Trapezoid ABCD

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2376 posts

#1 h Apr 30, 2009, 8:01 PM • 11 Y

Y by Davi-8191, tenplusten, Western_Spadefoot_Toad, Adventure10, Mango247, GoldenLisa, Rounak_iitr, HWenslawski, and 3 other users

Trapezoid $ABCD$ , with $\overline{AB}||\overline{CD}$ , is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$ . Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$ , respectively. Let the line through $G$ parallel to $\overline{AB}$ intersects $\overline{BD}$ and $\overline{BC}$ at points $R$ and $S$ , respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $\overline{BG}$ bisects $\angle CBD$ .

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not_trig

2049 posts

not_trig

#2 h Apr 30, 2009, 8:46 PM • 4 Y

Y by Western_Spadefoot_Toad, Adventure10, HWenslawski, and 1 other user

yay this problem made me happy

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Yongyi781

2142 posts

Yongyi781

#3 h Apr 30, 2009, 9:18 PM • 4 Y

Y by Western_Spadefoot_Toad, Adventure10, Mango247, HWenslawski

Who attached a diagram with lots of angle chasing like I did?

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greentreeroad

484 posts

greentreeroad

#4 h Apr 30, 2009, 9:34 PM • 1 Y

Y by Adventure10

huh this problem made me sad.

Another way for if part. Let BQ,DC intersect at T. Obviously DTB and QCB similar. Then this gives DGB and QSB similar since BG/BT=BS/BC. So <BDG=<BQS. Since <GPD=<BDC=<BRS so DRGP concyclic. So <BDG=<RPG. So<BQS=<RPG. similar <BQR=<SPG, adding gives <RQS=<RPS. So PQRS cyclic.

How to reverse it...

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lil hubeyG

109 posts

lil hubeyG

#5 h Apr 30, 2009, 9:48 PM • 9 Y

Y by antimonyarsenide, tim9099xxzz, MSTang, MathBoy2001, menpo, Adventure10, Mango247, and 2 other users

This solution uses pure angle chasing.

If

Only If

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cosinator

514 posts

cosinator

#6 h Apr 30, 2009, 9:50 PM • 2 Y

Y by Adventure10, Mango247

greentreeroad wrote:

How to reverse it...

I did it the same way. To reverse it I just showed $G$ was unique on $RS$ , (consider the necessary equality of $\angle GDB$ and $\angle SQG$ ).

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alt

261 posts

alt

#7 h Apr 30, 2009, 9:53 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

How many points do you think will be awarded for proving the "Only if" but not the "If"?

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chelseafc

18 posts

chelseafc

#8 h Apr 30, 2009, 9:56 PM • 2 Y

Y by Adventure10 and 1 other user

greentreeroad wrote:

huh this problem made me sad.

Another way for if part. Let BQ,DC intersect at T. Obviously DTB and QCB similar. Then this gives DGB and QSB similar since BG/BT=BS/BC. So <BDG=<BQS. Since <GPD=<BDC=<BRS so DRGP concyclic. So <BDG=<RPG. So<BQS=<RPG. similar <BQR=<SPG, adding gives <RQS=<RPS. So PQRS cyclic.

How to reverse it...

easy to know <AQR = <BPS since both ABPQ and RSPQ are cocyclic.
you can get <BQR = <SPG since <AQB = <APB. then <SPQ = <SCG since GSPC are cocyclic(because <GSC = <GPC).then <BQR = <BCG. Now suppose <DBQ bigger than <CBQ. then <BRQ is smaller than <BGC. since they are both bigger than 90 degree, sinBRQ > sinBGC.using sine law in triangle BRQ and BGC and simplify you can get BQ/BC > BD/BM. Now by proving (<BCQ) > (<BMD) > (180 - <BCQ), you know that sinBMD > sinBCQ. now using sine law in triangle BDM and BCQ you can get BQ/BC < BD/BM,which is a contradiction. The other way is similar(but not completely same though, the thinking is the same, just using sine law). Hence <DBQ cannot be bigger than or smaller than <CBQ.

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Alumen

157 posts

Alumen

#9 h May 1, 2009, 1:02 AM • 11 Y

Y by OlympusHero, Ultroid999OCPN, Lcz, Adventure10, Mango247, Math_legendno12, Alex-131, and 4 other users

Warning: slightly offtopic post as it contains no relevance to the problem itself

On the 'solution' sheet for this problem, all I did was draw a diagram and point out the grammar error:
"Let the line through G parallel to AB intersects"

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Altheman

6194 posts

Altheman

#10 h May 2, 2009, 7:33 PM • 16 Y

Y by TheStrangeCharm, Mobashereh, JasperL, e_plus_pi, DPS, MarkBcc168, Infinityfun, Quidditch, Adventure10, Mango247, and 6 other users

Since $RS$ and $DC$ are parallel, the circumcircle of $BSR$ , call it $\omega'$ , is tangent to $\omega$ at $B$ .

i) If $PQRS$ is cyclic:

$X\in RSG\cap BB\cap QP$ exists since it is the radical center of $\omega$ , $\omega'$ and $PQRS$ . By pascal on $BQQPAB$ , $G$ , $X$ , and $QQ\cap AB$ are collinear so $QQ\cap AB$ lies on $GX$ . But $GX=RS\parallel AB$ so those lines intersect at infinity so $QQ\parallel AB\parallel CD$ . It follows directly that $BQ$ bisects $\angle DBC$ .

ii) If $BG$ bisects $\angle DBC$ :

Let $X=BB\cap RS$ , $\{P', Q\}=\omega \cap XQ$ . Then $BX^2=XS\cdot XR=XP'\cdot XQ$ so $P'QRS$ is cyclic. Let $\{A', P'\}=P'G\cap \omega$ . By pascal on $BQQP'A'B$ , $G$ , $X$ , and $QQ\cap A'B$ are collinear. But the angle bisector implies that $QQ\parallel DC\parallel GX$ so $A'B\parallel DC$ so $A=A'$ , $P=P'$ so $PQRS$ is cyclic, as desired.

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Hamster1800

1189 posts

Hamster1800

#11 h May 2, 2009, 8:01 PM • 5 Y

Y by rashah76, Adventure10, Mango247, and 2 other users

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max221

107 posts

max221

#12 h May 2, 2009, 10:40 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

did anyone else try inverting around B? i think im a little crazy...

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Ubemaya

1560 posts

Ubemaya

#13 h May 2, 2009, 11:12 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

did you find a solution with it?

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jmerry

12096 posts

jmerry

#14 h May 3, 2009, 6:57 AM • 5 Y

Y by myh2910, Adventure10, and 3 other users

I got inversion to work. Like many other solutions the key was finding the right magic point to mark, after which stuff just works.

The inverted picture
The magic point

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not_trig

2049 posts

not_trig

#15 h May 3, 2009, 2:20 PM • 2 Y

Y by Adventure10 and 1 other user

So I saw the first line of my proof after inverting. I was like "oh my, $R'S'$ looks parallel to $P'Q'$ ."

and then I was like wait that may be trivial. and then I was happy.

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Mogmog8

1080 posts

Mogmog8

#39 h May 5, 2022, 3:15 AM • 1 Y

Y by centslordm

Invert about $B$ with arbitrary radius, denoting $X'$ is the inverse of $X.$ We see $C',D'$ are points on $\overline{S'B},\overline{R'B},$ such that $\overline{C'D'}\parallel\overline{S'R'}.$ Also, $A'=\overline{BB}\cap\overline{C'D'},$ $G'$ is on $(BR'S'),$ and $P'=(A'BG')\cap\overline{C'D'}.$ Since $P'Q'R'S'$ is a trapezoid, it suffices that $P'S'=Q'R'.$ Notice $\measuredangle A'P'G'=\measuredangle A'BG'=\measuredangle BS'G',$ so we can let $P'=(C'G'S')\cap\overline{C'D'}.$

Notice $\begin{align*}\measuredangle G'Q'P'=\measuredangle Q'P'G'&\iff \measuredangle Q'BR'+\measuredangle BD'Q'=\measuredangle C'S'G'\\&\iff \measuredangle G'BR'+\measuredangle BR'S'=\measuredangle S'BG'+\measuredangle BG'S'\\&\iff \measuredangle R'BG'=\measuredangle G'BS'\end{align*}$ so $\measuredangle R'BG'=\measuredangle G'BS'$ is equivalent to $G'R'=G'S'$ and $G'Q'=G'P'.$ Since $\measuredangle BGR=\measuredangle BSR=\measuredangle BCD=\measuredangle SCP=\measuredangle SGP,$ we have $G'R'=G'S'\land G'Q'=G'P'\iff\triangle Q'G'R'\cong\triangle P'G'S'\iff Q'R'=P'S'.$ $\square$

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kamatadu

466 posts

kamatadu

#40 h May 11, 2023, 10:49 AM • 1 Y

Y by HoripodoKrishno

Optx wrote:

Anyone tried to invert from point Q? I got some interesting things but cant solve it fully.

I gotchu homie :love:

. NOOO, why is no one Inverting at $Q$ in any of the earlier posts and Inverting with $\sqrt{bc}$ at $B$ T__T . Am I too dumb for not having thought about $\sqrt{bc}$ T__T .

Beautiful problem :omighty:

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Before starting, we prove that for any point $G$ , if we let the line parallel to $CD$ through $G$ meet $BD$ and $BC$ at $R$ and $S$ repectively and also let $P=\odot(QRS)\cap\odot(BCD)$ , then if $T=PQ\cap RS$ , then we will have that $T$ lies on the tangent through $B$ to $\odot(BCD)$ .

This simply follows by applying Radax to $\left\{\odot(BRS),\odot(BCD),\odot(PQRS)\right\}$ .

We first adress the if direction that is when $BG$ is the angle bisector.

Redefine $P=\odot(QRS)\cap\odot(BCD)$ . Let $X=CG\cap QR$ and $Y=DG\cap QS$ . Now we have,
$\begin{align*} \measuredangle GXQ &=\measuredangle CXQ=\measuredangle XQC+\measuredangle QCX\\ &=\measuredangle RQC+\measuredangle QCG\\ &=\measuredangle QRS+\measuredangle RSC+\measuredangle SCQ+\measuredangle QCG\\ &=\measuredangle QRS+\measuredangle DCG\\ &=\measuredangle QRG+\measuredangle RGX\\ &=\measuredangle RXG .\end{align*}$
So we get that $X\in\odot(BCD)$ and similarly we also get $Y\in\odot(BCD)$ . Now $\measuredangle GXR=\measuredangle CXQ=\measuredangle CBQ=\measuredangle QBD=\measuredangle GBR$ which gives that $XRGB$ is cyclic and similarly $BGSY$ is also cyclic. Thus we have the following final POP relations $QR\cdot QX=QG\cdot QB=QS\cdot QY$ .

Now we perform an Inversion with center $=Q$ and radius $=\sqrt{QG\cdot QB}$ which swaps $\left\{R,X\right\}$ , $\left\{G,B\right\}$ and $\left\{S,Y\right\}$ . Also notice that by Radax on $\left\{\odot(BCD),\odot(PQRS),\odot(XRSY)\right\}$ , we get that $T=PQ\cap RS\cap XY$ . Thus we finally derive that $\left\{RS,\odot(BCD)\right\}$ also get swapped. We also deduce that $\left\{P,T\right\}$ also get swapped under this Inversion. Thus now it just suffices to prove that $K=QA\cap RS$ , then $TBKQ$ is cyclic which simply follows from $\measuredangle TBQ=\measuredangle BDQ=\measuredangle BAQ=\measuredangle GKQ=\measuredangle TKQ$ and we are done by Inverting back :thumbup:

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Now for the other direction, we have that $\overline{A-G-P}$ are collinear. So now Reim's on the the lines $QG$ and $PG$ with the circles $\odot(PQR)$ and $\odot(BCD)$ , we get that the tangent at $G$ to $\odot(PGQ)$ is parallel to $AB\parallel CD$ . Now letting $T=RS\cap PQ$ , we get that $TG^2=TP\cdot TQ=TB^2$ from the initial relation we deduced at the beginning of the solution. This clearly gives $TG=TB$ which immediately finishes as then $\measuredangle DBQ=\measuredangle BDQ+\measuredangle DQB=\measuredangle TBG+\measuredangle DCB=\measuredangle BGS+\measuredangle GSB=\measuredangle GBS=\measuredangle QBC$ and we are finally done :stretcher:

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This post has been edited 4 times. Last edited by kamatadu, May 11, 2023, 10:58 AM
Reason: failed pre nursery, so made countably infinitely many grammatical errors :P

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Tafi_ak

309 posts

Tafi_ak

#41 h Jun 3, 2023, 9:17 AM • 1 Y

Y by CT17

Taking inversion at $B$ gives the following problem,

Inverted Problem wrote:

Let $\omega$ be the circumcircle of $\triangle ABC$ and $G$ be a point on arc $BC$ of $\omega$ not containing $A$ . Let $P\in AB$ and $Q\in AC$ such that $PQ\parallel BC$ . Let $T$ be the intersection $PQ$ and tangent at $A$ of $\omega$ . Let $R=PQ\cap (ATG)$ other than $T$ and $S=AG\cap PQ$ . Prove that $AG$ is the bisector of $\angle BAC$ if and only if $BCRS$ is cyclic.

Let $(ATG)=\gamma$ . Define the function $f(X)=P(X, \gamma)-P(X, \omega)$ It is known that $f$ is linear. So by linearity we have $\frac{ST}{SR}=\frac{f(S)-f(T)}{f(S)-f(R)}=\frac{TA^2}{P(R, \omega)} \implies P(R, \omega)=\frac{TA^2\cdot SR}{ST}$ Let $O$ be the center of $\omega$ . Since $PQ\parallel BC$ and $OB=OC$ therefore is sufficient to show $AG$ is the angle bisector of $\angle BAC$ iff $OR=OS$ or equivalently $P(R, \omega)=P(S, \omega)$ . Therefore we can say $AG$ is the angle bisector iff
$\begin{align*} ST\cdot SR=SA\cdot SG=P(S, \omega)=P(R, \omega)=\frac{TA^2\cdot SR}{ST}\iff TA=ST \end{align*}$ which is easy angle chase.

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HamstPan38825

8855 posts

HamstPan38825

#42 h Aug 14, 2023, 1:28 AM • 1 Y

Y by a_smart_alecks

Kind of an arbitrary geometry problem, but enjoyable nonetheless.

Invert about $B$ . In the resulting picture,

$\overline{AB}$ is tangent to $(BCD)$ and $(BRS)$ ;
$P = \overline{AD} \cap (ABG)$ , and $Q = \overline{AD} \cap \overline{BG}$ .

Observe that $\angle APG = \angle ABG = \angle BRG$ . For one direction, assume that $\overline{BG}$ is an angle bisector. Then $\angle GQP = \frac{\widehat{BS}+\widehat{RG}}2 = \angle BRG = \angle GPQ$ hence $GP = GQ$ , and the result follows by symmetry.

On the other direction, setting $G'$ to be the arc midpoint, notice that $\angle G'QP = \angle BQC$ by same computation, hence $G'$ lies on $\overline{BQ}$ .

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OronSH

1723 posts

OronSH

#43 h Nov 3, 2023, 9:32 PM

Y by

Invert about $B,$ and let $X'$ be the image of a point $X$ under this inversion. First, there is a homothety at $B$ taking $DC$ to $RS,$ so $(BCD)$ and $(BSR)$ are tangent so $C'D' \parallel R'S'.$ Next, let $R'S'$ intersect $G'P'$ at $Z.$ We get $\angle R'ZG'=\angle A'P'G'=\angle A'BG'=\angle BS'G'$ and $\angle G'R'Z=\angle G'BS',$ so $\angle R'G'B=\angle ZG'S',$ and letting $G'P'$ intersect $(BR'S'G)$ again at $K$ we get that $BK \parallel R'S'.$ We find that $P'Q'R'S'$ is cyclic iff it is an isosceles trapezoid. However, it is easy to see now that this is true iff $G$ is the midpoint of arc $R'S'$ by considering the perpendicular bisector of $R'S',$ so this is true iff $BG$ bisects $\angle CBD.$

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shendrew7

792 posts

shendrew7

#44 h Dec 27, 2023, 3:37 AM

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Invert at $B$ with arbitrary radius. We preserve $CE \parallel RS$ , but now we know $AB$ is tangent to $(BCD)$ and $(BRS)$ . Points $P$ , $Q$ , $R$ , and $S$ are mapped to the points of a trapezoid. Then
$\begin{align*} BG \text{ bisector} &\iff \measuredangle GSR = \measuredangle SRG = \measuredangle SBG = \measuredangle GBR \\ &\iff \measuredangle GSB = \measuredangle GBA = \measuredangle AQB = \measuredangle PQR = \measuredangle GPR \\ &\iff PQRS \text{ isosceles trapezoid.} \quad \blacksquare \end{align*}$

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asdf334

7586 posts

asdf334

#45 h Dec 27, 2023, 4:28 PM

Y by

nice and cute. it doesn't fall immediately

Let $\ell$ be the line through $G$ parallel to $AB$ . Also let $U$ and $V$ be the intersections of $QR$ and $PS$ with $\omega$ .

If $PQRS$ is cyclic, then by Reim's $UV\parallel AB\parallel CD$ . By Pascal's on $BQVPAC$ we then find $QV$ passes through $AC\cap \ell=T$ .

Here's the thing, though: $R$ and $T$ should be symmetric about the perpendicular bisector of $AB$ . Since $U$ and $V$ are also symmetric, it follows that $UR\cap VT=Q$ is on the perpendicular bisector, i.e., $BG$ bisects $\angle CBD$ .

If $BG$ bisects $\angle CBD$ , then $Q$ is on the perpendicular bisector of $AB$ . It suffices to have $UV\parallel AB$ (points are defined the same way). For the most part, everything here is identical to the previous. By Pascal's on $BQVPAC$ we again find $QV$ passes through $AC\cap \ell=T$ . Now, instead of $UR$ and $VT$ being given as symmetric, we instead have $QR$ and $QT$ symmetric. Thus $U=QR\cap \omega$ and $V=QT\cap \omega$ are clearly symmetric, so $UV\parallel AB$ .

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asdf334

7586 posts

asdf334

#46 h Dec 27, 2023, 4:37 PM

Y by

Also we can draw $PQ\cap RS=M$ and reflect $G$ across $M$ to $N$ . Then we need to show something like $(R,S;G,N)=-1$ which isn't hard if you notice the radical center of $(BRS),(BDC),(PQRS)$ (I didn't oops)

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atdaotlohbh

168 posts

atdaotlohbh

#47 h Mar 18, 2024, 4:19 PM • 1 Y

Y by kotmhn

Nice problem! I think I came up with a wonderful solution(not really humble there

). Here is a sketch of it:
Let $RS$ intersect $\omega$ and $PQ$ at $X, Y, Z$ . $CG$ intersect $\omega$ at $F$ . Then our idea is to use Desargues' Involution theorem on $FPCD$ and line $RS$ . It is not hard to verify that then $PQRS$ is concyclic is equivalent to $P, R, F$ lie on the same line, which is equivalent to tangent at $P$ being parallel to $CD$ by Pascal's theorem, which is exactly what we wanted.

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rstenetbg

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rstenetbg

#48 h May 19, 2024, 9:37 AM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -25.251594460355093, xmax = 24.927192375407017, ymin = -12.702874719263233, ymax = 7.819465053019219; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-1.3452036517273975,0.28788835564809817),0.5880326582315872,-39.87646176125591,-0.33902432445145547)--(-1.3452036517273975,0.28788835564809817)--cycle, linewidth(1) + qqwuqq); draw(arc((-4.069482163222376,2.56383778197193),0.5880326582315872,-39.87646176125591,-0.33902432445144914)--(-4.069482163222376,2.56383778197193)--cycle, linewidth(1) + qqwuqq); draw(arc((-0.7080729112973836,3.584814203364297),0.5880326582315872,-57.110838967275384,-17.57340153047094)--(-0.7080729112973836,3.584814203364297)--cycle, linewidth(1) + qqwuqq); draw(arc((3.2753562200269513,0.26054776469097185),0.5880326582315872,179.66097567554854,285.6547838267051)--(3.2753562200269513,0.26054776469097185)--cycle, linewidth(1) + blue); draw(arc((2.6410188261351224,2.5241306755260275),0.5880326582315872,179.66097567554854,285.6547838267051)--(2.6410188261351224,2.5241306755260275)--cycle, linewidth(1) + blue); draw(arc((-0.7080729112973836,3.584814203364297),0.5880326582315872,-163.10464711843196,-57.110838967275384)--(-0.7080729112973836,3.584814203364297)--cycle, linewidth(1) + blue); /* draw figures */ draw((-6.192115171650039,-4.679590254706536)--(4.677785160575875,-4.7439091915836125), linewidth(1)); draw((4.677785160575875,-4.7439091915836125)--(2.6410188261351224,2.5241306755260275), linewidth(1)); draw(circle((-0.7431525425540128,-2.343643479006383), 5.928561467353529), linewidth(1)); draw((-4.069482163222376,2.56383778197193)--(2.6410188261351224,2.5241306755260275), linewidth(1)); draw((-4.069482163222376,2.56383778197193)--(-6.192115171650039,-4.679590254706536), linewidth(1)); draw((-4.069482163222376,2.56383778197193)--(4.677785160575875,-4.7439091915836125), linewidth(1)); draw((-0.7080729112973836,3.584814203364297)--(4.677785160575875,-4.7439091915836125), linewidth(1)); draw((-6.192115171650039,-4.679590254706536)--(3.6084618748954225,1.6826815245216786), linewidth(1)); draw((-0.7080729112973836,3.584814203364297)--(9.912245917184492,0.2212762280214005), linewidth(1)); draw((-11.367451861549748,0.3471915994932009)--(-0.7080729112973836,3.584814203364297), linewidth(1)); draw((-11.367451861549748,0.3471915994932009)--(9.912245917184492,0.2212762280214005), linewidth(1)); draw((-11.367451861549748,0.3471915994932009)--(-6.192115171650039,-4.679590254706536), linewidth(1)); draw((4.677785160575875,-4.7439091915836125)--(9.912245917184492,0.2212762280214005), linewidth(1)); /* dots and labels */ dot((-6.192115171650039,-4.679590254706536),dotstyle); label("$A$", (-6.552155928590619,-5.215258871114335), NE * labelscalefactor); dot((4.677785160575875,-4.7439091915836125),dotstyle); label("$B$", (4.5812624005941,-5.293663225545214), NE * labelscalefactor); dot((2.6410188261351224,2.5241306755260275),dotstyle); label("$C$", (2.71915898286074,2.7231820150121155), NE * labelscalefactor); dot((-4.069482163222376,2.56383778197193),linewidth(4pt) + dotstyle); label("$D$", (-4.592047067818661,2.6447776605812368), NE * labelscalefactor); dot((-0.7080729112973836,3.584814203364297),linewidth(4pt) + dotstyle); label("$Q$", (-1.0442500298214181,3.7816407998289754), NE * labelscalefactor); dot((1.4345559109606933,0.27144007421207395),dotstyle); label("$G$", (1.072667539812296,-0.39339107361530595), NE * labelscalefactor); dot((3.6084618748954225,1.6826815245216786),linewidth(4pt) + dotstyle); label("$P$", (3.836421033500756,1.4295101669026198), NE * labelscalefactor); dot((-1.3452036517273975,0.28788835564809817),linewidth(4pt) + dotstyle); label("$R$", (-1.2598620045063336,0.44945573651663817), NE * labelscalefactor); dot((3.2753562200269513,0.26054776469097185),linewidth(4pt) + dotstyle); label("$S$", (3.346393818307767,0.4102535593011989), NE * labelscalefactor); dot((9.912245917184492,0.2212762280214005),linewidth(4pt) + dotstyle); label("$R'$", (9.991162856324703,0.37105138208575966), NE * labelscalefactor); dot((-11.367451861549748,0.3471915994932009),linewidth(4pt) + dotstyle); label("$S'$", (-11.922854207105782,0.48865791373207745), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

We consider inversion $\Psi$ centered at $G$ with radius $r=-\sqrt{GB\cdot GQ}.$ We know that $\Psi(Q)=B$ and $\Psi(P)=A.$

Let $QC\cap RS = R'$ and let $QD\cap RS = S'$ .

Claim: $R'\equiv \Psi(R).$

Proof: Angle chasing gives $\angle BQR' = \angle BQC = \angle BDC = \angle BRS = \angle BRR'.$ Hence, $BRQR'$ is cyclic, so $GR\cdot GR' = r^2$ and $G$ is on segment $RR'$ , so we are done.

In a similar fashion we can prove that $S'\equiv \Psi(S).$ Now $PQRS$ is cyclic iff $ABR'S'$ is cyclic. However, since $AB\parallel R'S'$ , we get that $ABR'S'$ is cyclic iff $AS'=BR'.$

Denote with $S_{XY}$ the perpendicular bisector of segment $XY$ . Clearly, we have that $S_{AB}\equiv S_{DC}.$

Direction 1: If $G$ lies on the angle bisector, then $Q\in S_{AB}$ and since $\hspace{2mm}$ $RS\perp S_{AB}$ , we get that $S'$ and $R'$ are symmetric with respect to $S_{AB}$ . However, $A$ and $B$ are also symmetric wrt to $S_{AB}$ , so $AS'=BR'$ and we are done.

Direction 2: If $PQRS$ is cyclic, then $ABR'S'$ is cyclic. Thus, $S'$ and $R'$ are symmetric wrt to $S_{AB}.$ Since $D$ and $C$ are also symmetric wrt to $S_{AB}$ , we get that $Q=DS'\cap CR'$ lies on $S_{AB}$ . Hence, $Q$ is the midpoint of the arc $CD$ and hence $G$ lies on the angle bisector of $\angle CBD$ .

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palindrome868

1174 posts

palindrome868

#49 h Jun 8, 2024, 2:08 AM

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after inversion, the key insight really was trying to remove points from the diagram lol

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kotmhn

56 posts

kotmhn

#50 h Dec 2, 2024, 8:06 PM

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Great problem. I don't think a solution similar to mine has been posted yet so here it is,

First the if condition.
Observe that as $RS \parallel CD$ $(BCD),(BSR)$ are tangent.
Claim: $PQ \cap RS \in BB$
Proof:
Let $X = BB \cap RS$ and let $XQ \cap \omega = P' \neq Q$
Observe
$\angle QBX = \angle QBC + \angle CBX = \angle DBQ + \angle ABD = \angle ABG \dots (1)$ Additionally by sine law in $\triangle BGX$
$\frac{BG}{BX} = \cos(\angle GBX) \dots (2)$ Similarly by applying the extended sine law on $\omega$ we get
$\frac{AB}{BQ} = \cos(\angle DBQ + \angle ABD) = \cos(\angle ABG) = \cos(\angle GBX) \dots (3)$ With $1,2,3$ we conclude
$\triangle ABG \sim \triangle BQX$ So
$\angle BAG = \angle BQX = \angle BQP'$ This implies $A-G-P'$ and hence $P' \equiv P$ .

Finally by pop we get that
$XP\cdot XQ = XB^2 = XR \cdot XS$ Hence $PQRS$ cyclic.

Now the only if part:
By radical axis we get $PQ \cap RS \in BB$ .
Next
$\begin{align*} \angle XGP &= \angle RGA \\ &= \angle GAB \\ &= \angle PQG \end{align*}$ Hence $XG$ is tangent to $(PQG)$ . So by many pops
$XG^2 = XP \cdot XQ = XB^2$ Hence $XB = XG$ .
Now as its well known that the tangent from a vertex, the opposing side and the perpendicular bisector of the angle bisector of the angle at vertex are concurrent, we are done.

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Ilikeminecraft

300 posts

Ilikeminecraft

#52 h Feb 20, 2025, 6:36 PM

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Invert about $B$ . Problem becomes:

inverted statement wrote:

Let $ABC$ be a triangle, $D$ lies on $AC$ such that $(BDC)$ is tangent to $AB,$ and $G$ is any point. $Q$ is the intersection of $BG$ and $AC,$ $R, S$ lie on $BD, BC$ such that $BRSG$ is tangent to $AB.$ Finally, $P = AC\cap (ABG).$ Show that $PQRS$ is cyclic if and only if $BG$ bisects $\angle RBS.$

First, note that inversion preserves parallel lines, so $PQ = DC\parallel RS.$ Note that $BG$ is bisector if and only if $RGS$ is isosceles, so it suffices to prove that $BG$ is bisector if and only if $GPQ$ is isosceles. However, $BG$ bisector $\iff \angle DBQ = \angle QBC \iff \angle BDQ + \angle ABD = \angle QBC + \angle ABD \iff \angle ABQ = \angle QBC + \angle BCD = \angle AQB\iff GPQ$ is isosceles.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(30.124370900642376cm); 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 = -13.141884669958879, xmax = 16.982486230683495, ymin = -1.941572657856366, ymax = 16.98538991910735; /* image dimensions */ /* draw figures */ draw((-5.449099046300786,11.855808032891977)--(1.1131748507888175,12.077324983344363), linewidth(1.2)); draw(circle((1.2130404729564792,9.118880655589328), 2.960129386178395), linewidth(0.8)); draw((-1.7438108774116874,9.258150446796272)--(1.1131748507888175,12.077324983344363), linewidth(1.2)); draw((1.1131748507888175,12.077324983344363)--(2.3520623455555123,6.386666323466413), linewidth(1.2)); draw((-0.07271075964964728,8.086596251773575)--(1.1131748507888175,12.077324983344363), linewidth(1.2)); draw(circle((1.2900851678349268,6.83648922198775), 5.2438208148059084), linewidth(0.8)); draw(circle((-2.028074943150312,7.822514229806174), 5.288748889515066), linewidth(0.8)); draw((-1.7201244339576545,2.54273854326088)--(-0.07271075964964728,8.086596251773575), linewidth(1.2)); draw((-3.947928660123173,7.083203380299419)--(-1.7438108774116874,9.258150446796272), linewidth(1.2)); draw((2.3520623455555123,6.386666323466413)--(3.3078437941783156,1.9964161994385508), linewidth(1.2)); draw((3.3078437941783156,1.9964161994385508)--(-3.947928660123173,7.083203380299419), linewidth(1.2)); draw((-3.947928660123173,7.083203380299419)--(-0.07271075964964728,8.086596251773575), linewidth(1.2)); draw((2.9299198470002734,5.981549135844178)--(3.3078437941783156,1.9964161994385508), linewidth(1.2)); draw((-0.07271075964964728,8.086596251773575)--(2.9299198470002734,5.981549135844178), linewidth(1.2)); draw((-5.449099046300786,11.855808032891977)--(-0.07271075964964728,8.086596251773575), linewidth(1.2)); draw((-1.7201244339576545,2.54273854326088)--(3.3078437941783156,1.9964161994385508), linewidth(1.2)); draw((-1.7201244339576545,2.54273854326088)--(-3.947928660123173,7.083203380299419), linewidth(1.2)); draw((-1.7201244339576545,2.54273854326088)--(2.9299198470002734,5.981549135844178), linewidth(1.2)); /* dots and labels */ dot((-5.449099046300786,11.855808032891977),linewidth(3.pt) + dotstyle); label("$A$", (-5.370549042331302,11.97162499805736), NE * labelscalefactor); dot((1.1131748507888175,12.077324983344363),linewidth(3.pt) + dotstyle); label("$B$", (1.0220012320075105,12.368548054307151), NE * labelscalefactor); dot((2.3520623455555123,6.386666323466413),linewidth(3.pt) + dotstyle); label("$C$", (2.4425679596383576,6.519155646415494), NE * labelscalefactor); dot((-1.7438108774116874,9.258150446796272),linewidth(3.pt) + dotstyle); label("$D$", (-1.6520067258858488,9.381179788848197), NE * labelscalefactor); dot((-0.07271075964964728,8.086596251773575),linewidth(3.pt) + dotstyle); label("$Q$", (0.019248247797500808,8.211301307269867), NE * labelscalefactor); dot((-1.7201244339576545,2.54273854326088),linewidth(3.pt) + dotstyle); label("$G$", (-1.631116038714807,2.6752692069438346), NE * labelscalefactor); dot((2.9299198470002734,5.981549135844178),linewidth(3.pt) + dotstyle); label("$P$", (3.006616513256488,6.101341902994662), NE * labelscalefactor); dot((-3.947928660123173,7.083203380299419),linewidth(3.pt) + dotstyle); label("$R$", (-3.866419566016287,7.208548323059868), NE * labelscalefactor); dot((3.3078437941783156,1.9964161994385508),linewidth(3.pt) + dotstyle); label("$S$", (3.382648882335242,2.132111340496752), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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Curious_Droid

34 posts

Curious_Droid

#53 h Wednesday at 5:15 AM

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Either I have a simple solution or I am trolling very hard.

Let $R', S'$ be intersections of $PR, QS$ with $(ABCD)$ . By Pascal on $APR'QBD$ , $QR'$ , $AD$ , and $RS$ concur. Let this point be $R''$ . By Reims, $PQRS$ is cyclic if and only if $R'S' \parallel RS \parallel BC$ . But since $R'', S$ are symmetric and $R' = QR'' \cap (ABCD)$ and $S' = QS \cap (ABCD)$ , this can only happen iff $Q$ is the arc midpoint. :thumbup:

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MathRook7817

615 posts

MathRook7817

#54 h Wednesday at 6:17 PM

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not_trig wrote:

yay this problem made me happy

i think u meant: "happy this problem made me"

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