Yay I don't think I suck at geo anymore D:

by shiningsunnyday, Jun 8, 2016, 2:20 PM

First thing first, today was officially the last day of school but I didn't go because 1) I won't have a place to do math, since a lot of people come to the library (where the antisocial nerd, me, mythically lives) to ask people to sign their yearbooks (which I think has little purpose, story for another day...) and 2) I drank milk tea at 9 PM last night, cause... YOLO, so I ended up not falling asleep until 4 AM, and 3) I'm weird. Anyways, as the bell rang for its final time this school year, laughter reverberating throughout the campus, foreshadowing a night of partying/sleepovers and the like. But me? I'll take some nice geo problems over all that any day.

...so here we go!

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.176238571853332cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.823380811362441, xmax = 14.352857760490894, ymin = -4.334778680755408, ymax = 6.36546447043235; /* image dimensions */ /* draw figures */ draw(circle((2.56,0.56), 3.2975475249186914)); draw((10.108740840048487,0.5643715298338)--(2.56,0.56)); draw((3.998762976580706,3.527116576125383)--(10.108740840048487,0.5643715298338)); draw((10.108740840048487,0.5643715298338)--(1.0829846426639882,3.508261405182885)); draw((0.203313076424267,2.8664792267316295)--(10.108740840048487,0.5643715298338)); draw((2.56,0.56)--(-0.7375469719776571,0.5580903669536675)); /* dots and labels */ dot((2.56,0.56),dotstyle); label("$O$", (2.6191587328882453,0.7212054530896249), NE * labelscalefactor); dot((10.108740840048487,0.5643715298338),dotstyle); label("$P$", (10.171336291304584,0.7212054530896249), NE * labelscalefactor); dot((1.0829846426639882,3.508261405182885),dotstyle); label("$B$", (1.1405218635562042,3.662579870578087), NE * labelscalefactor); dot((0.203313076424267,2.8664792267316295),dotstyle); label("$C$", (0.15476395066817678,3.0902043082560082), NE * labelscalefactor); dot((3.998762976580706,3.527116576125383),linewidth(3.pt) + dotstyle); label("$A$", (4.065996959869059,3.6148819070512475), NE * labelscalefactor); dot((-0.7375469719776571,0.5580903669536675),linewidth(3.pt) + dotstyle); label("$D$", (-1.0217858163272107,0.5304135989822653), NE * labelscalefactor); dot((5.49128518079106,2.0704260551156106),linewidth(3.pt) + dotstyle); label("$E$", (5.560533150376714,2.168043680070436), NE * labelscalefactor); dot((5.692297685331387,1.5907913899354293),linewidth(3.pt) + dotstyle); label("$F$", (5.751325004484074,1.6910640448020369), NE * labelscalefactor); dot((5.857546971977658,0.5619096330463327),linewidth(3.pt) + dotstyle); label("$G$", (5.926217537415821,0.6576081683871717), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Power is a powerful tool in geometry. Consider a point $P$ and circle $O.$ We all know that the quantity $PA^2=PE \cdot PB=PF \cdot PC$ is constant, independent of whichever secant you choose (consequence of power of a point), but things get interesting when we assign a value as to what that constant is, which we denote by the power of $P$ with respect to $O,$ and is defined as $PG\cdot PD=(PO+r)(PO-r)=PO^2-r^2.$ It's not hard to see this value is positive when $P$ is outside the circle, negative inside, and zero on the circle. The three main lemmas that arise out of this are:

The radical axis of two circles $w_1, w_2$ is defined as the locus of points in which the power of a point $P$ with respect to two circles is equal, is a perpendicular line through the two centers, and in the following case, through the intersection of the two circles, specifically as shown:

The radical lemma, which states that if there're two points $A, B$ on circle $w_1,$ two points on circle $w_2, C, D$ these four points are cyclic if and only if $AB$ and $CD$ intersect on the radical axis of $w_1$ and $w_2.$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.097334085533035cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.10964681802779, xmax = 18.987687267505247, ymin = -4.8664695616756015, ymax = 10.034638270919947; /* image dimensions */ /* draw figures */ draw(circle((-0.58,1.42), 5.2109500093553)); draw(circle((7.268639194831133,1.4816595514931006), 4.363096137421303)); draw((xmin, -127.28991704893257*xmin + 492.973281467901)--(xmax, -127.28991704893257*xmax + 492.973281467901)); /* line */ draw((xmin, 1.0193950051511613*xmin + 5.058630771270418)--(xmax, 1.0193950051511613*xmax + 5.058630771270418)); /* line */ draw((xmin, -4.39622452783308*xmin + 25.652304530725015)--(xmax, -4.39622452783308*xmax + 25.652304530725015)); /* line */ /* dots and labels */ dot((-0.58,1.42),dotstyle); dot((3.84,4.18),dotstyle); dot((7.268639194831133,1.4816595514931006),dotstyle); dot((3.8828173415990226,-1.270215860395413),linewidth(3.pt) + dotstyle); dot((1.2256922172331493,6.308095295370543),dotstyle); label("$B$", (1.0753600214877324,6.735581707685568), NE * labelscalefactor); dot((-5.432511733166115,-0.4792445549441988),dotstyle); label("$A$", (-6.164851362254823,-0.4382057184079812), NE * labelscalefactor); dot((3.8026441174508783,8.93502719096729),linewidth(3.pt) + dotstyle); dot((4.6963973911030195,5.005887127506636),dotstyle); label("$C$", (4.9722322035632365,5.761363662166691), NE * labelscalefactor); dot((6.4739075072571275,-2.8084464436014804),linewidth(3.pt) + dotstyle); label("$D$", (7.186364125197045,-3.6708383239933458), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Last but not least, the radical axes of three circles with non-collinear centers (Evan is extremely fussy about emphasizing this) are concurrent.

For now, take these three lemmas for granted. A full treatment is found in, not surprisingly, E.G.M.O!!!

...But first, here's two nice gems aren't specifically under this topic but which I liked:

Canada 1997 P4 wrote:

The point $O$ is situated inside the parallelogram $ABCD$ such that $\angle AOB+\angle COD=180^{\circ}$ . Prove that $\angle OBC=\angle ODC$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7.237281212948929cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.8686738152158329, xmax = 14.105955028164761, ymin = -3.7514588449191577, ymax = 3.2687856439024827; /* image dimensions */ /* draw figures */ draw((3.52,2.24)--(2.0617552409438877,-2.9596007520017023)); draw((2.0617552409438877,-2.9596007520017023)--(11.99430889316315,-2.9596007520017023)); draw((3.52,2.24)--(13.452553652219263,2.24)); draw((13.452553652219263,2.24)--(11.99430889316315,-2.9596007520017023)); draw((3.52,2.24)--(3.848864903635227,-0.8642373547872702)); draw((3.848864903635227,-0.8642373547872702)--(2.0617552409438877,-2.9596007520017023)); draw((3.848864903635227,-0.8642373547872702)--(13.452553652219263,2.24)); draw((3.848864903635227,-0.8642373547872702)--(11.99430889316315,-2.9596007520017023)); /* dots and labels */ dot((3.52,2.24),dotstyle); label("$A$", (3.4034723453549893,2.4238528005227606), NE * labelscalefactor); dot((2.0617552409438877,-2.9596007520017023),dotstyle); label("$D$", (1.8492131643231609,-3.2090328220087185), NE * labelscalefactor); dot((11.99430889316315,-2.9596007520017023),dotstyle); label("$C$", (12.103151251264688,-3.1986015523373643), NE * labelscalefactor); dot((13.452553652219263,2.24),linewidth(3.pt) + dotstyle); label("$B$", (13.532235196240261,2.371696452165988), NE * labelscalefactor); dot((3.848864903635227,-0.8642373547872702),dotstyle); label("$O$", (3.9354670985940716,-0.6533717525268438), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Solution

HS Forum wrote:

Let $ABC$ be a triangle with $AB = 8$ , $BC = 9$ , and $CA = 10$ . The line tangent to the circumcircle of $ABC$ at $A$ intersects the line $BC$ at $T$ , and the circle centered at $T$ passing through $A$ intersects the line $AC$ for a second time at $S$ . If the angle bisector of $\angle SBA$ intersects $SA$ at $P$ , compute the length of segment $SP$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15.55256211478783cm); 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 = -16.493925424082587, xmax = 36.05863669070524, ymin = -15.71165718472974, ymax = 12.159008144862254; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw(arc((11.059997386776805,1.1264661862241279),1.2423773549595232,108.25626751684904,180.04631065256604)--(11.059997386776805,1.1264661862241279)--cycle, qqwuqq); draw(arc((3.06,1.12),1.2423773549595232,58.79786939035357,130.58791252607057)--(3.06,1.12)--cycle, qqwuqq); draw(arc((3.06,1.12),1.2423773549595232,-121.20213060964645,-49.41208747392945)--(3.06,1.12)--cycle, qqwuqq); draw(arc((-3.4157352090193904,-9.571812451715854),1.2423773549595232,-12.992173745363443,58.79786939035356)--(-3.4157352090193904,-9.571812451715854)--cycle, qqwuqq); draw(arc((3.06,1.12),1.2423773549595232,0.04631065256602237,58.79786939035356)--(3.06,1.12)--cycle, blue); draw(arc((16.07228044377464,-14.068171636262198),1.2423773549595232,108.25626751684905,167.0078262546366)--(16.07228044377464,-14.068171636262198)--cycle, blue); /* draw figures */ draw((3.06,1.12)--(11.059997386776805,1.1264661862241279)); draw((8.24058816721552,9.673449961372697)--(11.059997386776805,1.1264661862241279)); draw((3.06,1.12)--(8.24058816721552,9.673449961372697)); draw(circle((7.057233308567444,4.544580789309355), 5.263613559678153)); draw((xmin, -1.167218282905096*xmin + 4.691687945689594)--(xmax, -1.167218282905096*xmax + 4.691687945689594)); /* line */ draw((-3.4157352090193904,-9.571812451715854)--(16.07228044377464,-14.068171636262198)); draw((-3.4157352090193904,-9.571812451715854)--(11.059997386776805,1.1264661862241279)); draw((3.06,1.12)--(-3.4157352090193904,-9.571812451715854)); draw((11.059997386776805,1.1264661862241279)--(16.07228044377464,-14.068171636262198)); draw((1.0674660895324941,-2.1697884466818023)--(11.059997386776805,1.1264661862241279)); draw((11.059997386776805,1.1264661862241279)--(1.0674660895324941,-2.1697884466818023)); /* dots and labels */ dot((3.06,1.12),dotstyle); label("$A$", (0.8993575453507378,0.68772390073598), NE * labelscalefactor); dot((11.059997386776805,1.1264661862241279),dotstyle); label("$B$", (12.494879524972953,0.8119616362319325), NE * labelscalefactor); dot((8.24058816721552,9.673449961372697),linewidth(3.pt) + dotstyle); label("$C$", (8.229383939611925,10.46109242641757), NE * labelscalefactor); dot((16.07228044377464,-14.068171636262198),linewidth(3.pt) + dotstyle); label("$T$", (17.754276994301602,-14.75916787926077), NE * labelscalefactor); dot((-3.4157352090193904,-9.571812451715854),linewidth(3.pt) + dotstyle); label("$S$", (-4.85699086596172,-10.245196822907833), NE * labelscalefactor); dot((1.0674660895324941,-2.1697884466818023),linewidth(3.pt) + dotstyle); label("$P$", (-0.42584496660608695,-2.293981751166878), NE * labelscalefactor); dot((-2.107643527157205,7.151768004434068),dotstyle); label("$E$", (-2.1651732635494194,8.266225765989077), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Nice problem -- features cyclic quads, PoP, and angle chasing.
Solution

Now let's power through some problems!

P50 of 106 wrote:

Let $ABC$ be a non-right triangle with orthocenter $H$ and let $M, N$ be points on its sides $AB$ and $AC.$ Prove that the common chords of circles with diameters $CM$ and $BN$ passes through $H.$

Diagram attached in solution

Solution

BAMO 2012 wrote:

Given a segment $AB$ in the plane, choose on it a point $M$ different from $A$ and $B$ . Two equilateral triangles $AMC$ and $BMD$ in the plane are constructed on the same side of segment $AB$ . The circumcircles of the two triangles intersect in point $M$ and another point $N$ . Prove that no matter where one chooses the point $M$ along segment $AB$ , all lines $MN$ will pass through some fixed point $K$ in the plane.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12.567908618748916cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.58965732228952, xmax = 27.978251296459398, ymin = -7.686932861499198, ymax = 11.17611087405477; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((1.92,-1.12)--(7.7,-1.12)--(4.81,3.8856268338740567)--cycle, zzttqq); draw((7.7,-1.12)--(15.04,-1.12)--(11.37,5.236626463777779)--cycle, zzttqq); /* draw figures */ draw((1.92,-1.12)--(15.04,-1.12)); draw((1.92,-1.12)--(7.7,-1.12), zzttqq); draw((7.7,-1.12)--(4.81,3.8856268338740567), zzttqq); draw((4.81,3.8856268338740567)--(1.92,-1.12), zzttqq); draw((7.7,-1.12)--(15.04,-1.12), zzttqq); draw((15.04,-1.12)--(11.37,5.236626463777779), zzttqq); draw((11.37,5.236626463777779)--(7.7,-1.12), zzttqq); draw(circle((4.81,0.5485422779580201), 3.3370845559160367)); draw(circle((11.37,0.9988754879259247), 4.237750975851852)); draw((xmin, -14.56699140724605*xmin + 111.0458338357946)--(xmax, -14.56699140724605*xmax + 111.0458338357946)); /* line */ /* dots and labels */ dot((1.92,-1.12),dotstyle); label("$A$", (2.024048474582725,-0.8480284462314892), NE * labelscalefactor); dot((15.04,-1.12),dotstyle); label("$B$", (15.141291369440482,-0.8480284462314892), NE * labelscalefactor); dot((7.7,-1.12),dotstyle); label("$M$", (7.8259059088467335,-0.8480284462314892), NE * labelscalefactor); dot((4.81,3.8856268338740567),dotstyle); dot((11.37,5.236626463777779),dotstyle); dot((7.44487811196122,2.5963583508613004),linewidth(3.pt) + dotstyle); label("$N$", (7.545622941007892,2.7676218388895535), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Solution

2012 JMO 1 wrote:

Given a triangle $ABC$ , let $P$ and $Q$ be points on segments $AB$ and $AC$ , respectively, such that $AP = AQ$ . Let $S$ and $R$ be distinct points on segment $BC$ such that $S$ lies between $B$ and $R$ , $\angle{BPS} = \angle{PRS},$ and $\angle{CQR} = \angle{QSR}.$ Prove that $P, Q, R, S$ are concyclic.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.267137976618258cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.449822481858106, xmax = 25.81731549476015, ymin = -6.375813475339232, ymax = 10.206364505956353; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw(arc((5.32,3.12),0.739175838690739,-126.35849197893974,-104.60336200894038)--(5.32,3.12)--cycle, qqwuqq); draw(arc((12.748538253708311,-3.744503025785759),0.739175838690739,137.25984643600262,180.00030493788594)--(12.748538253708311,-3.744503025785759)--cycle, qqwuqq); draw(arc((11.815291475204473,2.4354585951520624),0.739175838690739,-81.4125497884481,-65.67394149160917)--(11.815291475204473,2.4354585951520624)--cycle, blue); draw(arc((3.5314879795615175,-3.744552080495362),0.739175838690739,3.04937885967508e-4,36.7242417666089)--(3.5314879795615175,-3.744552080495362)--cycle, blue); /* draw figures */ draw((5.32,3.12)--(11.815291475204473,2.4354585951520624)); draw((xmin, 5.322170124254279e-6*xmin-3.744570875675181)--(xmax, 5.322170124254279e-6*xmax-3.744570875675181)); /* line */ draw((9.152380978789576,8.326007021082681)--(14.609031057690258,-3.7444931239265418)); draw((9.152380978789576,8.326007021082681)--(0.26667500337709027,-3.7445694563854452)); draw((5.32,3.12)--(3.5314879795615175,-3.744552080495362)); draw((11.815291475204473,2.4354585951520624)--(12.748538253708311,-3.744503025785759)); draw((12.748538253708311,-3.744503025785759)--(5.32,3.12)); draw((3.5314879795615175,-3.744552080495362)--(11.815291475204473,2.4354585951520624)); /* dots and labels */ dot((5.32,3.12),dotstyle); label("$P$", (4.553690535089895,3.356668400755501), NE * labelscalefactor); dot((11.815291475204473,2.4354585951520624),dotstyle); label("$Q$", (12.388954425211729,2.6914101459338355), NE * labelscalefactor); dot((3.5314879795615175,-3.744552080495362),dotstyle); label("$S$", (3.4695659716768112,-4.700348240973558), NE * labelscalefactor); dot((12.748538253708311,-3.744503025785759),dotstyle); label("$R$", (12.783181539180122,-4.6757090463505335), NE * labelscalefactor); dot((9.152380978789576,8.326007021082681),dotstyle); label("$A$", (9.062663151103404,8.703373633951848), NE * labelscalefactor); dot((0.26667500337709027,-3.7445694563854452),linewidth(3.pt) + dotstyle); label("$B$", (-0.22631322177688354,-4.6757090463505335), NE * labelscalefactor); dot((14.609031057690258,-3.7444931239265418),linewidth(3.pt) + dotstyle); label("$C$", (14.926791471383266,-4.651069851727509), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Wait what I solved this in less than 5 mins. What has EGMO done to me?

Solution

2008 IMO 1 wrote:

Let $H$ be the orthocenter of an acute-angled triangle $ABC$ . The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing through $H$ intersects the sideline $BC$ at points $A_{1}$ and $A_{2}$ . Similarly, define the points $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ .

Prove that the six points $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ are concyclic.

Diagram credit to user "fclvbfm934."
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.23024805366, xmax = 6.98560554737, ymin = -1.53508634754, ymax = 4.58657752606; /* image dimensions */ /* draw figures */ draw((1.07322932756,3.89331217285)--(-0.14,-0.18)); draw((-0.14,-0.18)--(5.22,0.32)); draw((5.22,0.32)--(1.07322932756,3.89331217285)); draw((0.466614663782,1.85665608642)--(2.54,0.07)); draw((2.54,0.07)--(3.14661466378,2.10665608642)); draw((3.14661466378,2.10665608642)--(0.466614663782,1.85665608642)); draw(circle((2.54,0.07), 1.88550997846)); draw(circle((0.466614663782,1.85665608642), 0.908953928174)); draw(circle((3.14661466378,2.10665608642), 1.94969602822)); draw((0.433547786318,1.74563691517)--(5.22,0.32)); draw((1.07322932756,3.89331217285)--(1.43945830558,-0.0326624714943)); draw((-0.14,-0.18)--(2.39129283322,2.75752417264)); /* dots and labels */ dot((1.07322932756,3.89331217285),dotstyle); label("$A$", (1.12555124101,3.97179504302), NE * labelscalefactor); dot((-0.14,-0.18),dotstyle); label("$B$", (-0.369462335575,-0.414004251654), NE * labelscalefactor * 1.02); dot((5.22,0.32),dotstyle); label("$C$", (5.27206288189,0.40082445009), NE * labelscalefactor); dot((2.54,0.07),dotstyle); label("$M$", (2.51208194743,-0.148555641131), NE * labelscalefactor); dot((3.14661466378,2.10665608642),dotstyle); label("$N$", (3.20534730063,2.17976950738), NE * labelscalefactor); dot((0.466614663782,1.85665608642),dotstyle); label("$P$", (0.314561582545,1.8919989834), NE * labelscalefactor); dot((1.2975787164,1.48828672451),dotstyle); label("$H$", (1.3740803299,1.31645793546), NE * labelscalefactor); dot((1.21510891674,2.37236297685),dotstyle); label("$X$", (1.32175841645,2.37597668281), NE * labelscalefactor); dot((0.95371411682,1.08923744814),dotstyle); label("$Y$", (1.00782693575,0.910963106223), NE * labelscalefactor); dot((4.35596906992,0.577350190659),dotstyle); label("$Z$", (4.42183178834,0.374663493365), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Solution

...and with that said here're three more good-looking problems, among others, I'll try to solve before moving to Chapter 3 EGMO!

USAMO 2010 P1 wrote:

Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$ . Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ,$ respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle{XOZ}$ , where $O$ is the midpoint of segment $AB.$

Canada 2007 P5 wrote:

Let the incircle of triangle $ABC$ touch sides $BC, CA,$ and $AB$ at $D, E,$ and $F,$ respectively. Let $w, w_1, w_2$ , and $w_3$ denote the circumcircles of triangles $ABC, AEF , BDF,$ and $CDE$ respectively. Let $w$ and $w_1$ intersect at $A$ and $P$ , $w$ and $w_2$ intersect at $B$ and $Q, w$ and $w_3$ intersect at $C$ and $R.$
(a) Prove that $w_1, w_2$ , and $w_3$ intersect in a common point.
(b) Show that lines $PD, QE,$ and $RF$ are concurrent.

IMO 2009 P2 wrote:

Let $ABC$ be a triangle with circumcenter $O$ . The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ , respectively. Let $K, L,$ and $M$ be the midpoints of the segments $BP, CQ,$ and $PQ$ , respectively... (darn character limit)

This post has been edited 2 times. Last edited by shiningsunnyday, Jun 8, 2016, 3:53 PM

power of a point

Angle Chasing

Cyclic Quadrilaterals

geometry

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do 2006 G2; it's a nice geometry problem that you'll enjoy

by AMN300, Jun 8, 2016, 2:54 PM

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also from what my parents tell me (parents grew up in China, I grew up in US) don't most people in Shanghai spend most of their summer+free time preparing for Gao kao or learning English or some other form of studying?

@first comment, thanks for the recommendation, I'll definitely look into it soon

I go to an American school (same system as everyone else in US), so I'm not associated with the Gaokao system, though what you said is true of the local kids around here.

This post has been edited 1 time. Last edited by shiningsunnyday, Jun 8, 2016, 3:04 PM

by AMN300, Jun 8, 2016, 2:56 PM

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The thing about geometry is that one gains an advantage by just knowing more stuff. And I'm not good at that.

Me 3 months ago:

I need to mem the angular momentum formulas. $L=I\omega$ . WTF IS WITH ALL THE GREEK LETTERS I need to mem all the angular momentum formulas.

Lol I prefer to call it developing one's geometric intuition as opposed to memorizing.

This post has been edited 1 time. Last edited by shiningsunnyday, Jun 9, 2016, 4:29 AM

by MathAwesome123, Jun 8, 2016, 3:15 PM

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hint for #1
hint for #2
hint for #3

Someone's been doing his homework. But thanks!

This post has been edited 1 time. Last edited by shiningsunnyday, Jun 9, 2016, 4:30 AM

by wu2481632, Jun 8, 2016, 3:27 PM

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The blog is locked right?
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:O a new background picture
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did you get into MIT?
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wait what new site?
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Yea, doing a bit of cleaning before migrating to new site
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Were there posts made in December 2017 for this blog and then deleted?

I ask because I was purging my thunderbird inbox and I found emails indicating new blog posts of yours.

email do not lie
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@below sorry not accepting contribs
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contrib plez?
also wow this blog is very popular
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first shout of december
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Darn, whenever what?

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