geometry
circumcircle
perpendicular bisector
convex quadrilateral
IMO Shortlist

We have that $\triangle{ADX}$ is similar with $\triangle{BCX}$, thus there exists 2 points $M,N$ on the perpendicular bisector of $(AB)$ such that $\triangle{AMN}$ is similar with $\triangle{ADX}$, and $\triangle{BMN}$ is similar with $\triangle{BCX}$.
So we have that $\frac{AD}{AM}=\frac{AX}{AN}$ and $\angle{DAM}=\angle{XAN}$, but $\triangle{ADM}$ beeing similar with $\triangle{AXN}$, we have $\frac{AD}{AX}=\frac{DM}{XN}$, and similar $\frac{BC}{BX}=\frac{CM}{XN}$, thus we have $CM=DM$, therefore $M$ lies on both perpendicular bisectors of $AB$ and $CD$, therefore $M \equiv Y$.
Now it follows easy, because $\angle{AYB}=2\angle{AYN}=2\angle{ADX}$.

[quote="Tiks"]Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.[/quote]
[hide="Solution"]
Let $ D'$ and $ C'$ be the points so that $ \triangle BC'A\sim \triangle BCX\sim \triangle ADX\sim \triangle AD'B$.  Let $ Z$ be the circumcenter of $ \triangle BD'A$.  Now, from the similarities, we have that $ \angle BD'A \equal{} \angle BC'A$, so $ ABC'D'$ is cyclic, which means that $ Z$ is the circumcenter of $ ABC'D'$.  Furthermore, since $ AB \equal{} AB$, we see that $ \triangle BD'A\cong \triangle AC'B$, so $ AD' \equal{} BC'$, so $ ABC'D'$ is an isosceles trapezoid.  This means that $ \angle ZD'A \equal{} \angle ZBC'$.  Now, we see that a spiral similarity maps $ \triangle ADD'\sim \triangle AXB\sim \triangle C'CB$.  From the fact that $ AD' \equal{} BC'$, we have that $ \triangle ADD'\cong \triangle C'CB$, so $ DD' \equal{} BC$ and $ \angle AD'D \equal{} \angle CBC'$.  We obtain that $ \angle ZD'D \equal{} \angle AD'Z \plus{} \angle AD'D \equal{} \angle ZBC' \plus{} \angle C'BC \equal{} \angle ZBC$.  Furthermore, $ ZD' \equal{} ZB$, so $ \triangle ZD'D\cong \triangle ZBC$, from which we observe that $ ZD \equal{} ZC$, so $ Z$ lies on the perpendicular bisector of $ CD$.  This means that since $ ZA\equal{}ZB$, $ Z$ is the intersection between the perpendicular bisectors of $ CD$ and $ AB$, so $ Z \equal{} Y$.  We conclude that $ \angle AYB \equal{} \angle AZB \equal{} 2\angle AD'B \equal{} 2\angle ADX$, as desired.  [/hide]

Does this solution work? I am really unsure as to whether it works because it seems like too simple a solution especially for an ISL #6. Does anyone think that this problem is a little similar to IMO 1975 #3?

[b]Solution[/b]

Define point $Y'$ to be on the perpendicular bisector of $AB$ and such that $\angle{AY'B}=2\angle{ADX}$. Define point $Z$ to be such that $\angle{AZY'}=\angle{BZY'}=\angle{AXD}=\angle{BXC}$. Note that since $Z$ and $Y'$ are on the perpendicular bisector of $AB$, $AY'=BY'$ and $AZ=BZ$.

By AA similarity, we have that $\triangle{AXD} \sim \triangle{AZY'}$ and $\triangle{BXC} \sim \triangle{BZY'}$. This yields by spiral similarity that $\triangle{AZX} \sim \triangle{AY'D}$ and $\triangle{BZX} \sim \triangle{BY'C}$. Therefore,

\[\frac{ZX}{DY'}=\frac{AZ}{AY'}=\frac{BZ}{BY'}=\frac{ZX}{CY'}\]
This yields that $DY'=CY'$ and therefore that $Y'=Y$ is the intersection of the perpendicular bisectors of $AB$ and $CD$. Therefore $\angle{AY'B}=\angle{AYB}=2\angle{ADX}$.

Let $M,N$ are midpoints of the sides $AB$ and $CD$ and $DX\cap AB=P(1)$, $AX\cap CD=Q(.)$. Then since $ \triangle AXD \sim \triangle BXC$ we get that  $ \angle BXP=\angle CXQ $, $ \angle AXP=\angle DXQ $ and \[ \frac{AX}{BX}=\frac{DX}{CX}. \] So \[ \frac{AP}{BP}=\frac{DQ}{CQ} \] and the parallel lines pass through $M$ and $N$ to $DX$ and $AX$ respectively, intersect on $AD$. Analoguosly, the parallel lines pass through $M$ and $N$ to $CX$ and $BX$ respectively, intersect on $BC$. Hence $ \angle AYB=2\angle ADX$.

Dear mathuz, i dont understand how the result follows. (rest is ok)
Could you explain a bit ?

My solution:\\
Let circumcerntres of $\triangle DAX$ and $\triangle BCX$ be $O_1,O_2$. Let $Y$ be point such that $AY=BY$ and $\angle AYB=2\angle ADX$. \\Since $\angle AO_1X=2\angle ADX$ we have $\triangle O_1AX$ ~$\triangle AYB$ so $\frac{AO_1}{AY}=\frac{AX}{AB}$ and since $\angle O_1AY=\angle XAB$ we have that $\triangle AO_1Y$~$\triangle AXB$. Similary we have $\triangle BO_2Y$~$\triangle AXB$.\\
Now we have $O_1Y=\frac{AY}{AB}BX$ and $O_2Y=\frac{BY}{AB}AX$ and $O_1X=\frac{AX}{AB}AY$ and $O_2X=\frac{BX}{AB}BY$.
\\Since $\angle AO_1Y=\angle YO_2B$ and $\angle DAX=\angle CBX$ we have $\angle DO_1Y=\angle CO_2Y$. Now by Law of Cosine we have\\
$DY^2=O_1D^2+O_1Y^2-2O_1D\cdot O_1Y\cos(DO_1Y)=AY^2(\frac{AX^2+BX^2-2AX\cdot BXcos(DO_1Y)}{AB^2})$ \\and
$CY^2=O_2Y^2+O_2C^2-2O_2Y\cdot O_2Ccos(CO_2Y)=AY^2(\frac{AX^2+BX^2-2AX\cdot BXcos(DO_1Y)}{AB^2})$ which implies\\
$DY=CY$ so $Y$ is the intersection of perpendicular bisectors of $AB,CD$.

Feels too convenient, so perhaps I missed something? :maybe:
[quote=Tiks]Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.[/quote]

WLOG let rays $\overrightarrow{BA}, \overrightarrow{CD}$ intersect. Let $M, N$ be midpoints of $\overline{AB}, \overline{CD}$ respectively. Let $P$ be the spiral center $\overline{AB} \mapsto \overline{DC}$. Construct $E, F$ outside $ABCD$ with $\triangle AEB \sim \triangle DYC$ and $\triangle AYB \sim \triangle DFC$. Observe that $\triangle AEY \sim \triangle DYF$ with spiral center $P$ again. Further, we see $k=\tfrac{EM}{MY}=\tfrac{YN}{NF}$. Let $L$ be the point on $\overline{DA}$ with $\tfrac{AL}{LD}=k$. Suppose $X$ is the point inside $ABCD$ with $\triangle LMN \sim \triangle XBC$.

[b]Lemma.[/b] $\overline{XL} \perp \overline{DA}$ and $\triangle XAD \sim \triangle XBC$.

(Proof) By linearity, $\triangle AEY \mapsto \triangle LMN \mapsto \triangle DYF$ under spiral similarity with pivot $P$. Apply $\triangle PDA \sim \triangle PNM \sim \triangle PYE$ hence $\angle PME=\angle PLA$. Combined with $\angle PMB=\angle PLX$ we obtain $\angle XLA=\angle BME=90^{\circ}$. Now reflect $X$ in $L$ to get $X'$. Again linearity gives $\triangle X'AD \sim \triangle XBC$ and we're done. $\blacksquare$

Finally, we see $\triangle XAD \sim \triangle LMN \sim \triangle DFY$ hence $\tfrac{1}{2}\angle AYB=\angle AYE=\angle ADX$ and we're done here.

Assume $AB,CD$ aren't parallel; otherwise the problem doesn't make sense.

Let $X_1,X_2$ be the reflections of $X$ over $BC,AD$ and $X_3,X_4$ be the projections of $X$ onto $BC,AD$. Let $M,N$ be the midpoints of $AB,CD$. Clearly $X_2AXD, XBX_1C$ are directly similar kites, so they're also similar to their direct average $X_4MX_3N$. Since $AX_4:X_4D=BX_3:X_3C$ and $AB,CD$ aren't parallel, this implies $X_4,X_3$ are the only such points on $AD,BC$ which divide segments $AD,BC$ in the same ratio with $X_3X_4\perp MN$.

Next construct $Y_1,Y_2$ with $DYC\sim AY_1B, DY_2C\sim AYB$. Once again $AY_1BY, DYCY_2$ are directly similar kites, so for any $r$ their weighted average $r(AY_1BY)+(1-r)(DYCY_2)$ is also a kite. Since $Y_1M:MY=YN:NY_2$ we can choose $r$ so that this resulting kite contains $M,N$; then the other two vertices of the kite must lie on $AD,BC$, divide those two segments in the same ratio, and create a segment perpendicular to $MN$. This implies these other two vertices are precisely $X_4,X_3$ by our earlier work, hence $MX_4NX_3\sim Y_1AYB$ and $\angle AYB =\angle X_2DX=2\angle ADX$; the other equality follows similarly.

Let us deal with non-trivial cases with $AB \not \parallel CD$.

Let $\odot (ADX)$ and $\odot (BCX)$ meet at $Z$, and $\odot (ABZ)$ meets $\odot (CDZ)$ at $W$. We claim that $W$ coincides with $Y$.

Simple angle chasing gives $\angle AYB=\angle AWB$, and $\angle CYD=\angle CWD$. Also line $ZX$ is the internal angle bisector of $\angle AZB, \angle CZD$, so it suffices to prove that $ZW \perp XZ$.

Now invert the diagram with centre $Z$. Denote by $T '$ by the image of any point $T$.
Note that $X'=A'D' \cap B'C'$, and $W'=A'B' \cap C'D'$, and $ZX'$ bisects both $\angle A'ZB'$ and $\angle C'ZD'$. Assume that line $ZX'$ meets $A'B', C'D'$ at $U, V$, and a line perpendicular to $ZX'$ passing $Z$ meets $A'B', C'D'$ at $P, Q$, respectively. Then,
$$ X'(A'B', UP)=Z(A'B',UP) = -1=Z(C'D', VQ)=X(C'D', VQ),$$ so $P$ and $Q$ coincides with $W'$. Thus $X'Z \perp WZ'$, which implies $W \equiv Y$. $\square$

First, a long phantom point argument: [list] [*]Fix $\triangle AXD$ first and let $O$ denote its circumcenter. [*]Let $B$ be as in the original problem. [*]Then we redefine $Y$ such that $\angle AYB = 2 \angle ADX = \angle AOX$ and $AY = BY$. [*]We then redefine $C$ such that $\angle DYC = 2 \angle DAX = \angle DOX$ and $YC = YD$. [*]We will prove $\triangle AXD \sim \triangle BXC$ (oppositely oriented). This will solve the problem. [/list]

[asy] pair A = dir(125); pair D = dir(275); pair X = dir(-5); pair O = origin; draw(unitcircle);

pair Y = 1.2*dir(-20); pair B = Y+(A-Y)*X/A;

pair C = Y+(D-Y)*X/D; filldraw(A--O--X--cycle, invisible, deepgreen); filldraw(A--B--Y--cycle, invisible, deepgreen); filldraw(D--O--X--cycle, invisible, blue); filldraw(D--C--Y--cycle, invisible, blue); filldraw(B--X--C--cycle, invisible, grey); draw(circumcircle(B, X, C), grey);

dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$O$", O, dir(-X)); dot("$Y$", Y, dir(Y)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C));

/* TSQ Source:

A = dir 125 D = dir 275 X = dir -5 O = origin R-X unitcircle

Y = 1.2*dir(-20) B = Y+(A-Y)*X/A

C = Y+(D-Y)*X/D A--O--X--cycle 0.1 lightgreen / deepgreen A--B--Y--cycle 0.1 lightgreen / deepgreen D--O--X--cycle 0.1 lightblue / blue D--C--Y--cycle 0.1 lightblue / blue B--X--C--cycle 0.1 yellow / grey circumcircle B X C grey

*/ [/asy]

Anyways, the main idea is that $\triangle AOX \sim \triangle AYB$ and $\triangle DOX \sim \triangle DYC$. We break symmetry now and set up as follows. Since we have a similarity $\triangle AOY \sim \triangle AXB$, there should be a single complex number $t \in {\mathbb C}$ such that \begin{align*} 	y-a &= t(0-a) \implies y=(1-t)a \\ 	b-a &= t(x-a) \implies b=t(x-a)+a. \end{align*} Next since $\frac{c-y}{d-y} = \frac{x-0}{d-0}$ we have \[ c = y + \frac{(d-y)x}{d} = \frac{d(x+y)-xy}{d}. \] Now, we simply calculate \begin{align*} 	\frac{c-b}{x-b} &= \frac{\frac{d(x+y)-xy}{d}-[t(x-a)+a]}{x-[t(x-a)+a]} \\ 	&= \frac{dx + (d-x)[(1-t)a] - dt(x-a)-da}{d(1-t)(x-a)} \\ 	&= \frac{(1-t)[(d-x)a+dx-da]}{d(1-t)(x-a)} \\ 	&= \frac{(1-t)x(d-a)}{(1-t)d(x-a)} = \frac{x(d-a)}{d(x-a)} \\ 	&= \frac{\frac1d-\frac1a}{\frac1x-\frac1a} 	= \overline{\left( \frac{d-a}{x-a} \right)} \end{align*} as desired.

hint : let $O$ be the circumcircle of $\triangle ADX$ then use rotation and similarity properties.  :jump:
also use this technique. you should began from verdict and arrive to assumptions.

Here's a pretty nice complex solution that nobody's posted yet.

Let $P, Q$, both inside $ABCD$, be points on the perp. bisector of $AB$ such that $\triangle{APQ}\sim \triangle{ADX}$ (note there is only one choice for $P$ and $Q$). For our setup, set $(ABP)$ be the unit circle, let $p=1$, and let $a, q, x$ be the free variables; note $b=\bar{a}=\frac{1}{a}$, and $q$ is real. We desire to prove $PC=PD$, or $|c-1|=|d-1|$.

Now, note $A$ is the spiral center of $PQ, DX$, by the spiral center formula we have $a=\frac{px-qd}{p+x-q-d}=\frac{x-qd}{1+x-q-d}$, which we can rearrange to get $d=\frac{a+ax-aq-x}{a-q}$. Thus, we have $d-1=\frac{(a-1)(x-q)}{a-q}$, so changing $a$ for $\frac{1}{a}$ in this expression, we get $c-1=\frac{(1-a)(x-q)}{1-aq}$. Thus, it suffices to prove that $|a-q|=|1-aq|$, or $(a-q)(\frac{1}{a}-q)=(1-aq)(1-\frac{q}{a})$. But multiplying both sides by $a$, we find the equation indeed holds, so we are done.

Noone has posted a short and neat sollutiion using complex weighted gliding lemma, so here it is :-).

[color=#00f][b]Complex Weighted Gliding Lemma:[/b][/color] Let $A_1B_1C_1$ and $A_2B_2C_2$ be two similar triangles. Construct points $X$, $Y$, $Z$ such that triangles $A_1XA_2$, $B_1YB_2$ and $C_1ZC_2$ are similar. Then $\triangle XYZ\sim\triangle A_1B_1C_1$.
[color=#f00][b]Proof sketch:[/b][/color] As $A_1B_1C_1\sim A_2B_2C_2$ there exist linear function $f$ of complex plane mapping $A_1B_1C_1\mapsto A_2B_2C_2$. Also consider identity function $i$ as a linear function of complex plane. Then to get to points $X$, $Y$, $Z$ you just take weighted average of these functions, but average of linear functions is linear, hence $XYZ\sim A_1B_1C_1$. $\square$

[color=#f00][b]Problem Proof:[/b][/color] Denote $X_1$ reflection of $X$ by $AD$ and $X_2$ reflection of $X$ by $BC$. Note that $X_1AX\sim XBX_2$, construct $Y'$ such that $DY'C\sim X_1AX$. Note that $X_1DX\sim XCX_2$, hence if we can complex glide them to $AY'B$, hence $AY'B\sim X_1DX$. Because triangles $AY'B$ and $DY'C$ are isosceles, we get $Y'=Y$ and from similarities we got the angle condition, that problem wanted. $\square$

Solved with [b]nukelauncher[/b].

[hide=Diagram (No geogebra, just Microsoft Paint!)]
[img]https://lh3.googleusercontent.com/-wEXeb24yCh0/X6JiZ56BeWI/AAAAAAAAG9k/JN5K3nx1zvAoPvi4VUmZP36NsC91cdlqgCLcBGAsYHQ/s0/ISL_2000_G6.PNG[/img]
[/hide]

Let $O$ be the cirumcenter of $\triangle ADX$. Redefine $Y$ such that $\triangle AOX\sim \triangle AYB$, hence $\angle AYB=\angle AOX=2\angle ADX$. We want to show $Y$ lies on the perpendicular bisector of $\overline{CD}$. Let $(ADX)$ be the unit circle, so $O=0$. We fix $A,X,D$ and $B$ as arbitrary points, and will calculate $Y$ and $C$ based on these points.

WLOG by rotation set $x=1$, so $OX$ is the real axis. From the spiral similarity sending $\triangle AOX$ to $\triangle AYB$, we have
 \[ \frac{y-a}{0-a} = \frac{b-a}{x-a} \implies y=\frac{-a(b-a)}{x-a}+a=a\cdot \frac{1-b}{1-a}\]
Let $\bullet'$ denote the reflection over real axis of $\bullet$. The angle conditions imply that there exists a spiral similarity sending $A'\mapsto B$ and $D'\mapsto C$. Hence
\begin{align*}
\frac{c-x}{\bar{d}-x} = \frac{b-x}{\bar{a}-x} \implies c&=\frac{(\bar d-x)(b-x)}{\bar a-x} +x \\
&=\frac{\left(\frac1d-1\right)(b-1)}{\frac1a-1} + 1 \\
&= \frac{b/d-1/d-b+1/a}{1/a-1} \\
&= \frac{ab-a-abd+d}{d(1-a)}.
\end{align*}
Since we wanted to show $Y$ is on the perpendicular bisector of $CD$, it suffices to prove that $y$ satisfies
\begin{align*}
|y-d|=|y-c|\iff (y-d)(\bar y-\bar d)=(y-c)(\bar y-\bar c).
\end{align*}
So if $Z=\tfrac{y-d}{y-c}$, we need to prove $Z\cdot \bar Z=1$. Magically, \begin{align*}     Z=\frac{y-d}{y-c}&=\frac{a\cdot\frac{1-b}{1-a}-d}{a\cdot\frac{1-b}{1-a}-\frac{ab-a-abd+d}{d(1-a)}}\\     &=\frac{a(1-b)-d(1-a)}{a(1-b)-\frac{ab-a-abd+d}d}\\     &=\frac{a-ab-d+ad}{a+\frac{a-ab}d-1}\\     &=\frac{ad-abd-d^2+ad^2}{ad+a-ab-d}\\     &=d. \end{align*}
Hence $Z\cdot\bar Z=d\cdot \bar d=1$. The end.

[hide=Motivational Remarks]Really hard to get a foothold on this problem. Since the quadrilateral is quite unwieldy, we started with $\triangle AXD$, and then a reflection over any line through $X$ followed by a spiral similarity at $X$ would get $\triangle XCB$. We tried to get $2\angle ADX$ by reflecting $X$ over $AD$, but couldn't get anywhere from there.

After a long time, we realized another way to get $2\angle AXD$ was to involve the circumcenter of $\triangle AXD$. Then we want to show $\triangle AOX$ spiral sims to $\triangle AYB$ at $A$. Note that $B$ can be any point in the plane by degrees of freedom ($\triangle XCB$ can be scaled and pivoted about $X$), so this suggested redefining $Y$ in terms of this spiral sim. After this redefinition, the complex value of $Y$ is pretty simple. To get $C$, we realized that if we set $X=1$, then we can just reflect $\triangle AXD$ over $OX$ -- the real axis -- to get the conjugates of $A$ and $D$, and then there exists a spiral sim to $\triangle CBX$! This used the idea from before, and since we have $A,X,D,B$, solving for $C$ is quite nice.

Finally, we want to show $Y$ lies on the perpendicular bisector of $CD$ by the redefinition, and magically, $(y-d)/(y-c)$ simplified to just $d$, which means $(y-d)/(y-c)$ is the reciprocal of its conjugate, proving $YD=YC$! The reason for this was that  \[ \frac{y-d}{y-c} = \frac{d-o}{x-o} = d\] since $\angle DYC=\angle DOX$, which is the symmetric equivalent of $\angle AYB=\angle AOX$ from our redefinition. In retrospect, we ``had to have'' $(y-d)/(y-c)$ cancel out nicely to $d$.[/hide]

complex solution without phantom points

[asy]
unitsize(2cm);
pair A, B, C, D, X, Y;
X=(0,0);
A=(-1,2);
B=(-2,-2);
C=(3,-1);
D=A*(2*foot(C/B,(0,0),(1,0))-C/B);
Y=extension((A+B)/2,circumcenter(A,B,C),(C+D)/2,circumcenter(B,C,D));
draw(A--B--C--D--A--X--D--C--X--B);
draw(A--Y--B--C--Y--D);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$X$", X, W);
label("$Y$", Y, S);
[/asy]

Since $\angle ADX=\angle BCX$ and $\angle DAX=\angle CBX$, this means that $\triangle ADX\sim\triangle BCX$, so $\frac{d-x}{a-x}=\frac{\overline c-\overline x}{\overline b-\overline x}$. since $X$ is inside $ABCD$. Assume without loss of generality $x=0$. Then, we have $\overline bd=a\overline c$.
Now, since $|y-a|=|y-b|$, we must have
\begin{align*}
(y-a)(\overline y-\overline a)&=(y-b)(\overline y-\overline b)\\
-a\overline y-y\overline a+|a|^2&=-b\overline y-y\overline b+|b|^2\\
y(\overline b-\overline a)+\overline y(b-a)&=|b|^2-|a|^2.
\end{align*}
Similarly, we have
$$y(\overline d-\overline c)+\overline y(d-c)=|d|^2-|c|^2.$$ Therefore, we have
$$y=\frac{(|b|^2-|a|^2)(d-c)-(b-a)(|d|^2-|c|^2)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$ We need to show that $\frac{(a-y)(d-x)^2}{(y-b)(a-d)^2}$ is real. We have $$y-b=\frac{\overline a(b-a)(d-c)-(b-a)(|d|^2-|c|^2)+(b-a)(b\overline d-b\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}=\frac{(b-a)(\overline ad-\overline ac-d\overline d+c\overline c+b\overline d-b\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$ Similarly, we have
$$y-a=\frac{(b-a)(\overline bd-\overline bc-d\overline d+c\overline c+a\overline d-a\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$ Therefore, we have $\frac{a-y}{y-b}=-\frac{\overline bd-\overline bc-d\overline d+c\overline c+a\overline d-a\overline c}{\overline ad-\overline ac-d\overline d+c\overline c+b\overline d-b\overline c}$. Since $\overline bd=a\overline c$, we also have $b\overline d=\overline ac$. Therefore, this means that
$$\frac{a-y}{y-b}=-\frac{-\overline bc-d\overline d+c\overline c+a\overline d}{\overline ad-d\overline d+c\overline c-b\overline c},$$ so $$\frac{(a-y)(d-x)^2}{(y-b)(a-d)^2}=\frac{(\overline bc+d\overline d-c\overline c-a\overline d)d^2}{(\overline ad-d\overline d+c\overline c-b\overline c)(a-d)^2}.$$ Therefore, it suffices to show that $$\frac{(\overline bc+d\overline d-c\overline c-a\overline d)d^2}{(\overline ad-d\overline d+c\overline c-b\overline c)(a-d)^2}=\frac{(b\overline c+d\overline d-c\overline c-\overline ad)\overline d^2}{(a\overline d-d\overline d+c\overline c-\overline bc)(\overline a-\overline d)^2}.$$
This is equivalent to
$$\frac{(\overline bc+d\overline d-c\overline c-a\overline d)^2}{(\overline ad-d\overline d+c\overline c-b\overline c)^2}=\frac{(a-d)^2\overline d^2}{d^2(\overline a-\overline d)^2}.$$ We will show that $\frac{\overline bc+d\overline d-c\overline c-a\overline d}{\overline ad-d\overline d+c\overline c-b\overline c}=\frac{(a-d)\overline d}{d(\overline d-\overline a)}$. Adding one to each side gives $$\frac{\overline ad-a\overline d+\overline bc-b\overline c}{\overline ad-d\overline d+c\overline c-b\overline c}=\frac{a\overline d-\overline ad}{d(\overline d-\overline a)},$$ which is equivalent to $$\frac{a\overline d-\overline ad+b\overline c-\overline bc}{d(\overline d-\overline a)+\overline c(b-c)}=\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}.$$ Therefore, it suffices to show $\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}=\frac{b\overline c-\overline bc}{\overline c(b-c)}$. Since $d=\frac{a\overline c}{\overline b}$, this means that we have
\begin{align*}
\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}&=\frac{a\frac{\overline ac}b-\overline a\frac{a\overline c}{\overline b}}{\frac{a\overline c}{\overline b}(\frac{\overline ac}b-\overline a)}\\
&=\frac{\frac cb-\frac{\overline c}{\overline b}}{\frac{\overline c}{\overline b}(\frac{c-b}b)}\\
&=\frac{\overline bc-b\overline c}{\overline c(c-b)}\\
&=\frac{b\overline c-\overline bc}{\overline c(b-c)}.
\end{align*}

Therefore, this means that $\angle AYB=2\angle ADX$.

Let $Y',Z$ be points on perpendicular bisector of $AB,$ such that $AY'Z\stackrel{+}{\sim} ADX,BY'Z\stackrel{+}{\sim} BCX.$
By spiral similarity $AXZ\stackrel{+}{\sim} ADY',BXZ\stackrel{+}{\sim} BCY',$ so $$\frac{|AY'|}{|Y'D|}=\frac{|AZ|}{|ZX|}=\frac{|BZ|}{|ZX|}=\frac{|BY'|}{|Y'C|}\implies |Y'C|=|Y'D|\implies Y'=Y.$$ Finally $\measuredangle AYX=2\measuredangle AYZ=2\measuredangle ADX,$ as desired.

Note that $\triangle AXD\sim \triangle BXC$. There exists a unique spiral similarity that takes $DX$ onto line $EY$. Let $D\to P$ and $E\to Q$. We have that $\triangle BQP\cong \triangle AQP$ so $\triangle BQP \sim \triangle BXC$. We also have $\triangle BQX\sim \triangle BPC$ and $\triangle AQX\sim\triangle APD$. Thus,
\[\frac{DP}{CP}=\frac{AP}{BP}\cdot \frac{QX}{AQ}\cdot \frac{BQ}{QX}=1\]
so $CP=DP$, implying $P=Y$ and the result follows.

Weird solution:
Let $O$ be the circumcenter of $\triangle AXD$, and denote by $K, Z$ the images of $D, O$ under the spiral similarity taking $AD$ to $AB$.
Now $DK/AD=BX/AX=BC/AD$ so $DK=BC$. Now in directed angles,
\[ \angle ZKD = \angle ZKA + \angle AKD = \angle ZKA + \angle ABX = \angle ZKA + \angle ABZ + \angle ZBX = \angle KAB + \angle ZBX = \angle XBC + \angle ZBX = \angle ZBC \], so \triangle ZBC \cong \triangle ZKD$, whence $Z = Y$. This finishes.


calculus
integration
function
derivative
limit
geometry
geometric transformation

Hello

I want to compute this integral: \[ \int_{\minus{}\infty }^{a}\exp (\minus{}itx)dx\]

can I compute it like an ordinary integral and say that it is equal to $ \frac {\minus{}1}{it} exp(\minus{}ita)$

or it is something more complicated than that?

thank you

Yes and no. Yes, in practical terms integration of complex-valued functions is no different from integration of real-valued functions: the Fundamental Theorem of Calculus is true for both (although the Mean Value Theorem fails for complex-valued functions). But no, the integral is not equal to $ \frac {\minus{}1}{it} \exp(\minus{}ita)$. It diverges because antiderivative does not have a limit at $ \minus{}\infty$.

(I assume that $ t$ is real.)

Thank you for your answer :-)

Actually, in a paper I found, the author is giving a function and its Fourier transform.

The initial function is:
\[ f(x) \equal{} (2l)^{ \minus{} p}\sum_{j \equal{} 0}^{p}\binom{p}{j}( \minus{} 1)^{j}F_{n}(x \plus{} (p \minus{} 2j)l)
\]
where
\[ F_{n}(u) \equal{} \frac {1}{n}\sum_{k \equal{} 1}^{n}1(X_{k} < u)
\]
$ F_{n}(u)$ is the empirical distribution function, based on an iid sample $ X_k$ of size $ n$
the function $ f(.)$is an estimator for the derivative of a distribution function

The author gives the Fourier transform of f:
\[ (2l)^{ \minus{} p}\sum_{j \equal{} 0}^{p}\binom{p}{j}( \minus{} 1)^{j}\frac {1}{n}\sum_{k \equal{} 1}^{n}\exp (it\left[ X_{k} \minus{} (p \minus{} 2j)l\right] )/( \minus{} it)
\]
What I don't understand is how did he find this result.
First, the function $ F_{n}(.)$ does not belong to L1 so how did he do in order to compute the FT.
I tried to ignore this and to compute the integral, but you told me that it does not converge. (I am not surprised by your answer)
And still, if I look at his result, it seems that he used something like the result that you said it does not converge.

On the other hand, the function f(x) belongs to L1, and it has a FT..
So I'm lost
:-)

Could you help me with that?
Thank you

$ F$ doesn't belong to $ L^1$, but $ f$ does.

The Fourier transform of $ F$ does exist in the sense of tempered distributions, and it blows up at zero; there's a multiple of the principal value of $ \frac1s$, and a multiple of the Dirac $ \delta$. To compute it, find the transform of $ f$ and recall the relation $ \widehat{g'}(s)\equal{}is\widehat{g}(s)$ (possibly with another constant, depending on your definition). That division by $ s$ gets you everything except the appropriate multiple of $ \delta$, which comes from $ \frac{F(\infty)\plus{}F(\minus{}\infty)}{2}$

[quote="andreea.b"]Hello

I want to compute this integral:
\[ \int_{ \minus{} \infty }^{a}\exp ( \minus{} itx)dx
\]
can I compute it like an ordinary integral and say that it is equal to $ \frac { \minus{} 1}{it} exp( \minus{} ita)$

or it is something more complicated than that?

thank you[/quote]

Evaluating complex integrals is a bit of a different beast than evaluating real integrals.  The integral which you gave is nonetheless evaluated straight forwardly. Though its limit as the lower limit of integration approaches minus infinity is not defined.

Actually, there's no difference here; a complex-valued function might as well be an $ \mathbb{R}^2$-valued function if it's a function of a real variable. The only reason to use complex numbers is algebraic convenience; it's easier to write $ \frac{\minus{}1}{it}e^{\minus{}itx}$ as an antiderivative than the real analogue $ \left(\frac1t\sin(tx),\frac1t\cos(tx)\right)$.
Also, this came from a Fourier transform, which is normally written with complex notation. That transform viewpoint allows us to give an answer despite the divergence of the integral.

[quote="andreea.b"]...What I don't understand is how did he find this result.
First, the function $ F_{n}(.)$ does not belong to L1 so how did he do in order to compute the FT.
I tried to ignore this and to compute the integral, but you told me that it does not converge. (I am not surprised by your answer)
And still, if I look at his result, it seems that he used something like the result that you said it does not converge.

On the other hand, the function f(x) belongs to L1, and it has a FT..
So I'm lost
:-)

Could you help me with that?
Thank you[/quote]

The FT is defined as $ F\left(\omega\right) \equal{}\int_{\minus{}\infty}^{\plus{}\infty} f\left(t\right) e^{\minus{}i\omega t}\, dt$.

Typically $ f\left(t\right)$ is bounded such that $ \lim_{|t|\to\infty} f\left(t\right) \equal{}0$ Therefore, the FT of such a function actually exists.

One of the techniques that often proves really helpful is that

$ \mathcal{F}\left\{\frac{d^nf\left(t\right)}{dt^n}\right\}\equal{}\left(i\omega\right)^n\mathcal{F}\left\{f\left(t\right)\right\}$

[quote="jmerry"]$ F$ doesn't belong to $ L^1$, but $ f$ does.

The Fourier transform of $ F$ does exist in the sense of tempered distributions, and it blows up at zero; there's a multiple of the principal value of $ \frac1s$, and a multiple of the Dirac $ \delta$. To compute it, find the transform of $ f$ and recall the relation $ \widehat{g'}(s) \equal{} is\widehat{g}(s)$ (possibly with another constant, depending on your definition). That division by $ s$ gets you everything except the appropriate multiple of $ \delta$, which comes from $ \frac {F(\infty) \plus{} F( \minus{} \infty)}{2}$[/quote]

$ f_{n}(x)\equal{}\frac{1}{n}\sum_{i\equal{}1}^{n}\delta _{\left( X_{i}\right) }(x)$

$ \widehat{f}_{n}(t)\equal{}\int \frac{1}{n}\sum_{i\equal{}1}^{n}\delta _{\left( X_{i}\right) }(x)e^{\minus{}itx}dx\equal{}\frac{1}{n}\sum_{i\equal{}1}^{n}e^{\minus{}itX_{i}}$

Using$ \widehat{f}_{n}(t)\equal{}it\widehat{F}_{n}(t)$
I can say that $ \widehat{F}_{n}(t)\equal{}(it)^{\minus{}1}\frac{1}{n}\sum_{i\equal{}1}^{n}e^{\minus{}itX_{i}}$

Is this right?
thank you

[quote="Dr. No"]The FT is defined as $ F\left(\omega\right) \equal{} \int_{ \minus{} \infty}^{ \plus{} \infty} f\left(t\right) e^{ \minus{} i\omega t}\, dt$.

Typically $ f\left(t\right)$ is bounded such that $ \lim_{|t|\to\infty} f\left(t\right) \equal{} 0$ Therefore, the FT of such a function actually exists.[/quote]

$ f\to 0$ as $ |t|\to\infty$ is not enough to ensure that the relevant integral is well defined.

for most purposes, you want $ f$ to be in some better class such as $ f\in L^1$ (so that the integral converges in the 'usual' sense), or $ f\in L^2$ (where you have to do some technical work to define the FT but in which it has better properties (the FT (with a normalizing factor of $ (2\pi)^{\minus{}1/2}$) is a unitary isomorphism of $ L^2$)) or $ f\in\mathcal{S}$ (the schwarz class, which has much better decay properties than merely $ f\to 0$ as $ |t|\to \infty$)

[quote="andreea.b"][quote="jmerry"]$ F$ doesn't belong to $ L^1$, but $ f$ does.

The Fourier transform of $ F$ does exist in the sense of tempered distributions, and it blows up at zero; there's a multiple of the principal value of $ \frac1s$, and a multiple of the Dirac $ \delta$. To compute it, find the transform of $ f$ and recall the relation $ \widehat{g'}(s) \equal{} is\widehat{g}(s)$ (possibly with another constant, depending on your definition). That division by $ s$ gets you everything except the appropriate multiple of $ \delta$, which comes from $ \frac {F(\infty) \plus{} F( \minus{} \infty)}{2}$[/quote]

$ f_{n}(x) \equal{} \frac {1}{n}\sum_{i \equal{} 1}^{n}\delta _{\left( X_{i}\right) }(x)$

$ \widehat{f}_{n}(t) \equal{} \int \frac {1}{n}\sum_{i \equal{} 1}^{n}\delta _{\left( X_{i}\right) }(x)e^{ \minus{} itx}dx \equal{} \frac {1}{n}\sum_{i \equal{} 1}^{n}e^{ \minus{} itX_{i}}$

Using$ \widehat{f}_{n}(t) \equal{} it\widehat{F}_{n}(t)$
I can say that $ \widehat{F}_{n}(t) \equal{} (it)^{ \minus{} 1}\frac {1}{n}\sum_{i \equal{} 1}^{n}e^{ \minus{} itX_{i}}$

Is this right?
thank you[/quote]

Hello,

Reading all these posts in this topic, I have the feeling that the answers "seem" to answer your question in different directions, but in fact after all, you seem to be so lost as to the start of this topic.

I don't want to polemicate here, and because my english is very poor (thanks Systranbox for the translation  :) ...), I will try to be as explicit as I can.

First, it seems that in all the answers you received, there is an aspect that is not well presented. You asked how to perform a Fourier Transform in a "classical" way, and the gentle people which gave answers to your topic viewed many aspects of the problem.

Yes, for the Fourier Transform we need functions in L1 space... Yes, the zero limit to the end of real axes is not suficient... Yes, we can work with L2 spaces and so on...

But for your problem, all the previous answers miss the fact that your 'f' is a density function, so is positive with integral over R equal to 1. So in L1 space, so in L2 space...

By the way, if someone rest too suspicious (from the intelectual point of view), in this way one can polemicate the definition of the characteristic function for a random variable (which is a well-defined function), and of course, totally reject the empirical counterpart obtained from a sample, by "plug-in", using the definiton you used for the 'fn' as mean of Diracs centred in the sample's points.

Second, all the key of the problem is the Dirac function as particular case of distributions. But the Dirac function is not a function ! One can define it as a 'function', but with "strange" properties (equal to zero almost everywhere, but with integral equal to one... not too strange if we try to assimilate it with a degenerate density function).

On the other hand, the Dirac "function" is a distribution (not in the sense of the probability theory, but in the sense of Schwarz: as "generalized function" acting on the space of the test functions), and a measure (because of this, your empirical counterpart of the density as mean of Diracs is a measure too).

So... now, if you want to compute your Fourier Transform in a "classical way" for your functionals, you will be in trouble. And all the answers you obtained before will keep you so lost as in the starting point of this topic. But if you accept some conventions which are not virtual, but are 'translations' of some physical facts (classical or quantum physics), you can compute the Fourier Transform of the constant function, of the Heaviside function (all these functions are not in L1 !), and of course of a Dirac distribution...

Try to prove that the Fourier Transform of a Dirac is equal to 1, in a classical way :-).... or that the Fourier Transform of  a constant is a Dirac... And however, the Fourier Transform exists in the space of the distributions.

The idea is that if you want to prove that something (noted by F) is a Dirac, you must prove that this "thing" acts like a Dirac, so you must compute the <F, g> in the space of temperated distributions (in particular for some "good" test function with a nice decay to infinity).

In conclusion, the last calculus you made is corect in the sense of the distributions, and your Fourier Transform exists :-)

Have a nice day !

There exist many types of Fourier Transform, following the value of two parameters a and b. As is pointed out at :
  http://mathworld.wolfram.com/FourierTransform.html

"Unfortunately, a number of other conventions are in widespread use. For example, (0,1) is used in modern physics, (1,-1) is used in pure mathematics and systems engineering, (1,1) is used in probability theory for the computation of the characteristic function, (-1,1) is used in classical physics, and (0,-2pi) is used in signal processing. In this work, following Bracewell (1999, pp. 6-7), it is always assumed that a=0 and b=-2pi unless otherwise stated."

 You can see the Fourier Transform of some typical functions, but a more exhaustive list you can find at:
  http://en.wikipedia.org/wiki/Fourier_transform

 So the constants of the integration will be different following the definition used for the Fourier Transform (which is not unique). If you want the unitary definition which allows the symmetry between the direct and the inverse transform, you must divide by sqrt(2*Pi).

Repeating my point from earlier- that last calculation is [b]not[/b] right, because its inverse transform is $ F_n\minus{}\frac12$. If there isn't a $ \delta$ at zero, we can't get a nonzero average value, and this cumulative distribution function has $ F_n(\minus{}\infty)\equal{}0,F_n(\infty)\equal{}1$.

[quote="jmerry"]Not quite- you can add something that will be zero after multiplying by $ it$, and you have to in this case. The exact quantity you need comes from the constant of integration.[/quote]

I don't understand; where can I add something?

[quote="jmerry"]Repeating my point from earlier- that last calculation is [b]not[/b] right, because its inverse transform is $ F_n \minus{} \frac12$. If there isn't a $ \delta$ at zero, we can't get a nonzero average value, and this cumulative distribution function has $ F_n( \minus{} \infty) \equal{} 0,F_n(\infty) \equal{} 1$.[/quote]

I tried to compute the inverse transform of my result, but I'm not able to find the solution. I split the integral in two parts, around 0.  I'm stuck at the part with $ \lim_{\epsilon\rightarrow 0}\int_{\minus{}\infty }^{\minus{}\epsilon }(it)^{\minus{}1}\exp (iat)dt$
It's a kind of exponential integral, but I don't know how to go further.

I saw in your previous post that you were saying something about $ \frac{F_n( \minus{} \infty) \plus{}F_n(\infty)}{2}$ which in this case is $ \frac12$ but I don't see where I can use this.

That limit diverges; it's comparable to $ \minus{}\ln(\epsilon)$ in absolute value. If you want anything convergent, you need to combine with the other side before taking limits.

While multiplication by $ t$ is invertible for smooth functions, it's not for distributions: $ t\delta(t)\equal{}0$. After the transform, that corresponds exactly by the ambiguity of the indefinite integral.
Thus, we add $ \delta$ (or a specific constant multiple, depending on the exact version of the Fourier transform you use) for each $ 1$ we add to the function globally. Which functions don't have a $ \delta$ in them? Functions with average value zero. The transform of an $ L^1$ function is finite everywhere, and $ L^1$ functions have average value zero.
Also important: you have to be careful defining $ \frac1t$ as a distribution. If you want it to make sense, it should be specified as a principal value.

could you recommend me a book/(s) /website that could help me understand all this things?
thank you

[quote="jmerry"]
Also important: you have to be careful defining $ \frac1t$ as a distribution. If you want it to make sense, it should be specified as a principal value.[/quote]

The Fourier Transform of the distribution 1/t is well-known and is equal to -i*Pi*signum(s) (following the definition that andreea used for its Fourier transform).

[quote="jmerry"]That limit diverges; it's comparable to $ \minus{} \ln(\epsilon)$ in absolute value. If you want anything convergent, you need to combine with the other side before taking limits.

While multiplication by $ t$ is invertible for smooth functions, it's not for distributions: $ t\delta(t) \equal{} 0$. After the transform, that corresponds exactly by the ambiguity of the indefinite integral.
Thus, we add $ \delta$ (or a specific constant multiple, depending on the exact version of the Fourier transform you use) for each $ 1$ we add to the function globally. Which functions don't have a $ \delta$ in them? Functions with average value zero. The transform of an $ L^1$ function is finite everywhere, and $ L^1$ functions have average value zero.
Also important: you have to be careful defining $ \frac1t$ as a distribution. If you want it to make sense, it should be specified as a principal value.[/quote]

I don't understand the logic of the purpose... Wanting to compute a Fourier transform for distributions, using classical integral calculus is not a funny work, at least for someone having little knowledge in the calculus of distributions (is visible that andreea don't understand all that you say).

Of course that the limit diverges... as classical integral calculus. But after that, adding complements from the distributions calculus (Dirac and so on) for proving a result started in a classical way, is a little bit vicious (in my opinion).

All that andreea needs can be found in the "classical" results about the distributions and their Fourier Transforms (results that one can try to prove by a classical integral calculus, but is better to work in the associated Hilbert space.

The limit I was referring to was the last line of the previous post; trying to decouple the two sides of a principal value is a bad idea.

Looking back at andreea's second post, there is a classical way to approach this- $ f$ is a linear combination of characteristic functions of [b]bounded[/b] intervals. Take the simple case $ p\equal{}1$ and $ n\equal{}2$, with the samples equal to $ \minus{}1$ and $ 1$.
We have $ F_n(x)\equal{}\begin{cases}0&x<\minus{}1\\ \frac12&\minus{}1<x<1\\ 1&1<x\end{cases}$. Then $ f(x)\equal{}\frac1{2l}\left(F_n(x\plus{}l)\minus{}F_n(x\minus{}l)\right)\equal{} \frac1{2l}\left(\frac12\chi(I_1)\plus{}\frac12\chi(I_2)\right)$, where $ \chi(I_1)$ and $ \chi(I_2)$ are the characteristic functions of $ (\minus{}1\minus{}l,\minus{}1\plus{}l)$ and $ (1\minus{}l,1\plus{}l)$ respectively. Now that's something you can take a transform of; finite intervals behave nicely.

In this light, the question that started the thread was the wrong question to ask. Probability densities have nice transforms (even for discrete distributions), and cumulative distribution functions don't. Most of my posts in this thread have really focused on that initial wrong question.


In what resolution do you usually prefere to work?

I work in the highest resolution my computer supports - 1920x1080 on windows, 1400x1050 on linux, though i only use half the width on them usually. A better guide is probably 800x600 or 1024x768

Does your monitor support those resolutions also? Or is it just your video card? :) They seems pretty huge resolutions ...

I use 'standard' [by looking at the poll] 1024x768 even though I could have higher one - I feel just .... hm ... confident with this one  :lol:

I just got Windows XP recently, and I kinda like 1024x768 better than Windows ME's 800x600.

Usually a larger resolution allows for higher details. There are some draw backs though:

1) you need a good graphics card to maintain a higher resolution
2) you need a good monitor, if it's a CRT you should look for refresh rates above 85, and if you are using a TFT, the native resolution is the best.
3) some times the windows icons seem very small. This can be easily managed by changing the font size in the advanced controls panel.

Depends on the operating system. If I am using Red Hat, I prefere 1280x1024, if I work in Windows, I prefer 1024x768!

Whenever I'm using my Windows ME I prefer to use an 800 x 600 screen resolution, and when I'm using a laptop, a 1024 x 768 screen resolution looks better!

Okay so now we should do a goodness of fit test for the data.. or maybe wait for a few more votes and then do it.  Does the student T-distrobution adequately describe the resolutions of monitors? Or maybe the tails are small enough that normal fits the data.

Then we will know the distrobution and be able to calculate things like the probability someone has a monitor resolution greater than 1024x768. Or the probability that someone has a resolution between 600x800 and 1280x1024.

Sound fun? :-D

1024x768... because my parents arent letting me buy out of this crappy monitor....... :(

I use an unusual setting on my laptop - 1400x1050!  On my desktop, it only allows me to use one setting - 1024x768, I think.

I'm using 1280x1024 on Linux, but that forces me to use a 60Hz refresh rate (my monitor can't use a 75 rate for any resolution above 1152x864 for some reason) and I'm worried that a rate that low could be harmful for eyes and/or head. Anyone know anything about that?

[quote="stupidkid"]I never understood why my ATI Radeon X800XT computer only supports 1280x1024 (on a 19in LCD monitor) max and my ATI Mobility Radeon on my laptop can go up to about 2048 something (only on a 15in monitor).[/quote]

There is no monitor to date that is 15" viewable that supports 2048xsomething resolution, be it laptop or not. You couldn't even imagine how small that would be.

Also, note that the resolutions you can choose from come from your graphics card drviers, not the graphics card itself or the monitor. Most drivers these days (if you have up-to-date drivers) will detect acceptable settings for your monitor and only give you those to choose from. Laptops monitors are sketchy business, and neither ATI nor Nvidia supports custom drivers for Mobile cards, because you can easily screw up your screen and they don't want to be liable.

I personally have a 15.4" WUXGA laptop and I run 1920x1200 resolution, which is the max for my monitor. Stuff is very small and that's the way I like it. When I'm at home during the summer I have another monitor that sits on my desk that I hook up for a dual monitor rig. It's a bit older and only supports 1600x1200 so that's what I run on that.

[quote=checkmatetang][quote=chessmaster22]Time to revive this old post :D[/quote]

Hi, why do you revive threads? You've seemed to do that a lot. And you don't offer any additional material- just that you want to revive the thread.[/quote]

This is the second thread I revived...

I personally, think that 1600 x 1200 is a good resolution to put the site in.  Its not too lagy yet it provides very detailed images.

[quote=chessmaster22][quote=checkmatetang]Okay, still, why is it necessary?[/quote]

Why isn't it?[/quote]

Uh... cause it's spammy?

im just bringing old threads that people created back to life

I just want people to be able to voice their opinion about the topic without them having to look to deep into AoPS' site


I'm reposting this from the "Round Table" section because I wasn't sure in which category it fit:

I'm currently reading Euclid's "The Elements," and I had some questions which might be useful to others who are also studying the book.

First of all, what is Euclid meant to teach me? I see that most of his proofs show concepts we learned in basic geometry, although I would never have thought of them myself. Is the importance of Euclid in the way he reached his solutions (proofs)?

Second, I was wondering what I could do to get the most out of the book. One of my friends said that he tried to independently prove each proposition before looking at the so-called "solution," but I was wondering whether this was efficient. I know many of you have found the solution of a problem obvious, but only after you see it. That happens to me a lot, so I'm kind of hesitant about reading the proofs straight.

Thank you for considering this post. Any suggestions would be greatly appreciated.

Actually, I'd say that reading that for geometry probably actually isn't that wise.

Many of the stuff in there you can "assume" in modern contests/proofs. Also, some more problem solving orientated geometry might be good.


trigonometry
geometry
geometric transformation
homothety
power of a point
radical axis
geometry proposed

In isosceles triangle $ABC(AC=BC)$ the point $M$ is in the segment $AB$ such that $AM=2MB,$ $F$ is the midpoint of $BC$ and $H$ is the orthogonal projection of $M$ in $AF.$ Prove that $\angle BHF=\angle ABC.$

We need only prove $BF^{2}=FH\cdot FA$, or $BF^{2}-FA^{2}=AH\cdot FA$(*). Put $CA=CB=x,BA=y$. Then $AH=AM\cdot \cos \angle{HAM}=\frac{2x}{3}\cdot \frac{AF^{2}+y^{2}-\frac{x^{2}}{4}}{2AF\cdot y}$. Now, I see $AF^{2}=\frac{x^{2}+y^{2}}{2}-\frac{x^{2}}{4}$ and $BF^{2}=\frac{x^{2}}{4}$. Therefore (*) is easy too.

Another proof:
[img]8274[/img]

Use a homotety with centre B and factor 2. You can easily see that the problem is equivalent to prove that $AF \perp C'K$ where $\frac{AK}{KB}= \frac{1}{2}$.

Now I'll use vectors. The origin is O, the circumcentre.
$(A-F)(C'-K) = (A-\frac{B+C}2)(\frac{2}{3}A+\frac{1}{3}B+C) = \ldots = (\frac{2}{3}A^{2}-\frac{1}{6}B^{2}-\frac{1}{2}C^{2})+\frac{2}{3}C(A-B) = 0$.

Let $E$ be a point on $BC$ such that $CE/BE=2$
We have $ME//AC$ so $EM/AC=EB/BC=1/3=EF/FC=DE/AC$ So $ME=ED$
Because $EM//AC$ and triangle $ABC$ is isosceles we have triangle $MEB$ is isosceles
so $ME=EB=ED$
Thus $\angle DBM=90=\angle MHD$
So $M,B,D,H$ are cyclic
This means $\angle BHF=\angle DMB=\angle ABC$ $q.e.d$

A simpler proof would be to observe that $ \tan \angle ABC \equal{} 3 \tan \angle FAB \equal{} \tan \angle BHF$

[quote="bilarev"] [color=darkred]In isosceles triangle $ ABC\ ,\ (CA \equal{} CB)$ consider the point $ M\in (AB)$ so that $ MA \equal{} 2\cdot MB$ , the midpoint $ F$ of the side $ [BC]$ and the orthogonal projection $ H$ of the point $ M$ on the line $ AF$ . Prove that $ \angle BHF \equal{} \angle ABC$ .[/color]  [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the midpoint $ L$ of the side $ [AB]$ , the centroid $ G\in AF\cap CL$ of the triangle $ ABC$ , the midpoint $ N$ of the segment $ [AM]$ and the point $ D\in AF$ for which $ DB\perp AB$ . Thus, $ LN\equal{}LM$ , $ LA\equal{}LB$ , $ AN \equal{}$ $ NM \equal{}$ $ MB\equal{}$ $ 2\cdot NL$ , $ AG \equal{} GD$ , $ GC\equal{}$ $ DB\equal{}$ $ 2\cdot GL$ , $ AC\parallel GN\parallel DM$ and the quadrilateral $ MHDB$ is cyclically. Therefore, $ \widehat {BHF}\equiv$ $ \widehat {BHD}\equiv$  $ \widehat {BMD}\equiv$ $ \widehat {BAC}\equiv$ $ \widehat {ABC}$ , i.e. $ \widehat {BHF}\equiv\widehat {ABC}$ .

[b][u]Remark.[/u][/b] Observe that $ \frac {MB}{ML} \equal{} \frac {AB}{AL} \equal{} 2$ , i.e. the points $ A$ , $ M$ are harmonically conjugate w.r.t. the pair $ \{L,B\}$ . Since $ HA\perp HM$ obtain $ \widehat {MHB}\equiv\widehat {MHL}$ . Thus, $ \widehat {LHA}\equiv\widehat {BHD}\equiv\widehat {FBL}$ , i.e. the quadrilateral $ LHFB$ is cyclically.[/color]

[size=117][color=darkred][b]Nice problem, Bilarev[/b] ! Thank you.[/color][/size]

It was much better (I think)  to take A’-symmetrical of A about F, see that D is the centroid of A’BC, hence AD/A’D=AM/MB and MD||A’B||AC, and $ \angle BMD \equal{} \angle ABC$, then BD and A’C are perpendicular, consequently $ BD\perp AB$ and BDHM cyclic,  $ \angle BHD \equal{} \angle BMD$, done.

Best regards,
sunken rock

[hide]
Let the circle through A,C tangent to BC be $\omega_1,$ and the circle through $A,B$ tangent to $CA,CB$ be $\omega_2.$ Let $\omega_1\cap \omega_2=H_1,$ where $H_1 \neq A.$ Let $M_1$ be the point on $AB$ such that $M_1H_1\perp AH_1.$ Because $AH_1$ is the radical axis of $\omega_1,\omega_2,$ it bisects segment $BC,$ so it suffices to show $\frac{AM_1} {AB} =\frac{2} {3}$

Invert about a circle centered at A.
Notice $\omega_2$ becomes a line through $B'$ parallel to $AC'.$ $\omega_1$ becomes a line through $C'$ intersecting $\omega '_2 $ at $H'_1.$
The image of $BC$ becomes a circle through $A,B',C'$ tangent to $\omega '_1,\omega '_2,$ at $C',B',$ repsectively. Thus, $H_1'C'=H_1'B'.$
Additionally, since $AC'||\omega_2',$ $AB=BC'.$ $M_1'$ becomes the foot of the perpendicular from $H_1'$ to $AB'.$
Since $A,B,M$ were originally colinear, it sufficies to show, $\frac{AM_1'}{AB'}=\frac{3} {2}.$
Let $\angle C'AB'=\theta$
Then, $\angle C'AB'=\angle H_1'B'M_1'=\angle B'C'A=\angle C'B'H_1'=\theta$
Thus, $AB'=B'C'=2B'H_1'\cos\theta,$ and $B'M_1'=B'H_1'\cos\theta=\frac{AB'} {2}$ and the result follows.

[/hide]

If you didn't see the cyclic quads...

Let $G$ be the centroid of $ABC$. It suffices to show that $HBM \sim CBG$, for then $\angle GCB = \angle MHB$ implies the result.

First, $\angle BGC = 180 - (90 - \angle GBA) = 90 + \angle GAB = \angle HMB$.

Second, let $h_C$ denote length of altitude from $C$, and let $N$ denote the midpoint of $AM$. We have
\begin{align*}
\frac{HM}{CG} &= \frac{3}{2}\cdot \frac{HM}{h_C} = \frac{3}{2} \cdot \frac{[AMF]}{[ABC]} \cdot \frac{AB}{AF}\\
&= \frac{AN}{AG} = \frac{GB}{MB}.
\end{align*}

Therefore $HBM \sim CBG$, so the problem is finished.

I think no one posted this simple solution:
Let $N$ be the midpoint of $AB$ and $G$ the centroid of $\triangle ABC$. We have $\dfrac{AG}{GF} = \dfrac{AM}{MB} = 2$ so $GM \parallel BC$. Note that $\angle MHG = 90^{\circ} = \angle GNM$ so $HNMG$ is cyclic. Furthermore $\angle ANH = \angle AGM = \angle AFB$ because $HNMG$ is cyclic and because $GM \parallel BC$. Thus $HNBF$ is cyclic and $\angle BHF = \angle BNF = \angle BAC = \angle ABC$.

[hide=Proof]
Let $FM$ meet $AC$ at $G$ and let $BG$ meet $AF$ at $K$. By Menelaus Theorem on triangle $ABC$ we have $\frac{AM}{MB}\cdot\frac{BF}{FC}\cdot\frac{CG}{GA}=1$ which gives $\frac{CG}{GA}=\frac{1}{2}$. Thus, $C$ is the midpoint of $GA$. Since $AC=BC$, $C$ is the circumcenter of triangle $ABG$ and thus we deduce that $\angle ABG=90^\circ$. But $\angle MHK=90^\circ$, so $KHMB$ is cyclic. Hence, $\angle BHF=\angle BMK$. Moreover, by Ceva's Theorem, we have $\frac{AM}{MB}\cdot\frac{BK}{KG}\cdot\frac{CG}{CA}=1$ which implies that $\frac{AM}{MB}=\frac{KG}{BK}$ so by Thales' Theorem, $KM$ is parallel to $GA$. Thus, $\angle BMK=\angle BAC$ which implies that $\angle BHF=\angle BAC=\angle ABC$.
[/hide]

Using cartesian coordinates, was only the logical way to do it . . . .
We want to show that $BF^2=AF.FH$.

Let $A(-3,0)$, $B(3,0)$, $C(0,c)$ meaning that $M(1,0)$ and $F\left( \frac{3}{2},\frac{c}{2} \right)$

We easily get that the line defined by the points $A$ and $F$ is the following $y=\frac{c}{9}x+\frac{c}{3}$.
With this we can easily calculate the line defined with the points $H$&$M$ (not sponsored), which is the following $y=-\frac{9}{c}(x+1)$.

We easily get that $H\left( -\frac{c^2+27}{c^2+81},\frac{9(c^2-54)}{c(c^2+81)} \right)$.

Now all that's left to do is to see if $BF^2=AF.FH$. Which is easily seen when we plug in what we got.


So the terms will be in the form +1/3^k sin^(3^k) 3^k * x???

I need you to explicitly state the pattern, I can't find enough terms.

use $\sin x=\frac{e^{i x}-e^{-ix}}{2i}$, it becomes a simple geometric series

Actually it doesn't. $\frac{e^{i3^{k}x}}{3^{k}}$ isn't a geometric series.

[hide="Solution (a)"]\begin{eqnarray*}\sin 3t &=& \sin 2t\cos t+\cos 2t\sin t\\ &=& 2\sin t(1-\sin^{2}t)+(1-2\sin^{2}t)\sin t\\ &=& 3\sin t-4\sin^{3}t\end{eqnarray*}

which gives

$\sin^{3}t={1\over 4}(3\sin t-\sin 3t)$

Therefore

${1\over 3^{k}}\sin^{3}3^{k}x={1\over 4}\left({1\over 3^{k-1}}\sin 3^{k}x-{1\over 3^{k}}\sin 3^{k+1}x\right)$

The expression in the parentheses is obviously telescoping when summed, hence the desired PARTIAL sum is

$S_{n}={1\over 4}\left(3\sin x-{1\over 3^{n}}\sin 3^{n+1}x\right)$

Now using the Two Policemen Theorem, we find $-{1\over 3^{n}}\leqslant{1\over 3^{n}}\sin 3^{n+1}x\leqslant{1\over 3^{n}}$, hence the second term in the parentheses tends to zero. Therefore

$S={3\over 4}\sin x$[/hide]

[hide="Solution (b)"]Similarly as above, we find

$\cos^{3}t={1\over 4}(3\cos t+\cos 3t)$

hence

$(-1)^{k}{1\over 3^{k}}\cos^{3}3^{k}x={1\over 4}\left((-1)^{k}{1\over 3^{k-1}}\cos 3^{k}x+(-1)^{k}{1\over 3^{k}}\cos 3^{k+1}x\right)$

The RHS is also telescoping when summed, by the pattern $A+B-B-C+C+D-\dots$, hence

$S_{n}={1\over 4}\left(3\cos x+(-1)^{n}{1\over 3^{n}}\cos 3^{n+1}x\right)$

Same as above, we find that the second term in the parentheses tends to zero, hence

$S={3\over 4}\cos x$[/hide]

[quote="Farenhajt"]Actually it doesn't. $\frac{e^{i3^{k}x}}{3^{k}}$ isn't a geometric series.
[/quote]

If you integrate it it is!

What is the Two Policeman Theorem? I tried a cursory google but it turns up nothing.

[quote="Farenhajt"]Quote from [url]http://www.math.jmu.edu/~vanwyk/courses/235/c2.html[/url] (gotten from googling on "two policemen theorem"):

"[b]Section 5.[/b] The pinching theorem (squeeze theorem, sandwich theorem, two policemen theorem)"[/quote]
Note to self: Do not misspell theor[b]e[/b]m.

[quote="Phelpedo"]I think the more common name is the "squeeze principle," though I suppose I can understand where the policemen come in :wink:[/quote]
Oooooooooooooooooooooooh.

That would have helped....

Darn you Serbians and your odd vernacular.

Hm...ironic, isn't it.

[quote="PenguinIntegral"]
I think we need some humoristic sticky at the top of this fourm where the questions consist of "I want to score X on competition Y" and "Where do I learn subject Z" and all the answers are "AoPS II".

Really, I do.

Wait, you say? Masspoints aren't covered by AoPS II? Has PenguinIntegral's superpowered liver gotten the best of him?

No, you fools. Learn to use google and Kedlaya's geometry packet.
[/quote]

[quote="Farenhajt"][quote="PenguinIntegral"]Darn you Serbians and your odd vernacular.[/quote]

Wow, we are squeezing in some national hatred here? And what are you - some "Way to go, Georgie, f*** the darned Iraqis brainless and take all their oil" kind of American?

Sorry, baby, but it isn't this "darned Serbian's" fault that you can't spell and/or use Google...[/quote]

No, no, no national hatred. I'm sorry if that was offensive or anything. Completely, and I mean completely, sarcastic. That google page was just an outline, so I skipped it.

I'm not racist, nor nationalist. You can check out my post history in the round table.

I'm sorry if I seem a little serious or apologetic in the above, but I don't want to be seen as racist.

 :maybe: <-Me

@Phelpedo

AoPS II doesn't talk about calculus. AMC contests are strictly precalc, and that was a post in the AMC forum.

When I googled, I found the same outline Farenhajt did. It's an outline, no actual subject matter on the page. Hence, the "ask more questions" bell went off in my head. Yes, I probably should have actually read the page, and typed squeeze theor[b]e[/b]m into google. But I'm a idiot. And while I may be an idiot, I'm not a hypocrite. :wink:

Also, the above is in good humor, in no way ment to offend or be rude, etc...

[quote="Farenhajt"][quote="PenguinIntegral"]
No, no, no national hatred. I'm sorry if that was offensive or anything. Completely, and I mean completely, sarcastic. That google page was just an outline, so I skipped it.

I'm not rascist, nor nationalist. You can check out my post history in the round table.
[/quote]

Apology accepted, and I'm also offering one for jumping into conclusion so hastily. Usually I'm not like that  :)[/quote]
I'm also usually not like a frat boy, especially on an international forum. Wording things without smilies really makes it sound mean :( .
Observe:
[quote]
I think we need some humoristic sticky at the top of this fourm where the questions consist of "I want to score X on competition Y" and "Where do I learn subject Z" and all the answers are "AoPS II".  :P

That would be a good joke.

Wait, you say? :ninja:  Masspoints aren't covered by AoPS II? Has PenguinIntegral's superpowered liver gotten the best of him? :huh:

No, you silly people :roll: . Learn to use google and Kedlaya's geometry packet.  :P
[/quote]
[quote]I think we need some humoristic sticky at the top of this fourm where the questions consist of "I want to score X on competition Y" and "Where do I learn subject Z" and all the answers are "AoPS II".

Really, I do.

Wait, you say? Masspoints aren't covered by AoPS II? Has PenguinIntegral's superpowered liver gotten the best of him?

No, you fools. Learn to use google and Kedlaya's geometry packet. [/quote]


geometry
parallelogram
vector
geometry proposed

Find the centre of gravity of a quadrilateral and justify your answer.

Let $ABCD$ the quadrilateral, and $G$ its center of gravity.

If $O$ is the origin, then $\overline{OG} = \dfrac{1}{4}(\overline{OA}+\overline{OB}+\overline{OC}+\overline{OD})$

And this equation is true for every position of $O$.

So we can choose the position of $O$ out of the lines $AB,CD$

Then we can construct a parallelogram with the vectors $\overline{OA},\overline{OB}$

The diagonals of this parallelogram intersect at $M$, where $M$ is the midpoint of $AB$. So we have

$\overline{OM}=\dfrac{1}{2}(\overline{OA}+\overline{OB})$

Similarly, if $N$ is the midpoint of $CD$, then we have $\overline{ON}=\dfrac{1}{2}(\overline{OC}+\overline{OD})$

And the result holds independently of the position of $O$.

So, if we choose the point $O$ out of the line $MN$, then we can construct a parallelogram with the vectors $\overline{OM},\overline{ON}$.

If $K$ is the midpoint of $MN$, then

$\overline{OK}=\dfrac{1}{2}(\overline{OM}+\overline{ON})$

$\overline{OK}=\dfrac{1}{2}(\dfrac{1}{2}(\overline{OA}+\overline{OB})+\dfrac{1}{2}(\overline{OC}+\overline{OD}))$

$\overline{OK}=\dfrac{1}{4}(\overline{OA}+\overline{OB}+\overline{OC}+\overline{OD})$

So $K \equiv G$

In other words the center of gravity is "the midpoint of opposite midpoints"

P.S.
Similarly, if we start with other pairs, (i.e. $\overline{OA}+\overline{OD}$), then we'll find that :

-$\ G$ is the midpoint of $PR$, where $P,R$ are the midpoints of $AD,BC$
-$\ G$ is the midpoint of $ST$, where $S,T$ are the midpoints of $AC,BD$


$k$ is an integer. We define the sequence $\{a_n\}_{n=0}^{\infty}$ like this:
\[a_0=0,\ \ \ a_1=1,\ \ \ a_n=2ka_{n-1}-(k^2+1)a_{n-2}\ \ (n \geq 2)\]
$p$ is a prime number that $p\equiv 3(\mbox{mod}\ 4)$
a) Prove that $a_{n+p^2-1}\equiv a_n(\mbox{mod}\ p)$
b) Prove that $a_{n+p^3-p}\equiv a_n(\mbox{mod}\ p^2)$

we can solve this problem by finding the general term of the sequance then use some knowledge in Z_i

Easily you see $a_n=\frac{i(k-i)^n-i(k+i)^n}2$. Now we must prove that for each $0\neq t\in\mathbb Z_p[i]$ we have $t^{p^2-1}=1$. This is beacuse of lagrange theorem in the multiplicative group $\mathbb Z_p[i]$ that has $p^2-1$ elements. After this the second statement is trivial

P.S. $p\equiv3$ is used to prove that $\mathbb Z_p[i]$ is a group


Since we have a official waterloo thread, why not have a official UofT thread.

First question: Who here is coming to UofT next year?

Get its a joke, no one goes to uoft!

[hide]

of cource I'm kiddining, there are like 40 bakjillionz students that go there[/hide]

Wow, U of T is so unpopular all of a sudden. Or did everyone applying to U of T get kidnapped by the fierce competitors? Oh-oh, I fear my fate ... But otherwise I'm going to U of T (math/physics). Is anybody ... *glances around*  .... else .... *looks around again* ... going?

Another question (I'm too lazy to start a separate thread) - what scholarships is everybody applying for?

isn't it just that alot of the canadian aops crowd is from toronto, and alot of them want to leave the city for university?

That and Uwaterloo has just as good a reputation as uft.

Actually I think there are quite a few strong math/physics students at UT: Roger, Tianyi (David), Robert, Jacob, Janos, Oleg, etc; so going to UT shouldn't be a bad dicision.
By the way, I will be extremely happy if some kind person goes to UT and help me out in my study.


algebra
polynomial
search
algebra unsolved

We find $ p(y)\equal{}0$ where $ p$ is a polynomial of degree $ 12$ which is irreducible over $ \mathbb{Q}$. Hopeless !

cf    http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1800808255&t=199338 for a much better exercise.


geometry
trigonometry
rotation
trig identities
Law of Cosines

[geogebra]eb0ef61e0827ed3fd7a0b03f0a86c07e536241df[/geogebra]

Suppose $ P$ is an inteior point of a square $ ABCD$.
If $ AP\equal{}1,BP\equal{}2,CP\equal{}3$, find the area of square $ ABCD$.

[hide="solution"]
Let $ \angle ABP \equal{} \alpha, \angle CBP \equal{} \beta$ and $ \alpha \plus{} \beta \equal{} \frac {\pi}{2}$

From the cosine laws;
$ 1 \equal{} x^2 \plus{} 4 \minus{} 4x\cos (\alpha) \qquad(1)$
$ 9 \equal{} x^2 \plus{} 4 \minus{} 4x\cos (\beta) \qquad(2)$

Since $ \alpha \plus{} \beta \equal{} \frac {\pi}{2} \longrightarrow \cos \beta \equal{} \sin \alpha$

$ 16x^2(\sin^2(\alpha) \plus{} \cos^2(\alpha)) \equal{} (x^2 \plus{} 3)^2 \plus{} (x^2 \minus{} 5)^2$

$ \Rightarrow 2x^4 \minus{} 20x^2 \plus{} 34 \equal{} 0$ since $ [ABCD] \equal{} x^2 \equal{} a$

$ \Rightarrow a^2 \plus{} 10a \plus{} 17 \equal{} 0 \Rightarrow a \equal{} 5\pm 2\sqrt {2}$

However $ 2(5\minus{}2\sqrt{2}) < 3^2$  ($ \because 1< 4\sqrt{2}$) So this cannot be the answer because the point P is $ 3$ unit from one corner

$ \therefore [ABCD] \equal{} \boxed{5 \plus{} 2\sqrt{2}}$
[/hide]

[quote="ocha"]
From the cosine laws;
$ 1 \equal{} x^2 \plus{} 4 \minus{} 4x\cos (\alpha) \qquad(1)$
$ 9 \equal{} x^2 \plus{} 4 \minus{} 4x\cos (\beta) \qquad(2)$

Since $ \alpha \plus{} \beta \equal{} \frac {\pi}{2} \longrightarrow \cos \beta \equal{} \sin \alpha$

$ 16x^2(\sin^2(\alpha) \plus{} \cos^2(\alpha)) \equal{} (x^2 \plus{} 3)^2 \plus{} (x^2 \minus{} 5)^2$
[/quote]

I believe you meant to square $ \sin(\alpha) \minus{} \cos(\alpha)$, but you forgot the term $ \minus{}2\sin(\alpha)\cos(\alpha)\equal{}\minus{}\sin(2\alpha)$

[quote="236factorial"][quote="ocha"]
From the cosine laws;
$ 1 \equal{} x^2 \plus{} 4 \minus{} 4x\cos (\alpha) \qquad(1)$
$ 9 \equal{} x^2 \plus{} 4 \minus{} 4x\cos (\beta) \qquad(2)$

Since $ \alpha \plus{} \beta \equal{} \frac {\pi}{2} \longrightarrow \cos \beta \equal{} \sin \alpha$

$ 16x^2(\sin^2(\alpha) \plus{} \cos^2(\alpha)) \equal{} (x^2 \plus{} 3)^2 \plus{} (x^2 \minus{} 5)^2$
[/quote]

I believe you meant to square $ \sin(\alpha) \minus{} \cos(\alpha)$, but you forgot the term $ \minus{} 2\sin(\alpha)\cos(\alpha) \equal{} \minus{} \sin(2\alpha)$[/quote]

I don't think so , because ocha's work is possibly

$ 4x\cos (\alpha)  \equal{} x^2 \plus{} 3  \qquad(3)$
$ 4x\cos (\beta) \equal{} x^2 \minus{} 5  \qquad(4)$

Sum of the squares of (3) and (4) !!

Rotate P about B 90 degrees to the other side of BC, call it P'. Then $ PP' \equal{} 2 \sqrt {2}$, and $ \angle BPP' \equal{} 45$. Also, by the law of cosines on $ \triangle PP'C$, $ 1 \equal{} 8 \plus{} 9 \minus{} 12\sqrt {2} \cdot cos CPP'$, so $ cos CPP' \equal{} \frac {2 \sqrt {2}}{3}$. So by the law of cosines, $ BC^2 \equal{} 4 \plus{} 9 \minus{} 12cos(45 \plus{} cos^{ \minus{} 1}\frac {2\sqrt {2}}{3}) \equal{} 5 \plus{} 2 \sqrt {2}$

For this problem, ocha's way is nicer but my method can be applied to many other problems, such as for equilateral triangles.


Let $ a$, $ b$, $ c$, $ d$ be four positive real numbers such that $ a\geq b \plus{} c \plus{} d$. Prove that

      $ \left(a \plus{} b\right)\left(b \plus{} c\right)\left(c \plus{} d\right)\left(d \plus{} a\right)\geq 20abcd$.

I think we can use mixing variables. If we replace $b,c$ by $t,t$ where t is their geometric mean, then the condition is fulfilled. The R.H.S. of the inequality remains unchanged while L.H.S. of the inequality decreased. So we can repeating the mixing process until b=c=d, then the rest is easy, we only need to prove $a^2+t^2 \ge 3at$ with $a \ge 3t$ which is obvious.


It give $xy=y^{2}-y+4-y^{3}$ and $y^{6}-y^{5}+3y^{4}-10y^{3}+5y^{2}-8y+16=0$. But last equation had not rational solutions. Can be find by computer.


geometry
trigonometry
trig identities
Law of Cosines

In the square ABCD, lines from B, C, and D meet at a point E inside the square, where <CDE = 15, <DEC = 90. What is <CBE?

I'm not sure what you find difficult about this problem. Have you used the fact that the angles of a triangle have measures that sum to 180 degrees?

edit. Sorry, I thought this problem was much easier than it is. Ignore the above.
I found a solution using trig.
[hide="hints"]Extend CE to the point E' so that BE' is perpendicular to CEE' at E'.
Note that triangle BE'C is congruent to triangle CED.
Calculate the area of triangle BEC in two different ways to get something involving sin(CBE).
Use the law of cosines on the same triangle.
Divide the sine and cosine expressions to get tan(CBE).
Calculate many lengths and plug them in.
[/hide]
Yeah, there's probably a better way.

That's the problem--Using that fact you don't have enough information to get the angle you need. You can get angle DCE to be 75, angle ECB to be 15 but cannot get angles CEB or angle CDE.

I don't think there is a very easy way to do it.  I'd do law of cosines twice.  Let $ s$ be the side length of the square.  Then $ EC \equal{} s \sin 15$ then do Law of cosines to get $ EB$ in terms of $ s$ (note $ \sin 15 \equal{} \frac{1}{4}(\sqrt{6} \minus{} \sqrt{2})$).  so $ EB \equal{} s\sqrt{1\plus{}sin15\minus{}sin30} \equal{} s/2 \sqrt{2 \plus{} \sqrt{6} \minus{} \sqrt{2}}$.  Then do law of cosines on $ \angle EBC$, which is not pretty...

I'm getting that $ \angle CBE \approx 13.81139335^{\circ}$.  Not something you could find by hand.

[hide="my solution"]
Let $ |CD|\equal{}|BC|\equal{}a$, $ |CE|\equal{}x$ and $ \angle CBE \equal{} \varphi$

From right triangle trigonometry we have: $ \sin 15^{ \circ} \equal{} \frac{x}{a}$.

$ \angle CEB \equal{} 165^{\circ}\minus{}\varphi$

From law of sine we have:

$ \frac{\sin \varphi}{\sin (165^{\circ}\minus{}\varphi)}\equal{} \frac{x}{a} \implies \frac{\sin \varphi}{\sin 15^{ \circ} \cos \varphi \plus{} \cos 15^{ \circ} \sin \varphi}\equal{}\sin 15^{ \circ}$

$ \sin \varphi \equal{} \sin^2 15^{ \circ} \cos \varphi \plus{} \frac{\sin 30^{\circ}}{2} \sin \varphi$

$ \frac{3}{4} \sin \varphi\equal{} \sin^2 15^{ \circ} \cos \varphi$

$ \tan \varphi\equal{} \frac{4}{3} \sin^2 15^{ \circ}$

$ \tan \varphi\equal{} \frac{2}{3} (2\sin^2 15^{ \circ})$

$ \tan \varphi\equal{} \frac{2}{3} (1\minus{} \cos 30 ^{\circ})$

$ \tan \varphi\equal{} \frac{2\minus{}\sqrt 3}{3}$

$ \boxed{\varphi \equal{} \arctan \left(\frac{2\minus{}\sqrt 3}{3}\right)}$[/hide]


Let $ a,b,c$ be the side lengths of a triangle of area $ S$. Prove that $ 4\sqrt{3}S\le 3\sqrt[3]{a^{2}b^{2}c^{2}}$.

[hide="my solution"]we need to prove that $ 4\sqrt{3}\cdot\frac{abc}{4R}\le 3\sqrt[3]{a^{2}b^{2}c^{2}}$. which is equivalent to $ R\ge \frac{1}{\sqrt{3}}\sqrt[3]{abc}$. but this is true since $ R\ge \frac{1}{\sqrt{3}}\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\ge \frac{1}{\sqrt{3}}\sqrt[3]{abc}$.[/hide]


geometry
geometric transformation
rotation
trigonometry

We put a rod of length $L$ to the edge of a frictionless table. The rod falls off the table, and arrives on the floor so that its ends are just opposite as they were on the table. How high is the table?

Seems impossible to solve... What's the solution?

Why do you think so? First present your thoughts they may be in the right track.

In my opinion it the problem could be solved perhaps if one considers the rotation
the rod gains from falling from rest and one could also determine the free fall time (as expression).
So in this time the rod should have done $(n+\frac12)$ turns, but probably one could also show
that only some $n$ is possible that suits the free fall time, which then could be solved.

Am I on the right track?

Well, I can solve it if I know the initial velocity of the center of mass, or something like that. But like this... well, I have no idea. I also tried to fit in an odd integer number of turns into the time of the free fall, but couldn't do it  :(

Well I wouldnt go on that track.  :wink:
Here is [hide="hint"] Find with dynamical assumptions in what position will the rod leave the table, and find the rods angular velocity at this moment[/hide] for now

I have assumed the [b]length of rod [/b]as $2l$ and of the [b]table [/b]as $L$
let me post the solution.
let initially the rod be perfectly vertical and then you give it a slight push and it starts moving the point of contact of the rod with the table be $O$ about $O$ the centre $C$ of the rod rotates in a circular path as do the other points.
$let \theta_{o}$ be the angle that vertical makes with the rod when it slips off.
now $v^{2}_{o}=gl \cos \theta_{o}$
where $v^{2}_{0}$ is the speed of c.m when it falls down
also, $\frac{(dv)(d \theta)}{(dt)(d \theta)}=g \sin \theta$
thsi gives $v^{2}_{0}=2gl(1-\cos \theta_{0})$
this gives $\cos \theta_{0}=\frac{2}{3}$
now speed $v_{0y}$ of cm now is $\sqrt{\frac{2gl\sqrt{5}}{9}}$
also from cm FOR there is no net torque after the rod moves off so all particles as seen from cm will have the same angular velocity as they had before falling off which is same as that of cm's as seen from our FOR.
so $\omega$ $=$ $\sqrt{\frac{2g}{3l}}$
now say the length of that place is $L$
now the angular disp. of any particle must be $(n+\frac{1}{2})2\pi-\cos^{-1}\frac{2}{3}$
  so time $t$ to fall down is $\frac{(n+\frac{1}{2})2\pi-\cos^{-1}\frac{2}{3}}{\sqrt{\frac{2g}{3l}}}$
now $L$ $=$ $(\sqrt{\frac{2gl\sqrt{5}}{9}})(\frac{(n+\frac{1}{2})2\pi-\cos^{-1}\frac{2}{3}}{\sqrt{\frac{2g}{3l}}})+\frac{g(\frac{(n+\frac{1}{2})2\pi-\cos^{-1}\frac{2}{3}}{\sqrt{\frac{2g}{3l}}})^{2}}{2}$
so the lengths are in general lots

More or less (More I think  :D ) what I and [b]Djole[/b] were discussing above, very good [b]pardesi.[/b]

Now , [b]Thaakisfox[/b] you did not recommend this track, how come?
I mean what did your solution look like? Please post it.


If $abc=1$, one can substitut $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$
Does this substitution have a name? (Like Ravi-Substitution for $a=x+y$ etc..) :maybe:

[quote="Yimin Ge"]If $abc=1$, one can substitut $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$[/quote]

This is famous [b]harazi's substitution[/b].

He had already reply this before [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7942&start=140]here[/url]:

[quote="mahbub"]I think i got what i wanted. Look:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=harazi%5C%27s+substitution&t=6198

from here i guessed that:
when abc=1 then substitution of a=x/y, b=y/z and c=z/x is called harazi's substitution! :)[/quote]

[quote="harazi"]Come on, I refuse to be consider the greatest idiot on the planet!  :P . THis substitution born in the same period as the inequalities. It would be shameful to call it harazi substitution.  ;)  ;)[/quote]


Hi.
This year I will be taking the AP exam in Chemistry and Euro History.
Do any of you know good test prep books that prepare you throughly for these AP tests?

Additionally, do you know any good PSAT prep books?

Thanks.

For AP Chemistry, I've heard Zumdahl is a good textbook, but I used Masterton (I think that's how you spell it). For the review book for AP Chemistry, I would suggest Princeton Review's Book.

For PSAT prep books, I would suggest that you get the College Board Official Study Guide for the SAT and disregard the SAT essay section. Also, old PSAT tests are helpful, they can be bought from the Collegeboard store for about 3 dollars per test.

Don't know about Euro History, the College Confidential Forum would be useful to research this

http://talk.collegeconfidential.com/

For AP Chemistry, I personally find REA useful. Lots of practice tests!!!
Same goes for AP Euro...

As for PSAT, just grab any SAT prep book although I prefer Barron's.
College Board is always good but I just like the Barron's.


geometry
circumcircle
trigonometry
geometric transformation
rotation
trig identities
Law of Cosines

Let $ ABC$ be an equilateral triangle and $ M$, $ N$ be the midpoints of $ AB$ and $ AC$. Let the intersection of $ MN$ and the arc $ AB$ of circumcircle of $ \triangle{ABC}$ be $ K$ and let $ P\equal{}AB \cap CK$. Show that $ AP^2 \equal{} BP \cdot BA$.

Let AB=s. Let the center of the circumscribed circle by O and the intersection of AO and MN be L. Let KJ be the chord of the circle of center O, formed by the extension of MN.
Then the radius of the circle is s/sqrt(3).
KJ=2KL=2sqrt(KO^2-OL^2)=2sqrt(s^2/3-s^2/48)=(sqrt(5)/2)*s
KM=(KJ-MN)/2=(KJ-s/2)/2=(sqrt(5)-1)/4*s
angles KPM and BPC are equal.
MN and MC are parallel so angles KMP and PBC are equal.
So triangles KMP and CBP are similar and KM/PC=MP/BP.

So BP=(3-sqrt(5))/2*s.
AP=AB-BP= (sqrt(5)-1)/2*s
Therefore AP^2=BP*AB

[hide="different solution outline"]

Let $ X$ be diametrically opposite from $ C$, and let $ MN$ meet the circumcircle $ \omega$ in $ L$ such that the order of points on the segment is $ K,M,N,L$.  We use power of a point about $ M$ to find $ KM$ (because $ KM \equal{} NL$ by symmetry).  We then use the Law of Cosines on triangle $ KNC$ to find $ CK$.  Note that $ \angle XKP$ is right because $ CX$ is a diameter, and $ \angle XMP \equal{} \angle CMP$ is right because $ ABC$ is equilateral.  Hence $ KXMP$ is cyclic.  Thus $ CM\cdot CX \equal{} CP\cdot CK$.  We can thus find $ CP$.  Now that we know $ CP$ and $ CK$, we use power of a point around $ P$ to get that $ PB \cdot PA \equal{} PK \cdot PC$.  From this we can find the value of $ \lambda \equal{} \frac{BP}{BA}$, because we know that $ \frac{BP}{PA} < \frac{BM}{BA} \equal{} \frac{1}{2}$.  Then we need only show that $ \lambda \equal{} (1\minus{}\lambda)^2$, and if that holds true then we are done.  The calculations involved would be nasty.

[/hide]

Thanks for the two nice solutions. I've come up with a more geometrical proof but it's quite complex.

We first show that $ AP^2 = BP \cdot BC \Longleftrightarrow AK^2= BK \cdot CK$. It is known that $ CK=AK+BK$, so:

$ \frac{AP}{BP}\cdot\frac{AP}{BC}=1 \Longleftrightarrow \frac{AK}{BK} \cdot \frac{1}{\frac{AB}{AP}}=1 \Longleftrightarrow \frac{AK}{BK} \cdot \frac{1}{1+\frac{BK}{AK}}=1 \Longleftrightarrow$

$ \frac{AK}{BK} \cdot \frac{AK}{BK+AK}=1 \Longleftrightarrow \frac{AK^2}{BK \cdot CK}=1$

Now we return to the problem. By $ R(A, 60^o)$, we have $ M \to N$. Suppose $ K \to K'$, then $ K'$ is on $ CK$. By the rotation we have $ \angle{KAB}=\angle{K'AN}$. Since $ \angle{ANK'}=\angle{CMK}=120^o$, so we have

$ \triangle{ANK'} \sim \triangle{AKB} \Longrightarrow AK^2 = AM \cdot AC \Longrightarrow \triangle{AK'M} \sim \triangle{ABK'}$

Hence $ \angle{ABK}=%Error. "ANGLE" is a bad command.
{AK'M}$. By $ \angle{AK'N} = \angle{ABK}$, we have $ \angle{NK'M} = \angle{KBK'}$.

By $ \triangle{AKM} \sim \triangle{KCN}$ we have $ MN^2= KM \cdot KN = K'N \cdot KN \Longrightarrow  \triangle{ANK} \sim \triangle{K'NM}$, yielding $ \angle{NK'M} =\angle{NAK} \Longrightarrow \angle{KBK'} = \angle{CAK}$. Hence $ \triangle{KAC} \sim \triangle{KBK'} \Longrightarrow KA^2 = KB \cdot KC$.

Let $\{\ E\ \}\in BC\cap AP$ clearly $PE=AP\ (\ 1\ )$. Since $\widehat{EPB}=\widehat{DAC}=60^\circ$ and $\widehat{PBE}=\widehat{ADC}=60^\circ+\widehat{DCB}$ we infer $\Delta EPB\sim\Delta CAD\implies\frac{PE}{PB}=\frac{AC}{AD}\ (\ 2\ )$; with $(1)$ we get $\frac{PA}{PB}=\frac{PE}{PB}\ (\ 3\ )$, from angle bisector theorem ($PD$ is the bisector of $\angle APB$) we have $\frac{AP}{PB}=\frac{AD}{BD}\ (\ 4\ )$. From $(2)$ and $(4)$ we get $\frac{AC}{AD}=\frac{AD}{BD}$, the required relation.

[color=#BF0000][b]Solution provided by my son.[/b][/color]

Best regards,
sunken rock


In how many ways can you choose 20 coins from 20 nickels and 20 dimes?
How about 20 coins from 20 pennies, 20 nickels and 20 dimes?

In the first case it would be 21, since given any number of nickels chosen (between 0 and 20, both included), the number of dimes is given by $ 20\minus{}$number of nickels.

In the second case, let us assume that $ k$ nickels are chosen ($ 0\leq k\leq20$).  Then, we have $ 20\minus{}k$ coins to choose between dimes and pennies.  Or, applying the same reasoning as above, we would have $ 21\minus{}k$ ways of choosing these $ 20\minus{}k$ coins, for a grand total of
\[ \sum_{k\equal{}0}^{20}\left(21\minus{}k\right)\equal{}21^{2}\minus{}\frac{20\cdot21}{2}\equal{}11\cdot21\equal{}231.\]


function
algebra
floor function
\/closed

how can a rounddown function be expressed in algebraic form?

e.g. 0.999 rounds down to 0; each number rounds to the greatest integer less than or equal to it.

thanks in advance.

The floor function (looks like this [img]http://mathworld.wolfram.com/fimg90.gif[/img]) I suppose, look [url=http://mathworld.wolfram.com/FloorFunction.html]here[/url].

My explanation or the image? The image was designed to be viewed in the LGBlackPrint style if you meant that.

I meant the image...but the font explains a lot...yes...


Ok, I'm not a lazy type, it's just my problem, that's all! I come to US during my freshmen year (legal immigration  :P ), and my english has been improved a lot during the 2 years I stay. Unfortunately, it's not good enough to overcome the SAT I  :ninja: . That's why I desperately need help from the experienced people. If anyone in here have gone to college successfully, can you share with me your experience on how to take a good score for the writing and cri reading part, like:
-Which books you use to practice
-What shoould we prepare on test day
-Any valuable strategies?
-Vocab list...
I'm asking you guys a savor, and I really appreciate it
I've done several practice test, and my lastest result is:
CR: 500
WR:510
Ma:780
I'm not really worrying about math, but that would make my total score 1790, which is not good at all
I see that some people in this forum score really high on the SAT, so I just want to ask. I hope I don't annoy you guys at all.
By the way, I'm a junior (05-06), I would like to ask when should I take the SAT 1 (I'm planning to take it twice).
Any advices on SAT II on these subjects would be greatly appreciate:
-MATH (IC IIC)
-CHEM
-PHYSICS
If you take both IC and IIC, does it count as 1 or 2 subject test?
Thanks so much
 :roll: Have a great day

I have used Kaplan, Barron, Princeton, Peterson and CollegeBoard SAT prep books. CollegeBoard is certainly best, because it writes the real SAT, and the next is Kaplan and Princeton. Do practices will help a lot. I plan to finish 60 SAT tests. But the most important part is to analyze how do you make mistakes after each test.   :ewpu:

Just take IIC, it's the only one worth anything. The curve is also better so it is easier to get 800. :)

Thanks you guys.
2aznphatso: isn't IIC the SATII math test? what do you mean by taking IIC, I heard about the curve, but I don't know how it works

there's two SAT 2 math tests, level 1 (IC) and level 2 (IIC).
There's really no point to taking IC if you plan to take IIC later because IIC scores are better, usually around -5 is still 800 while on IC you will need -1 or perfect to get an 800.

[quote="tokenadult"]This looks like a thread that would be very welcome in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=143]College Forum[/url] here on AoPS. Could the moderator here kindly move it over?[/quote]

Certainly. Moving now.

I agree with the advice posted already to take the SAT II Math Level 2 and not to bother taking the Level 1 test. You will do better on the Level 2 test, that test is more impressive to colleges, and you save money by not taking two tests in the same subject.

About your other questions, keep reading. Be sure to mention to college admissions officers when you apply and write essays that you have only lived in the United States very recently, but make sure to READ, READ, READ, and READ. Find some interesting magazine and subscribe to it. Read the newspaper every day. Find books you like and try to read those. It's possible to do TOO MUCH test preparation while not doing enough general reading. Don't make that mistake, but do test preparation with the genuine College Board books and spend the rest of your time reading a lot of English. That will help.

Good luck on your tests, and in your college applications.

Could/should piyota plan on also taking the TOEFL to demonstrate sufficient English mastery to succeed in college classes given in English, and also to drive home the fact that English is not his native language, so that his SAT scores are interpreted with that in mind?

Thanks u huys very much, I really appreciate it!
By the way, I plan to take 3 SAT II subjec test which are:
IIC Math
Chem
Phy
I heard that barron Physic prep book is good, how about the other 2? What should I cover for those three?

Barrons' IIC hands down. Nothing else comes close. WARNING THOUGH: I never got more than 40/50 right on the Barrons' tests and still got an 800. The real thing is MUCH easier than the Barrons' (but I guess they want you to work harder so you will be more prepared for the real thing)

Thank you, and can I have one more question:
Can you guys explain how the curve is calculated for me? it depends on what factor?
If I plan t take SAT I 2 times, and SAT II each 1 time, and I'm a junior right now, what time should I take each test?
And which level of math/chem/phy you supposed to comple in HS to take the SAT II:
I'm taking (AP Cal AB/ Ap CHem/ H phys) and I'm having no problem, the classes go pretty smooth  :roll:

so, i took most of the SAT II tests that i took 3 at a time(i took 7= 1+3+3)... if you don't mind the mental exhaustion that can result from taking 3 science-related tests in a row, then you can take them in june. else, which is what i really encourage, you should split up, taking the math earlier (chem could also be taken in may, as the AP test is about a week after the may test date.)the physics curve is also nice... :lol:
the SAT I curve is usually easier (i think) in january and march (i only took the test once, in a january sitting, and heard this statement somewhere- it might be completely false!)


function
calculus
derivative
vector
real analysis
real analysis solved

This might be a really easy problems for you, but i have been trying to solve it and could do it, so i hope you can give me a hand. It says: If f: R^3-->R has continuous partial derivatives(we call this class one function in spanish but i dont know if its de same in English) and L is the Subset of R^3 generated by (1,1,1) suppose f(x,y,z)=0 <--> (x,y,z) belongs to L.
Prove that the gradient of f in (0,0,0) if equal to (0,0,0).

I tried using the fact that the directional derivatives are equal to the gradient multiplied by the unity vector, but i couldnt prove that the gradient in (0,0,0) has to be always 0 or which is the same, that the unity vector and the gradient are never perpendicular.

I am really sorry for my awful English. If there is something you see confusing, please tell me and i will try to write it clearer.

Hint: $f$ preserves sign outside $L$.

If you do not see why [hide="click here."]
If you have two points outside $L$ such that $f$ is positive at one point and negative at the other point, you can join them by a curve in $R^3$ avoiding $L$. Then $f$ must vanish somewhere on that curve, which is impossible. [/hide]
If you see why but do not see what to do next [hide="click here."]
Since $f$ is always positive or always negative outside $L$, it has a global extremum at $(0,0,0)$. This means that its gradient must be zero. [/hide]


MATHCOUNTS
geometry
3D geometry
calculus
integration
function

anyone interested in doing the [u] most [b] amazing [/b] [/u] mock mathcounts ever?  :P   it was written by myself and pratik (ragnarok23).  we have been working to perfect it for a few months, and it is flaw-free (or so we hope).  it has some pretty good questions that appear on mathcounts alot.  the sprint questions range from fairly easy to very challenging.  the targets range from medium to very hard.

in my opionion, its like a state.  pratik says its [i]easy. [/i]  but he thinks everything is easy.

but seriously everyone, this is a very good test, and well give it if maybe 20 people will participate.  any takers?  :ninja:

so far we have:

Go Around the Tree
anirudh
bpms
13375P34K43V312
jhredsox
humblemastermind
sonny
funcia
SplashD
i_like_pie
goldendomer

if we have reasonable time frames accounting real MC(if it goes that long) finals and midterms

i think most of us have exams next week.  so we can wait to start until after then.  but wed like to know who would be interested.

I'll do it though I hope it lives up to your words because quite frankly, I waa a little disappointed by someperson's mock mc, no offense to her or anything.

[quote="i_like_pie"]I guess I'll do it, even though there is already a mock MC going on...[/quote]

i think were going to start it after exams, so the other one should be over...

and to splashD, i agree.  our test may not be the bestest problems ever (although i think they are extremely good), but it is verrry well written and presented.

[quote="funcia"]so if we are say going to be doing midterms for HS latin will you allow us time(if it goes that long?)[/quote]

yess.  :)  :starwars:  :pilot:

I'll do it if its as[b][i] [u]amazing[/u][/i][/b] as you say it is.

I'm in.

How do you guys have midterms already???

When did your school start???

-jorian

For many it's already been 4 months. Also, I'll do it.

[quote="13375P34K43V312"]I don't not want to do it.[/quote]
Not to be mean, but that is so not original...
chess64 beat you to it...
EDIT: This is going to be the third mock mathcounts?

Jhredsox, I was two away but I was also in the process of challenging a question.

Why does it matter to you? Why waste your time posting about it?

oh, so technically there were 2 you were arguing about?

this is fun, it isn't wasting time (yes bpms, i'm getting started in 2007 :P )

-jorian

I didn't actually get a 44, because Eli said that time constraints didn't matter for some reason, I actually got a 40 taking time constraints into account.

[quote="jhredsox"][quote="anirudh"][quote="jhredsox"](as anirudh says, he doesn't perform well under time pressure...but then again, he could have taken the whole week for his answers :P )[/quote]

What I meant when saying that is that I'm horrible at Countdowns.[/quote]

that's not what u said. specifically on neal wu's forum, you said you get 10's on chapters

-jorian[/quote]

Yeah. For target rounds. :wink:

[quote="jhredsox"]no, that's not what you said...i will go and quote you if you don't believe me

-jorian[/quote]

huh?

Well...I improved :P

[quote="jhredsox"]it'd be mean of me to quote you, but this is what he said (just to show):
[quote="AnirudhD"]I probably wont' get anywereh[/quote]

[quote="AnirudhD"]I'm not good under time pressure[/quote]
[url]http://nebula.s3.bizhat.com/viewtopic.php?t=95&start=15&mforum=nebula[/url]

this is only to show that anirudh said he didn't think he was bad at mathcounts...he said he would have trouble making the school team

-jorian[/quote]

As I said before, I may have improved. :P

[quote="jhredsox"]it'd be mean of me to quote you, but this is what he said (just to show):
[quote="AnirudhD"]I probably wont' get anywereh[/quote]

[quote="AnirudhD"]I'm not good under time pressure[/quote]
[url]http://nebula.s3.bizhat.com/viewtopic.php?t=95&start=15&mforum=nebula[/url]

this is only to show that anirudh said he didn't think he was bad at mathcounts...he said he would have trouble making the school team

-jorian[/quote]
Relax, do you get some personal satisfaction by showing that anirudh is a liar? Some people ought to chill out...

Yes.  Admin/mods, can you please lock this thread?

All other discussion of the test can take place in the already created forum called Mock Dispute/Discussion Thread or whatever.

ok, i was just trying to make a point
disregard what i have said
-jorian

[quote="Walk Around The River"]Yes.  Admin, can you please lock this forum?

All other discussion of the test can take place in the already created forum called Mock Dispute/Discussion Thread or whatever.[/quote]
I think you mean this thread, not the forum right?
Mods have that power too...

Ya, I would edit what I said, but too many people have quoted me!

-jorian


i was about to submit my solutions so i hit the browse button and all ym documents were there except for the compressed zipped folder with the solutions
so i checked my documents from the desktop icon that says computer and all the files were there including the compressed zipped folder containing the solutions
 :?: wut happened is there a speacial way to submit ur solutions besides hitting the upload button from ur accounts or something :?:

...you know, usamts site exists for a reason.

[url]http://www.usamts.org[/url]

[url]http://www.usamts.org/Rules/U_RulesSubmitting.php[/url]

rules about submitting solutions

Do not submit a folder or more than one file.

Do not submit a zip file.

You must submit a single PDF file.

If you can't figure out how to do this, then print your solutions out and mail them in.

I'm not sure what I can add to what I posted before:[quote="DPatrick"]You must submit a single PDF file.

If you can't figure out how to do this, then print your solutions out and mail them in.[/quote]

[quote="peregrinefalcon88"]are you saying that i should combine my five separate latex files into one latex file and then send it in[/quote]Yes. He is. Or if you're having trouble just mail them in.

or just write your solutions by hand like me  :rotfl:


calculus
integration
calculus computations

Please I need a lot of help in this one: How can i set it up?

A home buyer can afforsd to  spend no more than 800/month on mortgage payments. Suppose that the interest rate is 9% and that the term of the mortgage is 20 years. Assume that interest is compounded continuously and that payments are also made continuously.

a. Determine the maximum amount that this buyer can afford to  borrow.

b. Determine the total interest paid during the term of the mortgage.

Let $y$ be the balance due on the loan. How does the balance change?

The interest accrued (at interest rate $r$) causer the balance to increase by $ry$ per unit time.

The mortgage payments - let's say they are at a rate of $m$ per unit time - cause the balance to decrease.

So now we have the differential equation $\frac{dy}{dt}=ry-m.$

There's one complication here that will cause us to step carefully: what is the unit of time? Interest rates are conventionally given per year, while you see the payment quoted per month. We'd better agree on a common unit. Let's say we use the year as a unit, in which case $\$800$/month becomes $\$9600$/year.

So, $\frac{dy}{dt}=ry-m,$ where $r=.09$ and $m\le9600.$

I'll skip the details of solving this equation - if you're studying differential equations, you should know the technique. We get that $y=\frac{m}{r}+Ce^{rt}.$

Now, what do we have for an initial condition? At time zero, the balance must be the price of the house. Since we don't know that yet, let's give it a name, $P.$ If $y(0)=P,$ then $C=P-\frac{m}{r}$ and we have

$y=\frac{m}{r}+(P-\frac{m}{r})e^{rt}.$

Finally, what of that 20-year term? That means that $y(20)=0.$

One thing you notice right away: no matter what the term, we have to have $P<\frac{m}{r}.$ If not, then we have that $y\to\infty$ as $t\to\infty$ - not only do you never pay off the house but eventually you owe all the money in the world.

Let $T$ be the time to pay off the debt. In this case, we'll use $T=20$ years. That gives us

$0=\frac{m}{r}+(P-\frac{m}{r})e^{rT}$

$Pe^{rT}=\frac{m}{r}(e^{rT}-1)$

$P=\frac{m}{r}(1-e^{-rT}).$

Using $m=9600,\ r=.09,$ and $T=20,$ we get

$P=\frac{9600}{.09}(1-e^{-1.8})\approx\$89035.$

Changing 20 years to 40 years doesn't make a huge difference: in that case, $P$ would be approximately $\$103750.$ Setting $T$ to some extremely long time would, in the limit, get us $P=\frac{m}{r},$ which in this case would be about $\$106667.$

[No, that won't buy you a house - or even a condo - these days where I live, in Southern California.]

Computing the interest paid would involve an integral of $y(t).$ I'll let you work out the details.

One more detail: the actual mechanics of such financial transactions involve periodic compounding of interest and periodic payments rather than continuous compounding and continuous payments. In other words, the mathematics are those of a difference equation rather than a differential equation. But the difference equation and the differential equation are reasonable approximations of each other. Our conclusions will differ by only a few dollars.

[quote="Kent Merryfield"]Computing the interest paid would involve an integral of $y(t).$ I'll let you work out the details.[/quote]

Is that $\int_{0}^{20}r y(t) \, dt$? I think it's because of

[quote="Kent Merryfield"]The interest accrued (at interest rate $r$) causer the balance to increase by $ry$ per unit time.[/quote]

I'm asking this because I've never paid attention in my economics class and I want to be sure I got it right  :D

Yes that is right. On the other hand, it can also be calculated without the integral - it's just the total payments minus the principal.


Denote by ${\mathcal H}_{n} \; (3\le n \le 9 ) , $ the set of all     $ n\times n $ matrices  having all elements in the set $ \{-1,1\} . $ Suppose that there exists  $ H \in {\mathcal H}_{n} $ such that
\[
HH^{T}=\left(\begin{array}{cccccc}
b_{1,1} & 0        & b_{1,3}  &\ldots  & b_{1,n-1} & 0\\
b_{2,1} & b_{2,2}  & b_{2,3}  &\ldots  & b_{1,n-1} & 0\\
b_{3,1} & b_{3,2}  & b_{3,3}  &\ldots  & b_{3,n-1} & b_{3,n}\\
\vdots  &  \vdots  &   \vdots & \vdots & \vdots    &\vdots\\
b_{n,1} & b_{n,2}  & b_{n,3}  &\ldots  & b_{n,n-1} & b_{n,n}\\
\end{array}\right)\; .
\]
[b] Which are the possible  values of $ n $  ? [/b]

Those $0$ come from summing $\sum a_ib_i$, $\sum b_ic_i$, $\sum a_ic_i$, where $a_i$-s are the elements of the first row, $b_i$-s are the elements of the second row, and $c_i$-s are the elements of the last row. To have $\sum a_ib_i=0$ we must first have $n$ even. Now let's make all the $a_i$-s 1. When we make an $a_i$ which was $-1$, 1 we also change the sign of $b_i$ and $c_i$. Obvious the sums remain the same. So, half $b_i$-s and half $c_i$-s are $-1$ and the other halfs are $1$. There are $k$ $i$-s for which $c_i=b_i=1$. This means there are also $k$ $j$-s for which $b_j=c_j=-1$, and so we have $n-2k$ products $b_ic_i=-1$.  Now, to have $\sum b_ic_i=0$, we must have $2k=n-2k\rightarrow n=4k$.


"Find all the ellipses with eccentricity e = 2/3, directrix x = 9 and one of it's foci either c = 4 or c = -4"

I found the one with centre at the origin and c=±4 and a=±6 and b=±sqrt20.
Then I think I've got one with centre (72/13, 0) and c = 4 and 92/13, a = 30/13 and 114/13, b = ±(sqrt500)/13

I think there's one more but my head is swimming and I'm lost and confused...can anyone assist?


function
topology
algebra unsolved
algebra

Well i am kind of stuck with this problem :

1.Does there exist a surjective ,injective function from $R \longrightarrow R^2$?

I neither managed to prove there exist nor to prove that ther dont exist such a function ,
But i managed to generelize the question (ya,great deal  :( )

2.Does there exist a surjective ,injective funtion from $R^m \longrightarrow R^n$?

I think the answer should be no cause ,how shall we turn the line $R$ to a plane $R^2 : !:$

I do one. two should be similar.
Wlog I give an example of how to map [0,1] -> [0,1]^2
For each pair (x,y) in [0,1]^2 let f(x,y) = 0.a1b1a2b2.... in decimal expansion
where we know that x= 0.a1a2... and y=0.b1b2b3...
Clearly the function is injective and surjective.
For R^m to R^n you can use the same idea.

Such function exists!

Let's map $[0,1[$ to $[0,1[^2$:
Map $0,a_1a_2a_3...$ (decimal expansion) to $(0,a_1a_3a_5 ; 0,a_2a_4a_6)$. This is bijective, and the generalisation for $\mathbb{R} \mapsto \mathbb{R}^n$ is similar.

Now map bijectivly  $\mathbb{R}_0^+$ to $[0,1[$, e.g. by $x \mapsto \frac{x^2}{x^2+1}$

Now map bijectivly  $\mathbb{R}$ to $\mathbb{R}_0^+$, but at the moment I don't see a nice mapping, but some ugle ones are easy to construct.
E.g, take $x \mapsto e^x$ for x not being a nonnegative integer, $0 \mapsto 0$ and $x \mapsto e^{x-1}$ for positive integers $x$.

The question is, what type of function is allowed?   If continuous, there is no bijection between $R^m$ and $R^n$ (Brouwer's theorem in topology, on homeomorphism-invariance of dimension).  There are continuous non-injective surjective functions (space-filling curves of Peano, Hilbert, and others) from $R$ to $R^2$.  If discontinuous functions are allowed, anything is possible.

To zetax and ali :nice way of thinking ,i was  trying to prove that we dont have such function ,so i was wrong .

To fleeting-guest: wow,so we dont have a continuous function,ha ?   :) ,is there a elementry proof for $R \longrightarrow R^2$? would you give some idea`s about it ?(Wow man this problem is so beautiful :) ,and i was some how right werent I? ,we dont have a continuous, function from a line to a plane :lol: )

Well, we do know that $\mathbb{C}%Error. "isomorphic" is a bad command.
\mathbb{R^{2}}$ and since $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as gropus..... :)

There is no continous bijective function $f$ from $\mathbb{R}^2$ to $\mathbb{R}$:
$\mathbb{R}^2 \backslash \{ (0,0) \}$ is path-connected, $\mathbb{R} \backslash \{ f(0) \}$ is not, contradicting the continuity of $f$.

$R^1$ is distinguished by the fact that removing a point leaves a disconnected space, as you point out.
$R^2$ is distinguished by the fact that removing a point leaves a space with non-trivial fundamental group.
But I don't know an elementary way to prove that $R^3$ is topologically different from $R^4$.


Using Geometry for Probability Part 3

We use area to find the probability that I meet my wife at the airport.

This guy is smart.... He just taught me what I could never learn!!!!    :lol: