GeometryDifficulty 8.2Prove itTeam selection tests for BMO 2018 · Saudi Arabia · 2018
Let ABC be a triangle with M,N,P as midpoints of the segments BC,CA,AB respectively. Suppose that I is the intersection of angle bisectors of ∠BPM, ∠MNP and J is the intersection of angle bisectors of ∠CNM, ∠MPN. Denote (ω1) as the circle of center I and tangent to MP at D, (ω2) as the circle of center J and tangent to MN at E.
1. Prove that DE is parallel to BC.
2. Prove that the radical axis of two circles (ω1), (ω2) bisects the segment DE.
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Official solution
1) Note that ∠MNC=∠MPB=∠A then by angle chasing, we have IP∥JN. Denote K=PJ∩IN then K is the incenter of triangle MNP. Hence, MK is the angle bisector of ∠NMP, thus MK∥IP. Denote X=IN∩MP then MKIP=XMXP=NMNP. Similarly, MKJN=MPNP. Thus JNIP=MNMP. Since △IPD∼△JNE then JNIP=NEPD. Therefore, MNMP=NEPD which implies that DE∥NP∥BC.
2) Suppose that DE cuts (ω1), (ω2) at R,S respectively. We have ERDS=JE⋅cos∠JESID⋅cos∠IDR=NE⋅sin∠NESPD⋅sin∠PDR=AB⋅sinBAC⋅sinC=1. Hence DS=ER. Denote T as midpoint of DE then PT/(ω1)=TD⋅TR=TE⋅TS=PT/(ω2), which implies that T lies on the radical axis of (ω1),(ω2). □
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