circumcenter lies on circumcircle, intersecting circles

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circumcenter lies on circumcircle, intersecting circles

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Source: JBMO Shortlist 2005

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#1 h Jan 1, 2018, 11:22 AM • 2 Y

Y by Adventure10, Mango247

Let $C_1,C_2$ be two circles intersecting at points $A,P$ with centers $O,K$ respectively. Let $B,C$ be the symmetric of $A$ wrt $O,K$ in circles $C_1,C_2$ respectively. A random line passing through $A$ intersects circles $C_1,C_2$ at $D,E$ respectively. Prove that the center of circumcircle of triangle $DEP$ lies on the circumcircle of triangle $OKP$ .

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#2 h Jan 1, 2018, 12:49 PM • 1 Y

Y by Adventure10

Invert with center $P$ , radius $PA$ . Let $X'$ be the inverse of $X$ . $A$ is fixed, $O'$ is the reflection of $P$ in line $AD'$ , $K'$ is the reflection of $P$ in line $AE'$ . Since $D, A, E$ are collinear and the line $DAE$ does not pass through $P$ , thus $A, D', E', P$ are concyclic.
Hence, the statement can be rephrased as follows:
Given cyclic quadrilateral $AD'PE'$ , let $O'$ and $K'$ be the reflections of $P$ in $AD'$ and $AE'$ . Prove that the reflection of $P$ in $DE$ lies on $O'K'$ .
Now consider the feet of the perpendiculars $X, Y, Z$ from $P$ to $AD$ , $AE$ , $DE$ respectively. $X, Y, Z$ are collinear (this is the Simson line of $P$ with respect to $AD'E'$ ). Now consider the homothety with center $P$ , ratio $2$ . This maps $O'K'$ to $XY$ . $Z$ is mapped to the reflection of $P$ in $DE$ , and the conclusion follows.

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#3 h Jan 1, 2018, 1:59 PM • 3 Y

Y by Mathlover_1, Adventure10, Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.226824610143254, xmax = 14.05730020596206, ymin = -9.642381612521445, ymax = 19.108619982523095; /* image dimensions */pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */draw(circle((-8.38,-2.11), 5.9819060507500454), linewidth(2) + wrwrwr); draw(circle((3.14,1.49), 8.59902320034084), linewidth(2) + wrwrwr); draw((xmin, -0.05494193155701337*xmin + 2.737708924116689)--(xmax, -0.05494193155701337*xmax + 2.737708924116689), linewidth(2) + wrwrwr); /* line */draw((-10.858478404933466,3.334294701453851)--(-2.9377935943060507,-4.593060498220641), linewidth(2) + wrwrwr); draw((-2.9377935943060507,-4.593060498220641)--(11.717790604543337,2.0939108747224546), linewidth(2) + wrwrwr); draw(circle((-3.2911980878398563,1.8378338810875408), 6.440597585152131), linewidth(2) + wrwrwr); draw((-8.38,-2.11)--(0.6609181144409348,6.9233107296767775), linewidth(2) + wrwrwr); draw((0.6609181144409348,6.9233107296767775)--(3.14,1.49), linewidth(2) + wrwrwr); draw((0.6609181144409348,6.9233107296767775)--(-2.9377935943060507,-4.593060498220641), linewidth(2) + wrwrwr); draw((3.14,1.49)--(-2.9377935943060507,-4.593060498220641), linewidth(2) + wrwrwr); draw((-2.9377935943060507,-4.593060498220641)--(-8.38,-2.11), linewidth(2) + wrwrwr); /* dots and labels */dot((-8.38,-2.11),dotstyle); label("$O$", (-8.249853389524922,-1.804049661854995), NE * labelscalefactor); dot((-5.32,3.03),dotstyle); label("$A$", (-5.208580592666341,3.3378960977821968), NE * labelscalefactor); dot((3.14,1.49),dotstyle); label("$K$", (3.2568179140534212,1.8015830354515723), NE * labelscalefactor); dot((-2.9377935943060507,-4.593060498220641),linewidth(4pt) + dotstyle); label("$P$", (-2.825727679663742,-4.343669213870925), NE * labelscalefactor); dot((-11.44,-7.25),dotstyle); label("$B$", (-11.32247951418617,-6.945995421492187), NE * labelscalefactor); dot((11.6,-0.05),dotstyle); label("$C$", (11.722216420773185,0.265269973120948), NE * labelscalefactor); dot((-10.858478404933466,3.334294701453851),dotstyle); label("$D$", (-10.72676628593552,3.6514293758088545), NE * labelscalefactor); dot((11.717790604543337,2.0939108747224546),linewidth(4pt) + dotstyle); label("$E$", (11.847629731983847,2.334589608096891), NE * labelscalefactor); dot((0.6609181144409348,6.9233107296767775),linewidth(4pt) + dotstyle); label("$F$", (0.7799050176428238,7.163002089707424), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let F be the center of $(DPE)$
This means $FK\perp EP$ and $FO\perp DP$

We have $\angle EDP=\frac{1}{2}\angle AOP=\angle KOP$ and $\angle DEP=\frac{1}{2}\angle AKP=\angle OKP$ so $\angle OPK=\angle DPE$

Now, $\angle OFK=\angle OFP+\angle KFP=(90-\angle DPF)+(90-\angle EPF)=180-\angle DPE=180-\angle OPK$ which implies that $O, F K, P$ are cyclic.

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