of simple rules suitable for human play.

is there a website where you can go to play connect 4?


email
MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
number theory

Dr. Andreescu and Dr. Feng have been working hard on the curriculum for the upcoming summer. They have posted a .pdf summarizing the different topics and difficulty levels. Those of you who are already accepted into the camp should look at this .pdf carefully. It is worthwhile to think about what classes will interest you. Feel free to email Titu if you have any questions now (or post here or PM me).

Last year, they first day you signed up for a class with the guidance of Dr. Feng. So you don't have to make the choice immediately, but it would be best to have a good idea about what you want to take before you go.

Look here:

http://www.awesomemath.org/summer.shtml

http://www.awesomemath.org/AMSP_dailysch.pdf

Are they going to be enforcing the lights out this year???
Last year I really enjoyed the freedom to stay up and do math.

I enjoyed quite a bit of freedom having Oleg as a counselor (the Russian guy who stayed out until 4 A.M. every night).

I'm pretty sure all the mentors stayed up until 4 AM every night.

Math counts. Rofl what a great name.

(edit) oh dang...one of the sample problems (#12? listed as Mathcounts 2007 by Chengde Feng) isn't exactly Mathcounts 2007, it's sorta Mathcounts 2008...hmmmm I'm thinking that should be taken out?

also apparently there won't be contrived algebraic number theory this year, or at least they're not telling us there will be

Ha...I went to bed at 10:30 every night and got up at 6:30 to run...I suppose the last night everybody stayed up late haha

It is really better to get a consistent 8 hours of sleep that way you are awake. One or two hours more of math doesn't do much if you are too tired to concentrate.

[quote="Altheman"]I suppose the last night everybody stayed up late[/quote]Nope. I was too tired by the last night. :-D
[quote]One or two hours more of math doesn't do much if you are too tired to concentrate.[/quote]Yeah, that's why I didn't learn much. I recommend that everyone who's going this year go to sleep early most nights.

I think the problem is that people get riled up before they go to bed... Like right now I am not that tired, but I know that I am going to be tired tomorrow because I have to wake up early. However, when you are at the camp, I think that there is more motivation to use your time efficiently.

Dr. Andreescu didn't want the courses to be confused with school courses.

That's why they are 1.5, 2.5, 3.5, etc.

Also, you can sign up for different levels in different areas.

Just make sure you choose your level, and they will not force you into some level

because of your AIME or USAMO scores like last year.

DH

[quote="CatalystOfNostalgia"]also apparently there won't be contrived algebraic number theory this year, or at least they're not telling us there will be[/quote]
Contrived my ***, that was like the best part of the camp, it was a lot of fun, and not nearly as complicated as anyone made it out to be once you got the hang of it. Besides, solving diophantine equations is always a ton of fun.


modular arithmetic
number theory unsolved
number theory

Let $ m\ge 2$ ($ m\in N^*)$
$ A_m \equal{} \{ m \plus{} 1,3m \plus{} 2,5m \plus{} 3,...\}$

Prove that $ \forall n\geq 2$     $ \exists a\in N$ ( $ 1\leq a\leq m$)such that
$ 2^a\in A_m$ or $ 2^a \plus{} 1\in A_m$

We need to show that for every $ m\geq 2$, there exists $ a$, $ 1\leq a\leq m$, such that $ 2^a \equal{} (2k \minus{} 1)m \plus{} k$ or $ 2^a \plus{} 1 \equal{} (2k \minus{} 1)m \plus{} k$. This is equivalent to showing that there exists $ a$, $ 1\leq a \leq m$, such that $ 2^a \equiv m \plus{} 1 \pmod{2m \plus{} 1}$ or $ 2^a \equiv m \pmod{2m \plus{} 1}$. Note that because $ (2,2m \plus{} 1) \equal{} 1$,

(1) If there is a $ b\leq m$ such that $ 2^b \equiv 1 \pmod{2m \plus{} 1}$, then $ 2^{b \minus{} 1} \equiv m \plus{} 1 \pmod{2m \plus{} 1}$.

(2) If there is a $ b\leq m$ such that $ 2^b \equiv \minus{} 1 \pmod{2m \plus{} 1}$, then $ 2^{b \minus{} 1} \equiv m \pmod{2m \plus{} 1}$.

So it suffices to show that (1) or (2) must occur.

Let $ \phi(2m \plus{} 1) \equal{} d$, and let $ e$ be the order of 2 modulo $ 2m \plus{} 1$. By Euler's extension to FLT, we know $ 2^d \equiv 1 \pmod{2m \plus{} 1}$, and we know that $ e|d$.  so if $ e < d$, then $ e\leq \frac {d}{2}\leq m$ (because of the obvious fact that $ \phi(2m \plus{} 1) < 2m \plus{} 1$). So (1) is true for $ b \equal{} e$.

If $ e \equal{} d$, then $ 2, 2^2, ..., 2^d$ form a reduced residue system modulo $ 2m \plus{} 1$. Thus there exists $ j$, $ 1\leq j\leq d$, such that $ 2^j\equiv \minus{} 1 \pmod{2m \plus{} 1}$. Then $ 2^{2j}\equiv 1 \pmod{2m \plus{} 1}$, and it follows that $ 2j \equal{} e \equal{} d < 2m \plus{} 1$, so $ j\leq m$. So (2) is true for $ b \equal{} j$.


If $ a,b,c\in\mathbb{R}$, prove that $ (a^2\plus{}b^2\plus{}c^2)^2 \ge a^3b \plus{} b^3c \plus{} c^3a$.

[quote="Johan Gunardi"]If $ a,b,c\in\mathbb{R}$, prove that $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge a^3b \plus{} b^3c \plus{} c^3a$.[/quote]
Maybe  $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a^3b \plus{} b^3c \plus{} c^3a)$ $ ?$
If so, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6026

Yes it should be $ (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}\ge 3(a^{3}b\plus{}b^{3}c\plus{}c^{3}a)$.

My friend made a wonderful manipulation of this inequality into this form:
\[ (a^2\minus{}c^2\minus{}2ab\plus{}bc\plus{}ac)^2\plus{}(b^2\minus{}a^2\minus{}2bc\plus{}ab\plus{}ac)^2\plus{}(c^2\minus{}b^2\minus{}2ac\plus{}ab\plus{}bc)^2\ge0.\] :D


[quote="April"]For a positive integer $ n$, consider the equation $ \frac{1}{x\minus{}1}\plus{}\frac{1}{4x\minus{}1}\plus{}\cdots\plus{}\frac{1}{k^2x\minus{}1}\plus{}\cdots\plus{}\frac{1}{n^2x\minus{}1}\equal{}\frac{1}{2}$.
(a) Prove that, for every $ n$, this equation has a unique root greater than $ 1$, which is denoted by $ x_n$.
(b) Prove that the limit of sequence $ (x_n)$ is $ 4$ as $ n$ approaches infinity.[/quote]
see problem 4.2.50 page 201 at http://www.scribd.com/doc/63634668/Selected-Problems-of-the-Vietnamese-Mathematical-Olympiad-1962-2009

Let $f_n(x)=\sum_{k=1}^n\frac{1}{k^2x-1}$. Between $x\in (\frac{1}{(k+1)^2},\frac{1}{k^2}$ $f(x)$ decrease from $+\infty$ to $-\infty$. Therefore it had $n-1$ roots less than $1$. For $x>0$ it decrease from $+\infty$ to $0$. Therefore for $a>0$ it had one root $x>1$. Because this equation equavalent to polinomial with degree $n$, they are all roots $f(x)=a$. $f(4)=\sum_{k=1}^n \frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}=\frac{1}{2}-\frac{1}{2(n+1)}$.
Therefore $x_n=4-\frac{1}{2(2n+1)f'(4)}+O(n^{-2})\to 4.$


calculus
integration
logarithms
calculus computations

Evaluate $ \int \sqrt \frac{{e^x}\minus{}1}{{e^x}\plus{}1} d\ x$

Put $ t \equal{} \sqrt {\tanh (\tfrac{x}{2})} \equal{} \sqrt {\frac {e^x \minus{} 1}{e^x \plus{} 1}}$. Then $ \frac {dt}{dx} \equal{} \frac {1 \minus{} \tanh^2 (\tfrac{x}{4})}{4\sqrt {\tanh (\tfrac{x}{2})}}$, so we have $ \sqrt {\tanh (\tfrac{x}{2})} \, dx \equal{} \frac {4t^2}{1 \minus{} t^4} \, dt$. Then

$ \int \sqrt {\frac {e^x \minus{} 1}{e^x \plus{} 1}} \, dx \equal{} \int \frac {4t^2}{1 \minus{} t^4} \, dt \equal{} 2 \int \left( \frac {1}{1 \minus{} t^2} \minus{} \frac {1}{1 \plus{} t^2} \right) \, dt$

The rest of calculation is straightforward.

hello, by setting $ t^2=\frac{e^x-1}{e^x+1}$ we get ${ x=\ln(\frac{t^2+1}{1-t^2}})$ and $ dx=\frac{4t}{t^4-1}dt$ and your integral is
$ 4\int\frac{t^2}{t^4-1}dt$.This is easily to solve. Note that
$ \frac{t^2}{t^4-1}=\frac{1}{4(t-1)}-\frac{1}{4(1+t)}+\frac{1}{2(t^2+1)}$.
Sonnhard.


Let $ a,b,c,d>0$ s.t. $ abcd\equal{}1$.Prove that:
$ (1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)(1\plus{}d^2)\geq (a\plus{}b\plus{}c\plus{}d)^2$

[quote="Inequalities Master"]Let $ a,b,c,d > 0$ s.t. $ abcd \equal{} 1$.Prove that:
$ (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2)(1 \plus{} d^2)\geq (a \plus{} b \plus{} c \plus{} d)^2$[/quote]
$ abcd \equal{} 1\Rightarrow$ two in four number are $ \ge$ or $ \le$ 1

assume two number are b and d we have

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

we need prove that ($ c \plus{} a \plus{} bd \plus{} 1)^2\ge (a \plus{} b \plus{} c \plus{} d)^2$

$ \Leftrightarrow (b \minus{} 1)(d \minus{} 1)\ge 0$ which is true by assume :wink:

This problem can be prove by Mixing Variable :)

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84043


Answer this question and explain how you get the answer. Thank you.

Question: An oil change costs $16 and a cooling system flush is $61. Find the total costs for both services.

Note: Please disable BBCode in order to prevent "messy words."

[hide]I got $\$$77. How did I get that? I added them together.[/hide]

Very easy
[hide] the total cost is 77 dollars.[/hide]

thats trivial.  what do you mean by "messy words"?  $Do you mean this?$

Why would someone post a question like that?

trolling (doubtful) but honestly i can't figure it out either.

[quote="Anonymous52111"]Answer this question and explain how you get the answer. Thank you.

Question: An oil change costs $16 and a cooling system flush is$61. Find the total costs for both services.

Note: Please disable BBCode in order to prevent "messy words."[/quote]

[hide]trick question? becase it is 16+61=77[/hide]

Since so many people got the same answer, it has to be right. :D

[quote="benlowkh"][hide]Easy, 61+16 is 77   :lol:[/hide][/quote]
I don't suggest disabling the BBCode. I don't know why Anonymous52111 said that.

[quote="t0rajir0u"][quote="Anonymous52111"]Note: Please disable BBCode in order to prevent "messy words."[/quote]

Or you could just use \text{}.[/quote]
It isn't suggested that we use the \text command because it takes a while for the pages to load.

this should be reagarded as  :spam:  :spam:  :spam:
but anyway
$\$$77


Given two sequences of positive integers $ a_{1}, a_{2}, ... a_{n}, b_{1}, b_{2}, ... b_{n}$ such that $ a_{i}$ [url=http://en.wikipedia.org/wiki/Majorization]majorizes[/url] $ b_{i}$, prove or disprove that

$ \frac{\prod a_{i}!}{\prod b_{i}!}$

is an integer.

The relevant entry is [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]here[/url].  ;)

[b]Follow-up:[/b]  The fact that the Catalan numbers $ C_{n}\equal{}\frac{(2n!)}{(n\plus{}1)! n!}$ are integers cannot be proven from the above exercise.  Can you generalize the above exercise in such a way that it can?

I posted about this before:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=115296[/url]

It's not hard to come up with counter-examples.

Well...even though this conjecture is not true, you could still use it...

2n,1 majorizes n+1,n


Let a,b,c,d,p,q  the sides and diagonals of a quadrilateral.

Prove that

max(a,b,c,d,p,q)/min(a,b,c,d,p,q) \geq  \sqrt 2

We just use these 2 results:

(1) One of the angles A, B, C, D is >=90

(2) For a triangle XYZ s.t. <XYZ >=90 we have max(XZ/XY, XZ/YZ)>= \sqrt2


[quote="anirudh"][color=black]At the beginning of a study 24 fish are caught, tagged and released back into a large
pond containing fish. A few days later, 19 fish are caught from the pond and three of
them have tags. Using this ratio, how many total fish would you expect to be living in
the pond?[/size][/quote]

[hide]
3:16
3/16=24/x
3x=24(16)
x=8(16)=128
128+24 = 152?
[/hide]

[quote="anirudh"][size=150][b][color=black]At the beginning of a study 24 fish are caught, tagged and released back into a large
pond containing fish. A few days later, 19 fish are caught from the pond and three of
them have tags. Using this ratio, how many total fish would you expect to be living in
the pond?[/color][/b][/size][/quote]
[hide="solution"]3/19=24/x
                       24/3=8, so 8*19=152 fish :o [/hide]

[quote="math92"]oops, what was i thinking.  so 128 is my answer and all of you got...152[/quote]

You forgot to add back the 24 with the tags, which makes 152.

thanks ccy.  i was wondering what i did wrong :)


When the $n's$ dice are thrown at one time, find the probability such that the sum of the cast is $n+3.$


in a class of 20 students, five like to go camping and ten like to ride bikes. What is the greatest possible number of students that like to do neither? i thought the answer was 5, but apparently not~

I am not quite sure if this is right
[hide] Say that 5 of the students like to go camping [b]and[/b] ride bikes.  This means that no students like to [b]only[/b] go camping and 5 students like to [b]only[/b] ride bikes.  The left over people who don't like biking nor camping is 20-5-5-0=$10$ [/hide]

maybe it is the right answer. it makes sense.

yes, because you have 20 kids, and 10 ride bikes, now of that ten u can have 5, and there's 10 kids left

jorian


Fie p,q numere reale pozitive astfel incat p+q=1. Sa se arate ca:
a/(pb+qc) + b/(pc+qa) + c/(pa+qb) >=3 ,
oricare ar fi numerele pozitive a,b,c .

Cauchy: $\sum \frac{a}{pb+qc}\sum a (pb+qc) \geq \left( \sum a \right)^{2}$.

Stim ca $\left( \sum a \right)^{2}\geq 3 \left( ab+bc+ca \right)$, deci \[\sum \frac{a}{pb+qc}\geq \frac{3(ab+bc+ca)}{\sum a(pb+qc)}= 3 .\] Egalitatea se obtine pentru $a=b=c$.

[hide]
  Notam cu $\ LHS$ membrul stang al inegalitatii.Folosind Cauchy in forma lui Engel, obtinem ca $\ LHS\geq\frac{(a+b+c)^{2}}{ab(p+q)+bc(p+q)+ca(p+q)}$, care conform ipotezei e echivalenta cu $\ (a+b+c)^{2}\geq\ 3(ab+bc+ca)$, adica $\ a^{+}b^{2}+c^{2}\geq\ ab+bc+ca$.[/hide]

ai zis ca ai folosit Cauchy in forma lui Engel .eu aceeasi inegalitate o stiam sub numele de inegalitatea lui Titu Andreescu .

mai, stii cum sta treaba?Unii zic la inegalitatea asta Lema Titu, Bergstrom,Cauchy in Engel's form,...s-au furat unii pe altii.Inegalitatea asta e pur Cauchy, sau CBS..Mai e o inegalitate in situatia asta.Inegalitatea lui Karmata..Cezar Lupu a zis ca de fapt ea a fost descoperita de tripleta de aur $\text{Hardy-Littlewood-Polya}$.Mai apoi le-a furat-o Karamata asta.. :mad:
[i]Concluzie[/i]Inegalitatea asta se numeste $\text{Cauchy}$. :)

Pardon, cand am zis eu ca inegalitatea Hardy-Littlewood-Polya a fost furata de Karamata? :maybe:
Ea a fost prima oara publicata de cei trei, iar dupa un an a publicat-o si Karamata. Probabil ca el nu stia ca o descoperisera cei trei inainte. Oricum asta da inegalitate, nu Cauchy ( Engel, Titu, Bergstrom, cutare form). :wink:


Has anyone heard of this book? Is it any good?

Titu Andreescu has put together a lot of interesting books with Olympiad problems recently, and I've been wondering if Mathematical Olympiad Treasures might be the best "entry level" book for that level of math competitions. I, too, would be interested in hearing from readers of the book what they think of it.


[quote="Ceva Theorem"]Suppose that the population of a certain rabbit colony doubles every year. If you start with two rabbits, find and prove a formula that predicts the rabbit population after n years. Prove using induction.[/quote]

[hide]$1$st year= $2$ rabbits
$2$nd year= $4$ rabbits
$3$rd year= $8$ rabbits

number of rabbits$=2^n$

I have to go so I'll finish this post later[/hide]

[quote="Ceva Theorem"]Suppose that the population of a certain rabbit colony doubles every year. If you start with two rabbits, find and prove a formula that predicts the rabbit population after n years. Prove using induction.[/quote]
Uh...................................
[hide]$2^{n+1}$
$2^{0+1}=2$

...
$2^{n+1}\cdot 2=2^{n+2}$
$2^{(n+1)+1}=2^{n+2}$...does this work?[/hide]

[quote="mathgeniuse^ln(x)"][quote="Ceva Theorem"]Suppose that the population of a certain rabbit colony doubles every year. If you start with two rabbits, find and prove a formula that predicts the rabbit population after n years. Prove using induction.[/quote]

I can't use mathematical induction to prove this, but it is pretty trivial without induction.

For your signature, don't you need a beginning quote in your statement.[/quote]

The book I got it from said it was possible. I don't know how, but I'm still working on figuring it out.

For the signature, it was a typo. i'm too lazy to change it  :D

Well, it's simpler not to use induction, just try the first times and you'll see the answer......


let x,y be real and x,y such that : $xy(x+y)=x^{2}-xy+y^{2}$
  find maximum of expression$P =\frac{1}{x^{3}}+\frac{1}{y^{3}}$

\[A=\left(\frac{1}{a}+\frac{1}{b}\right)^{2}\] Dividing both sides by $a^{2}b^{2}$.

[quote="romano"]Complete your solutions !
 From $xy \geq \frac{x+y}{4}$ , we can't conclude $P \leq 16$[/quote]
    To Soarer :
  Because if $xy <0$ and $x+y <0$ , then divide two sides for $xy$ , the sign'll be reserved .

oh, romano says right .
    but i think this pro have many solution than only one solution


Horizontal beam AB is 30 meters long. Let D be a point on AB such that AD=10m and BD=20m. Let C be a point exactly 24m under D. There are cables AC and BC. There is a load of 5000N at C. Find the tension on cables AC and BC using vectors.

Perhaps this would be better suited in the physics forums.

[hide="Solution"]
The vertical component of the tension needed by the strings is 5000 N. Using the information given, find the angles that the strings make with the vertical, being $22.6^\circ, 39.8^\circ$ Then, since the horizontal component of their tensions must be equal, we'll call it $z$. Also, we'll define $x$ as the vertical component of the first string. Then we set up a system of equations as such:

$\tan(22.6)x = z$
$\tan(39.8)(5000-x) = z$

Solving this system we get $x = 3334.2 N$
Therefore, the vertical components are $3334.2, 1665.6$ and the horizontal components are $1387.9 N$ Using Pythagorous, we get the tensions in the string as:

$AC \approx 3611.5 N$
$BC \approx 2168.1 N$

Hopefully my arithmetic is correct.
[/hide]


Just a quick query, is each term in the top (and bottom will be the same) 1/(n! * (n,1)^n) or is it (1/n!)*(n,1)^n. I would guess without brackets it would have to be the second version?


logarithms
Euler
limit
real analysis
real analysis solved

Find the sum of the series

$ \sum_{n=2}^{\infty} (-1)^n \frac{\ln n}{n}$ !

[quote="Nené"]The sum is $\displaystyle \ln 2 \left( \gamma - \frac {\ln 2} {2}\right) $

$ \gamma $ is the Euler constant.[/quote]

That's very nice to know, but how do you prove this, that means, how do you deduce this result? Your answer appears to come from nowhere.


That's a standard way to do it. Another way is to get help of complex numbers, although it isn't needed here...

[quote="Slizzel"]Well i finally dit it :D!! Why you didn't told me what you used?! I finally descovered that $\sin\alpha\cos\beta=\frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]$   :roll: I use this for the first time :)[/quote]
Yes, product to sum. There are more formulas about it.
sin*sin
sin*cos
cos*cos


integration
real analysis
real analysis unsolved

how to prove this ?

$ 2\int_{\minus{}1}^{1}\sqrt{1\minus{}t^{2}}dt\equal{}\int_{\minus{}1}^{1}\frac{1}{\sqrt{1\minus{}t^{2}}}$

Integrate by parts: $ \int_{ \minus{} 1}^1 \sqrt {1 \minus{} t^2}\,dt \equal{} \int_{ \minus{} 1}^1 \frac {t^2}{\sqrt {1 \minus{} t^2}}\,dt$ and add to this the identity $ \int_{ \minus{} 1}^1 \sqrt {1 \minus{} t^2}\,dt \equal{} \int_{ \minus{} 1}^1 \frac {1 \minus{} t^2}{\sqrt {1 \minus{} t^2}}\,dt$.

[quote="luckman168"]how to prove this ?

$ 2\int_{ \minus{} 1}^{1}\sqrt {1 \minus{} t^{2}}dt \equal{} \int_{ \minus{} 1}^{1}\frac {1}{\sqrt {1 \minus{} t^{2}}}dt$[/quote]

$ 2\int_{ \minus{} 1}^{1}\sqrt {1 \minus{} t^{2}}dt \equal{} \int_{ \minus{} 1}^{1}\frac {1}{\sqrt {1 \minus{} t^{2}}}dt\equal{}\pi$


integration
calculus
real analysis
real analysis solved

Compute the following primitives
A= \int ( \sqrt (sin^4x+cos^4x) )/(sin^3x.cosx) dx
B= \int ( \sqrt (sin^4x+cos^4x) )/(sinx.cos^3x) dx
x \in (0; \pi /2)

A = \int  tgx/ \sqrt (1+tg^4x).dx/cos^2x +  \int cotg^4x/ \sqrt (1+cotg^4x).dx/sin^2x

In the first integral we do the change variable y = tgx
\int  tgx/ \sqrt (1+tg^4x).dx/cos^2x =  \int y/ \sqrt (1+y^4) dy = 1/2.ln(y^2 +  \sqrt (1+y^4) )+K = 1/2.ln( tg^2x + \sqrt (1+tg^4x) ) +K

In the second integral we do the change variable w = cotgx
 \int cotg^4x/ \sqrt (1+cotg^4x).dx/sin^2x = -  \int w^3/ \sqrt (1+w^4)dw = -1/2. \sqrt (1+w^4) + K = -1/2.\sqrt (1+cotg^4x) + K

A = 1/2.ln( tg^2x + \sqrt (1+tg^4x) ) -1/2.\sqrt (1+cotg^4x) + K

For B use the same idea.


Let ABC is a  triangle,prove that:
$ \frac {r_a}{r_b \plus{} r_c} \plus{} \frac {r_b}{r_c \plus{} r_a} \plus{} \frac {r_c}{r_a \plus{} r_b}$
$ \geq \frac {a \plus{} r_a}{b \plus{} c \plus{} r_b \plus{} r_c} \plus{} \frac {b \plus{} r_b}{c \plus{} a \plus{} r_c \plus{} r_a} \plus{} \frac {c \plus{} r_c}{a \plus{} b \plus{} r_a \plus{} r_b}$
$ \geq \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}.$

Note:

left<=>
$ ( \minus{} 8R \plus{} r)s^3 \plus{} ( \minus{} 32R^2 \minus{} 4Rr \plus{} r^2)s^2 \plus{} (32R^3 \plus{} r^3 \plus{} 32R^2r \plus{} 10Rr^2)s \plus{} 128R^4 \plus{} 160R^3r \plus{} 14Rr^3 \plus{} 72R^2r^2 \plus{} r^4\geq 0.$

as degree(s)=odd number,
Thus,Sense to solve this problem.
BQ

right<=>

$ \minus{}4s^4\minus{}5s^3r\plus{}(12Rr\plus{}16R^2\plus{}5r^2)s^2\plus{}3r(r\plus{}2R)(4R\plus{}r)s\plus{}r(r\plus{}2R)(4R\plus{}r)^2\geq0.$

The first to make such a preliminary judge:
Estimated to be able to solve this problem not many people,
But nice!
BQ


I hope i post it in the right section.
find thevalue of
\[ \sum_{n=0}^{\infty} \left ( \dfrac{(-1)^n}{n!} \left ( \sum_{m=1}^{\infty}\dfrac{2^{n-m}}{m} \right )^n \right )  \]
 ;)

EDIT: I forget the $(-1)^n$ so that this problem consider to be little hard. but now it's fixed. Now it is easy I think.(it is enough to post this in intermediate forum). sorry  :oops:

This appears when I'm doing my homework about multivariant discrete prob . the result seem reasonable and  true according to my homework exercise. But my teacher told me that I got too serious and complicating the problem. since it is just an ordinary hypergeometry  :P


calculus
integration
inequalities
function
derivative
real analysis
real analysis unsolved

Let $f$ is differentiable on $[-1,1]$,such taht $|f'(x)| \le M$ and $\int_{-a}^af(x)dx=0$,Prove:
\[ |\int_{-1}^1f(x)dx| \le M(a-a^2)  \]

[b]Remark[/b]I am not sure about this problem,the oringal problem in my book is to prove:
\[ |\int_{-1}^1f(x)dx| \le M(1-a^2)  \]
I prove it by $f(a) \le Ma$,but what about this problem.Thanks

$f$ does not have to be dofferetiable on the whole interval $[-1,1]$? M is the upper boun of $f'$ or of the $|f'|$? Can you post a proof of the original problem?  Maybe it will help us to find a soltuion or a counterexample.

Ok,I am sorry,and I edited the post.
My proof for:\[ |\int_{-1}^1f(x)dx| \le M(1-a^2)  \]
I proved that:$|f(a)| \le Ma$
otherwise:for example:$f(a)>Ma$ then we have $f(x) >Ma-(a-x), x \in [-a,a]$(because $|f'(x)| \le M$)
so,$\int_{-a}^a f(x)dx>0$,contradiction.
thus we proved that $|f(a)| \le Ma$.
and\[ |\int_a^1 f(x) dx| \le |\int_a^1 (Ma+M(x-a))dx| \le \frac 1 2M(1-a^2)  \]
similar:\[ |\int_{-1}^{-a} f(x) dx| \le \frac 1 2M(1-a^2)  \]
So we prove the inequality.
But the equality can't be hold.(or tend to)

[quote="zhaobin"]Let $f$ is differentiable on $[-1,1]$,such taht $|f'(x)| \le M$ and $\int_{-a}^af(x)dx=0$,Prove:
\[ |\int_{-1}^1f(x)dx| \le M(a-a^2)  \]
[/quote]

I think that your aproximation does't work. Think about $a=0$. Even if $a\neq 0$, try the function $f(x)=Mx$ on $[-a,1]$ and $f(x)= -Ma-Mx$ on $[-1,-a]$. We have that  $|\int_{-1}^1f(x)dx|$ goes to $M$ when $a$ is going to 0 and since the right side expresion is going to $0$ when $a$ is going to $0$. I know that this cfuntion is not differentiable in $-a$ but it can be slightly modified such that it becomes differentiable in $-a$ ant it's intgral do not differ by so much.

A better aproximation I think is this:  $|\int_{-1}^1f(x)dx| \le M(1-\frac{a^2}{2}-\frac{a}{2})\le M(1-a^2)$ I do not know if it holds but I can construct a sequence of functions which tends to this aproxiamtion. These function are looking like $g(x)=Mx-\frac{a}{2}$ on $[0,1]$ and $g(-x)=g(x)$. Actually my sequence of functions is $G_n$ which are the antiderivatives of the functions $g_n=-1$ on $[-1,-\frac 1n]$, $g_n(x)=1$ on $[\frac 1n,1]$ and linear on $[-\frac 1n,\frac 1n]$.

I hope that my remarks are usefull. I hope that you will get the ideea and complete the detailes by youself. If you have any questions I'll be around here.  ;)

Thanks very much.
(Sorry,$0<a<1$)
[quote="xirti"][quote="zhaobin"]Let $f$ is differentiable on $[-1,1]$,such taht $|f'(x)| \le M$ and $\int_{-a}^af(x)dx=0$,Prove:
\[ |\int_{-1}^1f(x)dx| \le M(a-a^2)  \]
[/quote]

I think that your aproximation does't work. Think about $a=0$. Even if $a\neq 0$, try the function $f(x)=Mx$ on $[-a,1]$ and $f(x)= -Ma-Mx$ on $[-1,-a]$. We have that  $|\int_{-1}^1f(x)dx|$ goes to $M$ when $a$ is going to 0 and since the right side expresion is going to $0$ when $a$ is going to $0$. I know that this cfuntion is not differentiable in $-a$ but it can be slightly modified such that it becomes differentiable in $-a$ ant it's intgral do not differ by so much.
;)[/quote]
But if you do like that,you can't satisfy the condition $|f'(a)| \le M$.

[quote="xirti"]

A better aproximation I think is this:  $|\int_{-1}^1f(x)dx| \le M(1-\frac{a^2}{2}-\frac{a}{2})\le M(1-a^2)$ I do not know if it holds but I can construct a sequence of functions which tends to this aproxiamtion. These function are looking like $g(x)=Mx-\frac{a}{2}$ on $[0,1]$ and $g(-x)=g(x)$. Actually my sequence of functions is $G_n$ which are the antiderivatives of the functions $g_n=-1$ on $[-1,-\frac 1n]$, $g_n(x)=1$ on $[\frac 1n,1]$ and linear on $[-\frac 1n,\frac 1n]$.

I hope that my remarks are usefull. I hope that you will get the ideea and complete the detailes by youself. If you have any questions I'll be around here.  ;)[/quote]
I think you may mean $g(x)=Mx-M\frac{a}{2}$ to satisfy:$\int_{-a}^a f(x)dx=0$.
then the value is what I want to desired :)
(Sorry,if I am wrong)

1) zhaobin  "But if you do like that,you can't satisfy the condition ..."

The change consists of soothing the differece between the dervative at left in $-a$ and the derivative at right in $-a$. This can be done by modifying $f$ on an interval $[-a-\epsilon,-a]$ such that on that interval $f'$ is increasing from $M$ to $-M$ and, so, it's modulus do not exceed the limit $M$. The method of modifying is the same with that presented later in my previous post. (I'm not very sure that this modification works, but someone more experienced certainly can fix the details :D ).

2) Yes, the function is actually $g$ is $g(x)=Mx-M\frac a 2$
But I think that $\int_{-1}^1g(x)dx=2\int_0^1g(x)dx=2\int_0^1Mxdx-2\int_0^1 M\frac a 2=M(1-a)$ :D
Neither $M(1-\frac a 2-\frac{a^2} 2)$ nor $M(a-a^2)$ but still better aproximation than that of the problem :lol:

oh,yes,I made a mistake when I was calculuting,sorry for boring you. :D
And our new problem is:
Let $f$ is differentiable on $[-1,1]$,such taht $|f'(x)| \le M$ and $\int_{-a}^af(x)dx=0$,Prove:
\[ |\int_{-1}^1f(x)dx| \le M(1-a)  \]
Hope somebody can help us,thanks :)

It was my pleasure :lol: For proveing the new problem you must modify your previous proof:

Prove that $|f(-a)+f(a)|\le Ma$, otherwise $\int_{-a}^af(x)dx\ge 0$ (I think that you can get this result). I'll let you the pleasure of proving your problem by yourself.  ;)

[quote="xirti"] $|f(-a)+f(a)|\le Ma$, [/quote]
Thanks very much  :)

[quote="xirti"]You are welcomed ;)  Post your solution to complete this topic.[/quote]
Ok,thanks,and sorry for late:
First I will prove:$|f(-a)+f(a)| \le Ma$ by your hint.(It is based on $|f(a)|,f(-a)| \le Ma$,see #3)
If $|f(-a)+f(a)| > Ma$,wlog $f(-a)+f(a)>Ma$ thus we get(because  $|f(-a)| \le Ma$:
\[ \begin{cases} f(x)>f(a)-(a-x)M,x \in \left[\frac{Ma-2f(a)}{M},a\right] \\ f(x)>f(-a)-(-a-x)M,x \in \left[-a,\frac{Ma-2f(a)}{M}\right] \end{cases}  \]
Thus we have:$\int_{-a}^a f(x)dx >0$,contradiction.thus $|f(-a)+f(a)| \le Ma$ .
then we have:$|f(x)+f(-x)| \le |f(a)+M (x-a)+f(-a)| \le Ma+M(x-a),x \in [a,1]$ thus:
\[ |\int_{-1}^1f(x)dx|=\int_a^1 |f(x)+f(-x)| dx \le M(1-a)  \]
So we are done. :)
(In fact,here we can prove  $|f(a)|,f(-a)| \le Ma$ directly,I just take it for convenience :) )

[quote="zhaobinFirst I will prove:$|f(-a)+f(a)| \le Ma$ by your hint.(It is based on $|f(a)|,f(-a)| \le Ma$,see #3)
If $|f(-a)+f(a)| > Ma$,wlog $f(-a)+f(a)>Ma$ thus we get(because  $|f(-a)| \le Ma$:
\[ \begin{cases} f(x)>f(a)-(a-x)M,x \in \left[\frac{Ma-2f(a)}{M},a\right] \\ f(x)>f(-a)-(-a-x)M,x \in \left[-a,\frac{Ma-2f(a)}{M}\right] \end{cases}  \]
Thus we have:$\int_{-a}^a f(x)dx >0$,contradiction.thus $|f(-a)+f(a)| \le Ma$ .
[/quote]
My aproach is using the fact that:
$\int_{-a}^af(x)dx\ge \int_{-a}^a \max(f(a)-(a-x)M,f(-a)-(-a-x)M)$.
Anyway it was an entertaining problem ;)


All right I'm gonna host a AIM math jam at 8:30 Eastern (michigan time) on Friday.
The chat room will open between 8:00-8:30, most likely 8:30. I'm too lazy to get everybody's sns so just log on at that time and get invited and invite everyone else.

I will do a regular math jam and a 2 person countdown, u must be ready to give explanations.
Everybody who doesn't have AIM go get it.

P.S. kyymathcaunt, [poopyloopy]  :rotfl:  :rotfl: are you serious??? :heli:

P.S. #2: any of u private site people want to take over the site?

P.S. #3: i kno easyaspi wanted to host one too so is there anyway for 2 moderaters if anyone wants to?

i would except I can't manage two at once ;)

Can you delay to 900? i have office hours till then :puppydogface:

[quote="tz"]
P.S. #3: i kno easyaspi wanted to host one too so is there anyway for 2 moderaters if anyone wants to?[/quote]
Not really... That's basically impossible.
I think its best to just host two separate ones. More practice!  :D

Anyways, I can't go, sorry. I have state the next morning and we have something that night. (We're leaving for Austin at 1)

is that 9 o'clock? sure biffanddoc
ALL RIGHT ITS RESCUDLED FOR 9 O'CLOCK!!!!

P.S. the "P.S." stands for "postscript". "post" means after and "script" means write.
So it means "after-write" or a little not "after" you "write" the main section. :D

[quote="Klebian"]
Anyways, I can't go, sorry. I have state the next morning and we have something that night. (We're leaving for Austin at 1)[/quote]

Same here, except its in Little Rock.

Lucky people get to leave the day before. I have to meet my coach at 5:30 to get down there.
Good Luck!

lucky people have theirs 10 minutes away :rotfl:

i'd be interested, given if I can find a computer at mathcounts states...

[b][size=200]PEOPLE IT WAS LAST FRIDAY!!![/size][/b]

LOL Thats funny. I want to go to a math jam that was last week also.  :D

Ha, klebian i submitted my message before you. :D

haha, two people posted the same thing at once. And the whole LAST FRIDAY thing is funny :D

Specific dates please!!!

This is too old and useless, locked.


algebra
polynomial
superior algebra
superior algebra unsolved

[b]Problem:[/b] Let E be an extension field of a field F and let $ \alpha \in E$ be transcendental over F. Show that every element of the simple extension $ F(\alpha)$ is also transcendental over F.

[b]Attempt at the Solution:[/b]
The elements of $ F(\alpha)$ will look like this: $ \frac {\sum_{i \equal{} 0}^{\infty}f_{i}\alpha^{ik}}{\sum_{j \equal{} 0}^{\infty}f_j\alpha^j}$. If something that looks like that is algebraic over F, the following must be true:

$ \sum_{k \equal{} 0}^{\infty}\left(\frac {\sum_{i \equal{} 0}^{\infty}f_{ik}\alpha^{ik}}{\sum_{j \equal{} 0}^{\infty}f_{jk}\alpha^{jk}}\right)^kf_k \equal{} 0$
I kind of see how to finish this, but it seems like things will get very very messy when I start indexing the coefficients of f_k that are nonzero and start convolving things. Can someone give me an instruction telling me what to do next?

Don't expand out everything, just write the hypothetical algebraic element as $ \frac{p(\alpha)}{q(\alpha)}$, with $ q$ of least possible degree. Then show $ q$ divides $ p^n$ where $ n$ is the degree of some nonzero polynomial with $ p/q$ as a root.

[quote="scorpius119"]Don't expand out everything, just write the hypothetical algebraic element as $ \frac {p(\alpha)}{q(\alpha)}$, with $ q$ of least possible degree. Then show $ q$ divides $ p^n$ where $ n$ is the degree of some nonzero polynomial with $ p/q$ as a root.[/quote]
I don't see why $ q(\alpha)$ has to divide $ p(\alpha)^n$ for some n. $ q(\alpha)$ and $ p(\alpha)$ are arbitrary polynomials. They could have no factors in common.

As Scorpius119 said, write out the algebraic equation:

$ \left(\frac {p\left(\alpha\right)}{q\left(\alpha\right)}\right)^n \plus{} \text{something}\cdot\left(\frac {p\left(\alpha\right)}{q\left(\alpha\right)}\right)^{n \minus{} 1} \plus{} \text{something}\cdot\left(\frac {p\left(\alpha\right)}{q\left(\alpha\right)}\right)^{n \minus{} 2} \plus{} ... \plus{} \text{something}$
$ \equal{} 0$
$ \Longleftrightarrow$
$ \left(p\left(\alpha\right)\right)^n \plus{} \text{something}\cdot\left(p\left(\alpha\right)\right)^{n \minus{} 1}\cdot\left(q\left(\alpha\right)\right) \plus{} \text{something}\cdot\left(p\left(\alpha\right)\right)^{n \minus{} 2}\cdot\left(q\left(\alpha\right)\right)^2$
$ \plus{} ... \plus{} \text{something}\cdot\left(q\left(\alpha\right)\right)^n \equal{} 0$.

Do you see the divisibility now?

  darij

Its pretty clear from that equation that if you convolve everything and collect the coefficients of common powers of $ \alpha$, you will get a polynomial in F[x] with $ \alpha$ as a zero. That will complete the proof. However, I still do not see what you are talking about with q dividing p.

EDIT: yes i do. take all the terms except the first to the RHS and factor out a q(x);  Hmm. we chose q of minimum degree and got a contradiction right? How does the contradiction rely on the fact that $ \alpha$ is transcendental.

EDIT EDIT: never mind; I figured it all out; thanks for the help

every of element of $ F(x)$, which is NOT in $ F$, is trancendental over $ F$. here an alternative easy proof for this:

first, very be the definition, $ x$ is trandencental over $ F$. now let $ b \in F(x) \backslash F$, and write $ b \equal{} f/g$ with coprime $ f,g \in F[x]$. then $ x$ is algebraic over $ F(b)$ because it is annihilated by $ f(t) \minus{} g(t) b \in F(b)[t]$. if $ b$ was algebraic over $ F$, $ x$ would be trandencental over  $ F$, a contradiction.


AMC
AIME
geometry
3D geometry
sphere
tetrahedron
function

[quote="PenguinIntegral"][quote="Karth"]Okay, there are like 28234 of these going on, but here's another one. None of the previous ones were really that recent, so to avoid bringing up an old thread, here's a new one  :D

So yea. Post problems around AIME-level here.

[b]Problem 1[/b]

If $\sec x+\tan x = 22/7,$ find $\csc x+\cot x.$[/quote]
Are you missing a condition, or is my knowledge of AIME problems failing?[/quote]

Not the former, and hopefully not the latter  :D

[quote="mustafa"][quote="PenguinIntegral"][quote="Karth"]Okay, there are like 28234 of these going on, but here's another one. None of the previous ones were really that recent, so to avoid bringing up an old thread, here's a new one  :D

So yea. Post problems around AIME-level here.

[b]Problem 1[/b]

If $\sec x+\tan x = 22/7,$ find $\csc x+\cot x.$[/quote]
Are you missing a condition, or is my knowledge of AIME problems failing?[/quote]

If you mean the integer between 0-999, I don't think that that is honesty really important. It's just there for grading convinence, and most problems can be easily changed to fit that.[/quote]

Oh yea, heheh. Good point. $x$ can be a fraction. But if you really have the urge to add the numerator and denominator, go ahead  :D

Note that $\sec^{2}x-\tan^{2}x=1\implies \sec x-\tan x=\frac{7}{22}$.  Solving a system for $\sec x, \tan x$, one eventually obtains the values of $\sin x$ and $\cos x$.  Now find the wanted value using $\sin x, \cos x$ to get $\frac{29}{15}$ if I did not miscalculate.

[quote="mustafa"][quote="PenguinIntegral"][quote="Karth"]Okay, there are like 28234 of these going on, but here's another one. None of the previous ones were really that recent, so to avoid bringing up an old thread, here's a new one  :D

So yea. Post problems around AIME-level here.

[b]Problem 1[/b]

If $\sec x+\tan x = 22/7,$ find $\csc x+\cot x.$[/quote]
Are you missing a condition, or is my knowledge of AIME problems failing?[/quote]

If you mean the integer between 0-999, I don't think that that is honesty really important. It's just there for grading convinence, and most problems can be easily changed to fit that.[/quote]
I know I know, it's just there is another problem like this that has another condition. Sorry...

[quote="DarkArchon"][quote="chess64"][b]Problem 2[/b]

(HMMT 2005) Find the sum of the absolute values of the roots of $x^{4}-4x^{3}-4x^{2}+16x-8=0$.[/quote]

[hide]
sum of roots = 4
product of roots = 8

only choice: 4, 2, 1, 1

sum of absolute values of roots = [b]8[/b]
(yeah, I know I completely disregarded the
signs, but since it only asks for the sum of the
absolute values, those don't matter...right?)
[/hide][/quote]

This is incorrect.

[hide="Big Hint"]
Try factoring by manipulating the quadratic term.[/hide]

[hide="Partial Solution"]
With the above hint, we can rewrite the equation as the following:

$(x^{4}-4x^{3}+4x^{2})-(8x^{2}-16x+8)=x^{2}(x-2)^{2}-8(x-1)^{2}=0$

Noticing a difference of squares, we can factor this into two quadratics, find the solutions, and sum the absolute values.[/hide]

[b]Problem 3[/b]

Compute the exact value of $\sum_{k=1}^{1001}\,\log_{8}\left(\cos\frac{k\pi}{2003}\right)$.

Is the answer.....
[hide]$\frac{1001}{3}$?[/hide]
I kind of saw a pattern, and don't really have a correct argument.

What's the justification for the second equality?

man why's I gotta do all this trigo

let x = cot theta, y = cot phi => them square root things are csc theta, csc phi, so adding we get cot theta + csc theta = (1 + cos theta) / sin theta and the same thing for phi right

well guess what

cot (theta/2) = (1 + cos theta) / sin theta

so BAM cot (theta/2) * cot (phi/2) = 1

cot (theta/2) = tan (phi/2)

man wtf is this

this is clearly sayin

theta/2 + phi/2 = pi/2 + some multiple of pi man

y'know what I'm getting bored it's triv now since cot (x) + cot (pi - x) = 0

[quote="JavaMan"][quote="mathgeniuse^ln(x)"]

[hide]And the answer by the way is $\frac{\sqrt{6}-1}{10}$ and this comes from $1=2r(\sqrt{6}+1)$[/hide][/quote]

I got the same answer, but my method took pretty long (involving creating an inner tetrahedron, finding the height of that, and other long computation). Did you use a simpler method?[/quote]

I didnt' show my work, sorry.

[quote="1337h4x"]Let x,y be real numbers such that $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$. Find $x+y$.[/quote]

[hide]Let $x+\sqrt{x^{2}+1}=c$ where $c$ is a real.  Subtracting both sides by $x$ and squaring gives:
$x^{2}+1=c^{2}-2cx+x^{2}. \\ \Rightarrow 2cx=c^{2}-1 \\ \Rightarrow x=\frac{c^{2}-1}{2c}.$

It follows that $y+\sqrt{y^{2}+1}=\frac{1}{c}.$  Subtracting both sides by $y$ and squaring yields: $y^{2}+1=\frac{1}{c^{2}}-\frac{2y}{c}+y^{2}\\ \Rightarrow \frac{2y}{c}=\frac{1}{c^{2}}-1 \\ \Rightarrow y=\frac{1}{2c}-\frac{c}{2}=\frac{1-c^{2}}{2c}.$

Finally, adding the expressions for x and y gives:
$x+y=\frac{c^{2}-1}{2c}+\frac{1-c^{2}}{2c}=0.$ [/hide]

[quote="zhou_qiang"]Find the last three digits of the product of the first 1003 odd positive integers. :)[/quote]

[hide] So we're trying to find $1 \cdot 3 \cdot 5 \cdot ... \cdot 2005 \mod{1000}$. We break it up into $\mod{125}$ and $\mod{8}$.

Case 1: $\mod{125}$: we can clearly see that $125$ is an odd integer between $1$ and $2005$, this number is $0 \mod{125}$.

Case 2: $\mod{8}$: We examine the product of the four odd residues $\mod{8}$, noting that because $8$ is even, every odd number is one of these residues. We have that $1 \cdot 3 \cdot 5 \cdot 7 = 105 \equiv 1 \mod{8}$. For each set of residues $\mod{8}$ (each corresponding to a particular multiple of $8$), we get a product of 1, so the product $1 \cdot 3 \cdot 5 \cdot ... \cdot 1999 \equiv 1$ as $2000$ is a multiple of $8$. The product is then $2001 \cdot 2003 \cdot 2005 \equiv 1 \cdot 3 \cdot 5 = 15 \equiv 7 \mod{8}$.

By the Chinese Remainder Theorem, this has one solution $\mod{125 \cdot 8}= \mod{1000}$, and checking multiples of $125$, we see that $375 \equiv 7 \mod{8}$, so the last three digits are $\boxed{375}$.[/hide]

If $a_{1}=100$ and $a_{n}= 2 a_{n-1}+\frac{100!}{(100-n)!(n-1)!}$, and $\sum^{100}_{k=1}a_{k}= a \cdot b^{c}-d \cdot e^{f}$, with $a, b, c, d, e, f$ positive integers and $c$ and $f$ as large as possible, find $a+b+c+d+e+f$.

[quote="calc rulz"][quote="1337h4x"]Let x,y be real numbers such that $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$. Find $x+y$.[/quote]

[hide="Problem Solving Note"]The second you see $x+\sqrt{x^{2}+1}$ and $y+\sqrt{y^{2}+1}$, think of the formula:

$arcsinh{x}= \ln{(x+\sqrt{x^{2}+1})}$
etc for y

Then you get $asinh{x}+asinh{y}=0$, giving you essentially the same thing as jackb's solution.[/hide][/quote]

the trig substitution is more natural in my opinion...i don't use those formulas that often...

[quote="1337h4x"]Let x,y be real numbers such that $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$. Find $x+y$.[/quote]

[hide="42342034905834th solution"]
it's equivalent to $x+\sqrt{x^{2}+1}=-y+\sqrt{y^{2}+1}$, or
\[x+y+\sqrt{x^{2}+1}-\sqrt{y^{2}+1}=0\]
Now use conjugates for the radical expression:
\[\sqrt{x^{2}+1}-\sqrt{y^{2}+1}=\frac{(x+y)(x-y)}{\sqrt{x^{2}+1}+\sqrt{y^{2}+1}}\]
so the original expression becomes
\[(x+y)\left( 1+\frac{x-y}{\sqrt{x^{2}+1}+\sqrt{y^{2}+1}}\right) =0\]
Since $-x\leq |x|<\sqrt{x^{2}+1}$ and $y\leq |y|<\sqrt{y^{2}+1}$, the second factor is positive, so the first factor must equal 0 and $\boxed{x+y=0}$.
[/hide]

calc rulz, was my solution correct? If so, can someone post another problem?

an original problem...[it might be a bit too difficult...]

Let $ABCD$ be a parallelogram, with $AB=6$ and $BC=4$. There is a point, $P$, is inside $ABCD$ such that $AP*CP=12+\pi$ and $BP*DP=12-\pi$. If $O_{1}$, $O_{2}$, $O_{3}$, and $O_{4}$ are the circumcenters of $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, and $\triangle DAP$, then find the maximum value of $(O_{1}O_{3})(O_{2}O_{4})$.

[hide="hint 1"]translate[/hide]

[hide="hint 2"]ptolemy inequality[/hide]

Find the number of 10 digit integers formed by some or all the five digits, namely, 0, 1, 2, 3, and 4, such that these integers contain none of the three blocks 22, 33,and 44.


n people vote for one of 27 candidates. Each candidate's percentage of the vote is at least 1 less than his number of votes. What is the smallest possible value of n? (So if a candidate gets m votes, then 100m/n ≤ m-1.)

help

[hide]Let m_1, m_2, ...m_27 be the number of votes each of the candidates got.  Notice that  :Sigma: m_i = n. Since 100 m_i /n  :le: -1+m_i, we sum for i between 1 and 27 and get 100 ( :Sigma: m_i)/n :le: -27 + ( :Sigma: m_i), i.e. 100 :le: -27+n. So 127 :le: n, and 127 is the minimum value for n.[/hide]

You're close, though.  Everything you've done so far is right.

[hide]Set up an inequality, we have p_1  :ge:  p_1(n)/100-1, where p_1 is the percentage of candidate 1. Same thing for candidate 2,3...27. Add all the inequality and then simplify it we have n  :ge:  127. 127 doesn't work since splitting it evenly, we have 4s, 4/127=3.14%. n has to be larger. When everything is 5 we have n=135 which works because 5/135=3.7% so 127 :le:   n   :le: 135. So n=134 which is the answer[/hide][/hide]