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

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.23.3" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $ABC$ be a triangle and $AD,BE,CF$ be cevians concurrent at a point $P$. Suppose each of the quadrilaterals $PDCE,PEAF$ and $PFBD$ has both circumcircle and incircle. Prove that $ABC$ is equilateral and $P$ coincides with the center of the triangle.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.23.3" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _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"]Indian team selection test 2018[/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"]Hard

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" 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"]Quadrilaterals having both an incircle and a circumcircle are called bicentric. Read more about them here and here.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"] Show that $BCEF$ is cyclic. Afterwards, study the consequences of this result.    [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]

As $\angle BFC+\angle BEC=\pi$, hint 1 means that $\angle BFC=\angle BEC =\frac{\pi}{2}$. Hence $P$ is the orthocentre of $ABC$. Let $T$ be the circumcenter of $AFPE$ and $L$ be its incentre. From the information given in the second link of hint 1, we see that $T,L,K$ are colinear.Also, as $\angle AEP=\frac{\pi}{2}$$T$ lies on $AP$.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

As $L$ lies on $AP$$AD$ is the internal bisector of $\angle A$. As $AD$ is also an altitude, this means that $AB=AC$. By symmetry, $BC=AB$ and $CA=BC$. Hence the triangle is equilateral.

# Similar Problems

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