All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Shared Vertex (Posted on 2013-09-25) Difficulty: 3 of 5

Let squares ABCD and AB'C'D' ( both labeled counter-
clockwise ) share vertex A.

Prove that the midpoints of line segments BD, B'D,
B'D', and BD' are the vertices of a square.

  Submitted by Bractals    
No Rating
Solution: (Hide)

Consider the vertices of the squares as complex
numbers. WOLOG let A = 0, B = 2, C = 2(1+i), and
D = 2i. For an arbitrary complex number z ≠ 1,
let B' = Bz, C' = Cz, and D' = Dz. The midpoints
are then

         P = (B + D)/2 = 1+i
         Q = (B' + D)/2 = i+z
         R = (B' + D')/2 = z+iz
         S = (B + D')/2 = 1+iz

If P+R = Q+S, then the diagonals of the quadrilateral
PQRS bisect each other. If (R - P)i = S - Q, then the
diagonals are perpendicular and of equal length.
If both, then PQRS is a square.

         P+R = (1+i) + (z+iz)
                  = (i+z) + (1+iz)
                  = Q+S

         (R - P)i = [(z+iz) - (1+i)]i
                      = iz-z-i+1
                      = (1+iz) - (i+z)
                      = S - Q

QED  

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionCoordinatized SolutionJer2013-09-27 10:16:15
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (11)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information