point set
combinatorics
Triangle
combinatorial geometry
IMO Shortlist

Please if someone can provide a solution !

Notice the bound $2n-5$ cannot be lowered, since a planar triangulation using $n$ vertices has precisely $2n-5$ (bounded) facets (triangles).

Let $S=\{S_1,S_2,\cdots, S_n\}.$ A triangle, henceforth, will be referred to as the union of its perimeter and interior.

The crucial lemma is as follows:

[b]Lemma:[/b] Given a segment $\overline{AB}$ and a set of points $X=\{X_1,X_2,\cdots, X_m\}$ such that all of the $X_i$ lie on the same half-plane WRT $AB,$ the intersection $\bigcap_{i=1}^{m}\triangle ABX_i$ is nonempty.

[b]Proof:[/b] We will actually show that the intersection is a nondegenerate triangle with $AB$ as a side. The case $m=3$ is trivial. Assume the claim is true for $m=k,$ and that the desired intersection is $\triangle ABP.$ Then from the convexity of $ABPX_{k+1},$ we have exactly one of $AP\cap BX_{k+1},BP\cap AX_{k+1}$ nonempty.

WLOG let the latter case be true. Then if $BP\cap AX_{k+1}=P',$ $\triangle ABP\cap \triangle ABX_{k+1}=\triangle ABP'$ which is clearly nondegenerate. Hence by induction, we may conclude.

[b]Back to the main problem:[/b]

Let the convex hull of $S$ have $k$ vertices. The constructions for $k=3,4$ are trivial. If $k>4,$ we can have a construction with $n$ points by taking a point in $\triangle S_iS_{i+1}S_{i+2}\cap \triangle S_{i+1}S_{i+2}S_{i+3}$ for all $1\leq i\leq n,$ where indices are taken $\mod n.$ Since $n<2n+5,$ this suffices.

Now the problem is trivial. For each point $X$ within the convex hull of $S,$ the line $XS_1$ partition the rest of the $S_i$ into two half planes, so by [b]Lemma[/b], at most $1$ new point is necessary for each half plane, for a total of $2$ new points for each point in the convex hull of $S,$ so we may conclude.

[quote=Generic_Username]...Let the convex hull of $S$ have $k$ vertices. The constructions for $k=3,4$ are trivial. If $k>4,$ we can have a construction with $n$ points by taking a point in $\triangle S_iS_{i+1}S_{i+2}\cap \triangle S_{i+1}S_{i+2}S_{i+3}$ for all $1\leq i\leq n,$ where indices are taken $\mod n.$ Since $n<2n+5,$ this suffices.

Now the problem is trivial. For each point $X$ within the convex hull of $S,$ the line $XS_1$ partition the rest of the $S_i$ into two half planes, so by [b]Lemma[/b], at most $1$ new point is necessary for each half plane, for a total of $2$ new points for each point in the convex hull of $S,$ so we may conclude.[/quote]
But apart from that triangles, there are many others, such as: $S_i S_j S_k$ (where i,j,k are not consequtive), $X_iX_jS_k$, $X_i X_jX_k$, where $X_i,X_j, X_k$ are inside the convex hull. You should also take care of them!?

[quote=orl]$ S$ be a set of $ n$ points in the plane. No three points of $ S$ are collinear. Prove that there exists a set $ P$ containing $ 2n \minus{} 5$ points satisfying the following condition: In the interior of every triangle whose three vertices are elements of $ S$ lies a point that is an element of $ P.$[/quote]

Let $\vec v$ be a vector with $\vec v\nparallel \vec{S_iS_j}$ for $S_i,S_j\in S$.

$\{\vec s\pm\varepsilon\vec v\mid \vec s\in S\}$ is a set of $2n$ points satisfying the condition for sufficent small $\varepsilon>0$. At least 5 points are not inside the convex hull and therefore not needed.

[quote=dgrozev][quote=pi_quadrat_sechstel]
Let $\vec v$ be a vector with $\vec v\nparallel \vec{S_iS_j}$ for $S_i,S_j\in S$.
$\{\vec s\pm\varepsilon\vec v\mid \vec s\in S\}$ is a set of $2n$ points satisfying the condition for sufficent small $\varepsilon>0$. At least 5 points are not inside the convex hull and therefore not needed.[/quote]
Let $s_1,s_2\in S$. Why are you sure that there is a point of $S$ in the [b]interior [/b]of the triangle determined by the points $\vec{s_1}+\varepsilon\vec{v}, \vec{s_1}-\varepsilon\vec{v}, \vec{s_2}+\varepsilon\vec{v}$?[/quote]

Of course, there is not necessarily a point of $S$ in a triangle with vertices in $P$.

The condition is that in the interior of any triangle whose vertices are in $S$ is a point of $P$. You switched $S$ and $P$.

My bad, I obviously forgot the statement! Nice solution!


calculus
integration
rotation
conics
parabola
real analysis
real analysis unsolved

[quote="Nea"]Find surface of the zone closed by $ox : y^{2}= 4ax$, $x+y=3a$[/quote]
Instead we can think of $x$ as $y$ and $x$ as $y$. So the picture is rotated 90 degress.

I think we need to know what $a$ is in sign. If it is zero then you have a strange case. So lets say $a>0$, so $x^{2}= 4ay$ is a parabola and $x+y=3a$ is a line. Just write it in standard form $y=\frac{x^{2}}{4a}$ and $y=-x+3a$.

Let us find the intersection points,
$\frac{x^{2}}{4a}=-x+3a$

$x^{2}=-4ax+12a^{2}$

$x^{2}+4ax-12a^{2}=0$

$x= \frac{-4a \pm \sqrt{16a^{2}+48a^{2}}}{2}= \frac{-4a \pm 8a}{2}=-2a \pm 4a =-6a, 2a$

The top curve is the line and the bottom is the parabola.
Hence,

$\int_{-2a}^{6a}\left[(-x+3a)+\frac{x^{2}}{4a}\right] dx$


If f is a linear function such that:   f(x+2) - f(x)= 6, find the value of

f^-1 (x+2) - f^-1(x)

f^-1 = Inverse of f

It's very easy :lol:
If $f: \mathbb R\to \mathbb R$ is linear then $f(x)=ax+b$,$a\neq0$ a,b $\in \mathbb R$ .$f(x+2)-f(x)=6$ $\leftrightarrow$ $2a=6$ $\rightarrow$ $a=3$
so $f(x)=3x+b$ $\rightarrow$ $f^{-1}(x)=\frac{x-b}{3}$ $\rightarrow$ $f^{-1}(x+2)-f^{-1}(x)=\frac{x+2-b-x+b}{3}=\frac{2}{3}$


I solved this question wrong, but I would like to see some feedback.

Problem:

Five balls are numbered 1 through 5 and placed in a bowl. Josh will randomly choose a ball from the bowl, look at its number and then put it back into the bowl. Then Josh will again randomly choose a ball from the bowl and look at its number. What is the probability that the product of the two numbers will be [b]EVEN and GREATER than 10[/b]? Express your answer as a common fraction.

When it says even and greater, does that mean the product itself is > and = or just an [b]EVEN NUMBER And GREATER than 10?[/b]

When it says "even and greater than 10", it means the number is even, and greater than 10 (so 10 itself would not work.)

[hide=" click to reveal content"]

$ 3 \times 4 \equal{} 12, 4 \times 3 \equal{} 12, 4 \times 4 \equal{} 16, 4 \times 5 \equal{} 20$ and $ 5 \times 4 \equal{} 20$ are the only $ 5$ products which are both even and greater than $ 10$.

There are $ 5 \times 5 \equal{} 25$ possibilities altogether, so the required probability is $ \frac{5}{25}\equal{}\frac{1}{5}$.

[/hide]


conics ellipse geometry rhombus ratio incenter rotation

Given five lines, suposse that it's popsible to draw an elipse such that The lines are tangente to it.... To construct the elipse

Denote the 5 lines a, b, c, d, e in such a way that $A \equiv d \cap a, B \equiv a \cap b, C \equiv b \cap c, D \equiv c \cap d$, the quadrilateral ABCD is convex, $E \equiv e \cap CD, F \equiv e \cap DA$ and the pentagon ABCEF is also convex. Let $P \equiv a \cap c, Q \equiv b \cap d$. Using a central projection, project the line PQ to infinity is such a way that the projected quadrilateral ABCD becomes a square A'B'C'D' ( a rhombus would be sufficient). That is, draw an arbitrary square A'B'C'D'. Using the invariance of cross ration of 4 points in central projection, find the projections E', F' of the points E, F

$\frac{E'C'}{E'D'} = \frac{EC}{ED} \cdot \frac{PD}{PC}$

$\frac{F'D'}{F'A'} = \frac{FD}{FA} \cdot \frac{QD}{QA}$

The center of any ellipse inscribed in the square A'B'C'D' is its diagonal intersection $I' \equiv A'C' \cap B'D'$ and its main axes are the square diagonals $A'C' \perp B'D'$. Let the line $e' \equiv E'F'$ meet the lines $a' \equiv A'B'$ and $b' \equiv B'C'$ at M', N', respectively. The ellipse is inscribed in the triangle $\triangle M'N'B'$. Using a parallel projection, project the triangle $\triangle M'N'B'$ into a triangle $\triangle M''N''B''$ and its inscribed ellipse into its incircle. That is, find the intersections X', Y', Z' of the cevians M'I', N'I', B'I' with the opposite triangle sides, which project to angle bisectors M''I'', N''I'', B''I'' and construct an arbitrary triangle $\triangle M_1N_1B_1$ with the sides in the ratio $\frac{M_0B_0}{M_0N_0} = \frac{X'B'}{X'N'},\ \frac{N_0B_0}{N_0M_0} = \frac{Y'B'}{Y'M'}$ and the incenter $I_0$. Translate it so that its incenter coincides with I' and rotate it so that $B_0 \in B'I'$. Draw parallels to $B_0M_0, B_0N_0$ through A', C' intersecting at $B'' \in B'I'$ and intersecting the lines $I'M_0, I'N_0$ at M'', N'', respectively. Find the tangency points S''. T'', W'' of the incircle (I') of the triangle $\triangle M''N''B''$ with the sides $B''M'', B''N'', M''N''$ and parallel-project them back to the ellipse tangency points S', T', W' with the lines $a' \equiv B'M', b' \equiv B'N, e' \equiv M'N'$. The ellipse tancency points $U', V'$ with the lines $c', d'$ can be found similarly. Using invariance of the cross-ratio of 4 points in central projection, project the tangency points $S' \in a', T' \in b', U' \in c', V' \in d', W' \in e'$ back to the lines a, b, c, d:

$\frac{SA}{SB} = \frac{S'A'}{S'B'} \cdot \frac{PA}{PB}$

$\frac{TB}{TC} = \frac{T'B'}{T'C'} \cdot \frac{QB}{QC}$

$\frac{UC}{UD} = \frac{U'C'}{U'D'} \cdot \frac{PC}{PD}$

$\frac{VD}{VA} = \frac{V'D'}{V'A'} \cdot \frac{QD}{QA}$

The 5 tangency points $S \in a, T \in b, U \in c, V \in d, W \in e$ define the desired ellipse.

The lines define a pentagon.  Let the vertices of this pentagon be $A_1$, $A_2$, $A_3$, $A_4$, and $A_5$, as shown.  Let $T_1$ be the point of tangency of the ellipse with $A_1 A_2$.

Then by Brianchon's theorem, $A_1 A_3$, $A_2 A_5$, and $A_4 T_1$ concur.  Thus, $T_1$ is uniquely defined, as are the other points of tangency.  Then all five points of tangency can be used to construct the ellipse.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/976d4a6309424c9117efda0d873fff4e.png[/img]


In acute triangle $ ABC$, the altitude from $ A$ meets the circumcircle $ O$ at $ D$. Let $ P$ be a point on $ O$, and let $ Q$ be the foot of the perpendicular from $ P$ to $ AB$. Prove that if $ Q$ is outside $ O$ and $ 2\angle QPB\equal{}\angle PBC$, then points $ D,P,Q$ are collinear.

Suppose the line PD is extended to meet AB at Q'

DPAB is a cyclic quarilateral ,  angle BPQ' = angle BAD ............. (1)

let angle BPQ = x  ,  angle PBC = 2x

=>  angle PBQ = 90-x
=>   angle ABC = 180 - (90-x) - (2x)
                        = 90 - x
=>    angle BAD = 90 - angle ABC
                         = x

therefore ,  angle BAD = x = angle BPQ ................(2)

Compare (1) and (2) , angle BPQ' = angle BPQ
we get D , P , Q  collinear

I prefer using inversion here,An inversion by center$ A$,and an arbitrary radius$ k$,turnes the problem into this one:

Problem: Triangle$ AB'C'$ is given,$ AK$,meet the circumcircle of this triangle at $ E'$(which $ K$ is the circumcenter).

$ Q'$ is a point on $ A'B'$,in which the quadrilateral$ B'Q'D'E'$ is cyclic.$ P'$ is a point on $ B'C'$ which $ \angle AP'Q'=\frac{\pi}
{2}$

and also $ 2\alpha=\angle P'AC'=2\angle Q'P'B'$,prove that$ AQ'P'D'$ is cyclic.

$ 2\alpha=\angle P'AC'=2\angle Q'P'B' \Longleftrightarrow \angle AP'C'=\frac{\pi}{2}-\alpha
\Longleftrightarrow\angle AC'P'=\frac{\pi}{2}-\alpha\Longleftrightarrow \angle B'C'E'=\alpha=\angle B'AD'$

it means($ AQ'P'D'$ is cyclic) :wink:


If $x,y,z>0$ and $a \geq 1 $ then we have:

$ \frac{x}{ax+y+z} + \frac{y}{x+ay+z} + \frac{z}{x+y+az}  \leq \frac{3}{a+2}$.

Since the desired inequality is homogenous, we may assume that x+y+z = 1.
Thus it may be rewritten as f(x)+f(x)+f(z)  \leq 3f(1/3) where f(x) = x/[(a-1)x+1], which holds since f is concave.

Pierre.

After a lot of computation you have

2 \sum x^3 +a \sum (x^2y +xy^2)  \geq 6axyz + \sum (x^2y+xy^2)

which is easy to check.

2 \sum x^3 \geq  \sum (x^2y+xy^2)

as x^3+y^3=(x+y)(x^2-xy+y^2) \geq xy(x+y)

and this is obvious by AM-GM

and

 \sum (x^2y+xy^2) \geq 6xyz

again by AM-GM


Let $a,b,c,d$ be real numbers such that $|a|\leqq2,|b|\leqq2,|c|\leqq2,|d|\leqq2,a+b=1,c+d=1$.
Determine the range of $ac+bd$ which can be valued.

The maximum is clearly $5$, achieved when $a=c=2, b=d=-1$.

I have a feeling the minimum is $-4$, achieved when $a=d=2, b=c=-1$.

For the lower bound, $ac$ cannot drop below $-2$ since one of $a$ and $c$ must be negative for $ac$ to be negative and the negative term cannot go below $-1$ (this would leave, say, $b=1-a > 2$). Similarly for $bd$.

For the upper bound, $ac+bd \leq |ac|+|bd|$ and the maxima of $|a|,|b|,|c|,|d|$ are $2, 2, 1,1$ in some order. Rearrangement gives this to be at most $5$.


calculus
derivative
function
integration
inequalities
induction
real analysis

[quote="perfect_radio"]Prove that $f^{\prime} (t) \geq g(0) f(t)$ for all $t \in \left[ 0,1 \right]$.

[/quote]

Could check this inequality ?

We have $f=f*g.$ That means $f'=f'*g=f*g'.$ Taking it as $f'=f*g',$ we note that $f$ is positive and $g'$ is negative, so $f'$ is negative.

That inequality can't be pointing in the right direction.

Maybe it was typo... I don't know. Does the reverse inequality always hold?

[quote="perfect_radio"]Let $g : \left[ 0,1 \right] \to \mathbb R$ be a decreasing function, differentiable with the derivative continuous, and $f : \left[ 0,1 \right] \to \left[ 0,\infty \right)$ be a continuous function such that \[ f(t) = \int_0^t g(s) f(t-s) \, ds , \, \forall t \in \left[ 0,1 \right] .  \]

Prove that $f^{\prime} (t) \geq g(0) f(t)$ for all $t \in \left[ 0,1 \right]$.
[/quote]In this case, $f = 0$, since $f$ is a fixed point of a Volterra integral operator which has spectral radius 0.

[quote="Kent Merryfield"]We have $f=f*g.$ That means $f'=f'*g=f*g'.$
[/quote]

Can you explain more? I have no ideea how do you differentiate $\int_0^tK(t,s)ds$ where $K$ is a two variable function, and nice enough (differentiable WRT any variable, if not less) ? :?

$f'(t)=g(t)f(0)+\int_{0}^{t}g(s)f'(t-s)ds$  now integrate by part

$f'(t)=g(t)f(0) + [-g(s)f(t-s)]_{s=0}^{s=t}+\int_{0}^{t}g'(s)f(t-s)ds$

${f'(t)=g(0)f(t)+}+\int_{0}^{t}g'(s)f(t-s)ds$

g is decreasing then $g'\leq 0$ , $f\geq 0$ so the integral term is negative

$f'(t) \leq g(0)f(t)$

In fact $f(t)\leq f(0)e^{tg(0)}$

Here's what i did. Correct my work.

\[ \int_0^t g(s)f(t-s) ds=\int_0^t g(s) \int_s^t f'(x-s)dxds + \int_0^t g(s)f(0)ds=  \]
\[ \int_0^t g(s)\int_0^t \chi([s,t])f'(x-s)dx ds+\int_0^t g(s)f(0)ds  \]

Now intercharging the integrals we get:

\[ \int_0^t\int_0^t \chi([s,t])g(s)f'(x-s)dsdx+\int_0^t g(s)f(0)ds=  \]
\[ \int_0^t\int_0^t g(s)f'(x-s)dsdx+\int_0^t g(s)f(0)ds  \]

Now differentiating WRT $t$ we get :

\[ \int_0^t g(s)f'(t-s)ds+g(t)f(0) \].

I did the same things as moubinool but since i'm a beginer i wrote every step. is it ok?

[quote="perfect_radio"]Let $g : \left[ 0,1 \right] \to \mathbb R$ be a decreasing function, differentiable with the derivative continuous, and $f : \left[ 0,1 \right] \to \left[ 0,\infty \right)$ be a continuous function such that \[ f(t) = \int_0^t g(s) f(t-s) \, ds , \, \forall t \in \left[ 0,1 \right] .  \]

Prove that $f^{\prime} (t) \geq g(0) f(t)$ for all $t \in \left[ 0,1 \right]$.

[i]Problem 110 of the Romanian MO Shortlist 2005, 12th grade[/i][/quote]
To make the assertion that under these conditions $f=0$ more explicit, let $M = \max_x |g(x)|$. Then
$0 \le f(x) \le \int_0^t |g(s)| f(t-s)ds \le M \int_0^tf(s) ds$.
Iterate this estimate:
$0 \le f(x) \le M \int_0^t M \int_0^s f(\tau) d\tau \, ds = M^2\int_0^t(t-s)f(s) ds$.
By induction, we obtain for all $n$
$0 \le f(x) \le M^n \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s) ds$.
Now send $n$ to $\infty$, and the result is $0 \le f(x) \le 0$.

The problem becomes meaningful if one assumes that
\[ f(t) = f(0) + \int_0^t g(s) f(t-s) \, ds , \, \forall t \in \left[ 0,1 \right] .  \]


No seriously, how do you deal with scary, new notation? Its' very intimidating...

I mean, would you feel really intimidated if you didn't know anything about equalities and then clicked on the below link?
http://activities.tjhsst.edu/vmt/pages/tjusamo/tjusamo2k8/InequalitiesList.pdf

I am.

Yeah that's most definitely calculus based.

Its combinatorics that gives me the heeby jeebies.

...Learn what it means?

The only notation in that pdf that might actually be unfamiliar to you is the notation $ \sum_{sym} f(x_1, x_2, ... x_n)$.  This is a [i]symmetric sum[/i], and it just means to take the sum over all permutations of the $ n$ variables $ x_1, x_2, ... x_n$.

It is easiest to get a grasp on in fewer variables.  For example,

$ \sum_{sym} f(x, y) \equal{} f(x, y) \plus{} f(y, x)$.
$ \sum_{sym} f(x, y, z) \equal{} f(x, y, z) \plus{} f(x, z, y) \plus{} f(y, x, z) \plus{} f(y, z, x) \plus{} f(z, x, y) \plus{} f(z, y, x)$.

This is shorthand to make expressing inequalities easier; many inequalities involve symmetric sums.  An $ n$-variable symmetric sum has $ n!$ terms.

To give you a heads up, many inequalities involve [i]cyclic sums[/i] $ \sum_{cyc} f(x_1, x_2, ... x_n)$, which means to cycle the order of the variables (instead of permuting them).  For example,

$ \sum_{cyc} f(x, y) \equal{} f(x, y) \plus{} f(y, x)$.
$ \sum_{cyc} f(x, y, z) \equal{} f(x, y, z) \plus{} f(y, z, x) \plus{} f(z, x, y)$.

Note that the cyclic and symmetric sums are the same only for two variables.  An $ n$-variable cyclic sum has $ n$ terms.  It's the three-variable or $ n$-variable cases that you'll see most often anyway.

Question about the .pdf: Is number 12 a typo? It looks like it should be $ \sum_{cyc}(a^5\minus{}a^2\plus{}3)\ge (a\plus{}b\plus{}c)^3$ instead of $ (a^5\minus{}a^2\plus{}3)(b^5\minus{}b^2\plus{}3)(c^5\minus{}c^2\plus{}c)\ge(a\plus{}b\plus{}c)^3$.


$ x,y,z \in \mathbb Z^\plus{}$
$ \frac{1}{x}\minus{}\frac{1}{y}\equal{}\frac{1}{z}$  and $ h\equal{}gcd(x,y,z)$

show that $ hxyz$  and $ h(y\minus{}x)$  are both perfect squares

From the given we know that
$ x<y , x<z$. Also, $ yz \minus{} xz \equal{} xy$.
Let $ x\equal{}ah$, $ y\equal{}bh$ and $ z\equal{}ch$, $ a<b$, $ a<c$.
Now the given equation becomes: $ bc \equal{} a(c\plus{}b)$.

Case 1: $ (a,b)$, $ (b,c)$ and $ (c,a)$ are all relatively prime.

Now our equation becomes:
Note that $ gcd(b,a) \equal{} 1$ and $ b>a$ which implies $ b$ must divide $ c\plus{}b$. Similarly, $ c$ must divide $ c\plus{}b$. However, also note that $ gcd(b, b\plus{}c)$ and $ gcd(c, b\plus{}c)$ = $ 1$. So a contradiction.

Case 2: $ (a,b)$ and $ (c,b)$ relatively prime but $ (a,c)$ are not. This implies that $ a$ divides $ c$ or $ a\equal{}1$. In the first case, Let $ c \equal{} ak$, $ 1<k$. Therefore, $ bk \equal{} c\plus{}b$, $ b(k\minus{}1) \equal{} c$, which implies that $ b$ divides $ c$, contradiction. If $ a\equal{}1$, we similarly have $ b(c\minus{}1) \equal{} c$ implying $ b$ divides $ c$, a contradiction.

Case 3: $ (a,b)$, $ (a,c)$, are not relatively prime but $ (b,c)$ are. Then note that $ c\plus{}b$ divides $ bc$. Yet $ gcd(c\plus{}b,b)$ and $ gcd(c\plus{}b,c)$ are $ 1$. So contradiction.

Case 4: $ (b,c)$ are not relatively prime but $ (a,b)$ and $ (a,c)$ are. Then $ bc$ divides $ b\plus{}c$ and $ b\plus{}c$ divides $ bc$ so $ b\plus{}c \equal{} bc$ Then $ b(c\minus{}1) \equal{} c$ implying that $ b$ divides $ c$ and $ c(b\minus{}1) \equal{} b$ implying that $ c$ divides $ b$ implying that $ b\equal{}c$. Then $ b^2 \equal{} 2b$ and $ b\equal{}c\equal{}2$ and $ a\equal{}1$.

Now, $ hxyz$ = $ h^4abc$ = $ 4h^4$, clearly a perfect square.
Also, $ h(y\minus{}x)$ = $ h^2(b\minus{}a)$ = $ h^2(1)$, clearly a perfect square.


What are the three least significants bits in the binary representation of $3^{101}+2$?

So, if $3^{101}+2 \equiv 5 \bmod 8$ the least three significant bits in binary are $101$?

I am sorry $1+2=3 =(011)_{2}$. I edit my post.


geometry
circumcircle
incenter
Euler
power of a point
radical axis
geometry unsolved

In $ \triangle ABC$ with $ \angle{ACB} \equal{} 60$ ,let $ AA_1$ and $ BB_1$,$ (A_1 \in BC,B_1 \in AC)$be the bisector of $ \angle{BAC}$ and $ \angle{ABC}$.The line $ A_1B_1$ meets the circumcircle of $ \triangle ABC$ at points $ A_2$ and $ B_2$ .If $ R$ is the midpoint of the arc $ AB$,not containing point $ C$,and $ P$ and $ Q$ are the midpoints of $ A_1B_1$ and $ A_2B_2$,respectively,prove that $ RP\equal{}RQ$

$ I \equiv AA_1 \cap BB_1$ is the incenter and $ I_a, I_b, I_c$ are the A-, B-, C-excenters of the $ \triangle ABC,$ $ O_c$ is the circumcenter of the $ \triangle I_aI_bI.$ Quadrilaterals $ ICI_aB,\ ICI_bA$ with diagonal intersections $ A_1,\ B_1$ are cyclic, having right angles at C, A, B. Then $ A_1C \cdot A_1B \equal{} A_1I \cdot A_1I_a,$ $ B_1C \cdot B_1A \equal{} B_1I \cdot B_1I_b$ and $ A_1B_1$ is the radical axis of the circles $ (O),\ (O_c),$ $ OO_c \perp A_1B_1,$ $ Q \equiv OO_c \cap A_2B_2 \equiv OO_c \cap A_1B_1.$ $ I_c$ is the orthocenter and O the 9-point circle center of the $ \triangle I_aI_bI,$ $ O_cOI_c$ its Euler line $ \Longrightarrow (I_cQ \equiv I_cOQO_c) \perp A_1B_1.$

$ \angle A_1IB_1 \equal{} \frac {\angle C}{2} \plus{} 90^\circ.$ $ \angle C \equal{} 60^\circ \Longrightarrow \angle A_1IB_1 \equal{} 120^\circ,$ $ CA_1IB_1$ is cyclic, $ \angle IA_1B_1 \equal{} \angle ICB_1 \equal{} \angle ICA_1 \equal{} \angle IB_1A_1,$ $ \triangle IA_1B_1$ is isosceles, $ IP \perp A_1B_1.$ As a result, $ IP \parallel I_cQ$ (both $ \perp (PQ \equiv A_1B_1)$).

$ IAI_cB$ is cyclic, because of the right angles at A, B. R is its circumcenter because the $ \triangle RAI$ with angles $ \angle IRA \equal{} \angle B,\ \angle RAI \equal{} \frac {\angle C \plus{} \angle A}{2}$ is isosceles, $ RA \equal{} RI \equal{} RI_c.$ R is on the midparallel of $ I_cQ \parallel IP,$ both $ \perp PQ\ \Longrightarrow RP \equal{} RQ.$


geometry
incenter
circumcircle
vector
angle bisector
power of a point
radical axis

Let $ I_{a}$ ,$ I_{c}$ be the $ A$, $ C$-excenters, respectively. It is clear that $ B$, $ B_{1}$ and $ B_{2}$ are collinear; therefore, the perpendicularity of $ B_{2}I$ and $ B_{1}I_{b}$ is equivalent with the fact that $ I$ is the orthocenter of triangle $ I_{b}B_{1}B_{2}$. Thus, it is suffice to show that $ IB \cdot BI_{b} \equal{} BB_{1} \cdot BB_{2}$ (the power of $ I$ wrt. the circumcircle of $ I_{b}B_{1}B_{2}$). But, on the other hand, we know that $ I$ is the orthocenter of $ I_{a}I_{b}I_{c}$ and so $ IB \cdot BI_{b} \equal{} BI_{a} \cdot BI_{c}$. In this case, the problem reduces to proving that $ BB_{1} \cdot BB_{2} \equal{} BI_{a} \cdot BI_{c}$. But this is just a consequence of $ (B_{2}, I_{a}, B, I_{c}) \equal{} \minus{} 1$ and $ B_{1}I_{a} \equal{} B_{1}I_{c}$ (since the circumcircle of $ ABC$ is the nine-point center of $ I_{a}I_{b}I_{c}$).

$ BI$ cuts the circumcircle $ (O)$ of $ \triangle ABC$ again at $ Y$ and $ (Y)$ is a circle with center $ Y$ and radius $ YA \equal{} YC \equal{} YI \equal{} YI_b.$ $ B_2I_b$ cuts $ (Y)$ again at $ Q.$ $ \overline{B_2Q} \cdot \overline{B_2I_b} \equal{} \overline{B_2A} \cdot \overline{B_2C} \equal{} \overline{B_2B} \cdot \overline{B_2B_1}$ $ \Longrightarrow$ $ BB_1I_bQ$ is cyclic and the angle $ \angle B_1QI_b \equal{} \angle B_1BI_b$ is right. Since $ II_b$ is a diameter of $ (Y),$ $ Q \in (Y)$ and $ B_1Q \perp I_bQ,$ $ B_1Q$ goes through $ I$ $ \Longrightarrow$ $ I$ is orthocenter of $ \triangle B_1B_2I_b.$

We notice that if we prove $B_1I$ is perpendicular to $B_2I_b$ ,we are done.
Let $ \odot AIC \cap B_2I_b=L$.As $\odot AIC$ and $\odot ABC$ have $AC$ as the radical axis and as $B_1B_2,AC,B_2I_b $ are concurrent, then $B_1BLI_b$ is cyclic. So $ B_1L$ is $\perp$ to $B_2I_b$.So $ I$ sholud lie on $B_1L$ as $II_b$ is the diameter of $ \odot AIC$. Hence done.

Let $\{I, X\}\in B_2I\cap\odot (AIC)$; from power of $B_2$ w.r.t. $\odot(ABC),\odot(AIC)$ we get (already proven $B_1-B-B_2$ are collinear): $B_2B\cdot B_2B_1=B_2A\cdot B_2C=B_2I\cdot B_2X$, hence $BB_1XI$ is cyclic, i.e. $B_2X\cap B_1X$. As $II_b$ is a diameter of $\odot (AIC)$, we infer $I_bX\bot IX$, meaning $B_1-X-I_b$ are collinear, and we are done.

Best regards,
sunken rock

$ B_{1} $ is the midpoint of the arc $AC$ containing $B$, in the circumcircle of $\triangle ABC\cdot \cdot \cdot (1)$
$I_{b}$ is the $B$-excircle's center$\cdot \cdot \cdot (2) $
Angle bisector of $\angle ABC$ intersects $AC$ at $B_{2}\cdot \cdot \cdot (3)$
$ I  $ is the incenter of $\triangle ABC\cdot \cdot \cdot (4)$

By $(3)$ and $(4)$, $\angle IBB_{2}=\angle R\cdot \cdot \cdot (5)$
By $(1)$ and $(4)$,
\[\angle B_{1}BI\\=\angle IBC+\angle B_{1}BC
\\=\angle IBC+\angle B_{1}AC
\\=\angle IBC+(\angle R-\frac{1}{2}\angle AB_{1}C)
\\=\angle R+(\angle IBC-\frac{1}{2}\angle ABC)
\\=\angle R\cdot \cdot \cdot (6)\]
By $ (5) $ and $ (6) $, $  B $ and $B_{1},B_{2}$ are collinear$\cdot \cdot \cdot (7)$
By,$ (2) $ and $ (4) $, $\angle IAI_{b}=\angle ICI_{b}=\angle R\cdot \cdot \cdot (8)$
Let $ H $ be a point such that $ H $ is in $ B_{1}I $ and $ BH\bot HI_{b}. \cdot \cdot \cdot (9) $
By $ (6) $ and $(9)$, $B$ and $B_{1},I_{b},H$ are concyclic.$\cdot \cdot \cdot (10)$
By $ (8) $ and $(9)$, $A$ and $I,C,I_{b},H$ are concyclic.$\cdot \cdot \cdot (11)$
By $ (1) $ , $ A $ and $B,C,B_{1}$ are concyclic.$\cdot \cdot \cdot (12)$
By $ (10) $ and $(11),(12)$, $ I_{b} $ and $ H,BB_{1}\cap AC(=B_{2}) $ are collinear.$\cdot \cdot \cdot (13)$($\because$ $BB_{1}\cap AC\cap I_{b}H$ is a radical ceneter)
By $(9)$ and $(13) $, $B_{1}I\bot I_{b}B_{2}.\cdot \cdot \cdot (14)$
By $ (5) $ and $ (7) $, $ B_{1}B_{2}\bot BI_{b}.\cdot \cdot \cdot (15) $
By $ (14) $ and $ (15) $, I is orthocenter of $\triangle B_{1}B_{2}I_{b}$.
So $ B_{2}I\bot B_{1}I_{b} $.
$ (Q,E,D,) $

Let $l$ be the line through $I$ perpendicular to $B_1I_B$. Define $P'=B_1I_B\cap l$. An easy angle chase shows $B_2,B,B_1$ are collinear on the external angle bisector of $B$ while $B_2,A,C$ are collinear by the definition. Now consider the circles $\Gamma_1=(ABC),\Gamma_2=(BB_1I),\Gamma_3=(AIC)$. Clearly $B_1\in\Gamma_1$ while $P'\in\Gamma_2$ since $\angle B_1P'I=\angle B_1BI=90^{\circ}$ and $P'\in\Gamma_3$ since $II_B$ is a diameter, where $I_B$ is the $B$-excenter, and $\angle IP'I_B=90^{\circ}$. It is easy to see that the Radical Axes of $\Gamma_1,\Gamma_2$ is $BB_1$, $\Gamma_2,\Gamma_3$ is $IP$, and $\Gamma_3,\Gamma_1$ is $AC$. By Radical Concurrence on $\Gamma_1,\Gamma_2,\Gamma_3$, these lines concur at $B_2$, which is enough to conclude that $B_2,I,P'$ are collinear, showing $P=P'$.

Complex numbers: vertices of triangle are $a^2,b^2,c^2$, and it's circumcircle is a unit circle. Then incenter is $-ab-bc-ca$, $B_2$ as intersection of lines $BB_1,AC$  has coordinates $\frac{(b^2+ac)ac-(a^2+c^2)b^2}{ac-b^2}$.
$\frac{B_2-I}{I_b-B_1}=\frac{(ab+bc+ca)(ac-b^2)+(b^2ac+a^2c^2-a^2b^2-b^2c^2)}{(ab+bc-2ac)(ac-b^2)}$
The conjugate of this number is $\frac{(a+b+c)(b^2-ac)+abc+b^3-bc^2-a^2b}{(c+a-2b)(b^2-ac)}$
Adding two last complex numbers we get $0$ (as you don't believe check here [hide]https://www.symbolab.com/solver/step-by-step/simplify%20%5Cfrac%7B%5Cleft(%5Cleft(ab%2Bbc%2Bca%5Cright)%5Cleft(ac-b%5E%7B2%7D%5Cright)%2B%5Cleft(b%5E%7B2%7Dac%2Ba%5E%7B2%7Dc%5E%7B2%7D-a%5E%7B2%7Db%5E%7B2%7D-b%5E%7B2%7Dc%5E%7B2%7D%5Cright)%5Cright)%7D%7B%5Cleft(ab%2Bbc-2ac%5Cright)%5Cleft(ac-b%5E%7B2%7D%5Cright)%7D%2B%5Cfrac%7B%5Cleft(a%2Bb%2Bc%5Cright)%5Cleft(b%5E%7B2%7D-ac%5Cright)%2Babc%2Bb%5E%7B3%7D-bc%5E%7B2%7D-a%5E%7B2%7Db%7D%7B%5Cleft(c%2Ba-2b%5Cright)%5Cleft(b%5E%7B2%7D-ac%5Cright)%7D[/hide])
Hence $B_2I\perp IbB_1$

Let $I_A,I_C$ be the $A$, $C-$ excenter, By Brokard's Theorem on Quadrilateral $I_CACI_A$ $\implies$ $I$ is the orthocenter of $\Delta B_2B_1I_B$

[quote=Erken]Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.[/quote]

Was this question bonus or $I_b$ and $I_B$ the same?

It's well-known that under $\sqrt{ac}$ inversion, $\{I,I_B\},\{B_1,B_2\}$ swap, so we're done because then we must have$$BI\cdot BI_B = AB\cdot BC = BB_1\cdot BB_2,$$which means that $I$ is the orthocenter of $\triangle B_1B_2I_B.$ $\blacksquare$

We will prove that infact $I$ is the orthocenter of $\Delta B_{1}B_{2}I_{B}$.
Now let $B_{1}I\cap I_{B}B_{2}=F$. By PoP
$B_{2}F\cdot B_{2}I_{B} =B_{2}A\cdot B_{2}C=B_{2}B\cdot B_{2}B_{1}$. Hence
quadrilateral $B_{1}BI_{B}F$ is cyclic.
Now let $IE\perp B_{1}I_{B}$, it is also seen that there exists a circle $(I_{B}CEIAF)$. Hence, as $\angle B_{1}FI_{B}=\angle B_{1}BI_{B}=90^{\circ} \Rightarrow BIEB_{1}$ is cyclic too. As $BB_{1}\cap FI_{B}=B_{2}$ Its consequence leads to the fact that $B-I-E$
$\blacksquare$

[quote=Jupiter_is_BIG]
Was this question bonus or $I_b$ and $I_B$ the same?[/quote]
It was a bonus

Notice that $\angle I_BBB_1 = 90$. (Indeed $\angle I_BBB_1 = \angle MBA + \angle B_1BA = \angle MB_1A + \angle B_1AC = 90$).

Denote by $P$ the intersection of line $B_2I_B$ with the circumcircle of $\triangle IAC$. It is immediate that \[B_2B \cdot B_2B_1 = B_2C \cdot B_2A = B_2P \cdot B_2I_B.\] Thus $B, B_1, P, I_B$ are concyclic. Now by the incenter/excenter lemma, we have that $II_B$ is the diameter of $(CAI_B)$. Using this, one obtains \[\angle I_BPI = 90 = \angle I_BBB_1 = \angle I_BPB_1.\] Therefore $P, I, B_1$  are collinear, and $I$ is the orthocenter of $B_2BI_B$, as desired. $\blacksquare$

We prove that $I$ is the orthocenter of $\triangle I_BB_1B_2$. In particular, it is equivalent to show that \[BI\cdot BI_B = BB_1\cdot BB_2.\] Let $D$ and $M$ be the intersections of line $BI$ with $AC$ and $(ABC)$. Since $(BD;II_B) = -1$, $BI\cdot BI_B = BD\cdot BM$. Then, observe that $MB\perp B_1B_2$ and $B_2D\perp B_1M$, so in fact $D$ is the orthocenter of $\triangle MB_1B_2$. Hence, \[BB_1\cdot BB_2 = BD\cdot BM = BI\cdot BI_B\] as desired.

Well, it is known that the external angle bisector of $\triangle ABC$ is simply $\overline{BB_1}$. Now, notice that,
\[\measuredangle IAI_b=\measuredangle ICI_b=90^\circ\]
and thus, $I$,$A$,$C$ and $I_b$ are concyclic. Now, let $B_3=(IBB_1)\cap \overline{B_1I_b}$. It is easy to see that since $\measuredangle IB_3B_1=\measuredangle IBB_1 90^\circ$, $B_3$ also lies on $(IAC)$. Now, let $B_3'=\overline{B_2I} \cap (IBB_1)$. Then,
\[B_2I\cdot B_2B_3' = B_2B \cdot B_2B_1 = B_2A\cdot B_2C\]
Thus, $B_3'$ must also lie on $(IAC)$ which implies that $B_3'=B_3$ and indeed, $B_2I \perp B_1I_b$ as required.


Let $x,a,b$ be postive integer such that $x^{a+b}=a^{b}b$. prove that $a=x$ and $b=x^{x}$

find all natural number $a^{m}+1=(1+b^{k})^{n}$ good luck :lol:

[quote="nayel"]$x^{a+b}=a^{b}b$
$\Rightarrow (\frac{x}{a})^{b}\cdot x^{a}=b$
so $a|x \Rightarrow x=ka$
$\Rightarrow k^{b}\cdot k^{a}\cdot a^{a}=b\Rightarrow k^{b}\le b$
but we can easily prove that $2^{b}>b, \forall b\in \mathbb N$(try yourself)
so $k=1$ and the conclusion follows...
tell me if i'm mistaken somewhere. :P

ps. i want more diophantine equations.[/quote]

yuo didn't prove alx

[quote="ali666"][quote="nayel"]
$\Rightarrow (\frac{x}{a})^{b}\cdot x^{a}=b$
so $a|x$
[/quote]I think this step is not true.can you exaplain it?[/quote]

i think its because $b$ is an integer (RHS), and so $\frac{x}{a}$ has to be a natural number, so $a|x$....

[quote="nayel"][quote="ali666"][quote="nayel"]
$\Rightarrow (\frac{x}{a})^{b}\cdot x^{a}=b$
so $a|x$
[/quote]I think this step is not true.can you exaplain it?[/quote]

if it's not true, can you give a counter example? what i have to explain is clear in kathi;)'s post. :P[/quote]
the result is true but you dont prove that a|x.
if a>b and $a=x^{2}$then $(\frac{x}{a})^{b}\cdot x^{a}$ is integer
but $a\not{|}x$

okey that's all i needed. :blush:  now let $(x,a)=g$ and... i'm stuck. someone please prove that $g=a$

Let $(x,a) = d, x = dp, a = dq$
We have $d^{a}\cdot p^{a+b}= q^{b}\cdot b$
Since $(p,q) = 1$, then $d^{a}= k \cdot q^{b}$ with $k \in \mathbb{N}$ and $b = k \cdot p^{a+b}$
If $\geq 2$, then $b = k \cdot p^{a+b}\geq 2^{b+1}> b$ contradicts the assumption.
Hence, $p = 1 \Rightarrow x = d, a = xq$

Now assume that $q \geq 2$, we have $x^{a}= q^{b}\cdot b$, then $b = q^{M \cdot a-b}\cdot R^{a}$ with $M \cdot a \geq b$ so that $x = q^{M}\cdot R$
But, $a = q^{M+1}\cdot R$ and $b = q^{M \cdot R \cdot q^{M+1}}\cdot R^{a}> q^{M+1}\cdot R \cdot M = M \cdot a \geq b$ contradicts the assumption.
Hence, $q = 1 \Rightarrow x = a, b = \frac{x^{a+b}}{a^{b}}=x^{x}$ as desired..  :wink:

17 years revive...

We know that $b=\left(\frac{x}{a}\right)^b\times x^a$
if $\frac{x}{a}\geq 2$ we have $b<\left(\frac{x}{a}\right)^b$, so $\frac{x}{a}=1$ and $a=x$
plug in to get $b=x^x$


superior algebra
superior algebra unsolved

Prove that k[x] / (x^4+x^3+x^2+x+1) is a field with 32 elements.

Please help!  :D

O yea. I forgot to add. k[x] = Z/2Z (Z integers)[/quote]

[quote="ZetaX"]Definitely wrong as stated, whatever $k$ is.[/quote]
Even if it is $\mathbb Z/2\mathbb Z$.

By the way, $k[x]$ is not $\mathbb Z/2\mathbb Z$, but $k$ probably is, you mixed up some symbols.

is it because the denominator is not irreducible?

im sorry..im new to the subject...

I haven't checked if it is irreducible, but even if it is: the quotient has just $|k|^{4}$ elements, so $16$ in your case. $16 \neq 32$.


Find all positive integers (m,n) such that m|n^2 + 1 and n|m^2 + 1. I think I have a solution to the problem, but I obtained it by making some good guesses rather than by rigor. The result is fairly interesting, so I'm looking for a method to solve the question.

Hi,
basically the idea is that if the pair (m,n) with n>m, then obviously (m,n)=1
and (n,(n^2+1)/m) and ((m^2+1/n),m) also work (easy to prove; one way is trivial and other way just involves some calculation and using original condition). further, m^2+1/n < m, since
mn-m^2 = m(n-m)>1. However this breaks down when m=1,n=2. So suppose some pair (m,n) works; then we use the above idea to keep generating numbers of the form ((m^2+1)/n, m). This would lead to infinitely many solutions, which is a contradiction since all 'm' values generated are strictly smaller than the original, and there are only finitely of these numbers. So we much reach a 'breaking point', which is when m=1,n=2. So thus all (m,n) values are generated by (1,2).
Now consider the Fibonacci sequence. [f(1)=1,f(2)=1,etc]. We note that numbers of the form
f(2k-1),f(2k+1) satisfy the above condition, because of the identity
f(2k-1)^2 + 1 = f(2k-3) * f(2k+1)
which can be proved by induction. the original value is (1,2), so these numbers are in fact the (m,n) values which are generated by (1,2); which are in turn the only values that work as proved above.
so finally, the only working values are: f(2k-1),f(2k+1) for some integer k.
Hope this helps.

Thank you--that, along with (1,1), is the answer I got, and its good to see how its done properly.


I think I have come up with a new formula to help solve the value of any exponent

Here it is:

$ (x^y) \equal{} (x^{(y \minus{} 1)} \plus{} x^{(y \minus{} 2)} \plus{} ... \plus{} x^0)(x \minus{} 1) \plus{} 1$

When $ x$ and $ y$ $ \not\equal{}$ $ 0$

I think it might be okay now, maybe...

You might want to set some limits on the values for $ x$ and $ y$. For example, what happens when we let $ x \equal{} 0$?

The only problem with the case $ x \equal{} 0$ is the term $ x^0$, which we can easily replace with $ 1$ with the assumption that $ y$ is a natural number.

What do you mean by "solve the value of any exponent"?  Your statement is true.  It's not particularly novel, nor does it help you much to actually compute the value of $ x^y$.  It's much more efficient to use

$ x^y \equal{} x \cdot x^{y \minus{} 1}$.

It works, but instead of using it to find exponents(which is much harder using the formula) use it to factor $ (x^n\minus{}1)$

Anything new sorry, that's just the formula for geometric progression. (I don't know if it's the correct English word, I mean just 1+x+x^2+...+x^n, assuming x^0=1).
Regards, Gil

could you do it by induction?
i read about this in mathematical toolchest, and i'm fairly sure that the proof using induction isn't too difficult

i hate to break it to you, but...subtract 1 both sides. you now have the factorization for $ x^n \minus{} 1^n$. :oops:

wow, I really cant believe we all missed that lol


geometry calculus integration MATHCOUNTS

Draw two concentric circles, one inside the other.  Draw a line segment tangent to the smaller circle with endpoints where it intersects the larger circle. The segment has a length of 24.  What is the area of the region outside the smaller circle but inside the larger circle? :D

Are you asking for help, or giving kids a problem to try??  I'll help, but don't want to give it away.

The sign shows that this is a challenge problem.

[color=indigo][hide]First you have to draw the radius of the smaller circle that is perpendicular to the line segment. Then you should draw a radius of the big circle that connects to one of the endpoints. x=radius of small circle y=radius of big circle. Becuase of pythagorean's theorem x2+122=y2 so 144=y2-x2. The answer is y2:pi:-x2:pi: which changes to :pi:(y2-x2) then you can use substitution to find the answer:[/color] [color=green]144 :pi: [/color] [/hide]

so basically, according to LadyKn 1 ght's explanation,

[hide]the smaller circle's radius is 5, and the larger circle's radius is 13.[/hide]

ajai, not necessarily -- there are lots of such pairs of circles other than the one you note.  (If we require that they have integral radii as well, the pair that you give is one of two: the other is radii of 9 and 15.)

Since no one has posted this  well known  funny way to solve this,  I will  post it here...

[hide](Since the  radius of  the inner circle is not given, and  we alll know  "oh_we_are_so_clever "  type attitude of those who post such problems, .... we  can trust that it will make NO difference  whatever  the radius of the inner circle would be... so why not assume it is zero. Now what do we have?  Just one circle with diameter 24 and area  $144\pi$  :))

Who needs mathematical arguments when we can trust the psycology of the probelm poser..:) [/hide]

[hide]If r is the radius of smaller circle and R is the radius of the next one. We have R 2 =r 2 +144. We have to find  (R 2 -r 2 )pi=144pi[/hide]

[hide]This is one the saab questions isn't it?  The trick is just the square of half the segment. So don't waste your time calculating it(especially in Mathcounts)[/hide]


Hi,

Is it true that a positive self-adjoint operator satisfies A^2 = A?
If not, when is the last condition true?

Thanks in advance.

[quote="DK"]Or is it maybe true that if (Ax|x) is positive then there is a B such that B^2 = A?[/quote]

true and unique if you require B to be positive definite as well.

In the finite-dimensional case you diagonalize $A$ and take square roots of diagonal elements, then go back to the original basis.

Infinite dimensions require the [url=http://en.wikipedia.org/wiki/Spectral_theorem]spectral theorem[/url] (I think).


[quote="iamhe"]How do you find the vertical  or horizontal asymptote of a function without graphing it?
Example:
How many vertical asymptotes doe sthe function y=  (x^2+6x+8)/(x^2-6x+8) have
4
3
2
1
0
Thanks a lot.[/quote]
[hide="approach"]
What values can x not equal?
you could find the horiz asymptotes of the inverse...
[/hide]

i'm not too sure if there are ANY vertical asymptotes...

then what is the difference between finding horizontal and vertical asymptotes?

If all you do is factor the denominator and solve for the zeros, wouldn't that one have 2?
[hide]The correct answer was 2.[/hide]

With rational functions, that's exactly what you do.

$ y \equal{} \frac {(x \plus{} 2)(x \plus{} 4)}{(x \minus{} 2)(x \minus{} 4)}$

At $ x \equal{} 2$ and $ x \equal{} 4$, the denominator is zero but the numerator is not, hence we have a vertical asymptote at each of these x values.

A [i]horizontal asymptote[/i] describes the value of $ y$ for large values of $ x$ -- in particular, a function has a horizontal asymptote if it eventually approaches that line for large $ x$ (either positive or negative -- not neces. both).

For example, in this case, we can divide the numerator and denominator by $ x^2$ to get:

$ y \equal{} \frac{ \left( 1 \plus{} \frac{2}{x} \right) \left( 1 \plus{} \frac{4}{x} \right)}{\left( 1 \minus{} \frac{2}{x} \right) \left( 1 \minus{} \frac{4}{x} \right)}$.

As $ x$ gets very large, this approaches $ \frac{1 \cdot 1}{1 \cdot 1} \equal{} 1$. As x becomes very negative, again, it approaches $ 1$. Thus, the only horizontal asymptote is $ y\equal{}1$.


The following is supposed to be part of a prelim exam but this is so easy I couldn't believe it ...

Given a finite ring $R$, show that there exist two positive integers $m \neq n$ such that $\forall x \in R, x^m=x^n$.

Consider $\mathbb{N} \to R^R , n \mapsto (x \mapsto x^n)$. This map can't be injective since it maps an infinite set into a finite set. The claim follows.


group theory
abstract algebra
superior algebra
superior algebra solved

Need some help!!!

Porve that if $N_1$ is a normal subgroup of $G_1$ and $N_2$ is a normal subgroup of $G_2$, then $G_1 \times G_2 / N_1 \times N_2 \simeq (G_1/N_1)\times(G_2/N_2)$

The kernell of the natural surjective morphism $G_1\times G_2\to (G_1/N_1)\times (G_2/N_2)$ is $N_1\times N_2$.
The Fundamental  Theorem of Isomorphism should blah blah blah


I know this was posted here before, but I don't want to revive an old, old thread.

Who wants an account? I could invite like 100 people.

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It's awesome.

Leave your name and email and someone will invite you. You will receive an link in your email box to sign up.

I also have about 100 invites, if anyone is interested.

My e-mail is alpha_gaurav_beta@yahoo.co.uk

My e-mail is sparklingstar_suntana@hotmail.com

I also have about 100 invites.  As if there arn't enough floating around.

It's really good, but htese days its a little slow, and its not just me, but also other ppl i know

:huh: sometimes it seems to me that Gmail is super super slow... sometimes I can log in on one account but not on another account.

you cant do both accounts at the same time, i dont think

Gmail is awesome for me. I don't ever remember it being slow.

Although sometimes it says the server's down and to "cross my fingers and try again later".  :P


Three circles of  radii 1 are mutually tangent. What is the area of the bounded region located in between the curves of the three circles?

set the centers of the circles to A B and C. we ave a right triangle ABC area $ \frac{\sqrt{3}} {4}$ and 3 60 degree sectors. do the area of the triangle minus the sectors and WHALAHHH!!


K guys, here's a toughie:

suppose x and y are two integers with 1 <x<y and x+y less than or equal to 100. kill knows their sum x+y and paul knowstheir product xy. the following conversation now takes place:

*Paul: I do not know hte two numbers.
*Jill: I knew that already.
*Paul: Now I know the two numbers.
*Jill: Now I know them also.

What are the two numbers?

Hint: [hide]Every even number from 4 to 100 can be written as the sum of two primes[/hide]

There are up to four solutions; Any hints ppl?

A similar thread was already posted. http://www.mathlinks.ro/Forum/viewtopic.php?t=146360


how do i proove this ?
$\displaystyle \sin6^\circ\cdot\cos12^\circ\cdot\sin42^\circ\cdot\cos24^\circ=\frac{1}{16}$

i've tried to use all the things and formulas in my manual and i
get to much too complicated stuff so i think i'm doing it wrong.
anyway at the answers part of the book it says that both
sides should be multiplied by $2\cdot cos(6^\circ)$

Multiply by $cos6^\circ$:

$cos6^\circ\cdot sin6^\circ\cdot\ cos12^\circ\cdot\ sin42^\circ\cdot\ cos24^\circ=$
$=\frac{1}{2} \cdot sin12^\circ\cdot\ cos12^\circ\cdot\ sin42^\circ\cdot\ cos24^\circ=$
$=\frac{1}{4} sin24^\circ\cdot\ cos24^\circ\cdot\ sin42^\circ=$
$=\frac{1}{8} sin48^\circ\cdot\ cos48^\circ=$
$=\frac{1}{16} sin96^\circ=\frac{1}{16} cos6^\circ$

So dividing it out again you get the desired result.


function
vector
inequalities
Functional Analysis
real analysis
real analysis unsolved

Let $ 0 < \alpha \le 1$. Let $ \Lambda_{\alpha}([0,1])$ be the space of Hölder continuous functions of exponent $ \alpha$ on $ [0,1]$. The space consists of all $ f$ satisfying $ \parallel{}f\parallel{}_\alpha < \infty$, where
\[ \parallel{}f\parallel{}_\alpha \equal{} |f(0)| \plus{} \sup_{x,y\in[0,1],\ x\neq y} \frac{|f(x)\minus{}f(y)|}{|x\minus{}y|^\alpha}\]
Prove that the space $ \Lambda_{\alpha}([0,1])$ is a Banach space with the norm $ \parallel{}\cdot\parallel{}_\alpha$.

[hide="hint"]$ \parallel{}f\parallel{} \le \parallel{}f\parallel{}_\alpha$ where $ \parallel{}\cdot\parallel{}$ is the supremum norm[/hide]

It is easy to show that it's a normed vector space.
To show completeness, we can show that any absolute convergence series converges. Suppose $ \sum |f_n|_{\alpha}$ exists. By the hint and the completeness of space of continuous functions under sup norm, we see that $ f \equal{} \sum f_n$ exists and is continuous. It remains to show that it's Holder continuous. Let $ S_n$ be the partial sums of $ f_n$, then
$ |f(x) \minus{} f(y)| \leq |f(x) \minus{} S_n(x)| \plus{} |S_n(x) \minus{} S_n(y)| \plus{} |S_n(y) \minus{} f(y)|$.
For each $ S_n$ there is $ B_n$ so that $ |S_n(x) \minus{} S_n(y)| \leq B_n|x\minus{}y|^{\alpha}$ for any $ x,y$. The absolute convergence of the series means $ B_n$ has a supremum, say $ B$. Choose large n so that $ |f(x) \minus{} S_n(x)|$ and $ |f(y) \minus{} S_n(y)|$ are all less than $ \frac{1}{2}|x\minus{}y|^{\alpha}$. Then as $ |S_n(x) \minus{} S_n(y)| \leq B_n|x\minus{}y|^{\alpha} \leq B|x\minus{}y|^{\alpha}$, we see that $ f$ is Holder continuous.

Now it remains to show $ \parallel{}f \minus{} S_n\parallel{}_\alpha \to 0$ in order to complete the proof.

We have the pointwise convergent sum $ f \equal{} \sum f_n$.
$ \parallel{} f \minus{} S_{n \minus{} 1} \parallel{}_\alpha \equal{} \parallel{}\sum_n^\infty f_k\parallel{}_\alpha \le \sum_n^\infty \parallel{}f_k\parallel{}_\alpha$.
The last inequality comes from the definition of $ \parallel{}\cdot\parallel{}_\alpha$ and the fact that $ |\sum_1^\infty a_k| \le \sum_1^\infty |a_k|$ for all convergent series $ \sum_1^\infty a_k$ of real numbers. But the last term goes to zero as $ n\to\infty$ because $ \sum_1^\infty \parallel{}f_k\parallel{}_{\alpha} < \infty$. The inequality shows at the same time that $ f\in\Lambda_\alpha([0,1])$ i.e. $ \parallel{}f\parallel{}_\alpha < \infty$ because $ \parallel{}f\parallel{}_\alpha \le \parallel{} f \minus{} S_n \parallel{}_\alpha \plus{} \parallel{}S_n\parallel{}_\alpha$.


Given $ a;b$ is positive real numbers satisfying $ ab|a^2\plus{}b^2\minus{}1$ and $ 1<a<b$
Find all value of $ A\equal{}\frac{a^2\plus{}b^2\minus{}1}{ab}$

A can get all natural values!
check $ (a,b)\equal{}(k,k^2\minus{}1)$ for any $ k$

[quote="Albanian Eagle"]A can get all natural values![/quote]

except $ 1$  :wink:


Prove that there exist infinitely many positive integers $a$
such that $(a^n + n^a)$is a prime for some positive integers $n$.

[quote="mps"]Prove that there exists infinitely positive integers $a$
such that $(a^n + n^a)$is a prime for some positive integers $n$.[/quote]
Are you sure about this problem? :?
If yes,what is the source of it?

Sorry,unknown source
I found  it  somewhere on other Math forum.
I try to solve it. But I can't.

Forget it,if it's useless.

Dear $mps$,I asked this question,becaus I don't know what should we prove from wich this problem will be done,usually problem of this kinde have this form
Prove that there exist infinitely many positive integers  $a$
such that $a^n+n^a$ [b]isn't[/b]  prime for some positive integers $n$.
If the above one isn't stupide problem(I havn't tried it),then mybe that is a correct version of your problem.

Isn't $a^1+1^a=a+1$ infinitely often prime ? :P

Yes, so I think the question is indeed with "isn't prime."
That would be a natural generalization for the well-known problem : $n^4 + 4^n \notin \mathbb {P}$. :)

n^4 +4^n =(n^2+2^n)^2 -2.2^n.n^2
 if n is even .what is happen?

If $n$ is even, then $n^4 + 4^n$ is even and cannot be prime.  :P

OK,here is a solution for problem with "isn't prime" :) .
we have that the numbers of this form$x^4+4y^4=(x^2+2y^2-2x^2y^2)(x^2+2y^2+2x^2y^2)$ isn't prime numbers.
Let's take $a=8k+6$
and $n=8m^4$=>
$a^n+n^a=(a^{2m^4})^4+4(2^{6k+4}m^a)^4=t^4+4l^4$isn't prime ;) .


quadratics
algebra
polynomial
combinatorics proposed
combinatorics

Let $ H$ be a set of $ 2000$ nonzero real numbers. How many negative elements should $ H$ have in order to maximize the number of four-element subsets of $ H$ with a negative product of elements?

Most of this solution goes away if one can use a calculator, but I don't have a way to avoid the calculations entirely.

If $ H$ has $ p$ positive elements and $ n$ negative elements, $ p+n = 2000 = s$, then there are $ \binom{p}{3}\cdot n+\binom{n}{3}\cdot p$ 4-element subsets of $ H$ with negative sum.  We can re-write this as $ \frac{pn}{6}(p^{2}+n^{2}-3p-3n+4) = \frac{pn}{6}((p+n)^{2}-2pn-3(p+n)+4) = \frac{pn}{6}(-2pn+(s^{2}-3s+4))$.  This is a quadratic polynomial in $ pn$ with negative leading coefficient, so it achieves its maximum when $ pn = \frac{s^{2}-3s+4}{4}$.  (Also note that because it's a quadratic, simply getting as close as possible to this value is good enough.)  This is itself just a quadratic in $ n$, $ 0 = n^{2}-sn+\frac{s^{2}-3s+4}{4}$ which has its roots at $ n = \frac{s \pm \sqrt{3s-4}}{2}= 1000 \pm \sqrt{1499}$.  So, we want to choose the integer values of $ n$ closest to $ 1000 \pm \sqrt{1499}$ -- whichever gives us a value of $ n(2000-n)$ closer to $ \frac{s^{2}-3s+4}{4}= 998501$, wins.  I don't see a better way out of this than just calculating: $ \sqrt{1499}$ lies between 38 and 39, so we're talking about $ n$ at $ 1038$ or $ 1039$ which lead to values of 998556 and 998479, the latter of which is better.  Thus, we want either 1039 or 961 negative elements in $ H$.


Let  $ M \equal{} \{ 1,2,.......,2048\}$ And $ X \subset M$ such that $ \left| X \right| \equal{} 15$
Show that there are two distinct subsets of $ X$ whose sum of elements is the same.

ie.    $ A,B \subset X\& A \cap B \equal{} \emptyset$

                         $ \sum_{\substack{a\in A}}a\equal{}\sum_{\substack{b\in B}}b$

Does this have something to do with the fact that 2^11=2048?

Use pigeon-hole principle here. The number of subsets of X is $ 2^{15}\minus{}1$. The largest sum of such subset is the sum of all elements of M which is $ 2^{12}\minus{}1$. So there are 2 subsets $ A$ and $ B$ with equal sums. If some number is in both A and B remove it from both. This doesn't change the fact that A and B have equal sums. The problem is solved.

Its a variation of [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1972]IMO1972[/url], which isn't difficult, that's why it's #1.


hello, from the second equation we get
$ y\equal{}\frac{3x}{x\minus{}3}$
plugging this in the first equation simplifying and factoring we get
$ \frac{(x^2\plus{}3x\minus{}9)(x\minus{}6)^2}{(x\minus{}3)^2}\equal{}0$
now it is easy to find all solutions of the system.
Sonnhard.

Multiply the equations to get $ \frac{x^2}{y^2}\plus{}\frac{y^2}{x^2}\plus{}\frac xy\plus{}\frac yx\equal{}4$.

By AM-GM, $ \frac{x^2}{y^2}\plus{}\frac{y^2}{x^2}\plus{}\frac xy\plus{}\frac yx\geq 4\sqrt{1}\equal{}4$ with equality iff $ x\equal{}y$.

Put $ x\equal{}y$ into either equation and you get $ 2x\equal{}12\implies \boxed{x\equal{}y\equal{}6}$.

Unfortunately, this doesn't get you the negative solutions...

hello, with your method you don't get the solutions
$ (\frac{3}{2}(\minus{}\sqrt{5}\minus{}1),\frac{3}{2}(\sqrt{5}\minus{}1))$
$ (\frac{3}{2}(\sqrt{5}\minus{}1),\frac{3}{2}(\minus{}\sqrt{5}\minus{}1))$
Sonnhard.

Yeah, I know, I mentioned that.
If this were a system in the positive reals, though, it would work nicely.

But yeah, for this problem I guess substitution is the best approach.

Hi Sonnhard and grn_trtle. I have the following method (which is much the same as Sonnhard's):

$ x, y \neq 0$.

On multiplying both sides by $ xy$, the first equation may be written as

$ x^{3} \plus{} y^{3} \equal{} 12xy$.

Similarly, the second equation is equivalent to

$ x \plus{} y \equal{} \frac {xy}{3}$.

Now,

$ (x \plus{} y)^{3} \equal{} x^{3} \plus{} y^{3} \plus{} 3xy(x \plus{} y)$.

Substituting $ x \plus{} y \equal{} \frac {xy}{3}$, we have

$ \frac {(xy)^{3}}{27} \equal{} 12(xy) \plus{} (xy)^{2}$,

$ (xy)^{3} \equal{} 324(xy) \plus{} 27(xy)^{2}$

$ (xy)[(xy)^{2} \minus{} 27(xy) \plus{} 324] \equal{} 0$

$ (xy \minus{} 36)(xy \plus{} 9) \equal{} 0$

$ xy \equal{} 36, x \plus{} y \equal{} 12 \implies x \equal{} 6, y \equal{} 6$

$ xy \equal{} \minus{} 9, x \plus{} y \equal{} 12 \implies x \equal{} \frac { \minus{} 3\pm3\sqrt {5}}{2}$

Therefore $ (x, y) \equal{} (6, 6), (\frac { \minus{} 3 \plus{} 3\sqrt {5}}{2}, \frac { \minus{} 3 \minus{} 3\sqrt {5}}{2}), (\frac { \minus{} 3 \minus{} 3\sqrt {5}}{2}, \frac { \minus{} 3 \plus{} 3\sqrt {5}}{2})$.


Simplify $ \left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4$; the result is:

$ \textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2$


complex numbers
superior algebra
superior algebra solved

Hi! I'm new to this field and I don't know if my problem is in the right place.

a) Show that set $A=\{z\in\mathbb{C} | z=a-b+i(a+b), a,b\in\mathbb{Z}\}$ is a group recording to the complex numbers addition.
b) Is the set $A$ a stabil part recording to the complex numbers multiplication?

Sorry for my bad english.

Start from $A=(1+i)Z[i]$. So $A$ is an ideal in $Z[i]$.


Let $a,b,c \in \mathbb{Z^{+}}$ be edge lengths of a rectangular parallelepiped. Prove (or disprove) that there exist no $(a,b,c)$ such that all face and space diagonals are integers.

[hide="Source"] PreCalc class. The teacher told us that he was not sure whether a "perfect box" was possible, but knew that 6 of the 7 lengths (edges, face diagonals, and main diagonal) could be made integers -- this was for extra credit :D  .  I'm not planning to use anyone's work other than for my own curiosity (dont worry :P).[/hide]

Maybe use the method for generating integer pythagorean triples? They can be made with $(a^{2}-b^{2},2ab,a^{2}+b^{2})$ for integers a and b. No time to do it right now though...


floor function
inequalities
function
algebra unsolved
algebra

The sum of several non-negative numbers is not greater than 200, while the sum of their squares is not less than 2500. Prove that among them there are four numbers whose sum is not less than 50.

[i]Proposed by A. Khabrov[/i]

Denote the given numbers $ a_1 \geq a_2 \geq \cdots \geq a_n \geq 0$, where $ n$ may be taken as large as wanted, by just appending zeroes.
Denote $ \sigma \equal{} \sum_{i\equal{}1}^n a_i$, so $ \sigma \leq 200$ and $ \sum_{i\equal{}1}^n a_i^2 \geq 2500$, and denote $ s \equal{} a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4$.
Take $ k \equal{} \left \lfloor \frac {\sigma \minus{} a_1 \minus{} a_2 \minus{} a_3} {a_4} \right \rfloor$ if $ a_4 \neq 0$, and $ k \equal{} 1$ otherwise. Then $ \sigma \equal{} a_1 \plus{} a_2 \plus{} a_3 \plus{} ka_4 \plus{} \varepsilon a_4$, for some $ 0 \leq \varepsilon < 1$.
The solution comes in two steps.

Step 1.
Build a decreasing sequence $ b_1 \equal{} a_1 \equal{} a$, $ b_2 \equal{} a_2 \equal{} b$, $ b_3 \equal{} a_3 \equal{} c$, $ b_i \equal{} a_4 \equal{} d$ for $ 4 \leq i \leq k\plus{}3$, $ b_{k\plus{}4} \equal{} \varepsilon a_4 \equal{} \varepsilon d$, $ b_j \equal{} 0$ for $ k\plus{}5 \leq j \leq n$.
Now, the sequence $ (b_1, b_2, b_3, b_4, \ldots, b_n)$ dominates (in the sense of $ \textsc{Karamata}$'s inequality) the sequence $ (a_1, a_2, a_3, a_4, \ldots, a_n)$, since
$ \sum_{i\equal{}1}^{\ell} b_i \geq \sum_{i\equal{}1}^{\ell} a_i$ for $ 1 \leq \ell \leq n\minus{}1$, and $ \sum_{i\equal{}1}^{n} b_i \equal{} \sum_{i\equal{}1}^{n} a_i \equal{} \sigma$. As $ f(x) \equal{} x^2$ is a convex function, $ \textsc{Karamata}$ yields $ \sum_{i\equal{}1}^n b_i^2 \geq \sum_{i\equal{}1}^n a_i^2 \geq 2500$
(this result can be obtained also in more elementary ways).

Step 2.
But $ \sum_{i\equal{}1}^n b_i^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} kd^2 \plus{} \varepsilon^2d^2$ = $ \sum a^2 \plus{}d(\sigma \minus{} \sum a) \minus{} \varepsilon d^2(1\minus{}\varepsilon) \leq \sum a^2 \plus{} d(200 \minus{} \sum a)$.
Take now $ a \equal{} d\plus{}\alpha$, $ b \equal{} d\plus{}\beta$, $ c \equal{} d\plus{}\gamma$, with $ \alpha, \beta, \gamma \geq 0$. Combining all these yields $ d \geq \frac {2500 \minus{} \sum \alpha^2} {200 \plus{} \sum \alpha}$.
Therefore $ s \minus{}50 \equal{} 4d \plus{} \sum \alpha \minus{} 50 \geq 4 \cdot \frac {2500 \minus{} \sum \alpha^2} {200 \plus{} \sum \alpha} \plus{} \sum \alpha \minus{} 50$ = $ \frac {3\sum \alpha(50\minus{}\alpha) \plus{} 2\sum \alpha\beta} {200 \plus{} \sum \alpha} \geq (50\minus{}s)\frac{3\sum \alpha} {200 \plus{} \sum \alpha}$, hence $ s \geq 50$.

Remark.
Similarly one can get that $ a_1 \geq 12.5$, $ a_1 \plus{} a_2 \geq 25$, and $ a_1 \plus{} a_2 \plus{} a_3 \geq 37.5$, all realized e.g. when $ a_i \equal{} 12.5$ for $ 1 \leq i \leq 16$.
However, one only has $ \sum_{i\equal{}1}^{\ell} a_i \geq 50$ for $ \ell \geq 5$, realized e.g. when $ a_1 \equal{} 50$ and $ a_i \equal{} 0$ for $ i \geq 2$.

This is my solution:

The problem does not change if we delete all the zeroes, so we can assume that all numbers are positive. Let $a_1 \geq a_2 \geq \cdots \geq a_n$ be those numbers, and suppose all sums of four are less than 50. So now we have three conditions:
1. $a_1 + a_2 + a_3 + a_4 < 50$
2. $a_1 + \cdots + a_n \leq 200$.
3. $a_1^2 + \cdots + a_n^2 \geq 2500$.

Extend the sequence by adding numbers $a_{n+1}, \cdots, a_m$ where $a_{n+1} = a_{n+2} = \cdots = a_{m-1} \geq a_m$ such that $a_1 + \cdots + a_m = 200$. This addition will not disrupt condition one and three, so we now have:

1. $a_1 + a_2 + a_3 + a_4 < 50$.
2. $a_1 + \cdots + a_m = 200$.
3. $a_1^2 + \cdots + a_m^2 \geq 2500$.

Let us now define a "transfer" operation as follows. Given $a,b$ with $a \geq b$, we replace them with $a+ \epsilon,b - \epsilon$ with $\epsilon > 0$. One can easily verify that a transfer on two numbers will not change their linear sums, but will only increase their sum of squares, because $(a+\epsilon)^2 + (b-\epsilon)^2 > a^2+b^2 \iff \epsilon(a-b)+\epsilon^2 > 0$

So now we can apply a series of transfers, beginning with $(a_1,a_m)$, followed by $(a_1, a_{m-1}), (a_1, a_{m-2}), ... $ and so on. In each step, we apply as much transfer as possible, possibly reducing the smaller number to zero in the process. We stop when $a_1 + a_2 + a_3 + a_4$ reach $50$, at which points our three conditions become:
1. $a_1 + a_2 + a_3 + a_4 = 50$.
2. $a_1 + \cdots + a_m = 200$.
3. $a_1^2 + \cdots + a_m^2 > 2500$.

(Note that the sign for condition 3 changed because equality is no longer possible. In order to reach condition 1, we must do at least one non-trivial transfer and that transfer strictly increase the sum of squares).

Now we apply another series of transfers similar to above. We start with the last numbers and apply the transfer, this time not to $a_1$ but to $a_5$. Our goal is such that $a_4 = a_5 = \cdots = a_{k-1} \geq a_k > a_{k+1} = \cdots = a_m = 0$ for some $k$. One can achieve this configuration by transferring from the smallest number each time, and raising the largest number that's less than $a_4$ to be up to par with $a_4$. Every transfer operation does not change our three conditions above.

After we are done, our sequence looks like this: $a, b, c, d, d, \cdots, d, e$.

Lastly, we transfer $(c-d)$ from $c$ to $a$, and $(b-d)$ from $b$ to $a$ to arrive at configuration $50-3d,d,d,d,\cdots,d,e$ where there are $N$ $d$'s. Again, these final transfers still do not change our three conditions, which can now be written as:

1. $(50-3d)+d+d+d = 50$.
2. $(50-3d) + Nd + e = 200$.
3. $(50-3d)^2 +  Nd^2 + e^2 > 2500$

But because $50-3d \geq d$, then $d \leq 12.5$, and thus $12d \leq 150$. Then
$(50-3d)^2 +  Nd^2 + e^2 = (50-3d)^2 + e^2 + d(200-e-(50-3d))$
$= 2500 - 150d + 12d^2 + e^2 -  ed$
$= 2500 + d(12d-150) + e(e-d) \leq 2500$ which contradicts condition 3.


In $ \triangle{ABC}$, let $ a$, $ b$, and $ c$ be the lengths of the sides opposite angles with measures $ A^{\circ}$, $ B^{\circ}$, and $ C^{\circ}$, respectively.  If $ B^{\circ} > 90^{\circ}$, $ b^2\equal{}c(c\plus{}a)$, and $ C^{\circ}\equal{}kB^{\circ}$, then find the set of all possible values of $ k$.

[quote="kunny"]Intuitively $ k \equal{} 2$.[/quote]

Could you please explain "intuitively?"  Also, the answer [hide="is"]$ 0.5$[/hide].  Thanks!

The problem is similar to that of [url=http://www.mathlinks.ro/viewtopic.php?t=58215]USAMO 1991 #1[/url].
Maybe I remember this problem. :lol:


geometry
3D geometry
tetrahedron
inequalities
trigonometry

In a tetrahedron $ABCD$, the medians of the faces $ABD$, $ACD$, $BCD$ from $D$ make equal angles with the corresponding edges $AB$, $AC$, $BC$. Prove that each of these faces has area less than or equal to the sum of the areas of the other two faces.

[hide="Comment"][i]Equivalent version of the problem:[/i] $ABCD$ is a tetrahedron. $DE$, $DF$, $DG$ are medians of triangles $DBC$, $DCA$, $DAB$. The angles between $DE$ and $BC$, between $DF$ and $CA$, and between $DG$ and $AB$ are equal. Show that: area $DBC$ $\leq$ area $DCA$ + area $DAB$. [/hide]

it's from the 48-th Polish Ol., they used the Ptholemy's theorem (check their solution), I remember solving it using only the cosine law.

Let $K, L, M$ be the midpoints of $BC, CA, AB$ respectively. We need to show that for example
$$DK\cdot BC\le DL\cdot CA+DM\cdot AB.$$
It suffices only to apply the Ptolemy's theorem for the spacial quadrilateral $DLKM$ whence the result follows immediately:
$$DK\cdot LM\le DL\cdot KM+DM\cdot KL.$$


Hi,

I am having trouble understanding orthonormal basis ...

say you have a space V spanned by vectors (v1, v2, v3)

when you construct the orthonormal basis (u1, u2, u3) using the gram-schmidt process ... what is the geometric significance of this new basis? and is it in V?

i guess my question is why do we need an orthonormal basis

hey k2imperial,

if you have a random set of vectors V that span the space, it is possible that v1, v2, v3 are not all mutually perpendicular. the gramm-schmidt process ensures the vectors u1, u2, u3 are to span the same space.

hmm, i guess you don't technically need it, but it makes things easier.

hope that helps!
k2imperial


calculus
function
geometry
3D geometry
sphere
algebra
polynomial

Maxima and Minima using trigonometric functions

1.) Find the largest conical tent that can be constructed having a given slant height?
2.) Find the altitude of the circular cone of maxima convex surface inscribed a sphere of radius a?

tnx in advance.

Question 1

I assume that the "slant height" is the hypotenuse of the cone.

Call this slant height $ l$.

Now we must use a variable to represent each possible cone: you could use the angle at the apex, but this involves more work. I will use the height $ h$. The radius of the cone by Pythagoras' Theorem is given by $ r^2 \equal{} l^2 \minus{} h^2$.

The volume of the cone is $ V \equal{} \frac{1}{3} \pi r^2 h$. Expressing in terms of $ h$ we have $ V(h) \equal{} \frac{1}{3} \pi (l^2h \minus{} h^3)$. This is a polynomial, which is why $ h$ instead of $ r$ is our prefered choice of variable. $ V(h)$ is a continuous function of $ h$, so on any closed interval it has maximum value at the endpoints or when $ V'(\theta) \equal{} 0$. Clearly $ 0 \leq h \leq l$, and $ V(0) \equal{} V(l) \equal{} 0$.

$ V'(\theta) \equal{} \frac{1}{3} (l^2 \minus{} 3h^2)$. Solving for $ V'(\theta) \equal{} 0$ yields $ h \equal{} \frac{l}{\sqrt{3}}$.

Substituting this value back into $ V$ gives $ V \equal{} \frac{1}{3} \pi (\frac{l^3}{\sqrt{3}} \minus{} \frac{l^3}{9\sqrt{3}}) \equal{} \frac{1}{3\sqrt{3}} \pi l^3 (1 \minus{} \frac{1}{9}) \equal{} \frac{8}{27\sqrt{3}} \pi l^3$

Question 2

The centres of the cylinder and sphere must coincide. The circular edges of the cylinder touch the sphere.

So the length of the line joining the centre of the cylinder to the circular edge is $ a$. We let our variable be $ \theta$: the angle between this line and the axis of the cylinder. A suitable diagram will show that the diameter of the cylinder is $ d \equal{} 2a\sin{\theta}$ and the height is $ h \equal{} 2a \cos{\theta}$.

The area of the curved surface is $ A \equal{} \pi dh$. Substituting our expressions gives $ A(\theta) \equal{} 4 \pi a^2 \sin{\theta}\cos{\theta}$. Using the double angle formulae, this is $ A(\theta) \equal{} 2 \pi a^2 \sin{2\theta}$. Note that $ A$ is maximum when $ \sin{2\theta} \equal{} 1$ which occurs for $ \theta \equal{} \frac{\pi}{4}$. Substituting this value into the expression for $ h$ gives $ h_{max} \equal{} 2a \cdot \frac{1}{\sqrt{2}} \equal{} \sqrt{2} a$.

Note that this was a calculus free solution!


1) Let
[A 0]
[ B ] = M
[0 C]
WHERE A, B, C are square matrices.
Prove that M is invertible if and only if A, B, and C are all invertible.

2) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
b) using only the definition of matrix multiplication , prove that B is upper triangular.

please hlep. thank you very much..

Could you possibly clarify that first matrix?  Here is the syntax:
[code]\begin{pmatrix}
a&b&c&\cdots&z\\
A&B&C&\cdots&Z\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&2&3&\cdots&4
\end{pmatrix}[/code]

This will yield:
$ \begin{pmatrix}
a&b&c&\cdots&z\\
A&B&C&\cdots&Z\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&2&3&\cdots&4
\end{pmatrix}$

So feel free to fill in the appropriate entries.  Once I get a better picture of what the matrix looks like, I should be able to help you.


[b]A)[/b] $1<a_1\leq a_2\leq \ldots\leq a_{12}<144$ are real numbers. Prove that there exist positive integers $1\leq i<j<k\leq 12$ that $a_i,a_j,a_k$ can be sides of triangle.

[b]B)[/b] Generalize.

Yes, you're right  :) . You can also generalize to quadrilaterals and m-gons... but the number instead of 144 will be given just by recurence.


1.a) Given concentric circles (circles that have the same centre), prove that two chords of the larger circle that are tangent to the smaller circle are congruent.

b)Given a circle with centre C, a point P (note that P can be inside, on, or outside of the circle), and length d, construct a line through P such that the length of the chord is d.

a) is evident: length of each hord equals $2\sqrt{R^2-r^2}$.
b) Let's construct circle, which is concentric our circle with the radius $R$, with the radius $r=\sqrt{R^2-\frac{d^2}{4}}$. Since $d=2\sqrt{R^2-r^2}$, the tangent to the second circle is the right hord. This method works, since $d\leq D=2R$ :arrow: $R^2-\frac{d^2}{4}\geq 0$.


Find the number of square units in the area of the quadrilateral. The dots are equally spaced vertically and horizontally. Express your answer as a mixed number.
[asy]size(70);
for(int i = 0; i < 5; ++i){
 for(int j = 0; j < 6;++j){
   dot((i,j));
 }
}
draw((2,0)--(1,3)--(2,5)--(4,4)--(2,0),linewidth(0.7));[/asy]

Blargh. BY picks (WoW ftw!)

6-1+5/2=7 1/2

For Pick's theorem, go [url=http://en.wikipedia.or/wiki/Pick%27s_theorem]here[/url].

[quote="ernie"]For Pick's theorem, go [url=http://en.wikipedia.org/wiki/Pick%27s_theorem]here[/url].[/quote]

Fixed your link :D


If a,b,c>0 prove the two ineqs
$(ab+bc+ca)[(a+b+c)^3+27abc]\geq 3(a+b+c)[3(a+b+c)abc+(ab+bc+ca)^2]$
and the second

$(a+b+c)[2(a+b+c)^4+11(a+b+c)abc+13(ab+bc+ca)^2]\geq (ab+bc+ca)[11(a+b+c)^3+15abc]$ :)

The first one is $((\sum a)^2 - 3(\sum ab))(\sum_{sym} a^2b - 6abc) \ge 0$ which is true by amgm.

By expanding the second ineq we get an inequality between multiindexes
$(5,0,0)$ and $(4,1,0)$ which is true by Muirhead.


[quote="Kondr"][quote="silouan"]If $a,b,c$ are positive reals such that $a^2+b^2+c^2=1$ prove that

$2abc(a+b+c)\leq 2(a+b+c)^2+1$[/quote]
I think this ineq is stronger
$18abc(a+b+c)\leq 2(a+b+c)^2$, which is equivalent to
$9abc\leq (a+b+c)(a^2+b^2+c^2)$
$\frac 3 2 [1,1,1]\leq \frac 1 2[3,0,0]+[2,1,0]$, that holds by Muirhead. Where is my mistake? :huh:[/quote]
no.

$1=(a^2+b^2+c^2)^2 \ge (ab+bc+ca)^2\ge 3abc(a+b+c)$ which is stronger.

silouan's inequality is very very weak.
$2(a+b+c)^2+1=(a^2+b^2+c^2)(3a^2+3b^2+3c^2+4ab+4ac+4bc)=\sum_{cyc}{3a^4+3a^2b^2+3a^2c^2+4a^3b+4a^3c+4a^2bc}\geq 2abc(a+b+c) \\\Leftrightarrow \sum_{cyc}{3a^4+3a^2b^2+3a^2c^2+4a^3b+4a^3c+2a^2bc}\geq0$... quite weak, huh? :| maybe it should be $2(a+b+c)^2+1\geq 21abc(a+b+c)$, but anyway it's too easy, just homogenize and apply muirhead.

in your original inequality, if $a,b,c>0$, then it's obvious that the equality would never hold. but if you mean the other one, as far as i know (in muirhead), the equality holds iff $a=b=c$. i hope i'm not wrong


inequalities
geometry
ratio
trigonometry
geometry proposed

We can go by this way:
Put $ CA=a;CB=b;CP=p \geq 0$
$ \Longrightarrow \frac{AB-PB}{AP} = \frac {\sqrt{a^2+b^2}-\sqrt{b^2+p^2}}{a-p}= f(p)$
$ \Longrightarrow f'(p)=\frac{\sqrt{a^2+b^2}.\sqrt{b^2+p^2}-(pa+b^2)}{\sqrt{b^2+p^2}.(a-p)^2}$
By Cauchy-Schwarts inequality: $ \sqrt{a^2+b^2}.\sqrt{b^2+p^2} \geq (pa+b^2)$
$ \Longrightarrow f'(p) \geq 0$
$ \Longrightarrow f(p) \geq f(0) = \frac{AB-BC}{AC}$
$ \Longrightarrow \frac{AB-PB}{AP} \geq \frac{AB-BC}{AC}$
It's an ugly solution but this method works for many problem. ;)

Actually, if you draw a nice picture, it becomes clear :). If you draw a picture, it could even be qualified as a proof without words :).

Let $I_P,T_P$ be the incetner of $APB$ and the point where the incircle touches $AP$, respectively, and let $I,T$ be the analogous points for $ABC$. We want to show that $\frac{AB+AP-BP}{AB+AC-BC}>\frac{AP}{AC}$. The first ratio is $\frac{AT_P}{AT}=\frac{AI_P}{AI}$. Now, since $\angle API_P>\angle ACI$, the conclusion is clear.

Indeed, why do we need nice geometric proofs if Mathematica (or Maple, depending on what you like better) does it as well? :D

Actually, Kunny's inequality

$\frac{AB-BP}{AP}>\frac{AB-BC}{AC}$

can be also proven as follows: By multiplying it with $AP\cdot AC$, we rewrite it in the form

$\left(AB-BP\right)\cdot AC > \left(AB-BC\right)\cdot AP$.

In other words,

$AB\cdot AC - BP\cdot AC > AB\cdot AP - BC\cdot AP$.

Now we add $BP\cdot AC - AB\cdot AP$ to both sides;

$AB\cdot AC - AB\cdot AP > BP\cdot AC - BC\cdot AP$.

But $AB\cdot AC - AB\cdot AP = AB\cdot \left(AC-AP\right) = AB\cdot CP$. Thus, it remains to prove

$AB\cdot CP > BP\cdot AC - BC\cdot AP$.

This is equivalent to

$AB\cdot CP + BC\cdot AP > BP\cdot AC$.

But this follows from the Ptolemy inequality, applied to the quadrilateral ABCP (and we have the strict > sign since this quadrilateral cannot be cyclic). Problem solved.

  Darij

[u][b]The author of this posting is : Charlesww[/b][/u]
____________________________________________________________________

[quote="kunny"]Given a right triangle $ABC$ with hypotenuse $AB$. We take an arbitrary point $P$ , which is different  from the vertices $A$ and $C$, on $AC$.
Prove that

\[\frac{AB-BP}{AP}>\frac{AB-BC}{AC}\][/quote]

Here is my solution:

Consider that
$\angle{ABP} = \alpha$, $\angle{CBP} = \beta$

All we need to prove is that
$PC\cdot BA + PA\cdot BC > AC\cdot PB$

the same as
$\frac{\sin{C}}{\sin{B}}\cdot\frac{\sin{\beta}}{\sin{C}}+\frac{\sin{A}}{\sin{B}}\cdot\frac{\sin{\alpha}}{\sin{A}}>1$

in fact, with $\alpha, \beta, B < \frac{\pi}{2}$, we get

$LHS=\frac{\sin{\alpha}+\sin{\beta}}{\sin{B}} > \frac{\sin{\alpha}\cdot\cos{\beta}+\sin{\beta}\cdot\cos{\alpha}}{\sin{B}}=1$

Q.e.d


and even without knowing the trick, it's really helpful to know the answer to 1!-10!. It helps speed stuff up a lot.

[quote="usahw1011"]1. If 5! *3!=n!, what is n?

Is there a trick to this?[/quote]
n=6 because 3!=6, so 6*5!=6!


geometry
trapezoid
AMC
AMC 10
rectangle
Pythagorean Theorem

2001 AMC10:

In trapezoid $ ABCD$, $ \overline{AB}$ and $ \overline{CD}$ are perpendicular to $ \overline{AD}$, with $ AB \plus{} CD \equal{} BC$, $ AB < CD$, and $ AD \equal{} 7$. What is $ AB\cdot{CD}$?

I think there can be multiple answers...

There can?

Well, here are the choices the test gave:

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

You typed the question wrong. It should be $ AB\plus{}CD\equal{}BC$, and $ AB<CD$.

oh yea, I did! Sorry. Can someone still give me a solution to the problem though?

[hide="Solution"]Let $ F$ be the foot of the altitude from $ B$ to $ \overline{CD}$. Since $ BEDA$ is a rectangle, we apply the Pythagorean Theorem:
\begin{align*}
AD^2=BF^2&=BC^2-CF^2 \\
&=\left(AB+CD\right)^2-\left(CD-AB\right)^2 \\
&=AB^2+2\cdot AB \cdot CD + CD^2-CD^2+2 \cdot AB \cdot CD - AB^2 \\
&=4 \cdot AB \cdot CD
\end{align*}Then, $ AB \cdot CD=\frac{AD^2}{4}=\frac{49}{4}=\boxed{12.25}$.[/hide]

I hope this helps.  New here to the forums...

The altitude of the trapezoid is 7, so the perpendicular line from segment DC and point B is 7.  Let's say the perpendicular intersection of point B and segment DC is E.  Now triangle BEC is a right triangle, with E being 90 degrees.  Say the length of segment AB is x and the length of segment DC is y.  The length of EC would be (y-x).  Since BC is AB+DC, BC is (x+y).  You may now use Pythagorean Theorem, producing the equation (y-x)^2 + (7)^2 = (x+y)^2.  Simplifying further gives us x^2 - 2xy + y^2 + 49 = x^2 + 2xy +y^2.  Subract x^2 and y^2 from both sides and add 2xy to both sides.  49 = 4xy.  Now divide by 4, giving us xy=12.25.  Recall x was length AB and y was length DC.  Therefore, the product is 12.25.  I hope it helps.  =D


[quote="marucho8d7"]A candle burns completely in one hour, there are two candles of equal size and length in a room. How can you burn out both candles in 45 minutes?[/quote]

Wait 40 minutes, and throw both in the fireplace

OR, just burn them normally. Nobody said that the candles being used in the problem was being burned in one hour.

Even if it was, cut 1/4 of it off, put the caps together, and burn all three at one time

You seriously expect that to be the answer? NOT!

NOTICES:

You have no way of telling time in the room (and you may NOT count the time)

The room has no fireplace; there is only a table with two candles and a box of matches. It is completely dark with no windows, but you can still see the table with the objects on it (don't ask why).

You MUST burn the candles completely. I don't care WHAT you do with them. Break them if you wish, but I want EXACTLY 45 minutes passed. Each candle burns for EXACTLY one hour, no more, no less.

ONE MORE THING. How do you know fore sure that you can cut one-fourth of it? I want a way that results in EXACTLY 45 minutes. And NO COUNTING!

ill post it in the intermediate section as well.

Now I hope you can think of a better answer.

[quote="marucho8d7"]A candle burns completely in one hour, there are two candles of equal size and length in a room. How can you burn out both candles in 45 minutes?[/quote]

 Here is my answer! We let both of them burnt out within 45 minutes. I know they don't burn completely. But you didn't tell we had to burn out them completely. OK?

hmm

spoiler

[color=white]light both, one at both ends and one at one end. when the one lit at both ends is finishes that's half an hour, and you know that half of the other one is left, light the second end of this and that will last 15 more minutes.[/color]

that took some tought :blush:

I really like Mikey's answer. I would have never thought of it. It's genius!! I can't believe it's so easy.

If you just let them both burn than it would take an hour. But if you break one in half and the other in forths. You can than light the halfs they burn for 30 minutes, than light the forths they burn in 15 minutes. Together that gives you 45 minutes.

How do you know
1) that you can break it into perfect halves?
2) that the candle burns at a constant rate?

Do the problem without relying on either of those two things.

I don't think Mikey's answer relies on either of those two assumptions...

[quote="themonster"]Mikey's answer does assume that the candles burn at constant rates, because after half an hour, he assumes that the candle lit from only one end has burned out half of the candle.[/quote]

No, he doesn't.  He assumes that, since it has been burned for half an hour, it has been burned through half of its life, which is perfectly O.K. to assume.

[quote="pakagawa"]Dear Marucho8d7:

Visit some other forums. like the games and fun factory.

You could interest us there. And, about the candle problem, the way you stated it, I could easily prove impossible be physics and time-counting.

So, tell me what is possible.[/quote]

what do you mean by that?

[quote="dts"]There are two aspects of being good at competitive mathematics; the Art of Problem Solving and the Art of Lucky Guessing.[/quote]

you bet

hey this problem is in the fun and games section! by the way my answer is in the funa dn games section but i'lll repeat it anyway.  get a robot. program it to hold both the candles and in exactly 45 mintes, it'll self destruct.  its the right answer.  everyone in the funa dn games agrees.

sorry. but for the last time: THERE IS NOTHING ELSE IN THAT ROOM AFTER YOU GO IN BUT THE CANDLES AND THE MATCHES! USING COMMON SENSE, WE MUST KNOW THAT YOU CANNOT TAKE ANYTHING WITH YOU INTO THE ROOM! GODDDDDD! WHATS WRONG WITH YOU? IM BEING SERIOUS HERE! sorry about not mentioning, but YOU HAVE TO LIGHT THE CANDLES AT ZERO MINUTES AND HAVE ALL THE CANDLES BURNED OUT BY 45 MINUTES! USE YOUR COMMON SENSE!

I am a stupid little seventh grader who does not fear my stupidity..

so I think this is how u solve it..

break one fourth of the both candles.
then burn them..
since the candle is 60 minutes long
3/4 would be 45 minutes
....
solution
even it might be the stupidest answer u heard don't insult me

[quote="shinwoo"]so I think this is how u solve it..

break one fourth of the both candles.
then burn them..
since the candle is 60 minutes long
3/4 would be 45 minutes
....
solution
even it might be the stupidest answer u heard don't insult me[/quote]

i do not say it, but i will.

WRONG. You have NO IDEA whether what you break is exact. ANd i mentioned NO BREAKING THE CANDLES and NOTHING BROUGHT WITH YOU INTO THE ROOM for a GAZILLION TIMES!

its a joy to see a problem like that!!
  a solution for that obviously follow  some assumptions
without them its not complete.


complex analysis
function
complex analysis unsolved

Let $f$ be a holomorphic function in the strip $x\in\mathbb R,-1<y<1$ with the property that
\[|f(z)|\le A(1+|z|)^{\eta}, \eta\in\mathbb R \textrm{ fixed}\]
for all $z$ in that strip.
Show that for each integer $n\ge 0$ there exists $A_{n}\ge 0$ so that
\[|f^{(n)}(x)|\le A_{n}(1+|x|)^\eta\]
for all $x\in\mathbb R$.


[quote="Kono"]Determine the triples of integers $ (x,y,z)$ satisfying the equation $ x^3 \plus{} y^3 \plus{} z^3 \equal{} (x \plus{} y \plus{} z)^3$[/quote]

easy
$ (x\plus{}y\plus{}z)^{3}\minus{}x^{3}\minus{}y^{3}\minus{}z^{3}\equal{}3(x\plus{}y)(y\plus{}z)(z\plus{}x)\equal{}0$ ...


A ship leaves port at 20 miles per hour with a bearing of S 46 E. There is a warning buoy located 5 miles directly north of the port. What is  the bearing of the warning buoy as seen from the ship after 4.5 hours?

I am having trouble understanding bearing problems.

I understand that the word bearing means location but how does one tackle such questions like the one here?

When you say "S 46 E", do you mean $ 46^{\circ}$ south of east?  East of south?  This problem seems like it should be tackled by using vectors.  Since the boat travels 20 mph for 4.5 hours, the length of the boat's position vector is 90.  You can use trig to plot the rest.


if  tan(a+b)=n tan(a-b)

show that (n+1) sin(2b)=(n-1)sin(2a)

without using componendo and dividendo

ayush, post these problems [url=http://joigroup.freezoka.com/index.php]here[/url], im sure you'd have more response! :)

I think just writing the given equation like sin(a+b) * cos(a-b)/cos(a+b) * sin(a-b) = n

And using sinacosb = (sin(a+b) + sin(a-b))/2 should get the result.


trigonometry
inequalities
inequalities proposed

[hide]Use $S=\frac{abc}{4R}=pr,\cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2bc}$, etc. And Denote $x=a+b,y=b+c,z=c+a$. We will have a algebraic inequality of $x,y,z$.[/hide]

[quote="Zamfirmihai"]prove that in any triangle holds:

cosA/a(b+c) + cosB/b(a+c) = cosC/c(a+b) <= 1/8Rr[/quote]
It's equivalent to $\sum_{cyc}(a-b)^{2}((a+b)^{2}-c^{2})\geq0,$ which triviality.


A pitcher contains 4 1/2  cups of water. A 1/  3 -cup scoop is used to  transfer water from the pitcher to a bowl. How many scoops are needed until the bowl has more water than the pitcher?

[hide=" Solution"]

We want $ x$ scoops to be more than half of $ 4\frac{1}{2}$, that is $ \frac{x}{3} > 2\frac{1}{4}$.

Therefore $ x > \frac{27}{4} \equal{} 6\frac{3}{4}$.

So $ x\equal{}7$.

[/hide]


number theory
greatest common divisor
number theory proposed

Prove that there is no integer $n \ge 2$ for which :
$\frac{3^n-2^n}{n}$ be an integer.

Obviously, $n$ must be odd. Let $p$ be the smallest prime divisor of $n$, then $p$ divides $3^n - 2^n$ and $p$ divides $3^{p - 1} - 2^{p - 1}$ (Fermat's little theorem), so $p$ divides $3^d - 2^d$ where $d$ is the greatest common divisor of $n$ and $p - 1$. However, since $p$ is the smallest prime divisor of $n$, $p - 1$ and $n$ must be coprime, so $d = 1$, and $p$ divides $3^1 - 2^1 = 1$, contradiction.

I'm pretty sure this was posted before but I couldn't find the right topic...


calculus
derivative
calculus computations

Okay - basically, what I'm trying to do is to learn the practical meaning behind the derivative.

For example, let's take a problem:

Suppose $C(r)$ is the total cost of paying off a car loan borrowed at an annual interest rate of r%. What is the practical meaning of $C'(r)?$

If the annual interest rate is $r$, $C'(r)$ shows how fast the costs increase/decrease if the annual interest rate is changed. Sorry, but in my opinion the derivative does not  make much sense here.

The typical example is:
Suppose $s(t)$ is the distance you have covered after time $t$. Then $s'(t)$ is your speed and $s''(t)$ acceleration.

Economists like to use the word "marginal". See if you can work that word into the description.


induction
trigonometry
inequalities
algebra proposed
algebra

Consider the sequences $(a_n)$ and $(b_n)$ defined by :
$a_0=\frac{\sqrt{2}}{2}$ and $a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^2}}$ for all $n \in \mathbb N$
$b_0=1$ and $b_{n+1}=\frac{\sqrt{1+b_{n}^2}-1}{b_n}$ for all $n \in \mathbb N$
Show that : $2^{n+2}a_n<\pi <2^{n+2}b_n$

It is easy to prove by induction that
$a_n = \sin(\frac {\pi} {2^{n+2}})$ and $b_n = \tan (\frac {\pi} {2^{n+2}})$

Using the well-known inequalities $\sin(x) < x$ and $\tan (x) > x$ for $0  < x < \frac {\pi} 2$ leads immediately to the desired conclusion.

Pierre.

Woow !!! You took too little time to solve it. Congratulations  :lol:


calculus
derivative
algebra
polynomial
algebra unsolved

Let $n \in \mathbb{N}^+, h \in \mathbb{R}$ and $f(x)$ be a polynomial of degree $n$ with real coefficients which does not have real roots.

Prove that

\[ F(x) = \sum^n_{k=1} h^k \cdot f^{(k)}(x)\]

also has no real roots. Note that $f^{(k)}(x)$ denotes the $k-th$ derivate of $f(x)$.

The degree $n$ hs to be even because of the real polynomial $f(x)$ hasn't real roots. Also the sign of $f(x)$ has to be constant, for ex. $f(x) > 0, \forall_{x\in R}$
Manifestly $F(x)$ has the same even degree $n$ than $f(x)$  and $F(-\infty)=f(-\infty)=f(+\infty)=F(+\infty)=+\infty$
So, there is $x_0\in R$ such that $ F(x_0) = \displaystyle  \min_{x\in R}(F(x)) \Longrightarrow  F'(x_0)=0 $
It's  easy to see that $F(x) =  \sum^n_{k=0} h^k \cdot f^{(k)}(x)=f(x)+h.F'(x)$, so $F(x_0)=f(x_0) > 0$, so $F(x) > 0 (\forall_{x\in R})$, so finally no real roots for $F(x)$.
 :cool: I prevent  the case $f(x) < 0$. In the real life I prefer the positive attitude ! :cool:


calculus
integration
function
derivative
complex analysis
complex analysis unsolved

Let $ z_0$ be an isolated point of a function $ f$ and suppose that $ f(z)\equal{}\frac{\phi(z)}{(z\minus{}z_0)^m}$, where $ m$ is a positive integer and $ \phi(z)$ is analytic and nonzero at $ z_0$. By applying the extended form of the Cauchy integral formula to the function $ \phi(z)$, show that $ \text{Res}_{z\equal{}z_0}\equal{}\frac{\phi^{(m\minus{}1)}(z_0)}{(m\minus{}1)!}$.

I do not see how to do this. Our book says:
Since there is a neighborhood $ |z\minus{}z_0|< \epsilon$ throughout which $ \phi(z)$ is analytic, the contour used in the extended Cauchy integral formula can be the positively oriented circle $ |z\minus{}z_0|< \frac{\epsilon}{2}$. How do I use this suggestion? I don't see that now. Thanks in advance.

Write down the Cauchy integral for the $ (m\minus{}1)$th derivative of $ \phi$. Compare it to the integral for the residue of $ f$.
What's that suggestion? It's just one possible choice for a curve that encloses $ z_0$ ad no other singularities. Many others would work as well.


Please don't double post, I answered on the other question.

[quote="williams"]show dat for all n; n(n+1)(n+2)(n+3)+1 is a perfect square.[/quote]

[hide="This is simple alternate"]$ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\plus{}1$
$ \equal{}n(n\plus{}3)(n\plus{}1)(n\plus{}2)\plus{}1$
$ \equal{}(n^2\plus{}3n)(n^2\plus{}3n\plus{}2)\plus{}1$
$ \equal{}A(A\plus{}2)\plus{}1$ by using $ A\equal{}(n^2\plus{}3n)$
$ \equal{}A^2\plus{}2A\plus{}1$
$ \equal{}(A\plus{}1)^2$
$ \equal{}(n^2\plus{}3n\plus{}1)^2$[/hide]


SEE ATTACHMENT Please find the value of x and y

Now isn't that simple ??
$ X = 4 ; Y = 2$
$4^5 - 2^5 = 992$
$4 - 2 = 2$
By the way , I think it would be better if you learnt $\LaTeX$

I wanted the solution and not any hit-and-trial method.

the solution is the harder thing try that!

Sorry, I don't have PPT so I didn't download it. But was the question:

$x^5-y^5=992$ Find all possible values of $(x,y)$

[quote="riddler"]x5-y5=992
x-y=2
i am sorry that i dont know how to use latex but that 5 is "raised to power 5"[/quote]

Just write the stuff, but put dollar signs in front of and at the end of whatever you write, it's that simple. And ^ means raised to.

x^5-y^5=992
x-y=2
try this
i want the solution

Hmmm..I'm sorry, but I can't find a solution to this by solving it, no matter how hard I tried, however, I believe there are more solutions to this equation than just $(4,2)$, however, I don't think I can get them all. Look for even numbers though, that do not divide by 3, because $992$ only divides by multiples of $2$, but not $3$. Yes, that would be just trial and error but I don't have a way of solving this.

I tried to break it down to:

$(x+y)(x-y)(x+y)(x-y)(x+y)=992$, and just substitute $2$ for $(x-y$ but I don't know if that's correct.  :(  :?  :help:

Your solution is wrong. It doesnt matter if you are unable because it is a very smart question and only sometimes does it click what to do.
I cannot give you the whole solution as I am unable to get latex but I can give you a hint
the equation2 should be succesively multiplied by itself to get the answer.

These problems are all the same and can be solved systematically as follows:

$(x-y)^5=x^5-y^5-5xy(x-y)((x-y)^2+3xy)+10x^2y^2(x-y)$

If you plug in the given values for $x-y$ and $x^5-y^5$, the above equation can be reduced to $(xy)^2+4(xy)-96=0$ from wich you get $xy=8 or xy=-12$

So now you have two systems of equations: $xy=8, x-y=2$ and $xy=-12, x-y=2$

The first of these has two solutions (4,2) and (-2,-4). The second doesn't have real solutions.

Ans: (4,2) and (-2,-4)

That was quite a bit of typing so I might have made a mistake or two :)

yup the answers are right bob, but I did the sum differently. :D

In a different way ? Could you show that method here ?  :? I worked out the sum using Bob's method......By the way , if you're not used to $\LaTeX$ , you could probably attach another powerpoint file... :)

sorry, i havent got time to do sums differently for you but i cna tell you that you have to sucesively multiply x-y by itself until you get x^5-y^5.

[quote="karthik_k"]But if if you multiply $(x-y)$ succesively by itself , you will not land up with $x^5 - y^5$.You'll end up with $(x-y)^5$ and then Bob's solution follows..
By the way , Bob , I think only the solution (4 , 2 ) is right for this particlular question.
$(-4)^5 - (-2)^5 \neq 992$
and
$(-4) - (-2) \neq 2$[/quote]

The second ordered pair in my answer was (-2,-4), not (-4,-2) - I think you got those mixed up.

Oops :oops:

But even $(-2 , -4 )$ satisfies only the equation $xy = 8 ; x -y = 2$
and not the actual question because -
$(-2)^2 - (-4)^2 \neq 992$

:oops_sign: That's my mistake.....
By the way , do you mind posing the complex solutions as well to this problem ?


Find a 7-digit number $ N$ satisfying these conditions:
a) $ N$ is a multiple of 7,
b) five of $ N$'s digits are 3's and the other two are 4 and 6,
c) the first digit is a 4.

N must be 4------, where 5 of the blanks are 3s. The remainder upon dividing 4333333 by 7 is 4, so you have to add three to get [b]4333336[/b].


Two heave right circular rollers of diameters D and d , respectively , rest on a rough horizontal surface as shown in the diagram. The large roller has a string wound around it to which a horizontal force P can be applied as shown. Assuming that the coefficient of friction is [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/a/4/3a4e56595df1d02f21e5c3ff7b0e9d80921858ff.gif[/img] has the same value for all the surfaces of contact , determine the necessary condition under which the large roller can be pulled over the small one.  :?:

Ans : [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/9/a/89a9ae539d55e40bafeb118b8d1dd5218af1fb2c.gif[/img]

[quote="rt_08"]hey i have done with ey frame-oi
that becomes easy[/quote]

Please explain a bit more  :|

That's still not a very enlightening response.  But here's a dumb solution (by "dumb," I mean you can get the answer by doing as little work as possible, which might actually be smart).

First of all, draw a right triangle of side lengths $ R \plus{} r$, $ R \minus{} r$, and $ 2\sqrt{Rr}$, two of whose vertices are the centers of each of the cylinders ($ R \equal{} D/2$ and $ r \equal{} d/2$).  Let $ \theta$ be the smallest angle (sorry I don't have a diagram).  Call $ F$ the normal force between the big cylinder and the small cylinder, $ W_1$ the weight of the big cylinder, and $ W_2$ the weight of the small cylinder.  The problem requires making the assumptions $ W_1 \gg W_2$ and $ R \gg r$.

Now I'm going to be cheap by not actually determining $ F$ by doing annoying and stupid geometry, but assuming that it has a factor of $ W_1$, or at least is some expression containing $ W_1$ (which is clear).  The only force to the right, on the small cylinder, is $ F\cos\theta$ (from the normal force), and the two forces to the left are $ \mu F\sin\theta$ (from friction between the cylinders) and $ (W_2 \plus{} F\sin\theta)\mu$ (from friction between the small cylinder and the ground).  Hence we have the condition
\[ \mu W_2 \plus{} 2\mu F\sin\theta \geq F\cos\theta.
\]
Assuming $ W_2$ is negligible compared to $ F$, we have
\[ \mu \geq \frac{1}{2}\cot\theta \equal{} \frac{\sqrt{Rr}}{R\minus{}r} \approx \sqrt{\frac{r}{R}} \equal{} \sqrt{\frac{d}{D}}.
\]

Thank you genericme ,  :)  however don't you think that would be a big assumption ? :?:

Ah, well, if you mean the assumption that $ W_1 \gg W_2$ and $ R \gg r$, I don't see how to get the answer otherwise.

If you mean the assumption that $ F \gg W_2$, you can determine that $ F \equal{} \frac {W_1}{\sin\theta \minus{} \mu\cos\theta}$ (and also $ F \equal{} \frac {P}{\cos\theta \plus{} \mu\sin\theta}$) by drawing a free-body diagram.  So indeed, since $ W_1 \gg W_2$, the assumption is justified.

hey i think the assumption is all right
and sorry i did'nt gave a very cheerful response

Can the moderator , Pardesi look into it? Someone told me that he is a wonderful  mathematician .....Thanks for the help. Waiting for some more replies . :)

Sushmiz, what was the answer you gave in the problem? It seems to me that the image representing the answer isn't on the server anymore (right click->show picture doesn't help  :) ). I'd like to analyze this problem further, but an answer would be extremely helpful...

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/9/a/89a9ae539d55e40bafeb118b8d1dd5218af1fb2c.gif[/img]

This is the given answer i,e
 coefficient of friction is greater than equal to the square root of the ratio of d/D .

[quote="hsiljak"]Thank you: so genericme's answer is OK- it's just one thing that bothered me there: D was taken as much bigger than d, so basically, if we want to be strict, we would have to take d/D=0... But nevertheless, the reasoning used in his/her solution seems ok.[/quote]

the idea when you say something is much bigger than another, or equivalently that some $ \lambda$ (here a certain ratio) is much smaller than one, is to take a taylor expansion in $ \lambda$. Usually we only keep the lowest nonzero term although in some contexts you need to go further.

Yes, just like in formulae for mathematical pendulum, or heat expansion... It's just that I had a more 'engineering' approach to it  :lol:


function
real analysis
real analysis unsolved

Let $f: [0,1] \to \mathbb{R}$ be a function and suppose that the set $\left\{(x,f(x))\left|\,x \in [0,1]\right\}\right. \subseteq \mathbb{R}^2$ is compact.
Prove that $f$ is continuous.

Take $x_{0} \in [0,1]$, and any sequence $(x_{n}) \rightarrow x_{0}$. Then you can find a convergent subsequence $(y_{n},f(y_{n}))$ of $(x_{n},f(x_{n}))$ (by compacity). This subsequence has limit $(x_{0},a)$, where $a$ can only be $f(x_{0})$ (by closedness). Whence $(f(x_{n}))$ converges to $f(x_{0})$.

To have a name for it, let $\Gamma= \{(x,f(x))|\,x \in [0,1]\}.$ This is the graph of $f,$ which we are presuming to be compact.

Now suppose $x_k\to x$ Consider the sequence in $\Gamma$ whose elements are $(x_n,f(x_n)).$ Since $\Gamma$ is compact, this has a convergent subsequence:

$(x_{n_k},f(x_{n_k}))\to (x,L).$

But since $\Gamma$ is compact, we must have $(x,L)\in\Gamma,$ and that implies that $L=f(x).$

Let's sharpen the same argument. Suppose $f$ is discontinuous at $x.$ Then $\exists\,\epsilon>0$ and $\exists$ a sequence $x_n$ such that $x_n\to x$ but $|f(x_n)-f(x)|\ge\epsilon.$

From the above argument, we must have a subsequence of $(x_{n_k},f(x_{n_k}))$ which tends to $(x,f(x)),$ which can't. So $f$ cannot be discontinuous.

does compact also imply closed and bounded and the other way around?
so if A is compact / closed and bounded, can you say A = A complement?

[quote="Arne"]How could we possibly have A = "A complement"?[/quote]
I guess he means A = A', set of all accumulation points. Answer = no

Compact <=> closed & bounded in euclidean spaces.

so basically if you have a compact set, then it is closed and bounded
so if X and Y are compact then X U Y is also compact?
how do you show that?

Take any open cover of $X \cup Y$. This is an open cover of $X$ which has a finite subcover (since $X$ is compact). The same holds for $Y$. So put these two finite subcovers together and you've got a finite subcover for $X \cup Y$, as desired.


induction
combinatorics proposed
combinatorics

Suppose $u_{1}, u_{2}, u_{3}, \ldots, u_{n}$ are [url=http://en.wikipedia.org/wiki/Automata_theory#Formal_description]words[/url] such that all $n$ cyclic concatenations $u_{1}u_{2}\ldots u_{n},\ u_{2}u_{3}\ldots u_{n}u_{1},\ \ldots,\ u_{n}u_{1}u_{2}\ldots u_{n-1}$ are equal. Prove that there exists a word $v$ such that all $u_{i}$ are concatenations of $v$, i.e. $\exists n_{i}\in \{0,1,2,\ldots\}\ u_{i}= v^{n_{i}}$.

For example, let $n=2, u_{1}=010101, u_{2}=0101$, then we have $u_{1}u_{2}= u_{2}u_{1}= 0101010101$, and we can find that $v=01, u_{1}= v^{3}, u_{2}= v^{2}$.

In the case of $n=2$, it's very easy.
Suppose that when $n\leq m$, the proposition is true.
Then when $n=m+1$, Let $u_{n+2}=u_{1}u_{2}$.
Then by assumption, $u_{3}, u_{4},....,u_{n+2}$ has $v$ s.t $u_{i}$s are concatenations of $v$. Let $v'$ be the smallest word among the all possible $v$s.
Assume that $u_{1}$ and $u_{2}$ isn't concatenations of $v'$, then $x$, $y$ exist satisfying $v'=xy=yx$. By assumption, $x,y$ has $v''$ s.t $x,y$ are concatenations of $v''$. Therefore, $v'$ is a concatenation of $v''$. It's contradiction to $v'$ is the smallest among the $v$s. Therefore proposition is true when $n=m+1$. Therefore by induction, the proposition is true. QED

that's a nice proof.

here's another proof using bi-infinite sequences.

Some notation:
If $y$ is a word, $y^\infty$ denotes a bi-infinite sequence repeating $y$ starting with 0-th coordinate. For example, if $y$ is 011, then $y^\infty$ is ...011011.011011011... $\in \{0,1\}^\mathbb{Z}$ where the first letter after the central dot is the 0-th term of the sequence. If $\mathcal{A}$ is an alphabet, [url=http://en.wikipedia.org/wiki/Shift_operator]the shift map[/url] $\sigma : \mathcal{A}^\mathbb{Z}\to \mathcal{A}^\mathbb{Z}$ is defined by $\sigma( \ldots lmn.opqr \ldots ) = (\ldots mno. pqr \ldots)$.

Proof:
Let $s_{i}= \sum_{k=1}^{i}|u_{k}|$ where $|u_{k}|$ denotes the length of the word $u_{k}$.
Let $w = u_{1}u_{2}\ldots u_{n}= u_{2}u_{3}\ldots u_{1}= \ldots$.
We have $w^\infty = (u_{i+1}u_{i+2}\ldots u_{i})^\infty = \sigma^{s_{i}}((u_{1}u_{2}\ldots u_{n})^\infty) = \sigma^{s_{i}}(w^\infty)$ for all i. Let $d = \gcd_{i=1}^{n}(s_{i})$. Then $w^\infty = \sigma^{d}(w^\infty)$  . it means that $w^\infty = v^\infty$ for some word $v$ with $|v| = d$. Since $d$ divides $|u_{i}| = s_{i}-s_{i-1}$, all of $u_{i}$ are concatenations of $v$.

This problem is in the first chapter from [url=http://www.math.washington.edu/SymbolicDynamics/]An Introduction to Symbolic Dynamics and Coding[/url]