Olympiad Maths Prep

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Problem 1759

IMO Shortlist mid-range; USAMO P2/P5
Geometry Difficulty 8.2 Prove it Team selection tests for BMO 2018 · Saudi Arabia · 2018

Let ABCABC be a triangle with M,N,PM, N, P as midpoints of the segments BC,CA,ABBC, CA, AB respectively. Suppose that II is the intersection of angle bisectors of BPM\angle BPM, MNP\angle MNP and JJ is the intersection of angle bisectors of CNM\angle CNM, MPN\angle MPN. Denote (ω1\omega_{1}) as the circle of center II and tangent to MPMP at DD, (ω2)\left(\omega_{2}\right) as the circle of center JJ and tangent to MNMN at EE.

1. Prove that DEDE is parallel to BCBC.

2. Prove that the radical axis of two circles (ω1\omega_{1}), (ω2\omega_{2}) bisects the segment DEDE.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1) Note that MNC=MPB=A\angle MNC = \angle MPB = \angle A then by angle chasing, we have IPJNIP \parallel JN. Denote K=PJINK = PJ \cap IN then KK is the incenter of triangle MNPMNP. Hence, MKMK is the angle bisector of NMP\angle NMP, thus MKIPMK \parallel IP. Denote X=INMPX = IN \cap MP then
IPMK=XPXM=NPNM. \frac{IP}{MK} = \frac{XP}{XM} = \frac{NP}{NM}.
Similarly, JNMK=NPMP\frac{JN}{MK} = \frac{NP}{MP}.
Thus IPJN=MPMN\frac{IP}{JN} = \frac{MP}{MN}. Since IPDJNE\triangle IPD \sim \triangle JNE then IPJN=PDNE\frac{IP}{JN} = \frac{PD}{NE}. Therefore, MPMN=PDNE\frac{MP}{MN} = \frac{PD}{NE} which implies that DENPBCDE \parallel NP \parallel BC.

2) Suppose that DEDE cuts (ω1\omega_{1}), (ω2\omega_{2}) at R,SR, S respectively. We have
DSER=IDcosIDRJEcosJES=PDsinPDRNEsinNES=ACsinCABsinB=1. \frac{DS}{ER} = \frac{ID \cdot \cos \angle IDR}{JE \cdot \cos \angle JES} = \frac{PD \cdot \sin \angle PDR}{NE \cdot \sin \angle NES} = \frac{AC \cdot \sin C}{AB \cdot \sin B} = 1.
Figure 1
Hence DS=ERDS = ER. Denote TT as midpoint of DEDE then
PT/(ω1)=TDTR=TETS=PT/(ω2), \mathscr{P}_{T/(\omega_{1})} = TD \cdot TR = TE \cdot TS = \mathscr{P}_{T/(\omega_{2})},
which implies that TT lies on the radical axis of (ω1),(ω2)\left(\omega_{1}\right), \left(\omega_{2}\right). \square

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.