CAYLEY BHHARR

by OronSH, May 7, 2024, 2:34 AM

Here's a proof of Simson Lines with Cayley-Bacharach.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.187120640720877, xmax = 17.980904953422378, ymin = -6.041596675626619, ymax = 5.164810647142181; /* image dimensions */ /* draw figures */ draw(circle((-1.612602379915067,-1.0418061079137126), 4.358088190074937), linewidth(1)); draw((-3.5764192790227765,3.9234333640026025)--(-5.34,-3.3), linewidth(1)); draw((-0.5286783829178066,3.179336088364934)--(-3.5764192790227765,3.9234333640026025), linewidth(1)); draw((-0.5286783829178066,3.179336088364934)--(-2.4229793237472355,1.24967704119482), linewidth(1)); draw((-5.34,-3.3)--(2.150494609122812,-3.239994844934419), linewidth(1)); draw((-0.5286783829178066,3.179336088364934)--(-0.47708552767717305,-3.261043969429335), linewidth(1)); draw((-3.88,2.68)--(2.150494609122812,-3.239994844934419), linewidth(1)); draw((-3.5764192790227765,3.9234333640026025)--(-0.47708552767717305,-3.261043969429335), linewidth(0.6) + linetype("2 2")); /* dots and labels */ dot((-3.88,2.68),dotstyle); label("$A$", (-4.431909692115152,2.7168090475082196), NE * labelscalefactor); dot((-5.34,-3.3),dotstyle); label("$B$", (-5.955110687442946,-3.484795004897815), NE * labelscalefactor); dot((2.150494609122812,-3.239994844934419),dotstyle); label("$C$", (2.4496948046336287,-3.5391950404452364), NE * labelscalefactor); dot((-0.5286783829178066,3.179336088364934),dotstyle); label("$P$", (-0.3791070438322734,3.342409456303565), NE * labelscalefactor); dot((-0.47708552767717305,-3.261043969429335),linewidth(4pt) + dotstyle); label("$X$", (-0.5151071327008263,-3.8655952537297646), NE * labelscalefactor); dot((-2.4229793237472355,1.24967704119482),linewidth(4pt) + dotstyle); label("$Y$", (-2.96310873233478,0.8128078033484722), NE * labelscalefactor); dot((-3.5764192790227765,3.9234333640026025),linewidth(4pt) + dotstyle); label("$Z$", (-3.6159091589038344,4.185610007288596), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Since $\measuredangle BXP=\measuredangle BZP=90^\circ,$ we have $BXPZ$ is cyclic. Similarly, $CXPY$ is cyclic.

Now the cubic $(BXPZ)\cup\overline{AYC}$ intersects the cubic $(CXPY)\cup\overline{AZB}$ at nine points: $A,B,C,X,Y,Z,P$ and the circle points $(1:\pm i:0).$

Now consider the cubic $(ABCP)\cup\overline{YZ}.$ It passes through the eight points $A,B,C,Y,Z,P$ and the circle points, so it passes through $X$ as well. Since $X$ lies on $BC$ it cannot lie on the circle, so it lies on line $YZ,$ finishing.

This post has been edited 1 time. Last edited by OronSH, May 7, 2024, 2:34 AM

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cmu do gonna have a lot of weird geo :'(

by peace09, May 7, 2024, 2:05 PM

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Sus weird geo

by bjump, May 31, 2024, 6:11 PM

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orz orz orz

by RaymondZhu, Jun 13, 2024, 12:57 AM

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CAYLEY BHHARR

by OronSH, May 7, 2024, 2:34 AM

3 Comments