Chords $A_1A_2$ and $B_1B_2$ meet at point $D$. Suppose $D'$ is the inversion image of $D$ and the line $A_1B_1$ meets the perpendicular bisector to $DD'$ at a point $C$. Prove that $CD\parallel A_2B_2$.
Problem
Source: Sharygin 2020 Correspondence Round Problem 17
Tags: geometry, Sharygin 2020
MP8148
04.03.2020 06:54
[asy][asy]
import olympiad;
size(7cm);
defaultpen(fontsize(10pt));
pair A1 = dir(145), B1 = dir(60), D = (3/5,0), D1 = (5/3,0), C = extension(A1,B1,(D+D1)/2,circumcenter(D,D1,A1)), A2 = intersectionpoint(A1+dir(A1--D)*0.0069--A1+dir(A1--D)*69,unitcircle), B2 = intersectionpoint(B1+dir(B1--D)*0.0069--B1+dir(B1--D)*69,unitcircle);
dot("$A_1$", A1, dir(145));
dot("$B_1$", B1, dir(95));
dot("$C$", C, dir(45));
dot("$D$", D, dir(225));
dot("$D'$", D1, dir(315));
dot("$A_2$", A2, dir(0));
dot("$B_2$", B2, dir(315));
draw(unitcircle);
draw(B1--B2--A2--A1--B1--C--D--D1--C);
[/asy][/asy]
We use complex numbers. Let lower case letters represent their respective uppercase points. Set $(A_1B_1A_2B_2)$ as the unit circle and $d$ be a real number.
First, we solve for $a_2$ and $b_2$. Since $A_1$, $D$, $A_2$ are collinear, we have $$\dfrac{a_1-d}{a_2-d} = \overline{\left(\dfrac{a_1-d}{a_2-d}\right)} = \dfrac{\tfrac{1}{a_1}-d}{\tfrac{1}{a_2}-d}.$$Solving, we get $$a_2^2\left(\dfrac{1}{a_1}-d\right) + a_2\left(da_1-\dfrac{d}{a_1}\right) + (d-a_1) = 0.$$We know that $a_1$ is also a root, so by Vieta's formulas: $$a_2 = \dfrac{-(da_1-\tfrac{d}{a_1})}{\tfrac{1}{a_1}-d} - a_1 = \dfrac{d-da_a^2}{1-da_1} - a_1 = \dfrac{d-a_1}{1-da_1}.$$By symmetry, we have $$b_2 = \dfrac{d-b_1}{1-db_1}.$$Now we solve for $c$. $C$ lies on $\overline{A_1B_1}$ implies $$\dfrac{a_1-c}{b_1-c} = \overline{\left(\dfrac{a_1-c}{b_1-c}\right)} = \dfrac{\tfrac{1}{a_1}-c}{\tfrac{1}{b_1}-c} \implies \overline{c} = \dfrac{a_1+b_1-c}{a_1b_1}.$$Recall that $D'$ is the inversion image of $D$ means that $$OD \cdot OD' = r^2 \implies d \cdot d' = 1 \implies d' = \dfrac{1}{d}.$$Thus, $C$ lies on the perpendicular bisector of $\overline{DD'}$ is equivalent to \begin{align*} (c-d)\overline{(c-d)} &= (c-d')\overline{(c-d')} \\ \implies (c-d)(\overline{c}-d) &= \left(c-\dfrac{1}{d}\right) \left(\overline{c}-\dfrac{1}{d}\right) \\ \implies -d(c+\overline{c})+d^2 &= -\dfrac{1}{d}(c+\overline{c}) + \dfrac{1}{d^2} \\ \implies \left(d-\dfrac{1}{d}\right)(c+\overline{c}) &= \left(d+\dfrac{1}{d}\right)\left(d-\dfrac{1}{d}\right) \\ \implies \overline{c} &= d + \dfrac{1}{d} - c.\end{align*}Therefore, \begin{align*} \overline{c} = \dfrac{a_1+b_1-c}{a_1b_1} &= d + \dfrac{1}{d} - c \\ \implies \left(\dfrac{1}{a_1b_1-1}\right) &= \dfrac{a_1+b_1}{a_1b_1} - \dfrac{1}{d} - d \\ \implies c &= \dfrac{\tfrac{1}{a_1} + \tfrac{1}{b_1} - \tfrac{1}{d} - d}{\tfrac{1}{a_1b_1} - 1} = \dfrac{d(a_1+b_1-a_1b_1d)-a_1b_1}{d(1-a_1b_1)}. \end{align*}It remains to show that $\overline{CD} \parallel \overline{A_2B_2}$, or $\dfrac{c-d}{a_2-b_2} \in \mathbb{R}$. We first calculate \begin{align*}c-d &= \dfrac{d(a_1+b_1-a_1b_1d)-a_1b_1 - d(d)(1-a_1b_1)}{d(1-a_1b_1)} \\ &= \dfrac{-(a_1b_1+d^2-a_1d-b_1d)}{d(1-a_1b_1)} = \dfrac{(a_1-d)(b_1-d)}{d(a_1b_1-1)},\end{align*}and \begin{align*}a_2 - b_2 &= \dfrac{(d-a_1)(1-db_1) - (d-b_1)(1-da_1)}{(1-da_1)(1-db_1)} \\ &= \dfrac{-a_1-d^2b_1+b_1+d^2a_1}{(1-da_1)(1-db_1)} = \dfrac{(a_1-b_1)(1-d^2)}{(1-da_1)(1-db_1)}.\end{align*}Then $$\dfrac{c-d}{a_2-b_2} = \dfrac{\tfrac{(a_1-d)(b_1-d)}{d(a_1b_1-1)}}{\tfrac{a_1-b_1)(1-d^2)}{(1-da_1)(1-db_1)}} = \dfrac{(a_1-d)(b_1-d)(1-da_1)(1-db_1)}{d(1-d^2)(a_1b_1-1)(a_1-b_1)}.$$Now we can verify that $$\overline{\left(\dfrac{c-d}{a_2-b_2}\right)} = \dfrac{(\tfrac{1}{a_1}-d)(\tfrac{1}{b_1}-d)(1-\tfrac{d}{a_1})(1-\tfrac{d}{b_1})}{d(1-d^2)(\tfrac{1}{a_1b_1}-1)(\tfrac{1}{a_1}-\tfrac{1}{b_1})} = \dfrac{c-d}{a_2-b_2},$$so $\overline{CD} \parallel \overline{A_2B_2}$ as desired. $\blacksquare$
Durga01
04.03.2020 06:56
I didnt use bash in ANY of my problems lol
mathsworm
04.03.2020 06:59
Given that you have a lot of time, this problem is easy using length chase though it is not very elegant. Menelaus theorem is the key. I will write out the complete solution when I get time. I did not register for sharygin though
GeoMetrix
04.03.2020 07:07
This was my favourite on the contest. [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 = -16.121969676410945, xmax = 32.96364784200301, ymin = -15.926203370293212, ymax = 16.48942845842549; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((10.68,-3.68), 6), linewidth(0.4)); draw((5.132974812280369,-1.3929688312565571)--(16.512489308673537,-2.272140467159853), linewidth(0.4) + rvwvcq); draw((4.686767354842838,-3.3951097386317093)--(14.460740273404182,0.9789701635725963), linewidth(0.4) + rvwvcq); draw((1.785042163954211,4.833084826605463)--(10.68,-3.68), linewidth(0.4) + blue); draw((7.957992458238445,11.282936285863038)--(4.686767354842838,-3.3951097386317093), linewidth(0.4) + dtsfsf); draw((7.957992458238445,11.282936285863038)--(16.512489308673537,-2.272140467159853), linewidth(0.4) + dtsfsf); draw((1.785042163954211,4.833084826605463)--(7.957992458238445,11.282936285863038), linewidth(0.4) + green); draw((1.785042163954211,4.833084826605463)--(5.985646938099579,2.4329863723724166), linewidth(0.4) + blue); draw((5.985646938099579,2.4329863723724166)--(8.567641788719527,-1.6583282879729353), linewidth(0.4) + red); draw((1.785042163954211,4.833084826605463)--(10.539987308858391,7.191621625517683), linewidth(0.4) + blue); draw((10.539987308858391,7.191621625517683)--(8.567641788719527,-1.6583282879729353), linewidth(0.4) + red); draw((5.176341976336869,1.5873782693162637)--(10.539987308858391,7.191621625517683), linewidth(0.4) + blue); draw((1.785042163954211,4.833084826605463)--(-7.167233142735727,-4.520764337544445), linewidth(0.4) + green); draw((-7.167233142735727,-4.520764337544445)--(14.460740273404182,0.9789701635725963), linewidth(0.4) + dtsfsf); draw((-7.167233142735727,-4.520764337544445)--(16.512489308673537,-2.272140467159853), linewidth(0.4) + dtsfsf); draw((7.957992458238445,11.282936285863038)--(8.567641788719527,-1.6583282879729353), linewidth(0.4) + green); /* dots and labels */ dot((10.68,-3.68),linewidth(3pt) + dotstyle); label("$O$", (10.87511995871673,-3.909789211564647), NE * labelscalefactor); dot((5.132974812280369,-1.3929688312565571),linewidth(3pt) + dotstyle); label("$A_1$", (4.117879105532472,-2.1886052206592295), NE * labelscalefactor); dot((16.512489308673537,-2.272140467159853),linewidth(3pt) + dotstyle); label("$A_2$", (16.64427370601084,-2.284226553487308), NE * labelscalefactor); dot((4.686767354842838,-3.3951097386317093),linewidth(3pt) + dotstyle); label("$B_1$", (3.7035199966107957,-4.356022098095681), NE * labelscalefactor); dot((14.460740273404182,0.9789701635725963),linewidth(3pt) + dotstyle); label("$B_2$", (14.572478161402458,0.9987725402767297), NE * labelscalefactor); dot((8.567641788719527,-1.6583282879729353),linewidth(3pt) + dotstyle); label("$D$", (8.70770308128027,-1.4555083356439587), NE * labelscalefactor); dot((1.785042163954211,4.833084826605463),linewidth(3pt) + dotstyle); label("$D'$", (1.2173653430807378,4.82362585339988), NE * labelscalefactor); dot((7.957992458238445,11.282936285863038),linewidth(3pt) + dotstyle); label("$X$", (8.006479973874358,11.485245373756035), NE * labelscalefactor); dot((5.985646938099579,2.4329863723724166),linewidth(3pt) + dotstyle); label("$C$", (6.2852959829689325,2.1143547566043153), NE * labelscalefactor); dot((5.176341976336869,1.5873782693162637),linewidth(3pt) + dotstyle); label("$M$", (5.010344878594544,1.7956169805107194), NE * labelscalefactor); dot((10.539987308858391,7.191621625517683),linewidth(3pt) + dotstyle); label("$E$", (10.652003515451213,7.309780506929928), NE * labelscalefactor); dot((-7.167233142735727,-4.520764337544445),linewidth(3pt) + dotstyle); label("$T$", (-7.037943057743428,-5.21661409354839), NE * labelscalefactor); dot((8.514631308168472,-0.5330541021454414),linewidth(3pt) + dotstyle); label("S", (8.006479973874358,-0.4036736745350923), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $M$ be the midpoint of $DD'$ and let $O$ be the centre of the circle. Let $A_1B_1 \cap A_2B_2=X$ and let $A_1B_2 \cap A_2B_1=T$. Finally let $S=XD \cap A_1B_2$ and let $CM \cap A_2B_2=E$ . We'll prove something stronger i.e. we'll prove that $XECD$ is a parallelogram . We begin with a few claims. Claim 1: $XD' \parallel CE$ Proof: Note that by brocards theorem (theorem 9.25 egmo page 179) on $A_1B_1A_2B_2$ we have that $X$ lies on the polar of $D$. But notice that since $D'$ is the inversion image of $D \implies D'$ lies on the polar of $D \implies XD'$ is precisely the polar of $D$. Now by definition of polar we have that $\angle OD'X=90^\circ$ but notice that since $CM$ is the perpendicular bisector of $DD' \implies \angle OMC=90^\circ \implies CM \parallel XD' \implies CE \parallel XD' $. Done .$\square$. Claim 2: $\{X,D',T\}$ are collinear. Proof: Notice that by brocards theorem on $A_1B_1A_2B_2$ we have that $XT$ is the polar of $D$. But we already proved in claim 1 that $XD'$ is the polar of $D \implies XD'\equiv XT$. Done. $\square$. Now back to the main problem. Notice that by lemma 9.12 egmo page 175 we have that $-1=(T,S;A_1B_2)$. Notice that claim1 and claim 2 together imply $XT \parallel CE \implies XT \cap CE= P_{\infty}$ where $P_{\infty}$ is the point at infinity along the line $CE$. Now projecting th harmonic bundle we had earlier onto the line $CE$ we have $-1=(T,S;A_1B_2) \overset{X}{=} (P_{\infty},XS\cap CE;C,E)$. Now note that by egmo lemma 9.8 page 173 we have that $XS\cap CE$ is the midpoint of $CE \implies XS$ bisects $CE\implies XD$ bisects $CE$. Now note that by thales theorem on $\Delta DD'X$ we have that $CE\cap XD$ is midpoint of $XD$. Now finally note that in quaderilaeral $XCDE$ we have that the diagnols bisect each other $\implies XCDE$ is a parallelogram as desired $\blacksquare$.
lminsl
04.03.2020 07:37
Construct points $E=A_1B_2 \cap B_1A_2, F=A_1A_2 \cap B_1B_2$, and let $M, N$ be the midpoints of $DE, DF$. Draw a line passing $D$ and parallel to $A_1A_2$, and let the line meet $B_1B_2$ at $X$, and similarly define $Y$. Now consider the radical axis $l$ of the degenerate circle $(D)$ and $\odot(A_1A_2B_1B_2)$. It is well-known that $M, N$ lies on $l$. Since $MN$ is the perpendicular bisector of $DD'$, it suffices to prove that $X,Y$ lies on $l$. Indeed, from $\angle B_2DX=\angle B_2A_2A_1=\angle B_2B_1A_1$, we have $$XD^2=XB_2 \cdot XB_1,$$thus $X$ lies on $l$, as desired.
amar_04
04.03.2020 10:46
One of my favourites from this year! Sharygin 2020 CR P17 wrote: Chords $A_1A_2$ and $B_1B_2$ meet at point $D$. Suppose $D'$ is the inversion image of $D$ and the line $A_1B_1$ meets the perpendicular bisector to $DD'$ at a point $C$. Prove that $CD\parallel A_2B_2$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -14.96, xmax = 7.94, ymin = -5.59, ymax = 6.63; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-1.5,1.23)--(-1.98,-3.17), linewidth(0.4) + wrwrwr); draw(circle((1.719926158758669,-1.3474464900464), 4.1244581313077795), linewidth(0.4) + red); draw((5.28,-3.43)--(-1.22,5.07), linewidth(0.4) + wrwrwr); draw((-1.22,5.07)--(-1.5,1.23), linewidth(0.4) + wrwrwr); draw((0.6428960486046713,2.633905167209277)--(-7.8,-2.97), linewidth(0.8) + wrwrwr); draw((-7.8,-2.97)--(5.28,-3.43), linewidth(0.4) + wrwrwr); draw((0.6428960486046713,2.633905167209277)--(-1.98,-3.17), linewidth(0.4) + wrwrwr); draw((-1.5,1.23)--(5.28,-3.43), linewidth(0.4) + wrwrwr); draw((-1.22,5.07)--(1.719926158758669,-1.3474464900464), linewidth(0.4) + wrwrwr); draw((-1.22,5.07)--(0.34048229401394065,-3.256286074560123), linewidth(0.4) + wrwrwr); draw((-7.8,-2.97)--(0.681288984765185,0.9197527954929827), linewidth(0.4) + wrwrwr); draw((-0.2693555076174075,2.9948763977464914)--(-13.28,-2.73), linewidth(0.4) + wrwrwr); draw((-7.8,-2.97)--(-13.28,-2.73), linewidth(0.8) + wrwrwr); draw((0.6428960486046713,2.633905167209277)--(4.96,5.13), linewidth(0.8) + wrwrwr); draw((4.96,5.13)--(-0.2693555076174075,2.9948763977464914), linewidth(0.4) + wrwrwr); draw((-1.22,5.07)--(4.96,5.13), linewidth(0.8) + wrwrwr); draw((-1.22,5.07)--(-13.28,-2.73), linewidth(0.8) + wrwrwr); draw((-7.8,-2.97)--(-1.22,5.07), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-1.5,1.23),dotstyle); label("$B_2$", (-2.12,1.49), NE * labelscalefactor); dot((-1.98,-3.17),dotstyle); label("$B_1$", (-2.34,-3.55), NE * labelscalefactor); dot((5.28,-3.43),dotstyle); label("$A_1$", (5.46,-3.73), NE * labelscalefactor); dot((-1.22,5.07),dotstyle); label("$D$", (-1.14,5.27), NE * labelscalefactor); dot((0.6428960486046713,2.633905167209277),linewidth(4pt) + dotstyle); label("$A_2$", (1.06,2.37), NE * labelscalefactor); dot((-7.8,-2.97),dotstyle); label("$E$", (-7.9,-3.39), NE * labelscalefactor); dot((1.719926158758669,-1.3474464900464),linewidth(4pt) + dotstyle); label("$O$", (1.72,-1.77), NE * labelscalefactor); dot((0.34048229401394065,-3.256286074560123),dotstyle); label("$Y$", (0.44,-3.69), NE * labelscalefactor); dot((-0.6078482148658891,1.8037336004187088),linewidth(4pt) + dotstyle); label("$X$", (-0.56,2.11), NE * labelscalefactor); dot((-0.34905265799788304,0.43893589767996094),linewidth(4pt) + dotstyle); label("$S$", (-0.2,-0.03), NE * labelscalefactor); dot((0.681288984765185,0.9197527954929827),dotstyle); label("$D'$", (0.76,1.11), NE * labelscalefactor); dot((-0.2693555076174075,2.9948763977464914),linewidth(4pt) + dotstyle); label("$P$", (-0.66,3.13), NE * labelscalefactor); dot((-13.28,-2.73),dotstyle); label("$C$", (-13.62,-3.11), NE * labelscalefactor); dot((4.96,5.13),dotstyle); label("$T$", (5.04,5.33), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let the circle be $\omega$. All Poles and Polars will be taken WRT $\omega$. So, as $D'$ is the inverse of $D$ we get that $D'\in\text{ Polar of } D$. Let $A_1B_1\cap A_2B_2=E$. So, $E\in\text{ Polar of }A$. So, $ED'$ is the polar of $D$. So, $DO\perp ED'$. Now we will rephrase the problem as follows. Rephrased Problem wrote: $A_1A_2B_2B_1$ is a cyclic quadrilateral with circumcenter $O$ and $A_1A_2\cap B_1B_2=D$ and $A_1B_1\cap A_2B_2=E$. Let $ED'\perp DO$ where $D'\in DO$. Let the perpendicular bisector of $DD'$ and $A_1B_1$ meet at a point $C$. Then $DC\|A_2B_2$. Let $P$ be the midpoint of $DD'$ and $A_1B_2\cap A_2B_1=S$. Notice that $S\in\text{Polar of } D$. Hence, $\overline{E-S-D'}$. Let $DS\cap A_2B_2=X$ and $DS\cap A_1B_1=Y$ and $CP\cap A_2B_2=T$. So, $$-1=(X,Y;D,S)\overset{E}{=}(T,C;TC\cap DE,\infty_{TC})$$So this implies that $DE$ bisects $TC$. Also notice that $PC\|D'E$ and $DP=PD'$. Hence $TC$ bisects $DE$. Hence, $TDCE$ is a parallelogram. Hence, $CD\|A_2B_2$. $\blacksquare$
greenturtle3141
04.03.2020 11:08
[asy][asy] size(14cm); draw(circle((0,0),1)); pair A1, A2, B1, B2, D, D1, X, Y, O; O = (0,0); A1 = dir(130); A2 = dir(-20); B1 = dir(-70); B2 = dir(45); X = extension(A1,B2,B1,A2); Y = extension(A1,B1,B2,A2); D = extension(A1,A2,B1,B2); D1 = extension(O,D,X,Y); draw(A1--X--Y--A1); draw(B2--Y); draw(B1--X); draw(A1--A2); draw(B1--B2); draw(O--D1,dashed); dot("$O$",O,NW); dot("$A_1$",A1,dir(A1)); dot("$A_2$",A2,dir(A2)); dot("$B_1$",B1,dir(B1)); dot("$B_2$",B2,dir(B2)); dot("$D$",D,dir(D)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D^\prime$",D1,dir(D1)); [/asy][/asy] Let $X = A_1B_2 \cap B_1A_2$, $Y = A_1B_1 \cap A_2B_2$. Let $Z$ be the infinity point along $A_2B_2$. $D'$ lies on the polar of $D$, which is $XY$. Further, we have $DD' \perp XY$. Now let the plane in which we are working on be $\rho$, and consider a projective transformation with singular line $XY$ that preserves the circle $O$. In particular, we consider the following central projection to another plane $\rho'$ from a point $P$, where $KL$ is the diameter parallel to $OD$: [asy][asy] size(8cm); import olympiad; pair A,B,C,D,E,F,W,X,Y,Z; A = (-3,1.5); B = (2.5,0); C = (-3,-1.5); draw(A--B--C); D = (0,4); E = (-2,-2); F = (2,-2); draw(E--D--F); dot("$P$",D,dir(D)); W = extension(A,B,D,E); X = extension(A,B,D,F); Y = extension(C,B,D,F); Z = extension(C,B,D,E); draw(anglemark(D,X,W,15)); draw(anglemark(Y,Z,W,15)); label("$\rho$",A,dir(A)); label("$\rho^\prime$",C,dir(C)); dot("$K$",W,dir(W)); dot("$L$",X,dir(X)); dot("$K^\prime$",Y,dir(Y)); dot("$L^\prime$",Z,dir(Z)); pair M = (Z+Y)/2; pair V = extension(D,M,W,X); dot("$D$",V,dir(V)); draw(D--M,dashed); [/asy][/asy] We note that by nature of this cross-section, the projection preserves parallelism to $XY$. Since $XY$ is the singular line, this maps $D$ to the center of the circle and maps $A_1B_2A_1B_1$ to a rectangle. [asy][asy] size(14cm); draw(circle((0,0),1)); pair A1, A2, B1, B2, D; D = (0,0); A1 = dir(-210); A2 = dir(-30); B1 = dir(110); B2 = dir(110+180); draw(A1--A2); draw(B1--B2); dot("$D*$",D,NE); dot("$A_1^*$",A1,dir(A1)); dot("$A_2^*$",A2,dir(A2)); dot("$B_1^*$",B1,dir(B1)); dot("$B_2^*$",B2,dir(B2)); pair X,Y; X = (2,2); Y = (2,-2); draw(X--Y); label("$i$",(2,0),E); pair Z = extension(A2,B2,X,Y); dot("$Z^*$",Z,dir(Z)); draw(B2--A2); draw(A2--Z,dashed); pair X = extension(A1,B1,D,Z); draw(B1--X); draw(X--Z); draw((-2,2)--(-2,-2),dashed); [/asy][/asy] Now we have a line $i$ that is the image of the infinite line. Notate the image of a point $S$ as $S^*$. Since $Z$ was the infinity point along $A_2B_2$, $Z^*$ is the intersection of $A_2^*B_2^*$ with line $i$. Now the image of the line through $D$ parallel to $A_2B_2$ is now the line $D^*Z^*$, which intersects $A_1^*B_1^*$ at the reflection of $Z^*$ over the center $D^*$. This in turn lies on the reflection of $i$ over $D^*$. Call this reflection $i^*$. It suffices to show that the pre-image of $i^*$ is the perpendicular bisector of $DD'$ since parallelism to the singular line is preserved. Consider the cross-section from before: [asy][asy] size(14cm); import olympiad; pair A,B,C,P,E,F,K,L,K1,L1; A = (-3,2); B = (2.5,0); C = (-3,-2); draw(A--B--C); P = (0,4); E = (-2,-2); F = (2,-2); draw(E--P--F); dot("$P$",P,dir(P)); K = extension(A,B,P,E); L = extension(A,B,P,F); K1 = extension(C,B,P,F); L1 = extension(C,B,P,E); draw(anglemark(P,L,K,15)); draw(anglemark(K1,L1,K,15)); label("$\rho$",A,N); label("$\rho^\prime$",C,S); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$K^\prime$",K1,dir(K1)); dot("$L^\prime$",L1,dir(L1)); pair M = (L1+K1)/2; pair D = extension(P,M,K,L); dot("$D$",D,dir(D)); draw(P--M,dashed); dot("$D^*$",M,dir(M)); pair i = extension(L1,K1,P,P+(K-L)); draw(B--i,dashed); draw(P--i,dashed); dot("$i$",i,dir(i)); pair i1 = 2*M-i; dot("$i^*$",i1,dir(i1)); draw(C--i1,dashed); pair D1 = extension(K,L,P,P+(K1-L1)); dot("$XY$",D1,dir(D1)); draw(P--D1--A,dashed); draw(P--i1,dotted); pair N = (D+D1)/2; dot("$M$",N,dir(N)); [/asy][/asy] We draw the parallel plane to $\rho$ intersecting $P$ to intersect $\rho'$ at the image $i$ of the infinity line. We reflect $i$ over $D^*$ to get the line $i^*$. We may also draw the plane parallel to $\rho'$ and passing through $P$ to intersect $\rho$ at line $XY$. We now show that the plane passing through $i^*$ and $P$ intersects $\rho$ at the midline between $XY$ and the line through $D$ parallel to $XY$. Now reinterpret this as a 2D geometrical problem on a cross-section! We need to show that $Pi^* \cap \rho = M$ is the midpoint of $(XY)D$. Let $\infty$ and $\infty'$ be the points of infinity along lines $\rho$ and $\rho'$, respectively. Then we can consider a projection from $\rho'$ to $\rho$ through $P$ and obtain: $$-1 = (i^*,i;D^*,\infty') = (M,\infty;D,XY)$$This completes the proof.
SHREYAS333
04.03.2020 12:00
Here is my solution Solution : Let $\omega$ be the given circle and $O$ be its centre and $r$ be its radius. Let $M$ be the midpoint of $DD'$. $\angle DB_2A_2=\angle B_1B_2A_2=\angle B_1AA_2=\angle B_1AD$. We need to prove that $CD\parallel A_2B_2$ , that is we need to prove $\angle CDB_1=\angle DB_2A_2$, that is we need to prove $\angle B_1AD=\angle CDB_1$. So,by converse of alternate segment theorem we need to prove that $CD$ is tangent to $\odot (AB_1D)$ or $(CB_1)(CA_1)=CD^{2}$. Now, $Pow_{\omega}(C)= (CB_1)(CA_1)=OC^{2}-r^{2}$. So ,we need to prove that $OC^{2}-r^{2}=CD^{2} \Longleftrightarrow OC^{2}-CD^{2}=r^{2}$. $\Longleftrightarrow (OM^{2}+MC^{2})-(DM^{2}+MC^{2})= OM^{2}-DM^{2}=(OM+DM)(OM-DM)=(OM+D'M)(OD)= (OD)(OD')= r^{2}$ . Thus , $(CB_1)(CA_1)=CD^{2} \implies CD\parallel A_2B_2$. $\blacksquare$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18.91cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -6.16, xmax = 31.66, ymin = -10.16, ymax = 7.46; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); /* draw figures */ draw(shift((11.96,0.56)) * scale(4.055268178554905, 4.055268178554905)*unitcircle, linewidth(2.) + red); draw((7.946321062254165,1.1392938673035227)--(13.335168996544018,-3.254984958154375), linewidth(2.)); draw((9.79022032130355,-2.865967913731234)--(16.006978247455294,0.8191660946646224), linewidth(2.)); draw((10.842051013905005,-6.185096130717064)--(11.96,0.56), linewidth(2.)); draw((xmin, -0.16574248378816628*xmin-2.141962380842888)--(xmax, -0.16574248378816628*xmax-2.141962380842888), linewidth(1.6) + dotted); /* line */ draw((10.238105981097865,-3.8388514954365283)--(11.566707678718961,-1.812911955293868), linewidth(2.) + blue); draw((13.335168996544018,-3.254984958154375)--(16.006978247455294,0.8191660946646224), linewidth(2.)); draw((10.238105981097865,-3.8388514954365283)--(10.842051013905005,-6.185096130717064), linewidth(2.)); draw((16.006978247455294,0.8191660946646224)--(13.335168996544018,-3.254984958154375), linewidth(2.)); draw(shift((9.602971903361077,-0.5251031963829035)) * scale(2.3483418820410797, 2.3483418820410797)*unitcircle, linewidth(2.) + blue); draw((10.238105981097865,-3.8388514954365283)--(11.96,0.56), linewidth(2.)); draw((7.946321062254165,1.1392938673035227)--(10.238105981097865,-3.8388514954365283), linewidth(2.)); /* dots and labels */ dot((11.96,0.56),linewidth(3.pt) + ududff); label("$O$", (12.04,0.68), NE * labelscalefactor,ududff); dot((7.946321062254165,1.1392938673035227),linewidth(3.pt) + xdxdff); label("$A_1$", (7.2,1.18), NE * labelscalefactor,xdxdff); dot((13.335168996544018,-3.254984958154375),linewidth(3.pt) + xdxdff); label("$A_2$", (13.46,-3.64), NE * labelscalefactor,xdxdff); dot((9.79022032130355,-2.865967913731234),linewidth(3.pt) + xdxdff); label("$B_1$", (9.34,-3.16), NE * labelscalefactor,xdxdff); dot((16.006978247455294,0.8191660946646224),linewidth(2.pt) + xdxdff); label("$B_2$", (16.08,0.9), NE * labelscalefactor,xdxdff); dot((11.566707678718961,-1.812911955293868),linewidth(4.pt) + uuuuuu); label("$D$", (11.82,-1.94), NE * labelscalefactor,uuuuuu); dot((10.842051013905005,-6.185096130717064),linewidth(2.pt) + uuuuuu); label("$D'$", (11.04,-6.56), NE * labelscalefactor,uuuuuu); dot((10.238105981097865,-3.8388514954365283),linewidth(2.pt) + uuuuuu); label("$C$", (9.72,-4.04), NE * labelscalefactor,uuuuuu); dot((11.204379346311987,-3.9990040430054674),linewidth(3.pt) + uuuuuu); label("$M$", (11.2,-4.58), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.5) * currentpicture; /* end of picture */ [/asy][/asy]
mueller.25
04.03.2020 16:56
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -9.336455861920932, xmax = 20.822741067203594, ymin = -7.934126922809146, ymax = 8.77703465431227; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-0.42507203935225474,-2.1872973988267157), 3.931602630175816), linewidth(0.6) + wrwrwr); draw((-3.799059128102572,-4.205640869733378)--(2.5587853680121975,0.372799125549208), linewidth(0.6) + wrwrwr); draw((-1.2289453758788529,1.6612460789616006)--(3.341980608682831,-3.3128255364302346), linewidth(0.6) + wrwrwr); draw((0.9919445364743702,-0.7555214450055447)--(4.5631842185762,-0.3090237829554982), linewidth(0.6) + wrwrwr); draw((1.1372121932086576,7.0625552422717295)--(10.69212983004632,-2.3938660895173167), linewidth(0.6)); draw((0.9919445364743702,-0.7555214450055447)--(4.972654260985429,3.266650602891781), linewidth(0.6) + wrwrwr); draw((0.9919445364743702,-0.7555214450055447)--(-0.42507203935225474,-2.1872973988267157), linewidth(0.6) + wrwrwr); draw((1.1372121932086576,7.0625552422717295)--(3.341980608682831,-3.3128255364302346), linewidth(0.6)); draw((3.341980608682831,-3.3128255364302346)--(-3.799059128102572,-4.205640869733378), linewidth(0.6)); draw((-3.799059128102572,-4.205640869733378)--(1.1372121932086576,7.0625552422717295), linewidth(0.6)); draw((3.341980608682831,-3.3128255364302346)--(10.69212983004632,-2.3938660895173167), linewidth(0.6)); draw((-1.2289453758788529,1.6612460789616006)--(10.69212983004632,-2.3938660895173167), linewidth(0.6)); draw((4.972654260985429,3.266650602891781)--(4.5631842185762,-0.3090237829554982), linewidth(0.6) + wrwrwr); draw(circle((-0.1751275056120044,-4.186438521132457), 3.6239824964642016), linewidth(0.4) + linetype("4 4") + wrwrwr); draw((-2.150978366969179,-1.148468391530618)--(0.9919445364743702,-0.7555214450055447), linewidth(0.6) + wrwrwr); /* dots and labels */ dot((-1.2289453758788529,1.6612460789616006),linewidth(3pt) + dotstyle); label("$A_1$", (-1.129198992683766,1.8058104461048161), NE * labelscalefactor); dot((3.341980608682831,-3.3128255364302346),linewidth(3pt) + dotstyle); label("$A_2$", (3.444121853126101,-3.1630408512345403), NE * labelscalefactor); dot((2.5587853680121975,0.372799125549208),linewidth(3pt) + dotstyle); label("$B_1$", (2.653060950067097,0.5203364786339377), NE * labelscalefactor); dot((-3.799059128102572,-4.205640869733378),linewidth(3pt) + dotstyle); label("$B_2$", (-4.515928483905127,-4.2754702461612615), NE * labelscalefactor); dot((0.9919445364743702,-0.7555214450055447),linewidth(3pt) + dotstyle); label("$D$", (0.6956597971696829,-0.6168135695133776), NE * labelscalefactor); dot((4.972654260985429,3.266650602891781),linewidth(3pt) + dotstyle); label("$D'$", (4.775684965685297,3.412652905443414), NE * labelscalefactor); dot((4.5631842185762,-0.3090237829554982),linewidth(3pt) + dotstyle); label("$C$", (4.655433860935201,-0.171841811542689), NE * labelscalefactor); dot((1.1372121932086576,7.0625552422717295),linewidth(3pt) + dotstyle); label("$E$", (1.2439837164932461,7.219633501414861), NE * labelscalefactor); dot((10.69212983004632,-2.3938660895173167),linewidth(3pt) + dotstyle); label("$F$", (10.786155859642482,-2.2483766820725695), NE * labelscalefactor); dot((-0.42507203935225474,-2.1872973988267157),linewidth(3pt) + dotstyle); label("$O$", (-0.1650935170806048,-2.6439071336020703), NE * labelscalefactor); dot((-2.150978366969179,-1.148468391530618),linewidth(3pt) + dotstyle); label("$C'$", (-2.0438631618457395,-1.0123440210428787), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the circumcircle of $A_1A_2B_1B_2$ be $\omega$. Let $O$ be the centre of $\omega$. Claim 1: $OA_1D'A_2$ is cyclic. Proof: $A_1,D,A_2$ are collinear. After an inversion with respect to $\omega$ , $A_1$ and $A_2$ do not change and $D$ goes to $D'$. Therefore, the line $A_1-D-A_2$ inverts to circle $\odot(OA_1D'A_2)$. We perform the transformation $\Gamma$ composed of an inversion centred at $D$ with radius $\sqrt{DA_1 \cdot DA_2}=\sqrt{DB_1 \cdot DB_2}$ and a homothety centred at $D$ with scale factor $-1$. $\Gamma$ swaps points $A_1$ and $A_2$. It also swaps $B_1$ and $B_2$.Therefore,$\Gamma$ transforms line $A_1B_1$ into the circumcircle of $DA_2B_2$ Claim 2: $\Gamma(D')\equiv O$. Proof: Since $OA_1D'A_2$ is cyclic, we get $DA_1 \cdot DA_2=DO \cdot DD'$. Therefore, $\Gamma$ swaps points $D'$ and $O$. Let $C'$ be the reflection of $D$ across the perpendicular bisector of $A_2B_2$. Claim 3: $\Gamma(C) \equiv C'$ Proof: We prove that $\Gamma(C') \equiv C$ which is equivalent to the claim. Assume that $\Gamma(C')\equiv K$. Note that $B_2A_2DC'$ is an isosceles trapezoid and is therefore cyclic. So, $C' \in \odot(DA_2B_2)$. This implies that $K$ lies on $A_1B_1$. Since $\Gamma(O)=D'$, we have $DOC' \sim DKD'$. But $OD=OC'$ since $D$ and $C'$ are reflections across the perpendicular bisector of $A_2B_2$ passing through $O$. Therefore, $KD=KD'$. So, $K$ lies on line $A_1B_1$ as well as the perpendicular bisector of $DD'$. Hence, $K \equiv C \implies \Gamma(C)=C'$. Main proof: As observed earlier, the quadrilateral $B_2A_2DC'$ is an isosceles trapezoid. So $C'D \parallel A_2B_2$. $\Gamma(C) \equiv C'$ implies that $C',D,C$ are collinear. $$\implies CD \parallel A_2B_2$$
Jupiter_is_BIG
04.03.2020 17:40
Short solution for lazy readers Problem 17. Chords $A_1A_2$ and $B_1B_2$ meet at point $D$. Suppose $D'$ is the inversion image of $D$ and the line $A_1B_1$ meets the perpendicular bisector to $DD'$ at a point $C$. Prove that $CD||A_2B_2$. Solution. Lemma. Given any circle, $\Gamma$ with center $O'$ and a point $X$ in the plane. Let $X'$ be the inversion image of $X$ with respect to $\Gamma$. Now, let $Y$ be any random point on the perpendicular bisector of $XX'$. Then, power of $Y$ with respect to $\Gamma$ is equal to $XY^2$.
Proof of lemma. We will use co-ordinate geometry for the proof. Without loss of generality, let the center of $\Gamma$ be $(0,0)$ and let $X$ lie on the $x$-axis. Thus, $D$ will have co-ordinate $(a,0)$ for some $a$. Since, $O',X,X'$ are collinear, $X'$ will have co-ordinate $(b,0)$. If $r$ denotes the radius of $\Gamma$, then $O'X\cdot O'X'=r^2\implies ab=r^2$. Now, obviously the perpendicular bisector of the line $XX'$ will be $x=(a+b)/2$. Therefore, let $Y$ have some co-ordinate $\left(\frac{a+b}{2},t\right)$ for some $t$. Thus, power of $Y$ will be $$O'Y^2-r^2=\left(\frac{a+b}{2}\right)^2+t^2-r^2=\frac{a^2}{4}+\frac{b^2}{4}+2\cdot\frac{ab}{4}+t^2-r^2+2\cdot\frac{ab}{4}-2\cdot\frac{ab}{4}=\left(\frac{a-b}{2}\right)^2+ab+t^2-r^2=\left(\frac{a-b}{2}\right)^2+t^2$$where the last equality holds because $ab=r^2$. Now obviously, by distance formula, we have $XY^2=\left(\frac{a-b}{2}\right)^2+t^2$ and thus, $O'Y^2-r^2=XY^2$ completing the proof of the lemma.
Proof. Note that power of $C$ with respect to the circle is $CB_1\times CA_1$ and thus, by our lemma, $CB_1\times CA_1=CD'^2=CD^2\implies CD$ is tangent to circumcirle of triangle $A_1B_1D$ at $D$. Thus, $\angle CDB_1=\angle B_1A_1D=\angle B_1A_1A_2=\angle B_1B_2A_2$. Hence, we have $\angle B_1DC=\angle B_1B_2A_2\implies CD||A_2B_2$. Feedback: This was really a very easy problem. However, the lemma was beautiful to observe.
GeoMetrix
04.03.2020 19:12
The official solution. Proof: Since $C$ lies on the radical axis of the given circle and the point $D$, we have $CD^2 = CB_1 \cdot CA_1$, therefore, $\angle CDB_1 = \angle DA_1C = \angle A_2B_2D$ $\blacksquare$.
aops29
04.03.2020 21:29
PUjnk wrote: I felt this was simple. Yup, it was. My solution is attached.
Attachments:
P17_diagram.pdf (2kb)
P17.pdf (82kb)
SPHS1234
28.11.2021 08:13
Since $CM$ is the radical axis of the zero circle at $D'$ and $\omega$, we have that $CD^2=CD'^2=CA_1.CB_1 \implies CD$ is tangent to $(A_1DB_1)$.Thus $\angle CDB_1 =\angle DA_1B_1= \angle DB_2A_2 \implies CD \parallel A_2B_2$
guptaamitu1
02.06.2022 23:06
Let $E = \overline{A_1B_2} \cap \overline{B_1A_2}$, $F=\overline{A_1B_1} \cap \overline{A_2B_2}$ ; $G \in \overline{A_1B_1}$ such that $\overline{DG} \parallel \overline{EF}$. By Brokard, we have points $D',E,F$ collinear. Redefine $C \in \overline{A_1B_1}$ such that $\overline{DC} \parallel \overline{A_2B_2}$. By projecting segment $DD'$ onto $\overline{A_1B_1}$ through the direction along line $EF$, we obtain our problem is equivalent to showing $C$ is the midpoint of segment $GF$. [asy][asy] size(250); pair B1=dir(20),B2=dir(100),A2=dir(130),A1=dir(160),D=extension(A1,A2,B1,B2),E=extension(A1,B2,A2,B1),F=extension(A1,B1,A2,B2),C=extension(D,foot(D,E,foot(E,A2,B2)),A1,B1),G=2*C-F; draw(arc((0,0),dir(0),dir(180))); dot("$A_1$",A1,dir(-60)); dot("$A_2$",A2,dir(160)); dot("$B_1$",B1,dir(-110)); dot("$B_2$",B2,dir(B2)); dot("$D$",D,dir(D)); dot("$E$",E,dir(-90)); dot("$F$",F,dir(-90)); dot("$C$",C,dir(-90)); dot("$G$",G,dir(-90)); draw(F--B2^^C--D,orange); draw(E--F^^D--G,fuchsia); draw(G--B1,purple); draw(D--F,royalblue); draw(B2--A1--D--B1--A2,grey); [/asy][/asy] Let $\infty$ be the point at infinity along line $A_1B_1$. We want to show $$ (G,F ; C, \infty) = -1 \iff (\overline{DG},\overline{DF} ; \overline{DC},\overline{D \infty}) = -1 $$But by definition we have that $$ \overline{DG} \parallel \overline{FE} ~~,~~ \overline{DF} \parallel \overline{FD} ~~,~~ \overline{GC} \parallel \overline{FA_2} ~~,~~ \overline{D \infty} \parallel \overline{FA_1} $$Hence we obtain (which can also be seen from color coding in the figure), $$ (\overline{DG}, \overline{DF} ; \overline{DC} , \overline{D \infty}) = (\overline{FE} , \overline{FD} ; \overline{FA_2} , \overline{FA_1}) $$So we just want $$ (\overline{FE} , \overline{FD} ; \overline{FA_2} , \overline{FA_1}) = -1$$But this is of course true, for example by: Projecting onto line $A_1A_2$ and using Ceva-Menelaus configuration on $\triangle DA_1B_1$ ; or Again projecting onto $\overline{A_1A_2}$ and using $\overline{EF}$ is the polar of $D$.