This is a fairly simple problem from Futility Closet, which is currently under a hiatus.
“Robert Bilinski proposed this problem in the April 2006 issue of Crux Mathematicorum. On square ABCD, two equilateral triangles are constructed, ABE internally and BCF externally, as shown. Prove that D, E, and F are collinear.”
See Line Work
I found this problem from the 1981 Canadian Math Society’s magazine, Crux Mathematicorum, to be quite challenging.
“Proposed by Kaidy Tan, Fukien Teachers’ University, Foochow, Fukien, China.
An isosceles triangle has vertex A and base BC. Through a point F on AB, a perpendicular to AB is drawn to meet AC in E and BC produced in D. Prove synthetically that
Area of AFE = 2 Area of CDE if and only if AF = CD.”
See the Linked Triangles Problem
(Update 2/22/2023) Alternative Solution
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This turned out to be a challenging puzzle from the 1980 Canadian Math Society’s magazine, Crux Mathematicorum.
“Proposed by Leon Bankoff, Los Angeles, California.
Professor Euclide Paracelso Bombasto Umbugio has once again retired to his tour d’ivoire where he is now delving into the supersophisticated intricacies of the works of Grassmann, as elucidated by Forder’s Calculus of Extension. His goal is to prove Neuberg’s Theorem:
If D, E, F are the centers of squares described externally on the sides of a triangle ABC, then the midpoints of these sides are the centers of squares described internally on the sides of triangle DEF. [The accompanying diagram shows only one internally described square.]
Help the dedicated professor emerge from his self-imposed confinement and enjoy the thrill of hyperventilation by showing how to solve his problem using only highschool, synthetic, Euclidean, ‘plain’ geometry.”
Alas, my plane geometry capability was inadequate to solve the puzzle that way, so I had to resort to the sledge hammer of analytic geometry, trigonometry, and complex variables.
See Neuberg’s Theorem
This is a somewhat challenging math cryptogram in a slightly different guise from the Canadian Math Society’s magazine, Crux Mathematicorum.
“But you can’t make arithmetic out of passion. Passion has no square root.” (Steve Shagan, City of Angels, G.P. Putnam’s Sons, New York, 1975, p. 16.)
On the contrary, show that in the decimal system
has a unique solution.
See the Passion Kiss Problem
This is an interesting problem from the 1977 Canadian Math Society’s magazine, Crux Mathematicorum.
“206. [1977: 10] Proposed by Dan Pedoe, University of Minnesota.
A circle intersects the sides BC, CA and AB of a triangle ABC in the pairs of points X, X’, Y, Y’ and Z, Z’ respectively. If the perpendiculars at X, Y and Z to the respective sides BC, CA and AB are concurrent at a point P, prove that the respective perpendiculars at X’, Y’ and Z’ to the sides BC, CA and AB are concurrent at a point P’.”
See the Twin Intersection Puzzle
Puzzles and Problems: plane geometry, Dan Pedoe, Crux Mathematicorum