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# 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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