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Trilinear coordinates and locus

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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 $ ABC$ be an equilateral triangle and $ P$ in its interior. The distances from $ P$ to the triangle's sides are denoted by $ a^2, b^2,c^2$respectively, where $ a,b,c>0$. Find the locus of the points $ P$ for which $ a,b,c$ can be the sides of a non-degenerate 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"]Romanian Master in Mathematics, 2008[/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"]Medium[/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"]This problem obviously points towards an application of trilinear coordinates (see more here). [/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"]

Start with hints

[/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"]If there exists a non-degenerate triangle with sides a,b,c then the area of the triangle must be positive. Make use of this fact employing Heron's formula.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Hint 1 will give you a locus in terms of trilinear coordinates. However, it is easier to work in cartesian coordinates as we are already familiar with many curves in them.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Take A=(1,0,0), B= (0,1,0), C= (0,0,1). There is a simple relationship between the cartesian and trilinear coordinates for this choice.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

.Note that all points in the plane of ABC satisfy x+y+z=1 (why?). For any P in the interior, Let Q be the foot of the perpendicular from P to AB and R be the foot of the perpendicular from P to the XY plane. It is possible to show that \frac{PR}{PQ}= \sqrt{\frac{2}{3}} (it is because the angle between the plane and the Z axis is \arccos \sqrt{\frac{2}{3}}). Hence, \frac{z}{c^2}=\sqrt{\frac{3}{2}}. By symmetry, \frac{x}{a^2}=\sqrt{\frac{2}{3}}=\frac{y}{b^2}. This relates the two coordinate systems. For a triangle with sides a,b,c, the square of the area is \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c). For this to be positive, we must have (after simplification) a^4+b^4+c^4< 2(a^2b^2+b^2c^2+c^2a^2). In cartesian coordinates, this translates to \frac{3}{2}(x^2+y^2+z^2)<3(xy+yz+zx), which is equivalent to (x+y+z)^2>2(x^2+y^2+z^2). As P lies on the plane x+y+z=1, this means that x^2+y^2+z^2<\frac{1}{2}. This last equation is that of the interior of a solid sphere. Hence, our desired locus is the intersection of this solid sphere with the plane x+y+z=1, which is precisely the interior of the circumcircle of ABC.  

 

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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%%" 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"]

Similar Problems

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