Let $ABC$ be a triangle, and let $T$ be a point on the extension of $AB$ beyond $B$, and $U$ a point on the extension of $AC$ beyond $C$, such that $BT = CU$. Moreover, let $R$ and $S$ be points on the extensions of $AB$ and $AC$ beyond $A$ such that $AS = AT$ and $AR = AU$. Prove that $R$, $S$, $T$, $U$ lie on a circle whose centre lies on the circumcircle of $ABC$.
Problem
Source: South African Mathematics Olympiad 2020, Problem 5
Tags: geometry, cyclic quadrilateral, circumcircle
MiltonMath12
17.01.2021 10:45
$\triangle RAT\equiv\triangle SAU$ since are isosceles and $\angle RAT=\angle SAU$ then $RTUS$ are isosceles trapezoid. Take $D$ a circumcenter of $RTUS$ and we have by angle chasing: $$\angle TDU=2\angle TRU=2\frac{\angle BAC}{2}=\angle BAC$$Suppose that $G=TD\cap\odot(ABC)$ and $F=UD\cap\odot(ABC)$ Then if $D\in \odot(ABC)$ we must have: $$\angle ABD=\angle ACD$$Clearly $\triangle DRU\simeq DST$ since $ST=RU$ Since $BT=CU$ by LAL $\triangle BTF\simeq\triangle CDU$ then $\angle DBT=\angle DCU$ then $\angle ABD=\angle ACD$ then $ABCD$ are cyclic and $D\in \odot(ABC)$ $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -27.914259227537215, xmax = 34.1400087404224, ymin = -23.20904681608514, ymax = 11.601883995209299; /* image dimensions */ pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-4.88,-3.33)--(-0.84,4.17)--(5.9,-3.11)--cycle, linewidth(0.4) + sqsqsq); /* draw figures */ draw((-4.88,-3.33)--(-0.84,4.17), linewidth(0.4) + sqsqsq); draw((-0.84,4.17)--(5.9,-3.11), linewidth(0.4) + sqsqsq); draw((5.9,-3.11)--(-4.88,-3.33), linewidth(0.4) + sqsqsq); draw((xmin, 1.8564356435643565*xmin + 5.729405940594059)--(xmax, 1.8564356435643565*xmax + 5.729405940594059), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, 0.020408163265306138*xmin-3.2304081632653054)--(xmax, 0.020408163265306138*xmax-3.2304081632653054), linewidth(0.4) + wrwrwr); /* line */ draw(circle((-3.7043248415442607,-5.253826724943427), 16.17937858721433), linewidth(2) + linetype("0 3 4 3") + blue); draw((-19.802467668518727,-3.634540156500382)--(2.198301935011469,9.810412008065846), linewidth(0.4) + wrwrwr); draw((12.314348222998323,-2.979094934224524)--(-13.035942502383131,-18.470982368285515), linewidth(0.4) + wrwrwr); draw(circle((0.4723474385076766,-1.375024486876135), 5.698206065018499), linewidth(0.4) + linetype("4 4") + red); draw((-3.7043248415442607,-5.253826724943427)--(12.314348222998323,-2.979094934224524), linewidth(0.4) + wrwrwr); draw((-3.7043248415442607,-5.253826724943427)--(-13.035942502383131,-18.470982368285515), linewidth(0.8) + wrwrwr); draw((12.314348222998323,-2.979094934224524)--(2.198301935011469,9.810412008065846), linewidth(0.4) + wrwrwr); draw((xmin, -1.6363590836353064*xmin-11.315432328140295)--(xmax, -1.6363590836353064*xmax-11.315432328140295), linewidth(0.4) + wrwrwr); /* line */ draw((-3.7043248415442607,-5.253826724943427)--(2.198301935011469,9.810412008065846), linewidth(0.4) + wrwrwr); draw((5.561612683330907,-3.9380171960440595)--(0.04185737731306283,4.306896901758722), linewidth(0.4) + wrwrwr); draw((-4.88,-3.33)--(5.561612683330907,-3.9380171960440595), linewidth(0.4) + wrwrwr); draw((-4.88,-3.33)--(0.04185737731306283,4.306896901758722), linewidth(0.4) + wrwrwr); draw((-0.84,4.17)--(0.04185737731306283,4.306896901758722), linewidth(0.4) + wrwrwr); draw((5.561612683330907,-3.9380171960440595)--(5.9,-3.11), linewidth(0.4) + wrwrwr); draw((-3.7043248415442607,-5.253826724943427)--(5.9,-3.11), linewidth(0.4) + wrwrwr); draw((-19.802467668518727,-3.634540156500382)--(-3.7043248415442607,-5.253826724943427), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-4.88,-3.33),dotstyle); label("$A$", (-4.670935859747463,-2.77654394858623), NE * labelscalefactor); dot((-0.84,4.17),dotstyle); label("$B$", (-0.6168678304818085,4.736995465652787), NE * labelscalefactor); dot((5.9,-3.11),dotstyle); label("$C$", (6.139912218294282,-2.5603269870253955), NE * labelscalefactor); dot((2.198301935011469,9.810412008065846),dotstyle); label("$T$", (2.41016963136988,10.358636466234497), NE * labelscalefactor); dot((12.314348222998323,-2.979094934224524),dotstyle); label("$U$", (12.518312584338911,-2.4522185062449777), NE * labelscalefactor); dot((-19.802467668518727,-3.634540156500382),linewidth(4pt) + dotstyle); label("$R$", (-19.589906207445072,-3.2089778717079005), NE * labelscalefactor); dot((-13.035942502383131,-18.470982368285515),linewidth(4pt) + dotstyle); label("$S$", (-12.83312615866898,-18.0198397386251), NE * labelscalefactor); dot((-3.7043248415442607,-5.253826724943427),linewidth(4pt) + dotstyle); label("$D$", (-3.481742571162871,-4.830605083414163), NE * labelscalefactor); dot((0.4723474385076766,-1.375024486876135),linewidth(4pt) + dotstyle); label("$E$", (0.6804339388832009,-0.9386997753191326), NE * labelscalefactor); dot((5.561612683330907,-3.9380171960440595),linewidth(4pt) + dotstyle); label("$F$", (5.761532535562821,-3.479249073658944), NE * labelscalefactor); dot((0.04185737731306283,4.306896901758722),linewidth(4pt) + dotstyle); label("$G$", (0.24800001576153108,4.736995465652787), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
rafaello
20.01.2021 13:18
Easy, but nice. By Power of a Point, $$AU\cdot AS=AR\cdot AT\implies RSTU\text{ is cyclic.}$$ Now we claim that the midpoint of arc $BAC$ is the centre of $(RSTU)$. Let that midpoint be $F$. We have $\triangle FCU\cong \triangle FBT$, since $$\measuredangle UCF=\measuredangle ACF=\measuredangle ABF=\measuredangle TBF$$and $$FB=FC \text{ and } CU=BT.$$ Also, $\triangle SFC\cong \triangle RFB$, since $$\measuredangle SCF=\measuredangle ACF=\measuredangle ABF=\measuredangle RBF$$and $$FB=FC \text{ and } CS=AC+AS=AU+AB=BR.$$ Thus, $F$ lies on the perpendicular bisectors of $UT$ and $SR$, meaning it is indeed the centre of $(RSTU)$.