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

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $O$ be the circumcenter and $G$ be the centroid of a triangle $ABC$. If $R$ and $r$ are the circumcenter and incenter of the triangle, respectively,
prove that$$OG \leq \sqrt{ R ( R - 2r ) } .$$

Balkan MO 1996 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Geometry [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Comments" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Let $I$ be the incentre. Euler's theorem says that $OI^2=R(R-2r)$. Hence the result actually proves that $OG\le OI$. [/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"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" 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!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]The distance $OG$ is easily computable from standard formulae. For example, one can use complex numbers by assuming that $O=0$ and $|A|=|B|=|C|=R$. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]The centroid is given by $\frac{A+B+C}{3}$. Hence $OG^2=\frac{|A+B+C|^2}{9}=\frac{(A+B+C)(\overline{A}+\overline{B}+\overline{C})}{9}=\frac{3R^2+A\overline{B}+A\overline{C}+B\overline{A}+B\overline{C}+C\overline{A}+C\overline{B}}{9}=R^2-\frac{(2R^2-A\overline{B}-B\overline{A})+(2R^2-A\overline{C}-C\overline{A})+(2R^2-B\overline{C}-C\overline{B})}{9}=R^2-\frac{|A-B|^2+|B-C|^2+|C-A|^2}{9}=R^2-\frac{a^2+b^2+c^2}{9}$.   [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Show that $R(R-2r)=R^2-\frac{abc}{a+b+c}$. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]Combining all the hints, the problem reduces to proving that $(a^2+b^2+c^2)(a+b+c)\ge 9abc$. This follows from      $\frac{a^2+b^2+c^2}{3}\ge 3(abc)^{\frac{2}{3}}$ and $\frac{a+b+c}{3}\ge 3(abc)^{\frac{1}{3}}$. [/et_pb_tab][/et_pb_tabs][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"]

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