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"]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 \( AA_1 \).
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"]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://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"]
