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Source: 2009 USAMO problem 5

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1080 posts

#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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466 posts

#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:

: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 .

https://i.imgur.com/H1bHDfP.png

Beautiful problem :omighty:

:omighty:

.

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:

:thumbup:

.

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:

:stretcher:

.

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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309 posts

#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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8857 posts

#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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1727 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 $\angle CBD.$

This post has been edited 1 time. Last edited by OronSH, Nov 3, 2023, 9:32 PM

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792 posts

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

Y by

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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7586 posts

#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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7586 posts

#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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170 posts

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

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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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1174 posts

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

Y by

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

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56 posts

kotmhn

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

Y by

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

Y by

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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#53 h Mar 19, 2025, 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:

:thumbup:

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#54 h Mar 19, 2025, 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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