1970 Canadian MO Problems/Problem 6

Problem

Given three non-collinear points $A,B,C,$ construct a circle with centre $C$ such that the tangents from $A$ and $B$ are parallel.

Solution

Construct segment $AB$ . Find the midpoint of $AB$ (denote the midpoint $M$ ), by constructing its perpendicular bisector, and the point where the perpendicular bisector and $AB$ meet is the midpoint of $AB$ . Join $M$ to the centre of the circle, point $C$ . Construct lines $l$ parallel to the line $MC$ through $A$ , and line $k$ parallel to $MC$ through $B$ . From $C$ , drop a perpendicular to line $k$ , and let this point be $D$ . Construct the circle with centre $C$ and radius $CD$ . Let this circle pass through E on line $l$ . Then this circle is tangent to lines $l$ and $m$ , and lines $l$ and $k$ are parallel.

Proof: We have line $l$ parallel to $CM$ by our construction, and line $k$ parallel to $CM$ by our construction. Thus, lines $l$ and $k$ are parallel. Thus, the line $CD$ is a transversal to $l$ and $k$ . Since $\angle BDC = 90^{\circ}$ , by the co-interior angle theorem for parallel lines, $\angle AED = 180 - \angle BDC = 180 - 90 = 90$ . Thus, the circle is indeed tangent to lines $l$ and $k$ 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

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1970 Canadian MO Problems/Problem 6

Problem

Solution