geometry
parallelogram
IMO Shortlist
homothety
incenter
exterior angle
Spiral Similarity

Here is the TST problem (notations changed):

[color=blue][b]Problem 3 of 3rd German TST 2006.[/b] Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.[/color]

[i]Note.[/i] A matter of notation: The [i]$A$-excenter[/i] of a triangle $ABC$ is defined as the center of the excircle of triangle $ABC$ which touches the interior of the side $BC$ and the extensions of the sides $CA$ and $AB$.

[color=green][See the attached graphic - click it to enlarge it.][/color]

[i]Solution.[/i] We WLOG assume that the point X lies on the extension of the segment BC beyound C, and that the point Y lies inside the segment CD. The other case can be treated analogously.

Let K' and L' be the Y-excenter and X-excenter of triangle XCY, respectively. Then, both of these points K' and L' lie on the exterior angle bisector of the angle XCY, what yields $\measuredangle XCL^{\prime}=90^{\circ}+\frac{\measuredangle XCY}{2}$ and $\measuredangle K^{\prime}CX=\measuredangle YCL^{\prime}=90^{\circ}-\frac{\measuredangle XCY}{2}$; also, the point K' lies on the exterior angle bisector of the angle CXY, what gives $\measuredangle K^{\prime}XC=90^{\circ}-\frac{\measuredangle CXY}{2}$, and the point L' lies on the exterior angle bisector of the angle CYX, leading to $\measuredangle CYL^{\prime}=90^{\circ}-\frac{\measuredangle CYX}{2}$. Thus, by the sum of angles in triangle CK'X, we get

$\measuredangle CK^{\prime}X=180^{\circ}-\measuredangle K^{\prime}XC-\measuredangle K^{\prime}CX=180^{\circ}-\left(90^{\circ}-\frac{\measuredangle CXY}{2}\right)-\left(90^{\circ}-\frac{\measuredangle XCY}{2}\right)$
$=\frac{\measuredangle CXY+\measuredangle XCY}{2}=\frac{180^{\circ}-\measuredangle CYX}{2}$          (since < CXY + < XCY = 180° - < CYX by the sum of angles in triangle XCY)
$=90^{\circ}-\frac{\measuredangle CYX}{2}$.

Thus, < CK'X = < CYL'. Combining with < K'CX = < YCL', we see that triangles K'CX and YCL' are similar.

Since ABCD is a parallelogram, we have AB || CD, so that Thales yields $\frac{XC}{XB}=\frac{XY}{XA}$. Now, the homothety with center X and ratio $\frac{XC}{XB}=\frac{XY}{XA}$ obviously maps the point B to the point C and the point A to the point Y, and leaves the point X fixed. Thus, this homothety must map the A-excenter of triangle XBA to the Y-excenter of triangle XCY; in other words, it maps the point K to the point K'. Since the center of this homothety is X and the ratio is $\frac{XY}{XA}$, it thus follows that the point K' lies on the line XK and satisfies $\frac{XK^{\prime}}{XK}=\frac{XY}{XA}$. This transforms into $\frac{KK^{\prime}}{XK}=\frac{XK^{\prime}-XK}{XK}=\frac{XK^{\prime}}{XK}-1=\frac{XY}{XA}-1=\frac{XY-XA}{XA}=\frac{AY}{XA}$. Similarly, the point L' lies on the line YL and we have $\frac{LL^{\prime}}{YL}=\frac{AX}{YA}$, so that $\frac{YL}{LL^{\prime}}=\frac{YA}{AX}$, what rewrites as $\frac{LY}{L^{\prime}L}=\frac{AY}{XA}$. Comparison with $\frac{KK^{\prime}}{XK}=\frac{AY}{XA}$ yields $\frac{KK^{\prime}}{XK}=\frac{LY}{L^{\prime}L}$. Thus, the points K and L are corresponding points in the similar triangles K'CX and YCL' (since they lie on the respective sidelines K'X and YL' and divide them in the same ratio $\frac{KK^{\prime}}{XK}=\frac{LY}{L^{\prime}L}$). Since corresponding points in similar triangles form equal angles, this implies < KCX = < LCL'. Thus,

$\measuredangle KCL=\measuredangle KCX+\measuredangle XCL=\measuredangle LCL^{\prime}+\measuredangle XCL=\measuredangle XCL^{\prime}$
$=90^{\circ}+\frac{\measuredangle XCY}{2}=90^{\circ}+\frac{180^{\circ}-\measuredangle BCD}{2}=180^{\circ}-\frac{\measuredangle BCD}{2}$.

Since this angle is independent of the line g, this solves the problem.

  Darij

[quote="madatmath"]Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.[/quote]

Another solution

1) A bit of angle chasing first $\angle BAK = \frac{\angle BAX}{2} = \frac{\angle AYD}{2} = \angle ALD$. This last equality is because $Y$, $L$, $D$ and the incenter, say $I$ of triangle $AYD$ are concyclic so $\angle IYD = \angle ILD$.
2) $\angle ADL = \angle ABK \Rightarrow \triangle ADL$ and $\triangle ABK$ similar
3) Hence $BK \cdot DL = AB \cdot AD = BC \cdot CD$ Hence $\frac{BK}{BC} = \frac{CD}{DL}$ but
4) $\angle CBK = \angle LDC$ hence $\triangle CBK$ is similar to the $\triangle CLD$. Hence the angle between lines $CK$ and $CL$ is the angle between lines $BC$ and $DL$. This last clearly do not depend on the location of point $X$ hence result.

Daniel

Changed form of this problem is [color=blue][b]Iran's TST[/b][/color],which is here [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=498848#[/url]

[hide="Without ratios"]
WLOG let $ Y$ be closer to $ A$ than $ X$ (same as in Darij's configuration).

Since $ K$ is the $ A$-excenter of $ ABX$, $ K$ lies on the external bisector of $ \angle ABC$ and the internal bisector of $ \angle BAX$. Similarly, $ L$ lies on the external bisector of $ \angle ADC$ and the internal bisector of $ \angle DAY$. For convenience, let $ \angle DAL \equal{} \alpha$ and $ \angle BAK \equal{} \beta$. Let the intersection of $ AK$ with $ DC$ be $ N$, and the intersection of $ AL$ with $ BC$ be $ M$.

Since $ \angle LDN \equal{} \frac{1}{2} \angle BAD \equal{} \angle LAK \equal{} \angle KBM$, we know that quadrilaterals $ ADLN$ and $ ABKM$ are cyclic. We know that $ \angle DNA \equal{} \angle NAB \equal{} \beta$, and $ \angle DNL \equal{} \angle DAL \equal{} \alpha$. Similarly with quadrilateral $ ABKM$, we know that $ \angle BMA \equal{} \alpha$ and $ \angle BMK \equal{} \beta$. Therefore $ \angle ANL \equal{} \angle AMK \equal{} \alpha \plus{} \beta$, so $ NKML$ is cyclic. However, $ \angle CNL \equal{} \angle LMC \equal{} \alpha$, so $ CNKML$ is cyclic. Therefore $ \angle KLC \equal{} 180 \minus{} \angle KML \equal{} 180 \minus{} \alpha \plus{} \beta \equal{} 180 \minus{} \frac{1}{2}\angle BAD$, which is constant, as desired.
[/hide]

Assume wlog that $ X$ lies between $ A$ and $ Y$. Let $ S$ be the point of intersection of the lines $ BC$ and $ AL$ and let $ T$ be the point of intersection of $ DC$ and $ AK$. Moreover, let $ \angle DAL \equal{} \angle LAX \equal{} \alpha$ and $ \angle XAK \equal{} KAB \equal{} \beta$. First, we claim that $ SLCKT$ is a cyclic quadrilateral. We have $ \alpha \plus{} \beta \equal{} \angle SBK \equal{} \angle SAK$ and $ \alpha \plus{} \beta \equal{} \angle LDT \equal{} \angle LAT$. Thus, quadrilaterals $ ATLD$ and $ ABKS$ are cyclic. A simple angle chase shows that $ \angle BKA \equal{} \alpha$ and $ \angle ALD \equal{} \beta$. Therefore, $ \angle KSL \equal{} \angle KSB \plus{} \angle BSA \equal{} \angle KAB \plus{} \angle BKA \equal{} \beta \plus{} \alpha$ and $ \angle KTL \equal{} \angle ATD \plus{} \angle DTL \equal{} \angle ALD \plus{} \angle DAL \equal{} \beta \plus{} \alpha$. Thus, quadrilateral $ SLKT$ is cyclic. Moreover, $ \angle SKT \equal{} 180 \minus{} \angle AKS \equal{} 180 \minus{} \angle ABS \equal{} 2(\alpha \plus{} \beta)$ and $ \angle SCT \equal{} \angle BCD \equal{} 2(\alpha \plus{} \beta)$. Thus, $ SLCKT$ is cylic. It follows that $ \angle KCL \equal{} 180 \minus{} \angle KTL \equal{} 180 \minus{} ( \alpha \plus{} \beta) \equal{} 180 \minus{} \angle DAB/2$ and that is clearly constant.

edit: Hmm, my solution is exactly the same as orznorz' solution…^^ it would be better to read the posts above before posting^^ orznorz even chose the same names for the angles $ \angle DAL$ and $ \angle KAB$...  :lol:

Seems too easy to be right...

By angle chasing, ABK~LDA, so AB/BK=LD/DA, also KB/BC=CD/DL and <KBC=<CDL, so KBC~CDL. Then by angle chasing again <KCL=180-<BAD/2, which is independent of g.

If you intersect CD with AK and BC with AL at S and T respectively, angle chasing yields that TKBA, SLDA, and KCTLS are all cyclic.  This gives $\angle KCL=\angle KTL=\pi - \dfrac{\angle BAD}{2}$ as desired.

you also get a bunch of similar triangles here, such as KBA ~ KCL ~ ADL.

Finally I came up with a solution with a geometric approach:(withuot any bash)

Note that $\angle{ADL}=\angle{ABK}=90+\frac{B}{2}$ and $\angle{DAL}=\angle{AKB} \Rightarrow \frac{BK}{BC}=\frac{BK}{AD}=\frac{AB}{LD}=\frac{DC}{DL}$
.Also note that $\angle{LAC}=\angle{CBK} \Rightarrow \triangle{CBK} \sim \triangle{LDC}$.
Hence $\angle{KCL}=360-\angle{KCB}-\angle{DCL}-\angle{DCB}=360-(\angle{KCB}+\angle{CKB})-\angle{A}=360-90+\frac{B}{2}-180+B=90+3\frac{B}{2}$ which is fixed,so we are done!!!

[hide]
Let $\angle BAD = \angle BCD = A,$ so $\angle ABD = \angle ADC = 180^{\circ} - A.$ Let $\angle BAY = \angle AYD = \theta . $ Since $K$ and $L$ lie on the internal angle bisectors of $\angle ABX$ and $\angle ASY$ respectively, $\angle BAK = \dfrac{\theta}{2}$ and $\angle YDL = 180^{\circ} - \left( 180^{\circ} - A + \dfrac{A}{2} \right) - \dfrac{A-\theta }{2} = \dfrac{\theta}{2},$ so $\angle BAK = \angle ALD.$ Also, $\angle ADL = 180^{\circ} - A + \dfrac{A}{2} = \angle ABK + \angle XBK = \angle ABK,$ so $\triangle ABK \sim \triangle ADL \implies \dfrac{AK}{AL} = \dfrac{BK}{AD} = \dfrac{AB}{DL}.$ Since $K$ and $L$ lie on the external bisectors of $\angle ABC$ and $\angle ADC$ respectively, $\angle CBK = \dfrac{A}{2}$ and $\angle YDL = \dfrac{A}{2},$ so $\angle CBK = \angle YDK.$ Using $AD=BC$ and $AB=CD,$ we get that $\dfrac{BK}{BC} = \dfrac{BC}{DL},$ so $\triangle BCK \sim \triangle LDC.$ Let $BC$ intersect $DL$ at $J.$ Considering the spiral transformation centered at $C$ which sends $D,L$ to $B,K$ respectively, we deduce that $\angle CKL = 180^{\circ} - \angle CJD = 180^{\circ} - \left(180^{\circ} - \dfrac{A}{2} - (180 - A) \right) = 180^{\circ} - \dfrac{A}{2},$ which is clearly independent of $\theta.$ $\blacksquare$
[/hide]

[hide=Solution]Let $\angle BAD = 2y$ and note $\angle KAL=y$ by the definition of excenters.  Let $\angle LAD=\theta$ and get $\angle KAB=y-\theta$.  Note that by excenter definitions we have $\angle KBC=\angle LDC=y$.  I claim that $KBA\sim KCL$, which is sufficient.

Let $AL\cap BC=E$ and $AK\cap CD=F$.  From $\angle KAE=\angle KBE=\angle CDL=y$ we have $KBAE$ and $AFLD$ cyclic, and by $KBAE$ and the well-known spiral similarity lemma, we have reduced the problem to showing $KCEL$'s concyclicity.

By the cyclic quadrilaterals, we have $\angle DFL=\theta$ and $\angle AEB=\theta$, so $FCEL$ is cyclic.  We also get $\angle AEK=180-\angle ABK=180-(180-y)=y$ and $\angle KFL=180-\angle ADL=y$, so $FKEL$ is also cyclic.  Thus, $FKCEL$ is a cyclic pentagon and we are done.[/hide]

[quote=ISL 2005 G3]Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.

[i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i][/quote]

Let $\ell_b$ and $\ell_d$ denote the external bisectors of $\measuredangle ADC$ and $\measuredangle ABC$, respectively. Consider the pencil of lines $\ell$ through $A$ and observe that $K, L$ move on ranges $\ell_b, \ell_d$ respectively, such that $\measuredangle KAL=\tfrac{1}{2}\measuredangle DAB$.

Let $\Gamma : \ell_b \mapsto \ell_d$ be the composition of rotation about $A$ with angle $\tfrac{1}{2}\measuredangle DAB$ followed by projection through $A$. Let $\Phi: \ell_b \mapsto \ell_d$ be the composition of rotation about $C$ with angle $180^{\circ}-\tfrac{1}{2}\measuredangle DAB$ followed by projection through $C$. Note that both $\Gamma$ and $\Phi$ provide a one-one projective map from $\ell_b$ to $\ell_d$.

Thus, in order to establish $\Gamma=\Phi$, it suffices to check this for three possibilities of line $\ell$. However, the result is immediate when line $\ell$ coincides with $\overline{AB}, \overline{AC},$ and $\overline{AD}$; so we are done! $\blacksquare$

All what you have to do is \(\Delta DLY \sim \Delta BKX\) and \(\Delta DLC \sim \Delta BCK\).

[hide=Solution][quote=ISL 2005 G3]Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.
[/quote]
[b][color=#000]Solution:[/color][/b] $$\begin{cases} \angle BXK=90^{\circ}- \tfrac{1}{2} \angle AXB =90^{\circ} -\tfrac{1}{2} \angle DAY =\angle DLY \\ \angle XBK=90^{\circ}-\tfrac{1}{2} \angle ABC=90^{\circ}-\tfrac{1}{2} \angle ADC=\angle LDY \end{cases} \implies \Delta LDY \sim \Delta XBK $$
Also, $\Delta DYA \sim \Delta CYX \implies \tfrac{DY}{CY}=\tfrac{AY}{YX}=\tfrac{BC}{CX}$ $\implies$ $BC \times CD= XB \times DY$. Also, $BK$ $\times$ $LD$ $=$ $XB$ $\times$ $DY$ $$\implies LD \times BK= XB \times DY= BC \times CD \implies \tfrac{BC}{BK}=\tfrac{DL}{DC} \implies \Delta LDC \sim \Delta CBK$$
Hence, as a result, $\angle KCL$ is constant $\qquad \blacksquare$
[/hide]

[quote=madatmath]Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.
[/quote]
Let $\ell_B,\ell_D$ be the external angle bisectors of $\angle B,\angle D$ respectively. We consider the general version of this problem where $L,K$ are defined as the intersections of the angle bisectors of $\measuredangle DAY, \measuredangle YAB$ with $\ell_D,\ell_B$ respectively. Thus, they are the excenters when $Y \in \vec{DC},$ and might correspond to incenters if $Y$ is outside the segment $CD.$

We claim that $\theta=\measuredangle KCL=\pi-\tfrac{1}{2}\angle C$ works. Fix $ABCD$ and move $L$ on $\ell_D.$ Clearly, $L \mapsto K$ is projective since $\angle KAL$ is fixed. Also, let the rotation at $C$ by angle $\theta$ counterclockwise send $L \in \ell_D$ to $K^\ast \in \ell_B.$

It suffices to check $K=K^\ast,$ for which we just need three positions of $L.$ Pick $L=D, \infty_{AD}$ and the $A$ excenter of $\triangle ACD.$ It is easy to check these cases so done. $\square$

Denote the external angle bisectors of $B$, $C$ as $\ell_B$, $\ell_D$, and define $L'$ to be the intersection of the clockwise rotation of $CK$ about $C$ by $180-\frac{\angle C}{2}$ with $\ell_B$. We wish to show that $L=L'$, which we will do with moving points. If $K$ varies with degree $1$, note that $\angle KAL$ is fixed, so $AL$ and $L$ also vary with degree $1$. As $CL'$ is just a rotation of $CK$, $\deg CL'=1\implies \deg L'=1\implies \deg(L=L')=2$.

The cases $g=AB, AC, AD$ are easily verified to work, so we are done.

@below Thanks

@above: It doesn't work since $XL$ is defined as bisector so it's "square-root" map. [hide=It's fixable though.]Note that $\angle KAL$ is fixed.[/hide]
-----
Through angle chasing, we see that $AK\perp YL$ and $AL\perp XK$. Thus let $K_1 = AK\cap YL$, $L_1=AL\cap XK$ and $P = XK\cap YL$, the $C$-excenter of $\triangle CXY$. Notice that $P$ is the orthocenter of $\triangle AKL$. The following claim will imply the problem.

[color=red][b]Claim:[/b][/color] $KCLP$ is cyclic.

[i]Proof:[/i] WLOG $P$ lie on the different side of $L$ w.r.t. $AK$.  Let $T = PC\cap KK_1$. Notice that
$$\angle XPT = 90^{\circ} - \angle XYP = \angle KAY$$
or $A,X,T,P$ are concyclic. Moreover, notice that
$$\angle XCT = 90^{\circ} - \tfrac{\angle A}{2} = \angle ALP = 180^{\circ}-\angle XKT$$
thus $X,C,K,T$ are concyclic. Hence
$$\angle KCP = \angle KXT = \angle KAP = \angle KLP$$ as desired. $\blacksquare$

The claim implies the problem as $\angle KCL = 180^{\circ} - \angle KPL = 180^{\circ} - \angle KAL$.

[quote=madatmath]Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.

[i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i][/quote]

Solution from [i]Twitch Solves ISL[/i]:

Let $\alpha = \angle BAK = \angle KAX$ and $\beta = \angle XAL = \angle LAD$. Since $\angle ABK = \angle ADL = 180^{\circ}-(\alpha+\beta)$, it follows that \[ \triangle ABK \sim \triangle LDA \implies 	\frac{AB}{LD} = \frac{BK}{DA} 	\implies \frac{CD}{LD} = \frac{BK}{BC} 	\implies \triangle LDC \sim \triangle CBK. \] Then, $\angle BCK + \angle LCD = 180^{\circ}-(\alpha+\beta)$, which is fixed. On the other hand, $\angle BCD = 2(\alpha+\beta)$ is fixed too. Therefore $\angle KCL$ is fixed as desired.

We use the method of moving points.

Let $K,L$ move along the exterior angle bisectors of $\angle ABC,\angle ADC$ respectively and note $\angle KAL$ is fixed. Thus the map $K\mapsto AK\mapsto AL\mapsto L$. Similarly the map $K\mapsto CK\mapsto CL'\mapsto L'=CL'\cap DL$ is projective where $\angle KCL'$ is fixed at $180^\circ-\angle BCD/2$. To show these maps are the same, it suffices to check three cases for $X$. At $X=B$, we have $K=B$ and $L$ is the point at infinity along the angle bisector of $\angle BCD$, so the maps must be the same. Similarly the result trivially holds when $Y=D$. For $X=C$, we get $X=Y=C$ so the result is obvious by excenter angles. Hence we are done.

We claim that $\angle KCL = 180^{\circ} - \frac{\angle BCD}{2},$ which is independent of $g.$(You can easily find this by considering the case $C=X=Y$)

Let $\angle KCA=\theta$. It is easy to then se that $\angle BAC=\angle CAD=2\theta$ and $\angle BAD=\angle BCD=4\theta$. By incenter excenter duality, we know that $\angle I_1CK=90^{\circ}$. It then follows from simple angle chasing that $\angle KCL$(reflex) is equal to $180^{\circ}+2\theta,$ so the other part is just $180^{\circ}-2\theta=180^{\circ}-\frac{\angle BCD}{2},$ as desired.

The key claim in this problem was really just the incenter-excenter duality.

Interesting how much G3 is easier than G2...
Let AL meet BC at P and AK meet DC at Q.
Step1 : QLDA and BKPA are cyclic.
∠LDQ = (180 - ∠ADC)/2 = ∠DAB/2 = ∠LAQ ---> QLDA is cyclic.
∠PBK = (180 - ∠ABC)/2 = ∠DAB/2 = ∠PAK ---> BKPA is cyclic.

Step2 : QKCPL is cyclic.
∠CQL = ∠DAL = ∠CPA ---> CQLP is cyclic.
∠KPC = ∠KAB = ∠KQC ---> KQPC is cyclic.

Now we have ∠KCL = ∠KPL = 180 - (∠KPA) = 180 - (∠KPC + ∠CPA) = 180 - ∠BAC/2 so ∠KCL is independent of line g.
we're Done.

As always, I have an overcomplicated solution, so I won't be posting it here. However, I do have a diagram which I think will help with understanding this configuration :).

[img width=60]https://cdn.discordapp.com/attachments/872490629714808853/947392676263776256/unknown.png[/img]

Since $L,D,Y,$ and the incenter of $ADY$ are concyclic, we have $$\angle ALD = \frac{1}{2}\angle AYD = \frac{1}{2}\angle BAX = \angle BAK.$$ Similarly $\angle BKA = \angle DAL$, so $\triangle BAK \sim \triangle DLA$. Therefore $\frac{BA}{BK} = \frac{DL}{DA}$. Since $AD=BC$ and $AB=CD$ we have $$\frac{DA}{BK}=\frac{DL}{BA}\Rightarrow \frac{BC}{BK}=\frac{DL}{CD}.$$ However, note that $\angle CBK = \angle CDL = \frac{1}{2}(180^{\circ}-\angle ABC)$, so $\triangle BKC \sim \triangle DCL$. Therefore $\angle BCK + \angle DCL = \angle BCK + \angle BKC = 180^{\circ} - \angle CBK = 180^{\circ}-\frac{1}{2}(180^{\circ}-\angle ABC)$ and is thus fixed. Now $$\angle KCL = 360^{\circ}-\angle BCD - (\angle BCK +\angle DCL)$$ which is fixed as $\angle BCD$ and $\angle BCK + \angle DCL$ are.

We have $\angle KAL=\frac{1}{2}\angle ABC=\angle KBC=\angle LDC$ so let $AK$ and $DC$ intersect at $E$ and $AL$ and $BC$ intersect at $F.$ $ADLE$ and $ABKF$ are cyclic. Now, $\angle AEL=\angle AFK=\angle KAL$ so $KFLE$ is cyclic. In particular, $\angle KEL$ and $\angle KFL$ are fixed. $\angle LEC=\angle LAD=\angle CFA$ so $ECFL$ is cyclic. Thus, $\angle KCL=\angle KFL$ is fixed.

Let $I$ and $L_1$ denote the incenter and $X$-excenter of $ABX$ respectively. It's easy to see $ABX$ and $YDA$ are homothetic. Now, if we let $C_1$ denote the image of $C$ under the homothety taking $YDA$ to $ABX$, $$\frac{BC_1}{C_1A} = \frac{DC}{CY} = \frac{AB}{CY} = \frac{BX}{XC}$$ which means $AC \parallel C_1X$.

Properties of internal and external bisectors imply $BIAL_1$ and $BIXK$ are cyclic with diameters $BA$ and $BX$ respectively. Because $I = AK \cap XL_1$, the Spiral Center Lemma yields $BAL_1 \overset{+}{\sim} BKX$. Now, combining this similarity with $AC \parallel C_1X$ gives $$BK \cdot BL_1 = BA \cdot BX = BC \cdot BC_1$$ so $\frac{BK}{BC_1} = \frac{BC}{BL_1}$. In addition, we have $$\angle KBC = \angle KBX = \angle ABL_1 = \angle C_1BL_1.$$ Thus, SAS Similarity implies $BKC \sim BC_1L_1$. Now, because the aforementioned homothety yields $BC_1L_1 \sim DCL$, we know $$\angle KCB + \angle BCD + \angle DCL = \angle BCD + \angle KCB + BC_1L_1$$ $$= \angle BCD + \angle KCB + \angle BKC = \angle BCD + (180^{\circ} - \angle CBK)$$ $$= \angle BCD + 180^{\circ} - \left( \frac{180^{\circ} - \angle ABC}{2} \right) = 180^{\circ} + \frac{\angle BCD}{2}$$ so $\angle KCL = 180^{\circ} - \frac{\angle BCD}{2}$, which finishes. $\blacksquare$

at this point the questions are getting ridiculous. i find the needed points to add in and then i dont add them in because im convincing myself that they dont work. jeez i need to hone my intuition more!1

Let $C'$ be the point on $BK$ such that
\[\frac{C'B}{C'K}=\frac{CD}{CY}=\frac{XA}{XY}=\frac{XB}{XC}\implies C'X\parallel CK.\]

Also notice that $\triangle BKX\sim \triangle DYL$.
Now notice that
\[\measuredangle DCL+\measuredangle KCB=\measuredangle BC'X+\measuredangle KCB=\measuredangle BC'X+\measuredangle C'XB=\measuredangle KBX\]
which is constant. Also notice that $\measuredangle BCD$ is constant, so $\measuredangle KCL$ is constant (since it is the last part of a $360^{\circ}$ angle).

Here is a sketch
Observe BLC is similar to DCK.
[hide=Proof]
Observe Angle CBL=KDC.
Then observe BL*DK=AB*AD(Can be found out using trigonometry)
[/hide]
We are done


search limit trigonometry function algorithm

I think there was a generalization of Ramanujan's formulas that lead to one which churns out 50 more accurate digits of pi per summand... its pretty crazy

http://en.wikipedia.org/wiki/Pi Quite a few formulas for it there, look for fast converging ones.

Or try

http://mathworld.wolfram.com/PiFormulas.html

I'd say C will probably compte the fastest, but I don't know TOO much about programming. Maybe search fast executing languages on Google?

does this work?   :maybe:

$\pi=\arcsin (1)-\arcsin (-1)$

edit: oh i get it. you want $\pi$ expressed as a series of numbers.

[quote="Franz Joseph"][quote="Hokkage"]All of those have pretty slow convergence. You might want to search the forum for pi formulas. Ramanujan's is one of the best as far as I know (one of my friends got up to about 10 million digits on his computer, but that was in C).

As far as excel is concerned... why not try using a language that's more powerful?[/quote]

Any language suggestions then?[/quote]

C is fast, you could mix in some assembler but only if you're skilled enough, or else just go with C.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=314393#314393]Here[/url] there is another way based on the formula ${\pi = \lim_{n \rightarrow \infty}n* \sin(\frac{180}{n}})$

i have a question off the topic: how does a calculator evaluate trigonometric functions? i mean, does $\pi$ appear in its algorithm?

Some do, if you put it in the right mode. @Franz Joseph: It is very hard and is never exact to find pi. Look at mathworld.

[quote="arsemesomething"] how does a calculator evaluate trigonometric functions? [/quote]

http://mathworld.wolfram.com/TaylorSeries.html

[quote="arsemesomething"] how does a calculator evaluate trigonometric functions? [/quote]

$\sin x = \frac{x^{1}}{1!}-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+...$
$\cos x = \frac{x^{0}}{0!}-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+...$

This is proven using Taylor and Maclaurin series.

Correct me if I'm wrong...

We already had a high amount of threads about this, search, everything you want to ask was probably already answerded ;)


inequalities
geometry
circumcircle
geometry unsolved

let incircle of triangle ABC touch sides AB, BC, CA at D, E, F respectively. prove that (ABC)[u]>[/u] 4(DEF)

[quote="28121941"]The inequality is no new, actually. And it still holds when DEF is any cevian triangle inscribed in ABC. Just express the area of ABC as sum of areas of 4 triangles, one DEF, the other with vertices in each of the vertices of ABC and two of the other three points D,E,F, and apply the inequality of the arithmetical-geometrical means,[/quote]

just joking

DEF is a cevian triangle inscribed in the triangle ABC (D belong to BC, E belong to CA, F belong to AB) if the lines AD, BE and CF are concurrent.

the area on the pedal triangle of the incircle is: $ A'\equal{}\frac{Ar}{2R}$ so the inequality become: $ 2R\ge 4r$ that is know

[quote="28121941"]The inequality is no new, actually. And it still holds when DEF is any cevian triangle inscribed in ABC. Just express the area of ABC as sum of areas of 4 triangles, one DEF, the other with vertices in each of the vertices of ABC and two of the other three points D,E,F, and apply the inequality of the arithmetical-geometrical means,[/quote]

i m still not clear...

[quote="28121941"]With your notations, call x = AD = AF, y=BD = BE, z= CE = CF (tangents to incircle from the vertices are of the same lenght).
Express in terms of x,y,z the areas of the triangles ABC, ADF, BDE and CEF.
Then you have (ABC) = (DEF) + (ADF) + (BDE) + (CEF) or equivalently
(DEF) = (ABC) - (ADF) - (BDE) - (CEF) and apply the inequality arithmetic geometric[/quote]

is that by using heron?

Yes, and then eliminate the sinuses of the angles using the extended theorem of sinus, using the circumradius (later it can be simplified)


If $ \overline{AB}\equal{}16, \overline{MA}\equal{}2$ and $ PC\perp AB$, how to prove that $ C$ is the midpoint of $ AO$? Find $ \overline{MC}$.

$ MC^2\equal{}ME*MP\equal{}MA*MB\equal{}2*18$ so $ MC\equal{}6$, $ AC\equal{}4\equal{}CO$...

Nice Altheman. I was wondering how to use $ ME$ and $ MP$ since it wasn't given. I knew that the solution will come with power of a point, i just couldn't see how...


group theory
abstract algebra
superior algebra
superior algebra unsolved

$\mathbb{Z}(p^{\infty}) = \mathbb{Z}[p^{-1}] / \mathbb{Z}$ is the $p$-sylow subgroup of $\mathbb{Q}/\mathbb{Z}$, and can be described as the colimit of the $\mathbb{Z}/p^{m}$ with the transition maps

$\forall m \leq n ~: ~ \mathbb{Z}/p^{m}\to \mathbb{Z}/p^{n}~,~ \overline{z}\mapsto p^{n-m}\overline{z}$

we have

$\hom(\mathbb{Z}/p^{n},\mathbb{Z}(p^{\infty})) \cong \text{Tor}_{p^{n}}(\mathbb{Z}(p^{\infty})) \cong \mathbb{Z}/p^{n}$

the last isomorphism is given by $\overline{z/p^{n}}\to \overline{z}$. we continue by taking limits:

$\text{End}(\mathbb{Z}(p^{\infty})) \cong \hom(\text{colim}_{n}~ \mathbb{Z}/p^{n},\mathbb{Z}(p^{\infty})) \cong \text{colim}_{n}~ \hom(\mathbb{Z}/p^{n},\mathbb{Z}(p^{\infty}))$
$\cong \text{lim}_{n}~ \mathbb{Z}/p^{n}\cong \mathbb{Z}_{p}$

one can easily check that this is even an isomorphism of rings. just shout if you want something without limits, i.e. which hides them ;).


floor function
number theory proposed
number theory

The number $ x_{n}$ is defined as the last digit in the decimal representation of the integer $ \left\lfloor\sqrt{2}^{n}\right\rfloor\;\;(n\equal{}1,2,3,\cdots)$. Determine wether the sequence $ x_{1},x_{2},\cdots,x_{n},\cdots$ is periodic.

The same problem
Find all m such that the last digit of $ \Large\[m^{\frac{n}{2}]$ is periodic


The $ 5\times 5$ grid shown contains a collection of squares with sizes from $ 1\times 1$ to $ 5\times 5$. How many of these squares contain the black center square?
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt));

for(int i=0; i<=5; ++i)
{
draw((0,i)--(5,i));
draw((i,0)--(i,5));
}
fill((2,2)--(2,3)--(3,3)--(3,2)--cycle);[/asy]$ \textbf{(A)}\ 12\qquad
\textbf{(B)}\ 15\qquad
\textbf{(C)}\ 17\qquad
\textbf{(D)}\ 19\qquad
\textbf{(E)}\ 20$

[hide]
the number of squares with the black squares are:
1x1:1
2x2:4
3x3:9
4x4:4
5x5:1
so 19, or D
[/hide]

[quote="#H34N1"]The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$. How many of these squares contain the black center square?
(A) 12 (B) 15 (C) 17 (D) 19 (E) 20[/quote]

[hide="solution"]There is one 1 by 1 square that contains the middle one.

There are 4 2 by 2 --------------------------------

There are 9 3 by 3 ---------------------------------

There are 4 4 by 4 ---------------------------------

There is 1 5 by 5 ----------------------------------

Adding them all together we get 19[/hide]

Well there's only 1 spot the center can be "in" the 1x1 square so 1.
There's 4 spots the center square can be "in" the 2x2 spot so 4.
Same with 3x3 for 9.
However, there's only the center 2x2 of the 4x4 square that the center can be "in" or else the 4x4 square won't be in the 5x5 grid anymore so 4.
etc.

[quote="Quevvy"][quote="rohit.is.the.lebron23"]I don't  see that picture "The 5x5 grid [b][u]shown ..................[/quote]
Can't you just imagine it?[/quote]No he can't because this person has too much time in his hand posting ridiculous replies. If you want to know, check his profile and all of his previous 13 posts.

[quote="10000th User"][quote="Quevvy"]Is there a formula to this? Maybe a pattern? so that you could solve this very quickly.

I know it can be done quickly, but is there a formula?[/quote]Sure, by induction, for any $n\in\mathbb{Z}^{+}$, in a $2n-1\times2n-1$ grid, the number of squares containing the black center square is
\[2\sum_{k=0}^{n-1}k^{2}+(n^{2})=\frac{(n-1)n(2n-1)}3+n^{2}=\frac{n(2n^{2}+1)}3 \]
In this problem, $n=3$ for the $2n-1\times2n-1=5\times5$ grid and you get 19 squares :thumbup:[/quote]
Sane as that one USAMTS problem...


hey

since spicy things are acidic, why do people tell you to drink milk, which is also slightly acidic.  why not just drink water, which is more basic than milk???

thanks,
Johnny

I actually pondered this question a while ago.

The taste of "spicy" does not come solely from the acidity of the food, but these foods activate your pain sensory signals. My guess as to why one should drink milk as opposed to water is that milk either subdues the pain or lessens the ability of spicy foods to induce pain.

Now, this may be all false, but I am pretty sure that this is true. The biology subforum might be a better place to ask this question.

and i also think that people drink milk because of its many nutrition and vitamins, right? [/b]

yea but i'm talking about why ppl eat it after peppers...

The reason people say to drink milk is that it coats the tastebuds and reduces the "pain" from the spiciness.  Actually drinking water is one of the worst things that can be done, as it allow the air to reach the tastebuds and enhances the "pain" from spiciness.

Actually the best thing I've found is to eat starch, such as cooked rice.  The starch coats the tastebuds and reduces the "pain" from spiciness, until it is wash away by drinking water, or something similar.

also ice cream or ice or something frozun work well since they numb ur tongue.  :rotfl:

I always thought that milk was helpful because the substances calling spiciness are fat-soluble (and hence water-insoluble).  Milk contains lots of fat-soluble substances, being a colloid of various biochemicals.  Not sure if this is a good explanation, though, but I'm sure it's nothing a little Googling wouldn't fix.


If m and n are roots of ax^2 + bx + c = 0, then what is the value of a^2 + b^2?

This one has choices, but I don't think you'll need them.

Oh yes, the answer is in terms of a, b, and c.

[hide]Since m and n are roots, (x-m)(x-n)=ax :^2: +bx+c, so

x :^2: -nx-mx+mn=ax :^2: +bx+c

Then we turn the left side into quadratic form:

x :^2: +(n+m)x+mn=0

so:

a=1

b=(-n-m)

c=mn

a :^2: +b :^2: =1+ n :^2: +2mn+m :^2:

Now to put it into terms of a,b, and c:

The product of the roots is c/a, and the sum of the roots is-b/a.  So the 2mn term equals 2c, and the n :^2: +m :^2: term is the sum of the roots, squared, minus 2mn. This is equal to b :^2:-2c.[/hide]

[hide]Simplified, a :^2: +b :^2: = 1+b :^2:

I think that is the answer. [/hide]

a doesn't have to equal 1.

You were going in the right direction, though.

the choices are here: (in spoiler)

[hide]A. (b^2 - 2c)/a

B. (b^2 - ac)/a^2

C. (b^2 - 2ac)/a^2

D. (b^2 + 2ac)/a^2[/hide]


I saw this from an old AMC/AJHSME contest from 1998:

A 4x4x4 cubical box contains 64 small identical cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?

A)48  B)52  C)60  D)64  E)80

HELP PLEASE AND EXPLAIN

I tried so many times but I can't get it... :wallbash_red:

B)$ 52$. All $ 16$ cubes of the lowest tier, and the $ 12$ cubes each on the contour of the other three tiers.

Let's use complementary counting.

The cubes not touching a side/bottom of the box is 2x2x3=12; so, there are 64-12=[b]52[/b] cubes that do, as previously gotten.


function
induction
algebra proposed
algebra

Find all functions $f:N\to N$ such that:  1) $0$ is on $f(N)$
  2)$f(f(n-1)+1)=n$ for any $n$ natural number without $0$. ;)

it's clear that f must be a bijection!
we have so $f(n+1)=f(f(f(n-1)+1)+1)=f(n-1)+2$ and that for all $n\geq1$
by induction we have:
$f(2k)=f(0)+2k$ and $f(2k+1)=f(1)+2k$ for all $k$ in $N$
let's note e the integer such that $f(e)=0$
NB: $f(1)=f(1+f(e))=e+1$ then $f(2k+1)\geq2$
and so $e=2p$ then $0=f(0)+2p$ then $e=0$ and $f(0)=0$
so finally $e=0$ and $f(1)=1$
$f=id_N$ is the solution

$f(2k+1)\geq2$ if we suppose that $e\geq1$
sorry ! :oops:


Mπορείτε να με βοηθήσετε να βρώ ένα φυσικό πρόβλημα για τα παρακάτω προβλήματα

[b]1. [/b] Δίνετε το πρόβλημα συνοριακών τιμών
Δu=0
u(α,θ)=1 με 0<θ<π
u(ρ,0)=u(ρ,π)=0 με 0<=ρ<=α

Να βρεθεί ένα φυσικό πρόβλημα που να έχει την παραπάνω μαθηματική διατύπωση

[b]2. [/b]Δίνετε το πρόβλημα συνοριακών τιμών
Δu(x,y)=0 με (χ,y) ανήκει στο (0,π)x(0,1)
$ u(x,0)\equal{}u_ocosx$
$ u(x,1)\equal{}u_osin^2{x}$
$ u_x(0,y)\equal{}u_x($π$ ,y)\equal{}0$

Να βρεθεί ένα φυσικό πρόβλημα που να έχει την παραπάνω μαθηματική διατύπωση

[b]3. [/b]Δίνετε το πρόβλημα συνοριακών τιμών

Δu(r,θ,φ)=0 με r>α ,θ ανήκει στο [0,π] και το φ ανήκει στο [0,2π) με
u(α,θ,φ)=0  με θ ανήκει στο [0,π] και το φ ανήκει στο [0,2π) και
το $ lim_{r\rightarrow {\infty}} u(r,$θ,φ$ )\equal{}\minus{}A\[ \cdot\] r\cdot$$ cos$θ με Α>0

Να βρεθεί ένα φυσικό πρόβλημα που να έχει την παραπάνω μαθηματική διατύπωση

Spinos, τα είχες ξαναβάλει αυτά μου φαίνεται. Aπλώς βάλε πυκνωτές κατάλληλου σχήματος στο δυναμικό των συνοριακών επιφανειών.

Cheerio,

Durandal 1707


Evaluate and express as a common fraction: $ (16)^{\minus{}\frac{1}{4}}\minus{}(8)^{\minus{}\frac{2}{3}}$

evaluating $ (16)^{\minus{}\frac{1}{4}}$ we end up with $ \frac{1}{2}$
evaluating $ (16)^{\minus{}\frac{2}{3}}$ we end up with $ \frac{1}{4}$
subtracting:
$ \frac{1}{2}\minus{}\frac{1}{4}\equal{}\boxed {\frac{1}{4}}$


Prove that for each positive integer number $ n$ , there exist a perfect square number stars with $ n$

For example, with $ n\equal{}1$ we can choose 16, $ n\equal{}2$ we can choose 25, $ n\equal{}10$ we can choose 100,etc.

Plz help me, thanks

For some positive integer $ n$ with $ k$ digits consider the set with all numbers starting with the digits of $ n$ but having $ 2(k\plus{}k^2)$ digits.

So the possible numbers whose square are included in this set are those with $ k \plus{} k^2$ digits or $ k \plus{} k^2 \minus{} 1$ digits. We need only to show that if we pick a number $ l$ with at most $ k\plus{} k^2$ digits, $ (l \plus{} 1)^2 \minus{} l^2 < 10^{k \plus{} 2k^2}$ which is easy to see. This way there can't be a gap between two consecutive squares that contain all numbers from the set.

PS: $ (2k\plus{}k^2)$ was just some random number sufficiently larger than $ k$

[quote="hungnsl"]Prove that for each positive integer number $ n$ , there exist a perfect square number stars with $ n$

For example, with $ n \equal{} 1$ we can choose 16, $ n \equal{} 2$ we can choose 25, $ n \equal{} 10$ we can choose 100,etc.

Plz help me, thanks[/quote]
It is an easy problem . We only need to show that there exist a positive integer $ m$ (large enough) such that ,exist an integer $ x$ satisfies  $ n.10^{m}<x^2<(n\plus{}1).10^m$ .This is obvious because the sequence $ \{\sqrt{(n\plus{}1).10^m}\minus{}\sqrt{n.10^m}\}$ tends to infinite ,hence in the interval $ (\sqrt{n.10^m},\sqrt{(n\plus{}1).10^m})$ can contain very many natural numbers . My solution also prove that for any positive $ k$ there exist $ k$ consecutive perfect square that begin with $ n$ and also works for any p th power.


Asymptote AMC USA(J)MO USAMO geometry incenter function

Usamo #6:

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair pa, pb, pc, qa, qb, qc; pair A = origin; pair B = (3,0); pair C = 2.5*dir(77); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B); pair ta = intersectionpoint(in,I--interp(I,(I+O-A),10)); pair tt = (-(O-A).y,(O-A).x); pair tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); pair o = circumcenter(A,ta,tb); dot(o); label("$P_A$",o,E); pa = o; ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); o = circumcenter(A,ta,tb); dot(o); label("$Q_A$",o,E); qa = o; tt = (-(O-B).y,(O-B).x); ta = intersectionpoint(in,I--interp(I,(I+O-B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$P_B$",o,E); pb = o; ta = intersectionpoint(in,I--interp(I,(I-O+B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$Q_B$",o,E); qb = o; E = intersectionpoint(in,B--C); tt = (-(O-C).y,(O-C).x); ta = intersectionpoint(in,I--interp(I,(I+O-C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); o = circumcenter(C,ta,tb); dot(o); label("$P_C$",o,E); pc = o; ta = intersectionpoint(in,I--interp(I,(I-O+C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); o = circumcenter(C,ta,tb); dot(o); label("$Q_C$",o,E); qc = o; dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5)); draw(pb--qb,linewidth(1.5)); draw(pc--qc,linewidth(1.5));[/asy]

It doesn't like it because the code is too long? How long can it be?

Edit: just one pair of circles:

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair pa, pb, pc, qa, qb, qc; pair A = origin; pair B = (3,0); pair C = 2.5*dir(77); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B); pair ta = intersectionpoint(in,I--interp(I,(I+O-A),10)); pair tt = (-(O-A).y,(O-A).x); pair tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); pair o = circumcenter(A,ta,tb); dot(o); label("$P_A$",o,E); pa = o; ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); o = circumcenter(A,ta,tb); dot(o); label("$Q_A$",o,E); qa = o; dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5));[/asy]

Edit 2: Can I do 2 pairs of circles?

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair pa, pb, pc, qa, qb, qc; pair A = origin; pair B = (3,0); pair C = 2.5*dir(77); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B); pair ta = intersectionpoint(in,I--interp(I,(I+O-A),10)); pair tt = (-(O-A).y,(O-A).x); pair tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); pair o = circumcenter(A,ta,tb); dot(o); label("$P_A$",o,E); pa = o; ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); o = circumcenter(A,ta,tb); dot(o); label("$Q_A$",o,E); qa = o; tt = (-(O-B).y,(O-B).x); ta = intersectionpoint(in,I--interp(I,(I+O-B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$P_B$",o,E); pb = o; ta = intersectionpoint(in,I--interp(I,(I-O+B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$Q_B$",o,E); qb = o; dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5)); draw(pb--qb,linewidth(1.5));[/asy]

Guess so.

Edit 3: Making my code fewer lines

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair o,pa, pb, pc, qa, qb, qc, A = origin, B = (3,0), C = 2.5*dir(77), O = circumcenter(A,B,C), I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C), cir=circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B), ta = intersectionpoint(in,I--interp(I,(I+O-A),10)), tt = (-(O-A).y,(O-A).x), tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); dot(pa = o = circumcenter(A,ta,tb)); label("$P_A$",o,E); ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); dot(qa = o = circumcenter(A,ta,tb)); label("$Q_A$",o,E); tt = (-(O-B).y,(O-B).x); ta = intersectionpoint(in,I--interp(I,(I+O-B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); dot(pb = o = circumcenter(B,ta,tb)); label("$P_B$",o,E); ta = intersectionpoint(in,I--interp(I,(I-O+B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); dot(qb = o = circumcenter(B,ta,tb)); label("$Q_B$",o,E); E = intersectionpoint(in,B--C); tt = (-(O-C).y,(O-C).x); ta = intersectionpoint(in,I--interp(I,(I+O-C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); dot(pc = o = circumcenter(C,ta,tb)); label("$P_C$",o,E); ta = intersectionpoint(in,I--interp(I,(I-O+C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); dot(qc = o = circumcenter(C,ta,tb)); label("$Q_C$",o,E); dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5)); draw(pb--qb,linewidth(1.5)); draw(pc--qc,linewidth(1.5));[/asy]

Still too long...I could abuse syntax more to make it shorter but it becomes unreadable then. Is there a reason for the limit to the length of code?

[quote="i_like_pie"]If the code of a certain picture is so long that the box you get can't contain it all (even when you maximize it), then clicking Ctrl+A and then pasting somewhere will give you the entire code, plus a "Close Window" line at the bottom.[/quote]
If you are a [b]true[/b] AoPSer, then "somewhere" will be TeXnicCenter (or the equivalent for your operating system) :wink: .

[quote="perfect_radio"]
How on Earth can I get the name of a point to go exactly to the right of the point? I tried E, but the name just gets over the point :ewpu:[/quote]
Your problem is that you defined a point with the name $E$ in your code. So, it is now used for positioning your labels when you write label(...,E). To make everything work right, you should now write [b]label("M",M,plain.E)[/b] :)

Thanks, fedja! It makes sense  :yup:

[quote="i_like_pie"]If the code of a certain picture is so long that the box you get can't contain it all (even when you maximize it), then clicking Ctrl+A and then pasting somewhere will give you the entire code, plus a "Close Window" line at the bottom.[/quote]
What's wrong with using [b]Page Down[/b]?

If you click on an asy image to make the box containing the code pop up, hitting page down does not do anything within that box.

It doesn't? That's strange. If I click on a box that contains text over the limit, then "Page Down" does as its name says.

Page Down [b]does[/b] work, after all.

I never use Page Down, so I didn't realize it could be used in this case. :oops:

[quote="Hamster1800"]Still too long...I could abuse syntax more to make it shorter but it becomes unreadable then. Is there a reason for the limit to the length of code? [/quote]

Is it what you were after? :? At least, your best code is equivalent to this one and I believe this one is easier to read. I hope, you've got the idea ;). As to restrictions on the code, it should be at most 2000 characters. That  is quite a lot if used with skill and imagination. Just don't try to construct a regular 65537-gon with straight edge and compass  :D .

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair A, B,C, O, I, E,ta,tt,tb; pen q=linewidth(1), r=linewidth(1.5); A = origin; B = (3,0); C = 2.5*dir(77); O = circumcenter(A,B,C); I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,q); draw(in,q); draw(cir,q); E = intersectionpoint(in,A--B); pair routine(pair A, real sign, pen p,string lab) { ta = intersectionpoint(in,I--interp(I,(I+sign*(O-A)),10)); tt = (0,1)*(O-A); tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),p+q); pair o = circumcenter(A,ta,tb); dot(o); label(lab,o,plain.E); return o; } draw(routine(A,-1,red,"$P_A$")--routine(A,1,red,"$Q_A$"), r); draw(routine(B,-1,blue,"$P_B$")--routine(B,1,blue,"$Q_B$"), r); E = intersectionpoint(in,B--C); draw(routine(C,-1,green,"$P_C$")--routine(C,1,green,"$Q_C$"), r); dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); void smile(pair center, real radius) { filldraw(circle(center,radius),yellow,brown); filldraw(circle(center+0.5*radius*dir(120),0.2*radius),blue,red); filldraw(circle(center+0.5*radius*dir(60),0.2*radius),blue,red); draw( (center+0.7*radius*dir(240))--(center+0.7*radius*dir(300)), red+linewidth(2) ); } smile((5,0),0.5); smile((5,1),0.3); // And so on, and so forth.[/asy]

Yes yes, that was what I was after...I'm rather used to programming in C++ so I felt that "unrolling" the routine would be better, but it seems the focus should be on saving memory. Thanks.

[quote="Hamster1800"]I'm rather used to programming in C++ so I felt that "unrolling" the routine would be better[/quote]
Better from what point of view? :? Written with routine, the code essentially says: "Here is what we do for $A$ and outer tangency; for inner tangency we need just one small change; and then we do exactly the same for B and C". Isn't it how you would explain your construction to a human being? What's the advantage of writing exactly the same lines again and again? Anybody who reads the code will understand what's going on after the first repetition and will just skip the next lines. Besides, little tricks like defining pens and declaring all variables in one line are also harmless for understanding: nobody really will read your variable declarations anyway and pens are just pens: who really cares about the exact colors and thicknesses you are going to use? As to the machine, you do not really do much more operations with declaring routine than without it. Actually, you do less, because your code now works from the first attempt once you have figured out a part of it and when you do manual unrolling you are almost certain to forget to change some pa to pb during the first time (assuming you've done cut and paste), so you have to run your code more times. Unrolling may have certain advantages when you are to run the same correct code again and again but you are in the situation when it is run as many times as you need to make it work and then never more (the picture is generated and stored; the code is retired). So, if you are used to C++, just program as you normally would; insert a few comments here and there about what you are doing but use full power of the language. If you really want to help sparing the server time, test run your code on your home PC first, make sure it works there, check its length, and post it once. If we all get into a few such good habits, it'll be much easier to convince Valentin to keep restrictions on programming to the bare minimum needed for safety.   :)

In C++, or java, or other languages, there is an overhead in calling a routine. There is no way around this, but as these images are only compiled once, there's no real point in worrying about it. Inlining a routine into your code, or unrolling a loop, or basically using functional C instead of objective C, will cause the program to run faster. That's what I meant by better. In this case, we have different priorities - namely to keep the program short, rather than fast.

You are perfectly right, of course :) but the overhead is the same for any routine call (basically, you have to store the Program State Word (or whatever it is called in English) in the stack and that's it) and, if you count the number of times you called incenter or intersectionpoint or whatever, which are also nothing but routines, you'll realize that a few more routine declarations will hardly make any difference. It is not like having this overhead against not having it but rather like having it 1001 times against 1000. The inlining really makes sense mostly in relatively simple programs at the bottom level or inside long loops when your call within some subroutine is multiplied by the number of times your subroutine is called by other subroutines or by the number of times the loop is executed but you tried to do it at the very top in a linear segment of the code, which never makes sense.
Anyway, this discussion is for the Computer science subforum. I just wanted to say that using routines is not [i]an abuse of syntax[/i] and it doesn't [i]make the code unreadable[/i]. Rather it is a normal programming technique that makes the code more clean and clear even if a tiny bit slower :lol:.

Right, by abuse of syntax, I meant:

Instead of something like
a = [i]expr[/i];
o = a;

use

o = a = [i]expr[/i];

By combining this with methods, you can do assignments inside calls to routines, and it quickly becomes unreadable.

How about this? (I hope Valentin won't kill me for advertising such constructs :lol:)
[asy]path f(path p) { draw(p); p=rotate(7)*scale(1.2)*p; return p; } path g(path p) { draw(p); p=shift((1,0))*p; return p; } path p=(0,0)--(1,0); g(f(g(f(f(f(g(f(p))))))));[/asy]
Also, you might be interested in looking at this code ;):
[asy]size(200); pen w=red+linewidth(2), n=blue+linewidth(1), t; path operator +(path p) { draw(p, t); return p; } pair operator +(pair p) { dot(p,linewidth(7)); return p; } path p,q; pair A,B; t=w; p=+circle((1.0),1); q=+(shift(0.5,0)*p); A=+intersectionpoint(p,q); B=+(intersectionpoints(p,q))[1]; p=+(A--B); t=n; p=+circle(A,1.7); q=+circle(B,1.7); A=intersectionpoint(p,q); B=(intersectionpoints(p,q))[1]; +((1.3*A-0.3*B)--(1.3*B-0.3*A)); /*and so on; the advantage of overloading an operator compared to */ /*declaring a function is that it allows you constructs like */ +(+(2,3)..(2.5,2.5)..+(3,3)); /*without extra parentheses. The disadvantage is that you have */ /*to adjust your pen in a separate operator while the function would */ /*allow you a second optional argument. Anyway, I do not see why */ /*this drawing-en-passe should be considered a bad style or why it */ //may be unsafe or hard to read.[/asy]

worse would be something like this:

[asy]path f(path p) { draw(p); p=rotate(7)*scale(1.2)*p; return p; } path g(path p) { draw(p); p=shift((1,0))*p; return p; } path p=(0,0)--(1,0); g(p=(p--f(p=(p--g(p=(p--f(p=(p--f(p=(p--f(p=(p--g(p=(p--f(p)))))))))))))));[/asy]

And this should be equivalent to your second:

[asy]size(200); pen w=red+linewidth(2), n=blue+linewidth(1), t; path operator +(path p) { draw(p, t); return p; } pair operator +(pair p) { dot(p,linewidth(7)); return p; } path p,q; pair A,B; t=w; p=+((A=+intersectionpoint((p=+circle((1,0),1)),(q=+(shift(0.5,0)*p))))--(B=+(intersectionpoints(p,q))[1])); /*Note the above line particularly */ t=n; A=intersectionpoint((p=+circle(A,1.7)),(q=+circle(B,1.7))); B=(intersectionpoints(p,q))[1]; +((1.3*A-0.3*B)--(1.3*B-0.3*A)); /*I tried to mess with this line but it resisted... */ /*and so on; the advantage of overloading an operator compared to */ /*declaring a function is that it allows you constructs like */ +(+(2,3)..(2.5,2.5)..+(3,3)); /*without extra parentheses. The disadvantage is that you have */ /*to adjust your pen in a separate operator while the function would */ /*allow you a second optional argument. Anyway, I do not see why */ /*this drawing-en-passe should be considered a bad style or why it */ //may be unsafe or hard to read.[/asy]

Actually let's just make super unreadable code:

[asy]path p; real i; for(p = (0,0)--(1,1), i = 0; i< 10; p = p..shift((2,0))*scale(1.2)*p, ++i) draw(p);[/asy]

Hmm..that one wasn't too bad. Let's try..

[asy]size(100); path p=unitcircle, q=p..shift(1,0)*p; real i; for(i=0; i< 6; ++i) { q=q..scale(1.5)*(p=q..p); } draw(q);[/asy]

Yeah, you are completely right about some readability issues for the first code :). Of course, all such tricks should be applied only when you need to create a complicated picture and usual explicit assignment style doesn't work. My point was just that up to a certain level, the readability is not affected much (what, indeed, is wrong with the convention that objects with + in front of them are drawn and without it aren't?) and that sometimes such tricks allow to spare quite a lot of space. Anyway, with the standard black-and-white pictures I'm in favor of things like drawing-en-passe but not very  fond of the assignments within function calls; here I agree with you wholeheartedly. The real question is whether we can come up with some constructs that allow to write a code that is simultaneously legible and terse. For instance, it doesn't make sense to define pair cc(pair A, pair B, pair C){return circumcenter(A,B,C);}  unless you have less than 7 circumcenters in your code. Starting with 7, it is beneficial (if I counted the letters right :P). Drawing-en-passe also makes sense only if you have sufficiently many draw commands. On the other hand, why should you define a lot of things you won't draw? Commands like interp never make sense because you have full vector calculus (actually, complex number arithmetic) on pairs. Predefining pens, especially those with linewidth() makes sense if you use them at least twice, and so on. And it almost always makes sense looking for "and this is done similarly" things and defining subroutines for them. Anyway, the thing I wanted to attract attention to was that the linear code with repetitions may not be the best solution in many cases and it seems to be quite widespread here so far. So, I intentionally mentioned a few weird and not so weird ways of doing things just to sparkle some interest in the programming aspects of Asymptote. I should confess that my first impression from your question was that you just didn't know of existence   of subroutines. It turned out that you knew them perfectly well but were strongly opposed to them for some reasons. But I bet that many people here who wrote long repetitive codes just thought  that it was the only possible way. So, now, I hope some of them start asking what else is possible and that is the main thing I really wanted to achieve by my constructs from the previous post. The only thing I want to add is that if somebody wants to play with some new programming things, he'd better do it on his/her home PC first or we'll all be in danger that Valentin will come and just turn them off for "abuse". ;)

[quote="i_like_pie"]Page Down [b]does[/b] work, after all.[/quote]
What are you doing that allows you to use page down in the asymptote window?  It definitely does not work on any computer that I have used (they all use windows, in case that makes a difference).

Nothing special... :huh:

Click the drawing, make sure you are in the window (it might switch back to the main window automatically), and "scroll" with Page Up/Down all you want. I'm not sure what could restrict you from doing so. If those keys still don't work, try [b]Home[/b] and [b]End[/b], which do the same thing (for me, at least).

I'd like to finish with what Hamster1800 will call an ultimate abuse: full compass-straight edge geometry. We'll construct the circumcircle of a given triangle. Note that the fixed cost is quite high, about 1/3 of total limit, but it is a FIXED cost in every sense ot this word. The variable cost is quite low: about 5 characters per elementary operation.
[asy]size(400); path D(path p, pen q=blue) { path s=p; draw(p,q); return s; } pair D(pair p, pen q=red) { dot(p,q); return p; } path C(pair A, real R) { return circle(A,R); } path C(pair A, pair B) { return circle(A,sqrt((A.x-B.x)^2+(A.y-B.y)^2)); } path L(pair A, pair B, real a=0.2) { return (a*(B-A)+B)--(a*(A-B)+A); } pair IP(path A, path B) { return intersectionpoint(A,B); } pair oIP(path A, path B) { return (intersectionpoints(A,B))[1]; } pair MP(string s, pair A, pair B=plain.S) { label("$"+s+"$",A,B); return A; } pair A=(0,0), B=(0,3), C=(1,2), D, E, F, G; path X,Y; D(MP("A",A)--MP("B",B)--MP("C",C)--A,black); X=D(L(MP("D",D(IP(D(C(A,B)),D(C(B,A))))),MP("E",D(oIP(C(A,B),C(B,A))))),green); Y=D(L(MP("F",D(IP(D(C(A,C)),D(C(C,A))))),MP("G",D(oIP(C(A,C),C(C,A))))),green); D(C(MP("O",D(IP(X,Y))),A),red);[/asy]

[quote="fedja"]I'd like to finish with what Hamster1800 will call an ultimate abuse: full compass-straight edge geometry. We'll construct the circumcircle of a given triangle. Note that the fixed cost is quite high, about 1/3 of total limit, but it is a FIXED cost in every sense ot this word. The variable cost is quite low: about 5 characters per elementary operation.
[asy]size(400); path D(path p, pen q=blue) { path s=p; draw(p,q); return s; } pair D(pair p, pen q=red) { dot(p,q); return p; } path C(pair A, real R) { return circle(A,R); } path C(pair A, pair B) { return circle(A,sqrt((A.x-B.x)^2+(A.y-B.y)^2)); } path L(pair A, pair B, real a=0.2) { return (a*(B-A)+B)--(a*(A-B)+A); } pair IP(path A, path B) { return intersectionpoint(A,B); } pair oIP(path A, path B) { return (intersectionpoints(A,B))[1]; } pair MP(string s, pair A, pair B=plain.S) { label("$"+s+"$",A,B); return A; } pair A=(0,0), B=(0,3), C=(1,2), D, E, F, G; path X,Y,Z; D(MP("A",A)--MP("B",B)--MP("C",C)--A,black); X=D(L(MP("D",D(IP(D(C(A,B)),D(C(B,A))))),MP("E",D(oIP(C(A,B),C(B,A))))),green); Y=D(L(MP("F",D(IP(D(C(A,C)),D(C(C,A))))),MP("G",D(oIP(C(A,C),C(C,A))))),green); Z=D(L(MP("H",D(IP(D(C(B,C)),D(C(C,B))))),MP("I",D(oIP(C(B,C),C(C,B))))),green); D(C(MP("O",D(IP(X,Y))),A),red);[/asy][/quote]

Might as well put in the third perpendicular bisector. It's only 79 more characters if I counted correctly.

[quote="Hamster1800"]
Might as well put in the third perpendicular bisector. It's only 79 more characters if I counted correctly.[/quote]
Actually, much less than that: starting with 3, the cost per bisector is just "B(A,B,"G","H")"; ,i.e. 17 characters because it now makes sense to define a subroutine that constructs a perpendicular bisector for an interval, which would be a good style anyway. This cost is almost optimal, because  you need to specify the endpoints and the labels. The only thing that can be improved is removing the quotation marks on string definitions, so we are down to 13 ;).


Let $ P$ be a polynomial of degree $ d$,is it possible to find an open, bounded subset $ U$ in $ C$ such that for every $ s\in P(u)$, $ P(x)\equal{}s$ exactly has a $ d$ solution in $ U$ .

[quote="HilbertThm90"]Could you please try to rephrase your question. I have no idea what it is asking.  :maybe:[/quote]
 :blush:
A polynomial-like map of degree $ d$ is a holomorphic map $ f: U' \to U$ such that every point in $ U$ has exactly d preimages in $ U'$, where $ U$, $ U'$ are open sets isomorphic to disc and $ U$ contains $ U'$ in its interier.

Show that every polynomial is  polynomial-like map.


[hide="1"]1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4. If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2? [/hide]

[hide="2"]2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can. If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls. Express your answer in terms of $ \pi$. [/hide]

[hide="3"]3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass. The rectangle has dimensions of 4 feet by 3.5 feet. How many feet of weather-stripping will be needed to go around the glass? Express your answer to the nearest integer. [/hide]

[hide="4"]4.
The numbers from 1 through 8 are separated into two sets, x and y. The numbers in x are multiplied together to get $ a$. The numbers in y are multiplied together to get $ b$. The larger of the two numbers $ a$ and $ b$ is written down. What is the smallest possible number that can be written down using this procedure? [/hide]

oops i got it from a different website lol... :blush:


It's known that there is always a prime between $n$ and $2n-7$ for all $n \ge 10$. Prove that, with the exception of $1$, $4$, and $6$, every natural number can be written as the sum of distinct primes.

For small $n$ the statement can be verified instantly.

Proof by induction that every $n\geq 7$ can be written as the sum of distinct primes.
For basic cases $n=7,8,\dots,20$ the statement trivially holds.
For $n\geq 20$ suppose that the statement holds for all integers in the interval $[7,n-1].$
Let $p$ be prime in the interval $[\lfloor n/2\rfloor+1, 2\lfloor n/2\rfloor - 8].$ Then $n=p+(n-p)$ where $7\leq n-p< p.$ Applying induction to $n-p$ completes the proof.

He has shown that $ p>n\minus{}p \geq 7$, thus $ p$ does not occure in the decomposition of $ n\minus{}p$ and the ones summing to $ n\minus{}p$ are distinct by induction.


don't worry too much for RMO as it is quite easy though i myself haven't qualified it yet( Xth grader now)

in my opinion, u shud complete upto NCERT XI syllabus alongwith few other books

take special care in classical geometry , permutations and combinations and equations and inequalities for RMO

also go thru past RMO papers

u can find them in resource section of aops
or
goto http://www.hbcse.tifr.res.in

I sat for RMO last time and did nothin other than bench warming :P :D
Anyways.....number theory is a hot favourite of the RMO paper setters,u may xpect 1-2 questions out of the 6/7 of them.So is geometry.[hide]And I''m bad at both :(  :( [/hide]


:wallbash_red:

because it epicly phails, just like me.

:surrender:
I surrender. Can you tell me my favorite emoticon?

:wallbash_red:

I forgot. =D

No really, thats not my fav emoticon. I have 2 favorites:

:D <-- you'll know why if you've ever taken an AoPS class with me.

:yoda: <-- Is this really a smiley?

Se my avatar, my signature, and my blog.

Hmm... I like:

:harhar:   :yankchain:  :heli:

lol randomness

lol yay. consecutive posts again!

So far, two people are obssessed with  :D !

Maybe I'll keep a tally here of WHO LIKES  :D !!! SPREAD THE  :D ON MY BLOG!

lol those smilies totally represent emo aopsers :D

I like a lot of them, but :rotfl: is my favorite.  Yay for uncontrollable laughter!

:yinyang: 'cause it's pretty.

 :D simply 'cause it's awesome.  :D

:stretcher: Lol, this one is never used. I like it because you would only see it in the forum if somebody was doing a really random post.

:D
[color=white]
The message is too small. Please make the message longer before submitting.[/color]

Yay for your avatar! Spread the word of  :D !!!

SPAM SUCKS!!!!

[quote="mathelete"]Click on the picture for a better image.[/quote]
ROFL ROFL ROFL :D

[quote="eggylv999"]:rotfl:  :rotfl:  :rotfl:[/quote]
Nnnnnnnnnuuuuuuuu  :D is better than rofl and spam

mine is: [img]http://www.planetsmilies.com/smilies/fighting/fighting0020.gif[/img]

awesome right?

^^ That is not from Aops.
Mine favorite emoticons!!!
[img]http://www.artofproblemsolving.com/Forum/images/smilies/trampoline.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/jump.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/heli.gif[/img]
I don't use this one much, but here it is: [img]http://www.artofproblemsolving.com/Forum/images/smilies/icecream.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/juggle%5B1%5D.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/wow.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/plane.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/play_ball.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smilies/starwars.gif[/img]
And the last one: :rotfl:

:fool: This is how i feel after every problem i get wrong on ftw

:D He has an ugly smile. He should get braces.

Mine is  :ddr:

^^ spammer.

My favorite emoticons are  :wow:, because he seems so excited,  :wallbash_red:, I find myself doing it every time the Vikings :surrender:, (they seem to do it a lot); or :rotfl:, because I find myself doing it a lot.

:yinyang:  :yinyang:  :yinyang:  :yinyang:
too much kung fu panda

Thai is from January dude, please stop reviving!


Twin primes are prime numbers that differ by 2. Given that $ a$ and $ b$ are the greatest twin primes with $ a < 100$, evaluate the value of $ a\plus{}b$.

Why is the sum of 2 twin primes divisible by 12 ? (except 3,5 of course)

Because primes besides 3 cannot be 3 mod 12 or any multiple of 2 mod 12. The only possible mods are 1 mod 12, 5 mod 12, 7 mod 12, and 11 mod 12. All of the possible twin prime pairs (11+1 mod 12, 5+7 mod 12) add up to 0 mod 12.


geometry
perimeter
symmetry
geometry open

What propierties has a convex figure such that if you divide it's perimeter by the half it divides it's area by the half too?

Any figure with rotational symmetry order 2


the answer for ur question is

My answer is based on the fact that $ {T^2}$ proportional to $ \frac{1}{M}$...

[hide]7:1[/hide]

[quote="hash_include"][hide="W"][hide="R"][hide="O"]wrong answer da  :D [/hide][/hide][/hide][/quote]
nice shrids. :rotfl:

thanks guys :D
got the idea from some guy at the calculus forum

PS is any1 bothered to answer this qsn? :D

actually i thought you were feigning da  :D  By saying W fer wrong n then R fer right  :D

Carcul often does that  :rotfl:

Shrids that was brilliant da!  :thumbup:

And I didn't even check this, sorry since ranga had answered, I assumed it must have been obviously RIGHT.

Anyway
[hide]answer is 1:1.

T is independent of mass of satellite, depends only on mass of planet. [/hide]

yes. perfect :D

i thought ppl would get this earlier
and i suspect that was a dig at ranga :P

see da.. the main wrong nv that u did was u took $ M$ to be mass of satellite
but actually it's mass of EARTH.
so time period is not related to mass of satellite at all :)

nice one mate...

got confused a bit :oops:

thats y i gave choices.. most of u will just see 49 is a square number n rite 7:1 or 1:7
:D


Find all polynomials $ P(x,y)$ such that for all reals $ x$ and $y$,
\[P(x^{2},y^{2}) =P\left(\frac {(x + y)^{2}}{2},\frac {(x - y)^{2}}{2}\right).\]

[quote="khashi70"]Find All Polynomials $ P(x,y)$ such that for all real $ x,y$ we have :
$ P(x^{2},y^{2}) = P(\frac {(x + y)^{2}}{2},\frac {(x - y)^{2}}{2})$[/quote]

The equation may be written $ P(\frac {x^2 + y^2}{2} + \frac {x^2 - y^2}{2},$ $ \frac {x^2 + y^2}{2} - \frac {x^2 - y^2}{2})$ $ = P(\frac {x^2 + y^2}{2} + xy,$ $ \frac {x^2 + y^2}{2} - xy)$

And so : $ P(x + y,x - y) = P(x + \sqrt {x^2 - y^2},x - \sqrt {x^2 - y^2)}$ $ \forall x,y$ such that $ x\geq |y|$

Let then $ Q(x,y) = P(x + y,x - y)$ We have $ Q(x,y) = Q(x,\sqrt {x^2 - y^2})$ $ \forall x,y$ such that $ x\geq |y|$

So $ Q(x,y) = Q(x, - y)$ $ \forall x,y$ such that $ x\geq |y|$ and so $ \forall x,y$ ($ Q$ is a polynomial). So $ \exists$ $ R(x,y)$ such that $ Q(x,y) = R(x,y^2)$ And :

$ R(x,y^2) = R(x,x^2 - y^2)$ $ \forall x,y$ such that $ x\geq |y|$ and so $ \forall x,y$ ($ R$ is a polynomial).

And a general solution of such an equation is $ R(x,y) = H(x,y) + H(x,x^2 - y)$ With $ H(x,y)$ any polynomial.

And so a general solution of initial equation : $ P(x,y) = H(\frac {x + y}{2},(\frac {x - y}{2})^2) + H(\frac {x + y}{2},(\frac {x + y}{2})^2 - (\frac {x - y}{2})^2)$

Which may be written in a simplier manner : $ P(x,y) = H(x + y,(x - y)^2) + H(x + y,4xy)$ (for any polynomial $ H(x,y)$)

we have $ P(x^{2},(-y)^{2})=P(x^{2},y^{2})=P(\frac{(x+y)^{2}}{2},\frac{(x-y)^{2}}{2})=P(\frac{(x-y)^{2}}{2},\frac{(x+y)^{2}}{2}) $. so $ P(x,y) $ is symmetric.
then we can write $ P $ in the form $ P(x,y)=Q(x+y)+xyR(x,y) $.
then we obtain that $ R(x,y)=\frac{(x-y)^{2}}{4}S(x,y) $ with $ S $ satisfying the initial equation.
how can I finish the problem now?

[quote="pco"][quote="khashi70"]Find All Polynomials $ P(x,y)$ such that for all real $ x,y$ we have :
$ P(x^{2},y^{2}) = P(\frac {(x + y)^{2}}{2},\frac {(x - y)^{2}}{2})$[/quote]
And a general solution of such an equation is $ R(x,y) = H(x,y) + H(x,x^2 - y)$ With $ H(x,y)$ any polynomial.

And so a general solution of initial equation : $ P(x,y) = H(\frac {x + y}{2},(\frac {x - y}{2})^2) + H(\frac {x + y}{2},(\frac {x + y}{2})^2 - (\frac {x - y}{2})^2)$

Which may be written in a simplier manner : $ P(x,y) = H(x + y,(x - y)^2) + H(x + y,4xy)$ (for any polynomial $ H(x,y)$)[/quote]
how can we get the general solution ?????

[quote=TheBeatlesVN]how can we get the general solution ?????[/quote]
We have to solve $R(x,y)=R(x,x^2-y)$ $\forall x,y$

1) Any $R(x,y)=H(x,y)+H(x,x^2-y)$ is a solution, whatever is $H(x,y)$
Trivial result : just check

2) any such polynomial may be written as $R(x,y)=H(x,y)+H(x,x^2-y)$ for some $H(x,y)$
Trivial result : just set $H(x,y)=\frac 12R(x,y)$

And so $R(x,y)=H(x,y)+H(x,x^2-y)$ indeed is a general solution for such an equation.
Q.E.D.

The answer is
$$ P(x,y) \equiv Q \left( x+y, xy(x-y)^2 \right) $$
for any $Q(x,y) \in \mathbb R[x,y]$. First we prove all these solutions work. Note that both
$$ P(x) \equiv x+y \qquad , \qquad P(x) \equiv xy(x-y)^2 $$
satisfy the given assertion. It follows all specified solutions indeed work.

Now we prove the converse direction. Changing $y$ to $-y$, it isn't hard to get $f$ is symmetric, i.e.
$$ f(x,y) = f(y,x) ~~ \forall ~ x,y \in \mathbb R $$ Note if we subtract from $P$ any polynomial $A(x+y)$, then $P$ would still satisfy the assertion. By doing this, we may WLOG assume $P(x,0) \equiv 0$. As $P$ is symmetric, hence $xy$ divides $P(x,y)$. Now putting $y=x$ gives
$$ P(x^2,x^2) = P(2x^2,0) = 0 $$
It follows $P$ is divisible by $(x-y)$. As $P$ is symmetric, thus $P$ is divisible by $(x-y)^2$. So $xy(x-y)^2$ divides $P$. Again, we cancel a factor of $xy(x-y)^2$ from $P$. Then $P$ still satisfies the given assertion. We keep proceeding like this, and we are done. $\blacksquare$

[color=#f00][b][color=#f00]Remark: [/color][/b][/color] I am not exactly sure how this solution set is same as the one mentioned by [b]pco[/b] , though I cannot find any flaw in my solution (maybe someone else can comment on this).

[quote=guptaamitu1]I am not exactly sure how this solution set is same as the one mentioned by [b]pco[/b] , though I cannot find any flaw in my solution (maybe someone else can comment on this).[/quote]
I did not check your proof/result but it is very common that many totally equivalent general form for a given F.E. exist.
As soon as each is a general  solution (any function in this form is a solution and any solution may be written in this form), they are quite equivalent.

Example with the "trivial" FE : "find all the functions from $\mathbb R\to\mathbb R$ such that $f(x)=f(-x)\quad\forall x$"
Different general solution :
1) $g(x)$ whatever is even function $g(x)$ from $\mathbb R\to\mathbb R$  :D
2) $g(x)+g(-x)$ whatever is function $g(x)$ from $\mathbb R\to\mathbb R$
3) $g(|x|)$ whatever is function $g(x)$ from $\mathbb R_{\ge 0}\to\mathbb R$
4) $\forall x\ge 0$ $f(x)=g(x)$ and $\forall x<0$ $f(x)=g(-x)$ whatever is function $g(x)$ from $\mathbb R_{\ge 0}\to\mathbb R$

But for example $g(x)g(-x)$ whatever is $g(x)$ from $\mathbb R\to\mathbb R$ is not a general solution (some solutions can not be written in this form)


How many 7 digit palindromic are there if the
(a) first digit cannot be zero
(b) first digit cannot be zero and digits cannot be repeated

a. 9*10*10*10=9000
b. clearly 0, 7 digit palindromic number are in the form abcdcba so cleary there must be 3 pairs of digits repeated.  but you said we couldnt repeat.

I have a feeling you mistyped b. because unless it was some trick problem.  Maybe you mean how many 7 digit palindromic numbers are there such that the first 4 digits are all distinct.

yup, i mistyped it. it was a different question.


I'm having to withdraw from a class I just started, because I realized that the course was going way too slow, and lots of it is review. It's not one of those withdrawal's where you are about to fail something (those have another designation like WF or something).

How does that look on a transcript though? Does it show up on your transcript? I'm not sure if the rules are similar at every college, or if they change radically from place to place.

It shows on your transcript. Transcripts record history - you can't choose which part of it is shown. One or two W's on a transcript really doesn't mean anything and will generally be ignored by whoever is reading it. A semester with W's in all courses taken records some dramatic life event; an explanation might be in order. A dozen W's on a transcript looks bad.


I am reading a paper, and it says let $ q \equal{} f((1\minus{}2p)^{\minus{}2a},b)$, where $ f(\cdot )$ is a polynomial function.
What does this mean?

I don't think the computer section would be suitable for this topic...
And a polynomial function is just a function that's a polynomial, isn't it?

a pol' function is a usuall pol'

http://www.מורה-פרטי-למתמטיקה.com/


In the last student council election, 3000 ballots were cast. Candidate A received 55% of the votes and candidate Z received 45%. How many more votes did the winner receive?

Obviously, Candidate A was the winner, so we just need to perform $ 0.55 \times 3000 \equal{} \boxed{1650}$.

Multiply and figure out that the winner received 3000*.55=1650 votes, while the 2nd place guy received 3000*.45 or 1350 votes.
1650-1350.
The winner received 300 more votes.
Note: Izzy, it says how many MORE votes did the winner receive.

Alternatively, the winner recieved 55-45=10% of 3000 votes, a much simpler calclation which yields the answer of 300.


If my school does not offer the AMC 12 can I go to a nearby school to take it?  Are there other options if this is not one?  And would I still be eligible to qualify for AIME, i.e would it effectively be the same as if I took it at my own high school?

It doesn't matter where you take it, just remember to put down the school you attend (not the proctor) when you're filling out the answer sheet.

I think you have to ask someone for that? Like your school, perhaps.

Pretty much the same topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=322254]here[/url].


inequalities
search
inequalities unsolved

Prove that for any $a,b,c$ positive reals we have:

\[\frac{1}{a^2+ab+b^2}+\frac{1}{b^2+bc+c^2}+\frac{1}{c^2+ca+a^2} \geq \frac{9}{(a+b+c)^2}.\]

[b]Vialli[/b] posted this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13280&highlight=cauchy]problem[/url] on AoPS before. [b]zabelman[/b] provided a "brute force" (straight-up multiplying in his own language) solution.

Has this been posted before on Mathlinks? Is there any other solution other than "straight-up multiplying"?

Using the identity:

\[ \displaystyle x^{2}+y^{2}+xy=( \sum x)^{2}- \sum xy - (\sum x)z\] you get that:

\[ \displaystyle \frac{ (\sum x)^{2}}{ x^{2}+y^{2}+xy}= \frac{1}{1-p- \sum mn}\] with $\displaystyle m=\frac{x}{x+y+z}$  etc

The inequality becomes: \[ \sum \frac{1}{1-m- \sum mn} \geq 9\] with $m+n+p=1$ Clearing the denominators the ineq becomes: \[[1-4(\sum mp)+9mnp
]+[\sum mn(3 \sum mn-1)^{2}] \geq 0.\] which is obvious, as it is Schur.

[quote="galois"]well i believe the source is gazeta and harazi gave a similar solution to this problem quite sumtime bak[/quote]

I searched gazeta and harazi, didn't come up with this inequality. Could you post a link if you knew it? Thanks.

Fuzzylogic, the topic you search is "inq8" posted by Gil and you can find it in Solved Problems -Section Inequalities.

You can also use option "search"  with "9" in Inequalities - Solved Problems to find it at once.


Metro results:

1. Alan Chou (Clarke)
2. Jason Li (Clarke)
3. Kevin Wen (Diamond)
4. Shi An (Boston Latin)
5. Allison Bukys (Clarke)

OK, I may first mention that 3 of the top 5 are from Clarke.

6. ?
7. Michael Mc... (McCall)
8. Amy Zhang (Diamond)
9. ?
10. Peijin Zhang (Clarke Individual)

IIRC, the countdown round went something like, 9>10,8,7, 6>9,5, 4>6,3,2, 1>4.

Teams:

1. Diamond
2. Clarke
3. Bigelow
4. McCall
5. Brown
6. ?
7. Roxbury Latin

The top 6 teams made it. I just threw in RL because it's SAML's school.

I'll take People Who ________ for 800.

"carl"

What are People Who Suck?

Trebek: No, that's incorrect.

What is people who h4x0r?

Yes. We would also have accepted, "What is people who can make it to nationals twice?"

arg postponed until feb 13 cuz of inclement weather... :mad:

So who are these people who got first?
I know one is alanchou.

Wow, there are only 6 chapters...
Either a lot of people are lying or our chapter is the only one left to go...

I said first because it didn't say for which year this poll was.

I said first because it didn't say from where.

[quote="frost13"]There are at least 3 chapters left.  People ALWAYS lie on these polls!!!

Maybe 4 kids tied and all checked off 1st?[/quote]

I remember how 14 people clicked 46 as their national score last year

Me too!

At home...
1998...
2nd time I did it...

First from way back when  :P

Congratulations to all the high placers from this year's chapter mathcounts, and good luck in states!

I remember I did well in states in Connecticut, but then came to Massachusetts and got owned by Sway Chen and Stephen Xu from Lexington over and over  :D

1 day until states!!!!!
too bad I have to wake up at like 7:30... but whatever....

MAN I AM SO PUMPED FOR STATES

wait i'm not going, dangit. i'm pretty pumped for my carnegie audition though. sort of.

gah, I got 5th written, 1st countdown, but Allison gave up her spot since she was going to Mexico, so I'm going to nats :)

Why would someone give up their spot at Nats?

Alright, you have some sort of a point. Who is Carl though?

[quote="CatalystOfNostalgia"]I don't know but I hear he can't read very well[/quote]
QFAT

(Quoted for Absolute Truth)

[quote="ZhangPeijin"]Lets see.
On one side, you could go to Nats, probably not do that great since Carl could only get 16th, and come back home feeling sad and being called a nerd by other people at your school...
On the other side, you could go to Mexico, have a good time looking at sights, learn some spanish, and have fun with friends...
Although going to Nats would give you a bunch of free stuff, like a TI-89...[/quote]

last year, I hear they gave out some version of a Ti-84, but it's still free :)

Dude, the Nationals trip last year was probably the most fun four days of my life. and the second most fun three days of my life was the Nationals trip from the year before.

I've forged lifelong friendships with people from MathCounts, and the type of people you meet encourages you to work harder. Before I went to nationals in 7th grade, my attitude was basically "man I'm a loser who does math for fun." But then I went to nationals. Then I went to MathPath.

So try experiencing going to Nationals before saying that it's not worth it or whatever.

I have to agree. It was well worth the experience.

Yeah we all love free stuff.
That's legal of course.
We wouldn't take anything free if it was illegal...


The least common multiple of 51x^2 y z ^3 and 34 x^3 z is:

The lcm is such that we must consider the multiple of oith numbers accordingly. In practical terms with variables, this means we take the highest power of each variable. Thus the lcm is $ \boxed{102x^3yz^3}$.


function
calculus
derivative
real analysis
real analysis unsolved

Show that the function $ f(x,y) \equal{} |xy|$ is differentiable at $ 0$, but is not of class $ C^1$ on any open neighborhood of $ 0$.

I don't get it: isn't it necessary that the partials exist for the total derivative to exist?

Note that the partials do exist at $ (0,0)$ and are both zero, and the total differential is zero at that point. This is because $ f(x,y)\minus{}f(0,0)\minus{}0\equal{}|xy|\equal{}o(\|(x,y)\|)$ as $ \|(x,y)\|\to0.$

I'm not familiar with little-$ o$ notation (although I've heard of it), but I assume from context that it means $ \frac{|xy|}{\|(x,y)\|} \rightarrow 0$ since that's what we need here. I've fooled around with that quite a bit and can't seem to get it: could you point in me in the right direction?

Also interesting: this seems to be a case where the "Calculus III method" of computing partials by holding one variable constant and differentiating in the usual way (i.e. without $ \epsilon$-$ \delta$ methods) fails.

No, you can compute the partials at $ (0,0)$ directly. These involve approaching the origin along one axis or the other - but on either axis, the function is identically zero and hence has zero derivative.

As for that limit: note that $ |xy|\le\frac12(x^2\plus{}y^2)$ (Call it AM-GM or prove it directly.) Hence $ \frac{|xy|}{\|(x,y)\|}\le\frac12\cdot\frac{\|(x,y)\|^2}{\|(x,y)\|}\equal{}\frac12\|(x,y)\|,$ which tends to zero.

Oh, I'm with you 100% now. It's been too long since I saw partial derivatives defined properly.


there are three different non-negative integer x, y, z.

Prove that $ \frac {1}{(x \minus{} y)^2} \plus{} \frac{1}{(y \minus{} z)^2} \plus{} \frac{1}{(z \minus{} x)^2} \ge \frac{4}{xy \plus{} yz \plus{} zx}$.

Thanks.

2008 Vitnam ...

I'm sorry I don't know how to use Latex...It's so hard for me...


In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $BC$. Let $F$ be the intersection of $AE$ with $CD$. $AD=3, DB=2, BE=3, CE=4$. Find $EF: FA$ in lowest terms.

By mass points, we can put a weight of 8 on A, 12 on B, and 9 on C, therefore E has a weight of 21, and EF:FA = 8:21

Just use Menelaos' theorem, $\frac{3}{2}\cdot \frac{7}{4}\cdot \frac{EF}{FA}=1.$ :D

$\frac{AD}{DB}\cdot \frac{BC}{CE}\cdot \frac{EF}{FA}=1.$  :)


How would you solve:

$ 10^x \minus{} x \minus{}1 \equal{} 0$

I know one of the answers is 0, but how could you find that out algebraically?
( I think there are actually 2 answers)

Short answer: There isn't a good way.

Long answer: We can try to determine the number of roots and bound them, but we can't get to them without root approximation algorithms.

[hide="Shield your eyes from the CALCULUS :o"]For example, if $ f(x)\equal{}10^x\minus{}x\minus{}1$, $ f'(x)\equal{}10^x\ln 10\minus{}1$ and $ f''(x)\equal{}10^x\ln^2 10$ which is clearly positive for all $ x$. Therefore, $ f$ is convex. $ f$ is decreasing when $ 10^x<\frac{1}{\ln 10}$, or $ x<\minus{}\ln\ln 10$. At that point, we have a minimum by the Second Derivative Test, but for negative $ x$ a little before $ \minus{}\ln\ln 10$, it goes positive because $ \minus{}x$ outweighs $ 10^x$. So there's another root in $ (\minus{}1,\minus{}\ln\ln 10)$.[/hide]

move -x-1 to the other side so you get 10^x = x +1

from there i would graph y=10^x and y=x+1

i have no idea how you'd solve it using algebra

[quote="Ronnicus"]How would you solve:

$ 10^x \minus{} x \minus{} 1 \equal{} 0$

I know one of the answers is 0, but how could you find that out algebraically?
( I think there are actually 2 answers)[/quote]

This equation (or an equivalent one to it) has already been asked
and tackled in this board on a previous date.

The other root is around $ \minus{}0.85$
Source: Wolfram

hello, we get
$ x_1\equal{}0$
or
$ x_2\equal{}\minus{}\frac{\text{LambertW}\left(\minus{}\frac{1}{10}\ln(10)\right)\plus{}\ln(10)}{\ln(10)}$
Sonnhard.


calculus
integration
trigonometry
calculus computations

Can someone help me set up these integral questions with partial fractions. (I don't know how to divide them to make them integrable in convenient form):

1. indefinite integral of (5x 2)/(x 2 +x+1) dx.

2. Indefinite integral of (x 3 +2x 2 +2)/[(x 2 +1) 2 ] dx.

Help is very appreciated. Thanks.

1. This is not a proper fraction. Do long division first to write it as $5+\frac{\text{something}}{x^2+x+1}.$ Then complete the square in that denominator.

2. You need four unknown coefficients to deal with that fourth degree denominator.

The form is $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$.

Once you've done that, use the trigonometric substitution $x=\tan\theta$ to integrate $\frac{D\,dx}{(x^2+1)^2}.$


[hide="Hint 1"]
$\log_{a} b = \frac{ \log b}{\log a}$
[/hide]

[hide="Hint 2"]
$\log_{3^n} 3^m = \frac{m}{n}$
[/hide]

[quote="lkryptonicl"]The solution of $\log_{9}x+\log_{27}x=15$ can be written in the form  $3^a$.  Determine the  value of a.[/quote]

[hide]
$\frac{\log_3x}{\log_3{9}}+\frac{\log_3x}{\log_3{27}} = 15$

$\frac{\log_3x}{2} + \frac{\log_3x}{3} = 15$

$5\log_3x = 90$

$\log_3x = 18$

$x = 3^{18} \implies a = 18$
[/hide]

[quote="lkryptonicl"]The solution of $log_{9}x+log_{27}x=15$ can be written in the form  $3^a$.  Determine the  value of a.[/quote][hide]
$\log_{9} x =\frac{\log_{3} x}{2}$
$\log_{27} x = \frac{\log_{3} x}{3}$
Thus, we have:
$15=\frac{\log_{3} x}{2}+\frac{\log_{3} x}{3}$
$15=\frac{3\log_{3} x +2\log_{3} x}{6}$
$15=\frac{\log_{3} x^5}{6}$
$90=5\log_{3} x$
$18=\log_{3} x$
$x=3^18$
 :w00t: big number!

[/hide]

a = 18 its vrey simple just put the values


AMC
AIME
trigonometry
MATHCOUNTS
AMC 12
AMC 10
LaTeX

I am considering starting a Mock AMC 10/AIME set.

It will take some time before more is posted because of Mock Mathcounts.

So if you consider joining me in this in a few weeks, signup here or \pm me.

Thanks.
More details later.

[quote="veezbo"]Hopefully it'll be an AIME. If it gets created, send me a PM to remind because I'd be interested.[/quote]

Both AMC 10 and AIME.

I will post here when created.

I join.

But I'm only taking the AMC 10.  AIME's just too hard for me....... :blush:

I'll take both I guess. It'll be good training.

I'll try out the AIME. I need the practice.

I shall take both, but I make no promises. I may drop out at any time due to time contraints and other events conflicting with my time.

[quote="Bacteria"]I would like to join in on the AIME please.[/quote]

Please make clear on whether to take or make the test. Also, you will have to QUALIFY for AIME through AMC 10.

[quote="athunder"]Any helpers? it is pretty challenging to set up AIME questions. so..ya are you looking for any helpers?

In aime you must strategically plan out the questions.(especially problems 10-15).

AMC 10 is similar it a bit like a Mock State MC, the difference is that trig may be in it.[/quote]

No trig in AMC 10. If I have enough helpers, [i]possibly[/i] an AMC 12.

Again, you need to [b]QUALIFY[/b] for AIME.

I'll sign up to take the test. Thank you for making it.

[quote="PowerOfPi"]\sign up for any test there is going to be, but I am willing to be a problem maker if needed.[/quote]

Problem maker please :blush:

I understand that you are running Mock Mathcounts but this wont be very soon. :)

Thanks everyone:

[hide="Signups (8)"]veezbo
Bacteria
Maybach
Ihatepie
zapi2007
cheeseyicecream
NuncChaos
bokagadha[/hide]
[hide="Helpers (2)"]athunder
PowerOfPi[/hide]

Alright yes I am looking for helpers. The more the better...

I'll join.  Best of luck to all contestants!

[quote="abcak"]I have received liked 5.  The vast majority of them have very high scores.  I guess that means this was very easy?  :P[/quote]

Yeah, see my post :P

1. MCLegend-138
2. Thunder365-138
3. Tenniskidperson3-138
4. kamallohia-126.5
5. Jason27603-120
6. brightzhu-118.5
7. pascal12-91.5

Ugh, I'm embarrassed about my score >.<
Oh, and also, this was before the new problems were updated

Should the 126.5 be a 136.5? 126.5 is not divisible by 1.5...

In any case, not to rush, but is the AIME done, and regardless of its status, what is the expected date for the release of the AIME?

Hm...that's embarrasing.

Mock AMC 10 probably comes before Mock AIME.

[quote="abcak"]Hey other helpers, It'd be nice if I weren't the only one doing problems :P[/quote]

Sorry I should be on there more now...
I was busy

rts2007: 42
kamallolhia: 127.5

Helpers, send me scores. I've received 2 so far.

[quote="abcak"]1. MCLegend-138
2. Thunder365-138
3. Tenniskidperson3-138
4. kamallohia-127.5
5. Jason27603-120
6. brightzhu-118.5
7. pascal12-91.5
8. rts2007: 42

Bump.  I think deadline is up...[/quote]

What happened to the other 80 people that signed up?

I PMed my answers to limac. I think you should get in contact with the other helpers.

I got a few too. I'll post up the results tomorrow. Sorry for the delay.

Ok, here are the ones I have recevied.

kamallohia - 127.5
Bacteria - 120
bokagadha - 118.5
pascal12 - 88.5
darkdieuguerre - 124.5

[hide="Specifics, if courious"]
kamallohia, all 20 right and 5 blanks.
Bacteria, 6, 10 incorrect, 19 correct, and 4 blanks.
bokagadha, 18 correct, 7 blank.
pascal12, 3,7,10,11,16 incorrect, 13 correct, 7 blank.
darkdieuguerre - 10,23 incorrect, 20 correct, 3 blank[/hide]

edit: sorry dark, your's was a bit on top of the others, so I didn't catch it straight away.

1. MCLegend-138
1. Thunder365-138
1. Tenniskidperson3-138
4. kamallohia-127.5
5. dardieuguerre - 124.5
6. Bacteria - 120
6. Jason27603-120
8. brightzhu-118.5
8. bokagadha - 118.5
9. pascal12-91.5
10. rts2007: 42


function
calculus
derivative
induction
algebra
polynomial
calculus computations

Prove:
Let f be a function that is differentiable n times on a real interval I. If $ x_1  < ... < x_p$are points of I and $ n_i$, $ 1 \le i \le p$ are natural numbers such that $ n_1  \plus{} ... \plus{} n_p  \equal{} n$ and $ f^{(k)} (x_i ) \equal{} 0$ for $ 0 \le k \le n_i  \minus{} 1$, then there exists a point $ \xi$ in the closed interval $ [x_1 ,x_p ]$ at which $ f^{(n \minus{} 1)} (\xi ) \equal{} 0$.

feels like we can use some form of interpolation and use the error to show the desired derivative is zero .

I only know about the Lagrange kind. I know nothing about the Hermite kind.

Let's do this by induction on $ n,$ for $ n\ge 2.$

For $ n\equal{}2,$ there are two cases. One is the trivial case of $ f$ having a double root at $ x_1,$ in which case $ f'(x_1)\equal{}0$ is already built in. The other has $ f(x_1)\equal{}f(x_2)\equal{}0,$ in which case our result, namely that there is a $ \xi\in(x_1,x_2)$ such that $ f'(\xi)\equal{}0,$ is the original Rolle's Theorem.

Now, suppose the theorem is true for $ n,$ and we have a function $ f$ that is $ n\plus{}1$ times differentiable and has $ n\plus{}1$ zeros (counting multiplicities) as in the hypotheses of the problem. That is, $ f$ has a zero of order $ n_i$ at $ x_i,$ and $ n_1\plus{}n_2\plus{}\cdots\plus{}n_p\equal{}n\plus{}1.$

Consider $ g(x)\equal{}f'(x).$ $ g$ is $ n$ times differentiable.

By Rolle's theorem, there exist points $ y_i\in(x_i,x_{i\plus{}1})$ such that $ f'(y_i)\equal{}g(y_i)\equal{}0.$ We have $ p\minus{}1$ such zeros of $ g.$

In addition, if $ f$ has a zero of order $ n_i$ at $ x_i,$ then $ g(x)$ has a zero of order $ n_i\minus{}1$ at $ x_i.$ (If $ n_i\minus{}1\equal{}0,$ then that's simply not a zero at all.)

Hence, $ g$ has zeros at all of the $ y_i$ and some of the $ x_i,$ and the total order of these zeros is $ (n_1\minus{}1)\plus{}(n_2\minus{}1)\plus{}\cdots\plus{}(n_p\minus{}1)\plus{}p\minus{}1\equal{}n.$ The interval on which these zeros occur could be as large as $ [x_1,x_p]$ or as small as $ [y_1,y_{p\minus{}1}],$ but in any case is contained in $ [x_1,x_p].$

The inductive hypothesis applies to $ g$ and gives us that there exists $ \xi$ in our interval such that $ g^{(n\minus{}1)}(\xi)\equal{}0.$ That gives us that $ f^{(n)}(\xi)\equal{}0,$ which completes our proof.

In effect, the proof is the repeated use of Rolle's Theorem.

That is an awesome proof KM. My thinking was along those lines but I did not try to use induction and so came up with immense difficulties trying to count the intermittent zeros.

One last question, since we're at it. How can we construct the Hermite polynomial given this data about points where f vanishes and the order of these points?


Prove that :
[img]http://nhanh.vinaanh.com/ph/b6edc1c4807668c63f30.JPG[/img]

EDIT: Actually, the identity is wrong: $ (C_{4016}^{2008})^2\equal{}\sum_{k \equal{} 0}^{2008}\frac{4016!}{(k!)^2[(2008\minus{}k)!]^2}$

This is a case of the generalization $ \binom{2n}{n} \equal{} \sum_{k \equal{} 0}^{n}\binom{n}{k}^2$. Prove this. Also, this topic should be in Intermediate Topics. Please read guidelines before posting problems randomly.

[quote="10000th User"]EDIT: Actually, the identity is wrong: $ (C_{4016}^{2008})^2 \equal{} \sum_{k \equal{} 0}^{2008}\frac {4016!}{(k!)^2[(2008 \minus{} k)!]^2}$

[/quote]

I don't think so
(my teacher 4 me)
Prove that :
 $ (C_{2n}^{n})^2 \equal{} \sum_{k \equal{} 0}^{n}\frac {(2n)!}{(k!)^2[(n \minus{} k)!]^2}$

Please help me !

Here's my thinking:
You have $ 2n$ people, then you divide them into 2 groups of $ n$ each, group A and group B. Then you choose $ n$ people to form a committee.
The right hand side represents choosing $ n$ people out of $ 2n$ people twice; the right hand side represents choosing, to include in the committee, $ k$ people from group A and $ n\minus{}k$ people from group B.
Here's a visualization for $ n\equal{}4$:
AAAA|NNNN
MAAA|BNNN
MMAA|BBNN
MMMA|BBBN
MMMM|BBBB

This is generalized by [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's identity[/url] (it's the case $ n \equal{} m \equal{} r$).  minsoens has already given you a combinatorial proof (that is generalizable).

[hide="Second proof"] First we let the notation $ [x^n] P(x)$ denote the coefficient of $ x^n$ in $ P(x)$.  In particular, $ [x^k] (1 \plus{} x)^n \equal{} {n \choose k}$.  Then

$ (1 \plus{} x)^{n \plus{} m} \equal{} (1 \plus{} x)^n (1 \plus{} x)^m \implies$
$ [x^r] (1 \plus{} x)^{n \plus{} m} \equal{} \sum_{k \equal{} 0}^{\infty} \left( [x^k] (1 \plus{} x)^n \right) \left( [ x^{r \minus{} k}] (1 \plus{} x)^m \right)$. [/hide]
There is also a path-counting solution.  Can you find it?


group theory
abstract algebra
superior algebra
superior algebra solved

Hi.  I'm trying to self-learn some group theory.

How can I show the following:

If [tex]|G|[/tex] and [tex]|H|[/tex] are finite then [tex]|f(G)|[/tex] is finite and divides both [tex]|G|[/tex] and [tex]|H|[/tex].

Thank you,
Bailey

I think you assume $f$ being a homomorphism from $G$ to $H$.
Let be $I= \ker f$. Then $f(a)=f(b) \iff f(ab^{-1})=e \iff ab^{-1} \in I$ ($e$ neutral element), so there are $|f(G)|= \frac{ |G| }{|I|}=( G : I )$ right residue classes of $I$ and this number obviously divides $|G|$. The other way $f(G)$ is a subgroup of $H$ and Langrange's theorem gives the second result.

Thank you for your  reply....I'm trying to digest what you posted.  Hopefully I will understand it.

Have a nice day.