Just for fun I thought I’d share a few interesting geometric theorems that I came across recently.
In 1899 Frank Morley, a professor at Haverford, discovered the following remarkable theorem.
The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle.
I’ve made a Geogebra applet illustrating this theorem. You can find several proofs of Morley’s miracle at this website.
The Pascal line
When he was sixteen years old Blaise Pascal discovered the following theorem.
If any hexagon (convex or not) is inscribed in a conic section and opposite sides are extended until they meet, then the three points of intersection will be collinear.
The line is now called the Pascal line.
I’ve made a Geogebra applet illustrating the Pascal line in the case where the conic section is a circle. When you try the applet, do not forget to try the nonconvex configurations!
In fact, given a hexagon, we could keep the vertices fixed and permute their order to obtain other hexagons. A little combinatorics shows that there are 60 different hexagons for each collection of six points. Each configuration has its own Pascal line. There is a lot known about these Pascal lines and their intersections.
This last theorem is remarkable, not for what it says, but because of the difficulty of the proof. In 1840 C. L. Lehmus asked for a purely geometric proof of the following elementary-looking theorem.
Any triangle with two angle bisectors of equal lengths is isosceles.
For example, suppose we have the triangle shown below with angle bisectors and of the same length. Prove that and are the same length.
Steiner gave the first purely geometric proof. Now there are many geometric (and trigonometric) proofs, but they are all tricky and are all proofs by contradiction. In 1852 Sylvester asked whether there exists a direct proof of this theorem. It appears that this is still an open problem. (From what I understand, there have been direct proofs, which were followed by arguments why the proofs really weren’t direct proofs. Moreover, there are some convincing arguments why a direct proof cannot exist.)
Bonus: Archimedean spiral in Adobe Illustrator
This isn’t a geometric theorem. But I thought I’d share it with you because it is cool. Earlier this week I needed to draw an Archimedean spiral () using Adobe Illustrator. Unfortunately Illustrator does not have that capability. However, after a little searching I found this ingenious hack on the Adobe forum by jpro2007. Basically, draw concentric circles separated by a constant radial distance . Then create a brush which is a sloped line of any length, but rises a distance of . Apply the brush to the circles to get the desired spiral. Amazingly clever! (I tweaked this a little by giving the circles radii and a vertically-aligned brush so that the spiral would begin at the origin and start in the first quadrant.)