Maths Olympiad Prep

Track / Stage 5 / 182 of 400 #1262 of 2444

Problem 1262

AIME late
Geometry Difficulty 5.3 Prove it Hong Kong competition problems · Hong Kong · 2016

Let OO be the circumcentre of a triangle ABCABC, and let \ell be the line going through the midpoint of the side BCBC and is perpendicular to the bisector of BAC\angle BAC. Determine the value of BAC\angle BAC if the line \ell goes through the midpoint of the line segment AOAO.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Next problem →

Official solution

We have BAC=120\angle BAC = 120^\circ.
Let MM be the midpoint of BCBC. Suppose ll meets the altitude from AA, the angle bisector of BAC\angle BAC, and the line AOAO at EE, PP, NN respectively.
Note that AE//MOAE // MO. In order that EMEM bisects AOAO, the quadrilateral AEOMAEOM must be convex. Therefore, BAC\angle BAC must be obtuse.
Now, since AE//MOAE // MO and AN=ONAN = ON, we have AENOMN\triangle AEN \cong \triangle OMN. It is well-known that AEAE and AOAO are isogonal lines with respect to BAC\angle BAC. Therefore, APAP bisects EAN\angle EAN. As we also have APENAP \perp EN, this shows AEN\triangle AEN is isosceles, with AE=ANAE = AN. Therefore,
OM=AE=AN=12OA=12OB. OM = AE = AN = \frac{1}{2} OA = \frac{1}{2} OB.
As BMO=90\angle BMO = 90^\circ, this yields OBM=30\angle OBM = 30^\circ. Thus, BAC=120\angle BAC = 120^\circ.
Figure 1

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