In this assignment, we look at a fairly complex answer to a seemingly simple question. This question was first proposed in 1775, around the American Revolutionary War, by the Italian I. F. Fagnano.
Given an acute triangle, inscribe the triangle of minimum perimeter.
Like Assignment 6, which dealt with viewing a picture, there are multiple ways to solve this problem, including a measurement approach, calculus approach, and several geometric approaches. In this problem, the calculus solution appeared first, by Fagnano himself. He used a result from his more famous father, Giulio Carlo de' Toschi di Fagnano. (The father earned a mention in the Catholic Encyclopedia, but not the son.) We're not going to talk about calculus this time, instead focusing on measurement and geometry.
This is a little more difficult to measure than the picture viewing, because we have three points that can move, not just one. I created a GeoGebra applet for the situation. Click through to fagnano.html to try it out. Given the outside orange triangle ABC, your challenge is to move blue points X, Y, and Z so that one lies on each side of the triangle and the perimeter is as small as possible.
For the initial triangle, with vertices at (0, 0), (5, 8), and (9, 1), the minimum perimeter is roughly 13.04. After you find that, you can move points A, B, and C and try another triangle if you like. You might even try a non-acute triangle. This wasn't part of Fagnano's Problem, but later, we'll briefly discuss it.
There are several geometric proofs of this problem. We will work through Fejer's approach. First, we need a preliminary theorem from almost 2000 years ago, proposed by the Greek Heron. Let's fix points Y and Z, which aren't on segment AB, and consider what point X on line AB minimizes the distance YX + XZ. To find this, we refect points Y and Z across AB, like a mirror, forming reflection points Y' and Z'. The line AB is the perpendicular bisector of Y'Y. The distance from Y to any point on line AB equals the distance from Y' to that point. For example, segment XY has length 4.54, just like segment XY'.
Given this, we can figure out the shortest path from Y to AB to Z by going from Y' across AB to Z. The shortest distance between two points is a straight line, so we draw from Y' to Z. The desired point X is at the intersection. Because we used reflections, we also discover that angles AXY and BXZ are equal.
The minimum triangle will need to satisfy this equal angle criterion on all three sides. I colored the three sets of equal angles in green, orange, and purple respectively.
We now have small triangles with two colored angles and one vertex. The lower left corner has a triangle with a green angle, a purple angle, and angle A, for instance. If we call the green angle G, purple angle P, and orange angle O, along with the known values of A, B, and C we have three equations with three unknowns.
Each triangle must have one of each color of angle. This means the green angle is C, the purple angle is B, and the orange angle is A. We formed three triangles similar to big triangle ABC.
We can manipulate the angles to make them equal, but can we build such a triangle without measurements? We can. We will begin with three lines. Each line passes through a vertex and is perpendicular to the opposite side of the triangle (or the extension of the side into a line). If the line consists of a portion inside the triangle, that segment inside the triangle is called an altitude. The three lines meet at a point called the Orthocenter.
Each of the vertex perpendicular lines crosses the opposite side extended at one point. When we connect these points, we form a triangle usually called the Orthic Triangle. In this diagram, the vertex perpendiculars are black lines, point O is the Orthocenter, and blue triangle XYZ is the Orthic Triangle.
The Orthic Triangle is the one we desire, with equal side angles. I'll demonstrate equal side angles for one side; the other two sides work the same way. Examine the four-sided quadrilateral OZCY. Angle OYC is a right angle, since it comes from the intersection of an altitude. So is angle OZC. Because we have two right angles inside the quadrilateral, we can circumscribe a circle around the figure.
Angle OYZ has its base point Y on the circle, and sits opposite circle arc OZ. Angle OCZ has its base point C on the circle, and sits opposite circle arc OZ. This means the two angles are equal. In the figure, angles OYZ and OCZ are in bright green.
Triangle BXC includes angle OCZ and a right angle at X. Therefore, angle B is complementary to OCZ: OCZ + B = 90 degrees.
Also, angle OYC is a right angle, from the altitude. Therefore, OYZ + ZYC = 90 degrees.
Substituting OYZ = OCZ means OCZ + B = 90 = OCZ + ZYC and thus angle B equals angle ZYC.
By taking the circle around quadrilateral XOYA, we can similarly show that angle B equals angle XYA. This is what we wanted.
We can repeat this with the other two sides to get the property we desire. When big triangle ABC is acute, sides of the orthic triangle makes three smaller similar triangles with equal side angles. Because it satisfies the minimum triangle criterion that we need, the orthic triangle is the inscribed triangle of minimum perimeter.
Unfortunately, the orthic triangle construction fails for non-acute triangles, because the Orthocenter lies outside the main triangle. (Yes, I know it's strange to think of a "center" outside a shape.) While there is an orthic triangle, it is not inscribed.
When the base triangle is not acute (right or obtuse), there is no exact solution. We can approximate it by finding the shortest vertex perpendicular, then taking the other two points as close as possible to that line's vertex. An approximate example is below, where B has moved to (-1, 6) making the triangle obtuse.
Much of this presentation was adapted from Courant's classic text What is Mathematics?, though that book refers to the altitude triangle, not the orthic triangle. Plus it calls vertex perpendiculars altitudes. I double-checked Fejer's construction, though.