RMO 2018 Tamil Nadu Problem 1 is from Geometry. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.

Problem

Let ABC be an acute-angled triangle and let D be an interior point of the line segment BC. Let the circumcircle of triangle ACD intersect AB at E (E between A and B) and let the circumcircle of triangle ABD intersect AC at F (F between A and C). Let O be the circumcenter triangle AEF. Prove that OD bisects \( \angle EDF \)

Key ideas you will need to solve this problem

  • Properties of Cyclic Quadrilaterals:
    • Opposite angles of a cyclic quadrilateral adds up to \( 180^o \)
    • Angle at the center is twice the angle at the circumference.
    • Angles subtended by the same arc at circumference are equal.

Also see

Advanced Math Olympiad Program


Hint 1: Draw a picture

RMO 2018 Tamil Nadu Problem 1


Hint 2: A construction would be nice

Join OE and OF. What kind of quadrilateral is OEDF? (Pause here. Try it yourself.)

RMO 2018 Tamil Nadu Problem 1 Solution


Hint 3: Some angle chasing

Note that \( EDCA \) is a cyclic quadrilateral (all of its four vertices are on a circle). Hence \( \angle ODB = \angle EAC = \angle A \) .

Similarly \( AFDB \) is cyclic. Hence \(\angle CDF = \angle FAB = \angle A \)

This implies \( \angle EDF = 180^o – 2 \angle A \). Since O is the center of \( \Delta AEF \), \( \angle EOF = 2\times \angle EAF = 2 \times \angle A\).

Therefore \( \angle EDF + \angle EAF = 180^o – 2 \times \angle A + 2 \times \angle A = 180^o \). This implies quadrilateral OEDF is cyclic.


Hint 4: OE = OF

OE = OF because O is the center and E, F are at the circumference of the circle passing through AEF. Therefore both are radii of the same circle.

This implies \( \angle OEF = \angle OFE \)


Hint 5: Final Lap; Some more angle chasing

Since OEDF is cyclic, hence \( \angle OEF = \angle ODF \) (angle subtended by the arc OF at the circumference).

Similarly \( \angle OFE = \angle ODE \).

But \( \angle OEF = \angle OFE \) (see hint 4).

Hence \( \angle ODE = \angle ODF \). The proof is complete,


Reference:

  • These ideas are usually discussed in the Geometry I module of Cheenta Math Olympiad Program.
  • Challenges and Thrills of Pre -College Mathematics by Venkatchala is a good reference for some these ideas.