geometry
circumcircle
IMO Shortlist
homothety
triangle -incenter

Given a triangle $ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle.

[i]Proposed by Dimitris Kontogiannis, Greece[/i]

Let $C'$ be a symmetrical point to $C$ with respect to $I$, $A'$ lie on line $CA$ such that $CA' = 2AB$ and $B'$ lie on line $CB$ such that $CB' = 2AB$.

$CA + CB = 3AB$ imply $CD = AB$, $BB' = DA$ and $AA' = EB$. We see, that $C'B' = C'A' = 2IE$ and $C'K = C'L = CE = AB$. Triangle $C'B'B$ is congruent to $ADK$ and $C'A'A$ to $BLE$. Thus $C'B = AK$ and $C'A = LB$. So triangle $C'BK$ is congurent to $ABK$ and $C'AL$ to $ABL$. Thus $\angle KC'B = \angle KAB$ and $\angle LC'A = \angle LBA$. It means that $AKBC'$ and $ALBC'$ are cyclic quadrilaterals thus $ALKB$ is cyclic quadrilateral.

[quote="xirti"]Do you infer that $C'$ is the excircle of $ABC$??

Edit: Yes. I see why $C'$ is the excircle of $ABC$. From the similar triangles $AIE$ and $AC'A'$ we find that $C'A'\perp AC$. Similarly for $C'B'\perp BC$. But, $A'$ and $B'$ are the poits of tangency of the excircle. ($CA'=CB'=\frac p 2$)
Sorry for my useless post.[/quote]

i think here we can make it better.
from CA' = 2AB = $\frac{AB + BC + CA}{2}$, then C' is the external center of triangle ABC. thus I, A, B, C' are lie on circle with diameter IC'.
another side, we have K, L, C' are the symmetrycal points of D, E, C then C'K, C'L are the tangents of (I), then K, L also lie on circle with diameter IC'.

I think Michal Marcinkowski has some typos on his notatations or my figure is wrong....... :?

This problem is certainly on the ISL from 2005. Question G1 I believe.

[color=darkred][b][u]PP[/u].[/b] Let $ABC$ be an acute  triangle. Its incircle $w=C(I,r)$ touches it at the points $D\in BC$ , $E\in CA$ , $F\in AB$ . Denote:
$\blacktriangleright$ the centroid $G$ of the triangle $ABC$ ;
$\blacktriangleright$ the [b]A[/b]-exincircle $w_a=C(I_a,r_a)$ , the [b]B[/b]-exincircle $w_b=C(I_b,r_b)$ , the [b]C[/b]-exincircle $w_c=C(I_c,r_c)$ ;
$\blacktriangleright$ the [u][b]Nagel[/b]'s point[/u] $N\in AD_1\cap BE_1\cap CF_1$ ; $n_a=AD_1$ , $n_b=BE_1$ , $n_c=CF_1$ ;
$\blacktriangleright$ the points $D_1\in BC\cap w_a$, $E_1\in CA\cap w_b$ , $F_1\in AB\cap w_c$ ;
$\blacktriangleright$ the reflections $D'$ , $E'$ , $F'$ of the points $D$ , $E$ , $F$ with respect to the point $I$ .

[b][u]Lemma 1.[/u][/b] $IG\perp BC\Longleftrightarrow b=c\ \ \vee\ \ b+c=3a$ .[/color]

[color=darkblue][u]Proof.[/u] $IG\perp BC$ $\Longleftrightarrow$ $IB^2-IC^2=GB^2-GC^2$ $\Longleftrightarrow$ $\frac{ac(p-b)}{p}-\frac{ab(p-c)}{p}=$

$\frac 19\left\{\left[2\left(a^2+c^2\right)-b^2\right]-\left[2\left(a^2+b^2\right)-c^2\right]\right\}$ $\Longleftrightarrow$ $3a(c-b)=c^2-b^2$ $\Longleftrightarrow$ $b=c\ \ \vee\ \ b+c=3a$ .[/color]

[color=darkblue][u]Remark.[/u] $N\in (IG)$ and $GN=2\cdot GI$ . Therefore, $IN\perp BC$ $\Longleftrightarrow$ $b+c=3a$ .[/color]

[color=darkred][b][u]Lemma 2.[/u][/b] $\boxed {a n^2_a+a(p-b)(p-c)=p(b^2+c^2)-bc(b+c)}\ \ (1)$ ;

$\frac{NA}{a}=\frac{ND_1}{p-a}=$ $\frac{n_a}{p}\ \ (2)$ ; $D'\in AD_1$ and $\frac{D'A}{p-a}=\frac{D'D_1}{a}=\frac{n_a}{p}\ \ (3)$ a.s.o.[/color]

[color=darkblue][u]Proof.[/u] Apply the [u][b]Stewart[/b]'s relation[/u] for the [u][b]Nagel[/b]-ray[/u] $[AD_1$ in the triangle $ABC$ and obtain the relation $(1)$ .

Apply the [u][b]van Aubel[/b]'s relation[/u] $\frac{NA}{ND_1}=\frac{F_1A}{F_1B}+\frac{E_1A}{E_1C}$ and obtain the relation $(2)$ . $D'\in AD_1\cap BC$ $\Longleftrightarrow$

$\frac{D_1D'}{D_1A}=\frac{2r}{h_a}$ $\Longleftrightarrow$ $\frac{D_1D'}{D_1A}=\frac ap$ $\Longleftrightarrow$ $\frac{D_1D'}{a}=\frac{D_1A}{p}=$ $\frac{D'A}{p-a}=\frac{n_a}{p}$, i.e. the relations $(3)$ .

[u]Remark.[/u] $\frac{AD'}{p-a}=\frac{ND_1}{p-a}=\boxed {\frac{ND'}{2a-p}=\frac{NA}{a}=\frac{n_a}{p}}\ \ (4)$.[/color]

[color=darkred][b][u]An equivalent and complete enunciation of the proposed problem[/u].[/b] Let $ABC$ be an acute triangle for which $b\ne c$ . Then

$B,C,E',F'$ are cyclically $\Longleftrightarrow$ $b+c=3a$ $\Longleftrightarrow$ $IN\perp BC$ $\Longleftrightarrow$ $N\equiv D'$ and in this case $I,B,C,E',F'$ belong to

the circle $C(M,MI)$ , where the point $M$ is the second intersection between the line $AI$ and the circumcircle of the triangle $ABC$ .[/color]

[color=darkblue][b][u]Proof (indication)[/u].[/b] The points $B,C,E',F'$ are cyclically $\Longleftrightarrow$ $NE'\cdot NB=NF'\cdot NC$ a.s.o.

[u]Example.[/u] $a=3$, $b=4$, $c=5$.[/color]

[color=darkred][u]Remark.[/u] $N\in w$ $\Longleftrightarrow$ $(b+c-3a)(c+a-3b)(a+b-3c)=0$ $\Longleftrightarrow$$N\in \{D',E',F'\}$ $\Longleftrightarrow$ $p^2+4r^2=16Rr$ .[/color]

:-)

We draw a circumcircle of the triangle $ABC$. Now let $BI$ meet the circumcircle at the point $M$ (for $M$ is different from $B$), so we have $MC=MA$ since $\angle MAC = \angle MCA= \angle MBC = \angle MBA$. Thus we have $MA=MI=MC$, since $\angle MIA= \angle IAB + \angle IBA=\angle IAM= \angle IAC \angle CAM$. Therefore we have $\angle MIA = \angle MAI$  similarly we have $\angle MIC = \angle MCI$. So resulting we have $MI=MC=MA$. Now lets draw a parallel line to the line $DB$ from point $M$ and let the line meet $IL$ at $N$ so $DB||MN$. Then we obtain a congrunecy between the triangles $MNL$ and $CPM$ ( $P$ is a midpoint of the side $AC$) these facts imply to $ML=MC$. Similarly we get $MK=MA$. Then we can easily say that the points $A,K,I,L,C$ all lie in a circle with centre $M$.

davron

:arrow: See the my first reply where I offered a equivalent and complete enunciation of the proposed problem ! :?

my solution is a "little" ugly because i'm not too good at geometry as to define "miraculous" points, however i consider it a little more natural than any of the above  :P i think it is normal to consider most of the times one's solution as the most natural. it is as follows...

let F be the tangency point of the incircle with AB. a simple angle chasing shows that $\angle EFK=\angle AEK=\frac{\theta}{2}$ and $\angle FEK=\angle AFK=\frac{\beta}{2}$. since KL=DE and they are parallel it follows that CI is the perpendicular bisector of KL, which meets the perpendicular bisector of AB at the midpoint of the minor arc AB. but it is a well-known fact that this point is the circumcenter of AIB, thus by symmetry it is enough to prove that AKIB is cyclic.

we'll have to prove that $\angle IKM=\frac{\beta}{2}$, or it is equivalent to prove that $\angle AMB=90-\frac{\beta}{2}=\angle FDB$, so we have to prove that $AK$ is parallel to $DF$, which is equivalent to prove that $\angle KAF=90-\frac{\beta}{2}$.

let $\angle KAC=\frac{\alpha-\rho}{2}$ and $\angle KAF=\frac{\alpha+\rho}{2}$. we want to prove that $\theta=\rho$.

by trig form of ceva's theorem in triangle AEF we have that
$\frac{\sin^2(\frac{\beta}{2})\cdot \sin (\frac{\alpha-\rho}{2})}{\sin^2(\frac{\theta}{2})\cdot \sin (\frac{\alpha+\rho}{2})}=1$.

a little algebra, the help of briggs formulas, and the condition help us proving that
$\frac{\sin^2(\frac{\beta}{2})\cdot \sin (\frac{\alpha-\theta}{2})}{\sin^2(\frac{\theta}{2})\cdot \sin (\frac{\alpha+\theta}{2})}=1$.

so we have that
$\frac{\sin (\frac{\alpha-\rho}{2})}{\sin (\frac{\alpha+\rho}{2})}=\frac{\sin (\frac{\alpha-\theta}{2})}{\sin (\frac{\alpha+\theta}{2})}$
and it is easy to conclude that $\theta=\rho$ as desired.  :D

Check this out [b][color=blue]campos[/color][/b]!!! I bet you'll like this :D :P

Drawing a good figure I conjectured that $K$ and $L$ lie on the circumcircle of triangle $ABI$. I'll just prove that $L$ lies on this circle, and the result will follow analogously for $K$.

The first proof uses some trigonometry. We want to prove that quadrilateral $ABLI$ is cyclic, and since $\angle ABI=B/2$, it suffices to show that $\angle ALI=B/2$.

Observe that in triangle $AEL$ we have \[\tan ALE=\frac{AE}{EL}=\frac{s-a}{2r}\] where $s$ is the semiperimeter of the triangle and $r$ is the inradius. Now we use the condition $a+b=3c$ for finding r: \[r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{(s-a)(s-b)c}{2c}}=\sqrt{\frac{(s-a)(s-b)}{2}}\] whence \[\tan ALE=\frac{s-a}{2}\sqrt{\frac{2}{(s-a)(s-b)}}=\sqrt{\frac{2(s-a)^{2}}{4(s-a)(s-b)}}=\sqrt{\frac{s-a}{2(s-b)}}\] But we know that \[\tan \frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}=\sqrt{\frac{c(s-a)}{2c(s-b)}}=\sqrt{\frac{s-a}{2(s-b)}}\] Thus $\tan ALI=\tan B/2$, that is, $\angle ALI=B/2$, and we are done.

Now I continue with a second synthetic proof I found. Let $P$ be a point on the segment $AC$ such that $AB=AP$. Note that $EP=c-(s-a)=c+a-s=s-b$. We claim that triangles $AEI$ and $LEP$ are similar. Indeed, we have $\angle AEI = \angle LEP= \pi/2$ and \[\frac{AE}{LE}=\frac{EI}{EX}\Leftrightarrow \frac{s-a}{2r}=\frac{r}{s-b}\Leftrightarrow r=\sqrt{\frac{(s-a)(s-b)}{2}}\] Hence $\angle APL=\pi/2-A/2$. But since triangle $ABP$ is isosceles, $\angle APB= \pi/2-A/2$; so $\angle APB=\angle APL$ and therefore $B$, $L$ and $P$ are collinear. Now we have $\angle BAI=\angle ILP=A/2$, that is, quadrilateral $ABLI$ is cyclic, and we are done.  ;)

nice, gabriel! congrats, and i must admit it is much more natural and nice than mine!  :thumbup:

Sketch-proof of method using polars.

Let $F$ be the point of tangency with the incircle and $AB$. It can be shown that $EF$ is bisected by $KL$ (a short sine-rule bash). But the midpoint of $EF$ is the inverse of $A$ wrt the incircle, and similarly with $DF$. Since lines invert to circles, with $K$,$L$ self inverse, we have $A$,$B$,$K$,$L$ concyclic as required. :)

Here is a sketch of a solution:

Let $BL\cap AC=E''$, $AK\cap BC=D''$, line $l$ touches the incircle at $L$, $l\cap AB=F'$ and $l\cap BC=D'$. $AK\cap BL=X$.

$l//AC$ since $\angle AEI=\angle F'LI=90$.

The $B$-excircle of $ABC$ touches $AC$ at $E''$. Consider a homothety center $B$ that maps $l$ to $AC$.

$\frac{BL}{BE''}=1-\frac{br}{A}=\frac{d(B,AC)-2r}{d(B,AC)}$.

Use mass points to determine $\frac{E''X}{XB}$ in terms of the sides. Use: $E''C=s_{ABC}-a$, $D''C=s_{ABC}-b$

Use stewart's theorem to determine the length of $BE''$ and $D''A$.

Compute the product, $XK*XA-XL*XB$. [use $b+a=3c$]

If this turns out to be $0$, then $ADLB$ is cyclic by the converse of the power of a point theorem.

Note: Perhaps the excircle can be used with angle information. I will look into that.

Let $A'$ and $B'$ be the feet of the perpendiculars from $C$ to $AI$ and $BI$, respectively. Note $CA'DIEB'$ is a cyclic hexagon and $CI$ is a diameter. We have $\angle A'B'I=\angle A'CI=90-\angle CIA'=\frac{B}2=\angle IBA$. So, $A'B'$ is parallel to $AB$.

We have $\angle CA'B'=CIB'=90-\frac{A}2=\frac{C}2+\frac{B}2=\angle ECI+\angle ICA'=\angle ECA'=\angle EB'A'$. So, $EB'$ is parallel to $CA'$. So, since $EB'CA'$ is cyclic, it is an iscosceles trapezoid. So, $CE=A'B'$.

Because $AC+BC=3\cdot AB$, we can find that $CE=\frac{AC+BC-AB}2=AB$. So, $A'B'=AB$. So, $A'B'AB$ is a parallelogram. Since $I$ is the intersection of the diagonals, we find $A'I=AI$ and $B'I=BI$.

We know $B'EDA'$ is cyclic. Applying a $-1$ homothety around $I$ brings this quadrilateral to $BLKA$. So, $BLKA$ is cyclic.

Finally got to this after procrastinating for months  :P And not gonna lie, I'm surprised I got this done without trigonometry  :-D

[i]Solution[/i] Let $K'$ denote the ex-touch point of $A-$excircle on $BC$. Define $L'$ similarly. Let $X$ be the reflection of $B$ across $B$ and $Y$ be the reflection of $A$ across $B$. By a well-known lemma, $A-K-K'$ and $B-L-L'$ are collinear.

By the given condition we have
    \begin{align*}
        a+b     = 3c \iff s  = 2c
    \end{align*}
It is well known fact that length of tangent to $A-$excircle from $A$ is always $s$. Therefore $Y$ and $X$ both lie on the $A$ and $C$ excircles respectively.

As $\triangle AK'B$ and $\triangle ABL'$ are isosceles, we get
    \begin{align*}
        \angle KAB = 90^\circ - \frac{B}{2} \\
        \angle ABL = 90^\circ - \frac{A}{2}
    \end{align*}

Observe that
    \begin{align*}
        \angle KLL' & = \angle KLE + \angle ELK \\
                    & = \angle KDE + 90^\circ - \angle AL'K \\
                    & = \angle IDE + 90^\circ - \angle ABL' \\
                    & = \frac{C}{2} + 90^\circ - \left( 90^\circ - \frac{A}{2} \right) \\
                    & = \frac{A+C}{2} \\
                    & = 90^\circ - \frac{B}{2} \\
                    & = \angle KAB
    \end{align*}
which finishes the solution. $\blacksquare$

[b]Remark. [/b] $I$ lies on the circle $\odot(ABKL)$. This can be proven by showing $\angle K'AB = \angle ABI$. I didn't claim but I rather saw it on the thread. Proof came in swiftly. Another thing I read on thread by Ineis1 #73, $AK$ and $BL$ meet on the incircle! The proof is again by simple angle chasing.

Beautiful problem!

Let $AK \cap BC=D_1$ and $BL\cap CA=E_1$
It is well known that $D_1$ and $E_1$ are extouch points. Also by trivial angle chase $DE \vert \vert KL$
$$BC + CA=3AB \implies s=2c \implies s-c=c \implies AE_1=BD_1=AB$$

Note that $\angle BAD_1=\angle BD_1A= \left(\frac{C+A}{2}\right)$ and $\angle ABE_1=\angle AE_1B= \left(\frac{B+C}{2}\right)$
Hence $AK$ and $BL$ meet on the incircle.
Now it suffices to show that $CI$ is the perpendicular bisector of $KL$, but it is true since $KL \vert \vert DE$

Notice that $CD=CE=AB$. Let $AK$ hit $BC$ at $X$. Let $BL$ hit $AC$ at $Y$. Then $X$ and $Y$ are the contact points of the $A$ and $B$-excircles with $BC$ and $AC$. Then, notice that $\triangle ABX$ is isosceles and $\triangle ABY$ is isosceles. It suffices to show that $\angle YLK=\angle XAB=\frac{180^{\circ}-\angle B}{2}$. WLOG, suppose that $AC\le BC$. Then if $DE$ intersects $AB$ at some point $M$ to the left of $AB$ and $BY$ intersects $DE$ at $P$ then we have that
\begin{align*}
\angle YLK&=\angle YPE \\
&= \angle EMA+\angle YBA\\
&=\angle CDE-\angle B+\frac{180^{\circ}-\angle A}{2} \\
&=\frac{180^{\circ}-\angle C}{2}-\angle B+\frac{180^{\circ}-\angle A}{2} \\
&=\frac{180^{\circ}-\angle B}{2}
\end{align*}
so we are done. $\blacksquare$

:love:  :love:
-----
We use barycentric coordinates, with $ABC$ as the reference triangle. We have the following points:
\begin{align*}
I &= (3a:3b:a+b) \\
D &= (0:a+b:2a-b) \\
E &= (a+b:0:2b-a) \\
\end{align*}
Now, we need to find $K$. Since $I$ is the midpoint of $DK$, we know that $K=2I-D$. Some computation gives $$K =\left(9a^2 : 5ab - 2a^2 - 2b^2 : ab - a^2+2b^2\right).$$ Similarly, we have $$L = \left(5ab - 2a^2-2b^2 : 9b^2 : ab + 2a^2 - b^2\right).$$ Next, consider $(ABK)$. If we can show that $L$ also lies on this circle, we are done. First, we evidently have $u=v=0$. We need $w$, and this is just more computation:
$$(b-2a)\left(-a^2+2ab+a^2+ab\right)-9ab^2=-w(6a+6b) \rightarrow 6ab(a+b)=w(6a+6b) \rightarrow w=ab.$$ Clearly, $L$ also lies on this circle, which finishes the problem. $\blacksquare$

Let $M$ the midpoint of the smaller arc $\widehat{AB}$. It is well-known that $MA=MB=MI$. From Ptomely's theorem on quadrilateral $CAMB$ we have:
\begin{align*}
 MA \cdot BC + MB \cdot CA = AB \cdot CM \\
MI (BC+CA) = AB \cdot CM \\
3MI = CM
\end{align*}
Define $T$ to be a reflection of $M$ over $I$, then from before the established result, we have $CT=TI=IM$. Note that $T$ is the centre of $\odot(IDCE)$, which is cyclic deltoid. Therefore by symmetry, we have that $\triangle DTI$ and $\triangle ITE$ are congruent isosceles triangles.

The key observation is that $\angle DIT = \angle TIE = \angle LIM = \angle KIM$ and $TI=IM$ combined with $ID =IE =IK =IL$. This implies that $\triangle DTI =\triangle ETI =\triangle LMI = \triangle KMI$, which gives us that $LM=MI=MK=MA=MB$. This means that points $A,L,I,K,M$ all lie on the same circle centred at the point $M$ with radius $MI$. This solves the problem.

Solved with support from [b]Taco12[/b].

We use barycentric coordinates with reference triangle $ABC$. It is well known that
$$I=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)=\left(\frac{3a}{4(a+b)},\frac{3b}{4(a+b)},\frac{1}{4}\right),$$
$$D=\left(0,\frac{a+b-c}{2a},\frac{a-b+c}{2a}\right)=\left(0,\frac{a+b}{3a},\frac{2a-b}{3a}\right),$$
$$E=\left(\frac{a+b-c}{2b},0,\frac{-a+b+c}{2b}\right)=\left(\frac{a+b}{3b},0,\frac{2b-a}{3b}\right).$$
$K$, being the reflection of $D$ over $I$, is equivalent to $$2I-D=\left(\frac{3a}{2(a+b)},\frac{3b}{2(a+b)}-\frac{a+b}{3a},\frac{1}{2}-\frac{2a-b}{3a}\right).$$
$L$, being the reflection of $E$ over $I$, is equivalent to $$2I-E=\left(\frac{3a}{2(a+b)}-\frac{a+b}{3b},\frac{3b}{2(a+b)},\frac{1}{2}-\frac{2b-a}{3b}\right).$$
We can unhomogenize $K$ to be $$\left(9a^2 : 9ab-2(a+b)^2 : (a+b)(2b-a)\right).$$
We can unhomogenize $L$ to be $$\left(9ab-2(a+b)^2 : 9b^2 : (a+b)(2a-b)\right).$$
The equation of a circle is $$-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0,$$ and since $A$ and $B$ are on it, we have that
$$0+(u(1)+0+0)(1)=0\Rightarrow u=0,$$
$$0+(0+v(1)+0)(1)=0\Rightarrow v=0.$$
Therefore, if we can show that both $K$ and $L$ satisfy the equation with the same $w$, we are done.
We can see that $$w=\frac{a^2yz+b^2zx+c^2xy}{z(x+y+z)},$$ so all that's left is verifying that $w$ is indeed equal for both points. This turns out to be true, and I will include the computation below. QED.

$$\frac{a^2(9ab-2(a+b)^2)(a+b)(2b-a)+b^2(9a^2)(a+b)(2b-a)+c^2(9a^2)(9ab-2(a+b)^2)}{(a+b)(2b-a)\left(9a^2+9ab-2(a+b)^2+(a+b)(2b-a)\right)}=\frac{a^2(9b^2)(a+b)(2a-b)+b^2(9ab-2(a+b)^2)(a+b)(2a-b)+c^2(9ab-2(a+b)^2)(9b^2)}{(a+b)(2a-b)\left(9ab-2(a+b)^2+9b^2+(a+b)(2a-b)\right)}$$
$$\frac{a^2(7b-2a)(a+b)^2(2b-a)+(a+b)^2(a^2)(9ab-2(a+b)^2)}{(a+b)(2b-a)(6a)(a+b)}=\frac{b^2(7a-2b)(a+b)^2(2a-b)+(a+b)^2(9ab-2(a+b)^2)(b^2)}{(a+b)(2a-b)(6b)(a+b)}$$
$$\frac{a^2(a+b)^2(6b)(2b-a)}{a(2b-a)}=\frac{b^2(a+b)^2(6a)(2a-b)}{b(2a-b)}$$
$$\frac{a^2b(2b-a)}{a(2b-a)}=\frac{b^2a(2a-b)}{b(2a-b)}$$
$$ab=ab$$

Replace $K$ and $L$ with $D'$ and $E'$. Add $F$, the $C$-intouch point, as well as its antipode $F'$ wrt the incircle. Also add the midpoints $M,N$ of $\overline{BC},\overline{AC}$ resp. and let $P,Q$ be the $A$-extouch and $B$-extouch points. Also let $s,a,b,c$ be defined as usual. It is easy to see that $c=AB=AQ=BP$. Furthermore, $\overline{MN}$ is tangent to the incircle by Pitot, in fact at $F'$. Finally, it is well-known that $D'$ lies on $\overline{AP}$ and $E'$ lies on $\overline{BQ}$. We now have the following key claim.

[b][color=#00f]Proof:[/color][/b] $F'$ lies on $\overline{AP}$ and $\overline{BQ}$.
[color=#00f]Claim:[/color] Symmetry: $D$ and $F$ are symmetric with respect to $B$, which implies that $D'$ and $F'$ are, as are $A$ and $P$, so $F'$ lies on $\overline{AP}$ follows from the fact that $D'$ lies on it. Similarly, $F'$ lies on $\overline{BQ}$. [hide=Note]It's also possible to prove this by proving that $QN/NF'=QA/AB$ through direct usage of lengths[/hide]

To finish we employ an angle chase:
$$180^\circ-\angle AD'E'=\angle F'D'E'=\angle FDE=\frac{\angle FIE}{2}=90^\circ-\frac{\angle FAE}{2}=\angle ABE',$$
where the last line follows from $AB=AQ$, implying the desired result. $\blacksquare$

The length condition is just $s=2c$. Let $D'$ be the excircle touchpoint to $BC$ (or the reflection of $D$ across the midpoint of $BC$). It is well known that $A,K,D'$ are collinear. However, $$BD'=s-c=c,$$ so $\angle BAD'=\angle BD'A$. Let $\angle ABC=2\beta$. Then, $\angle BAD'=\angle BD'A=90-\beta$. However, $\angle BID=90-\beta$ as well, so $AKIB$ is cyclic. Similarly, $AILB$ is cyclic. Hence, $ABLK$ is cyclic

Let $D'$ be the $A$-excentral touchpoint, thus the length condition implies $AB=BD'$; also, $A, K, D'$ are collinear.

The rest is just an angle chase. Observe
\begin{align*}
\angle DKL &= \angle D'KD + \angle LKD \\
&= 90^\circ - \angle KD'D + 90^\circ - \angle ELD \\
&= 90^\circ - (90^\circ - \frac B2) + 90^\circ - 90^\circ - \frac C2 \\
&= \frac B2 + \frac C2
\end{align*}
On the other hand, $\angle BAD' = 90^\circ - \frac B2$, implying $\angle ABL = 90^\circ - \frac A2$ symmetrically (the length condition still applies.) This is enough to prove the concyclicity.

Let the incircle touch side $AB$ at point $F,$ and let $M$ be the reflection of $F$ with respect to $I.$ We further claim that $ABKLI$ is cyclic.

The first step is to invert with respect to the incircle. The point $A$ gets mapped to the midpoint $X$ of $EF,$ and the point $B$ gets mapped to the midpoint $Y$ of $DF.$ Thus it suffices to prove that $K,L,X,Y$ are collinear. For now, we will only focus on the point $X$; the analysis for the point $Y$ is symmetrical to that of $X.$ A rephrasing is to prove that if $EF$ intersects $KL$ at $X,$ then $X$ is the midpoint of $EF.$

To do so, we will use projective geometry. If $X$ were indeed the midpoint of $EF,$ then we'd have $(E,F;X,P_{\infty}) = -1.$ Projecting through the point  $L$ and noting that $LM \parallel EF,$ we get $(E,F;K,M) = -1.$ Thus it suffices to prove that $EFKM$ is harmonic. To do so, since the tangents at $E,F$ intersect at $A,$ we must prove that $A,K,M$ are collinear. It turns out that $M$ is the Nagel point of $\triangle ABC,$ and this can easily be proven using barycentric coordinates and the given condition on the triangle.

With a good diagram, we see that I lies on this as well. Indeed, BID=90-B/2 $\implies$ KIB=90+B/2. On the other hand, the fact that 3AB=AE+BD+2CD $\implies$ 4AB=2AB+2CD=2s leads to CD=s-c=c. Since we can't do much with BAK, it's well known that extending this intersects at D', the reflection of D over midpoint of BC (motivating because in these concise problems, there will probably be some construction). Then BD'=BA and we get that BAD'=BDA', whence BAK=BAD'=90-B/2, so ABIK is cyclic, and analogously so is ABLI, so ABLK is cyclic, as desired. $\blacksquare$

$(CI)$ and $(II_C)$ are reflections about $I$, so $CIDE$ and $II_CAB$ cyclic implies $ABKL$ cyclic.

We invoke barycentric coordinates with $\triangle ABC$ as our reference triangle.

We can see that our length condition can be used in that $I = (3a:3b:a+b)$.

Notice now that $D = (0:a+b:2a-b)$ and $E = (a+b:0:2b-a)$ it can now be computed that $K = (9a^2:2a^2+5ab+2b^2:-a^2+ab-2b^2)$ and similarly $L = (2a^2 + 5ab+2b^2:9b^2:-2a^2+ab-b^2)$. Bash the rest from here. I believe in you. It's 12:13 so I must go to bed.

Let $M$ be the midpoint of $\overline{CI}$ and let $P$ be the midpoint of arc $AB$. Then, by ASA congruence we have $\triangle CME \cong \triangle APB$. In particular, we have $PI = IM = EM = DM$. Reflecting segments $\overline{EM}$ and $\overline{DM}$ over the perpendicular bisector of $\overline{MK}$ gives us segments $\overline{KP}$ and $\overline{LP}$ respectively, so
\[ PI = IM = EM = DM = PK = PL,\]
hence the result follows.


The length of a rectangle is $ 5$ inches and its width is less than $ 4$ inches.  The rectangle is folded so that two diagonally opposite vertices coincide.  If the length of the crease is $ \sqrt {6}$, then the width is:

$ \textbf{(A)}\ \sqrt {2} \qquad \textbf{(B)}\ \sqrt {3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \sqrt {5} \qquad \textbf{(E)}\ \sqrt {\frac {11}{2}}$

[geogebra]e006386445a926291c8231117e97e4983de47a77[/geogebra]

[hide]
Let the width $ \equal{}x$
After that construction, we have two similar triangles, therefore

$ \frac{5}{x} \equal{} \frac{\sqrt{25\plus{}x^2}}{\sqrt{6}}$

$ Rightarrow x^4 \plus{} 25x^2 \minus{} 25(6) \equal{} 0 \Leftrightarrow (x^2\minus{}5)(x^2\plus{}30)\equal{}0$

$ \Rightarrow x\equal{}\sqrt{5} \longrightarrow \boxed{D}$[/hide]


Well it depends...

#1 Comedy

a tie for number 1: Drama...only watch a handful thou

#2 Sports

#3 Informational (Nova, etc.)

dude, drama all the way! have u seen the show "House"? it's good. sports is like last (unless it's tennis or the olympics or something, but NO football) actually, the superbowl game wasn't that intolerable to watch...indiana rocks everyone's socks

[quote="mwpl11"][quote="kstan013"]<<BUMP>>[/quote]

hey, what do you mean by that?

BTW, comedy really gets to me sometimes.... it can get so aggravating![/quote]
you bumb a topic to get it to the top of the topics list so more people see it. I like sports.

Tendentially drama (ie House), but Malcolm in the Middle is the best thing ever seen on TV.

Here's the series I currently watch on TV: Family Guy, American Dad, South Park, Scrubs, Stargate(s), Desperate Housewives, Lost, October Road.

I tend to watch comedies and game shows more. Comedies, because, well, who doesn't like comedy? And about game shows, I like them because it shows real people like us rather than actors play roles. Plus it's fun to answer trivia questions before they do. :D

[quote="Arvind_sn"] And about game shows, I like them because it shows real people like us rather than actors play roles. [/quote]
I can see you don't watch people a lot. Most game shows seem like they are well....
Just start to watch people. That's what I think is the best way to watch TV.
I like TV on the radio if that music is counted as TV show.
I also used to watch Full House but it is now unavailable to me(plus extremely hard to understand.)

I don't watch TV all that much...but I like drama types the best.

I like comedy - laughing is good for you.

comedy i guess... i dont really watch tv at all.. but i like laughing  :D

I really only like comedy and informational, maybe the others I could watch, but that's about it.

Youtube [size=200]>[/size] [size=59]TV[/size]

Has music, comedy, the truth, math, etc.

[quote="Valentin Vornicu"]Stargate(s)[/quote]

Which do you like better, Atlantis, or SG-1?

I only watch a handful of TV, besides news. And my hand isn't even that big. Maybe 1 or 2 shows, such as Reality (i.e. Duel?)

Last time I watched TV was about two months ago at a friends house. Hehe...


[quote="SnowStorm"]Source: AHSME 1973 #4

Two congruent $30^\circ - 60^\circ - 90^\circ$ triangles are placed so that they overlap partly and their hypotenuses coincide.  If the hypotenuse of each triangle is 12, the area common to both triangles is

$\text{(A)} \ 6\sqrt{3} \qquad \text{(B)} \ 8\sqrt{3} \qquad \text{(C)} \ 9\sqrt{3} \qquad \text{(D)} \ 12\sqrt{3} \qquad \text{(E)} \ 24$[/quote]
[hide]
I got
$\text{(C)} 9\sqrt{3}$
THe length of the triangle that shares the area has lengths $6,6,\text{ and }6\sqrt{3}$. Dividing it into two triangles, the area of the triangle is $9\sqrt{3}$[/hide]

[hide]I probably got this wrong but for some reason I got D.  I found that the side length of the trangle that was in the overlap was $12, 4\sqrt(3), 4\sqrt(3)$  I got this because the angles were 30, 30 120 in the overlapping triangle.  I calculated the total area to be $12\sqrt(3)$  Not sure what I did wrong... [/hide]

There is nothing wrong with [b]Apprentice[/b]'s solution :)

Construct a semicircle whose diameter is hypotenuse say $AB$ of those triangles, then there are exactly two points say $C$ and $D$ on perimeter, such that triangles $\Delta ABC$ and $\Delta ABD$satisfy $\angle C=\angle D=90^\circ$, $\angle ABC=\angle BAD=30^\circ$ and $\angle BAC=\angle ABD=60^\circ$.

[hide]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=634[/img]

Sorry they dont look congruent, but you get the idea.
[/hide]


Let $x, y, z \in (0,\infty)$ such that $x + y + z = 1$. Prove that:
\[\frac{x}{xy + 1} + \frac{y}{yz + 1} + \frac{z}{zx + 1} \geq \frac{36xyz}{13xyz + 1}\]

Let $x, y, z \in (0,\infty)$ such that $x + y + z = 1$. Prove that:
\[\frac{x}{xy + 1} + \frac{y}{yz + 1} + \frac{z}{zx + 1} \geq \frac{729xyz}{783xyz + 1}\]

[quote=harazi]This one is a simple consequence of Schur inequality. In fact, we have
\[\sum_{cyc} \frac{x}{xy+1}=\sum_{cyc} \frac1{y+\tfrac1x} \ge \frac9{\sum\limits_{cyc} (y+\tfrac1x)}=\frac{9xyz}{xyz+xy+yz+zx}\].
All you have to do is to combine this inequality with $xy+yz+zx \le \frac{9xyz+1}4$, which is true because it can be writen as $(x+y-z)(y+z-x)(z+x-y) \le xyz$.[/quote]

Please explain $\sum_{cyc} \frac1{y+\tfrac1x} \ge \frac9{\sum\limits_{cyc} (y+\tfrac1x)}$

That follows from Cauchy-Schwarz inequality.


Take $ b=c=1 $ and $ a=\frac{\sqrt[3]{2}}{1-\sqrt[3]{2}} $. If my computations aren't wrong, then the problem is wrong. Anyway, I feel that this inequality cannot be true.  ;)

I just adore your posts! You can understand everything from them!

[quote="hxtung"]Let $a,b,c \in R$. Prove that: $\sum_{cyclic}(a+b)^4-a^4 \geq 0$[/quote]

Are you sure there is no typo here? Cause your inequality (at least, in the form you posted it), rewrites as

  $\left(b+c\right)^4 + \left(c+a\right)^4 + \left(a+b\right)^4 \geq a^4 + b^4 + c^4$,

what is [b]a)[/b] not cyclic at all and [b]b)[/b] incorrect, as Harazi showed...

  Darij

It is incorrect, but why is this not cyclic? :?

Well, of course, it [i]is[/i] cyclic, but more importantly, it is symmetric, so if Hxtung really had meant this inequality, then he would have written "symmetric" rather than "cyclic".

  Darij

Sorry :( . I'll be carefull next time. :D


hameye $n$hayi ra biyabid ke :$(a,b)=1 , a\mid n \wedge b\mid n \Longrightarrow a+b-1\mid n$

javabash $n=1 , n=12 , n=p^{k}$ hastesh . rahe hallam toolanie . age kasi rahe kootah balade bege :)


superior algebra
superior algebra unsolved

[b]Problem[/b]. Let $H$ be a real semigroup (a subset of real
numbers that is closed with respect to the additive operation).
If there exist non-empty sets $X$, $Y$ such that
a) $H=X\cup Y$,
b) If $x\in X$, then $x+k\in H$ for all integers $k$,
c) For every $y\in Y$ there exists an integer $k_{0}$ such that
$y+k_{0}\notin H$,
Then $Y$ (and $X$) are  real semigroups.

[b]More explanations[/b]:
You can use the following properties (I proved them):

$X+H\subseteq X$ (so clearly $X$ is semigroup), $X\subseteq \mathbb{Q}^{c}$ (so $Y\cap \mathbb{Q}=H\cap \mathbb{Q}$).
The set $Y$ can be a subset of $\mathbb{Q}$ or not. Consider the following examples:

\[H_{1}=\{ m\sqrt{2}+n\sqrt{3}+k|m,n\in \mathbb{N}^{*},k\in \mathbb{Z}\}\cup \{ m'\sqrt{2}+n'|m'\in \mathbb{N},n'\in \mathbb{N}^{*}\},\]
\[H_{2}=\{ m\sqrt{2}+r|m\in \mathbb{N}^{*},r\in \mathbb{Q}\}\cup \mathbb{Q}_{+},\]
Where $\mathbb{N}^{*}$ is the set of positive integers, $\mathbb{N}=\mathbb{N}^{*}\cup \{ 0\}$ and $\mathbb{Q}_{+}$ is the set of all positive rational numbers.

moderators i request u to take this post to the superior algebra forum in college playground :maybe:

First of all we can say that H is the disjoint union of X and Y because of b and c. Especially the conditions in b and c are not only necessary but sufficent:

b.' $h\in X \iff \forall k\in\mathbb{Z}: h+k\in H$
c.' $h\in Y \iff \exists k\in\mathbb{Z}: h+k\notin H$

Now, if h is in H and x is in X, then for all $k\in\mathbb{Z}$ (h+x)+k=h+(x+k) is in H, so h+x is in X. Especially is X a semigroup.


It´s possible to prove that for all positive integer n exists one prime in the form m^2 + n ?

I don't think you get an answer here. It is highly possible that it is an open question.


Find all functions $f: (0,1) \to \mathbb R$ such that
\[f(xyz) \equal{} xf(x) \plus{} yf(y) \plus{} zf(z)\]
for all $x,y,z \in (0,1)$.

[quote="caubetoanhoc94"]Find all function:$ (0,1) \to R$ such that:
$ f(xyz) \equal{} xf(x) \plus{} yf(y) \plus{} zf(z)$[/quote]

$ f(xyztu)\equal{}f((xyz)tu)\equal{}xyzf(xyz)\plus{}tf(t)\plus{}uf(u)$ $ \equal{}x^2yzf(x)\plus{}xy^2zf(y)\plus{}xyz^2f(z)\plus{}tf(t)\plus{}uf(u)$

$ f(xyztu)\equal{}f(x(yzt)u)\equal{}xf(x)\plus{}yztf(yzt)\plus{}uf(u)$ $ \equal{}xf(x)\plus{}y^2ztf(y)\plus{}yz^2tf(z)\plus{}yzt^2f(t)\plus{}uf(u)$

Equating implies $ (x^2yz\minus{}x)f(x)\plus{}(xy^2z\minus{}y^2zt)f(y)$ $ \plus{}(xyz^2\minus{}yz^2t)f(z)\plus{}(t\minus{}yzt^2)f(t)\equal{}0$

Setting $ y\equal{}z\equal{}t\equal{}\frac 12$ in this equation, we get $ (x^2\minus{}4x)f(x)\plus{}(x\plus{}\frac 54)f(\frac 12)\equal{}0$

And so $ f(x)\equal{}a\frac{4x\plus{}5}{x^2\minus{}4x}$

Plugging this back in the equation, we get $ a\equal{}0$

Hence the unique solution : $ f(x)\equal{}0$ $ \forall x\in(0,1)$


Show that if $a,b,c,d>0$ then we have:

$ \frac{a}{3b+c} + \frac{b}{3c+d} + \frac{c}{3d+a} + \frac{d}{3a+b} \geq 1$ .

Guess that this works using Jensen, defining $ f(x)= \frac{1}{x}$ and applying the ineq with  WLOG $a+b+c+d=1$ and with weights $a,b,c,d$.

DO you have different solutions?

What about grandpa Cauchy?
   We have sum a/(3b+c)>=(sum a)^2/[3ab+3bc+3cd+3da+2ac+2bc] and the last one is at least 1, since it reduces to a^2+b^2+c^2+d^2>=ab+bc+cd+da, true.


1. What will happen to the number of moles of $ SO_{3}$ in equilibrium with $ SO_{2}$ and $ O_{2}$ in the reaction

$ 2SO_{3}(g) \equal{} 2SO_{2}(g) \plus{} O_2(g)$ (= just is the same thing as the double arrow, i dont know how to type that)

a. Oxygen gas is added
b. The pressure is increase by decreasing the volume of the reaction container
c. The pressure is increased by adding argon gas
d. The temperature is decreased
e. Gaseous sulfur dioxide is removed

2. At a certian temperature, 4 mol $ NH_{3}$ is introduced into a $ 2L$ container, and the $ NH_{3}$ partially dissociates by the reaction $ 2NH_{3}(g) \equal{} N_{2}(g) \plus{} 3H_{2}(g)$

At eqilibrium, 2 mol $ NH_{3}$ remains.  What is the value of $ K$ for this reaction

3. Ammonia is produced by the Haber process, in which notrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst.  For the reaction

$ N_{2}(g) \plus{} 3H_{2}(g) \equal{} 2NH_{3}(g)$ equalibirum constants as a function of temperature are

$ 300 C, 4.34*10^{ \minus{} 5}$
$ 500 C, 1.45*10^ { \minus{} 5}$
$ 600 C, 2.25*10^{ \minus{} 6}$

Is the reaction exothermic or endothermic?

Im trying to study for the local USNCO exam, so explanations would be nice.

Thanks

3. From Le Chatelier's principle, the increasing temperature is met with a shift in the equilibrium that counters the temperature change, that is, the direction that is endothermic is now spontaneous. Because the equilibrium constant K is decreasing, it becomes less than the reaction quotient Q, thus, the reverse reaction is spontaneous, and therefore, the forward reaction is exothermic. I think.

hello

ya for the first question

this is only le chateliers principle

1.a.Oxygen gas added shifts the equilibrium to the backward direction therby increasing the number of moles of SO3 formed

1.b.As the volume fo the container is decreased,the equilibrium shits to the direction to minimize the presuure increase....so greater the number of moles of the species--greater is the presuure...so what happens the equilibrium shitf to a direction in which there are lesser number of molecules to minimise the pressure ...so amt of SO3 increases.

1.c.same as above----amt of so3 increases

1.d.First to decide that we need to whther the given reaction is exothermic or endothermic....Since it involves decompostion it must be a endothermic reaction i.e.

forward reaction---endothermic
backward reaction-exothermic

so as temperature increases,the reaction think of a way of increasing the temperature to compensate the decrease according to lechatliers principle...ok

So.exothermic reaction is favoured therby increasing the amount of SO3 in the equilibrium mixture...

1.e.when SO2 is removed.the reaction thinks of a way to increase the SO2 according to le chatilers principle.So amot of So3 at equilibrium  decreases as forward reaction is favoured

I think these are the answers for the first question...
plz tell me if they are wrong

Thanks

These questions are from Zumdalh Chemistry Book.
Unlike the Atkins one that I have, I could not fine an answer book for the even problems of the Zumdahl book, so I do not know if they are right.


Is there the quarter penalty in AIME, like that in AMC?

Well there is no penalty for AMC in a way because you get 6 points for getting it right and 1.5 for not answering and 0 for wrong answers. To answer your questions there is no penalty for AIME or any points for not answered questions because it's pretty hard to guess from 1-1000.


Prove that in every triangle we have:
ma*mb>=ra*rb*rc.

bye~

Now .. is the inequality correct?
Cause I think it is:
ma*mb*mc>=ra*rb*rc :?

tschuess!

I am sorry !
That is the ineq!
ma*mb*mc>=ra*rb*rc!

bye~

This is not very hard!

ma=sqrt((b^2+c^2)/2-a^2/4)>=sqrt((b+c)^2-a^)/4)=sqrt(p(p-a))
do the same with mb and mc

now you multiply them all and you get:

ma*mb*mc>=psqrt(p(p-a)(p-b)(p-c))=pS=ra*rb*rc

bye! hope it's correct! :D

Okay!

I have something to add!

ma*mb*mc>=ra*rb*rc>=la*lb*lc

this could be useful!

We talked about the product (a relation between them)!
here is a nice inequality:
la+lb+lc<=ma+mb+mc<=ra+rb+rc :D

We could use the fact that ra+rb+rc=4R+r

Take da, db, dc the distances from O to the sides, ma<=OA+da=R+da, so
ma+mb+mc<=3R+da+db+dc=3R+R(cosA+cosB+cosC)=3R+R(1+r/R)=4R+r

As for the first one, we can see that ma<=la and so on

Could sombody explain the notations?

What are ma,mb,mc? (medians??)
What are ra,rb,rc?
What are la,lb,lc?

Thanks!

$m_a$ medians

$l_a$ bisectors

$r_a$ exradii

[quote="georg"]If $ \triangle ABC$ with medians $ m_a,m_b,m_c$ and radii of excircles $ r_a,r_b,r_c$, then

$ m_am_bm_c\geq r_ar_br_c$.[/quote]Actually, the following refined inequality chain holds

$ 3r_ar_br_c\leq r_am_aw_a \plus{} r_bm_bw_b \plus{} r_cm_cw_c\leq 3m_am_bm_c$,

where $ w_a,w_b,w_c$ are the angle-bisectors of $ \triangle ABC$.

The first inequality can be obtained from

$ m_aw_a\geq r_br_c$,

which is equivalent to an old result $ \frac {m_a}{w_a}\geq\frac {(b \plus{} c)^2}{4bc}$, see

G. Tsintsifas, L. Bankoff, L. Goldstone, Problem E 2471, [url=http://www.maa.org/pubs/monthly.html]Amer. Math. Monthly[/url], 81 (1974), 406 and 82 (1975), 523 - 524.

The second inequality can be obtained from

$ m_bm_c\geq r_aw_a$,

with equality if and only if $ \triangle ABC$ is equilateral.

[b]Proof[/b] $ m_b^2m_c^2 \minus{} r_a^2w_a^2$

$ \equal{} \frac {\left(2c^2 \plus{} 2a^2 \minus{} b^2\right)\left(2a^2 \plus{} 2b^2 \minus{} c^2\right)}{16} \minus{} \frac {bc(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)(a \plus{} b \plus{} c)^2}{4(b \plus{} c)^2}$

$ \equal{} \frac {\left(a^2 \plus{} b^2 \plus{} c^2 \minus{} 3bc\right)^2}{16} \plus{} \frac {(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)\left[2(2bc \minus{} ca \minus{} ab)^2 \plus{} (b \minus{} c)^2\left(a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 8bc\right)\right]}{16(b \plus{} c)^2}$

$ \geq 0$.


Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q\minus{}q^p\equal{}7$

You already posted this here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219599


geometry
geometric transformation
homothety
geometry proposed

Let $ w_1,w_2$ with centers $ O_1,O_2$ be two circles not intersecting and no one is inside the other. Let $ \ell_1$ be tangent to $ w_1$ in $ A_1$ and tangent to $ w_2$ in $ A_2$. $ \ell_2$ is tangent to $ w_1$ in $ B_1$ and tangent to $ w_2$ in $ B_2$.

Prove that  $ A_1B_1 \parallel{}A_2B_2$ or $ A_1B_1, A_2B_2,O_1O_2$ are concurrent.

If the circles have the same radi, we have $ A_1A_2 \parallel{}B_1B_2$.

If the circles have different radi, the tangency points on each are simmetric respect to $ O_1O_2$ and so the tangents, they intersect over $ O_1O_2$

 :)

[quote="ElChapin"]If the circles have the same radi, we have $ A_1A_2 \parallel{}B_1B_2$.

If the circles have different radi, the tangency points on each are simmetric respect to $ O_1O_2$ and so the tangents, they intersect over $ O_1O_2$[/quote]

Sorry, it was a typo. It wasn't intended to be a so easy...
The question is:

[b]Prove that  $ A_1B_1 \parallel{}A_2B_2$ or $ A_1B_1, A_2B_2,O_1O_2$ are concurrent.[/b]

[i][size=75]*Note: I corrected the original post.[/size][/i]

as mentioned in previous post, there are 4 common tangents to the circles.
if the tangents are transverse common, then the three lines are obviously concurrent.
if they are direct common, draw the lines connecting the centres to the points of contact and prove the similarity of $ \Delta O_1A_1B_1$ and $ \Delta O_2A_2B_2$ and use it to prove that $ A_1B_1$ and $ A_2B_2$ are parallel

[quote="BaBaK Ghalebi"]but such two circles have 4 common tangents,which pair do you mean? :maybe:

(your statement doesn't hold for two external common tangents,so you should define which pairs you mean...)[/quote]

Well in fact the idea is to consider all possible cases and the statement holds for all of them. Here are pictures of three of the possible cases, obviously the interesting one is the third one (Whene you have an internal and external tangent). If the tangents are both external or internal the lines $ A_1B_1,A_2B_2$ are related with an homothecy so they are parallel. What about the other cases?


ARML geometry probability trigonometry algebra polynomial Vieta

Thanks to everyone who came to the practice today. I hope you all had a good time and came away with a little bit of mathematical knowledge.

Here are the problems that I used for practice and their solutions (sometimes just answers for now). Also the answers to the geometry problems are given in this thread. Feel free to start a discussion about any of the problems talked about today or to ask for problems like any of the ones you saw.

[hide="Answers to Geometry problems:"]

1. 30
2. 15/4 or 3.75
3. 2
4. 10
5. 7
----------------
1. $ 21^o$
2. $ \frac{5}{16}$
3. $ \sqrt {11}$

1. $ 216 \minus{} 36\pi$
2. $ 95^o$
3. $ \frac{7}{60}$
------------------
1. 8
2. $ 26: 7: 27: 13$
3. $ \frac{49}{1250}$

1. $ 32\sqrt {2} \plus{} 32$
2. $ 132^o$
3. $ \frac{1}{7}$[/hide]

For those of you who are interested in the probability questions we were discussing about during the break, here are two threads where those problems are discussed and solved.

The drunken man problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198960

The "two heads before three tails" problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200580

[hide="Geometry Worksheet Problem 5"]
Let $ FB \equal{} x$ and $ EB\equal{}y$.  Let $ \angle ADF \equal{} \angle FDE \equal{} \theta$.

We know $ \tan 2\theta \equal{} \frac{2\tan\theta}{1\minus{}\tan^2 \theta}$

From triangle ADF, we get $ \tan \theta \equal{} \frac{5}{2\plus{}y}\equal{}\frac{5}{5\plus{}x}$
Draw in a line from E to side DA.  Where this line intersect DA, call it point P.  From triangle DPE, we have $ \tan 2\theta \equal{} \frac{5\plus{}x}{2}$.

Plugging these values in, we get $ \frac{5\plus{}x}{2} \equal{} \frac{2(\frac{5}{5\plus{}x})}{1\minus{}(\frac{5}{5\plus{}x})^2}$

Cross multiply and simplify:

$ (5\plus{}x)\minus{}\frac{25}{5\plus{}x}\equal{}\frac{20}{5\plus{}x}$
$ (5\plus{}x)^2\minus{}25\equal{}20$
$ x^2\plus{}10x\minus{}20\equal{}0$
$ x\equal{}\minus{}5\plus{}3\sqrt{5}$

Since a side length of the original square is $ 5\plus{}x$, the side length is $ 3\sqrt{5}$.  Now we can find $ DE^2 \equal{} (3\sqrt{5})^2\plus{}2^2$ so $ DE\equal{}7$.

[/hide]

[hide=" Vieta vs. Viete"]
[url=http://mathworld.wolfram.com/VietesTheorem.html]MathWorld[/url] likes Vieta and [url=http://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te]Wikipedia[/url] says either way.
[/hide]


probability
calculus
derivative
expected value
probability and stats

Suppose that a random sample of three observations is drawn from a distribution with the following pdf:

f(x)= Bx^(B-1) for 0<x<1 & 0 otherwise.
B>0.

Determine the pdf of the sample median.

For one copy of the random variable, $ P(X\le x)\equal{}x^B.$ (At least that's true for $ 0\le x\le 1;$ we don't really need to spend any energy thinking about what happens outside that interval.)

The probability that the median of three is $ \le x$ is the sum of the probability that exactly two of the three are $ \le x$ plus the probability that all three are $ \le x.$ Those, in turn, are essentially binomial probabilities.

The probability that all three are $ \le x:$ $ x^{3B}.$

The probability that 2 out of 3 are $ \le x:$ $ 3x^{2B}(1\minus{}x^B).$

So if $ M$ is the median, $ P(M\le x)\equal{}3x^{2B}(1\minus{}x^B)\plus{}x^{3B}\equal{}3x^{2B}\minus{}2x^{3B}.$

Hence the pdf that you're looking for is the derivative of this:
\[ \boxed{6Bx^{2B\minus{}1}\minus{}6Bx^{3B\minus{}1}}\]
(It's that formula on $ [0,1],$ and zero otherwise.)

We can readily find the expected value, getting
\[ E(M)\equal{}\frac{6B^2}{6B^2\plus{}5B\plus{}1}\]
As $ B\to\infty,$ we get $ E(M)\to 1,$ as expected. If $ B\equal{}1,$ we get the uniform distribution and $ E(M)\equal{}\frac12,$ which is what we should get. In that particular case of the uniform distribution, the pdf of the median is $ 6x\minus{}6x^2\equal{}6x(1\minus{}x).$


algebra
polynomial
function
domain
complex analysis
complex analysis unsolved

Give explicitly a sequence of complex polynomials whose simple limit is the function $f(0)=1$ and $\forall z \neq 0, f(z)=1$. (the existence is guaranteed by the complex Stone-Weierstrass theorem applied to continuous functions $(f_{n})$ such that $f_{n} \rightarrow f$ pointwise, on compact domains $K_{n}$ such that $(K_{n})$ is increasing with limit $\mathbb{C}$).

[quote="julien_santini"]Give explicitly a sequence of complex polynomials whose simple limit is the function $f(0)=1$ and $\forall z \neq 0, f(z)=1$. [/quote]
Are you sure about the function $f$, is it constant $1$-in this case the problem becomes trivial?

[quote="Chun"]I think $f(z)=0\mbox{ }\forall z\neq 0$ . The existence is not difficult to show . We can use Runge's Approximation .
But I think it's hard to find such explicit polynomials[/quote]
I must misunderstand something about Runge's theorem, but it is neccessary for the function to be approximated by polynomials to be holomorphic, but $f(z)=0 \forall z\neq 0$ and $f(0)=1$ isn't even continious.

[quote="julien_santini"]Give explicitly a sequence of complex polynomials whose [b]simple limit [/b]is the function $f(0)=1$ and $\forall z \neq 0, f(z)=1$. (the existence is guaranteed by the complex Stone-Weierstrass theorem applied to continuous functions $(f_{n})$ such that $f_{n} \rightarrow f$ [b]pointwise[/b], on compact domains $K_{n}$ such that $(K_{n})$ is increasing with limit $\mathbb{C}$).[/quote]
To eugene : I think you're misunderstanding him . The convergence here is pointwise , not locally uniform :) .


limit
calculus
derivative
function
real analysis
real analysis unsolved

Having trouble with this practice problem on parts c & d. Not 100% on my work for b (hidden).

Let $ f(x_1,x_2) \equal{} 0$ if $ x_1 \cdot x_2 \not \equal{} 0$ and $ \equal{} 1$ if $ x_1 \cdot x_2 \equal{} 0$.
a) Determine if $ f$ is continuous at $ (0,0)$.
b) Determine if $ f$ is directionally continuous at $ (0,0)$.
c) Determine all $ u \not \equal{} (0,0)$ such that $ \text{D}f(0,0)u$ exist & give its value.
d) Determine if $ f$ is differentiable at $ (0,0)$.

Note: $ \text{D}f(z)u \equal{} \lim_{h \to 0} \frac {f(z \plus{} hu) \minus{} f(z)}{h}$

[hide]
a) I said f was not continuous and disproved using $ \delta \minus{} \epsilon$.
b) I'm not sure about this, but using the def'n of dir. cont., I found $ \lim_{h \to 0} f((0,0) \plus{} h(u_1,u_2)) \equal{} \lim_{h \to 0} f(h \cdot u_1, h \cdot u_2)$ and noted that $ (h \cdot u_1)(h \cdot u_2) \to 0$ as $ h \to 0 \implies \lim_{h \to 0} f(h \cdot u_1, h \cdot u_2) \to 1$ as $ h \to 0$.
c) I'm not sure about this one. I think it works for all nonzero $ u$.
d) I don't know how to do this one at all.[/hide]

[quote="skimnc"]$ (h \cdot u_1)(h \cdot u_2) \to 0$ as $ h \to 0 \implies \lim_{h \to 0} f(h \cdot u_1, h \cdot u_2) \to 1$ as $ h \to 0$.[/quote]  This line is wrong.  You should think about what "directionally continuous" means, and also what the graph of $ f$ looks like.

For part (c), you need to have done part (b) correctly and to know something about the relationship between differentiability and continuity and then to do a little bit more work.

For part (d), you need to have done part (c) correctly and know something about the relationship between the existence of directional derivatives and differentiability.

This any better?

[b]b)[/b] So if we think of approaching along, for example, the line $ x_1 \equal{} x_2 \not \equal{} 0$, we get $ \lim_{h \to 0} f((0,0) \plus{} h(u_1,u_1)) \equal{} f(h \cdot u_1, h \cdot u_1) \equal{} 0$ since $ h^2 \cdot u_1^2 \not \equal{} 0 \forall h > 0$.  This value is not $ 1$, our value of the function at this point.  Therefore not dir. cont.

[b]c)[/b] $ \text{D}f(0,0)u$ exists $ \iff$ $ (u_1,u_2) \not \equal{} (0,0)$ has either $ u_1 \equal{} 0$ or $ u_2 \equal{} 0$.

[b]d)[/b] $ f$ is not diff. at $ (0,0)$ since $ f$ is not cont. at $ (0,0)$.

Again, thanks a bunch already.


function
geometry
geometric transformation
reflection
logarithms
algebra
domain

Solve the system (in real numbers)
$\begin{cases}x^3+y=2\vspace{0.2cm}\\y^3+x=2\end{cases}$

$f: R\rightarrow R,\ f(x)=x^3+\sqrt [3] {2-x}\ ,\ f\nearrow.$
$f(x)=2\Longleftrightarrow x=1\ (y=1).$

[quote="Farenhajt"][quote="Virgil Nicula"]$f(x)=x^3+\sqrt [3] {2-x}\ ,\ f\nearrow.$[/quote]

Please, can you explain how did you get that this function is always increasing? Since $f'(x)=3x^2-\frac{1}{3\sqrt[3]{(2-x)^2}}$ and, for example, $f'(0)<0, f'({1\over 9})<0, f'({1\over 4})<0, f'(-{1\over 4})<0, f'(-{1\over 9})<0$ etc. etc.[/quote]
Sorry, Farenhajt, I see wrong $\boxed {\ +\ }$ instead of $\boxed {\ -\ }\ .$

[quote="The Zuton Force"][quote="joml88"][quote="fourierseries"]If you subtract the equations, you get:

$x(x^2-1)=y(y^2-1)$ which implies that $x=y$

That makes it easier![/quote]

Why does that imply that $x=y$?  What if you had $x=1$ and $y=-1$?  That still works...[/quote]

It doesn't work
Plug it into the original equation.

Since the two original equations are inverse functions, y=x for them to be equal, right?

That would leave the cubic $x^3+x-2=0$, no?[/quote]

I didn't say that $x\neq y$.  I was trying to point out that I don't see how you can logically conclude from what he said that $x=y$.

Your reasoning looks better :)

I think this argument is valid:

Since the curves $x^3+y=2$ and $y^3+x=2$ are mutually inverse, all their meeting points must lie on the line $y=x$.

X^3+Y=Y^3+X
X^3-X=Y^3-Y
X(X^2-1)=Y(Y^2-1), <=> X=Y.

In the original ecuation, X^3+X-2=0.

The only solution is X=Y=1.

[quote="Virgil Nicula"][i][size=125]Fourierseries[/size][/i] (who didn't detail) and [i][size=125]Farenhajt[/size][/i] are right ! Denote $f: R\rightarrow R$, where $f(x)=2-x^3$. This function is inversely and the its inverse function is $f: R\rightarrow R$, where $f^{-1} (x)=\sqrt [3] {2-x}$. The our system becomes: $f(x)=f^{-1}(x),\ x\in R\Longleftrightarrow f(x)=x,\ x\in R\Longleftrightarrow$ $x^3+x-2=0\Longleftrightarrow x=y=1\ .$
[b]Remark.[/b] I used the fact that the graphic $G_{f^{-1}}$ of the inverse function $f^{-1}$ is reflection of the graphic $G_f$ of the direct function $f$ w.r.t. the angle-bisector of the first quadrant (with the equation $y=x,\ x\ge 0$), i.e.
$\boxed {\ (\alpha, \beta )\in G_f\Longleftrightarrow(\beta ,\alpha )\in G_{f^{-1}}\ }\ .$ I will come back soon.

[b][size=100]Another examples (and with another ideas).[/size][/b][hide="Here are eight exercises."]

$\blacktriangleright\ 1^{\circ}.$ Ascertain $0<a\ne 1$ so that the graphics of the functions $f: R\rightarrow (0,\infty ),\ f(x)=a^x$ and $g: R\rightarrow (0,\infty )\rightarrow R,\ g(x)=\log_a x$ are tangent between them.
[u]Answer:[/u] $a=e^{\frac 1e}$ and the common point of the both graphics is $(e,e)\ .$

$\blacktriangleright\ 2^{\circ}.$ Solve the following equations:

$2.1.\ \ 2^x+2^{\sqrt{1-x^2}}=3$ ([u]Marcel Chirita[/u]).
$2.2.\ \ \log_a (a+x)=\log_b (b+x)$, where $a>1,\ b>1,\ a\ne b$ ( [u]V.N.[/u] ).
$2.3.\ \ \log_{x^2+3x+4} |x+2|=\log_{x^2+5x+9} |x+3|$ ( [u]V.N.[/u] ).

$\blacktriangleright\ 3^{\circ}.$ Solve the following systems of equations:

$3.1.\ \ e^x+e^y=2\ ;\ x+y=xy$ ( [u]V.N.[/u] ).
$3.2.\ \ x^5+y^5=33\ ;\ x^3-y^3=7\ .$
$3.3.\ \ x_1+x_2+\ldots +x_{n-1}+x_n=1,\ x_k\ge 0,\ k\in \overline {1,n},\ n>1\ .$
$.\ \ \ \ \ \ x_1\sqrt{1-x_2}=x_2\sqrt{1-x_3}=\ldots =x_{n-1}\sqrt{1-x_n}=x_n\sqrt{1-x_1}\ .$
$3.4.\ \ ax+by=(x-y)^2\ ;\ by+cz=(y-z)^2\ ;\ cz+ax=(z-x)^2\ ,$
where $\ a>0\ ,\ b>0\ ,\ c>0$ ([u]Laurentiu Panaitopol[/u]).[/hide][/quote]

See here another exercises with the same or another ideas.

[hide="1"]
The graphs are, again, symmetric over $x=y$, so they are tangent at a point on $x=y$, and are thus tangent to $x=y$.  So, let's say $x=y$, $a^x=y$, and $\log_a{x}=y$ intersect at some point.  Combining these, we have \[ a^y=y=\log_a{y} \implies a=\sqrt[y]{y} \]  So now, we have the function $a(y)=\sqrt[y]{y}$.  We know the maximum is this function is $\sqrt[e]{e}$, and for every other value of $a$ greater than or equal to $1$, there are two possible $y$ such that $\sqrt[y]{y}=a$.  Thus, \[ \boxed{a=\sqrt[e]{e}=e^{\frac{1}{e}}} \] is the only number which gives exactly one intersection.[/hide]

[hide="2.1"]
It seems that the answers are $x=0, 1$.  Clearly, the domain of $f(x)=2^x+2^{\sqrt{1-x^2}}$ is $[-1, 1]$, and in that domain, the value of $2^{\sqrt{1-x^2}} \in [1, 2]$.  Meanwhile, $2^x \in [\frac{1}{2}, 2]$.  Below $0$, the value of $2^x$ is below $1$, so the relation cannot hold.  I don't see a good solution for this without using calculus, but all I can see is that this can be reduced to \[ 2^{\sin{\theta}}+2^{\cos{\theta}}=3 \] where $\theta=\sin^{-1}{x}$[/hide]

[hide="2.2"]The only solution is $\boxed{x=0}$.  WLOG, let $a<b$.  Then \[ \log_a{(a+x)}=\log_b{(b+x)} \iff \log_a{(a+x)}-\log_a{a}=\log_b{(b+x)}-\log_b{b} \] \[ \implies \log_a{(1+\frac{x}{a})}=\log_b{(1+\frac{x}{b})} \] \[ \implies \frac{\log{(1+\frac{x}{a})}}{\log{a}}=\frac{\log{(1+\frac{x}{b})}}{\log{b}} \]
We know $a<b, x>0 \implies 1+\frac{x}{a} > 1+\frac{x}{b} > 1$.  Thus, \[ \log{(1+\frac{x}{a})} > \log{(1+\frac{x}{b})} \implies \log{a} > \log{b} \] \[ \implies a > b \]  which is a contradiction.

Now assume $x<0$.  We then have $1+\frac{x}{a} < 1+\frac{x}{b} < 1$.  Thus, \[ \log{(1+\frac{x}{a})} < \log{(1+\frac{x}{b})} < 0 \implies |\log{(1+\frac{x}{a})}| > |\log{(1+\frac{x}{b})}| \] \[ \implies |\log{a}| > |\log{b}| \implies \log{a} > \log{b} \] \[ \implies a>b \]

Finally, if we assume $x=0$, we get \[ \frac{\log{(1)}}{\log{a}}=\frac{\log{(1)}}{\log{b}} \implies 0=0 \] [/hide]

Oh, uh... :blush:  I'm gonna edit that out now.


MIT
college
Princeton
Harvard
CMU
Carnegie Mellon

What college/university do you as an undergrad plan to go to?
What do you plan to major in as an undergrad?
What do you want to minor in as an undergrad?
Why?
What college/university do you as an grad student plan to go to?
What do you plan to major in as an grad student?
What do you want to minor in as an grad student?
Why?

My Plan:
MIT
Physics
CS
They are my top interest and i believe the future of science!
Princeton
Mathematics or Theoretical Physics
I plan to become a professor at Cal Tech and teach Math or Theoretical Physics! :D

then shouldn't you be majoring in math if you want to be a mathematician?

Its good to have plans but remember that on some level, plans should also be fluid (as contradictory as that may sound). Plans always need to be concrete enough to give you direction, but not so concrete as to give you tunnel vision. Be open to the possibility of liking subjects that you didn't think you would like, or didn't even know anything about. A good plan should always be based on ideology, but open to reassessment and revision, if necessary.

[b]esmalls[/b], how much thought have you really put into your 'plan'? It basically says, "I like computers, but I also want to be like Richard Feynman". I suggest that, before declaring a life plan, you:

1. Figure out what being a researcher and professor entails
2. Develop an actual understanding of what it means to study math, physics and computer science in depth
3. Don't think about grad school much until you at least start your undergrad, or else it's counterproductive (for the reason gauss202 mentioned in the above post -- tunnel vision is not a good thing).

Items 1 and 2 above take time.

mit is ranked first in physics based on the polls of us news and warld report and first in cs.

You should study the methodology of those polls. You may find that they don't mean what they're advertised to mean.

i believe gauss202 is either a teacher or prof correct me if im wrong.

this is my plan like top #1
here's my whole plan either study cs at carnegie mellon or mit go to cal tech and study cosmology and physics go to harvard or princeton and study in mathematics maybe minor in something new that relates to the field i chose when i get there. what ya think?

I think you need some more exposure to the real world before you start making plans like this.  Having ambition is certainly not a bad thing, and it's good to have goals, but honestly planning this far in advance before having even applied to college yet is just silly.  Try to enjoy the rest of high school.

My plan is to survive my upcoming junior year at Caltech in chemical engineering, and graduate the year after that.  I don't even know what I want to do after that.


[quote="csbonny"]Let x,y be real numbers such that

$ \left( {x \plus{} \sqrt {1 \plus{} y^2} } \right)\left( {y \plus{} \sqrt {1 \plus{} x^2} } \right)$ = 1

show that $ \left( {x \plus{} \sqrt {1 \plus{} x^2} } \right)\left( {y \plus{} \sqrt {1 \plus{} y^2} } \right)$ = 1[/quote]

It's such an easy problem. :blush:

Use bruce force with ur time. :P

[hide="Big Hint for csbonny"]WLOG;$ x\geq y$[/hide] :rotfl:

ข้อนี้กระจอกมาก  :(


Evaluate $ \frac{1}{2!\plus{}1}\plus{}\frac{2}{3!\minus{}1}\plus{}\frac{3}{4!\plus{}1}$. Express your answer as a common fraction.


calculus
integration
trigonometry
function
calculus computations

how to find the integral of e^tanx w.r.t. x ?

please help...

hello, do you mean $ \int e^{\tan(x)}\,dx$?
Sonnhard.

yes i mean the same .how do we evaluate this one .

Maybe we can't  calculate the integral as basic function.

so u say we cannot evaluate this integaral

you can evaluate the integral. but you have to do it as an infinite series:

$ e^{\tan x} = 1 + \tan x + \frac {\tan^2 x}{2!} + ....
= \sum_{k=0}^{\infty} \frac {\tan^k x}{k!}$

and then integrate term by term. but even that is not easy :maybe:.

ok, going for the final check on all integration problems, that is, Mathematica Online Integrator  :wink: , the integral is not a simple one. it involves the imaginary unit and the exponential integral. so that settles it.

hello, mathematica gives the following answer
$ \int e^{\tan(x)}\,dx\equal{}\minus{}\frac{1}{2} i e^{\minus{}i} \left(e^{2 i} \text{Ei}(\tan
   (x)\minus{}i)\minus{}\text{Ei}(\tan (x)\plus{}i)\right)$
Sonnhard.

still i don get this .please help me someone.please

bascially what they meant was that the  given integral is not  some basic function but can be rather written  as  a sum of infinite series.    u dont get a well dressed answer for this :)

but is there any way to evaluate this???

No, not in the sense that you are probably thinking. The answer cannot be expressed using [url=http://en.wikipedia.org/wiki/Elementary_function]elementary functions[/url].

There is no room to discuss the integral more than this.

but y is int there any way..any explanation,. please help

See here for an explanation why: http://en.wikipedia.org/wiki/Differential_Galois_theory


[hide="Hint"]$ f(x)\equal{}(1\minus{}x)(1\minus{}x_1)...(1\minus{}x_{n\minus{}1})\minus{}1\plus{}(x\plus{}x_1\plus{}...\plus{}x_{n\minus{}1})$ is a linear function.[/hide]

this problem doesn 't have another solution?

[quote="onlymath"]Consider $ x_1 , x_1 ,..., x_n$  $ \in$ $ [0,1]$ prove that:
$ (1 \minus{} x_1)(1 \minus{} x_2)...(1 \minus{} x_n)$ $ \geq$ $ 1 \minus{} (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)$[/quote]
\[ \begin{array}{l}
 f\left( {x_1 ,x_2 ,...,x_n } \right) \minus{} f\left( {0,x_1  \plus{} x_2 ,...,x_n } \right) \\
  \equal{} \left[ {\left( {1 \minus{} x_1 } \right)\left( {1 \minus{} x_2 } \right) \minus{} \left( {1 \minus{} x_1  \minus{} x_2 } \right)} \right]\left( {1 \minus{} x_3 } \right)...\left( {1 \minus{} x_n } \right) \\
  \equal{} x_1 x_2 \left( {1 \minus{} x_3 } \right)...\left( {1 \minus{} x_n } \right) \ge 0 \\
 \end{array}
\]
therefor we have
\[ f\left( {x_1 ,x_2 ,...,x_n } \right) \ge f\left( {0,x_1  \plus{} x_2 ,...,x_n } \right)
\]
\[ \Rightarrow f\left( {x_1 ,x_2 ,...,x_n } \right) \ge f\left( {0,0,...,0,x_1  \plus{} x_2  \plus{} ... \plus{} x_n } \right)
\]
\[ \Leftrightarrow \left( {1 \minus{} x_1 } \right)\left( {1 \minus{} x_2 } \right)\left( {1 \minus{} x_3 } \right)...\left( {1 \minus{} x_n } \right) \ge 1 \minus{} \left( {x_1  \plus{} x_2  \plus{} ... \plus{} x_n } \right)
\]

Another solution is note that for $ n\equal{}2$ then we have
\[ \left( {1 \minus{} x_1 } \right)\left( {1 \minus{} x_2 } \right) \ge 1 \minus{} \left( {x_1  \plus{} x_2 } \right)
\]
\[ \Leftrightarrow x_1 x_2  \ge 0
\]
Now we suppose that we have
\[ \left( {1 \minus{} x_1 } \right)\left( {1 \minus{} x_2 } \right)\left( {1 \minus{} x_3 } \right)...\left( {1 \minus{} x_n } \right) \ge 1 \minus{} \left( {x_1  \plus{} x_2  \plus{} ... \plus{} x_n } \right)
\]
then we by apply the case $ n\equal{}2$ and $ n\equal{}n$ then we got

\[ \left( {1 \minus{} x_1 } \right)\left( {1 \minus{} x_2 } \right)\left( {1 \minus{} x_3 } \right)...\left( {1 \minus{} x_n } \right)\left( {1 \minus{} x_{n \plus{} 1} } \right) \ge \left[ {1 \minus{} \left( {x_1  \plus{} x_2  \plus{} ... \plus{} x_n } \right)} \right]\left( {1 \minus{} x_{n \plus{} 1} } \right)
\]
\[ \left[ {1 \minus{} \left( {x_1  \plus{} x_2  \plus{} ... \plus{} x_n } \right)} \right]\left( {1 \minus{} x_{n \plus{} 1} } \right) \ge 1 \minus{} \left( {x_1  \plus{} x_2  \plus{} ... \plus{} x_n  \plus{} x_{n \plus{} 1} } \right)
\]
It's proved.


If $ x,y,z\in N$ and $ xy\equal{}z^{2}\plus{}1$, prove that there exist integers $ a,b,c,d$ such that $ x\equal{}a^{2}\plus{}b^{2},y\equal{}c^{2}\plus{}d^{2},z\equal{}ac\plus{}bd$.

[hide="Using Gaussian Integers"]If $ \minus{}1$ is a quadratic residue mod $ xy$, then $ \minus{}1$ is a quadratic residue mod $ x$ and mod $ y$.  Then $ x$ and $ y$ can be written as sums of squares.  It now suffices to show (after some algebra) this can be done in such a way such that
$ x\equal{}a^{2}\plus{}b^{2}, y\equal{}c^{2}\plus{}d^{2}$
$ ad\minus{}bc\equal{}\pm1$

However, the given statement tells us moreover that (by factorization of the first equation in $ \mathbb{Z}[i]$)

$ \exists m, n\in\mathbb{Z}[i]$ such that $ N(m)\equal{}x, N(n)\equal{}y, mn\equal{}z\plus{}i$.
Let $ m\equal{}a\minus{}bi$, $ n\equal{}c\plus{}di$.  Then $ z\plus{}i\equal{}(ac\plus{}bd)\plus{}(ad\minus{}bc)i\implies ad\minus{}bc\equal{}1$, as desired.[/hide]

The problem almost certainly requires at least the theory of quadratic residues.  There are other, more general ways of solving quadratic Diophantines, but they are more complicated.


Let be ABC a triangle such that  $\angle A = 40\textdegree$ and $\angle C = 20\textdegree$. Let be point D on AC such that $AD = BC$. Find $\angle CBD$.

a trigonometry solution
from $AD=BC$ we have $\frac{AD}{BD}=\frac{BC}{BD}$. Let $\angle ADB =x$ from Sine Theoreme $\frac{AD}{BD}=\frac{sin(x+40)}{sin40}$ and $\frac{BC}{BD}=\frac{sinx}{sin20}$. We obtain $sin(x+40)=2sinxcos20$

$sin(x+40)=sin(x+20)+sin(x-20)$
$sin(x+40)-sin(x-20)=sin(x+20)$
$2sin30cos(x+10)=sin(x+20)$ $cos(x+10)=cos(70-x)$ ,$x=30$, $\angle DBC=30-20=10$

And geometrical: let the point $L$ on $AC$: $CL=BC$. Then $\angle ALB=100^\circ$ and $AL=BL$. But $AL=AC-CL=AC-BC=AC-AD=CD$. Consider the point $K$: $\triangle CKD=\triangle BCL$ and $\angle KCB=60^\circ$. Then
the triangle $BCK$ is equilateral. Thus $KB=KD$ and $\angle BKD=40^\circ$ and $\angle DBC=\angle DBK-\angle CBK=70^\circ-60^\circ=10^\circ$.

Point E is on CA such that $EB=BC$, also we have $EA=AB$.
Over EA we construct an equilateral triangle EFA.
$EA=EF=AF=AB$. Then it's easy to see angle $\angle FBA = 50^\circ$, so $\angle FBE=30^\circ$

Then $EF=AB; EB=AD, \angle FEB=\angle BAD=40^\circ$ so the triangles are congruents and $\angle FBE=\angle BDA=30^\circ$

$\angle DBC + 20^\circ= 30^\circ$
$\angle DBC= 10^\circ$


If $ a,b,c \ge 0$ then prove

$ \frac{a^3}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}ca\plus{}a^2} \ge \frac{a\plus{}b\plus{}c}{3}$.

$ \frac{a^3\plus{}b^3}{a^2\plus{}ab\plus{}b^2} \ge \frac{a\plus{}b}{3} \Leftrightarrow 2(a\plus{}b)(a\minus{}b)^2 \ge 0$.
Sum them up and note that $ \sum \frac{a^3}{a^2\plus{}ab\plus{}b^2} \equal{} \frac{b^3}{a^2\plus{}ab\plus{}b^2}$ because their difference is $ a\minus{}b\plus{}b\minus{}c\plus{}c\minus{}a \equal{} 0$

or we can prove the following inequality
$ \frac{a^3}{a^2\plus{}ab\plus{}b^2} \geq \frac{2a\minus{}b}{3}$
in other words , $ a^3\plus{}b^3 \geq a^2b\plus{}ab^2$ which is true
make 2 other similar and sum them up

[quote="soruz"]If $ a,b,c \ge 0$ then prove

$ \frac {a^3}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^3}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^3}{c^2 \plus{} ca \plus{} a^2} \ge \frac {a \plus{} b \plus{} c}{3}$.[/quote]

From Cauchy, $ \sum\frac{a^3}{a^2\plus{}ab\plus{}b^2}\ge \frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^3\plus{}a^2b\plus{}ab^2)}\equal{}\frac{(a^2\plus{}b^2\plus{}c^2)^2} {(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}\ge \frac{a\plus{}b\plus{}c}{3}$


Το γινόμενο ισούται με την πιθανότητα να έχουμε ακριβώς 50 "κορώνες" αν ρίξουμε ένα νόμισμα 100 φορές. Είναι γνωστό ότι αυτό είναι 8% περίπου (βλ. π.χ. Α' τόμο του Feller).

Lance

Κι εγώ έτσι το σκέφτηκα..
Απλώς μου άρεσε και είπα να την γράψω... :lol:

δες και αυτο, σαν λημμα (ειναι ουσιαστικα η λυση) http://www.mathlinks.ro/Forum/viewtopic.php?t=128779

Poli wraia,

Mia omws kai anaferthike i pithanothewritiki apodeiksi vriskw aformi na anaferw oti genika ean exete ena peirama sto opoio iparxoun dio endexomena ta opoia xaraktirizontai ws "epitixia" (me pithanotita na simvei $p$) kai "apotixia" (me pithanotita na simvei $1-p$), tote i pithanotita stis $n$ dokimes tou peiramatos na paroume akrivws $k$ "epitixies" (opote $n-k$ apotixies) einai
\[B(n,p,k)={n\choose k}p^{k}(1-p)^{n-k}\]
Prokeitai gia tis legomenes dokimes Bernoulli (ipirxan palia stin eksetastea ili tis 1hs Desmis).

Alexandros

[color=red]EDIT:[/color] Kwsta eksairetiki i apodeiksi stin opoia parapempeis!!


AMC
AIME
ARML
USAMTS
Mandelbrot Competition

I'm looking for as many as possible math contests.
The only ones i know is amc, aime, and usamo.
If anyone knows other contests post it up please.

- by the way i live in New Jersey- So no state contests other than
NJ.

New Jersey Math League (NJML)
American Scholastic Mathematics Association (ASMA) Contests
Mandelbrot Competition
Atlantic-Pacific (AT-PAC) Contests
American Mathematics Teachers of New Jersery (AMTNJ) Contest
Tournament of the Towns

errr.. dats all i can think of at the moment

where in NJ do you live? I'm in Bergen County... JW.. =)

oh.. and of course theres Harvard/MIT Math Tournament and ARML..

thanx
do u know if i can take any one of them any soon?
i'm from Bergen county, teaneck.

oh cool.. im in ramsey

ummm.. NJML, ASMA, Mandelbrot all happen every couple of months.. im not sure when the next one is.. but i'll be sure to tell you when i find out.. i'll ask my coach tomorrow

Tournament of the Towns is a four hour essay proof test that is held every couple of months too.. i'll update you on that by tomorrow

the AMTNJ is annual, and that has already past, (it was in the winter)

Harvard/MIT is at the end of every february.. (very fun)
and ARML is coming up on June 4-5th at PennState.. find yourself a team and go =) thats really fun

Here are some more, all online.

Cody Bowl - [url]http://www.codybowl.com[/url]
Ole Miss POW - [url]http://www.olemiss.edu/mathed/contest/[/url]
USAMTS - [url]http://www.nsa.gov/usamts/[/url]

Since this seems like a common question, here is a [url=http://goodlymaths.0catch.com/Contests/directory.htm]website[/url] that lists a lot of competitions.

aiighty... NJML and AT-PAC were yesterday.. so you missed that.. (so did i actually) but you can ask around and see if you can still take it...

You have to sign up for those contests at the beginning of the school year, anyhow.


[quote="Differ"]I am sorry; I do not understand the question. Could you rephrase it?[/quote]

I think the OP asked to show that the set of all $ 120$ five digits numbers made of $ 1,2,3,4,5$ in any order may be split in two subsets (one with $ n$ elements, the other with the $ 120\minus{}n$ others) such that the sum of squares of numbers in each of these two subsets are equal.

[quote="mathservant"]Using  1,2,3,4,5  numbers  only   one time  obtain five digit numbers .Show that  all the  five digit numbers  can be divide two  groups  that sum  of  the  square  of  the  numbers  in  two  groups  equal.[/quote]

$ \begin{cases}(\overline{abc})^2\plus{}(\overline{bca})^2\plus{}(\overline{cab})^2\equal{}(\overline{acb})^2\plus{}(\overline{bac})^2\plus{}(\overline{cba})^2 \\ \overline{abc}\plus{}\overline{bca}\plus{}\overline{cab}\equal{}\overline{acb}\plus{}\overline{bac}\plus{}\overline{cba} \end{cases}$ for any $ a,b,c$

$ \Rightarrow (\overline{deabc})^2\plus{}(\overline{debca})^2\plus{}(\overline{decab})^2\equal{}(\overline{deacb})^2\plus{}(\overline{debac})^2\plus{}(\overline{decba})^2$ for any $ a,b,c,d,e$.

So for any subset $ A_{(d,e)}$ which contains all such kind of 5-digit numbers of the form $ \overline{dek_1k_2k_3}$, we can always divide elements of $ A_{(d,e)}$ in two groups $ B_{(d,e)}$, $ C_{(d,e)}$ such that sum  of  the  squares  of  the  numbers  in  two group  equal, done.

[quote="stephencheng"][quote="mathservant"]Using  1,2,3,4,5  numbers  only   one time  obtain five digit numbers .Show that  all the  five digit numbers  can be divide two  groups  that sum  of  the  square  of  the  numbers  in  two  groups  equal.[/quote]

$ \begin{cases}(\overline{abc})^2 \plus{} (\overline{bca})^2 \plus{} (\overline{cab})^2 \equal{} (\overline{acb})^2 \plus{} (\overline{bac})^2 \plus{} (\overline{cba})^2 \\
\overline{abc} \plus{} \overline{bca} \plus{} \overline{cab} \equal{} \overline{acb} \plus{} \overline{bac} \plus{} \overline{cba} \end{cases}$ for any $ a,b,c$

$ \Rightarrow (\overline{deabc})^2 \plus{} (\overline{debca})^2 \plus{} (\overline{decab})^2 \equal{} (\overline{deacb})^2 \plus{} (\overline{debac})^2 \plus{} (\overline{decba})^2$ for any $ a,b,c,d,e$.

So for any subset $ A_{(d,e)}$ which contains all such kind of 5-digit numbers of the form $ \overline{dek_1k_2k_3}$, we can always divide elements of $ A_{(d,e)}$ in two groups $ B_{(d,e)}$, $ C_{(d,e)}$ such that sum  of  the  squares  of  the  numbers  in  two group  equal, done.[/quote]

Nice :)

This is a problem from PEN, and here you cand find a solution

http://projectpen.wordpress.com/page/9/

What is the  meaning  of  up lines?(_____)Thanks.

$ \overline{abc}$ denotes $ 100a \plus{} 10b \plus{} c$. For instance, $ \overline{123} \equal{} 123$. This is not stupid because $ abc$ could be understood as the product $ a.b.c$.


vector
linear algebra
matrix
geometry
absolute value

Find, if exists, the value of $m \in \Re$ so that the vectors $u=\langle1,1,-1\rangle$, $v=\langle2,-1,1\rangle$ and $w=\langle m,-1,m\rangle$ are in the same plane.

You're asking for these three vectors to be linearly dependent. Let's do this by putting a matrix into row echelon form:

$\begin{bmatrix}1&2&m\\1&-1&-1\\-1&1&m\end{bmatrix}\longrightarrow \begin{bmatrix}1&2&m\\0&-3&-1-m\\0&3&2m\end{bmatrix}\longrightarrow \begin{bmatrix}1&2&m\\0&-3&-1-m\\0&0&-1+m\end{bmatrix}$

If $m=1$ the matrix is singular and the three original vectors are dependent and thus lie in a plane. For any other value of $m,$ the vectors are independent and do not lie in a plane.

Does $w$ have to be of the form $\alpha u+\beta v$ to lie in the same plane. Because I also get to $m=1$. ??

[quote="NaZAf"]Does $w$ have to be of the form $\alpha u+\beta v$ to lie in the same plane.[/quote]
In this case, yes - because $u$ and $v$ are not linearly dependent themselves. In general, no. The vectors $u=\langle 1,-2,3\rangle$, $v=\langle-2,4,-6\rangle$, and $w=\langle 1,0,1\rangle$ lie in the same plane, but $w$ is not of the form $\alpha u+\beta v$.

Try using $\text{\TeX}$ commands \langle and \rangle for angle brackets.

[color=blue][I just edited NaZAf's post so he can see what \langle and \rangle look like - Moderator.][/color]

Also, if you compute the volume of the parallelpiped formed by these three vectors, the volume must be zero. (In order for the three vectors to lie in the same plane):

$w \cdot (u \times v)=\begin{bmatrix}m&-1&m\\1&1&-1\\2&-1&1\end{bmatrix}=$
$m(1-1)+1(1+2)+m(-1-2)=0 \Longleftrightarrow 3-3m=0 \Longleftrightarrow \boxed{m=1}$

NaZAf, you just computed the determinant of the same $3\times 3$ matrix that I was using. (Well, a permuation of the transpose of the same matrix - it is really the same thing.) The columns (or rows) of a square matrix are linearly dependent if and only if the determinant is zero. I used row elimination to row echelon form, but in fact I did exactly the work needed to find that determinant.

[hide="No longer relevant"]Except for the typo.

If you follow my work, you'll see that I computed a determinant of $3(1-m).$ You computed a determinant of $2(1-m).$ As an exercise in de-bugging: can you find your mistake?[/hide]

[quote="Kent Merryfield"]NaZAf, you just computed the determinant of the same $3\times 3$ matrix that I was using. (Well, a permuation of the transpose of the same matrix - it is really the same thing.) The columns (or rows) of a square matrix are linearly dependent if and only if the determinant is zero. I used row elimination to row echelon form, but in fact I did exactly the work needed to find that determinant.

Except for the typo.

If you follow my work, you'll see that I computed a determinant of $3(1-m).$ You computed a determinant of $2(1-m).$ As an exercise in de-bugging: can you find your mistake?[/quote]

I am sorry. I typed $1$ instead of $2$ in the matrix. I have just edited the matrix.

You were thinking in terms of that 3-dimensional peculiarity, the "vector triple product" $u\cdot (v\times w).$

A better way to encode and remember that knowlege:

Suppose we have $n$ vectors $\{v_{1},v_{2},\dots,v_{n}\}$ in $\mathbb{R}^{n}.$ These vectors determine a "parallelopiped" which consists of everything of the form $\sum_{k=1}^{n}t_{k}v_{k}$ where $0\le t_{k}\le 1$ for each $k.$ That parallelopiped has an $n$-dimensional measure, where 2-dimensional measure is area in the plane and 3-dimensional measure is volume in space.

Form those vectors into an $n\times n$ matrix. (I would by habit make them column vectors; coming from the "triple product" side you're more likely to make them row vectors at first. It doesn't really matter which.) The absolute value of the the determinant of this matrix is the $n$-dimensional measure of the parallelopiped.

The $\pm$ sign of the determinant indicates orientation. A plus sign in 2 dimensions means we've written the two vectors is counterclockwise order; a plus sign in three dimensions indicates that we've written the three vectors in right-handed order. If you permute the order, you may change the sign of the determinant but not the absolute value, which is the "volume."

You and I both relied on the observation that this "volume" is zero if and only if the set of vectors is linearly dependent, which is in turn equivalent to all of the vectors lying in some lower-dimensional subspace ([i]e.g[/i], a plane in $\mathbb{R}^{3}.$)

[This is all related to what that Jacobian determinant is doing in the change of variables formula for multiple integration.]


For simplicity, let's assume that there are two kinds of drivers. The safe drivers, who are 70% of the population, have probability $ 0.1$ of causing an accident in a year. The rest of the population are accident makers, who have a probablity $ 0.5$ of causing an accident in a year. The insurance premium is 400 dollars multiplied by one's probablity of causing an accident in the following year. A new subscriber has an accident during the first year. What should be his insurance premium for the next year?

This is essentially a Bayes formula computation.

The probability (based on the information we have) that this is a safe driver is $ \frac{(.1)(.7)}{(.1)(.7)\plus{}(.5)(.3)}\equal{}\frac{7}{22}.$ The probability that this is an accident maker is the complement of that, namely $ \frac{15}{22}.$

From that, we estimate the probability that this driver causes an accident in a year is

$ (.1)\,\frac{7}{22}\plus{}(.5)\,\frac{15}{22}\equal{}\frac{41}{110}\approxeq .3727$

Multiply that by $ \$400$ to get a premium of $ \$149.09.$


I don't really use a picture because I almost never have time. I pay too much attention to detail, so I normally just rewrite the picture algebraically and solve for the desired value analytically.

Suggestions?

it depends on a problem..
but mostly, i do it freehand.
unless it's something complicated...
like for triangles, i sometimes draw them out freehand, i draw them with a straightedge, or i don't draw them at all.
it depends on the type of person (visual learner, etc),
level of difficulty of problem (like, "what's the area of a triangle bounded by lines y=x+1, the x axis, and y=-2x-3?" and "what is the area of a triagle with base 5 and height 7?")

soo there are a lot of factors.

The geometers we have at school are useful (they're like a collection of tracing utencils with geo stuff attached) for drawing accurate figures.

if you want exact detail, you need something that will do it for you

Well more recently problems out of AOPS II and problems with a difficulty level similar the geometry problems in the intermediate section.

For Olympiad problems I find drawing a large, clean, accurate diagram helps a lot, because then I can effectively make conjectures...

But it seems the kinds of geometry problems you guys are talking about are more analytical...


calculus
integration
function
logarithms
limit
calculus computations

For $ p > 0$ let $ f_p(x) \equal{} x^{ \minus{} 1}(\log{x})^{ \minus{} p}$
a)  Given $ p > 0$ and $ \epsilon > 0$, show that $ x^{ \minus{} 1 \minus{} \epsilon} < f_p(x) < x^{ \minus{} 1}$ for sufficiently large $ x$.

b)  for which $ p$ does $ \int_{2}^{\infty}f_p(x)dx$ converge?

a)  Follows from the fact that $ \lim_{x\to\infty} \frac {\log x}{x^{\alpha}} \equal{} 0$ for any $ \alpha > 0$.

b)  For $ p\not \equal{} 1$, $ \int f_p(x)dx \equal{} \minus{} \frac {1}{(p \minus{} 1)(\log x)^{p \minus{} 1}}$.

What happens when $ p \equal{} 1$ (that is, what is $ \int \frac {dx}{x\log x}$?)

Or you could just use the [url=http://en.wikipedia.org/wiki/Integral_test_for_convergence]Integral Test[/url] combined with the [url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Condensation Test[/url].


inequalities integration calculus induction

Suppose $\{ x_n \}_{n\geq 1}$ is a sequence of positive real numbers such that $x_1 \geq x_2 \geq x_3 \ldots \geq x_n \ldots$, and for all $n$ \[ \frac{x_1}{1} + \frac{x_4}{2} + \frac{x_9}{3} + \ldots + \frac{x_{n^2}}{n} \leq 1 .  \] Show that for all $k$ \[ \frac{x_1}{1} + \frac{x_2}{2} +\ldots + \frac{x_k}{k} \leq 3.  \]

Regrouping terms and using the fact that $u_n$ is decreasing it suffices to prove that

$\sum_{n^2}^{(n+1)^2-1} 1/k \leq 3/n$

Comparing with $\int 1/t dt$ we get $\sum_{n^2}^{(n+1)^2-1} 1/k \leq ln (1+ 2n/((n^2-1))$

Then we use $ln(1+x) \leq x$ and we get our result.

[quote="Rijul saini"][quote="tµtµ"]Regrouping terms and using the fact that $ u_n$ is decreasing it suffices to prove that

$ \sum_{n^2}^{(n \plus{} 1)^2 \minus{} 1} 1/k \leq 3/n$

Comparing with $ \int 1/t dt$ we get $ \sum_{n^2}^{(n \plus{} 1)^2 \minus{} 1} 1/k \leq ln (1 \plus{} 2n/((n^2 \minus{} 1))$

Then we use $ ln(1 \plus{} x) \leq x$ and we get our result.[/quote]

[b]Hey isnt there any better solution which DOES NOT use calculus as it spoils an inequality......and is not easily understandable to 9th graders like me.....[/b][/quote]

Yeah, I actually had logged in now to post this problem, I too do not know ANYTHING about Calculus! :o ( I am going to be in 9th standard after my exams end). :(

[hide="hint only"]Try to use induction to establish the inequality. It is easy as it is only from RMO.[/hide]


abstract algebra
superior algebra
superior algebra unsolved

Given $ g_1,g_2,...,g_n$ are representatives of the conjugacy classes of a finite group $ G$. Show that if the $ g_i$ commute then the group is abelian. I like this problem we discussed it in class last week.

Since the conjugacy class of $ g_i$ has cardinality $ K_i \equal{} \left[ G : C_G \left( g_i \right) \right]$ and $ C_G \left( g_i \right) \geq n$, it follows that $ K_i \leq \frac {m}{n}$, where $ m \equal{} |G|$.
Summing over $ i, \, m \equal{} \sum K_i \leq m$, so $ K_i \equal{} \frac {m}{n}, \, \forall i \in \overline{1,n}$.
Now, WLOG $ g_n \equal{} e$, so that $ K_n \equal{} 1$. That's it.


Show that for real numbers $x_1, x_2, ... , x_n$ we have:
\[ \sum\limits_{i=1}^n \sum\limits_{j=1}^n \dfrac{x_ix_j}{i+j} \geq 0  \]
When do we have equality?

Generalizition  :wink: :

Let $p_1 , p_2 ,... , p_n , c$ be positive real numbers , then

$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\frac{x_{i}x_{j}}{(p_i +p_j)^c} \geq 0 $

Also see [url=http://mathworld.wolfram.com/HilbertsInequality.html]Hilbert`s Inequality[/url]

[quote="Forward"]Define function $\displaystyle f(y)= \sum_{i=1}^n \sum_{j=1}^n \dfrac{x_ix_jy^{i+j}}{i+j}$
Then,$f(0)=0$ and $\displaystyle f(1)=\sum_{i=1}^n \sum_{j=1}^n \dfrac{x_ix_j}{i+j}$
Since $\displaystyle f'(y)=\sum_{i=1}^n \sum_{j=1}^n x_ix_jy^{i+j-1}=\dfrac{(\sum_{i=1}^n x_iy^i)^2}{y} \geq 0$
We have $f(1) \geq f(0)=0$
Thus the Ineq we need is proven.[/quote]

nice prove!!

define$\ F(x)= (\sum_{i=1}^{n}a_{i}x^{i-\frac{1}{2}})(\sum_{j=1}^{n}a_{i}x^{i-\frac{1}{2}}) $ for $ 0<x \leq 1 $
$\ F(x)=g(x)^{2} $
integrate F(x) between 0 and 1 to get \[ \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\frac{x_{i}x_{j}}{i+j}\]
but $ \ F(x)=g(x)^{2}$
so it is $ \geq 0$
also equality occurs iff g(x) is identically 0.
so equality iff all $ \ a_{i}$ s are =0

For the generalized statement, see [url=http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A493&l=en]Problem A.493. of KöMaL[/url] ;-)

[quote=sumanguha]There are some notational mistakes in the proof which I have written years ago. Let me correct it. We define
$g(x) = \sum_{i=1}^{n} a_{i} x^{i - \frac{1}{2}}$.
Hence,  $F(x) = (g(x))^{2} = (\sum_{i=1}^{n} a_{i} x^{i - \frac{1}{2}}) (\sum_{j=1}^{n} a_{j} x^{i- \frac{1}{2}}) = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j}x^{i + j - 1}$.
So, $0 \leq \int_{0}^{1}  (g(x))^{2}  dx = \int_{0}^{1} \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j}x^{i + j - 1} dx = \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{a_{i} a_{j}}{i+j}$.

Equality holds [b]iff [/b] $g(x) \equiv 0$ for $0<x\leq 1$, which holds  [b]iff[/b] $a_{i} = 0 \forall i$.[/quote]

12 year bump to fix old mistakes orz

Is there a proof that doesn’t use derivatives and integration?


Let $ f(x) = x^{3}-3x+1$, where $ x$ is real. Find the number of distinct real roots of the equation $ f(f(x)) = 0$.


geometry
geometric transformation
reflection
group theory
abstract algebra
topology

classify all the 2-folded covering space of the Klein bottle ?

I cannot use the idea which is similar to the classification of covering space of the figure eight in the Hatcher's book, it seems to me that I need some thing ... could you guy give me some ideas? thanks

The universal cover is $ \mathbb{R}^2$, with a group $ G$ of symmetries described [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172547]here[/url]. A presentation of that group: generators $ X,Y$, relation $ XY\equal{}YX^{\minus{}1}$. ($ X$ is translation in $ X$ and $ Y$ is translation in $ Y$ together with reflection in $ X$)
What do the two-sheeted covers come from? They must come from index-2 subgroups of $ G$, and those subgroups must contain the subset of squares $ G^2$. $ G^2$ is abelian, generated by $ X^2$ and $ Y^2$, and has index $ 4$.
As it happens, there are three subgroups of index $ 2$, generated by the pairs $ (Y^2,X)$, $ (X^2,Y)$, and $ (X^2,XY)$. The first is a covering by a torus. The second and third are both coverings by Klein bottles.

Dear Jmerry,

If I use tha fact that if $ X$ is 2-fold covering space of the Klein bottle $ K$, then I have $ \chi(X) \equal{} 2\chi(K) \equal{} 0$

if $ X$ is orientable then $ \chi(X) \equal{} 2 \minus{} 2g \equal{} 0$ so $ g \equal{} 1$ then $ X$ must be a Torus
if $ X$ is non-orientable then $ \chi(X) \equal{} 2 \minus{} k \equal{} 0$  then $ X \equal{} \mathbb{R}P^2$ # $ \mathbb{R}P^2\equal{}K$
so I have only two connected- covering spaces,

and I still don't get your ideas : why the groups generated $ (Y^2,X)$, $ (X^2,Y)$, and $ (X^2,XY)$  corresponds to a covering by a torus,  and  Klein bottles respectively, could you please explain for me why ? thanks

All of those covering spaces can be described as $ \mathbb{R}^2/H$, where $ H$ is the associated group of symmetries. One of those groups is free abelian on two generators, and the other two are isomorphic to the group for the Klein bottle.
Draw the pictures. Look at a fundamental region, and what we can do to it.


logarithms
limit
function
calculus
real analysis
real analysis unsolved

Prove that ([u]through a elementary method[/u]) the sequence $a_n=\frac{\ln n}{n},\ n\in N^*$ is convergently and $\lim_{n\to \infty} a_n=0.$

suppose n=e^u for some u, we also notice that $\lim n \rightarrow \infty \implies \lim u \rightarrow \infty$

so $\lim_{u \rightarrow \infty} \frac{u}{e^u}$ and obviously exponential functions with a base>1 increase faster as we approach infinity, so it approaches 0

...i don't really think that is that rigirous, but my assumptions have been proven...

Consider $f(x)=x-ln(x)$. One can prove that $f$ is positive for all $x\geq 1$, ands so we have that our sequence ${a_n}$ is bounded and we can prove that it´s monotonic and so convergent. But we have $a_{2n}=\frac{ln(2n)}{2n}$ and so $a_{2n}=\frac{ln2}{2n}+\frac{1}{2}a_n$ and making $n\to +\infty$ we´ll have $L=\frac{L}{2}$ and so $L=0$, where $L=\lim_{n\to +\infty}a_n$

How show you that the sequence is monotonic and bounded ? The rest it is nicely and O.K.

It´s just you see the function $f$ that I said. You have to use a little calculus, of course.

[quote="Arne"]Could a moderator move this to the right forum?[/quote]

Yep...I just regained my access to the internet (was away for a few days) :D

[color=red]Moved from Intermediate.[/color]

Let $f(x)=\sqrt{x}-\ln x\ (x>0).$ We have $f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}=\frac{x-4}{2x(\sqrt{x}+2)}.$

Thus $f(x)$ is mimnimal and minimum at $x=4,$ so $f(x)\geqq f(4)=2-\ln 4=2(1-\ln 2)>0\ (\because e>2).$

From this, for $x>1,$ we obtain ${0<\frac{\ln x}{x}< \frac{1}{\sqrt{x}}.\ \therefore \lim_{x\to\infty} \frac{\ln x}{x}}=0.$


calculus
integration
logarithms
limit
calculus computations

Regrettably your answer is incorrect, but you could almost solve it! :)

kunny

Sorry , just a small mistake.. :D
I edited my post.


[quote="asterday1"]Does anybody know what the PA cutoff was last year?
On chapters I got a score of *BEEP*
...
a score of *BEEP*
.......
*BEEP* *BEEP* *BEEP* *BEEP* *BEEP*
I'll tell you when I'm allowed.
I want to see if I have a chance.
[size=59][color=red]Stupid website and their rules about posting scores grumble grumble...[/color][/size][/quote]

Perhaps post it in the PA forum? :P

[quote="asterday1"]Stupid website and their rules about posting scores grumble grumble...[/quote]

I would just like to comment on this.

The website isn't stupid, nor is this a rule that AoPS cooked up. It's a MATHCOUNTS rule, because if people say stuff like "I got a [insert low score here] and got [insert high place here]", or people just know that person is smart/not so smart, etc., one can determine with pretty good accuracy the difficultly level of the test.

I know it's not a stupid rule.
I was just having fun.

For a purely indicative purpose, I'd ask it for last year as well.


Does anyone know of any summer programs that end by June 26, and that teacher math/CS/Physics/Astronomy/Something along those lines?

The closest I can find quickly is
[url]http://www.math.utah.edu/hsp/[/url]
but as you'll notice, it ends June 30.

I would imagine that applications would be closed (or about to be!) for programs beginning and ending in June.  Because many high schools don't let out for the summer until the beginning of June, few programs would be scheduled then.

A useful list of mathematics summer programs can be found at
[url]http://www.ams.org/employment/mathcamps.html[/url]
and some of these programs do sciences as well.

Either way, the program that smbelcas provided says on its website that preference will be given to the applications submitted by April 1st, 2005....

yea.. they would have to collect the applications before anybody else in order to select poeple... and get the camp going


geometry
trapezoid
trigonometry
geometry proposed

Given isoceles trapezium $ABCD,AB//CD$ let $AC=BD=d,S_{ABD}=S_{ABC}=S_{1},S_{ACD}=S_{BCD}=S_{2}$ prove that
$\frac1{S_{1}}+\frac1{S_{2}}\ge\frac8{d^{2}}$

Draw perpendiculars of AC from B and D and the feet are E and F respectively.
Let $DF=h_{1}, BE=h_{2}, \angle DBE=\angle BDF=\theta$
Then $d=\frac{h_{1}+h_{2}}{\cos{\theta}}, 2S_{2}=dh_{1}, 2S_{1}=dh_{2}$
Then it's to prove $\frac{1}{h_{1}}+\frac{1}{h_{2}}\geq \frac{\cos{\theta}}{h_{1}+h_{2}}\iff 4\geq 4\cos{\theta}$
The equality holds when ABCD is a square :)


A friend proposed this problem to me, i wonder how you guys will approach it.

A octagonal pyramid has curved side lengths, find the octagonal pyramids volume.

Radius = 20cms
Heigth from middle of pyramid to highest point in middle (at top) = 20cms

Edit : Assume the arcs are like that of a circles.

Does middle pyramid mean middle of the base or simply middle of the pyramid as in somewhere suspended?

it means the base. so if i had one of the pyramids it would be touching the desk or something.

curved side lengths, what are those?  :?

the side lengths, are curved... so its like an arc instead of a straight line to the top point were they all meet.


[quote="wxydx00"]prove that:

$\frac{a+b+\sqrt{ab}}{3}\leq\sqrt{(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{2})^{3}}$[/quote]
$a,b>0$

Messy, but I think it works.
Square and compare.

[quote="minsoens"][quote="wxydx00"]prove that:

$\frac{a+b+\sqrt{ab}}{3}\leq\sqrt{(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{2})^{3}}$[/quote]isn't it possible to do
$\sqrt{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{2}\right)^{3}}\geq \frac{a+b}{2}$ by Power Mean Inequality
$\frac{a+b}{2}\geq \frac{a+b+\sqrt{ab}}{3}\Longrightarrow a+b\geq 2\sqrt{ab}$[/quote]

I think it should be $\frac{1}{2}\sqrt{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{\frac{2}{3}}\right)^{3}}\geq \frac{a+b}{2}$ by Power Mean Inequality
not  $\sqrt{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{2}\right)^{3}}\geq \frac{a+b}{2}$ by Power Mean Inequality

IT is posted by ZUMING FENG, so I don't think it is a wrong problem..

My math is very poor, so if I am wrong, Point it to me clearly.OK?

[quote="wxydx00"]I think it should be $\frac{1}{2}\sqrt{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{\frac{2}{3}}\right)^{3}}\geq \frac{a+b}{2}$ by Power Mean Inequality[/quote]
It is true, however not straightforward by the power mean inequality. Its particular case you are probably talking about is $\sqrt{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{2}\right)^{3}}\leqslant \frac{a+b}{2}$.

The problem is true, but ridiculously sharp.
[hide]
So if we let $a=x^{6},b=y^{6}$, square, and expand, it reduces to
\[(x-y)^{4}((x^{4}-y^{4})^{2}+4xy(x^{6}+y^{6})+10x^{2}y^{2}(x^{4}+y^{4})+4x^{3}y^{3}(x^{2}+y^{2}))\geq 0\]
if i didn't make a stupid mistake  :whistling:

[i](The fourth power makes it very difficult to deal with any other way.)[/i]
[/hide]

[quote="Rzeszut"][quote="minsoens"]$\sqrt{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}}{2}\right)^{3}}\geq \frac{a+b}{2}$ by Power Mean Inequality[/quote]
It is wrong. The sign is flipped.[/quote]oh not as easy as i thought. i am humbled :|


Hey guys, I have been working on this problem, but I cannot think of the solution.

The question is

Take an angle between 2 rays L1,and L2 with vertex O. Take a point A inside the angle. Consider all triangles with vertex O two sides of which belong to L1 and L2 and third side of which pass through the point A. Prove that the area of the triangle will be the smallest if the point A is a middle of the third side.

What I was thinking is  that the location of point A is independent on the size of the triangle as long as A's are on the same line because they all have the same height and base.
Probably I am thinking in wrong direction.

Please help me out sloving this one.
Thank you so much in advance.

See [url]http://www.mathlinks.ro/viewtopic.php?t=194335[/url] and follow the links.


complex analysis
complex analysis theorems

if f is map from D = C\{z_1, z_2, ...} to C\[0,inf) and g is map from C\[0,inf) to D, which is possible by Riemann Mapping Theorem then h = g(f(z)) is bounded.

Hence, there is z_0 in D s.t. |h(z_0)| >= |h(z)| for all z in D.

By Maximum Principle, h is constant c.

Therefore, g(f(z)) = c, or f(z) = g^-1 (c), which is constant.

This proof will, of course, fail if C does not contain a point of infinity. Does C have a point of infinity?

[quote="ksun"]Hence, there is z_0 in D s.t. |h(z_0)| >= |h(z)| for all z in D.[/quote]
Not obvious, since $D$ is not compact.

$\mathbb C$ does not include $\infty$. This is why there is notation $\overline{\mathbb C}=\mathbb C\cup\{\infty\}$.


If $ P$ be at the maximum distance and $ Q$ be at the minimum distance of the point $ A( \minus{} 7,2)$ from the points on circumference of the circle $ x^{2} \plus{} y^{2} \minus{} 10x \minus{} 14y \minus{} 151 \equal{} 0$.Then find $ |AP\minus{}AQ|$

[hide]$ x^2 \plus{} y^2 \minus{} 10x \minus{} 14y \minus{} 151 \equal{} 0$ is equivalent to $ (x\minus{}5)^2 \plus{} (y\minus{}7)^2 \equal{} 225$.

The point lies on the circumference of the circle, so $ AQ \equal{} 0$. Therefore, $ AP$ is the length of the diameter, or $ 30$, so the desired answer is $ 30$.[/hide]

The point does not lie on the circumference.Check it,it does not satisfies the equation of circle.
[hide]
$ (x\minus{}5)^{2}\plus{}(y\minus{}7)^{2}\equal{} 225$ From here the coordinates of the centre will be $ (5,7)$.For min and max case,points$ P$ and $ Q$ must lie on the line joining $ A$ and the centre of the circle whose equation can be easily found out.Now solve for $ x$ and $ y$,since it is quadratic equation we will get two values for x and y and then use distance formula to get the answer.[/hide]