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# 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
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!

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

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

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