IMO 2011 Problem 6
by liberator, Jul 21, 2015, 4:57 PM
Problem: Let 
be an acute triangle with circumcircle 
. Let 
be a tangent line to , and let 
and 
be the lines obtained by reflecting in the lines 
, 
and 
, respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .
Proposed by Japan
be an acute triangle with circumcircle 
. Let 
be a tangent line to , and let 
and 
be the lines obtained by reflecting in the lines 
, 
and 
, respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .Proposed by Japan
![[asy]
import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(9); pointfontpen=fp; real xmin=2.5,xmax=14,ymin=-2,ymax=8.8;
pen wwwwww=rgb(0.4,0.4,0.4), wwzzcc=rgb(0.4,0.6,0.8), yqqqyq=rgb(0.5019607843137255,0.,0.5019607843137255), qqcccc=rgb(0,0.8,0.8), cqcqcq=rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
pair A=(9.,3.2), B=(7.6,1.), C=(11.06,1.), P=(10.9037267872876,2.2894742445215552), A_1=(4.225302839837132,4.713218228557546), B_1=(13.318947812589142,4.209009023926156), C_1=(11.23751359726076,0.33222222307412996), M=(9.699442104579619,-0.3030337832920753), P_a=(10.9037267872876,-0.28947424452155524), P_b=(9.783568496159925,1.2405987537383683), P_c=(7.368886499833012,4.538918063810838), H=(9.,2.310909090909089), I_1=(10.610893030990937,2.689468626599);
D(A--B--C--cycle,wwwwww+linewidth(1)); D(A_1--B_1--C_1--cycle,wwzzcc+linewidth(1)); D(A--A_1,linetype("4 4")+wwzzcc);
D(CR((9.33,1.4445454545454548),1.786202861143444),green); D((xmin,-1.8625555265220795*xmin+22.598270831870913)--(xmax,-1.8625555265220795*xmax+22.598270831870913),wwzzcc); D(CR((8.755537154895718,4.161938324953828),4.563653413817887),yqqqyq); D(CR((9.33,0.5554545454545452),1.786202861143444),red); D(CR((10.73,2.7554545454545445),1.786202861143444),red); D(CR((7.27,2.7554545454545463),1.786202861143444),red); D(P_a--P_c,linetype("4 4")+wwzzcc); D(CR((5.579166076673129,0.695625707960863),4.239574899162703),linetype("0 2")+qqcccc+linewidth(1)); D(CR((10.26502172862502,2.9510482106080516),3.3028669264989134),linetype("0 2")+qqcccc+linewidth(1)); D(CR((10.293309185404135,0.43870973889632964),0.9501902769422019),linetype("0 2")+qqcccc+linewidth(1)); D(M--A_1,wwzzcc); D(M--B_1,wwzzcc);
D(A); MP("A",(9.078267017098327,3.3218831364461052),NE*lsf); D(B); MP("B",(7.208843909967847,0.5368242217415146),NE*lsf); D(C); MP("C",(11.233826314095717,0.7657331736350426),NE*lsf); D(P); MP("P",(10.985841616211061,2.406247328871993),NE*lsf); D(C_1); MP("C_1",(11.367356536033608,0.11715780993671321),NE*lsf); D(A_1); MP("A_1",(3.698906647600416,4.981473037674183),NE*lsf); D(B_1); MP("B_1",(13.484764341048743,4.046761484108944),NE*lsf); D(M); MP("M",(9.688690888814401,-0.7030992676817621),NE*lsf); D(P_a); MP("P_a",(10.966765870219934,-0.5886447917349982),NE*lsf); D(P_b); MP("P_b",(9.898524094716802,1.1663238394487165),NE*lsf); D(P_c); MP("P_c",(7.11346518001221,4.752564085780656),NE*lsf); D(H); MP("H",(9.078267017098327,2.4253230748631207),NE*lsf); D(I_1); MP("I_1",(10.680629680353023,2.8068379946856674),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* Labels, by liberator */
defaultpen(fontsize(9pt));
label("$\Gamma_a$",(10,-1.2),S*lsf);
label("$\Gamma_b$",(12,4),dir(30)*lsf);
label("$\Gamma_c$",(6,1),N*lsf);
label("$\Gamma_1$",(5,7),N*lsf);
[/asy]](http://latex.artofproblemsolving.com/f/b/8/fb89592e2b7f740039a8d16e7184e085a07a93d6.png)
Let 
,
analogously; let 
be the circumcircle of 
.
,
.
is tangent to 
,
,
is the Steiner line of 
w.r.t 
,
.
lie on 
concur at 
be the incenter of 
are the centers of 
then 
.
;
.
:
.
,
is the internal bisector of 
.
are the internal bisectors of 
respectively. Hence 
concur at 
as the Miquel point of the line 
.
is a 
![[asy]
import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(9); pointfontpen=fp; real xmin=2.5,xmax=14,ymin=-2,ymax=8.8;
pen wwwwww=rgb(0.4,0.4,0.4), wwzzcc=rgb(0.4,0.6,0.8), yqqqyq=rgb(0.5019607843137255,0.,0.5019607843137255), cqcqcq=rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
pair A=(9.,3.2), B=(7.6,1.), C=(11.06,1.), P=(10.9037267872876,2.2894742445215552), A_1=(4.225302839837132,4.713218228557546), B_1=(13.318947812589142,4.209009023926156), C_1=(11.23751359726076,0.33222222307412996), M=(9.699442104579619,-0.3030337832920753), P_a=(10.9037267872876,-0.28947424452155524), P_b=(9.783568496159925,1.2405987537383683), P_c=(7.368886499833012,4.538918063810838), H=(9.,2.310909090909089), I_1=(10.610893030990937,2.689468626599), A_2=(7.556877292936604,1.6603150697678517), B_2=(11.116107846985981,1.4629688117894197), K=(12.028150798496817,7.3426508899567695);
D(A--B--C--cycle,wwwwww+linewidth(1)); D(A_1--B_1--C_1--cycle,wwzzcc+linewidth(1)); D(I_1--A_2--B_2--cycle,cqcqcq); D(A_1--K--B_1--cycle,wwzzcc); D(A--A_1,linetype("4 4")+wwzzcc);
D(CR((9.33,1.4445454545454548),1.786202861143444),green); D((xmin,-1.8625555265220795*xmin+22.598270831870913)--(xmax,-1.8625555265220795*xmax+22.598270831870913),wwzzcc); D(CR((8.755537154895718,4.161938324953828),4.563653413817887),yqqqyq); D(P_a--P_c,linetype("4 4")+wwzzcc); D(CR((5.579166076673129,0.695625707960863),4.239574899162703),linetype("0 2")+red+linewidth(1)); D(CR((10.26502172862502,2.9510482106080516),3.3028669264989134),linetype("0 2")+red+linewidth(1)); D(CR((10.293309185404135,0.43870973889632964),0.9501902769422019),linetype("0 2")+red+linewidth(1)); D(M--A_1,wwzzcc); D(M--B_1,wwzzcc); D((xmin,0.33698371098991503*xmin+3.289359997533001)--(xmax,0.33698371098991503*xmax+3.289359997533001),wwzzcc); D((xmin,-2.4276798224809775*xmin+36.54314988522595)--(xmax,-2.4276798224809775*xmax+36.54314988522595),wwzzcc); D(I_1--A_2--B_2--cycle,wwzzcc);
D(A); MP("A",(9.078267017098327,3.3218831364461052),NE*lsf); D(B); MP("B",(7.208843909967847,0.5368242217415146),NE*lsf); D(C); MP("C",(11.233826314095717,0.7657331736350426),NE*lsf); D(P); MP("P",(10.985841616211061,2.406247328871993),NE*lsf); D(C_1); MP("C_1",(11.367356536033608,0.11715780993671321),NE*lsf); D(A_1); MP("A_1",(3.698906647600416,4.981473037674183),NE*lsf); D(B_1); MP("B_1",(13.484764341048743,4.046761484108944),NE*lsf); D(M); MP("M",(9.688690888814401,-0.7030992676817621),NE*lsf); D(P_a); MP("P_a",(10.966765870219934,-0.5886447917349982),NE*lsf); D(P_b); MP("P_b",(9.898524094716802,1.1663238394487165),NE*lsf); D(P_c); MP("P_c",(7.11346518001221,4.752564085780656),NE*lsf); D(H); MP("H",(9.078267017098327,2.4253230748631207),NE*lsf); D(I_1); MP("I_1",(10.680629680353023,2.8068379946856674),NE*lsf); D(A_2); MP("A_2",(7.266071147941229,1.9484294250849374),NE*lsf); D(B_2); MP("B_2",(11.386432282024735,1.5096872672890087),NE*lsf); D(K); MP("K",(12.07315913770532,7.5376230004852465),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* Labels, by liberator */
defaultpen(fontsize(9pt));
label("$\Gamma_1$",(5,7),N*lsf);
[/asy]](http://latex.artofproblemsolving.com/3/0/c/30cdce277f8b9fc56322d3d0a0baa20ca56c49e3.png)
Suppose the tangents to 
at 
respectively intersect at 
.
;
.
,
.
.
,
are collinear.
,
and 
are homothetic.
By a converse of Reim's theorem, 
are tangent at This post has been edited 2 times. Last edited by liberator, Jul 22, 2015, 1:35 PM
lying on 
I have a neat proof using dynamic geometry.
'
October 2018
April 2018