Interesting construction problems

Here are some interesting construction problems that I stumbled upon.

In the following problems, constructions can only be done using an unmarked straightedge and compass.

I came across the first problem accidentally while watching my friend fold origami. However, it might have been posed long ago by some mathematician. The third problem is easier than the others.

  1. Given a line segment of length ll and any positive integer nn. Show that, using straightedge and compass, it is always possible to divide the line into nn equal segments, no matter what nn is.

  2. Given a line segment of length 2+3+5\sqrt{2}+\sqrt{3}+\sqrt{5}, is it possible to construct a line segment of length 11?

  3. Given triangle ABC, construct the circumcircle, and the incircle.

  4. Given the perpendicular from A and two medians from A, B onto BC, AC respectively, reconstruct triangle ABC.

Note by Joel Tan
6 years, 9 months ago

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2: Follows directly from the definition of a constructible number.

Constructively, show that n\sqrt{n} is always constructible when nZ+n \in \mathbb{Z}^{+} and use compass equivalence theorem to add them up.

3: Let OO be the circumcenter of ΔABC\Delta ABC. It's pretty obvious that ΔAOB\Delta AOB is isoceles, and that the angle bisector of AOB\angle AOB is the perpendicular bisector of ABAB; hence, by similar argument for the other sides, OO is the intersection of the perpendicular bisectors of the sides of ΔABC\Delta ABC. Circumcircle construction is just a corollary.

Let OO^{\prime} be the incenter of ΔABC\Delta ABC. Let the shortest distance from the incenter to ABAB intersect the latter at XX; do similarly for ACAC, with the intersection YY. OXOYO^{\prime}X \equiv O^{\prime}Y (radii), and AXAYAX \equiv AY (convergent tangents). It's easy to see the angle bisector of ABC\angle ABC passes through the incenter; thus, by similar argument for the other angles, OO^{\prime} is the intersection of the angle bisectors of the interior angles of ΔABC\Delta ABC. Again, incircle is just a corollary (albeit a slightly more complicated one).

Jake Lai - 6 years, 5 months ago

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I'm new to compass and straightedge, so I'm sorry if I used any theorems incorrectly.

As for the incircle corollary, my method is to pick any side of ΔABC\Delta ABC and construct any circle with center OO^{\prime} that intersects that side. The median of the two intersections on the side will be a point on the circumference of the incircle.

Jake Lai - 6 years, 5 months ago

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