$\omega$ is circumcirlce of triangle $ABC$. We draw a line parallel to $BC$ that intersects $AB,AC$ at $E,F$ and intersects $\omega$ at $U,V$. Assume that $M$ is midpoint of $BC$. Let $\omega'$ be circumcircle of $UMV$. We know that $R(ABC)=R(UMV)$. $ME$ and $\omega'$ intersect at $T$, and $FT$ intersects $\omega'$ at $S$. Prove that $EF$ is tangent to circumcircle of $MCS$.
Problem
Source: Iranian National Olympiad (3rd Round) 2002
Tags: geometry, circumcircle, power of a point, radical axis, perpendicular bisector, geometry proposed
yetti
02.10.2006 18:11
Let (O), (P) be the congruent circumcircles of $\triangle ABC,\ \triangle UMV.$ Let their center line OP meet (O) inside of (P) at N. As $UV \parallel BC$ is radical axis of 2 congruent circles (O), (P) intersecting their center line at M, N inside of (O), (P), UV is the perpendicular bisector of MN. A tangent to (O) at N is parallel to UV. A circle (O') with the chord MC and tangent to $EF \equiv UV$ must be centrally similar to (O) with similarity center C and coefficient 1/2 and such circle is unique. Let S be the intersection of (O'), (P) other than M. Let CS meet (O), (P) again at $A_{0}, T_{0}.$ Let CS meet UV at $F_{0}$ and let $A_{0}B$ meet $MT_{0}$ at $E_{0}.$ Because of the central similarity of (O'), (O), $SM \parallel A_{0}B$ is a midline of $\triangle A_{0}BC$ and $\angle A_{0}BC = \angle SMC.$ BC is a tangent of (P) at M, hence $\angle ST_{0}M = \angle SMC = \angle A_{0}BC.$ As a result, the triangles $\triangle E_{0}BM \sim \triangle E_{0}T_{0}A_{0}$ are oppositely similar ($A_{0}T_{0}BM$ with the diagonal intersection $E_{0}$ is cyclic) and $\frac{E_{0}A_{0}}{E_{0}M}= \frac{E_{0}T_{0}}{E_{0}B},$ or $E_{0}A_{0}\cdot E_{0}B = E_{0}T_{0}\cdot E_{0}M.$ As $A_{0}B$ is a chord of (O) and $MT_{0}$ a chord of (P), $E_{0}$ lies on their radical axis UV. Let $A \in (O)$ be arbitrary, different from $A_{0}, B, C,$ let AC meet the radical axis at F, let SM meet (P) at T, and let AB meet MT at E. Since AC is a chord of (O) and ST a chord of (P) and their intersection F is on the radical axis of these 2 circles, $FA \cdot FC = FS \cdot FT,$ ATCS is cyclic and $\angle FCS = \angle FTA \equiv \angle STA.$ From the circle (O), $\angle ABA_{0}= \angle ACA_{0}\equiv \angle FCS.$ Since the chord MS of (P) is fixed, the angle $\angle STM = \angle ST_{0}M = \angle A_{0}BC$ is also fixed. Therefore, $\angle EBM \equiv \angle ABC = \angle ABA_{0}+\angle A_{0}BC =$ $= \angle ATS+\angle STM \equiv \angle ATF+\angle FTE = \angle ATE$ This means that ATBM with the diagonal intersection E is also cyclic and $EA \cdot EB = ET \cdot EM.$ Since AB is a chord of (O), MT a chord of (P), their intersecion E lies on the radical axis UV of these 2 circles. QED. The really interesting fact is that $S \in (P)$ is fixed and this does not depend on the circles (O), (P) being congruent. Of course, if they are not congruent, their radical axis UV is not the perpendicular bisector of MN. The circumcircle (O') of $\triangle MCS,$ still centrally similar to the circumcircle (O) of $\triangle ABC$ with similarity center C and coefficient 1/2, then does not touch this radical axis.
Attachments:
TheDarkPrince
27.02.2019 19:20
Sorry for the revive Omid Hatami wrote: $\omega$ is circumcirlce of triangle $ABC$. We draw a line parallel to $BC$ that intersects $AB,AC$ at $E,F$ and intersects $\omega$ at $U,V$. Assume that $M$ is midpoint of $BC$. Let $\omega'$ be circumcircle of $UMV$. We know that $R(ABC)=R(UMV)$. $ME$ and $\omega'$ intersect at $T$, and $FT$ intersects $\omega'$ at $S$. Prove that $EF$ is tangent to circumcircle of $MCS$.
Let $O$ be the circumcenter of $\triangle ABC$.
Claim. $M,C,O,S$ are concyclic even if $\omega$ and $\omega'$ are not congruent.
(Proof): [asy][asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
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pen dotstyle = black; /* point style */
real xmin = -5.918843077909076, xmax = 12.182145147119765, ymin = -3.1606227295920664, ymax = 6.498183282288042; /* image dimensions */
pair A = (0.,5.525228786209073), B = (-2.,0.), C = (6.02,0.), F = (2.7242635502210915,3.024866761474571), U = (-2.119393306614767,3.024866761474571), V = (6.139393306614767,3.024866761474571), M = (2.01,0.), O = (2.01,1.6730667502945462), T = (-2.3180930155914403,4.491110145115853), X = (4.015,0.8365333751472731);
draw(A--B--C--cycle, linewidth(2.));
draw(circle(O, 4.345026162285004), green);
draw(U--V, linewidth(1.));
draw(circle((2.01,4.3310515919411285), 4.331051591941129), red);
draw(M--T, linewidth(1.));
draw(T--(5.749051936119805,2.145302633104703), linewidth(1.));
draw(M--(5.749051936119805,2.145302633104703), linewidth(1.));
draw(O--M, linewidth(1.));
draw(O--C, linewidth(1.));
draw(O--(5.749051936119805,2.145302633104703), linewidth(1.));
draw((5.749051936119805,2.145302633104703)--C, linewidth(1.));
draw(U--T, linewidth(1.));
draw(T--A, linewidth(1.));
draw(O--M, linewidth(1.));
draw(O--C, linewidth(1.));
draw(circle((0.005,2.3988285522327626), 3.1264042321822574), purple);
draw(circle((0.8786611024178059,0.44041912121461413), 5.160168084677393), blue);
draw(circle((4.015,0.836533375147273), 2.172513081142502), dashed);
/* dots and labels */
dot(A,dotstyle);
label("$A$", (-0.10278784494898931,5.6821551246023185), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (-2.4,-0.3), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (6.039678651085388,-0.3), NE * labelscalefactor);
dot((-0.9050709469173063,3.024866761474571),dotstyle);
label("$E$", (-1.2,3.1747231491680044), NE * labelscalefactor);
dot(F,linewidth(4.pt) + dotstyle);
label("$F$", (2.6,3.1450493979794327), NE * labelscalefactor);
dot(U,linewidth(4.pt) + dotstyle);
label("$U$", (-2.55,3), NE * labelscalefactor);
dot(V,linewidth(4.pt) + dotstyle);
label("$V$", (6.262231784999678,2.981843766442288), NE * labelscalefactor);
dot(M,linewidth(4.pt) + dotstyle);
label("$M$", (1.7814953555253246,-0.5), NE * labelscalefactor);
dot(O,linewidth(4.pt) + dotstyle);
label("$O$", (1.7814953555253246,1.75038309211656), NE * labelscalefactor);
dot(T,linewidth(4.pt) + dotstyle);
label("$T$", (-2.65,4.554552579436591), NE * labelscalefactor);
dot((5.749051936119805,2.145302633104703),linewidth(4.pt) + dotstyle);
label("$S$", (5.876473019548243,2.076794355190849), NE * labelscalefactor);
dot(X,linewidth(4.pt) + dotstyle);
label("$X$", (4.081211072639644,0.9), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy][/asy]
As $EF$ is the radical axis of $\omega$ and $\omega'$, \[A,T,B,M \text{ are concyclic and } A,T,C,S\text{ are concyclic}.\]Also $BC$ is clearly (as $MU = MV$) tangent to $\omega'$ so \[\angle MOC =\angle BAC = \angle TAC- \angle TAB = \angle TSC - \angle TMB = \angle TSC - \angle TSM = \angle MSC\]giving $M,C,O,S$ are concyclic. $~\square$
Main Problem.
[asy][asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(7cm);
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 = -5.918843077909076, xmax = 12.182145147119765, ymin = -3.1606227295920664, ymax = 6.498183282288042; /* image dimensions */
pair A = (0.,5.525228786209073), B = (-2.,0.), C = (6.02,0.), F = (2.7242635502210915,3.024866761474571), U = (-2.119393306614767,3.024866761474571), V = (6.139393306614767,3.024866761474571), M = (2.01,0.), O = (2.01,1.6730667502945462), X = (4.015,0.8365333751472731);
draw(A--B--C--cycle, linewidth(2.));
/* draw figures */
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draw(U--V, linewidth(1.));
draw(X--(4.015,3.024866761474571), linewidth(1.));
draw(O--U,red);
draw(O--V,red);
draw(O--C,red);
draw(U--(2.01,4.3310515919411285), red);
draw(V--(2.01,4.3310515919411285), red);
draw(M--(2.01,4.3310515919411285), red);
/* dots and labels */
dot(A,dotstyle);
label("$A$", (-0.10278784494898931,5.6821551246023185), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (-2.5,-0.2674319887063197), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (6.1,-0.2822688643006056), NE * labelscalefactor);
dot((-0.9050709469173063,3.024866761474571),dotstyle);
label("$E$", (-1.3,3.1747231491680044), NE * labelscalefactor);
dot(F,linewidth(4.pt) + dotstyle);
label("$F$", (2.7,3.1450493979794327), NE * labelscalefactor);
dot(U,linewidth(4.pt) + dotstyle);
label("$U$", (-2.461851064440453,3.0411912688194316), NE * labelscalefactor);
dot(V,linewidth(4.pt) + dotstyle);
label("$V$", (6.262231784999678,2.981843766442288), NE * labelscalefactor);
dot(M,linewidth(4.pt) + dotstyle);
label("$M$", (1.7814953555253246,-0.5), NE * labelscalefactor);
dot(O,linewidth(4.pt) + dotstyle);
label("$O$", (1.5,1.75038309211656), NE * labelscalefactor);
dot(X,linewidth(4.pt) + dotstyle);
label("$X$", (4.081211072639644,0.9491918100251224), NE * labelscalefactor);
dot((2.01,4.3310515919411285),linewidth(4.pt) + dotstyle);
label("$O'$", (1.7963322311196104,4.480368201465162), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy][/asy]
We need to show that $EF$ is tangent to $\odot(MOC)$ if $\omega$ and $\omega'$ are congruent. Let $O'$ and $X$ be the circumcenters of $\triangle MUV$ and $\triangle MOC$ respectively. Clearly $M,O,O'$ are collinear (lies on the perpendicular bisector of $BC$) and $X$ is the midpoint of $OC$. Let $|OA| = R$ and $|OO'| = 2d$. We have $O'M = R$ (as $\omega$ and $\omega'$ are congruent). Therefore, \[OM = O'M - OO' = R - 2d.\]Now $O'VOU$ is a rhombus, so distance from $X$ to $EF$ is \[\frac{OM}{2} + \text{ distance from O to EF } = \frac{R-2d}{2} + d = \frac{R}{2} = \frac{OC}{2} = XO = \text{ radius of } \odot(MOC).\]Therefore $EF$ is tangent to $\odot(MOC)$ or $\odot(MCS)$ and we are done. $~\blacksquare$