# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Given a circle $\Gamma$, let $P$ be a point in its interior, and let $l$ be a line passing through $P$.  Construct with proof using a ruler and compass, all circles which pass through $P$, are tangent to $\Gamma$, and whose centres lie on $l$.

RMO 2019 Maharashtra and Goa region

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.23.3" open="off"]Geometry[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.23.3" open="off"]

Easy

# Try these problems first before watching the video or reading the hints:

(Send it to support@cheenta.com. Our priority response is for internal students, however we occasionally try to respond to external students as well). 1. How do you infer that a parallel line needs to be drawn through the center (to the given line AB (L)?  2. Can you find any isosceles triangle in the picture (once one of the little circles is drawn)? 3. How is the second small circle drawn? [/et_pb_text][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"]

# Watch the video

[/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="3.23.3"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]

Consider an inversion with respect to a circle with centre $P$. Call this map $f$. Note that, given any point $X$, $f(X)$ is constructible using ruler and compass. Construct the circle $f(\Gamma)$.

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Suppose $\Gamma'$ is one of our solutions. Then $f(\Gamma')$ is a line perpendicular to $l=f(l)$ and tangent to $f(\Gamma)$

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There can be no more than two lines perpendicular to $l$ and tangent to $f(\Gamma)$. Thus these two lines are the images of our solution circles.

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Invert the lines back to get the solution circles.

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# Similar Problems

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### One comment on “RMO 2019 Maharashtra and Goa Problem 2 Geometry”

1. Trishan Mondal says:

Sir check the mail I sent there a new solution which is with out inversion

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