tangent bash woo

area is $sr_A$ where $s$ is semiperimeter and $r_A$ is radius of circle

[asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair I,WW,X,Y,Z,T,A,B,C,D,P,Q,R; I=(0,0); WW=dir(150); X=reflect( (0,0),(0,1))*WW; Y=-WW; Z=-X; T=dir(110); A=2*WW*X/(WW+X); B=2*X*Y/(X+Y); C=2*Y*Z/(Y+Z); D=2*Z*WW/(Z+WW); P=2*WW*T/(WW+T); Q=2*T*X/(X+T); R=reflect(A,C)*P; draw(P--C--Q,gray); draw(circle(I,1),gray); draw(P--R,dashed); draw(circumcircle(P,Q,R)); draw(B--D); draw(A--B--C--D--cycle); draw(P--Q); dot("$I$",I,S); dot("$A$",A,N); dot("$D$",B,E); dot("$C$",C,S); dot("$B$",D,W); dot("$P$",P,NW); dot("$Q$",Q,NE); dot("$R$",R,E); [/asy][/asy] Claim: $BP\cdot DQ$ is fixed. Proof. In fact it equals $(BD/2)^2$. Let $I$ be the center of $\omega$, and let $R$ lie on $\overline{AD}$ such that $\overline{PR}\parallel\overline{BD}$. Then $IP=IR$ and $\overline{DI}$ bisects $\angle PQR$, so $PQRI$ is cyclic. It follows that $BP\cdot DQ=DR\cdot DQ=DI^2$, as desired. $\blacksquare$ Let $x=BP$ and $y=DQ$. Without loss of generality $\operatorname{Area}(ABCD)=2$. Then $\operatorname{Area}(\triangle APQ)=(1-x)(1-y)$, $\operatorname{Area}(\triangle BPC)=x$, $\operatorname{Area}(\triangle DQC)=y$, so \[\operatorname{Area}(\triangle CPQ)=2-(1-x)(1-y)-x-y=1-xy,\]which is fixed.

Toss on the complex plane with the incircle $\omega$ as the unit circle. Let $AB$ touch $\omega$ at $x$. The tangency points of the incircle form a rectangle, so the other tangengy points are $\frac{1}{x}, -x, -\frac{1}{x}$. Let $PQ$ touch $\omega$ at $y$, so we want to show that $[CPQ]$ is independant of $y$. By standard formulas we have $p=\frac{2xy}{x+y}, q=\frac{2y}{xy+1}, c=-\frac{2x}{x^2+1}$. We want to show that $$\begin{vmatrix} -\frac{2x}{x^2+1} & -\frac{2x}{x^2+1} & 1 \\ \frac{2xy}{x+y} & \frac{2}{x+y} & 1 \\ \frac{2}{xy+1} & \frac{2}{xy+1} & 1 \end{vmatrix}$$is independant of $y$. By expansion, we can verify that this simplifies to $4 \cdot\frac{x^2-1}{x^2+1}$, which is independant of $y$ as desired.

clean length bash

Let $O$ be the center of the circle, and $X, Y, Z$ be the tangency points on $AB, AD, PQ$. Note that $[CPQ] = [COP]+[OPZ]+[OQZ]+[COQ] = [AOP]+[AOQ]+[OXP]+[OYQ] = [AXOY]$, which is constant.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.72, xmax = 20.18, ymin = -16.92, ymax = 8.5; /* image dimensions */ pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw((-1.42,0.82)--(1.84,-4.96), linewidth(1) + wvvxds); draw((-1.42,0.82)--(5.18896745165881,1.4179541980072257), linewidth(1) + wvvxds); draw((5.18896745165881,1.4179541980072257)--(8.44896745165881,-4.362045801992775), linewidth(1) + wvvxds); draw((8.44896745165881,-4.362045801992775)--(1.84,-4.96), linewidth(1) + wvvxds); draw(circle((3.514483725829405,-1.771022900996387), 3.0277830214153028), linewidth(1) + sexdts); draw((2.116573626324194,1.1399757090483795)--(-0.23006034688701962,-1.2897703052125846), linewidth(1) + rvwvcq); draw((-0.23006034688701962,-1.2897703052125846)--(8.44896745165881,-4.362045801992775), linewidth(1) + rvwvcq); draw((8.44896745165881,-4.362045801992775)--(2.116573626324194,1.1399757090483795), linewidth(1) + rvwvcq); /* dots and labels */ dot((-1.42,0.82),dotstyle); label("$A$", (-1.34,1.02), NE * labelscalefactor); dot((1.84,-4.96),dotstyle); label("D", (1.92,-4.76), NE * labelscalefactor); dot((5.18896745165881,1.4179541980072257),dotstyle); label("$B$", (5.26,1.62), NE * labelscalefactor); dot((8.44896745165881,-4.362045801992775),linewidth(4pt) + dotstyle); label("C", (8.52,-4.2), NE * labelscalefactor); dot((3.514483725829405,-1.771022900996387),linewidth(4pt) + dotstyle); label("$E$", (3.6,-1.62), NE * labelscalefactor); dot((2.116573626324194,1.1399757090483795),dotstyle); label("$P$", (2.2,1.34), NE * labelscalefactor); dot((-0.23006034688701962,-1.2897703052125846),linewidth(4pt) + dotstyle); dot((3.6606203189312176,-4.795277209239557),linewidth(4pt) + dotstyle); label("$Q$", (-0.16,-1.12), NE * labelscalefactor); label("$G$", (3.74,-4.64), NE * labelscalefactor); dot((6.2117724085506145,-0.3954852654389809),linewidth(4pt) + dotstyle); label("$H$", (6.3,-0.24), NE * labelscalefactor); dot((3.368347132727606,1.2532314072467834),linewidth(4pt) + dotstyle); label("$I$", (3.44,1.42), NE * labelscalefactor); dot((1.3365933429403611,0.3323704300024052),linewidth(4pt) + dotstyle); label("$J$", (1.42,0.5), NE * labelscalefactor); dot((0.817195043108206,-3.1465605365538134),linewidth(4pt) + dotstyle); label("$K$", (0.9,-2.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] It is a well-known fact that $E$ is the midpoint of $AC$. Therefore: $$\frac{[APE]}{[PEC]} = \frac{AE}{EC} = 1$$$$\frac{[AQE]}{[QEC]} = \frac{AE}{EC} = 1$$It follows that: $$[QPC] = [PEC] + [QEC] + [QPE] = [APE] + [AQE] + [QPE]$$Notice that $[QPE] = [JPE] + [QJE]$, which by power of a point, is equal to $[QEK] + [PIE]$. Thus, $$[QPC] = [APE] + [AQE] + [QEK] + [PIE] = [AIEK]$$which is constant. $\blacksquare$

Complex bash FTW

Just came up with this trig bash : Let the radius of the incircle be $r$. Let $T$ be the point of tangency of Q and the incircle. Let $\angle TOQ = \alpha$ Let $\angle TOP = \beta$ Now, it can easily be determined from angle chasing that $\angle DBA = \angle BDA = \alpha + \beta$. So we have that $OC = \frac{r}{\cos(\alpha+\beta)}$ $OQ = \frac{r}{\cos(\alpha)}$ $OP = \frac{r}{\cos(\beta)}$ Thus we obtain \begin{align*} [CPQ] =& [OPQ] + [OQC] + [OCP]\\ &= \frac{r^2}{2}\left(\frac{\sin(\alpha+\beta)}{\cos(\alpha)\cos(\beta)}+\frac{\sin(\beta)}{\cos(\alpha+\beta)\cos(\alpha)}+\frac{\sin(\alpha)}{\cos(\alpha+\beta)\cos(\beta)}\right)\\ &= \frac{r^2}{2}\left(2\cdot \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\right)\\ &= \frac{r^2}{2}\left(2\cdot \frac{\sin(\angle DBA)}{\cos(\angle DBA)}\right)\\ \end{align*} Which is fixed, as desired.

[asy][asy] size(250); defaultpen(linewidth(0.6)+fontsize(11)); pair C = (0,0), B = (3,4), A = (8,4), D = (5,0), O = (4,2), P = (4.9,4); pair X = reflect(O,P)*B, Q = extension(P,X,A,D); draw(A--B--C--D--A,rgb(0.1,0.1,0.7)); draw(circle(O,2),rgb(0.1,0.6,0.1)); draw(P--Q,rgb(0.8,0.1,0.1)); dot("$A$",A,NE,linewidth(3.3)); dot("$C$",C,SW,linewidth(3.3)); dot("$B$",B,NW,linewidth(3.3)); dot("$D$",D,SE,linewidth(3.3)); dot("$P$",P,N,linewidth(3.3)); dot("$Q$",Q,SE,linewidth(3.3)); dot("$T$",(O.x,4),N,linewidth(3.3)); [/asy][/asy] Let $r$ denote the radius of the incircle $\omega$ of $ABCD$, and let $T$ be the tangency point of $\omega$ with $\overline{AB}$. Compute \[ [CPQ] = [ABCD] - [APQ] - [BPC] - [CDQ]. \]Now observe that $[APQ] = r(AT - PQ)$ since $\omega$ is the $A$-excircle of $\triangle APQ$. Furthermore, since the height of $ABCD$ equals $2r$, \[ [BPC] + [CDQ] = r(BP + QD) = 2rBT + rPQ. \]Miraculously, adding these two terms cancels out the common $rPQ$ term, and what's left is an expression that is fixed as $P$ and $Q$ vary. That is to say, the area of $\triangle CPQ$ is constant, which is what we wanted.

Quick trig bash lol. Let $\angle BAD = 2\theta$, $AC = 2h$, $AP=x$ and $AQ=y$. Let $R, S, T$ be the points of tangency between $AB$, $PQ$, and $AD$, respectively with the incircle, and let $O$ be the center of the incircle. We compute \[ PQ = PS+QS = PR + QT = h\cos\theta - x + h\cos\theta - y = 2h\cos\theta - (x+y),\]using the definition of cosine on $\triangle ARO$ and $\triangle ATO$. Thus by the law of cosines we have \[ (2h\cos\theta - (x+y))^2 = x^2 + y^2 - 2xy\cos 2\theta.\]After expansion this is \[ 4h^2 \cos^2 \theta = 4h(x+y)\cos \theta - 2xy (\cos 2\theta + 1) = 4\cos\theta (hx+hy - xy \cos \theta),\]hence $hx+hy-xy\cos\theta$ is constant. Now we can compute that \[ [CPQ] = [APC] + [AQC] - [APQ] = \frac12 \cdot 2hx \sin \theta + \frac12 \cdot 2hy \sin \theta - \frac12 xy \sin 2\theta = \sin \theta (hx + hy - xy \cos \theta),\]so it is constant, as desired. $\blacksquare$

Let $I$ be the center of the incircle of $ABCD$ and let the foot from $I$ to $\overline{AB}$ and $\overline{PQ}$ be $W$ and $T,$ respectively. Notice $$[CITP]=[IPT]+[CPI]=[IPW]+[API]=[AWI]$$since $\triangle IPT\cong\triangle IPW$ and $I$ is the midpoint of $\overline{AC}.$ Similarly, $[CITQ]$ is constant so $[CPQ]=[CITP]+[CITQ]$ is constant. $\square$

Let $AC$ and $BD$ meet at $I$, it's well known that $I$ is the center of circle. Let us call the circle $\omega$. Let $\omega$ meet $AD$ at $X$, $DC$ at $Y$, $CB$ at $K$, $BA$ at $T$ and $PQ$ at $S$. Note that $[CPQ] = [CIQ] + [CIP] + [IPQ] = [AQI] + [API] + [IQS]+ [IPS] = [AQIP] + [QXI] + [PTI] = [AXIP]$ witch $[AXIP]$ is fixed. we're Done.