Maths Olympiad Prep

Library / /248 of 462

Geometry Difficulty 5.9 AIME, harder Prove it Ireland

Triangle ABCABC is inscribed in a circle Γ\Gamma. The points X,YX, Y and ZZ are the midpoints of those arcs BC,CABC, CA and ABAB, respectively, on Γ\Gamma which do not contain the third point of the triangle. The intersection of the triangles ABCABC and XYZXYZ form a hexagon DEFGHKDEFGHK. Prove that the diagonals DG,EHDG, EH and FKFK are concurrent.

Solution

Join AXAX, BYBY and CZCZ. These lines are the bisectors of the angles of ABC\triangle ABC and intersect at its incentre II. Label the vertices of the hexagon as shown below and join KIKI and ZAZA.

Figure 1

We have KZI=XZC=XAC=KAI\angle KZI = \angle XZC = \angle XAC = \angle KAI, hence AZKIAZKI is a cyclic quadrilateral. This implies ZIK=ZAK=ZAB=ZCB\angle ZIK = \angle ZAK = \angle ZAB = \angle ZCB from which we infer that KIKI and BCBC are parallel. Similarly it follows that IFIF and BCBC are parallel. This shows that K,IK, I and FF are collinear, i.e. the diagonal KFKF passes through II.

The same argument shows that the other two diagonals, DGDG and EHEH, also pass through II, thus the three diagonals of the hexagon are concurrent.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.