How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]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.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" _builder_version="4.0" hover_enabled="0" open="on"]Regional Math Olympiad, 2019 Problem 2 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Geometry

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5/10

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

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Draw a diagram. Construction: Complete the hexagon. Can you find some parallelograms in the picture? Particularly, can you prove $BGCA_1$ is a parallelogram? [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]

(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.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"] 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? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"] 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![/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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