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!

Watch video

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

Geometry of AM GM Inequality

AM GM Inequality has a geometric interpretation. Watch the video discussion on it and try some hint problems to sharpen your skills.

Geometry of Cauchy Schwarz Inequality

Cauchy Schwarz Inequality is a powerful tool in Algebra. However it also has a geometric meaning. We provide video and problem sequence to explore that.

RMO 2019 Maharashtra and Goa Problem 2 Geometry

Understand the problemGiven a circle $latex \Gamma$, let $latex P$ be a point in its interior, and let $latex l$ be a line passing through $latex P$. Construct with proof using a ruler and compass, all circles which pass through $latex P$, are tangent to $latex...

RMO 2019 (Maharashtra Goa) Adding GCDs

Can you add GCDs? This problem from RMO 2019 (Maharashtra region) has a beautiful solution. We also give some bonus questions for you to try.

Number Theory, Ireland MO 2018, Problem 9

This problem in number theory is an elegant applications of the ideas of quadratic and cubic residues of a number. Try with our sequential hints.

Number Theory, France IMO TST 2012, Problem 3

This problem is an advanced number theory problem using the ideas of lifting the exponents. Try with our sequential hints.

Algebra, Austria MO 2016, Problem 4

This algebra problem is an elegant application of culminating the ideas of polynomials to give a simple proof of an inequality. Try with our sequential hints.

Number Theory, Cyprus IMO TST 2018, Problem 1

This problem is a beautiful and simple application of the ideas of inequality and bounds in number theory. Try with our sequential hints.

Number Theory, South Africa 2019, Problem 6

This problem in number theory is an elegant applciations of the modulo technique used in the diophantine equations. Try with our sequential hints

Number Theory, Korea Junior MO 2015, Problem 7

This problem in number theory is an elegant application of the ideas of the proof of infinitude of primes from Korea. Try with our sequential hints.