MathDB

8

Part of 2012 Canadian Mathematical Olympiad Qualification Repechage

Problems(1)

Show that <I1MI2=<O1MO2

Source: Canadian Repêchage 2012: Problem 8

5/19/2014
Suppose circles W1\mathit{W}_1 and W2\mathit{W}2, with centres O1\mathit{O}_1 and O2\mathit{O}_2 respectively, intersect at points M\mathit{M} and N\mathit{N}. Let the tangent on W2\mathit{W}_2 at point N\mathit{N} intersect W1\mathit{W}_1 for the second time at B1\mathit{B}_1. Similarly, let the tangent on W1\mathit{W}_1 at point N\mathit{N} intersect W2\mathit{W}_2 for the second time at B2\mathit{B}_2. Let A1\mathit{A}_1 be a point on W1\mathit{W}_1 which is on arc B1N\mathit{B}_1\mathit{N} not containing M\mathit{M} and suppose line A1N\mathit{A}_1\mathit{N} intersects W2\mathit{W}_2 at point A2\mathit{A}_2. Denote the incentres of triangles B1A1N\mathit{B}_1\mathit{A}_1\mathit{N} and B2A2N\mathit{B}_2\mathit{A}_2\mathit{N} by I1\mathit{I}_1 and I2\mathit{I}_2, respectively.*
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.1cm); 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 = -0.9748626324969808, xmax = 13.38440254515721, ymin = 0.5680051903627492, ymax = 10.99430986899034; /* image dimensions */
pair O_2 = (7.682929606970993,6.084708172218866), O_1 = (2.180000000000002,6.760000000000007), M = (4.560858774883258,8.585242858926296), B_2 = (10.07334553576748,9.291873850408265), A_2 = (11.49301008867042,4.866805580476367), B_1 = (2.113311869970955,9.759258690628950), A_1 = (0.2203184186713625,4.488514120712773); /* draw figures */ draw(circle(O_2, 4.000000000000000)); draw(circle(O_1, 3.000000000000000)); draw((4.048892687647541,4.413249028538064)--B_2); draw(B_2--A_2); draw(A_2--(4.048892687647541,4.413249028538064)); draw((4.048892687647541,4.413249028538064)--B_1); draw(B_1--A_1); draw(A_1--(4.048892687647541,4.413249028538064)); /* dots and labels */ dot(O_2,dotstyle); label("O2O_2", (7.788512439159622,6.243082420501817), NE * labelscalefactor); dot(O_1,dotstyle); label("O1O_1", (2.298205165350667,6.929370829727937), NE * labelscalefactor); dot(M,dotstyle); label("MM", (4.383466101076183,8.935444641311980), NE * labelscalefactor); dot((4.048892687647541,4.413249028538064),dotstyle); label("NN", (3.855551940133015,3.761885864068922), NE * labelscalefactor); dot(B_2,dotstyle); label("B2B_2", (10.19052187145104,9.463358802255147), NE * labelscalefactor); dot(A_2,dotstyle); label("A2A_2", (11.80066006232771,4.659339937672310), NE * labelscalefactor); dot(B_1,dotstyle); label("B1B_1", (1.981456668784765,10.09685579538695), NE * labelscalefactor); dot(A_1,dotstyle); label("A1A_1", (0.08096568938935705,3.973051528446190), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]
Show that ∠I1MI2=∠O1MO2.\angle\mathit{I}_1\mathit{MI}_2=\angle\mathit{O}_1\mathit{MO}_2.
*Given a triangle ABC, the incentre of the triangle is defined to be the intersection of the angle bisectors of A, B, and C. To avoid cluttering, the incentre is omitted in the provided diagram. Note also that the diagram serves only as an aid and is not necessarily drawn to scale.
Asymptotegeometryincentergeometric transformationrotationgeometry proposed