# Understand the problem

##### Source of the problem

##### Topic

##### Difficulty Level

5/10

##### Suggested Book

# Start with hints

**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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