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.

Given a line segment of length \(l\) and any positive integer \(n\). Show that, using straightedge and compass, it is always possible to divide the line into \(n\) equal segments, no matter what \(n\) is.

Given a line segment of length \(\sqrt{2}+\sqrt{3}+\sqrt{5}\), is it possible to construct a line segment of length \(1\)?

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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest2: Follows directly from the definition of a constructible number.

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

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

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

Log in to reply

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 \(\Delta ABC\) and construct any circle with center \(O^{\prime}\) that intersects that side. The median of the two intersections on the side will be a point on the circumference of the incircle.

Log in to reply