A Problem in an Equilateral Triangle II
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
- R. Honsberger, In Pólya's Footsteps, MAA, 1999, pp. 125-126
Copyright © 1996-2009 Alexander Bogomolny
A Problem in an Equilateral Triangle II
Construct CQN congruent to BMP as in the diagram. Let F be the center of CQN. Then E is the midpoint of DF. Indeed, CF||DP and CF = DP by the construction. And since the quadrilateral CFPD is a parallelogram, its diagonals bisect each other. Thus, E being the midpoint of CP is also the midpoint of DF.
Next observe, that triangles ABD and ACF are congurent. (BD = CF, AB = AC,  ABD = ACF = 30o.) Thus, AD = AF. Now, AC is obtained from AB by a rotation through 60o around A. The same then is also true of AF and AD. We see that ADF is equilateral. Its "half" - ADE - has angles 30o, 60o, 90o.
This proof is due to Narcyz R. Babnis who posted it at the CTK Exchange.
Copyright © 1996-2009 Alexander Bogomolny
|
