Kürschák Competition 2018[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.6" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.6" open="off"]Hard[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.22.7" open="off"][/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]
Note that $latex p^2| |\overline{v_j}-\ell\cdot\overline{v_k}|^2$. This implies that $latex p^2|\langle \overline{v_j},\overline{v_k}\rangle$. Also, from the Cauchy-Schwarz inequality we get $latex |\overline{v_j}|\cdot |\overline{v_k}|\ge |\langle\overline{v_j}.\overline{v_k}\rangle|$ hence $latex -|\overline{v_j}|\cdot |\overline{v_k}|\le\langle\overline{v_j}.\overline{v_k}\rangle\le |\overline{v_j}|\cdot |\overline{v_k}|$. Also, $latex | \overline{v_j}|\cdot |\overline{v_k}|=p^2$. As the dot product is also divisible by $latex p^2$, it has to be equal to $latex \pm p^2$ or 0. It cannot be $latex p^2$ because the vectors are distinct, hence it is either $latex -p^2$ or 0. Thus the two vectors are either perpendicular or colinear (adding to 0).
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.6"]Let us plot the vectors in $latex \mathbb{R}^3$ and identify them with their tips (as the tails are at the origin). We join two tips by a straight line segment if they correspond to perpendicular vectors. Show that the number of straight lines is at least $latex \frac{n(n-2)}{2}$. Also prove that, there do not exist 4 vectors such that every possible pair of tips is joined by a line segment.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.6"]From the last assertion, show that the number of straight line segments in the figure is at most $latex \frac{n^2}{3}$ (this is a special case of a result called Turan's theorem). Thus, $latex \frac{n^2}{3}\ge \frac{n(n-2)}{2}$. Hence $latex n\le 6$. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.26.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]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%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]