Perpendiculars (Euclidea series)

Beta pack is the only one for which we have received all of the stars. Even so, a couple of the puzzles are worth discussing because they illustrate some interesting ideas. In particular, we like Drop a Perpendicular (2.6) and Erect a Perpendicular (2.7).

Perpendicular bisectors
One of the ideas lurking is around the perpendicular bisector, which we saw way back in alpha pack (1.2). One of the things I always found interesting about the perpendicular bisector was that it was easier to construct an object with more conditions than either constructing a perpendicular alone or finding a midpoint alone.

The other great thing about the perpendicular bisector is that it can be defined in an entirely different way: it is the locus of all points that are equidistant from our two starting points. In case that isn't clear, assume we start with two distinct points A and B. The perpendicular bisector of the segment AB is also the locus of all points C such that distance AC is equal to distance BC.

Of course, the fact that this is a line means that we only have to find two such points to construct the perpendicular bisector (which is how you solved 1.2, right?)

Drop a perpendicular
For the 2 move solution using tools, I'll let you find a solution on your own. In case you need a hint: how many combinations of 2 moves are there anyway? You could just try them all and see what you find.

Way back in my HS geometry class, I learned a construction for dropping a perpendicular that uses 4 elementary moves. For the E start, though, that's not good enough.  The cool idea that breaks through here is to choose two totally arbitrary points on the line as centers of circles that we draw. For some reason, I get a kick out of the idea that arbitrary points can be helpful ("if the point we choose doesn't matter, how can it help to choose a point anyway?")

In this case, while the points on the line we choose don't matter, the circle we draw with those points as centers need to have the right radius. Click the button below if you want to see how it is done and a bit more explanation.

The key here is that the circle radii are equidistant from our target point and the new intersection point of the two circles. That means they are on the perpendicular bisector of the segment between those points. Another observation is that the new intersection of the two circles is the reflection of our target point through the line. Kind of cool that, no matter which two center points we choose for the two circles, the new intersection point will always be the same!

Erect a perpendicular
Erecting a perpendicular also uses the idea of choosing an arbitrary point, but goes a step farther. This time, we choose any point we want that is not on the line already! Well, it also fails if we happen to choose a point that is already on the perpendicular, but that should be impossible.... In any case, just choose a point somewhere off to the side.

This construction uses Thales' Theorem. I don't know exactly why I find this result to be so cool, since the proof isn't hard. For me, it transforms a circle into a family of right triangles. Given a length for the hypotenuse, all the right triangles with that hypotenuse are living right there on the arc of the circle with that length as the diameter.

Incidentally, Thales' Theorem and the two defining properties of perpendicular bisectors will come up a lot in other Euclidea Constructions.

V stars
If you want to know where the V stars are, click below:

Angle of 30 degrees (2.3) and Double Angle (2.4) both have two solutions.