 # 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
Geometry

5/10

##### Suggested Book
Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!

Draw a diagram. Construction: Complete the hexagon. Can you find some parallelograms in the picture? Particularly, can you prove $BGCA_1$ is a parallelogram? (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: $\angle C_1 A B = \angle A_1AC$ 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. 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. 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? 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!

# Connected Program at Cheenta

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

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.