Introduction
When the Danish mathematics professor Julius Petersen (1839-1910) published Metoder og teorier til løsning af geometriske konstruktionsopgaver (Methods and Theories for the Solution of Problems of Geometrical Constructions, Petersen, 1866), his handbook on geometrical constructions initially did not draw much attention. However, when a second addition appeared in 1879, it was translated into at least seven other languages within the next four years (for English, see Petersen, 1879). Ever since, the book has been considered a classic among aficionados of construction problems in Euclidean geometry. Part of the appeal of the book is a set of 400+ problems of increasing difficulty with often extremely succinct solutions. To this day, it is not uncommon to find geometry lovers musing on the kind of solution Petersen may have envisioned to one of his problems. A case in point is the Brazilian-Canadian math educator Luís Lopes, who recently appealed for help with a reconstruction of Petersen’s solution to Problem 327: “Construct a triangle, which shall have its vertex in a given line, having a given base and a given difference of the angles at the base” (see Lopes, 2020; Petersen, 1879, p.59).
A First Solution
When I first saw Luís’ message, I did not think the problem was a hard one. In fact, let AB be the given line segment and let ℓ be the given line. Now, let C be any point in the plane such that ∠BAC exceeds ∠ABC by a fixed angle. If we assume these angles to be oriented, this angle condition defines a projective relation between the pencil of lines centered at A and the pencil centered at B. For any such relation, the locus of all points of intersection of corresponding lines is a conic section C. Therefore, the vertex C of any solution triangle ABC has to lie on C as well as on ℓ. However, there were a few issues with this solution. First of all, since ℓ and C generally meet in two points, which point is the right C? Could there perhaps be two solutions to Problem 327? In fact, might it be that ℓ and C do not meet at all, in which case there would not be a solution to Problem 327? More importantly, clearly this was not the kind of solution that Petersen had in mind. All of his construction problems are elementary, in that they can be solved with the use of ruler and compass only.
After some thought, I found a way to overcome the last issue (see Figure 1). Note that without loss of generality we may assume that all solution triangles ABC are oriented in the same way. In other words, we may assume that all candidates for C lie on the same side of the line AB. In fact, to find a solution with opposite orientation, we just reflect ℓ in AB, solve the problem for that configuration and reflect the solution triangle in AB again. Thus, with A to the left of B, we may assume that C lies “above” the line segment AB, i.e. on the same side of AB as the top of the page. Now, let C be such that ∠BAC exceeds ∠ABC by a fixed angle γ and that C∗ on AC is such that BC∗ is the reflection of BC in AB. Then, by construction, ∠AC∗B equals γ. That is, C∗ lies on a circular arc ΣAB. Conversely, any point on ΣAB comes from a point C such that ∠BAC exceeds ∠ABC by γ. Next, note that the procedure mapping C to C∗ defines a transformation π on the whole plane. Therefore, the points of intersection of C with ℓ that we are interested in now correspond to the points of intersection of ΣAB with ℓ∗ = π(ℓ). What is more, since it can be shown that π is projective, ℓ∗ is a straight line as well. By construction, if K and L on ℓ are such that KA ⊥ AB and LB ⊥ AB, ℓ∗ passes through L and K′ , the reflection in AB of K. Therefore, we can now easily find the point(s) of intersection of ℓ∗ with ΣAB. Finally, let C∗ be such a point of intersection. Then, the corresponding point C is just the point of intersection of ℓ with the line C∗A. In other words, we now have a way of constructing C by means of ruler and compass only. Unlike my previous solution, this approach provides some insight into the number of solutions to Petersen’s problem. Clearly, if the line ℓ is such that both K and L lie “above” AB, there is only one solution to Problem 327. Given that C is assumed to lie above AB, the same is true in case K and L lie on opposite sides of the line segment AB. This only leaves the solution to the case that K and L both lie below AB unclear (an issue that the scope of this note does not allow us to explore). Still, even though the construction is elementary, its derivation is not. All in all, it seemed unlikely that this was the solution of Problem 327 that Petersen had in mind and it was no surprise that Luís did not care much for my construction.
A More Elementary Approach
As it turned out, Petersen’s problem comes with a hint. In fact, he suggests to switch the end points of the given line segment (possibly referring to a reflection) and then to use the “method of similitude” (basically the notion that non-congruent figures can be obtained from one another by means of a single dilation if and only if they are directly similar and similarly situated). With this in mind, it is not much of a leap to a more “natural” solution to Problem 327. The idea is as follows (see Figure 2)
Let ABC with C on ℓ be a solution to Problem 327. Now, let m be the reflection of ℓ in the perpendicular bisector of AB and let D be the reflection of C in the same line. Then ABCD is an isosceles trapezoid and both ∠CAD and ∠CBD equal the fixed angle γ. Thus, A and B lie on a circular arc ΣCD that is directly similar to the arc ΣAB defined above. Now, let C′ and D′ be any two points on ℓ and m respectively such that C′ D′ is parallel to CD and let δ be the dilation centered at I = ℓ ∩ m such C′ = δ(C), D′ = δ(D), and ΣC′D′ = δ(ΣCD). By construction, A′ = δ(A)=ΣC′D′ ∩ IA and B′ = δ(B)=ΣC′D′ ∩ IB. But that means that we can construct A′ and B′ such that A′ B′ C′ is similar to the triangle ABC we are looking for. All that is left to do is to construct ABC as a triangle on AB that is directly similar to the triangle A′B′C′ . Alternatively, we can construct C as the point of intersection of ℓ with the line through A and parallel to A′C′.
Reconciling the Two Solutions
The preceding construction appears to take Petersen’s hints into account and I think Luís was perfectly happy with this solution. To me, however, the whole procedure remained somewhat unsatisfactory in that it is not fully deterministic: even though the outcome is completely determined by our choice of ℓ and AB, the intermediate steps are not. So, what if my earlier elementary construction could be derived from this second elementary construction? Actually, this is not too hard to do (see Figure 3). Indeed, if we retain all notations of Figure 2, K′L is by construction parallel to IA and has twice its length. Now, let F be the point of intersection of K′L with the perpendicular bisector of AB. Since I is the midpoint of KL and IF is parallel to KK′ , it follows that F is the midpoint of K′L. Therefore, since A is the midpoint of KK′ , AF is parallel to ℓ. By a similar argument, BF is parallel to m. Since C′D′ (in Figure 2) and AB are by construction parallel as well, there is a dilation δ′ that transforms ABF into C′D′I by Petersen’s Method of Similitude. As before, let C∗ = ΣAB ∩ K′L. Since ΣAB is similar to ΣC′D′ and the lines K′L and IA are parallel, it follows that δ (ΣAB)=ΣC′D′ and δ′ (K′ L) = IA. Consequently, δ′ (C∗) = δ′ (ΣAB ∩ K′ L) = δ′ (ΣAB) ∩ δ′ (K′ L) = ΣC′D′ ∩IA, which is A′ in Figure 2. We conclude that C∗A and C′A′ = δ′(C∗A) are parallel, which proves that the first elementary construction always leads to the same result as the second.
Final Thoughts
Thus, we have a deterministic construction for Petersen’s Problem 327 for which we also have an elementary derivation. Nonetheless, is this construction really a “better” or more “elegant” solution than what Petersen probably had in mind? More importantly, in any teaching context, would the extra effort to get to the deterministic solution really be justifiable? It was a fun exercise to bring my original, more advanced approach down to the more elementary level that Petersen envisioned. On balance, however, I feel that Peterson’s own approach has more of a natural flow to it and that bringing in more advanced mathematics was just not the right way to go. On the other hand, verification that my initial elementary construction always gives a solution to Problem 327 might make for a good classroom problem. Of course, there is also the issue of what exactly happens when ℓ lies “below” AB.
Eisso J. Atzema is a Lecturer at the University of Maine.
References
Lopes, Luís (January 20, 2020). Re: Triangle construction [electronic list message]. Retrieved from: groups.io/g/euclid/message/546
Petersen, Julius (1866). Metoder og teorier til løsning af geometriske konstruktionsopgaver (…). Copenhagen:Schønberg.
Petersen, Julius (1879). Methods and Theories for the Solution of Problems of Geometrical Constructions (…). Copenhagen: Andr. Fred. Høst & Son. (also accessible on Google Books)
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