geometry
projective geometry
geometry theorems

A circle is incribed in a quadrilateral $ABCD$ and touches $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$. $AB$ and $CD$ meets at $E$, $AD$ and $BC$ meets at $F$. Show that:
(1): $AC$, $BD$, $PR$ and $QS$ are concurrent.
(2): $AC$, $EF$, $PQ$ and $RS$ are concurrent.
(3): $BD$, $EF$, $PS$ and $QR$ are concurrent.

[color=darkred][b]Remark.[/b] [i]Shobber[/i], this remarkable property of a circumscribed quadrilateral
must be posted at Section >> Geometry >> Geometry Theorems and Formulas.
Maybe [i]V. Vornicu[/i] or [i]D. Grinberg[/i] will be able to move it there.[/color][hide="Here is a elementary proof."]
$1^{\circ}.\blacktriangleright\ X\in PR\cap QS\ ,\ X_1\in PR\cap AC\ ,\ X_2\in QS\cap AC\ .$

$\widehat {AX_1P}\equiv \widehat {CX_1R}\ ,\ m(\widehat {APX_1})+m(\widehat {CRX_1})=180^{\circ}\Longrightarrow \frac{X_1A}{X_1C}=\frac{AP}{CR}\ \ \ (1)\ .$

$\widehat {CX_2Q}\equiv \widehat {AX_2S}\ ,\ m(\widehat {ASX_2})+m(\widehat {CQX_2})=180^{\circ}\Longrightarrow \frac{X_2A}{X_2C}=\frac{AS}{CQ}\ \ \ (2)\ .$

$AP=AS\ ,\ CR=CQ\ ;\ (1)\ \wedge \ (2)\Longrightarrow \frac{X_1A}{X_1C}=\frac{X_2A}{X_2C}\Longrightarrow$ $X_1\equiv X_2\Longrightarrow X\in AC\ .$

Prove similarly that $X\in AD\ .$ Therefore, the lines $AC$, $BD$, $PR$, $QS$ are concurrently.

$2^{\circ}.\ -\ 3^{\circ}.\blacktriangleright\ Y\in PQ\cap SR\ ,\ Z\in SP\cap RQ\ ,\ Y_1\in PQ\cap AC\ ,\ Y_2\in SR\cap AC\ .$

[u]Menelaus' theorem[/u] to the transversals $\overline {PQY_1}$, $\overline {SRY_2}$  in $\triangle ABC$, $\triangle ADC$ respectively $\Longrightarrow$

$\frac{Y_1A}{Y_1C}\cdot \frac{QC}{QB}\cdot \frac{PB}{PA}=1\Longrightarrow \frac{Y_1A}{Y_1C}=\frac{AP}{CQ}\ \ (3)\ .$

$\frac{Y_2A}{Y_2C}\cdot \frac{RC}{RD}\cdot \frac{SD}{SA}=1\Longrightarrow \frac{Y_2A}{Y_2C}=\frac{AS}{CR}\ \ (4)\ .$

$AP=AS\ ,\ CQ=CR\ ; (3)\ \wedge\ (4)\Longrightarrow \frac{Y_1A}{Y_1C}=\frac{Y_2A}{Y_2C}\Longrightarrow Y_1\equiv$$Y_2\Longrightarrow Y\in AC\ .$

Prove similarly that $Z\in BD$. [u]Pascal's theorem[/u] for $PPQRRS$, $QQRSSP$ $\Longrightarrow$

$E\in PP\cap RR\ ,\ Y\in PQ\cap RS\ ,\ Z\in QR\cap SP\Longrightarrow E\in YZ\ .$

$F\in QQ\cap SS\ ,\ Z\in QR\cap SP\ ,\ Y\in RS\cap PQ\Longrightarrow F\in YZ\ .$

Therefore, the points $E\ ,\ F\ ,\ X\ ,\ Y$ are collinearly , i.e. $\{X,Y\}\subset EF\ .$[/hide]


Let be given 2004 pupils divided into 3 groups so that each pupil is in no more than 2 groups, 2 any pupils are in the same group. Prove that there is a group that has no less than 1336 pupils.

If some pupil belongs to only one group then all the other pupils have to belong to the same group and the result is obvious. So, we assume that no pupil belongs to only one group. From the statement it follows that each pupil belongs to exactly two groups.
Let $A,B,C$ be the groups and $x_x, \cdots, x_{2004}$ be the pupils.
For all $i$, let $a_i = 0$ if $a_i \notin A$ and $a_i = 1$ otherwise.
Define $a_i,c_i$ in a similar way.

Thus, for all $i$, we have $a_i+b_i+c_i=2$.
Summing over $i$, it leads to $\dsp \Sum_{i=1}^{2004} a_i + \Sum_{i=1}^{2004} b_i + \Sum_{i=1}^{2004} c_i  = 2 \times 2004$
that is $|A|+|B|+|C| = 4008$.
It follows that $\max \{ |A|,|B|,|C|  \} \geq \frac {4008} 3 = 1336$, as desired.

Note that this is optimal in the sense that if we divide the pupils in three groups according to their non-residu modulo 3 (that is if $x_i \neq j \mod [3]$ then $x_i \in G_j$ where $j=1,2,3$) we fulfill the statements of the problem with three groups of 1336 pupils.

Pierre.


geometry unsolved
geometry
U sviti matematyky

Point $ P$ is chosen inside triangle $ ABC$. Let $ X,Y,Z$ be the intersections of $ AP,BP,CP$ with $ BC,CA,AB$ respectively. Let $ M_1,M_2,M_3$ be the midpoints of $ AC,AB,BC$ and $ N_1,N_2,N_3$ the midpoints of $ XZ,XY,YZ$ respectively.

Prove that $ M_1N_1,M_2N_2,M_3N_3$ concur.

This problem is quite nice, but rather easy.
[quote="mr.danh"]
A well-known lemma: Let a quadrilateral ABCD. Denote E=AB∩CD, F=AD∩BC. Then, the midpoints of segments AC,BD,EF are collinear.[/quote]
In our problem, denote P1,P2,P3 be the midpoints of PB,PC,PA.
Apply the lemma to three quadrilaterals BXPZ,CYPX,AZPY, we have that (M1,N1,P1),(M2,N2,P2),(M3,N3,P3) are three lines. Otherwise, it is easy to show that M1P1,M2P2,M3P3 are concurrent at the common midpoint of them. Hence result.

Nice solution,dear mr.danh!
If I am right , the line is called "Newton Line"


search
topology
real analysis
real analysis unsolved

1.   Can  you  show   a counterexample   that   a  map  f : R to R sends
open subsets to open subsets,   but   it is [color=red][b]not [/b][/color] strictly monotone, i.e. either
f(x) < f(y) for all x < y or f(x) > f(y) for all x < y.

2.     Suppose that a map f : R to R sends  open subsets to open subsets, and    it is strictly monotone, i.e. either
f(x) < f(y) for all x < y or f(x) > f(y) for all x < y.
then   the    map  is  continuous ?

Thank you !

1. See [url=http://www.mathlinks.ro/viewtopic.php?search_id=1710509765&t=308786]this thread[/url].

2. Yes. A monotone map can only have jump discontinuities and if $ f$ has a jump discontinuity at a point $ x_0$, the image of the interval $ (x_0 \minus{}1, x_0 \plus{}1)$ under $ f$ clearly won't be an open set (since $ f(x_0)$ isn't an interior point).


Let $a\in[-1;1],b\in[0;4], \alpha\in[0;10^o]$ and
         $\sqrt{9-5sin50^o}=a+bsin\alpha$
Find $a,b,\alpha$

Hey, there's one value of $a, b, \alpha$ satified the condition, only one. Why don't you try? :(


can someone explain to me why the answer is C.  I put C on the test, but when i later checked the problem i keep on getting B, so I dont really know how to do this.  Can someone explain to me how you get C?

[hide]#19 was not that hard IMO. Assume the paint brush length is 1, the center is 2, and then let's say the side of the triangle is x. So you have $4x+2 = 2x^{2}$. Quad gives you it's $1%Error. "pmsqrt" is a bad command.
{2}$. So that means the length of the nonshaded area on the square is $2+rt(2)$. Then, you have to add on $\sqrt{2}$ for the shaded regions and you get C.[/hide]

Could someone make a picture? I don't really get the explanation.

This was the square with the x right?

(Lol dang, and I thought I had the right answers to all of them... this is making me nervous)

[quote="johnz"]what were the answer choices.  I get an equation that when you solve it is 3sqr2[/quote]

I gots that too.

So let's say that you can understand my picture...

AB is the width of the paintbrush, and let that be 1.
Let the side, XY be $S$.

So XB is $\frac{\sqrt(2)}{2}$, as is CY.

Thus, BC is $S-\sqrt(2)$, so DB = DC = $\frac{S\sqrt(2)}{2}-1$.

Thus, the area of BCD is $\frac{1}{2}\left(\frac{S\sqrt(2)}{2}-1\right)^{2}$.

The total area of 4 of these triangles is equal to have of the square's area, which is $\frac{S^{2}}{2}$.

Thus, \[\frac{S^{2}}{2}= 2\left(\frac{S\sqrt(2)}{2}-1\right)^{2}\].

Multiplying through by 2 and taking the squareroot, $S = \frac{S\sqrt(2)}{2}-1$.

Solving for $S$, we get $S = 2\sqrt{2}+2$.


What is the probability that, given a class of 30 students, 2 or more of them have the same birthday?

Is this the classic birthday paradox problem?

Yes, but the number of students is variable.

There will be a big wrinkle if we include the possiblity of a birth of 2/29 that is 1/4 the possibility of a birth on all other dates because of leap years.
However, if we assume noone was born on 2/29...
We just figure out what the probablity of everyone being born on different dates and subtract from 1.
[tex]\ 1-(365!/335!/365^{30})[/tex]
Which comes out to [u]0.7063[/u]

But what happens if we incorporate 2/29?


I've always wondered that...

\[
\frac{1}{7} = 0.\overline{142857}
\]

Note that you would use this for a line segment too.


geometry
3D geometry
sphere
parameterization
topology
advanced fields
advanced fields unsolved

Let $ \gamma: [0, l] \longrightarrow S^2$ such that $ \gamma(0) \equal{} \gamma(l)$ be a natural parametrization of a smooth curve lying on the unit sphere $ S^2$. Prove $ l < 2\pi$ implies that the curve $ \gamma([0, l])$ lies on a hemisphere.

Look how far the midpoint of that curve can be away (measured in terms of the sphere) from its endpoints.

Do you mean that 'midpoint' of the curve is a point $ \gamma(\frac{l}{2})$ and endpoints are $ \gamma(0) \equal{} \gamma(l)$? Could you post an exact solution if you see any. Thanks a lot.

By the way, you should have $ l \leq \pi$ (otherwise, there are counterexamples).

If you know the geodetics on the sphere, it's easy:
The minimal distance (defined as the minimum/infimum of all possible lengths of curves connecting them) of two points on the sphere is length of the arc of a (great) circle between them. Thus the endpoints can be at most $ \frac \pi 2$ away from the midpoint $ M$, meaning that they (and similar all of the curve) is in the hemissphere with "center"/"pol" $ M$.

Please, notice that the given curve is closed ($ \gamma(0) \equal{} \gamma(l)$) and I am sure that the thesis is true under the assumption that $ l < 2\pi$. Could you give any counterexample that it is false?

I believe the $ 2\pi$ condition as well. Think about a closed curve maximized along a great arc of length $ 2\pi$. It then makes a single loop around (marking off two hemispheres). If it is not allowed to be of length $ 2\pi$ by even some small $ \varepsilon$, then it is forced into a single hemisphere to remain closed.

You can use
http://en.wikipedia.org/wiki/Crofton_formula  (notice the extension to Riemannian manifolds, our case=sphere)
to prove that "if any big circle intersects the curve (the closed one) then its length is at least $ 2\pi$" which is the same as the original problem.

I solved this problem some time ago with some simple and elementary argument (similar with ZetaX's) but I can't figure it now  :( .


Some bats were in a cave. Two bats could see out of their right eye, three could see out of their left eye, four could not see out of their left eye, and five could not see out of their right eye. What is the minimum possible number of bats in the cave?

You can have the following:

2 bats can see out their right eye, but not their left
3 bats can see out their left eye, but not their right
2 bats can't see out either eye

$ 2\plus{}3\plus{}2\equal{}\boxed{7}$

[quote=lcsmart]Scientifically Impossible! ;)[/quote]

no


the second part is easier, becuase for the infinity case we just have k=(continued fraction)=1+1/k, so k^2=k+1 so easy solve.
then, I believe that f_n(x)=the ratio between consecutive fibonacci's (I can prove this...)

work seems right, answer seems wrong

[hide]
$f_{n+1}(x) = 1 + \frac 1 {f_n(x)}$
Define $f_n(x) = \frac{ N_n(x) }{ D_n(x) }$.  From the recursive $f$, we have $N_{n+1}(x) = N_n(x) + D_n(x), D_{n+1}(x) = N_n(x)$  We have $N_1(x) = x+1, D_1(x) = x$.
By noticing we are adding the functions together, we see we'll have a fibonacci sequence of coefficients.  The denominator is the previous numerator, so the denominator's coefficients will be one fibonacci term before the numerator's.
$F_0 = 0, F_1 = 1, F_{n+2} = F_n + F_{n+1}, n \ge 0$.
Then $N_n(x) = F_{n+1}x + F_n, D_n(x) = F_nx + F_{n-1}$ which gives us $f_n(x) = \frac{ F_{n+1}x + F_n }{ F_nx + F_{n-1} }$
$k = \frac{ F_{n+1}k + F_n }{ F_nk + F_{n-1} }$
$F_nk^2 + (F_{n-1} - F_{n+1})k - F_n = 0$
$F_nk^2 - F_nk - F_n = 0$
$F_n(k^2 - k - 1) = 0$
Thus $k = \frac{1 + \sqrt{5}}2$ and is independent of $n$.
[/hide]

[quote="drunner2007"]Let $f_n(x)=1+\frac{1}{1+\frac{1}{1+\ldots+\frac{1}{x}}}$ (basically $f_n(x)$ has $n$ "+" signs). For example, $f_1(x)=1+\frac{1}{x}$ and $f_2(x)=1+\frac{1}{1+\frac{1}{x}}$. Find $k$ in terms of $n$ if $k$ is the positive solution to $k=f_n(k)$. Then find $\displaystyle\lim_{n\to\infty}k$.[/quote]
[hide="first part"] I first solved for $k=f_1(k)$. $k=1+\frac{1}{k} \longrightarrow k^2=k+1$. The solution is the famous golden ratio: $\frac{1+\sqrt5}{2}$. Then, looking at $k=f_2(k)$, we can see that $1+\frac{1}{1+\frac{1}{k}}=1+\frac{1}{f_1(k)}=f_1(f_1(k))$. Therefore, the positive solution to $k=f_n(k)$ is $k=\boxed{\frac{1+\sqrt5}{2}}$. [/hide]


A game is played with a special fair cubic die which has one red side, two blue sides, and three green sides. The [b]result[/b] is the color of the top side after the die has been rolled. If the die is rolled repeatedly, find the probability that the second blue [b]result[/b] occurs on or before the tenth roll. Express your answer as a decimal rounded to 4 significant digits.

[hide="Solution"]

This is equivalent to the probability that you will roll blue, which has a probability of $\frac{1}{3}$, at least twice in 10 rolls.

There is a probability of $\left(\frac{2}{3}\right)^{10}$ that you will get 0 blues, and $\binom{10}{1}\times \left(\frac{1}{3}\right)\times \left(\frac{2}{3}\right)^{9}$ probability that you will get exactly 1 blue.  Then the total probability that you will get at least 2 blues is
$1-\left(\frac{12}{3}\right)\times \left(\frac{2}{3}\right)^{9}=\frac{3^{9}-2^{11}}{3^{9}}\approx 0.8960$.[/hide]

[quote="jeez123"][quote="samath"][hide="Solution"]
Then the total probability that you will get at least 2 blues is
$1-\left(\frac{12}{3}\right)\times \left(\frac{2}{3}\right)^{9}=\frac{3^{9}-2^{11}}{3^{9}}\approx 0.8960$.[/hide][/quote]

I got everything up to there. Where does the 12/3 comes from and could you explain what you did there? thanks![/quote]

He just factored:
$\left(\frac{2}{3}\right)^{10}+10\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{9}$
Factor out a $\left(\frac{2}{3}\right)^{9}$....
$\left(\frac{2}{3}\right)^{9}\left(\frac{2}{3}+\frac{10}{3}\right)$

I hope this helps.

Yeah ... sorry I wasn't more clear  :blush:


The following problem i proposed to our National Trainning Camp 2005,

Problem: Find all $f:\mathbb{Z}_+\longrightarrow\mathbb{Z}_+$ such that for any $m,n\in\mathbb{Z}_+$ there exist a number $k\in\mathbb{Z}_+$ such that
\[3f^4(k)=f^4(m)+f^2(m)f^2(n)+f^4(n)\]
where $\mathbb{Z}_+$ the set of all positive integers, $f^r(x)=(f(x))^r$ for any $r\in\mathbb{Z}_+$.

...I think there is only one root: $f(x)=const$.
if not , take $m,n$ such that $f(m)<f(n)$ and $f(m),f(n) min$
-> not exist $k$.


Consider the polynomial :
  $ T(x) = x^3+17x^2 - 1239x +2001 $
  Put $ T_1(x) = T(x) ; T_{n+1}(x)= T(T_n(x)) $ for every n =1,2,3...
  Prove that : There exists an integer n >1 such that $ T_n(x)-x $ is divisible by 2003 for every integer x

the only thing that has to be proved is that for every integer $l$ from $[0,1,...,2002]$ there exists a number
$S(l)$ that $T_{S(l)}(l)=l (mod 2003)$, then the $n$ we are looking for is
$n=\prod_{i=0}^{2002}S(i)$.

$(T_n(l))$ is periodic of period $p$ less than 2003, if $(T_n(l))$ is periodicity starts from $n=0$ then $T_p(l)=l$ and it's done, if periodicity starts from $n>0$ then we have a sequence like $(a;b;...;d;x;...;z;x;...;z;...)$ so $d\rightarrow x$ and $z\rightarrow x$ but it's not possible since $T(x)$ is injective in $\mathbb{Z}/2003\mathbb{Z}$ (I have not yet the proof)


algebra
polynomial
vector
trigonometry
algebra unsolved

Let $ a;b_1;b_2;..;b_n;c_1;c_2;...;c_n$ be real number such that:
      $ x^{2n}\plus{}ax^{2n\minus{}1}\plus{}ax^{2n\minus{}2}\plus{}...\plus{}ax\plus{}1\equal{}(x^2\plus{}b_1x\plus{}c_1)(x^2\plus{}b_2x\plus{}c_2)....(x^2\plus{}b_nx\plus{}c_n)$
for all real number $ x$ .Prove that:  $ c_1\equal{}c_2\equal{}...\equal{}c_n\equal{}1$

I think you forget? that polinom had only comlex roots? or only 2 real roots.
For example $ (x\minus{}1)(x\minus{}2)(x\minus{}1)(x\minus{}0.5)\equal{}(x^2\minus{}3x\plus{}2)(x^2\minus{}1.5x\plus{}0.5)$.

I do not understand what you say
what you say does not affect the results.Because any real number $ a$ also there exit $ b_i;c_i$ such that conditions
of problem and we need prove $ c_1\equal{}c_2\equal{}..\equal{}c_n$.

Consider another polinom $ (x\minus{}0.5)^3(x\minus{}8)\not \equal{}(x^2\plus{}b_1x\plus{}1)(x^2\plus{}b_2x\plus{}1)$.

[quote="Rust"]Consider another polinom $ (x \minus{} 0.5)^3(x \minus{} 8)\not \equal{} (x^2 \plus{} b_1x \plus{} 1)(x^2 \plus{} b_2x \plus{} 1)$.[/quote]

Your polynomial does not satisfy the conditions. I will attempt this problem tomorrow.

the polynomial $ (x\minus{}0.5)^{3}(x\minus{}8)\not\equal{} x^4\plus{}ax^3\plus{}ax^2\plus{}ax\plus{}1$
In case polynomial can write: $ (x\minus{}m)^{3}(x\minus{}n)$ it also can write:$ (x\minus{}m)^2(x\minus{}m)(x\minus{}n)$
but I think this case can not occur(because coefficient is $ 1,a,a,a,1$

can Differ post your solution?
I really want to see solutions

Let $ y\equal{}x\plus{}\frac 1x\equal{}2cos2t$, then equation $ P(x)\equal{}0$ equavalent to $ 2cos(2nt)\plus{}a(1\plus{}2cos(2t)\plus{}...\plus{}2cos(2n\minus{}2)t)\equal{}0$
or $ f(t)\equal{}sin(2n\plus{}1)t)\plus{}(a\minus{}1)sin(2n\minus{}1)t\equal{}0$
If $ t\equal{}t_k\equal{}\pi \frac{2k\plus{}1}{2(2n\minus{}1)}$, then $ f(t_k)\equal{}(a\minus{}1)(\minus{}1)^k\plus{}sin(2n\plus{}1)t_k$.
If $ |a\minus{}1|\ge 1$ we had at least (n-1) pairs complex roots $ x_{2j\minus{}1}\equal{}e^{is_j},x_{2i}\equal{}e^{\minus{}is_j}$.

[quote="Rust"]Let $ y \equal{} x \plus{} \frac 1x \equal{} 2cos2t$, then equation $ P(x) \equal{} 0$ equavalent to $ 2cos(2nt) \plus{} a(1 \plus{} 2cos(2t) \plus{} ... \plus{} 2cos(2n \minus{} 2)t) \equal{} 0$[/quote]

I do not understand this part.

Here's my solution, which looks a lot less elegant:

First, what I am trying to prove is that $ r$ in $ x \equal{} rcis{\alpha}$ is always $ 1$ for $ \alpha > 0$. Taking care of the real solutions is easy - there are only two cases. So if there is a $ r < 1$, there must be a $ r > 1$ because the products of all the $ r$'s is 1. Assume there exists a $ r > 1$.

After multiplying our original equation by $ x \minus{} 1$ and doing a small amount of manipulation, we get the equation $ \frac {x^{2n \plus{} 1} \plus{} 1}{x^{2n} \minus{} x} \equal{} k$, where k is a real number (more specifically, $ \minus{} a \plus{} 1$).

Setting $ x^{2n} \equal{} y \equal{} a \plus{} bi$ and $ x \equal{} c \plus{} di$, we get the equation $ \frac {(a \plus{} bi)(c \plus{} di) \minus{} 1}{a \plus{} bi \minus{} c \minus{} di} \equal{} k$. That means the vectors on the numerator and denominator have the same polar angle, so $ \frac {bc \plus{} ad}{b \minus{} d} \equal{} \frac {ac \minus{} bd \minus{} 1}{a \minus{} c}$. Multiplying out gives $ a^2d \minus{} d \plus{} db^2 \equal{} c^2b \minus{} b \plus{} bd^2$. We divide and get $ \frac {a^2 \plus{} b^2 \minus{} 1}{c^2 \plus{} d^2 \minus{} 1} \equal{} \frac {b}{d}$. Converting to polar form gives $ \frac {r^{4n} \minus{} 1}{r^2 \minus{} 1} \equal{} \frac {r^{2n}\sin(2n\alpha)}{r\sin{\alpha}}$.

$ \frac {r^{4n} \minus{} 1}{r^2 \minus{} 1} \equal{} r^{4n \minus{} 2} \plus{} r^{4n \minus{} 4} \plus{} r^{4n \minus{} 6} \plus{} \cdots \plus{} 1 > 2nr^{2n \minus{} 1} > r^{2n \minus{} 1}\frac {\sin(2n\alpha)}{\sin{\alpha}}$

We have a contradiction, so $ r \equal{} 1$. Noting that $ b$ must be real finishes the proof.


geometry
trapezoid
function
perimeter
logarithms
ratio
AMC 10

The rule is same as AMC-10 and AMC-12. Anything that's acceptable to use on the AMC Competitions are allowed in this competition and I want you to send me the answers by 5:15. I'll grade them within today and hopefully have the result out soon.
All questions are exactly from the AHSME or AMC so hopefully, I don't have to answer any questions. And most of all, good luck. Answers must be turned to me by private message.
----------------------------------------------------------------------------------
1. Such sequence [a,b,c] mean $\frac {a+b}{c}$, where $c \neq 0$. What is the value of following:

[[60,30,90],[2,1,3],[10,5,15]]

A) 0
B) 0.5
C) 1
D) 1.5
E) 2

2. If M is 30% of Q, Q is 20% of P, and N is 50% of P, then $\frac {M}{N}$ =

A) $\frac {3}{250}$
B) $\frac {3}{25}$
C) 1
D) $\frac {6}{5}$
E) $\frac {4}{3}$

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

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

Then what does x+y+z =?

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

4. Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is:

A) 10
B) 12
C) 14
D) 16
E) 18

5. Imagine sequence

1,-2,3,-4,5,-6...

where nth term is $(-1)^{n+1} * n$. What is the average of the first 200 terms of the sequence?

A) -1
B) -0.5
C) 0
D) 0.5
E) 1

6. Mientka Publishing Company prices its best seller Where's Walter? as follows:

C(n) = 12n $\text {if}$ $1 \leq n \leq 24$,
C(n) = 11n $\text {if}$ $25 \leq n \leq 48$,
C(n) = 10n $\text {if}$ $49 \leq n$,

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

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

7. For how many integers n between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?

A) 0
B) 1
C) 2
D) 9
E) 10

8. In $\triangle ABC$, $\angle C = 90$ degrees, AC = 6 and BC = 8. Point D and E are on $\overline {AB}$ and $\overline {BC}$, respectively, and $\angle BED = 90$ degrees. If DE = 4, then BD =

A) 5
B) $\frac {16}{3}$
C) $\frac {20}{3}$
D) $\frac {15}{2}$
E) 8

9. Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with one penny. Which of the following amounts could Boris have after using the machine repeatedly?

A) 5.13 dollars
B) 3.63 dollars
C) 6.30 dollars
D) 7.45 dollars
E) 9.07 dollars

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

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

11. In trapezoid ABCD, $\overline {AB}$ and $\overline {CD}$ are perpendicular to $\overline {AD}$, with AB+CD = BC, AB<CD, and AD = 7. What is AB * CD?

A) 12
B) 12.25
C) 12.5
D) 12.75
E) 13

12. Let f(x) be a function with the two properties:

(a) for any two real numbers x and y, $f(x+y) = x + f(y)$, and
(b) f(0) = 2.

What is the value of $f(1998)$?

A) 0
B) 2
C) 1996
D) 1998
E) 2000

13. Five points around a circle are numbered 1,2,3,4, and 5 are in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point:

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

14. Mr. and Mrs. Zeta want to name baby Zeta so that its mongram (the first, middle, and last initial) will be in alphabetical order with no letter repeated. How many such monograms are possible?

A) 276
B) 300
C) 552
D) 600
E) 15600

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

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

16. The set of all real numbers x for which:

$x + \sqrt {x^2+1} - \frac {1}{x+\sqrt {x^2+1}}$

Is a rational number is the set of what?

A) integers x
B) rational x
C) real x
D) x for which $\sqrt {x^2+1}$ is rational
E) x for which $x+\sqrt {x^2+1}$ is rational

17. The perimeter of an equilateral triangle exceeds the perimeter of a square by 1989 cm. The length of each side of the triangle exceeds the length of each side of the square by [i]d[/i] cm. The square has perimeter greater than 0. How many positive integers are [u]not[/u] possible values for [i]d[/i]?

A) 0
B) 9
C) 221
D) 663
E) Not finite

18. For any positive integers n, let this function describes it:

$f(n) = \log_8 n$, if $\log_8 n$ is rational
$f(n) = 0$ otherwise.

What is then $\sum^{1997}_{n=1} f(n)$?

A) $\log_8 2047$
B) 6
C) $\frac {55}{3}$
D) $\frac {58}{3}$
E) 585

19. Two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6. The chord parallel to these chords and midway between them is of length $\sqrt {a}$ where a is:

A) 144
B) 156
C) 168
D) 176
E) 184

20. The number of pairs of positive integers (x,y) which satisfy the equation $x^2+y^2 = x^3$ is:

A) 0
B) 1
C) 2
D) not finite
E) none of these

21. A list of five positive integers has mean 12 and range 18. The mode and median are both 8. How many different values are possible for the second largest element of the list?

A) 4
B) 6
C) 8
D) 10
E) 12

22. Call a 7-digit telephone number $d_1 d_2 d_3 - d_4 d_5 d_6 d_7$ memorable if the prefix sequence $d_1 d_2 d_3$ is exactly the same as either of the sequences $d_4 d_5 d_6$ or $d_5 d_6 d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digts 0,1,2,.....9, the number of different memorable telephone numbers is:

A) 19810
B) 19910
C) 19990
D) 20000
E) 20100

23. A child has a set of 96 distinct blocks. Each block is one of 2 materials (plastic, wood), 3 sizes (small, medium, large), 4 colors (blue, green, red, yellow), 4 shapes (circle, hexagon, square, triangle). How many blocks in the set different from the "plastic medium red circle" in exactly two ways? (The "wood medium red square" is such a block.)

A) 29
B) 39
C) 48
D) 56
E) 62

24. In $\triangle {ABC}$, $\angle C$ is a right angle and $CB>CA$. Point D is located on $\overline {BC}$ so that $\angle CAD$ is twice $\angle DAB$. If $\frac {AC}{AD} = \frac {2}{3}$, then $\frac {CD}{BD} = \frac {m}{n}$, where m and n are relatively prime positive integers. Find m+n

A) 10
B) 14
C) 18
D) 22
E) 26

25. Evaluate this:

$\sum^{49}_{k=0} (-1)^{k} \binom {99}{2k}$, where $\binom {n}{j} = \frac {n!}{j!(n-j)!}$

A) $-2^{50}$
B) $-2^{49}$
C) 0
D) $2^{49}$
E) $2^{50}$

All right.

I need the answers send to me now.

Send me within 5 minutes. I'll grade them as soon as possible.

that was good, considering I started the test ten to twelve minutes late.... I was watching topics for replies... but I think I did well.
how far back to some of these questions go? ie from AHSME's and AMC10's and 12's

that wasnt too bad, though i didnt have a single a as an answer :? :worries:

lol, neither did I, but I only got to 16.

[quote="Silverfalcon"]15. In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored 23,14,11, and 20 points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than 18, what is the least number of points she could have scored in the tenth game?
[/quote]

[hide="My Complaint"]Let x=the average of the first five games:

x<(23+14+11+20+5x)/9

4x<68

x<17

Let y=the number of points in the tenth game:

(68+5x+y)/10>18

5x+y>112

To minimize y, 5x must be maximized:

80+y>112

y>32[/hide]

[quote="Silverfalcon"]14. Mr. and Mrs. Zeta want to name baby Zeta so that its mongram (the first, middle, and last initial) will be in alphabetical order with no letter repeated. How many such monograms are possible?
[/quote]

[hide="My Complaint"]
Any set of three letters will work, because we can arrange them in alphabetical order. Assuming that we are working with an alphabet of 26 letters, the answer would be C(26,3).[/hide]

I PMed you my complaint so others do not see the answer I chose incase they have the same complaint.

[quote="RC-7th"]
[quote="Silverfalcon"]14. Mr. and Mrs. Zeta want to name baby Zeta so that its mongram (the first, middle, and last initial) will be in alphabetical order with no letter repeated. How many such monograms are possible?
[/quote]

[hide="My Complaint"]
Any set of three letters will work, because we can arrange them in alphabetical order. Assuming that we are working with an alphabet of 26 letters, the answer would be C(26,3).[/hide][/quote]

You have to realize that: [hide] baby Zeta has to have a last name beginning with Z  :).[/hide]
 American Culture.

Actually number 1 was A...

no jkjk but i did get A for 23 i think

and RC-7th, i still have to look over your solution for 15, but for 14, the last name has to be Zeta, so the answer is C(25, 2)

If I'm not mistaken Silverfalcon posted it 10 minutes late... and i shouldn't be mistaken unless something is wrong with my comp because I was refreshing for 10 minutes :(

I have a complaint about #4 - I got >18 triangles, did anyone else?

As for RC-7th's complaints. I can field those.
[hide]
Nevermind, 14 has been fielded

For the basketball question, x<17 implies that 84 points = most that could've been scored in first 5 games. 84+68+a>180 and we get x=29[/hide]

... and I also got A for 23

EDIT: Sorry, used variable "x" twice... was confusing

What kind of place do they live in that doesn't allow them to name their baby what they want? :D  But Silverfalcon probably meant the answer to be C(25,2)

[quote="Lucky707"]
[hide]
Nevermind, 14 has been fielded

For the basketball question, x<17 implies that 84 points = most that could've been scored in first 5 games. 84+68+x>180 and we get x=29[/hide]

... and I also got A for 23[/quote]

Why is x<17?

[quote="RC-7th"][quote="Silverfalcon"]15. In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored 23,14,11, and 20 points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than 18, what is the least number of points she could have scored in the tenth game?
[/quote]

[hide="My Complaint"]

5x+y>112

To minimize y, 5x must be maximized:

80+y>112

y>32[/hide]

[/quote]

x can be greater than 16, since it's an average, and all you defined was that it was less than 17.

[quote="EFuzzy"]x can be greater than 16, since it's an average.[/quote]

Let x=the average of the first five games:

x<(23+14+11+20+5x)/9

4x<68

x<17

#4 is 16. If you got more than 18 then you probably connected the midpoints of all of the sides of the square and not the OPPOSITE sides.

#4 -

[hide]  Ah, opposite sides, gotcha[/hide]

I believe the problem implied that the triangles had to be made of lines that were already drawn in the diagram (e.g. ABM isn't drawn):

anyone other than silverfalcon know about how many people participated?

i got 29 for the basketball problem also

[quote="shake9991"]for the basketball question:
i got 29...
call the sum of the total points in her firs five games f.
so: (f+68)/9 >f/5
so:f<85
call the number points she scored in tenth game t.
so:
(f+68+t)/10> 18
so:
f+t>112
since we want t to be smallest, we want f to be the largest... i.e. f=84
and f+t to be smallest, so f+t = 113, t=29.... i think i went wrong somewhere[/quote]

I got 29 also. I think it's correct.

[quote="EFuzzy"]To RC-7th:
why not 16.8?[/quote]

Good Point.

i think answer to monogram question is [hide]B[/hide] (assuming that the last letter in the monogram is a Z)
l[hide]ets start listing the monograms:
A_Z: there are 24 possible letters for the _ , therefore 24 monograms
B_Z: similar idea, but 23 monograms... looking at this we see that our monorgrams go down to:
X_Z whihc has only one possible one... therefore, Mr. and Mrs. Zeta have 1+2+3+...+24 = 300 choices for possible monograms.[/hide]

lol....i missed #1....omfg *cuts throat*....omfg...and yes i'm still typing....126 here

103.5 here. I'm betting for lowest score  :lol:

138. Missed 2 questions, dumb mistakes for both.

here's number 15:

the highest average that the first five games could be was 16.8 (only five games) because it had to be less than seventeen.  {16.8(5)+17(4)+k}/10=18.1 (18.1 is a little less higher than 18, but again, it was out of 10 games so it couldnt be (18.0<x<18.1) so 29 is the answer.

Please discuss the problems and solutions on the Solution Thread.

Thank you.

can any1 show me the solution for qns 24?? The one on m+n?

This one was tough one. If you count the level of this problem as what might appear on AMC-10, I'm pretty sure it's #25. Anyway, it came from old AHSME (not that old).

Since the ratio of AC:AD is 2/3, we can reset their length to be:

2 :sqrt: 5 and 3 :sqrt: 5

Using Pythagorean Theorem, we can get CD = 5 and applying bisector theorem, say the midpoint of CD is E, then CE = 2 and ED = 3. Now, apply Pythagorean Theorem to the triangle ACE and get AE = :sqrt: 24. Since the EAB has same measure, apply bisector theorem and if we call the length of DB = x, then AB must equal :sqrt: 24 *x/3. Using Pythagorean Theorem of ABC,

$(2 \sqrt 5)^2 + (x+5)^2 = (\frac {x \sqrt 24}{3})^2$

Then it follows that BD = 9 and since CD = 5, it's 9+5 = 14.

This looks like a solid contest. Where are the answers?

whats the solution to number 25?

i just did the contest, i didn't answer that one and missed number 16.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be an infinite group containing element $ x\ne e$ having only finitely many conjugates. Prove that G is not simple.

let $ C$ be the finite conjugacy class of $ x$. then $ G\to Sym(C) , g\mapsto (y\mapsto g^{\minus{}1}y g)$ is a homomorphism, which is nontrivial (otherwise $ x$ lies in the center of $ G$, i.e. $ C\equal{}G$ is not finite) and not injective (since $ G$ is infinite and $ Sym(C)$ finite). thus it's kernel is a nontrivial proper normal subgroup of $ G$.

@-oo-:
The homomorphism can be trivial. Consider the group $ O_{3}(\mathbb{R})$. The element $ \minus{}id$ lies in the center and hence its conjugacy class has only one element.

One must change your proof a little bit:

Case 1: $ |C|\equal{}1$
Then x lies in the center of G and the subgroup $ \langle x\rangle\leq Z(G)$ is normal.

Case 2: $ |C|>1$
Now your proof is applicable.

thank you. $ x\in Z(G)\Rightarrow C \equal{} G$ was nonsense :D. in case 1, it may happen $ \langle x\rangle \equal{} G$ - then we have to take e.g. $ \langle x^{2}\rangle$.


limit
logarithms
Euler
inequalities
function
calculus
calculus computations

$\textbf{Show : }$

$\lim_{n\to1}\left[\zeta{(n)}-\frac{1}{n-1}\right]=\boxed{\gamma}$

$\lim_{n\to\infty}\frac{1}{\ln p_{n}}\prod_{i=1}^{n}\frac{p_{i}}{p_{i}-1}=\boxed{e^\gamma}$

Where $\gamma$ is the Euler-Mascheroni constant , $p_{n}$ is the n-th prime .

1) The limit can be written into an equivalent form as:

$\lim_{x\rightarrow 0}\zeta(x+1)-\frac{1}{x}=c$. This is a well-known result which is based on the following inequality involving the Zeta function:

$1+\frac{1}{2^{1+x}}+\cdots+\frac{1}{n^{1+x}}+\frac{1}{x}\left(\frac{1}{(1+n)^{x}}-1\right)<\zeta(x+1)-\frac{1}{x}<$

$<1+\frac{1}{2^{x+1}}+\cdots+\frac{1}{n^{x+1}}+\frac{1}{x}\left(\frac{1}{n^{x}}-1\right)$.

It follows that

$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln(n+1)\leq \underline{\lim}(\zeta(x+1)-\frac{1}{x})\leq \overline{\lim}(\zeta(x+1)-\frac{1}{x})$

$\leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n$.

The result follows.


"A point, P is located on the line $ y\equal{}2x$, it is 24 length units from the origin, what coordinates does it have?"


for numbers x, y,z satisfy: $ x \leq y \leq z$ and $ x\plus{}y\plus{}z\equal{}  \frac{1}{x} \plus{} \frac{1}{y} \plus{}\frac{1}{z}$
find smallest value of :$ xy^2z^3$

hello, do you mean real numbers or positive real numbers $ x,y,z$?
Sonnhard.

I assume you meant $ x,y,z>0$

[hide="Solution"]Using the conditions and AM-GM : $ 3z\equal{}z\plus{}z\plus{}z\ge x\plus{}y\plus{}z\equal{}\frac{1}{x} \plus{}\frac{1}{y}\plus{}\frac{1}{z}\ge\frac{3}{z}\Rightarrow
 z\ge 1$

In the same way : $ 3x\equal{}x\plus{}x\plus{}x\le x\plus{}y\plus{}z\equal{}\frac{1}{x} \plus{}\frac{1}{y}\plus{}\frac{1}{z}\le\frac{3}{x}$ so $ x\le 1$

Use the condition again : $ x\minus{}\frac{1}{x}\equal{}(y\plus{}z)\left(\frac{1}{yz}\minus{}1\right)\Rightarrow yz\ge1 \ (1)$

$ z\minus{}\frac{1}{z}\equal{}(x\plus{}y)\left(\frac{1}{xy}\minus{}1\right)\Rightarrow xy\le 1$

Also $ z\minus{}\frac{1}{z}\equal{}(x\plus{}y)\left(\frac{1}{xy}\minus{}1\right)\ge 2\left(\frac{1}{\sqrt{xy}}\minus{}\sqrt{xy}\right)\ge\frac{1}{\sqrt{xy}}\minus{}\sqrt{xy}\Rightarrow z\ge\frac{1}{ \sqrt{xy}}$ $ \Leftrightarrow xyz^2\ge 1 \ (2)$

From $ (1)$ and $ (2)$ we have $ xy^2z^3\ge 1$ .[/hide]

if x, y, z not positive  numbers ,resolving???

I sure x, y, z not positive numbers
they be just condition as previously

If i remember well this is an inequality by mr. Vasile Cartoaje and he said $ x,y,z>0$ .

[quote="bigzero"][quote="alex2008"]If i remember well this is an inequality by mr. Vasile Cartoaje and he said $ x,y,z > 0$ .[/quote]
i don't know this inequality by mr. Vasile Cartoaje but my problem not have x,y,z>0[/quote]

What's the source of your problem ?


function
algebra
polynomial
system of equations

Write $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$ as a function of $ \sigma_1\equal{}a\plus{}b\plus{}c$, $ \sigma_2\equal{}ab\plus{}bc\plus{}ca$, and $ \sigma_3\equal{}abc$.

[hide="Motivation"] This is the [url=http://en.wikipedia.org/wiki/Discriminant]polynomial discriminant[/url] of a general cubic $ (x \minus{} a)(x \minus{} b)(x \minus{} c)$.  In the two-variable case the discriminant of $ (x \minus{} a)(x \minus{} b) \equal{} x^2 \minus{} \sigma_1 x \plus{} \sigma_2$ is $ (a \minus{} b)^2 \equal{} \sigma_1^2 \minus{} 4 \sigma_2$, which should look familiar. [/hide]

I get an answer of
[hide]\[ (\sigma_1^2 \minus{} 2\sigma_2)((\sigma_1^2 \minus{} 2\sigma_2)^2 \minus{} 2\sigma_1\sigma_3) \minus{} 2\sigma_2(\sigma_1^2 \minus{} 2\sigma_2)((\sigma_1^2 \minus{} 2\sigma_2)^2 \minus{} 2\sigma_1\sigma_3) \\\plus{} 2\sigma_2(\sigma_1^2 \minus{} 2\sigma_2) \minus{} \sigma_1\sigma_3 \plus{} 4\sigma_3(\sigma_1\sigma_2 \minus{} 3\sigma_3) \minus{} 8\sigma_3^2
\] [/hide]Might be grossly wrong though.. t0rajir0u, how do you extend to the two-variable case to the three-variable case? I can't seem to do it.

[quote="aidan"]I get an answer of
[hide]
\[ (\sigma_1^2 \minus{} 2\sigma_2)((\sigma_1^2 \minus{} 2\sigma_2)^2 \minus{} 2\sigma_1\sigma_3) \minus{} 2\sigma_2(\sigma_1^2 \minus{} 2\sigma_2)((\sigma_1^2 \minus{} 2\sigma_2)^2 \minus{} 2\sigma_1\sigma_3) \\
\plus{} 2\sigma_2(\sigma_1^2 \minus{} 2\sigma_2) \minus{} \sigma_1\sigma_3 \plus{} 4\sigma_3(\sigma_1\sigma_2 \minus{} 3\sigma_3) \minus{} 8\sigma_3^2
\]
[/hide]Might be grossly wrong though.. t0rajir0u, how do you extend to the two-variable case to the three-variable case? I can't seem to do it.[/quote]
I am not sure if this works, but if you put $ a\equal{}b\equal{}0$, then $ \sigma_1\equal{}c, \sigma_2\equal{}\sigma_3\equal{}0$. Then solve for your thing and get $ c^6$, when in fact it should equal $ 0$... does that mean it's wrong?

[quote="aidan"]t0rajir0u, how do you extend to the two-variable case to the three-variable case? I can't seem to do it.[/quote]
Try a recursion.  I'm not sure if it works, but note that if $ P(x) \equal{} (x \minus{} a)(x \minus{} b)$ and $ \Delta$ is the discriminant of $ P(x)$ then the discriminant of $ P(x)(x \minus{} c)$ is $ \Delta P(c)^2$.  How are the coefficients of $ P(x)(x \minus{} c)$ determined by the coefficients of $ P(x)$?

That doesn't help writing $ \Delta$ in terms of the new symmetric sums though...

This isn't that hard a problem but it might be a bit computational...

[quote="t0rajir0u"]Try a recursion.  I'm not sure if it works, but note that if $ P(x) \equal{} (x \minus{} a)(x \minus{} b)$ and $ \Delta$ is the discriminant of $ P(x)$ then the discriminant of $ P(x)(x \minus{} c)$ is $ \Delta P(c)^2$.  How are the coefficients of $ P(x)(x \minus{} c)$ determined by the coefficients of $ P(x)$?[/quote]

How does $ \Delta P(c)^2$ come in?

Also, is it possible to write $ (a\minus{}b)(b\minus{}c)(c\minus{}a)$ in terms of elementary symmetric polynomials? I couldn't get that one  :P

It is of course impossible, since the expression is not symmetric. :)

[hide="Brute-Forcing"]

Note that the term has to be of the form $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} k_1\sigma_1^6 \plus{} k_2\sigma_1^4\sigma_2 \plus{} k_3\sigma_1^3\sigma_3 \plus{} k_4\sigma_1^2\sigma_2^2 \plus{} k_5\sigma_1\sigma_2\sigma_3 \plus{} k_6\sigma_2^3 \plus{} k_7\sigma_3^3$ with $ \sigma_1 \equal{} a \plus{} b \plus{} c$, $ \sigma_2 \equal{} ab \plus{} bc \plus{} ca$ and $ \sigma_3 \equal{} abc$. Now, setting $ a \equal{} b \equal{} 1\implies \sigma_1 \equal{} 2 \plus{} c, \sigma_2 \equal{} 1 \plus{} 2c, \sigma_3 \equal{} c$, the LHS is equal to zero, while the RHS is a polynomial of degree 6. Thus all coefficients of that polynomial have to be zero, which implies that $ k_1$ and $ k_2$ are zero (since the polynomial is of the form $ k_1c^6 \plus{} (12k_1 \plus{} 2k_2)c^5 \plus{} \mathcal{O}(c^4)$). The polynomial then turns out to be
\[ (k_3 \plus{} 4k_4)c^4 \plus{} (6k_3 \plus{} 20k_4 \plus{} 2k_5 \plus{} 8k_6)c^3 \plus{} (12k_3 \plus{} 33k_4 \plus{} 5k_5 \plus{} 12k_6 \plus{} k_7)c^2 \plus{} (8k_3 \plus{} 20k_4 \plus{} 2k_5 \plus{} 6k_6)c \plus{} (4k_4 \plus{} k_6)
\]
which gives the following system of equations:
\[ \begin{array}{rcc} k_3 \plus{} 4k_4 & \equal{} & 0 \\
6k_3 \plus{} 20k_4 \plus{} 2k_5 \plus{} 8k_6 & \equal{} & 0 \\
12k_3 \plus{} 33k_4 \plus{} 5k_5 \plus{} 12k_6 \plus{} k_7 & \equal{} & 0 \\
8k_3 \plus{} 20k_4 \plus{} 2k_5 \plus{} 6k_6 & \equal{} & 0 \\
4k_4 \plus{} k_6 & \equal{} & 0 \end{array} \\
[12pt] \implies (k_1,k_2,k_3,k_4,k_5,k_6,k_7) \equal{} \lambda(0,0, \minus{} 4,1,18, \minus{} 4, \minus{} 27)
\]
by standard Gaussian elimination. Now setting $ (a,b,c) \equal{} (1, \minus{} 1,0)$, we have $ \sigma_1 \equal{} \sigma_3 \equal{} 0$ and $ \sigma_2 \equal{} \minus{} 1$ and $ 4 \equal{} (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} k_6\sigma_2^3 \equal{} 4\lambda$ which means $ \lambda \equal{} 1$ and thus
\[ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} \sigma_1^2\sigma_2^2 \plus{} 18\sigma_1\sigma_2\sigma_3 \minus{} 4\sigma_1^3\sigma_3 \minus{} 4\sigma_2^3 \minus{} 27\sigma_3^2.
\]
[/hide]

I believe that is one of the best methods for doing this type of problem (I had a pretty ridiculous other solution that uses a whole bunch of random things but I considered this solution as well).

Also you can choose your values very carefully to get information that you want. Ex. Let $ f(t)$ be the polynomial with roots $ a,b,c$. We choose the coefficients so that things cancel.

$ t^3\minus{}1\equal{}0$,  then $ a,b,c\equal{}1,\omega,\omega^2$. This gives $ \minus{}27$ for the $ (abc)^2$ coefficient.

$ t^3\minus{}t\equal{}0$, then $ a,b,c\equal{}0,1,\minus{}1$. This gives $ \minus{}4$ for the $ (ab\plus{}bc\plus{}ca)^3$ coefficient.

Now we know that $ [(a\minus{}b)(b\minus{}c)(c\minus{}a)]^2\plus{}27 (a b c)^2\plus{}4(ab\plus{}bc\plus{}ca)^3$ is divisible by $ a\plus{}b\plus{}c$.

Then we can divide that out from both sides, and we choose $ t^3\minus{}3t\minus{}2$ to get $ k_5$.

Then everything is divisible by $ (a\plus{}b\plus{}c)^2$, and such we keep repeating, eliminating the terms with not every type of symmetric sum (If we can do this?).

I suppose this isn't better than what you did, but its more of an algorithm.


Se considera un triunghi $\ ABC$ in care luam punctele $\ M,N$ pe segmentul $\ BC$ astfel incat $\ m(<BAM)=2m(<CAN)$, $\ M,N$ diferite si $\ AB<AC$.Demonstrati ca exista un unic punct $\ X$ pe $\ (AC)$ cu proprietatea ca avem relatia $\ AB\cdot\ AM\cdot\ NC\cdot\ (AN^2-XC^2+AC^2-XN^2)=BM\cdot\ AN^2\cdot\ AC^2$.

As vrea sa stiu macar,..e grea sau usoara?De ce n-o-ncearca nimeni? :(

probabil ca n-o incearca nimeni , pt ca e prea lunga chestia de demonstrat :)
Nu neaparat ca ar fi prea usoara sau prea grea :) De obicei ..probleme frumoase au cerintze concise si clare :) Asta nu prea atrage  :blush:

Pai ce nu e concis?Uita-te din nou la problema..am mai adaugat ceva.

si punem pariu ca daca era la vreun concurs punea probleme participantilor?Trebuie sa ai vointa de fier ca sa rezolvi problemele astea <lungi>.Macar, ai o idee de rezolvare?Eu am solutionat-o in 8-9 randuri, nu stiu daca e chiar asa lunga...

salut !

nici mie nu-mi place sa ma apuc de o problema care nu-mi pare "eleganta"...cea de la olimpiada rusa mi-a placut si de aceea am "séché" pe ea vreme de doua ceasuri ! placerea pe care am simtit-o e numai cistig pentru mine ! :roll: pot chiar sa-tsi multsumesc ca ai pus-o pe forum !

bleaaaaaah
ideea imediat, a ramas doar un calcul inutil, care a dat frumos la sfarsit si m`a scapat de functii si de continuitate.
bleah
notam $k=\frac{BC\cdot AN^{2}\cdot AC^{2}}{AB\cdot AM\cdot NC}$.
sa il calculam pe $k$.
$k=\frac{BM}{AM}\cdot\frac{AC}{AB}\cdot\frac{AN}{NC}\cdot AN\cdot AC$.
din teorema sinusurilor, obtinem
$k=\frac{\sin\angle BAM}{\sin B}\cdot\frac{\sin B}{\sin C}\cdot\frac{\sin C}{\sin \angle NAC}\cdot AN\cdot AC$.
din ipoteza, $\sin\angle BAM=2\sin\angle NAC\cos \angle NAC$.
deci $k=2\cdot AN\cdot AC\cdot \angle NAC$, si din teorema cosinusului $k=AN^{2}+AC^{2}-NC^{2}$.
deci $XN^{2}+XC^{2}=NC^{2}$.
.......
naspa calculu asta..
acum sunt si eu curios :) doar intreb, nu dau cu parul.
ce solutie la nivelul clasei a VIII`a ai tu sorine ? pentru ca la multe probleme de pe forum zici ca nu e de clasa a VIII`a, deci nu ai cum sa o faci. acuma...functiile trigonometrice se invata intr`a IX`a...

Lucian, solutia ta e corecta, si in mare masura si eu am rezolvat-o la fel. :) Poate ca ar pune probleme destul de mari celor de-a VIII-a...


pifinity is ze best
centslordm is our idol
ur mom

A set of 25 square blocks is arranged into a $ 5\times5$ square.  How
many different combinations of 3 blocks can be selected from that
set so that no two are in the same row or column?

I used the solution given first, but forgot to divide by a 2 somewhere and got 1200.  Not realizing, I looked for another way.

Here is what I found:
[hide]Number the rows 1-5 and the columns A-E.  There are 5c3 ways to choose the rows and columns.  For example, ABE and 125 can be selected.  There are 3x2x1 ways to pair the letters and numbers (A can be paired with 1, 2, or 5.  Be can be paired with one of the two remaining, C gets the last number).  (5c3)(5c3)6 = 10x10x6 = 600.[/hide]

Alternatively:

You pick the first block. Looking at a pic, it becomes obvious that there are only 16 choices left (block you chose, 4 in same row, 4 in same column eliminated). Doing the same thing with the next there are 9 choices left (eliminate 7) Thus its 25x16x9=(3x4x5)^2=60^2=3600. However, we must divide by 3!=6 because order doesn't matter. It is 3600/6=600.

Consider the rows and columns separately. For rows, there are $5\cdot 4\cdot 3=60$ ways to choose $3$ distinct rows. Therefore, we have a total of $60^2=3600$. However, we need to divide by $6$ because we counted rearrangements of the three blocks, giving an answer of $\boxed{600}$.

Each possibility is determined by the row and the column of which the square is chosen from. For any possibility of rows and columns, we have $3\cdot 2=6$ ways of choosing the squares from the selection of rows and columns. Hence, the sol is $\binom 5 3\cdot \binom 5 3\cdot 6=\boxed{600}$.

[hide=Solution]
There are 25 choices for the first block. After selecting that block, there are nine blocks that we cannot choose next, one being the first block and the other eight being the blocks that were in the same row as the first block. So, there are 25-9=16 choice for the second block. Following similar reasoning, there are 9 choices for the third. Since there are 3! ways/orders to choose the same combination of blocks, the answer is (25*16*9)/6=[b]600[/b].
[/hide]


I spent about 35min per day on math over the summer holidays. :)

[quote="nutz_for2.718281828"]I'm planning to do 8 hours a day on summer holidays seriously.[/quote]

[i]do[/i] you spend 8 hours a day?
Sure, I'm planning to practice my cello 4 hours a day, seriously too, but I don't. Yet...

Bah, I only spend like an hour a day these days...
On math, that is

I spent different amounts of time per day. Some days 8 hrs, others only about 30 min. :(

I'm moving this topic to The Round Table, which should be a better forum for it.

Hm, 3 - 3.5 hrs.

Its mostly piano - 5 hrs.

[quote="conceive"][quote="rem"]Usually 5-7 hrs every day.
Hope it helps :)[/quote]

@rem: If the all other not sleeping time is spend how I wrote u via PM, then it is not so bad.. :wink:[/quote]
Good. :rotfl:

[quote="joml88"][quote="conceive"]Math during vacations? You are drunk, aren't you?[/quote]

Hopefully not.  Drunk deriving is dangerous ;)[/quote]

 :rotfl:

I usually do about 2 hours a day if I'm lucky...

And for people who are saying 8-10 hours, I highly doubt that :wink:

Some days, I can spend up to 5 hours on this website. It's so addicting. :blush: It's an average of about 4 hours of math, total.

I would say almost 24 hours if spending time on math means thinking about it.

wouldn't it be easier to measure the amount of time not spent on math?  :rotfl:

if "spending time on math" means factoring all numbers you happen to see... wow i do that all the time.

[quote="perfectnumber628"]if "spending time on math" means factoring all numbers you happen to see... wow i do that all the time.[/quote]

"We all use math every day (echo)... to predict weather, to tell time, to handle money. Math is more than formulas or equations; it’s logic, it’s rationality, it’s using your mind to solve the biggest mysteries we know..." Bleh.

Couple hours.

Right now 0 hrs but i never time my self or anything when i do math,like i'm satisfied only if i do the work i planned to do n stuff.

some times i have been spent $18$,

but, now, nothing,

Right now I'm spending about 6+ hours on math every day, due to being at [url=http://math.bu.edu/people/promys/]PROMYS[/url] and everything :P

Eh bien, there is french,writing, latin, math, flute, swimming, greek ,spelling, and reading...I switch on what I spend the most time depending on my goals...right now it's either math or french...

How could someone have so much time for math???!?!? I wish I had 50 hour days. That would be cool. :huuh: [/i]

Last night is the last night I stay up until five in the morning doing math.   :|

[quote="D.P.L"]I mean disincluding the time spent on AoPs. :lol:[/quote]

oh in that case maybe 15 minutes :wink:

a couple minutes? that might go up to an hour eventually...nah, I won't kid myself

I spend about 1 hour on this website each day, while posting problems I don't understand or just having fun. I get on and off really often to ask questions or discuss certain things.

As for math excluding this website, I spend about maybe 5 hours on weekends and 2 hours on weekdays. seriously

On thanksgiving holiday I will probably spend 6 hours at least because it's the last holiday before Vestavia Tournament.

*gets a sharpie*

*writes differential equations on bedsheets*

Now I spend 5-13 hrs a day on math!

Hmm. None, actually. I've been trying to figure out whether I want to or not.

other then this website I spend about 30 min on math...


probability
calculus
derivative
algebra unsolved
algebra

$p^m \sum_{k=m-1}^{+\infty}\binom k {m-1} q^{k-m+1}=1$
where $p+q=1,p>0,$,m is a positive interger.


Prove, that if lengths of all sides of the triangle and all its bisectors are rational numbers sines of all angles of this triangle are also rational numbers.

$ l_a \cdot l_b \cdot l_c  \equal{} \frac{8 abcpA}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

so $ \sin {\alpha} \equal{} \frac{l_a \cdot l_b \cdot l_c \cdot (a\plus{}b)(b\plus{}c)(c\plus{}a)}{2ab^2c^2(a\plus{}b\plus{}c)}$ that is rational and cyclic

ok, thanks,
but how prove, that:

$ l_a \cdot l_b \cdot l_c \equal{} \frac{8 abcpA}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

it's simple. Prove at first that $ l_a\equal{}\frac{\sqrt{bc(a\plus{}b\plus{}c)(b\plus{}c\minus{}a)}}{b\plus{}c}$, and so on.


[hide]Substituting the coordinates into the equation,
$y_1 = a + b + c\\
y_2 = a - b + c\\
y_1 - y_2 = 2b = -6\\
b = -3$ A.[/hide]

[hide]
since we know the x coordinates are 1 and -1 we just plug them in the functino to get:
(a+b+c)-(a-b+c)=6
so:
2b=6 and [b]b=3 (C is answer)[/b]

Edit: oops y1-y2=-6, so [b]b=-3[/b] not 3

[/hide]

This must have been one of the easier ones on that year's AHSME!

I think posting the answer again would be overkill


Find x if $ lne^{2cos^{2}x \minus{} 1} \equal{} cosx$, for $ 0 < x\leq\pi$.
A) $ 0$
B) $ \frac {\pi}{6}$
C) $ \frac {5\pi}{6}$
D) $ \minus{} \frac {\pi}{6}$
E) None of the Above

[quote="crazypianist1116"]Um, could you possibly have meant $ Lne$ not $ Ine$?[/quote]
Eh..its supposed to be a lowercase L >_>

OK then:
$ lne^{2cos^2x \minus{} 1} \equal{} cosx$
$ 2cos^2x \minus{} 1 \equal{} cosx$
$ 2cos^2x \minus{} cosx \minus{} 1 \equal{} 0$
$ (2cosx \plus{} 1)(cosx \minus{} 1) \equal{} 0$
$ cosx \equal{} \minus{} \dfrac{1}{2},1$
We throw out the negative root since a log can't be negative:
$ cosx \equal{} 1$
$ x \equal{} 0\to\boxed{A}$

Your answer is incorrect. Remark $ 0<x\leq 2\pi$.

hello it is given 0<x<=pi.
crazypianist1116's ans is right.

It was not $ 2\pi$ but $ \pi$.

The answer is $ E)$.

Whoops, I suppose that's what happens when you don't see the intricacies of the question...

$ ln e^{2\cos^2x \minus{} 1} \equal{} \cos{x}$
$ 2\cos^2x \minus{} 1 \equal{} \cos{x}$
$ \cos{2x} \equal{} \cos{x}$

There is no solution, is there? (in $ 0 < x \le \pi$, that is)

[quote="ffao"]$ ln e^{2cos^2x \minus{} 1} \equal{} cosx$
$ 2cos^2x \minus{} 1 \equal{} cosx$
$ cos 2x \equal{} cos x$

There is no solution, is there? (in $ 0 < x \le \pi$, that is)[/quote]
I think you're a bit confused. $ 2cos^2x \equal{} 2(cosx)^2$ not $ cos2x$.

No, ffao is correct. ($ \cos{2x}\equal{}2\cos^2x\minus{}1$)

I suppose I'll have to brush up on my trig identities then...

Just note that $ \cos(2x) \equal{} \cos(x \plus{} x) \equal{} \cos x\cos x \minus{} \sin x\sin x \equal{} \cos^2x \minus{} \sin^2x$, and since

$ \sin^2x \plus{} \cos^2x \equal{} 1$, $ \sin^2x \equal{} 1 \minus{} \cos^2x$ and $ \cos^2x\equal{}1\minus{}\sin^2x$, so plugging this in, we have

$ \cos(2x) \equal{} \cos^2x \minus{} \sin^2x \equal{} 2\cos^2x \minus{} 1 \equal{} 1 \minus{} 2\sin^2x$.


What is the product of all the positive integer divisors of 12?

The factors of 12 are $ 1, 2, 3, 4, 6, 12$. There are 3 pairs of them. The product is $ 1728$.

The factors of 12 are 1, 12, 2, 6, 3, and 4, so their product is $ 12 \times 2 \times 6 \times 3 \times 4 \equal{} 12 \cdot 12 \cdot 12 \equal{} 12^3 \equal{} 1728$.


group theory
superior algebra
superior algebra unsolved

a) Let $S_n$ be the symmetric group on $n$ letters, and let $f(n)$ be the number of pairs $\left(u,v\right)\in S_n \times S_n$ such that $u^2=v^2$. Show that $f(n)=p(n)n!$, where $p(n)$ denotes the number of partitions of $n$.
b) Generalize to other finite groups.

We are seeking the number of pairs $ u,v \in S_{n} $ such that $
u^{2} = v^{2} $. Write $ u = xy, v = y $, so we have $ xyxy = yy
$, so $ yxy^{-1} = x^{-1} $. Take now some $ x \in S_{n} $. Then
it is easy to see that $ x $ and $ x^{-1} $ are conjugate one to
another, because they have the same list of nonintersecting
cycles. Therefore for given $ x $, if we take some $ y_{0} \in
S_{n} $, such that $ y_{0}xy_{0}^{-1} = x^{-1} $, then we have a
bijection between the set of $ y $, such that $ yxy^{-1} = x^{-1}
$ and the set of $ z $, such that $ zxz^{-1} = x $, by $ z =
y_{0}^{-1}y $. Denote by $ g(x) $ the number of $ z $, which
commute with $ x $.
 So we have to find number of pairs $ x,z \in S_{n}
$, such that they commute : $ zxz^{-1} = x $, which is equal to $
\sum_{x \in S_{n}} g(x) $. Denote now by $ C_{1},C_{2}, \ldots ,
C_{p} \subset S_{n} $ the conjugation classes of the group $ S_{n}
$, where $ p = p(n) $. We have $ S_{n} $ acting by conjugation on
$ S_{n} $, so if we take some $ x \in C_{j} $, then the orbit of $
x $ is exactly $ C_{j} $, and number of elements in stabilizer of
$ x $ is exactly $ g(n) $, so we have that $ | C_{j} | \cdot g(x)
= n! $ for every $ x \in C_{j} $. Therefore $ f(n) = \sum_{x \in
S_{n} } g(x) = \sum_{j=1}^{p} \sum_{x \in C_{j}} g(x) =
\sum_{j=1}^{p} n! = p(n)n! $.

I hope it is correct.


function
inequalities
algebra proposed
algebra

Find all functions $ f: \left[0,\infty\right) \rightarrow \left[0,\infty\right)$ satisfying the conditions:
$ 4f(x) \geq 3x$ and $ f(4f(x)\minus{}3x)\equal{}x$ for all $ x \geq 0.$

The first inequality implies $ f(x)\geq\frac{3}{4}x$. Substitution into the second identity gives $ x\geq \frac{3}{4}(4f(x)\minus{}3x)\Longrightarrow f(x)\leq \frac{13}{12}x$. More, generally define two sequences via $ b_0\equal{}3,a_0\equal{}12$ and $ b_{n\plus{}1}\equal{}4a_n\plus{}3,a_{n\plus{}1}\equal{}4b_n$. Then $ \frac{b_n}{b_n\plus{}1}x\leq f(x)\leq \frac{a_n\plus{}1}{a_n}x$ for some $ n$ implies $ \frac{b_{n\plus{}1}}{b_{n\plus{}1}\plus{}1}x\leq f(x)\leq \frac{a_{n\plus{}1}\plus{}1}{a_{n\plus{}1}}x$. Taking limits $ n\rightarrow\infty$ we conclude $ f(x)\equal{}x$.


[geogebra]551d895dab53cb37d1fdcb0fb655ed202c6c2186[/geogebra]

If each of the three sides of a triangle are divided in four equal parts, and from each vetex is drawn a line which  divides the opposite side 3:1, such that the smaller part is one fourth of the total side lenght as shown in the  figure.....Prove that the ratio between the bigger triangle ($ IGH$) and the one made by the intersections of the lines drawn ($ JKL$), is $ \frac {4}{13}$, for every triangle. Generalize the ratio for the case when every side is divided in $ n$ parts.

Please help me solving it. Thanks! :D

[hide="Solution"]I'll do the general case:

Let $ GH \equal{} nGN$, $ HI \equal{} nHO$ and $ IG \equal{} nIM$

Note that $ [JKL] \equal{} [IGH] \minus{} ([GNM] \plus{} [HON] \plus{} [IMO]) \minus{} ([MJN] \plus{} [NKO] \plus{} [OML])$

Since the height of $ GNM$ is $ \frac {n \minus{} 1}{n}$ of $ IGH$ and its base is $ \frac {1}{n}$ of $ IGH$ (similarly for $ HON$ and $ IMO$) then its area is:

$ \frac {n \minus{} 1}{n} \cdot \frac {1}{n} \equal{} \frac {n \minus{} 1}{n^2}$ of $ [IGH]$.

So $ [GNM] \plus{} [HON] \plus{} [IMO] \equal{} \frac {3(n \minus{} 1)}{n^2}$ of $ [IGH]$

Now by Menelaus in triangle $ IGN$ we have that $ \frac {HN}{HG} \cdot \frac {IJ}{JN} \cdot \frac {GM}{MI} \equal{} 1$

But since $ \frac {HN}{HG} \equal{} \frac {n \minus{} 1}{n}$ and $ \frac {GM}{MI} \equal{} n \minus{} 1$ then we have:

$ \frac {IJ}{JN} \equal{} \frac {n}{(n \minus{} 1)^2} \Rightarrow \frac {JN}{IN} \equal{} \frac {(n \minus{} 1)^2}{n \plus{} (n \minus{} 1)^2}$

Now note that:

$ [IGN] \equal{} \frac {1}{n}[IGH]$

$ [IMN] \equal{} \frac {1}{n}[IGN]$

$ [MJN] \equal{} \frac {(n \minus{} 1)^2}{n \plus{} (n \minus{} 1)^2}[IMN]$

So putting these together we have: $ [MJN] \equal{} \frac {(n \minus{} 1)^2}{n^2(n \plus{} (n \minus{} 1)^2)}[IGH]$

Similarly for $ [NKO]$ and $ [OML]$

So $ [MJN] \plus{} [NKO] \plus{} [OML] \equal{} \frac {3(n \minus{} 1)^2}{n^2(n \plus{} (n \minus{} 1)^2)}[IGH]$

Putting all of the information we have together:

$ [JKL] \equal{} [IGH] \minus{} ([GNM] \plus{} [HON] \plus{} [IMO]) \minus{} ([MJN] \plus{} [NKO] \plus{} [OML])$

$ [JKL] \equal{} [IGH] \minus{} \frac {3(n \minus{} 1)}{n^2}[IGH] \minus{} \frac {3(n \minus{} 1)^2}{n^2(n \plus{} (n \minus{} 1)^2)}[IGH]$

$ \frac {[JKL]}{[IGH]} \equal{} 1 \minus{} \frac {3(n \minus{} 1)}{n^2} \minus{} \frac {3(n \minus{} 1)^2}{n^2(n \plus{} (n \minus{} 1)^2)}$

This is really ugly, but it works, and plugging in $ n \equal{} 4$ gives your desired $ \frac {4}{13}$[/hide]

Also, take a look at this: [url]http://en.wikipedia.org/wiki/Routh%27s_theorem[/url]

There is not any triangle GNM. And why its height is $ \frac{n\minus{}1}{n}$???

You'll need to draw in some extra lines to make the triangles I mentioned. Its height of $ \frac{n\minus{}1}{n}$ comes immediately from the ratios given at the beginning of the solution.


Okay.  So the countdown round will be an aim chatroom, beginning at[u][b] 7 PM EST, TOMORROW SATURDAY THE 6TH[/b][/u].   It will be a ladder competition between the top 6 competitiors, and more details will be giving at the beginning of the competition.

The top 6 written finishers were:

1. Ubemaya
2. SplashD
3. davidli
4. Anirudh
5. Fanatic
6. bpms

one the winner of 2-6 is decided, they will take on Ubemaya at 9 PM EST [due to scheduling problems for him].

BUMP as a reminder so there's a higher chance that if someone forgot, they'd log into AOPS and see this.

I thought it was a good idea :(

To Walk Around The River: What's your AIM?

eliross2

he's on right now (3:37 pm est)
-jorian

It's on in 15 minutes. IM me or eli to get an invite.

Results so far:
6. bpms
5. anirudh
4. fanatic
3. davidli
2. ?
1. ?

SplashD will go against Ubemaya for championship at 9 PM.

CD Scores:
fanatic vs bpms-fanatic 2-1
fanatic vs anirudh- fanatic 2-0
fanatic vs davidli- davidli 2-1
davidli vs splashD- splashD 2-0
splashD vs Ubemaya- ????

Not that it matters, but I thought I beat anirudh 2-0.

uhh...i dont think Ubemaya showed up, when should it be rescheduled

whenever you guys can. schedule it with each other and then let me or pratik know.  we are always around except on the saturday the 13th.

the countdown rounds were only 3 questions, and first person to 2 wins? wow...

yeah I won't be on aops january 13th. I got a math tournament too  :lol:


In A.D. 1990, an animal bone estimated to be 3000 yrs old was found. In about what year did this animal die?

Ah, It belongs to how long that animal lived?
If it died when it was 2999 years old, then it died in about A.D 1989

It should be 1011 B.C.

Although, that doesn't really matter, since 3,000 years is only an estimate.

Thank you for the answer to my math question. I thought it would be 1010 b.c. too, but what threw me off was the  no 0 year. But 1010 would be the only answer that fits. Thank you so much. By the way I have lots more, so I may be bugging some of you again for some answers, but you guys are math genius's so it shouldn't bother you right? :lol:


There are 5 red balls, 3 white balls, and 4 black balls in the bag. If we take one ball out of the bag in each turn without putting it back, what's the probability that the 4 black balls are all taken out with other balls left in the bag?

[hide="My Attempt"]Suppose we draw out each ball in turn and form a sequence of drawn balls. Then the restriction is "how many such sequences are there such that the last ball is not a black ball". (Since in this problem, the red balls are indistinguishable from the white balls, lets refer to them as "non-black" balls.) There are $ \binom{11}{4}$ desirable sequences.

To make the probability easier to calculate, we assume all the balls are distinct. Then we have $ \binom{11}{4} \cdot 4! \cdot 8!$ sequences of drawings, and each drawing happens with probability $ \frac{1}{12!}$, so our probability is $ \frac{330 \cdot 4! \cdot 8!}{12!}\equal{}\frac{2}{3}$.  :maybe:

(Is there an easier way to do this?)[/hide]

[quote="ThinkFlow"](Is there an easier way to do this?)[/quote]  There are 12 balls, 4 of which are black.  Therefore the last ball is black exactly $ \frac {4}{12}$ths of the time in a random ordering.

If I understood correctly the question, it's asking for the probability of drawing all $ 4$ black balls in $ 4$ turns. Then the probability is:

$ \frac{4}{12}\cdot \frac{3}{11}\cdot \frac{2}{10}\cdot \frac{1}{9}\equal{}\frac{1}{495}$

I'm surely missing the point :), could you please clarify this jelly119?

@Flame: It says with others left, not with ALL others left. That is, even if there is only one ball left after the last black ball is drawn, its ok.

[quote="Flame"]If I understood correctly the question, it's asking for the probability of drawing all $ 4$ black balls in $ 4$ turns. Then the probability is:

$ \frac {4}{12}\cdot \frac {3}{11}\cdot \frac {2}{10}\cdot \frac {1}{9} \equal{} \frac {1}{495}$

I'm surely missing the point :), could you please clarify this jelly119?[/quote]

Sorry for my ambiguity. :oops:   I mean, all the 4 black balls are taken out while there are still other balls (not necessarily all the other balls) left in the bag.
A possible exapmle,  [b]w b r r b r b r w b w r[/b].
By the way, I don't have the answer. :wink:

I get the same answer as ThinkFlow.

[hide]First we'll consider the case where balls are distinguishable in color but not between the ones of the same color. Arrange drawn balls in a sequence. We can have $ \frac {12!}{5!3!4!}$ different arrangements. Now consider sequences where the last ball is black. There are $ \frac {11!}{5!3!3!}$ of them. Using complementary probability gives us

$ 1 \minus{} \frac {\frac {11!}{5!3!3!}}{\frac {12!}{5!3!4!}} \equal{} \frac {2}{3}$.

Now we will do the same but we'll treat each ball as distinguishable object. There are $ 12!$ different arrangements in total and $ 4\cdot 11!$ different arrangements with black ball being last to be drawn (4 is here since there are 4 black balls). Again complementary probability gives us

$ 1 \minus{} \frac {4\cdot 11!}{12!} \equal{} \frac {2}{3}$[/hide]

[quote="JBL"] There are 12 balls, 4 of which are black.  Therefore the last ball is black exactly $ \frac {4}{12}$ths of the time in a random ordering.[/quote]

I think JBL's idea is perfectly neat. :coolspeak:

One more question: What's the probability that all the 4 black balls are drawn while there are both white and red balls left in the bag?

[quote="jelly119"]

One more question: What's the probability that all the 4 black balls are drawn while there are [b]both[/b] white and red balls left in the bag?[/quote]

This problem is exacly the original one I would like to propose. That's really challenging. But I've got some clever clues thanks to all your input.

[hide]Let
$ A$: The case where all black balls are drawn before red balls are.
     $ B$: The case where all balck balls are drawn before white balls are.
Then
$ P(A)\equal{}\frac{5}{5\plus{}4}\equal{}\frac59$
       $ P(B)\equal{}\frac{3}{3\plus{}4}\equal{}\frac37$
       $ P(A\cup B)\equal{}\frac{5\plus{}3}{3\plus{}4\plus{}5}\equal{}\frac23$ (JBL's idea! :) )
$ \Rightarrow P(A\cap B)\equal{}P(A)\plus{}P(B)\minus{}P(A\cup B)\equal{}\frac59\plus{}\frac37\minus{}\frac23\equal{}\frac{20}{63}$[/hide]


limit
integration
real analysis
real analysis unsolved

Evaluate \[ \lim_{n\rightarrow \infty} \sum _{k\equal{}1}^n \left( \sqrt{1\plus{}\frac{k}{n^2}}\minus{}1\right)\]

[hide="I got"] $ \frac{5}{24}$ , but i m not sure , that's why i posted[/hide]

I don't agree with that. Using the binomial series and the fact that for $ k\le n,\ \frac{k^2}{n^4}\equal{}O(n^{\minus{}2})$ I get:

$ \lim_{n\to\infty}\sum_{k\equal{}1}^n\left(\frac{k}{2n^2}\plus{}O(n^{\minus{}2})\right)\equal{}
\lim_{n\to\infty}\sum_{k\equal{}1}^n\frac12\cdot\frac{k}{n}\cdot\frac{1}{n}$

$ \equal{}\int_0^1\frac12\,x\,dx\equal{}\frac14.$

Yes you are right .I ve spotted my mistake .I had taken into acount that coefficient $ \minus{}\frac{1}{8}$ when I used series expansion.Thank you.


Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z).
Thanks in advance.

T is not injective and, as such, it does not have an inverse. For an example of the lack of injectivity, see $ (1,0,1)\rightarrow (2,1,1)$ and $ (2,\minus{}1,1)\rightarrow (2,1,1)$.

yes the linear application is not injective because kerT=Vect{(-1;1;0)}

_________________________
lahoucine


$ \emph{Palindromic primes}$ are numbers that are prime and which read the same backwards and forwards: for example, 11 is a palindromic prime. What is the sum of all palindromic primes between 100 and 200.

Look here, you're right.

http://primes.utm.edu/lists/small/1000.txt


As you read in the title, this is a Mock MathCounts test. It's not an ACTUAL MathCounts test, and it's not a contest.

I made it, and you can take a look and solve the questions if you're interested. You may also have seen some posts of mine in the Classroom math section, which I thought were Middle School Level. Whatever.

 :ninja:

Note: If you PM me your answers and I'm feeling goodhearted, I might grade them.

EDIT: Never mind, no edit.

This is why we have a MIDDLE SCHOOL TOURNAMENTS forum... :wink:

BTW: Make it PDF....

This is good.

However, #19 is a bit ambiguous. You can't tell whether 000020 would work if you use only what the problem says.

And some of the target problems wont show.

Still good nonetheless.

Eh...

Sorry it's not PDF - I don't have any converter or anything, and the online ones tend to mess up the borders and such.

Also, it's NOT a CONTEST, as I mentioned multiple times  :roll:

Oh and I forgot to mention - I ran out of ideas for the target round, so ... there's no problem or anything - some of the questions are blank.


I have submitted my first round solution online more than a week ago.

The site said "When we have received your solutions, a [check] will appear next to the round in the table above. Do not panic if you do not see a [check] right after mailing your solutions. It may be several days after the submission deadline until we log our receiving them."

Instead of the check that I'm suppose to see. I have a magnifying glass-ish symbol that has a link to a random, non-working page.

is that okay? Does that mean USAMTS received my solution?

[quote="007math"]As I remember, the magnifying glass will go to your solutions which you submitted (if online).  I believe if you do it online, the check mark will come up immediately, but your solution should show when you click the magnifying glass.  You should ask DPatrick (but I'm sure he's gonna reply to this post in the next 12 hours).[/quote]
The link doesn't work=.= It does look like something to links to my solution.


For HMMT, what are the time limits for the Subject tests? I have tried very hard to find them, but with no success.

[quote="Rules"]The Subject Tests (Algebra, Geometry, Calculus, and Combinatorics) are 50-minute tests each with 10 short-answer questions of increasing difficulty. Each test is out of a total of 50 points, with problems weighted according to difficulty. Individuals may take any two of the four tests. The problems are generally comparable to those of the Mandelbrot or ARML.[/quote]

If you were too lazy to read that, 50 minutes.


geometry
perimeter
geometric transformation
homothety
ratio
geometry proposed

Given a $n$_polygon on the plane. A point $O$ in the polygon is called Special point if every line through $O$ meets the perimeter of the polygon at two point $A$ and $B$, furthermore $\frac{OA}{OB}\in[\frac{1}{2},2]$. Prove that for all $n$_Polygon there exists at least one special point in the Polygon.

[quote="grobber"]I remember solving it before, but it's not worth the effort of looking for it :).

For each point $P$ on the border of the (I assume convex? you didn't mention that) polygon, let $S_P$ be the image of the polygon through the homothety centered at $P$ of ratio $\frac 23$. We want to prove that all $S_P$ have a common point. Since they're compact, this is equivalent to showing that any finite system has such sets has a non-empty intersection, and sicne they're convex, by Helly's Theorem, all we need to do is show that any three of them have a common point. This, however, is easy: if $P_1,P_2,P_3$ are points on the border of the polygon, the centroid of $P_1P_2P_3$ belongs to all $S_{P_i},\ i=\overline{1,3}$.[/quote]
you solved it yourself, didn't you? :P  :D  :lol:


can anyone tell me the links of

"Principles and Techniques in Combinatorics" by Chen Chuan-Chong and Koh Khee-Meng

i would be very grateful

I never saw this on the internet, but you can buy at Amazon.


AMC
AIME
ratio
geometry
inequalities
AoPS Books

Hello. I find that whenever a holiday begins and I plan on doing some math, I end up spending the majority of the time reviewing AOPS V1 and get very little of the other books done. Do you recommend this type of review or should I just move on to other books and rely on the fact that they will include some review. Also, I probably review AOPS V1 so much because I see an AMC problem which I cannot solve immediately (i.e. not in the time that the contest will permit) and conclude that I need to review the basics. Would you guys recommend just moving on to other books and not becoming fussed with the introduction series?

Thanks. Help is greatly appreciated. :)

I've found that it's impossible to really master the AMC without spending a lot of time practicing the AIME, and similarly it's impossible to master the AIME without spending a lot of time practicing the USAMO.  The same way, it's impossible to master AoPS Volume 1 without going on and doing harder stuff.  Don't convince yourself that you need to master AoPS Volume 1 before going on to more difficult math.  Actually, I'd argue that you should lose the mentality of mastering any aspect of math, since it will never happen - there will always be problems you can't solve.  Instead, when you want to spend time doing math, realize that your time will be best spent learning a new idea, studying a new section of a book, or doing problems you haven't done before.  Just try to allocate your time in the way that will yield the maximum possible results (and fun, of course).

I wouldn't worry so much when you can't solve an AMC problem.  You'll probably never reach the point where you know with absolute certainty how to solve every AMC problem immediately upon seeing the problem - I don't think anyone does.  If you do find a problem that you can't solve, spend an hour or so trying to solve it.  You'll learn a lot this way, whether you eventually solve the problem or you have to look at a solution.  This (especially if you solve the problem in the end) is how you really internalize problem solving techniques.

Thanks a lot. That is [b]really[/b] helpful. This will really help me balance pressure and fun when doing math. :)

I don't know if I can give any advice better than that, but I would just like to add that you should be spending time solving problems rather than reading about math. AoPS books are great, and that's the reason why, more time is spent doing problems, than doing the reading in AoPS books. Sometimes, you should also do practice competitions, so you can get used to real problems instead of just a bunch of problems that are in one field of math, then moving onto another.

Hope this helps  :D

Thanks. Sometimes what happens to me is I don't notice a certain solution (e.g. if you guys have AOPS V1, then the ratio example in Chapter 11--I solved it with areas but I think that the way the book solved it is better) and then spend hours trying to categorize the ideas used in the solution in my head. Do you guys do this? Or do you just move on but still acknowledge the solution as a valid method in this case?

Hmm, I am wondering the same thing. Personally, I think it is probably a good idea to think through the solutions and their ideas like that since it builds a stronger foundation but then again, it is rather time-consumptive.

But I think that although AoPS volume 1 is not that hard a book, there are definitely going to be problems with solutions that present new ideas and methods that would be beneficial to know. So I think it is a good idea to maybe visit volume 1 every once in a while like you do, but definitely move on. I'm sure similar methods will show up in more advanced math, and then maybe when you return to volume 1, the problems will be trivial and the solutions well-understood.

Above all, I think just doing problems help you to expand your thought process so that you are more able to see and develop these methods yourself.

I actually recommend doing volume 2. If you have a pretty solid (not necessarily complete mastery) in V1, you are ready to do V2. Now, I don't mean ALL of it.. especially not graph theory. But a lot of the topics in there are sometimes more interesting than the ones in V1 (especially inequalities and the proofs on diophantine equations). The first time, you might not understand 50% of V2. Do loads of problems. And go back to V2. You'll see everything in a different light, and improve A LOT. The first time I read through V2, I pretty much couldn't understand the harder proofs and only bothered with surface stuff; i couldn't solve  90% of the problems. Second time through, after doing loads of problems, I understand probaly 50%. Third time through, more concentrated, i could do almost all the problems, and I remembered everything much better (or at least, could derive most of the stuff fairly easily)


vector
linear algebra
linear algebra unsolved

For a n-dimension Vector Space $ V$, I choose n-linear independent vectors:

$ \alpha_1, \alpha_2, ..., \alpha_n$, for $ \forall$ $ \alpha, \beta \in V$

If $ \alpha \equal{} x_1\alpha_1 \plus{} x_2\alpha_2 \plus{} ... \plus{} x_n\alpha_n$ and $ \beta \equal{} y_1\beta_1 \plus{} y_2\beta_2 \plus{} ... \plus{} y_n\beta_n$

I define an inner product on $ V$

$ (\alpha, \beta) \equal{} \sum_{k \equal{} 1}^{n}x_ky_k$

Obviously $ \alpha_1, \alpha_2,...,\alpha_n$ is a orthogonal basis on $ V$.

so ,[color=red]can we say[/color]: If we define the inner product as $ \sum_{k \equal{} 1}^{n}x_ky_k$, then every basis is a orthogonal basis?

Under suitable choice of inner product, yes... so unfortunately orthogonality means little unless you know some more facts about the inner product you are using.


calculus
derivative
logarithms
calculus computations

I read this one somewhere and figured it would be an easy chain rule solution... It seems simple enough but I can't get the right answer.

$\frac{d}{dx}x^{x}$
Anybody care to help me out?  :P

Write $x^{x}=e^{x\ln x}.$ Either that or use logarithmic differentiation.

[hide="The answer."]$\frac{d}{dx}x^{x}=x^{x}(1+\ln x).$[/hide]


Let $x_{1}$ and $x_{2}$ be natural roots of ecuation $x^{2}+2mx+p^{2}+1=0$, $p \in \mathbb{Z}$, $m \neq 0$.

Show that:
a) $m^{2}\ge 2p$
b) $p^{4}+4m^{2}$ is not prime

Ioan Popa (Cluj Local Round, 2002)

a)Since $x_{1}$ and $x_{2}$ are natural roots then
 $m^{2}\ge{p^{2}+1}$  (discriminant)
$p^{2}+1\ge2p$    (Cauchy)
then
$m^{2}\ge2p$


function
integration
real analysis
real analysis theorems

Does anybody know how to proove the following series is or is not uniformly convergent on R(all real numbers). The Weierstrass M-Test doesn't work here, any help would be appreciated.
$ \sum_{n\equal{}1}^{\infty} \frac{1}{n\plus{}n(n\minus{}x)^2}$

For any $ x \in \mathbb{N}, \, x > 1$ you have $ \frac{1}{0}$ form in your sum for $ n \equal{} x\minus{}1$.

This is uniformly convergent.

Start with this: for all $ x\in\mathbb{R}$

$ \sum_{k\equal{}\minus{}\infty}^{\infty}\frac1{1\plus{}(x\minus{}k)^2}\le \sum_{k\equal{}\minus{}\infty}^{\infty}\frac1{1\plus{}k^2}\equal{}C<\infty.$

Then $ \sum_{n\equal{}N}^{\infty}\frac{1}{n(1\plus{}(x\minus{}n)^2)}\le \frac1N\sum_{n\equal{}N}^{\infty}\frac1{1\plus{}(x\minus{}n)^2}\le \frac{C}{N}.$

Since this can be made arbitrarily small for sufficiently large $ N,$ no matter what $ x$ is, the convergence is uniform.

============

The generalization looks like this: suppose $ g(x)$ is a "nice enough" function on $ \mathbb{R}$ such that $ \int_{\mathbb{R}}|g(x)|\,dx<\infty.$ (I'm not quite sure what "nice enough" should be; $ g$ continuous, even, and decreasing on $ [0,\infty)$ is sufficient.) Let $ (b_n)$ be a decreasing sequence that tends to zero. Then $ \sum_{n\equal{}1}^{\infty}b_ng(x\minus{}n)$ converges uniformly on $ \mathbb{R}.$

[quote="GrandFlee"]Why ? When you plug in n = x - 1 you get 1/2n term. I guess you didn't notice that (n-x) in denominator is squared.[/quote]

Ups.. Yes, that's true.. Sorry..


For $ x,y,z$ positive real numbers

$ \sum \frac{x^2y}{z(x^2\plus{}xy\plus{}y^2)} \geq \frac{\sum xy}{\sum x^2}$

It is here:
http://ddbdt.tk/forum/ca-c-cua-c-thi-trasn-vimf/136-ae-thi-cha-nh-tha-c-cua-c-thi-vicfj.html

[quote="manlio"]For $ x,y,z$ positive real numbers

$ \sum \frac {x^2y}{z(x^2 \plus{} xy \plus{} y^2)} \geq \frac {\sum xy}{\sum x^2}$

It is here:
http://ddbdt.tk/forum/ca-c-cua-c-thi-trasn-vimf/136-ae-thi-cha-nh-tha-c-cua-c-thi-vicfj.html[/quote]
Using the Cauchy-Schwarz Inequality, we have
$ \left[ \sum \frac{x^2y}{z(x^2\plus{}xy\plus{}y^2)}\right] \left[\sum yz(x^2\plus{}xy\plus{}y^2)\right] \ge \left(\sum xy\right)^2.$
Therefore, it suffices to prove that
$ \left(\sum xy\right)\left(\sum x^2\right) \ge \sum yz(x^2\plus{}xy\plus{}y^2),$
which is equivalent to the obvious inequality
$ xy^3\plus{}yz^3\plus{}zx^3 \ge xyz(x\plus{}y\plus{}z).$


i found this problem on ahsme,

show that if a is different from c, a^x=c^q and c^y=a^z, then xy=qz

i took the ecuation a^x=c^q and multiplied by a^z and by c^y
here is what results, a^(x+z) c^y=a^z c^(q+y), then i asumed that (x+z).y=z(q+y), and if we solve, it results that xy=qz.

i dont know if what i assumed is right, and if it is, i dont exactly know why, can someone explain please

I already answered your question in your other thread.  Please don't post questions twice.

Sorry, i considered the high school forum more appropiate for my problem


probability
videos
Alcumus
expected value

[quote="GameBot"]At a carnival game, Katya chooses a number from 1 through 6 and rolls two ordinary dice.  If her number comes up on one die, she wins three tickets.  If her number comes up on both dice, she wins six tickets.  If she plays all morning, how many tickets would she expect to win, on average, per turn?[/quote]

This should be $ 3 \times \frac{10}{36} \plus{} 6 \times \frac{1}{36} \equal{} 1$. The reasoning behind is this is that there is a $ \frac{1}{6} \cdot \frac{5}{6}$ probability of her number coming up on one die, but we must multiply by 2 because it could be either dice. Similar reasoning goes for the second term in the LHS of the equation. The fact that she plays all morning is irrelevant, as we are looking for expected value. To read/learn more about expected value, I suggest the videos and problems in Alcumus, as well as the AoPS Wiki article.

why in the world does it say that the answer is 1?

I literally put 13/12  and it said i was wrong

@rightysportsfan1 because the answer is 1.  There are 6*6=36 combinations of rolling the 2 dice.  Lets pretend Katya picked the number 1.  When she rolls 2 ones, then she will earn 6 tickets.  If she rolls a 1 on only one die, she earns 3 tickets.  There is a 5*2=10 combinations of rolling 1 on one die.  Therefore, 10*3=30 tickets.  The rest of the combinations will result in no tickets so we add 30+6=36.  Then 36 tickets/36 combinations=1 ticket per combination so the avg. tickets she would expect to earn is 1.


In the USAMTS Rules/FAQ, it is stated that computer solutions are acceptable. So isn't a solution justified with the help of a graphing utility (not just from a computer, but from a calculator) and just saying or explaining that the "graphing utility said so"?

You have to explain the program used and why it would produce the solution.

I was asking this question because it would seem justified that one uses a calculator to graph [two equations of an equality] and use that to justify one's answers. This is obviously trivial, but the ambiguity of the rules seem to allow this. I mean, any graphing utility is considered a "program."

EDIT: Sorry. I was conscious about referring to specific questions. I'm satisfied with zanttrang's response. That justifies it. Haha. Justifies.

Firstly, you're not allowed to discuss possible solutions to specific problems, so you should modify your post accordingly.  Secondly, graphing curves and examining their features visually is never rigorous - your solution is only as good as the graphing program that provided it, and you have not proven that your program is giving a true representation of the curves being graphed.  As an example of what I mean, try using a graphing program to find me all solutions to

$x = \sin\left(\frac{1}{x}\right).$

All USAMTS problems require you to give a proof of your answer.  Did you prove that the program gives you the EXACT right answer?


A convex pentagon $ABCDE$ has rational sides and equal angles. Show that it is regular.

See
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33920

  darij


geometry
analytic geometry
trigonometry
geometry unsolved

Given a triangle ABC,let A1,B1,and C1 be the midpoints of BC,CA,and AB respectively.
Let P,Q,and R be the points of tangency of the incircle k with the sides BC,CA,and AB.
Let P1,Q1,and R1 be the midpoints of the arc QR,RP,and PQ on which the points P,Q,and R divide the circle k,and
let P2,Q2,and R2 be the midpoints of arcs QPR,RQP,and PRQ respectively.
Prove that the line A1P1,B1Q1,and C1R1 are concurrent,as well as the lines A1P1,B1Q2,and C1R2.

I long for your answers.Thank you. :blush:

The first point of concurrency is known as the second mid-arc point ($ X_{178}$ in the [url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html]ETC[/url]).  This point has barycentric coordinates
\[ \left( \cos \frac{B}{2} \plus{} \cos \frac{C}{2}, \cos \frac{C}{2} \plus{} \cos \frac{A}{2}, \cos \frac{A}{2} \plus{} \cos \frac{B}{2} \right),\]
and is mentioned in these posts:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20543[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22225[/url]
In the second post, Darij mentions that even he does not know of a synthetic solution, but it can be proven using barycentric coordinates.

I haven't gone through all the calculations, but the second point of concurrency seems to have barycentric coordinates
\[ \left( \cos \frac{B}{2} \plus{} \cos \frac{C}{2}, \cos \frac{C}{2} \minus{} \cos \frac{A}{2}, \minus{}\cos \frac{A}{2} \plus{} \cos \frac{B}{2} \right).\]
(Note that these coordinates are the same as the coordinates above, except for the sign of $ \cos (A/2)$.)

[quote="pigfly"]Suppose $ I$ is incenter of triangle $ ABC$ , $ AI$ meet $ (I)$ in $ A_1$, such that $ A_1$ in segment $ AI$. $ A_2$ is midpoint of $ BC$, $ B_1$, $ B_2$, $ C_1$, $ C_2$ similar. prove that $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ are concurrent.[/quote]
Jean Louis Ayme (jayme) also asked me two weeks ago about a synthetic solution for this problem. Unfortunately, I have just a half-synthetic approach by making use of a result of Daneels, combined with the converse of Steinbart's theorem. Let me explain:

According to the converse of Steinbart's theorem, since the triangles $ ABC$ and $ A_{1}B_{1}C_{1}$ are perspective at $ I$ (and after seeing that the collinearity from the Steinbart condition does not hold), we have that the triangles $ A_{1}B_{1}C_{1}$ and $ DEF$ are perspective as well, where $ D$, $ E$, $ F$ are the tangency points of the incircle with $ BC$, $ CA$, and $ AB$, respectively. On other hand, according to [url=http://forumgeom.fau.edu/FG2006volume6/FG200629index.html]Eric Danneels, On triangles with vertices on the angle bisectors, Forum Geometricorum, 6 (2006) 247--253[/url], we conclude that $ A_{1}B_{1}C_{1}$ and $ A_{2}B_{2}C_{2}$ are perspective.

It would be even nicer to see a synthetic proof of Daneels' theorem.

I appreciate your answers.It being too hard,I will give up the problem. :( TOT

For a proof of Steinbart's theorem (including its converse) and some other references, see: Darij Grinberg, Variations of the Steinbart Theorem, available at [url]http://www.cip.ifi.lmu.de/~grinberg/[/url].

Dear Mathlinkers,
you can also se my proof (in French, but with a lot of figures) on
http://perso.orange.fr/jl.ayme   vol. 3  les points de Steinbart et de Rabinowitz
Sincerely
Jean-Louis


geometry
geometric transformation
reflection
angle bisector
geometry proposed

Let $ ABC$ be a $ A$ - right triangle. Define the points : $ \|\begin{array}{c} D\in (BC)\ ,\ \widehat {DAB}\equiv\widehat {DAC} \\
 \\
N\in (AB)\ ,\ NB = 2\cdot NA \\
 \\
E\in AD\cap CN\end{array}$

Suppose that $ \widehat {ABE}\equiv\widehat {ACB}$ . Ascertain $ m(\widehat {ACB})$.

[hide="Solution"]
Let $ BE$ meet $ AC$ at $ X$ and let the reflection of $ X$ across $ AB$ be $ Y$.  Since $ \angle XAB\equal{}90$, we have that $ \angle YAB\equal{}90$, so $ \angle XAB\plus{}\angle YAB\equal{}180$, which implies that $ X$, $ A$, and $ Y$ are collinear.  Thus, $ Y$ is a reflection of $ X$ over $ A$ as well.  Now, letting $ \angle ABE\equal{}\angle ACB\equal{}\alpha$, we have that $ \angle ABC\equal{}180\minus{}\angle ACB\minus{}\angle BAC\equal{}90\minus{}\alpha$.  Also, \[ \angle YBC\equal{}\angle ABC\plus{}\angle YBA\equal{}\angle EBA\plus{}90\minus{}\alpha\equal{}\alpha\plus{}90\minus{}\alpha\equal{}90\]  Ceva's Theorem on concurrent cevians $ AD$, $ CN$, and $ BX$ yields that $ \frac{CD}{BD}\cdot \frac{BN}{AN}\cdot \frac{AX}{CX}\equal{}1$.  Since $ \frac{BN}{AN}\equal{}2$ and $ \frac{CD}{BD}\equal{}\frac{AC}{AB}$ because $ AD$ is an angle bisector, we can see that $ \frac{AX}{CX}\equal{}\frac{AB}{2AC}$.  Since $ AX\equal{}\frac{XY}{2}$, we have that $ \frac{XY}{XC}\equal{}\frac{2AX}{CX}\equal{}\frac{AB}{AC}\equal{}\frac{BD}{CD}$, so $ XD\parallel BY$.  Also, since $ \angle YBC\equal{}90$, we have that $ XD\perp BC$.  Since $ \angle XDB\plus{}\angle XAB\equal{}90\plus{}90\equal{}180$, we have that $ XDBA$ is cyclic.  Therefore, $ \angle XDA\equal{}\angle XBA\equal{}\angle EBA\equal{}\alpha$.  Also, notice that $ \angle CAD\equal{}\angle BAE$ and $ \angle EBA\equal{}\angle ACB$, so \[ \triangle CDA\sim \triangle BEA\implies \frac{CD}{BD}\equal{}\frac{AC}{AB}\equal{}\frac{CD}{BE}\] and so $ BD\equal{}BE$.  Since $ \angle XBC\equal{}\angle ABC\minus{}\angle EBA\equal{}90\minus{}\alpha\minus{}\alpha\equal{}90\minus{}2\alpha$, we have that $ \angle EDB\equal{}90\minus{}\frac{\angle EBD}{2}\equal{}45\plus{}\alpha$.  It follows that $ 90\equal{}\angle XDB\equal{}\angle XDA\plus{}\angle ADB\equal{}45\plus{}\alpha\plus{}\alpha\implies \angle ACB\equal{}\alpha\equal{}22.5$ [/hide]

Remark regarding [b]QuattoMaster 6000[/b] solution:
After having seen $DX \perp BC$ and $BY %Error. "parralel" is a bad command.
DX$, take $Z$ the reflection of $B$ in $A$, the triangle $BDZ$ is right-angled at $D$, $AD=AZ$, so $2\alpha=m(\widehat{BAD})=45^\circ$.

[b]A less attractive solution:[/b]
Let's note $c=AB$, $b=AC$ and $\widehat {ACB}=\alpha$.
If $\{F\} \in BE \cap AC$, then $\triangle ABF \sim \triangle ACB$ and $c^2=b\cdot AF$ $(1)$; subsequently $CF=\frac{b^2-c^2}{b}$, $(2)$.
Applying Ceva for $ABC$ with concurrent cevians $AD, BF, CN$ we shall get $2bc=b^2-c^2$ or $\frac{2bc}{b^2-c^2}=1$ $(3)$, with equivalent writing: $\frac{\frac {2c}{b}}{1-\frac{c^2}{b^2}}=1$, $(4)$. But $(4)$ is equivalent to: $\frac{2\cdot tan (\alpha)}{1-tan^2 (\alpha)}=1$, i.e. $tan (2 \alpha)=1$, or $\alpha=22.5^\circ$, done.

Best regards,
sunken rock


If $ a^2\equal{}5a\minus{}3$ and $ b^2\equal{}5b\minus{}3$,do a and b represent roots of an equation of form $ x^2\equal{}5x\minus{}3$?
If yes,how?
Thanks. :)

Yes. By definition, if $ r$ is a root of $ f(x)$, then $ f(r) \equal{} 0$. In this case we have $ f(x) \equal{} x^2 \minus{} 5x \plus{} 3$. Because $ a$ and $ b$ are roots, we have:

$ f(a) \equal{} a^2 \minus{} 5a \plus{} 3 \equal{} 0\implies a^2 \equal{} 5a \minus{} 3\\
f(b) \equal{} b^2 \minus{} 5a \plus{} 3 \equal{} 0\implies b^2 \equal{} 5b \minus{} 3.$

The root of an equation is when the equation intersects the $ x$ axis.  In another words, it is when $ f(x)\equal{}0$.

Answering your question, yes.

Basically what TachyonPulse said.


function
linear algebra
matrix
calculus
derivative
topology
algebra

Let $ W_{\varepsilon}\equal{}\{(x,y,z): x^2\plus{}y^2<\varepsilon^2\}$ and $ A\equal{}\{(x,y,z): \textrm{ there exists a solution } (u,v) \textrm{ of the system of equations } (R)\}$,
where $ (R)$:
$ 1\plus{}u^3\equal{}x\plus{}uxy\minus{}v y^2, \\
\minus{}1\plus{}v^2\equal{}y\plus{}u^3 x\minus{}v^2 y^2 , \\
(1\plus{}u)^3\plus{}(1\plus{}v)^3\equal{}z\plus{}u^2 x\minus{} v^3 y.$

Prove that for sufficiently small $ \varepsilon>0$ the set $ W_{\varepsilon} \cap A$ is a sum of disjoint, connected $ \mathcal{C}^{\infty}$ submanifolds. How many such submanifolds are there? Determine the spaces tangent of these submanifolds at points of the set $ S\equal{}(W_{\varepsilon} \cap A) \cap (\{0\} \times \{0\} \times \mathbb{R})$.

$ A\cap (\{0\}\times\{0\}\times\mathbb{R}) \equal{}\{(0,0,0),(0,0,8)\}$.
1) The 3 equations can be written $ F(u,v,x,y,z)\equal{}0$. We obtain 2 local $ C^\infty$ implicit functions $ (x,y,z)\equal{}\phi(u,v),(x,y,z)\equal{}\psi(u,v)$ s.t. $ \phi(\minus{}1,\minus{}1)\equal{}(0,0,0),\psi(\minus{}1,1)\equal{}(0,0,8)$ if the matrix of partial derivative of $ F$ with respect to $ x,y,z$ is invertible:
$ det\begin{pmatrix}1\plus{}uy&u^3&u^2\\ux\minus{}2vy&1\minus{}2v^2y&\minus{}v^3\\0&0&1\end{pmatrix}\equal{}1$ if $ x\equal{}0,y\equal{}0$.
2) Moreover the first 2 equations can be written $ G(u,v,x,y)\equal{}0$. We obtain a local $ C^\infty$ diffeomorphism $ (x,y)\equal{}\theta(u,v)$ between the neighborhoods of $ (\minus{}1,\minus{}1)$ and $ (0,0)$ or between the neighborhoods of $ (\minus{}1,1)$ and $ (0,0)$ if the matrix of partial derivative of $ G$ with respect to $ u,v$ is invertible:
$ det\begin{pmatrix}3u^2\minus{}xy&\minus{}3u^2x\\y^2&2v\plus{}2vy^2\end{pmatrix}\equal{}6u^2v$ if $ x\equal{}0,y\equal{}0$.
3) Thus if $ \epsilon$ is small then we obtain 2 $ C^\infty$ disjoint manifolds of dimension $ 2$. Moreover $ (x,y)$ is in a connected domain then $ (u,v)$ also and the manifolds are connected.