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Geometry from RMO 2019 Problem 2 Solution

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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_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" open="off"]Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][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"]

Start with hints

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Draw a diagram. RMO 2019 Problem 2 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"]

RMO 2019 Problem 2 equal angles (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: \( \angle C_1 A B = \angle  A_1AC\) RMO 2019 Problem 4 equal angles 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.  RMO 2019 Problem 5 equal angles 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"]RMO 2019 Problem 5 equal angles 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"]RMO 2019 Problem 5 equal angles 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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Connected Program at Cheenta

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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.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Similar Problems

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