A Problem in an Equilateral Triangle

The following problem has been offered at the 1980 All-Union Russian Olympiad and published in Crux Mathematicorum a decade later (1990, 33 and 70):

A line parallel to the side AC of equilateral ABC intersects BC at M and AB at P, thus making BMP equilateral as well. D is the center of BMP and E is the midpoint of CP. Determine the angles of ADE.

I came across this problem in one of R. Honsberger's books, where it was accompanied by an interesting commentary:

While the complete specification of D inside BMP is well known, with E situated at the midpoint of CP, ADE is cast obliquely in ABC, and with the medians not meeting the sides of a triangle at distinguished angles, it is doubtful that comparing the directions of the sides of ADE with those of ABC will uncover much useful information. Of course, there is always the chance that there exists an elusive construction line which will make everything clear. Finding Euclidean geometry so attractive, I am always reluctant to throw in the towel on the synthetic approach; however, after several looks at the problem from this point of view, I turned to the analytic approach.

The applet is supposed to suggest what that "elusive construction line" might be which would make everything clear. A different solution appears elsewhere.

References

1. R. Honsberger, In Pólya's Footsteps, MAA, 1999, pp. 125-126

Copyright © 1996-2009 Alexander Bogomolny

The key observation here is that, as P travels over AB, the points E and D each trace a straight line. E stays on the midline E'E'' of ABC parallel to the base AB while D is always located on the angle bisector of angle B. With this realization at hand, the problem appears as a particular case of a more general problem of three similar triangles.

In two extreme positions ABE' and AOE'', where O is the center of ABC, we obviously have two right triangles with 30^o, 60^o, 90^o angles. If we could show that points D = D(P) and E = E(P) divide BO and, respectively, E'E'' in the same proportion, the theorem of three similar triangles would apply directly: the triangle ADE would be a linear combination of triangles ABE' and AOE'' and thus be similar to either. It would then have the same 30^o - 60^o - 90^o angles.

But the required condition is easily obtained from comparison of two configurations of similar triangles. E'E/EE'' = BP/PA from the similarity of triangles ABC and E"E'C, that of PBC and EE'C, and also the similarity of triangles APC and E"EC. On the other hand, BD/DO = BP/PA from the similarity of PBD and ABO.

Copyright © 1996-2009 Alexander Bogomolny