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Explore the Back-StoryRMO 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.

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 )

- 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

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

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.

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 $

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,**

- 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.

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.

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 )

- 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

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

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.

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 $

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,**

- 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.

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