Carl is given three distinct non-parallel lines $\ell_1, \ell_2, \ell_3$ and a circle $\omega$ in the plane. In addition to a normal straightedge, Carl has a special straightedge which, given a line $\ell$ and a point $P$, constructs a new line passing through $P$ parallel to $\ell$. (Carl does not have a compass.) Show that Carl can construct a triangle with circumcircle $\omega$ whose sides are parallel to $\ell_1,\ell_2,\ell_3$ in some order.

Attempt: To build a triangle with a circle, just have $ 3 $ points in that circle, at each point draw a line tangent to the circle and parallel to one of the given lines.

Am I right?

  • $\begingroup$ well, no. for one thing, the triangle must be inside the circle. $\endgroup$ – Will Jagy Jan 8 at 22:29
  • $\begingroup$ @WillJagy Why not? $\endgroup$ – Meulu Elisson Jan 8 at 22:32
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    $\begingroup$ First we need to construct the midpoint of the circle, then I would construct a triangle with sides parallel to the given lines, construct its circumcenter, and use parallels to make a similar triangle attached to the given circle. $\endgroup$ – Berci Jan 8 at 22:41

Alright, the Poncelet-Steiner theorem says we can do anything a compass and straightedge can do, if only we can construct the center of the given circle. That turns out to be fairly easy. Given any line, we can construct a diameter of the fixed circle that is perpendicular to the line. Now that I think of it (it's not in the diagram) we might as well make that first diameter, then simply make the diameter perpendicular to that. The intersection of the two (perpendicular) diameters is the center of the circle.

See attached diagram. The fixed circle is in red, so is the initial line being used.

Given two diameters that are not identical, the center is just where the diameters intersect.

So, the original problem ( inscribed triangle) can be solved. We know that by


enter image description here

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