MathDB

3

Part of 2012 Sharygin Geometry Olympiad

Problems(4)

Incenter and equal angles

Source: Sharygin Geometry Olympiad 2012 - Problem 3

4/28/2012
A circle with center II touches sides AB,BC,CAAB,BC,CA of triangle ABCABC in points C1,A1,B1C_{1},A_{1},B_{1}. Lines AI,CI,B1IAI, CI, B_{1}I meet A1C1A_{1}C_{1} in points X,Y,ZX, Y, Z respectively. Prove that ∠YB1Z=∠XB1Z\angle Y B_{1}Z = \angle XB_{1}Z.
geometryincenterperpendicular bisectorangle bisectorgeometry unsolved
paperfolding in a square, 3 incircles so that with one's radius = sum of other 2

Source: 2012 Sharygin Geometry Olympiad Final Round 8.3

8/3/2018
A paper square was bent by a line in such way that one vertex came to a side not containing this vertex. Three circles are inscribed into three obtained triangles (see Figure). Prove that one of their radii is equal to the sum of the two remaining ones.
(L.Steingarts)
geometryincircleinradiussquare
triangle construction by 4 points,C,L of bisector + M,N touchpoints of incircle

Source: 2012 Sharygin Geometry Olympiad Final Round 9.3

8/3/2018
In triangle ABCABC, the bisector CLCL was drawn. The incircles of triangles CALCAL and CBLCBL touch ABAB at points MM and NN respectively. Points MM and NN are marked on the picture, and then the whole picture except the points A,L,MA, L, M, and NN is erased. Restore the triangle using a compass and a ruler.
(V.Protasov)
geometryangle bisectorconstruction
MI = r/3 iff MI _|_ to a side, where M,I centroid , incenter of scalene

Source: 2012 Sharygin Geometry Olympiad Final Round 10.3

8/3/2018
Let MM and II be the centroid and the incenter of a scalene triangle ABCABC, and let rr be its inradius. Prove that MI=r/3MI = r/3 if and only if MIMI is perpendicular to one of the sides of the triangle.
(A.Karlyuchenko)
geometryCentroidincenterperpendicular