https://artofproblemsolving.com/wiki/index.php?title=1970_Canadian_MO_Problems/Problem_6&feed=atom&action=history1970 Canadian MO Problems/Problem 6 - Revision history2024-03-28T12:19:04ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1970_Canadian_MO_Problems/Problem_6&diff=43827&oldid=prevAirplanes1: Created page with "== Problem == Given three non-collinear points <math>A,B,C,</math> construct a circle with centre <math>C</math> such that the tangents from <math>A</math> and <math>B</math> are..."2011-12-22T01:24:30Z<p>Created page with "== Problem == Given three non-collinear points <math>A,B,C,</math> construct a circle with centre <math>C</math> such that the tangents from <math>A</math> and <math>B</math> are..."</p>
<p><b>New page</b></p><div>== Problem ==<br />
Given three non-collinear points <math>A,B,C,</math> construct a circle with centre <math>C</math> such that the tangents from <math>A</math> and <math>B</math> are parallel. <br />
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== Solution ==<br />
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Construct segment <math>AB</math>. Find the midpoint of <math>AB</math> (denote the midpoint <math>M</math>), by constructing its perpendicular bisector, and the point where the perpendicular bisector and <math>AB</math> meet is the midpoint of <math>AB</math>. Join <math>M</math> to the centre of the circle, point <math>C</math>. <br />
Construct lines <math>l</math> parallel to the line <math>MC</math> through <math>A</math>, and line <math>k</math> parallel to <math>MC</math> through <math>B</math>. <br />
From <math>C</math>, drop a perpendicular to line <math>k</math>, and let this point be <math>D</math>. Construct the circle with centre <math>C</math> and radius <math>CD</math>. Let this circle pass through E on line <math>l</math>. Then this circle is tangent to lines <math>l</math> and <math>m</math>, and lines <math>l</math> and <math>k</math> are parallel. <br />
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Proof: We have line <math>l</math> parallel to <math>CM</math> by our construction, and line <math>k</math> parallel to <math>CM</math> by our construction. Thus, lines <math>l</math> and <math>k</math> are parallel. Thus, the line <math>CD</math> is a transversal to <math>l</math> and <math>k</math>. Since <math>\angle BDC = 90^{\circ}</math>, by the co-interior angle theorem for parallel lines, <math>\angle AED = 180 - \angle BDC = 180 - 90 = 90</math>. Thus, the circle is indeed tangent to lines <math>l</math> and <math>k</math> and these two lines are parallel.<br />
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{{Old CanadaMO box|num-b=5|num-a=7|year=1970}}</div>Airplanes1