Difference between revisions of "1970 Canadian MO Problems/Problem 6"
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Latest revision as of 20:24, 21 December 2011
Problem
Given three non-collinear points construct a circle with centre
such that the tangents from
and
are parallel.
Solution
Construct segment . Find the midpoint of
(denote the midpoint
), by constructing its perpendicular bisector, and the point where the perpendicular bisector and
meet is the midpoint of
. Join
to the centre of the circle, point
.
Construct lines
parallel to the line
through
, and line
parallel to
through
.
From
, drop a perpendicular to line
, and let this point be
. Construct the circle with centre
and radius
. Let this circle pass through E on line
. Then this circle is tangent to lines
and
, and lines
and
are parallel.
Proof: We have line parallel to
by our construction, and line
parallel to
by our construction. Thus, lines
and
are parallel. Thus, the line
is a transversal to
and
. Since
, by the co-interior angle theorem for parallel lines,
. Thus, the circle is indeed tangent to lines
and
and these two lines are parallel.
1970 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |