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Math Olympiad

Geometry from RMO 2019 Problem 2 Solution

This beautiful application from Regional Math Olympiad 2019, Problem 2 is based on the concepts of Euclidean Geometry. Sequential hints are given to work the problem accordingly.

Understand the problem

Let ABC be a triangle with circumcircle \( \Omega \) and let G be the centroid of the triangle ABC. Extend AG, BG, and CG to meet \( \Omega \) again at \( A_1, B_1 \) and \(C_1\) respectively. Suppose \( \angle BAC = \angle A_1B_1C_1 , \angle ABC = \angle A_1 C_1 B_1 \) and \( \angle ACB = \angle B_1 A_1 C_1 \). Prove that ABC and \(A_1B_1C_1 \) are equilateral triangles.

Source of the problem
Regional Math Olympiad, 2019 Problem 2
Topic
Geometry

Difficulty Level

5/10

Suggested Book
Challenges and Thrills in Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Draw a diagram. RMO 2019 Problem 2 Construction: Complete the hexagon. Can you find some parallelograms in the picture? Particularly, can you prove \( BGCA_1 \) is a parallelogram? 

RMO 2019 Problem 2 equal angles (We will prove that the marked blue angles are equal) Notice that \( \angle ABC = A_1 C_1 B_1 \) (given) Also \( \angle A_1 C_1 B_1 =  \angle A_1 B B_1 \) (subtended by the chord \(A_1 B_1\) at the circumference. Hence \( \angle ABC = \angle A_1 B B_1 \) Now substracting \( \angle CBB_1 \) from both side we have IMPORTANT: \( \angle A_1 B C = \angle  ABB_1 \) Similarly notice that \( \angle BAC =\ C_1 B_1 A_1 \) (given) Also \( \angle  C_1 B_1 A_1 =  \angle C_1 A A_1 \) (subtended by the chord \(C_1 A_1 \) at the circumference. Hence \( \angle BAC = \angle C_1 A A_1 \) Now substracting \( \angle BAA_1 \) from both side we have IMPORTANT: \( \angle C_1 A B = \angle  A_1AC\) RMO 2019 Problem 4 equal angles Now \( \angle A_1 B C = \angle  A_1 A C \)  (subtended by the same segment \(A_1 C \) at the circumference.) And \( \angle C_1 A B = \angle C_1 C B \) ( subtended by the same segment \( C_1 B \) ) Thus \( \angle C_1 C B = \angle A_1 B C \)  Thus alternate angles are equal making \( BA_1 \) parallel to GC.  RMO 2019 Problem 5 equal angles Thus in \( \Delta A_1MB \) and \( \Delta GMC \) we have BM = CM, \( \angle BMA_1 = \angle GMC \) and \( \angle C_1 C B = \angle A_1 B C \)   Hence the triangles are congruent, making \( BA_1 = GC \)

Since one pair of opposite sides are equal and parallel hence the quadrilateral \(BGCA_1\) is a parallelogram.

RMO 2019 Problem 5 equal angles Similarly, we have \( CGAB_1, AGBC_1 \) to be parallelogram.  Therefore \( GM = MA_1 \) (since diagonals of a parallelogram bisect each other).  Since G is the centroid, hence \( GM = \frac{1}{2} AG \) implying \( MA_1 = \frac{1}{2} AG \) implying G is the midpoint of \( AA_1 \). 
Similarly G is the midpoint of \( BB_1 , CC_1 \)  Can you conclude that G is the center? 
RMO 2019 Problem 5 equal angles If G is not center then let O be the center. Hence OG is perpendicular to both \( AA_1, BB_1 \) at G as line from center hits the midpoint of a chord of a circle perpendicularly. Thus contradiction.  Now notice that \( \angle BGC = 2 \angle A \) (angle at center is twice at the circumference). But \( \angle BGC =  \angle BA_1 C \) (opposite angles of parallelogram) and \( \angle BA_1C = 180 – \angle A \) since \( BACA_1 \) is cyclic Hence \( 180 – \angle A = 2 \angle A \) implying \( \angle A  = 60^o\)  Similarly \( \angle B, \angle C \) are also 60 degrees.  Hence proved!

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By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

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