# Viewing a Square

By Jonathan Osters @callmejosters & Chris Robinson @isomorphic2crob, The Blake School

This week, we were taking a look through How to Solve It, and came across this problem that intrigued us. The problem is as follows:

Point P is a point outside a square. Define the “viewing angle” as an angle that has P as a vertex and vertices of the square as points on its rays, as if P was on the side of a sculpture and viewing it by walking around it (Figure 1). What is the locus of points for which the viewing angle is 90˚? What is the locus for which the viewing angle is 45˚? (Paraphrased from p. 234 of How to Solve It)

This problem is really fun! Give it a try before you continue reading.

Solution for 90º viewing angle:

We began by creating convenient points in the locus. We reasoned that a point “half of a side length” from the midpoint of a side could create a 45-45-90 triangle.

Then we remembered Thales’ Theorem, that says that any point on a circle forms a right angle with endpoints of a diameter. Therefore, any point “half of a side length” from the midpoint of a side would create a right angle.

Thanks, Thales! Repeat that with all four sides, and you will get the following locus:

Solution for 45º viewing angle:

We knew a few things more about this one, having solved the 90˚ version of the problem. We knew that in this scenario, we would have to take into account that you might be able to see one or two sides, rather than just one, like in the 90˚ version. Not being able to see the solution with our mind’s eye, we began by calculating a few special points in the locus.

Special point 1: On a side (extended)

On the extension of one side, you will only be able to see one side, but you will be on the cusp of seeing two sides. This gave us the impression that this was an important point to investigate.

The point that forms a 45º viewing angle on the extended side forms an isosceles triangle which means that the point is a side length away from the vertex of the square. So the following 8 points are included in the locus.

Special Point 2: On an diagonal (extended)

The points along the extended diagonal would include two sides in their viewing angle. By symmetry we know that the diagonal is the bisector of the 45º angle and that the diagonal also bisects the 270º outside of the square.

Nicely, this leaves 22.5º, which means this a set of two isosceles triangles and that the point is a side length away from the vertex of the square. At that point we realized that to keep a 45˚ angle at the viewing angle, we could have the angle intercept a 90˚ arc on a circle. This circle:

We can see that any point on this quarter circle creates a 45˚ viewing angle.

There is such a quarter circle surrounding each vertex of the square.

This leaves us with the task of discovering what the locus of points is when only one side is visible. After playing around with it some more, and trying more things, we saw that the endpoints of the arcs and the closest vertices of the square, form a square.

If we create a circle at the center of that “side square,” then points on the outer edge will intercept a quarter of the circle at the viewing angle, thus making the viewing angle a 45˚ inscribed angle again!

There are these “bulbous bumpouts” on each of the sides, so the complete locus gives this “cloud” shape:

After completing this problem the other day, Chris and I started discussing the “cosmic significance” of the problem. This problem could be a great problem for an honors geometry class working on loci or circle properties, but more importantly, this problem is a reminder of what Chris and I like so much about working at Blake. We will often find ourselves on flights of mathematical fancy where we toss out some random problem like this, and will work to solve it together. We are all eager and excited mathematicians, and are always looking for problems that energize us and help us “recharge our mathematical batteries.” This is one such problem.